Graph Theory in Memory of G.A. Dirac (Annals of Discrete Mathematics)

GRAPH THEORY IN MEMORY OF G.A. DIRAC

ANNALS OF DISCRETE MATHEMATICS

General Editor: Peter L. HAMMER Rutgers University,New Brunswick, NJ, U.S.A.

Advisory Editors: C. BERGE, Universitede Paris, France M. A. HARRISON, Universityof California,Berkeley, CA, U.S.A. V. KLEE, Universityof Washington,Seattle, WA, U.S.A. J.-H. VAN LINT CaliforniaInstitute of Technology,Pasadena, CA, U.S.A. G.-C.ROTA, MassachusettsInstitute of Technology, Cambridge, MA, U.S.A.

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NORTH-HOLLAND -AMSTERDAM

NEW YORK

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GRAPH THEORY IN MEMORY OF GmAm DIRAC Edited by

Lars DOVLING ANDERSEN Department of Mathematics and Computer Science Institute of Electronic Systems Aalborg University, Denmark

IvanTAFTEBERG JAKOBSEN SkanderborgAmtsg ymnasium, Denmark

CarstenTHOMASSEN Mathematical Institute The Technical University of Denmark

BjarneTOFT Mathematical Institute Odense University, Denmark

Preben Dahl VESTERGAARD Department of Mathematicsand Computer Science Institute of Electronic Systems Aalborg University, Denmark

1989

NORTH-HOLLAND-AMSTERDAM

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NEW YORK

OXFORD TOKYO

Elsevier Science Publishers R,V., 1989 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmittedin any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior written permission of the publishers, Elsevier Science Publishers R. V. (Physical Sciences and Engineering Division/, F! 0. Box 103, 1000 AC Amsterdam, The Netherlands.

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Preface

From 2 to 7 June 1985, a meeting was held at Sandbjerg, Denmark, in memory of the graph theorist Gabriel Andrew Dirac who died the year before. Attendance was by invitation only, and 55 mathematicians from 14 countries participated in lectures and discussions on graph theory related to the work of Dirac. This volume contains contributions in honour of the memory of Dirac from participants and others, and should not be seen only as proceedings from the meeting. All the papers have been refereed, and we wish to thank all the referees who have devoted their time to the project. We also thank the people who have helped typing the manuscripts: Susanne Albzk, Karin B. Andersen, Martin Hare Hansen, Allan L. Jensen, Frank Jensen, Susanne Kzseler and Astrid Pedersen. Professor W. T. Trotter Jr. is thanked for his careful editing of the contributions to the problem session. The meeting at Sandbjerg was made possible through donations and contributions from 0 0

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The Danish Natural Science Research Council The Carlsberg Foundation The Ministry of Education The Tuborg Foundation The Otto Mmsted Foundation The Thomas B. Thrige Foundation The University of Aarhus The University of Odense Aalborg University The Technical University of Denmark

The editors express their gratitude to all these sources for their interest and support.

V

vi Finally, we thank the Institute of Electronic Systems, Department of Mathematics and Computer Science, Aalborg University, for financial and technical assistance in preparing the volume.

Denmark, July 1988 Lars Dmling Andersen Ivan Tafteberg Jakobsen Carsten Thomassen Bjarne Toft Preben Dahl Vestergaard

Contents

Preface

Gabriel Andrew Dirac THEEDITORS Hamilton Cycles in Metacirculant Graphs with Prime Power Cardinal Blocks

7

B. ALSPACH The Edge-Distinguishing Chromatic Number of Paths and Cycles K. AL-WAHABI, R. BARI,F. HARARYand D. ULLMAN

17

On the 2-Linkage Problem for Semicomplete Digraphs J . BANG-JENSEN

23

The Fascination of Minimal Graphs R. BODENDIEK and K. WAGNER

39

Optimal Paths and Cycles in Weighted Graphs

53

J. A. BONDYand G. FAN A Note on Hamiltonian Graphs H. J. BROERSMA

71

Uniqueness of the Biggs-Smith Graph

75

A. E. BROUWER Some Complete Bipartite Graph - Tree Ramsey Numbers

S. B U R RP. , ERDOS,R. J. FAUDREE, C. C. ROUSSEAU and R. H . SCHELP vii

79

viii The Edge-Chromatic Class of Graphs with Maximum Degree at Least IVI - 3 A. G. CHETWYND and A. J. W. HILTON

91

On some Aspects of my Work with Gabriel Dirac P. ERDOS

111

Bandwidth versus Bandsize P. ERDOS,P. HELLand P. WINKLER

117

Circumference and Hamiltonism in K 1 ,s-Free Graphs E. F L A N D R II.NFOURNIER , and A. GERMA

131

The Prism of a 2-Connected, Planar, Cubic Graph is Hamiltonian ( a Proof Independent of the Four Colour Theorem) H . FLEISCHNER

141

A Note Concerning some Conjectures on Cyclically 4-Edge Connected 3-Regular Graphs H . FLEISCHNER and B. JACKSON

171

On Connectivity Properties of Eulerian Digraphs A . FRANK

179

Some Problems and Results on Infinite Graphs

195

R. HALIN On a Problem Concerning Longest Circuits in Polyhedral Graphs J . HARANTand H. WALTHER Interpolation Theorems for the Independence and Domination Numbers of Spanning Trees F. HARARYand S. SCHUSTER

211

22 1

Ein zum Vierfarbensatz Aquivalenter Satz der Panisochromie H. HEESCH

229

The Existence Problem for Graph Homomorphisms P. HELLand J . N E S E T ~ ~ I L

255

ix On Edge-Colorings of Cubic Graphs and a Formula of Roger Penrose 267 F. JAEGER Longest &Paths in Regular Graphs H. A. JUNG On Independent Circuits in Finite Graphs and a Conjecture of Erdos and P6sa. P. JUSTESEN

281

299

Contractions to Complete Graphs L. K. JORGENSEN

307

Triangulated Graphs with Marked Vertices H.-G. LEIMER

311

On a Problem upon Configurations Contained in Graphs with Given Chromatic Number F. LESCURE and H . MEYNIEL On Disjoint Paths in Graphs W. MADER Conjecture de Hadwiger: Un Graphe Ic-Chromatique Contraction-Critique n’est pas Ic-RQgulier J. MAYER

325

333

341

A Theorem on Matchings in the Plane M. D. PLUMMER

347

Removing Monotone Cycles from Orientations 0. R. L. PRETZEL

355

Disentangling Pairings in Trees G. SABIDUSSI

363

Colour-Critical Graphs with Vertices of Low Valency H. SACHSand M. STIEBITZ

371

About the Chromatic Number of 1-Embeddable Graphs H . SCHUMACHER

397

A

Problems and Conjectures in the Combinatorid Theory of Ordered Sets W. T. TROTTER A Proof of Kuratowski’s Theorem H. TVERBERG

40 1

417

Finite and Infinite Graphs whose Spanning Trees are Pairwise Isomorphic P. D. VESTERGAARD

42 1

Bridges of Longest Circuits Applied to the Circumference of Regular Graphs H.-J. VOSS

437

On a Standard Method Concerning Embeddings of Graphs K. WAGNER and R. BODENDIEK

453

Construction of Critical Graphs by Replacing Edges W. WESSEL

473

A Brief History of Hamiltonian Graphs R. J. WILSON

487

Erinnerungen an Gabriel Dirac in Hamburg E. WITT

497

Hamilton Paths in Multipartite Oriented Graphs

499

Problems Edited by W. T. TROTTER

515

C.-Q.ZHANG

This Page Intentionally Left Blank

Drawing by J. NeLtEl

Annals of Discrete Mathematics 41 (1989) 1-6 0 Elsevier Science Publishers B.V. (North-Holland)

Gabriel Andrew Dirac The editors

With the passing of Gabriel Andrew Dirac on July 20, 1984, at the age of 59, the international community of graph theorists lost one of its main figures, and Danish university life lost a most stimulating and generous colleague and teacher. Denmark had an early tradition in graph theory through the work of Julius Petersen, but after his death in 1910 graph theory was soon forgotten in this country (however Petersen’s highschool textbooks were used as late as in the early 1960’ies). With Dirac’s move to Aarhus University graph theory was again taken up at a serious level in Denmark. At Aarhus University, in 1966-67 and from 1970, Dirac attracted a large number of students. His lectures were catching. He presented graph theory as a general and important mathematical theory and did so with rigour and meticulous care. Dirac’s thorough style, influenced by the book of Konig (of which he had a very high opinion), was a delight for his students. Dirac was a fascinating person. He had an unconventional view of many matters, he had a penetrating and sharp mind and did not take very much for granted. This held not only for mathematics, but also in everyday political and social life. Therefore it was very enriching to know him and learn from him, even if one did not always agree fully with him. Dirac was a truly international figure. He was born on March 13, 1925, in Budapest, moved to England in 1937 when his mother married the great physicist and Nobel prize laureate P. A. M. Dirac. After wartime service in the aircraft industry and mathematical studies at the universities of Cambridge and London (obtaining his doctorate in 1951 under Richard Rado), he held university positions in London, Toronto, Vienna, Hamburg, Ilmenau, Dublin, Swansea and finally Aarhus. He was a British subject, but also knew Hungarian, German, French and Danish cultures and languages very well. Besides mathematics and a happy family life with his wife Rosemari Dirac and four children Meike, Barbara, Holger and Annette, Dirac’s great

The editors

2

passion was fine art. He took an interest in numerous other things and was a very knowledgeable and multi-faceted person. Dirac is among the most quoted graph theorists. He developed methods of great originality and made many fundamental discoveries. A survey of his main contributions is contained in the obituary by C. Thomassen [J.Graph Theory 0 (1985), 303-3181. By arranging the meeting at Sandbjerg in South Jutland, Denmark, in the summer of 1985, we wished to honour Dirac as a pioneer of graph theory, as an inventive and deep researcher for 35 years, as the most excellent teacher we have known, and last but not least as a good friend with a never ending interest in his students and friends. Dirac continues to be a rich source of inspiration, in mathematical research and presentation, as well as in the way we think about many other aspects of life. We wish to thank Dirac’s family and all participants a t Sandbjerg for making the meeting a worthy memory of G. A. Dirac and his achievements.

The publications of Gabriel Andrew Dirac [11 “The explicit determination of orifice parameters in shock absorbers,” Aircraft Engineering XIX (1947), 258-262. [2] “The vibration of propeller blades,” Aircraft Engineering XX (1948),

322-329, 343. [3] “Note on the colouring of graphs,” Math. 2. 5 4 (1951), 347-353. [4] “Collinearity properties of sets of points,” Quarterly J. Math. (2) 2 (1951), 221-227. [5] “Note on a problem in additive number theory,” J. London Math. SOC.26 (1951), 312-313.

[6] “A property of 4-chromatic graphs and some remarks on critical graphs,” J. London Math. SOC.27 (1952), 85-92. [7] “Ovals with equichordal points,” J. London Math. SOC.27 (1952)) 429-437. [8] “Connectivity theorems for graphs,” Quarterly J. Math. (2) 3 (1952)) 171-1 74. [9] “Some theorems on abstract graphs,” Proc. London Math. SOC.(3) 2 (1952), 69-81.

Gabriel Andrew Uirac

3

[lo] “Map colour theorems,” Canad. J. Math. 4 (1952), 480-490.

[ll] “The structure of k-chromatic graphs,” Fund. Math. XL (1953), 4255. [12] “The colouring of maps,” J. London Math. SOC.28 (1953), 476-480. [13] “Theorems related to the four colour conjecture,” J. London Math. SOC.29 (1954), 143-149. [14] With S. Schuster, “A theorem of Kuratowski,” Indagationes Math. 10 (1954), 343-348. [15] “Circuits in critical graphs,” Monatsh. f. Math. 59 (1955), 178-187. [16] “Map colour theorems related to the Heawood colour formula,” J. London Math. SOC.31 (1956), 460-471. [17] “Map colour theorems related to Heawood colour formula (II),” J. London Math. SOC.32 (1957), 436-455. [18] “A theorem of R. L. Brooks and a conjecture of H. Hadwiger,” Proc. London Math. SOC.(3) 7 (1957), 161-195. [19] “Short proof of a map colour theorem,” Canad. J. Math. 9 (1957), 225-226. [20] “Paths and circuits in graphs: extreme cases,” Acta Math. Acad. Sci. Hungar. 10 (1959), 357-362. [21] “GQnQralisationsdu thQor6me du Menger,” Comptes Rendus Paris 250 20 (1960), 4252-4253. [22] “Un theoreme de rkduction,” Comptes Rendus Paris 251 1 (1960), 24-25. [23] With M. Stojakovic, Problem cetiri Boje (The Four Colour Problem), Matematicka Biblioteka 10,Belgrade (1960).

[241 “Trennende Knotenpunktmengen und Reduzibili tat abstrakter Graphen mit Anwendung auf das Vierfarbenproblem,” J. reine angewandte Math. 204 (1960), 116-131. [25] “bchrome Graphen und vollstandige 4-Graphen,” Math. Nachr. 22 (1960), 51-60.

4

The editors

[26] “In abstrakten Graphen vorhandene vollstandige 4-Graphen und ihre Unterteilungen,” Math. Nachr. 22 (1960), 61-85. [27] “Connectedness and structure in graphs,” Rendiconti del Circolo Matematico di Palerrno Ser. II 9 (1960), 1-11. [28] “On rigid circuit graphs,” Abh. Math. Sem. Univ. Hamb. 25 (1961), 7 1-76. [29] “A contraction theorem for abstract graphs,” Math. Ann. 144 (1961), 93-96. [30] “Note on the structure of graphs,” Canad. Math. Bulletin 5 (1962), 22 1-227. [31] “On the maximal number of independent triangles in a graph,” Abh. Math. Sem. Univ.Hamb. 26 (1963), 78-82. [32] With P. Erdos, “On the maximal number of independent circuits in a graph,” Acta Math. Acad. Sci. Hungar. 14 (1963), 79-94. [33] “On the four-colour conjecture,” Proc. London Math. SOC. (3) 13 (1963), 193-218. [34] “Extensions of Menger’s theorem,” J. London Math. SOC.38 (1963), 148-161. [35] “Some results concerning the structure of graphs,” Canad. Math. Bull. 6 (1963), 183-210. [36] “On complete graphs and complete stars contained as subgraphs in graphs,” Math. Scand. 12 (1963), 39-46. [37] “Extensions of Turh’s theorem on graphs,” Acta Math. Acad. Sci. Hungar. 14 (1963), 417-422. [38] “Percy John Heawood,” J. London Math. SOC.38 (1963), 263-277. [39] “Homomorphism theorems for graphs,” Math. Ann. 153 (1964), 6980. [40] “Valency-variety and chromatic number of abstract graphs,” Wiss. Zeitschrift Univ. Halle 13 (1964), 59-63. [41] “On the structure of 5- and 6-chromatic abstract graphs,” J. reine angewandte Math. 214-215 (1964), 43-52.

Gabriel Andrew Dirac

5

[42] “Graph union and chromatic number,” J. London Math. SOC.30 (1964), 451-454. [43] “Extensions of the theorems of Tur&n and Zarankiewicz,” Proceedings of the Symposium in Smolenice June 1968, Czechoslovak Academy of Sciences (1965), 127-132. [44] “Generalizations of the five colour Theorem,” Proceedings of the Symposium in Smolenice June 1968, Czechoslovac Academy of Sciences (1965), 21-27. [45] “Chromatic number and topological complete subgraphs,” Canad. Math. Bull. 8 (1965), 711-715. [46] “Short proof of Menger’s graph theorem,” Mathematika 13 (1966), 42-44. [47] “Minimally 2-connected graphs,” J. reine angewandte Math. 228 (1967), 204-216. [48] “On the structure of &chromatic graphs,” Proc. Cambridge Phil. SOC.63 (1967), 683-691. [49] “Extension of Konig’s Lemma,” Math. Nachr. 40 (1969), 43-49. [50] “On Hamilton circuits and Hamilton paths,” Math. Ann. 197 (1972), 57-70. [51] “On arbitrarily traceable graphs,’’ Math. Scand. 31 (1972), 319-378. [52] “On systems of paths and circuits in graphs,” Math. Ann. 201 (1973), 133-144. [53] “Uber minimale KP-anfdlige Graphen,” Matematisk Institut, Aarhus Universitet, Preprint Series 20 (1970/71). [54] With C. Thomassen, “Graphs in which every finite path is contained in a circuit,” Math. Ann. 203 (1973), 65-75. [55] “Note on Hamilton circuits and Hamilton paths,” Math. Ann. 206 (1973), 139-147. [56] “On separating sets and Menger’s theorem,” Indagationes Math. 35 (1973), 49-62.

6

The editors

[57] With B. Aagaard Sprrensen and B. Toft, “An extremal result for graphs with an application to their colourings” J . reine angewandte Math. 268-269 (1974), 216-221. [58] “The number of edges in critical graphs,” J. reine angewandte Math. 268-269 (1974), 150-164. [59] With C. Thomassen, “On the existence of certain subgraphs in infinite graphs,” Indagationes Math. 36 (1974), 406-410. [60] “Terminable and interminable paths and trails in infinite graphs,” Discrete Math. 9 (1974), 19-34. [61] “Note on infinite graphs,” Discrete Math. 11 (1975), 107-118. [62] “Remarks on Ic-chromatic graphs and on terminable and interminable trails and paths,” Atti dei Convegni Lincei 17 (1976), 407-421. [63] “Structural properties and circuits in graphs,” Proceedings of the Fifth British Combinatorial Conference 1975, Congressus Numerantium X V (1976), 135-140. [64] With O.S. Nielsen, “On path-amenable graphs” Math. Nachr. 74 (1976), 263-268. [65] “Cardinal-determining subgraphs of infinite graphs,” Contribu ti del Centro Linceo Interdisciplinare di Sci. Mat. 34 (1977), 1-14. [66] “Hamilton circuits and long circuits,” Annals of Discrete Math. 3 (1978), 75-92. [67] “Interminable paths and trails: extreme cases,” Math. Nachr. 82 (1978), 7-19. [68] “Simplicial decompositions of infinite graphs,” Istituto Nazionale di Alta Matematica Francesco Severi, Symposia Mathernatica XXVIII (1987), 159-196.

Annals of Discrete Mathematics 41 (1989) 7-16 0 Elsevier Science Publishers B.V. (North-Holland)

Hamilton Cycles in Metacirculant Graphs with Prime Power Cardinal Blocks B. Alspach Department of Mathematics and Statistics Simon Fraser University Burnaby, B.C., Canada Dedicated to the memory of G. A . Dirac It is shown that every connected metacirculant graph with prime power cardinal blocks, other than the Petersen graph, has a Hamilton cycle.

1

Introduction

Metacirculant graphs were introduced in [2] as a logical generalization of the Petersen graph for the primary reason of providing a class of vertextransitive graphs in which thergmight be some new non-hamiltonian, connected vertex-transitive graphs. The reader is referred to [2] for their basic properties although their construction is now described. Let m and n be two positive integers and a be a unit in the ring of integers modulo n. Let Bi = {u;,o,ui,l, . , u ~ , ~ -for I } i = 0,1, . . . , m- 1 be m mutually disjoint sets of vertices each of cardinal n and called the blocks of the graph being defined. Choose sets of integers Sj {0,1,. . .,n - 1) for j = 0, 1, ..., 1Y.J with the property that amSj = Sj for all j = 0, 1, . . . , [fl, where by definition amSj = { amb: b E Sj }. The set SOhas the additional property that 0 SOand b E SOimplies n - b E SO.If m is even, then Sm12has the additional property that am12Sm/2= -Sm/2 where the negative is taken modulo n. We place edges in the graph as follows. The vertex uoIois adjacent t o uj,k for k E Sj, j = 0, 1, . . . , The remaining edges are obtained by letting the group generated by the two permutations e = (uo,ouo,1 . . U O , ~ - I ) ( u i l o ~ i l i u 1 , n - i ) . . . ( ~ m - i l O u m - l , l . . .urn-1,n-1) and T , where T ( u i , k ) =

..

LfJ.

.

.-

7

B . Alspach

8 ~ ; + ~ , ~act k ,

on the edges incident with UQO. The resulting graph is a metacirculant and is designated by G(m,n,a,SO,S1,. ..,S1,/2~). The Petersen graph is G(2,5,2,{ l, 4}, (0)). It was shown in [3] that if m is odd and n is a prime, then G ( m ,n , a,SO, S1,. ,s[,/2J) is hamiltonian as long as it is connected. The proof depends heavily on D. Marugit’s result [8] which states that every connected Cayley graph on a group which is a semidirect product of a prime order group by an odd order abelian group is hamiltonian. It was shown in [l]that if m is even and n is a prime, then G(m, n , a , SO,S1,. ..,S,/2) is hamiltonian as long as it is connected and not the Petersen graph. This time the proof depends heavily on E. Durnberger’s generalization [6] of MaruBiE’s result in which the former is able to replace odd order abelian by abelian. Recently, D. Witte and K. Keating [7] have generalized the MaruBiE and Durnberger results to obtain the following theorem which is crucial for the proof of the main theorem below.

..

Theorem 1. If G is a connected Cayley graph on a group whose commutator subgroup is cyclic and of prime power order, then G has a Hamilton cycle.

2

Main result

The purpose of this section is to provide a proof of the following result.

..

Theorem 2, Let G = G(m,n, a,SO,S1,. ,SLm/2]) be a metacirculant graph. If n is a prime power, G is connected, and G is not the Petersen graph, then G possesses a Hamilton cycle.

Proof. Let G be a metacirculant graph as described in the hypotheses. The proof proceeds by a sequence of reductions. One may assume n = p “ , p a prime, and e 2 2 .

(1)

The result has already been proved for n a prime in [l]and [3]. Thus, assumption (1) is valid. One may assume amf 1 (mod p e ) ,

(2)

It is shown in [2] that am 1 1 (mod p e ) implies that G is a Cayley graph on the group H generated by Q and r which were described in the introduction. If a = 1, then H is abelian and G has a Hamilton cycle as shown by several people independently (see, for example, [ 5 ] ) . If H is

Hamilton Cycles in Metacirculant Graphs

9

not abelian, then the group (e) generated by e contains the commutator subgroup of H and G has a Hamilton cycle by Theorem 1. Let d be an integer, 1 5 d 5 e , and define a quotient graph G/ p das follows. It has vertices wi,j for 0 5 i 5 m and 0 5 j 5 pe-d - 1 where one can think of wi,j as corresponding to the set of vertices { ui,t : t = j (mod pew') and ui,t E G}. Then wi,j is adjacent to w,.,~if and only if some vertex ui,t in G is adjacent to some vertex uv,uin G where t E j (mod p"-') and u = s (mod pep').

The quotient graph G/pd is isomorphic to the metacirculant graph G ( m , ~ ~ - ~ 5'6, , a 'Si,. , . ,Sim,2j) where

.

a' = a (mod p e P d ) and 1 5 a'

5 pe-d - 1, and

(3)

S i = { j : O < j < ~ ~ - ~ ja =nkd( m o d ~ ~ - ~k )E ,S ; } . It is easy to verify that (3) holds. The quotient graph G/pe was called G / e in [l, 31. If SO= 8, then G has a Hamilton cycle. (4) Suppose first that am f 1 (mod p ) . The quotient graph G/pe is a circulant graph with symbol S = { j : Sj # 8 or S m - j # 0 for 1 5 j 5 m - 1). It is connected because G is connected and, accordingly, Chen and Quimpo's main theorem in [5] implies that G/pe is edge-hamiltonian for m 2 3. Let Gj denote the subgraph induced on Bi by G for i = 0, 1, . . . , m - 1. Since SO = 8, each Gi = the complement of the complete graph with pe vertices. This implies that some S;, i # 0 , must contain an element b f 0 (mod p ) because G is connected. Thus, the edge wowj is in the quotient graph G / p e (the second subscript is dropped in this case as each block of G/pe is a singleton). Let WOW;. . .wo be a Hamilton cycle of G/pe containing the edge WOW; when m 2 3. The case m = 2 will be dealt with shortly. Now lift this cycle of G / pe to a path in G by starting at UO,O and taking the edge to U i , b . Then from ?hi$ take an edge to a vertex ~ jif wiwj , ~ is the next edge of the cycle in G/pe. Continue in this way until returning to a vertex U O , ~E Bo and call the resulting path P. If a f 0 (mod p ) , then the juxtaposition of the paths P,e a ( P ) ,e 2 " ( P ) , ..., e('+l)"(P) results in a Hamilton cycle. If a = 0 (mod p ) , the preceding juxtapositions will not work. Instead, the path P must be modified to ~ a' $ 0 (mod p ) . produce another path P' that returns to a vertex U O , with Since b f 0 (mod p ) and am $ 1 (mod p ) , then amb f b (mod p ) and

rp.,

B. Alspach

10 amb E Si by the property that amSi

= S;. Thus, start P' with the edge

UO,OUi,amb rather than UO,OUi,b and obtain all other edges of P' by taking an edge parallel t o each corresponding edge of P where parallel means the image under the appropriate power of e. The path P' will then terminate a t u ~ ,where ~ I a' = a - b amb. Now a' f 0 (mod p ) when a i 0 (mod p ) and P' gives rise t o a Hamilton cycle in G. In the case that m = 2, notice that -ab E S1 because as1 = -SI (recall that am/2Sm/2 = -Sm/, in a metacirculant graph when m is even). Thus, the 2-path Uo,OU1,bUo,b+(rb is in G. Now b 4-ab = 0 (mod p ) implies that a -1 (mod p ) which, in turn, implies that a2 E 1 (mod p ) . But this contradicts the present case that an f 1 (mod p ) and leads to the conclusion that b ab f 0 (mod p ) . Therefore, juxtapositions of this 2-path under powers of a! as above produces a Hamilton cycle in G. This completes the proof of (4) when am f 1(mod p ) . From (2) and the proof just completed, it may be assumed there is an integer d < e such that am t 1 (mod p d ) but am f 1 (mod pd+'). The proof now proceeds by induction on the exponent e since the result is true when n = p . Consider the quotient metacirculant G/pe-d whose vertices will be labelled q , j , 0 5 i < m and 0 5 j < p d , and whose twist multiplier is denoted a'. Since G is connected, G/pe-d is connected and, by the induction hypothesis, contains a Hamilton cycle H'. Now H' must contain an edge wi,jwrlSwhere s - j f 0 (mod p ) and by suitably operating on H' with powers of e and T , it may be assumed without loss of generality that WO,OWk,bI is an edge of H' with 6' f 0 (mod p ) . Now b' E SL corresponds to some b E Sk and b $ 0 (mod p ) . Now lift H' t o a path P in G that starts with the edge UO,OUk,b and terminates with a vertex ~ 0 where , ~ a E 0 (mod p d ) because H' returns to w o , ~ . Let WObe the set of vertices of Bo that we collapsed t o w o , ~ , that is, WO= { ~ 0 : c, I~ 0 (mod p d ) } . Now if a f 0 (mod #+I), then juxtapositions of P with images of P under powers of e0 will produce a Hamilton cycle. If, on the other hand, a = 0 (mod @I), then start P with UO,OUk,(rmb instead. Use parallel edges to replace the other edges of P to obtain P'. Then P' terminates at UO,+ where a' = a - b amb. Since am = 1 (mod p d ) , a' E WOas required. But a' f 0 (mod p d + l ) for if a' = 0 (mod pd+') held, then amb- b = 0 (mod p d + l ) would be true which would imply am i 1 (mod #+I). Hence, P' can be used t o produce a Hamilton cycle of G. This completes the proof of (4).

+

+

+

If IS01 = 1, then G has a Hamilton cycle. If 15'01 = 1, then the prime p must be 2 and n = 2e,

e

(5)

> 1. In addition,

Hamilton Cycles in Metacirculant Graphs

11

SO = {2"-l} must hold so that each Gj consists of 2e-1 disjoint edges. Therefore, the quotient metacirculant G/2 has 5'4 = 0 and has a Hamilton cycle H' because of (4). Now lift H' as above to a path P in G starting a t UO,O and terminating at either u0,o or ~ 0 , 2 c - 1 . If P terminates a t ~ 0 , 2 e - 1 , then juxtaposing P with a2e-1P yields a Hamilton cycle. If, on the other hand, P terminates at UO,O, let P, be the P. that P begins with U o , o U j , b . Then P, begins with path Q ~ ~ - ' Suppose ~ 0 ~ 2 eU - 1j , b + 2 e - l , Now remove the two preceding edges that begin P and P,, respectively, and insert the edges U O , O U O , ~ ~and -I Uj,bUj,b+2e-l (a's0 = s o as a must be odd). This amalgamates P and P, into a single cycle which must be a Hamilton cycle since P and P, each have length Ze-l. Hence, ( 5 ) is true. If ]Sol 2 3 , then G has a Hamilton cycle. (6)

Again recall the main result of Chen and Quimpo in [5] which states that a connected Cayley graph of degree a t least three on an abelian group is Hamilton connected if it is not bipartite and Hamilton laceable if it is bipartite. First consider the case that p is an odd prime so that each component of G;, i = 0, 1, . ., rn - 1, is not bipartite. Suppose that the components of Go, and thus each G; as well, have cardinal p d , d > 1. Then the quotient metacirculant G/pd has Sh = 0 and by (4) has a Hamilton cycle H'. Each vertex of H' corresponds to a component of some Gi and each edge of H' corresponds to a set of parallel edges joining the vertices of the two components that correspond to the endvertices of the edge. By the Chen-Quimpo Theorem, the subgraph induced on each component is Hamilton connected because the components are circulant graphs. It is then obvious that H' lifts to a Hamilton cycle of G. Now consider the case that p = 2. If the components of Go are Hamilton connected, then the preceding argument works. Suppose that the components of Go have cardinal 2d, d > 2, and induce bipartite graphs, that is, the graphs induced by the components are Hamilton laceable. Again the quotient metacirculant G / 2 d has a Hamilton cycle H' by (4). Each vertex of H' corresponds to a bipartite graph induced by a component of G. If w is a vertex of H', let A , and B,, respectively, denote the bipartition sets of the component corresponding to w. The cycle H' will be lifted in a different manner than either of the two earlier methods. Let WO,O be the vertex of H' whose corresponding component contains UO,O and let UO,O E Awo,o.Choose a vertex UO,b E Bwo,o. There exists a Hamilton path in the component joining UO,O and U 0 , b by the

.

B . Alspach

12

Chen-Quimpo Theorem. Now let w' be the next vertex of H' following w o , ~ . This means that UO,O is adjacent to a vertex 5 in either A,I or B,I, but whichever is the case, U0,b is then adjacent to a vertex y of the opposite bipartition set because of the action of ee-d+l. There is a Hamilton path joining x and y in the component D corresponding to 20'. Join 5 to q o , join y t o U0,b and remove from the Hamilton path in D an edge with endvertices x' and y' distinct from 5 and y. This is possible because ID1 2 4. Now take edges from x' and y' to vertices in opposite bipartition sets in the component corresponding to the next vertex of H'. Join them with a Hamilton path and remove an appropriate edge for extending the partial cycle in G into the next component. Continue working along H' in this way until reaching the last vertex of H' before w o , ~ .Do not remove an edge from the Hamilton path in this component. The result is a Hamilton cycle in G. The proof of ( 6 ) is complete. This leaves the case that [Sol = 2 . Since amSo = So, am f l (mod p e ) and ( 2 ) precludes am 3 1 (mod p"). It is not surprising that this remaining case requires the most intricate proof because the Petersen graph satisfies [Sol = 2 and a2 -1 (mod 5 ) where m = 2.

=

=

When /Sol = 2, G has a Hamilton cycle.

(7)

Let the components of Go have p d vertices, 1 5 d 5 e , so that they are cycles of length p d . The quotient metacirculant G / P " - has ~ Sh = 0 and must be connected. It follows from (4)that G / p e v d has a Hamilton cycle H'. Notice that each vertex of H' corresponds t o a cycle of length pd in G and an edge of H' corresponds to one or more parallel l-factors in G joining the two pd-cycles that correspond to the endvertices of the edge. Hence, the following observation is true.

The subgraph of G corresponding to an edge and its two endvertices in H' contains a graph isomorphic to a generalized Petersen graph.

(8)

The notation G(n,lc) will be used for the generalized Petersen graph with vertices UO, 211, . . ., ~ ~ - 1vo,, v1, . , vn-l and edge set { uivi,u ; u ~ + ~ , viv;+l : i = 0 , 1 , . , , n - 1 with subscripts reduced modulo n } . Since both cycles in the generalized Petersen graph referred to in (8) have length p d , gcd{n,k} = 1. Recall that K. Bannai [4]has proved that G ( n , k ) , when gcd{n, k} = 1, has a Hamilton cycle if and only if G ( n ,k) is not isomorphic to G ( n , 2 ) with n 5 5 (mod 6). But in the latter case there is a compensating result. Namely, in [l]it was shown that there are Hamilton paths in G ( n , 2 ) , when n = 5 (mod 6), from uo to every v;, i # 0. In fact, it is

.

..

13

Hamilton Cycles in Metacirculant Graphs

not hard to show that there is a Hamilton path joining any pair of distinct non-adjacent vertices, but this is not germane to the present proof. Consider lifting the Hamilton cycle H' in G / P " ~ to G. There are two cases t o consider. Let w i , j be the vertex following WO,O in H' and w r , s the vertex preceding w o , ~ . First, assume that the generalized Petersen graph G* induced by the edge Wr,8Wo,o is isomorphic to G(pd,2),pd = 5 (mod 6). In this case start H with u0,o and take an edge from it to a neighbor u ; , j ~in the cycle C ; , j corresponding to w i , j . Continue by taking one of the two Hamilton paths around C i , j starting a t u ; , j , and terminating a t some ui,h+jl. Then take an edge from u;,h+jl to a neighbor on the next cycle corresponding t o the vertex following W i , j on H'. Take a Hamilton path around this cycle starting at the neighbor and then jump to the next cycle using HI. Continue until the cycle C r , s is entered at vertex U ~ , ~From I . remarks above, there is a Hamilton path in G* joining and UO,O unless they happen to be adjacent in G'. Thus, if they happen to be not adjacent in G', H is a Hamilton cycle are adjacent in G*, then use the other Hamilton in G. If u0,o and u r , s ~ path when first travelling around C ; , j . This means that C j , j is exited at a different vertex so that if all subsequent Hamilton paths are given the same orientation ils before, the cycle Cr,s will be entered at a vertex different than U , . , ~ I . Thus, a Hamilton cycle can be constructed in G. It may be assumed that no edge of H' in the quotient graph induces a subgraph containing a generalized Petersen graph of the form G(pd,2 ) with pd 5 (mod 6) or else the argument above establishes the existence of a Hamilton cycle in G. This assumption leads to the second case. First, consider the subcase that H' has odd length. Each vertex of H' corresponds to a circulant subgraph of G (because ee-d is an automorphism of G ) and each is a pd-cycle. Thus, in G , H' corresponds to a sequence of circulant subgraphs each of which is a pd-cycle and such that successive circulants are joined by a 1-factor that is invariant under ee-d. Relabel their vertices as q , o , w q , .. , q P d - l , i = 0, 1, . ., pe-d - 1, so that the vertices whose first index is i 1 correspond to the vertex of IT' that follows the vertices of H' corresponding to those whose first index is i. Hence, e e - d ( v i , j ) = V i , j + l for all i and j . Without loss of generality, it may be assumed that the symbol of the - ~ that is, DO,; is adjacirculant graph on { D O , O , O ~ , ~ ., .. , ~ ~ , }~ isd {l,-1}, cent to v o , i + l for i = 0, 1, . . . , pd - 1. Construct two paths PI and P2 as follows. The initial vertex of PI is DO,O and the initial vertex of Pz is OOJ. Now DO,O is adjacent to v l , j for some j and thus DO,^ is adjacent to v l , j + l . Let ~ 1 , j - k and v l , j - k + l be the vertices immediately preceding the

=

.

+

.

B. Alspach

14

vertices w 1 , j and v l , j + l , respectively, on the cycle 2)1,02)1,kZ)1,2k . . . v1,o. Extend both PI and P 2 by the respective subpaths ' ~ i , j v l , j + k v l , j + 2 k . . .v l , j - k + i and Wllj+l211,j+k+ll)l,j+2k+l ., . v 1 , j - k . Notice that all the vertices in the set { v 1 , 0 , . . ,v l , p d - l } are used and that the second subscript of the last vertex of Pl immediately follows the second subscript of the last vertex of P 2 . That is, the last vertices of the partially constructed paths PI and P 2 have interchanged the role of who comes first. Continue extending PI and P 2 by working along the cycle H' until reaching . ., ~ ~ ~ e - d - ~ , ~ dSince - ~ } .there are an odd number of circulants, the terminal vertex of PI will have the form vmpe-d-l,j and the terminal vertex of P 2 will have the form Vmpe-d-l,j+l. Now Vmpe-d-l,j is adjacent to some vo,;and 2),pe-d-l,j+l is then adjacent to v o , ; + l . If i # 0, then v0,1 to PI and path 2)mpe-d-l,j+lV0,i+l add the path v,pe-d-l,jv0,j2)0,;-1.. 00,;+2. . .oo,o to P 2 thereby producing a Hamilton cycle in G. On the other hand, if i = 0, then at the first stage of the construction of PI and P 2 use the vertices q , j + k and v l , j + l ~ +instead ~ and go around the cycle in the opposite direction. Leave all other extensions unaltered so that the terminal vertices are shifted by 2k. Now complete the cycle as above. Now consider the subcase that H' has even length. The vertex v0,o is adjacent to some v 1 , j which in turn is adjacent to some v 2 , k and so on. Continue in this way constructing a path P that passes through each subcirculant until returning to some vertex W O , ~in {vo,o, v o , ~ , .. . ,v o , p d - l } . If T is relatively prime to p e , then it is easy to see that a Hamilton cycle can be found in G as done earlier. (It is a simple application of the factor group lemma mentioned in [9].) Thus, it may be assumed that T = 0 (mod p ) . In fact, even more can be assumed as is now shown. Let c; be t - j reduced modulo pd where w ; , j v i + l , t is the edge of P corresponding to the edge in H' from the ith vertex to the (i 1)st vertex. Now if any C i , i = 0, 1, . . , , mpe-d - 1 is relatively prime to p , then there is also an edge from vi,j to v i + ~ , j - ~because , am = -1 (mod p") and amS, = S, for all S,. Make this one change in P and the terminal vertex v0,+ will satisfy r' f 0 (mod p). Thus, G has a Hamilton cycle. Thus, it may be assumed that every c; is congruent to zero modulo p . If T = 0, a Hamilton cycle may be found as follows. Let v 2 i I j v 2 i + l , k be an edge of P . Because of (8), there is a generalized Petersen graph induced by the corresponding edge of H' so that it may be assumed that v 2 i , j v 2 ; + l , k is an edge of the geneia.lized Petersen graph. It contains a Hamilton cycle because of K . Bannai's Theorem [4]mentioned earlier. Because of the action of ee-d, there is a Hamilton cycle in the generalized Petersen graph

.

.

.

+

Hamilton Cycles in Metacirculant Graphs

15

. other words, the generalized Petersen graph using the edge v 2 i , j ~ 2 j + ~ , kIn has a Hamilton path from v 2 i , j t o v z ; + l , k . Find these Hamilton paths in the generalized Petersen graphs corresponding to alternate edges of P starting with vo,ovl,j and use the remaining edges of P to connect them together forming a Hamilton cycle in all of G. If each ci is zero, then r is zero and as was just shown this would imply the existence of a Hamilton cycle in G. Therefore, it may be assumed that some ci is nonzero. Let ps be the largest power of p that divides every c;, i = O , l , ..., mpe-d - 1. Clearly, 1 5 s < d. Let G' denote the subgraph of G obtained by replacing each vertex of H' with its corresponding circulant subgraph and replacing each edge of P with the 1-factor between the corresponding circulants where the second coordinates of the edges change by c;. Now @ - d is an automorphism of G' so that the quotient graph G ' / ( Q " - ~ )may " be formed. The edges going between the blocks of this quotient graph have zero change in the second coordinates and from above this implies that there is a Hamilton cycle in the quotient graph G'/(e"-d)s. Now lift this Hamilton cycle in the quotient graph to G' to see what it does there. Start a t the vertex v0,o and follow the Hamilton cycle until reaching a vertex of the form vO,ap~ for the first time. If a f 0 (mod p ) , then the factor group lemma [9] produces a Hamilton cycle in G' which is a subgraph of G. If a = 0 (mod p ) , then find a ci of the form bp' with b f 0 (mod p ) . Such a c; exists by the definition of s. Since am 3 -1 (mod p " ) , -bps is also in the same S j containing bp". Instead of using the edge with ci = bp", use the edge with c j = -bps instead. Make no other changes and the resulting path in G' reaches Wo,apL2bpU instead. Clearly, a - 2b f 0 (mod p ) and the factor group lemma gives a Hamilton cycle in G'. This completes all possible cases and the proof of (7) is complete. There0 fore, Theorem 2 is proved.

3

Conclusion and acknowledgements

The primary question suggested by the result in this paper is whether or not the Petersen graph is the only connected metacirculant without a Hamilton cycle. Indications are that the answer is affirmative. It is likely that the question will not be resolved until the question of whether or not every connected Cayley graph on a group with cyclic commutator subgroup has a Hamilton cycle is resolved. The author wishes to thank the Natural Sciences and Engineering Research Council of Canada whose support is greatly appreciated, the Department of Mathematics, University of Newcastle, Australia for its hospitality

16

B . Alspach

and assistance in the preparation of this paper, and David Witte for several interesting and productive conversations.

References [l] B. Alspach and T. D. Parsons, “On hamiltonian cycles in metacirculant graphs,” Annals of Discrete Math. 15 (1982), 1-7.

[2] B. Alspach and T. D. Parsons, “A construction for vertex-transitive graphs,” Canad. J. Math. 34 (1982), 307-318. [3] B. Alspach, E. Durnberger and T. D. Parsons, “Hamilton cycles in metacirculant graphs with prime cardinality blocks,” Annals of Discrete Math. 27 (1985), 27-34. [4] K. Bannai, “Hamiltonian cycles in generalized Petersen graphs,” J. Combin. Theory ( B ) 24 (1978), 181-188. [5] C. C. Chen and N. F. Quimpo, “On strongly hamiltonian abelian group graphs,” Combinatorial Mathematics VIII (K. McAvaney, ed.), Lecture Notes in Math. 884 (1980), 23-34. [6] E. Durnberger, “Every connected Cayley graph of a group with prime order commutator group has a Hamilton cycle,” Annals of Discrete Math. 27 (1985), 75-80. [7] K. Keating and D. Witte, “On Hamilton cycles in Cayley graphs in groups with cyclic commutator subgroup,” Annals of Discrete Math. 27 (1985), 89-102.

[8] D. MaruBiE, “Hamiltonian circuits in Cayley graphs,” Discrete Math. 46 (1983), 49-54. [9] D. Witte and J. A. Gallian, “A Survey: Hamiltonian cycles in Cayley graphs,” Discrete Math. 51 (1984), 293-304.

Annals of Discrete Mathematics 4 1 (1989) 17-22

o Elsevier Science Publishers B.V. (North-Holland)

The Edge-Distinguishing Chromatic Number of Paths and Cycles K. Al-Wahabi Department of Mathematics University of Qatar Doha, Qatar

R. Bari Department of Mathematics George Washington University Washington D. C., USA

F. Harary*

D. Ullman Department of Mathematics George Washington University Washington D. C., USA

Department of Mathematics University of Michigan Ann Arbor, Michigan, USA

Dedicated to the memory of Gabriel Dirac The edge-distinguishing chromatic number x l ( G ) of a graph G is defined as the minimum number n of colors { 1,2,... ,n} which can be assigned to the vertices V ( G )in such a way that when each edge e = UZI is assigned as its “color” the set of colors {c(u), c ( v ) } , all the edges of G have different colors. We derive a n exact formula for the edge-distinguishing chromatic number of a path and of a cycle. The formulas of Theorems 3.1 and 4.1 appeared in [l]in different but equivalent form. However, we have rederived them in order to provide not only better formulas but also clear, detailed, and complete proofs in this semi-expository note.

1

Definitions

Let G be a graph with vertex set V ( G )and edge set E ( G ) . An r-coloring of G is a mapping C#I from V ( G ) into the set N , = {1,2,. . .,r}. The color of an edge e = uw induced by the r-coloring 6, is q5E(e) = {q5(u),6,(w)}. Thus the coloring 6, induces a coloring q 5 of ~ the edges of G. *Present address: Department of Mathematical Sciences, New Mexico State University, Las Cruces, NM 88003 USA.

17

18

K. A1-Wahabi, R. Bari, F. Haxary and D. Ullman

The coloring 4 is edge-distinguishing, or more briefly admissible if 4~ is one-to-one. The edge-distinguishing chromatic numberX1 of G is the smallest integer r such that G has an admissible r-coloring. This invariant was introduced by Frank, Harary and Plantholt [l].

2

Homomorphisms from P, to I<;

In general, we follow the notation and terminology of the book [2]. Thus let P, be a path on n vertices, C, a cycle on n vertices, and let K;t be the complete graph on n vertices augmented with a loop at each vertex. A homomorphism from a graph G to a graph H is a mapping 7 from V(G)into V(H) such that q(u)q(v)E E ( H ) whenever uv E E(G). Thus a homomorphism 7 induces a map q~ from E ( G ) into E ( H ) by the rule ~ E ( U W ) = q(u)q(v). By analogy with colorings, a homomorphism q is admissible if q~ is one-to-one. Theorem 2.1. For a graph G, the following are equivalent: 1) X l G ) 5 r 2) there is an admissible homomorphism 7 : G

+

Kf.

Proof. By definition, if X1(G) 5 r , there is an admissible r-coloring 4 : V ( G ) + N , . Let f be any bijection from N , onto V(K,*). Define q by q(u) = f (+(u)). Then q is a mapping from V(G)to V ( K , * ) Since . any two (not necessarily distinct) vertices of I<,* are adjacent, q(u)q(v) E E ( K ; ) whenever uv E E(G),so q is a homomorphism. Let uv and u'v' be edges of G such that qE(uv) = r]E(u'v'). Then f(4(u))f(+(v)) = f (#(U'))f (4(v')), so that ~ E ( U W ) = 4,q(u'v'). Since 4 is an admissible coloring, q is also admissible. Conversely and similarly, if q is an admissible homomorphism from G into Kf, then 4(u) = f -1 (q( u ) ) is an admissible r-coloring of G. 0 Corollary 2.1. The edge-distinguishing chromatic number of a graph G is the smallest number r for which Kf contains an admissible homomorphic image of G. Corollary 2.2. Let q : V(G)---+ Ir'f be an admissible homomorphism from G into I<,* and let 6 be an admissible r-coloring of I<,*. Then the function 4 : V(G) + N,. defined by +(u) = qY(q(u)) is an admissible rcoloring of G. This coloring of G is called the r-coloring induced by the homomorphism q.

The Edge-Distinguishing Chromatic Number of Paths and Cycles

3

19

The edge-distinguishing chromatic number of a path

Obviously, the mapping 7 -, xl(Pn)is a non-decreasing integer function of n, since every admissible r-coloring of P, induces an admissible r-coloring of P, for m 5 n. Hence, to know this function, it suffices t o obtain a formula for M , = max{n : x1(Pn) = T } . Theorem 3.1. I f r is odd, .hfT

=

T ~ + T + ~

2

and if r is even, M T

=

r2

'

+4

-

2 . Proof. If T is odd, then every vertex of Kf has even degree, so K,*is eulerian. Let Pn = u1u2 u,, and let C n = w1v2 * vn be an eulerian trail of I<,*, so that w; E V ( K , * )for each i = 1 , 2 , . ..,n. The mapping q defined by q(ui) = vi is easily seen to be an admissible homomorphism from Pn to 117;. Hence

-

-

Since Kf cannot be an admissible homomorphic image of any longer Path, T 2 + T + l = r2+r+2 Mr = 2 2 If T is even, then every vertex of Kf has odd degree. Let G be a 1-factor of I<:. Then G has exactly $ edges. Let G' be any subgraph of G with f - 1 edges, and let H = I<,* - E(G'). Then H has exactly two vertices of odd degree, so H has an eulerian trail Pn of length n, and one can easily see that there is an admissible homomorphism q from P,,into H (and hence from Pn into Kf). Furthermore, no subgraph of Kf with more edges than H has an eulerian trail, since any such subgraph would have a t least four vertices of odd degree. Hence

-

so, for even r ,

r2

+4

M, = 2

.

K.Al-Wahabi, R. Bari, F. Harary and D. Ullman

20

Corollary 3.1. If either 1) r is odd, and

r2 - 2r 2

r2-rt2

2 ) r is even, and

4

+5 <

2

n l

T2+T+2

T2

< n l -

+

9

2 4

2

Of

'

The edge-distinguishing chromatic number of a cycle

The calculation of xl(Cn) is analogous to the calculation of xl(P,), except that it is not obvious that the mapping n -+ xl(C,) is a non-decreasing function of n. This fact will be proved below. First, we obtain for T 2 3 a formula for 7iTr = max{n : x,(C,) 5 T } . Theorem 4.1. I f r is odd,

T2 t T M, = 2

and if

T

is even,

'

T2 Mr = -

2' Proof. If T is odd, K: is eulerian, and as before, we see that any eulerian circuit of Kf induces the required coloring of C, with n = IE(K,*)I but not of any C, with n > lE(Kf)l. Hence

If r is even, then every vertex of KT has odd degree. Let G be a l-factor of I<:, and let H = I<: - E(G). Then every vertex of has even degree, and H is eulerian. Clearly, no subgraph of I<: with more edges than H is eulerian, so an eulerian circuit of H induces the required coloring of C, as above for odd T .

r r2 M , = I E ( H ) (= p(Iq1- 5 = -

2'

Theorem 4.2. The function n

+

xl(Cn), n 2 3, is non-decreasing.

The Edge-Distinguishing Chromatic Number of Paths and Cycles

21

Proof. Let n be an integer such that n > 3 and MT-1 < n < M,.We wish to prove that xl(Cn) = r . It is clear that xl(Cn) 2 r , since n > Mr-1.To prove that xl(Cn)5 r , we need only to prove that there is an admissible r-coloring of C,. It is sufficient, from Corollary 2.2, to find an eulerian subgraph of Iif with exactly n edges. We consider two cases: 1. Let X v- r 5 n < M,,and let H , be an eulerian subgraph of Kf of maximum size. Then we can find the required eulerian subgraph of Kf by removing M,- n 5 r loops from H,.

2. Let M,-1< n < M,- r . In this case, r is odd, since if r were even, it is easily seen that M,-1 = 2 > = M,- r , so our hypothesis is not satisfied. Also, this case does not arise when r = 3. Since r is odd, H , = K:. Let F be a hamiltonian cycle of K,, and let T be a 2-factor of K , - E ( F ) . Then G*= K ; - E ( T ) is connected, since it has a hamiltonian cycle, and every vertex of G* is even, so G*is eulerian. Also, IE(G*)I = M,- r. We can now construct an eulerian subgraph of G with exactly n edges by removing M, - r - n loops from G*. To do this, we must be certain that M,- r - n 5 r , the number of loops in G*.But this follows from the fact that M,- 2r = < = V,-1, since r is odd. 0

9

Corollary 4.1. If r 2 3 and either

2) r is even, and

r2 - r T2
then xi (Cn) = T .

A table of values of xl(Pn)and xl(Cn)follows.

n

1 1

2

3

4

5

6

7

8

910111

12 13 14 15 16 17 18 19 20 21 x1(Pn) 5 5 5 5 5 6 6 6 6 7 xl(Cn) 5 5 5 5 6 6 6 7 7 7 n

22

K. Al-Wahabi, R. Bari, F. Harary and D . Ullman

References [l] 0.Frank, F.Harary and M. J. Plantholt, “The line-distinguishing chromatic number of a graph,” Ars Cornbinatoria 14 (1982), 241-252. [2] F. Harary, Graph Theory, Addison-Wesley, Reading (1969).

Annals of Discrete Mathematics 41 (1989) 23-38 0 Elsevier Science Publishers B.V. (North-Holland)

On the 2-Linkage Problem for Semicomplete Digraphs J. Bang-Jensen Department of Mathematics and Computer Science Odense University Odense, Denmark Dedicated to the memory of G. A . Dirac We consider the problem of finding two disjoint directed paths with prescribed ends in a semicomplete digraph. The problem is NPcomplete for general digraphs as proved in [4]. We obtain best possible sufficient conditions in terms of connectivity for a semicomplete digraph to be 2-linked (i.e., it contains two disjoint paths with prescribed ends for any four given endvertices). We also consider the algorithmically equivalent problem of finding a cycle through two given disjoint edges in a semicomplete digraph. For this problem it is shown that if D is a 5-connected semicomplete digraph, then D has a cycle through any two disjoint edges, and this result is best possible in terms of the connectivity. In contrast to this we prove that if T is a 3-connected tournament, then T has a cycle through any two disjoint edges. This is best possible, too. Finally we give best possible sufficient conditions in terms of local connectivities for a tournament to contain a cycle through af given pair of disjoint edges.

1 Introduction and summary In this paper we consider the following problem known as the 2-linkage problem: Given af digraph

D

and four different vertices

21, 2 2 ,

y1,

312.

Decide whether or not D has two disjoint paths which start in 51 resp. 22 and end i n y1 resp. y2.

This problem is very difficult for general digraphs. In fact, in [4] it is proved that for general digraphs, the problem is NP-complete. 23

J. Bang-Jensen

24

Let us call af digraph 2-linked iff it contains two disjoint paths with given endvertices for any choice of the endvertices. In [9]the problem is solved for acyclic digraphs. For more general classes of digraphs it is not even known whether there is a sufficient condition in terms of the connectivity of the digraph. This should be compared t o the fact that the corresponding problem for undirected graphs is completely solved in [6] and [7]. A corollary of the result in these papers is that if G is a 6-connected graph, then G has a 2-linkage with any given endvertices. This is best possible in terms of connectivity and was first shown in [5]. From the 5-connected non 2-linked graphs one easily obtains 5-connected non 2-linked digraphs. It is easy to see that there can be no sufficient condition in terms of the number of internally disjoint paths between 21 and y1 respectively 22 and yz for general digraphs. This is seen as follows: Let k be any given natural number, k 2 2 and let D be the digraph with vertex set 21, 2 2 , y1, yz and the k 2 vertices z ; j , i, j E {1,2,, . . ,k}. Let the edges be 21

22 Zij

-+

-+

---t

zij+l,

Zil, Zik

-+

y1,

i = 1,2,. . . ,k,

zlj,zkj

---t

y2,

j = 1,2,. . . ,k,

i.= 1,2,. . . ,k, j = 1,2,. . . , k - 1,

.Zij-tZi+lj,j=1,2,

...,k, i = 1 , 2 ,...,k - 1 .

Then the local connectivity from 5; to yi is k for i = 1 , 2 and it is easy to see that D does not contain a 2-linkage with endvertices q , z Z , y1, yz. The fact that the problem seems so difficult for digraphs in general suggests that one should try to look at it for a smaller class of digraphs. In [8] it is shown that if T is a 5-connected tournament, then T has a 2-linkage with given endvertices (i.e., T is 2-linked). In this paper we consider semicomplete digraphs, which constitute a class of digraphs containing the tournaments as a subclass. We show that every 5-connected semicomplete digraph is 2-linked. It is shown that this result is best possible in terms of the connectivity. Then we look at the slightly more specialized problem of finding a cycle through two disjoint edges in a semicomplete digraph. (From an algorithmic point of view this problem is equivalent to the 2-linkage problem for semicomplete digraphs). We show that a 5-connected semicomplete digraph contains a cycle through any two disjoint edges. This result is best possible, too. Then we restrict ourselves to tournaments and show that a 3-connected tournament contains a cycle through any two disjoint edges, and we show

On the 2-Linkage Problem for Semicomplete Digraphs

25

that this also is best possible. The corresponding problem concerning edgedisjoint paths linking two given vertices s ~x2, t o two given vertices y 1 , y2 in a tournament (the weak linkage problem) was completely solved in [l].

2

Terminology and preliminaries

Most of the notion is the same as in [l],but for the sake of completeness we will include all the needed notation here, except for the very standard notation, for which we refer to [2]. A digraph D consists of a pair V ( D ) , E ( D ) ,where V(D)is a finite set of vertices and E(D)is a set of ordered pairs sy of vertices called edges. In our definiton of a digraph we do not allow multiple edges in the same direction be tween two vertices. An oriented graph is a digraph with no cycle of length 2 . A semicomplete digraph is a digraph with no non adjacent vertices. A tournament is an oriented graph with no non adjacent vertices. Thus tournaments are a special subclass of the semicomplete digraphs. If zy is an edge of the digraph D then we say that x dominates y and we will use the notation x + y to denote this. For any subset A of V ( D )U E ( D ) , D - A denotes the subgraph obtained by deleting all vertices of A and their incident edges and then deleting the edges of A still present. We write D - s instead of D - { s } when x E V ( D )U E ( D ) . The subgraph induced by a vertex set A of D is defined as D - ( V ( D )\ A ) and is denoted by D ( A ) . We will often write x E D instead of x E V ( D )or s E E ( D ) , but the meaning will always be clear. A path is a digraph with vertex set { s 1 , 2 2 , . .. ,xn} and with edge set ( 2 1 + x 2 , x 2 + 2 3 , . .,zn-l -, sn}. We call such a path an ( X I ,xn)-path and denote it by s 1 + 2 2 + . . . + 2,. When we just want to write that P is an (3,y)-path we sometimes write this as P(z,y). If P is a path containing vertices s , y in that order then we let P [ s ,y] denote the part of P from x to y. A component D' of a digraph D is a maximal subdigraph, such that for any two vertices x,y E D',D' contains an (5,y)-path and a (y,x)-path. A digraph D is strong or strongly connected if it has only one component. D is k-connected if for any set A of a t most k - 1 vertices, D - A is strong. If a digraph is not strong, then we can label its (strong) components D 1 , D2,. . ,Dk,k 2 2, such that no vertex of D;dominates any vertex of Dj if j < i . For general digraphs this labelling is not unique, but it is so for semicomplete digraphs. Thus here we can speak of the initial component of D

.

.

J . Bang-Jensen

26

namely D1 and the terminal component Dk. The remaining components (if any) are called intermediate components. If A and B are two problems about digraphs with the property that problem A can be transformed into problem B by reversing all the edges in the corresponding digraph, and renaming some vertices and B can be similarly transformed into A , then we say that there is a directional duality between these problems. The following lemma is a special case of Proposition 3.1 in [8],but we prove it here for the sake of completeness.

L e m m a 2.1. Let x , y , u, v be distinct vertices in a semicomplete digraph T . Suppose that T has three internally disjoint (u,v)-paths, and that P I , P2, P3 are internally disjoint ( 2 , y)-paths, which together induce a semicomplete digraph with no (x,y)-path of length 3. Then T has a (u,v)-path Q , which intersects at most two of the paths P I ,P2, P3.

Proof. Let Q be a (u,v)-path in T - { x , y } . We may assume that Q intersects all the paths Pi, i = 1,2,3. Let z1 be the first vertex in the set V ( Q )n V(P1 U P2 U P3) as we go along Q from u towards v , and let zz be the last one. Let i , j E {1,2,3} be chosen such that Z I E P;,z2 E Pj. If i = j and ~1 can reach 2 2 on Pi (i.e., it does not lie after z2 on Pi) then we Put

la’ = Q[u,X I ] U Pi[zl,221 U Q [ ~ z , v ] Otherwise, let x‘ be the immediate successor of x on Pj and let y’ be the immediate predecessor of y on Pi. Then x‘ # y‘ and the only edge between x‘ and y’ is y‘ + x’, since otherwise x -, x‘ --t y‘ --t y would be an (z,y)-path of length 3 in T ( V ( P 1 )U V ( P 2 )U V ( P 3 ) )contradicting the assumption in the lemma. Let

Q ’ = Q [ ~ , ~ ~ ] U P ~ [ ~ ~ , Y ’ ] U { Y ’ UPj[x’,~2] +X’} U&[z2,~] Then Q’is a ( u ,v)-path and it intersects at most two of the paths PI, Pz,P3. 0 Thus the lemma is proved.

Corollary 2.2. Let T be a semicomplete digraph and let x1,x2, y1,y2 be distinct vertices of T , such that T - {xi,yi} has three internally disjoint (~3-i,ys-i)-wthsfor i = 1,2. Then T has a pair of disjoint ( q , y l ) - , ( 2 2 , Yz)-paths.

Proof. By lemma 1 we may assume that T has an (xi,yi)-path of length 3, i=1,2. Let X i + a + b -+ yi be such a path. Since {a,b} does not separate 0 2 3 4 from y3-; in T - {x;,y;} the result follows.

On the ZLinkage Problem for Semicomplete Digraphs

3

27

A best possible sufficient condition for 2-linkages in terms of local connectivity

Theorem 3.1. Let T be a semicomplete digraph and let x1,x2, ~ 1 , 3 1 2be different vertices of T. If T-{xi, yi} has three internally disjoint ( 2 3 4 , y3-i); , has two internally disjoint (xi,yi)-paths, for i = 1 paths and T - { X ~ - y3-i} or 2, then T has a pair of disjoint (XI,y1)-,(52, y2)-pths. Proof. Without loss of generality we may assume that i = 1. Let PI,P2,P3 be internally disjoint (xz,yz)-paths in T - {xl,yl}, chosen such that they are minimal, i.e if x and y are vertices on Pj, which have distance greater than 1 and x is before y on Pj, then the only edge between x and y in T is y + x, j E {1,2,3}. Let P,Q be internally disjoint (zl,yl)-paths in T - ( 2 2 , yz}, chosen such that they are minimal. We may assume that each of the paths P and Q intersects all the paths Pi, i = 1,2,3. Let S = V(P1) U V(P2) U V(P3). Let {a,b} denote the set of the first vertices of P,Q respectively, which are in S. Similarly let {c, d } denote the set of the last vertices of P,Q, which are in S. (We do not decide whether a E P, b E Q or vice versa). If a,b,c,d all lie on the same path Pj, j = 1 , 2 or 3, then it is easy to see from the minimality of Pj,that T has disjoint (z1, yl)-, ( 2 2 , yz)-paths (since then T has a n (51, y1)-path, which intersects only Pj). Thus we may assume that a E P;and c E Pj for some i , j E {1,2,3}, with i # j. Let 'P denote the last of the three (22, y2)-paths (i.e k # i, j). Let x denote the vertex immediately succeeding 2 2 on Pj and let y denote the vertex immediately preceding y2 on Pi. If y -, x then T has an (XI,y1)path, which does not intersect Pk,namely

where a E R1, c E R2 and { R l , & } c {P,Q}. Thus we may assume that the only edge between x and y is x + y. We may assume, by renaming P and Q if necessary, that x E P,y E Q, since otherwise either P or Q does not intersect 2 2 + x + y + y2. Let z denote the vertex of V(P') n (V(P) U V(Q)), which lies closest to y2. Let w E { a , b } be the first vertex of P in S. Then we may assume that x --+ w, since otherwise we easily see that T has the desired paths by arguing analogously to the case when y -+ x. We may also assume that P[w,x] intersects Pk. Let u denote the first vertex on P[w,x], which lies on Pk. Suppose J E P. Then Q and the (x2, yz)-path ( 2 2 -, x + w} U P[w,z] U Pk[z,y2] are disjoint, proving that T has the desired paths. Thus we may assume that J E Q. If y precedes z on Q,then P [ x l ,u] U Pk[u,z] U Q [ J , y1] and 2 2 -, x y + y2 ---f

J . Bang- Jensen

28

are disjoint. Thus we may assume that x precedes y on Q. If Q[y,y1] does not intersect Pk, then R [ z l ,a] u Pi[a,y] u Q[y, yi] does not intersect pk. Here R denotes that of P,Q, which contains a. 1.e we may assume that Q[y,yl] intersects pk in some vertex v. By the minimdity of Q and Pk, v is the vertex immediately preceding z on Pk. Now we can repeat the argument above with v instead of z (y precedes v on Q), and we conclude that T has the desired paths, This completes the proof of the theorem. 0 Theorem 3.1 is best possible as is seen from the semicomplete digraph

T in Figure 1 (all edges not shown go from left to right). The incidence matrix of T is given on the next page.

Figure 1. All double edges are shown. The digraph T is 4-connected, and it does not have disjoint (q,y1)-, (zz,yz)-paths. We leave it to the reader to check this. If we delete the edges y2 --+ q , y 1 -, 52 then we get a tournament TI, which is 4-connected too, and does not have disjoint (21,yl)-, ( 2 2 , yz)-paths. If we reverse the directions of the edges x1 --+ y2 and 52 -,y1 in T', we get a 2-connected tournament T". There are two internally disjoint (xi,y;)-paths in T"- {23-i, y3-i) for i = 1,2, but T" does not have disjoint (21,y1)-, (22,yz)-paths. T' can be modified to give an infinite family of 4-connected tournaments without

On the ZLinkage Problem for Semicomplete Digraphs

29

the desired paths, by substituting a &connected tournament for the vertex bg and choosing the edges appropriately. We leave it to the reader to check this.

51 2

0 0 Y 1 Y1 1 a1 1 a2 1 a3 1 a4 1 a5 1 Y2 1 bl 0 b2 1 b3 1 b4 1 bs 1 52 0

4

1 0 0 1 1 1 1 0 0 0 1 0 1 0 0 0

Y

Y1 a1 a2 a3 a4 a5 Y2 bl b2 b3 b4 b5 5 2

0 1 0 0 0 1 1 1 1 1 0 0 0 1 1 1

0 0 1 0 1 0 0 0 0 1 0 0 0 0 0 1

0 0 1 0 0 0 0

1 1 1 0 0 0 0 0 0

0 0 0 1 1 0 0 0 1 1 0 0 0 0 0 0

0 0 0 1 1 1 0 0 0 1 0 0 0 0 0 0

0 1 0 1 0 1 1 0 0 0 0 0

0 0 0 0

0 1 0 1 0 0 1 1 0 0 0 0 0 0 0

1 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0

1 0 1 1 1 1 1 1 1 1 0 0 0 1 1 0

0 1 1 1 1 1 1 1 1 1 1 0 0 0 1 0

0 0 1 1 1 1 1 1 1 1 1 1 0 0 0 0

0 1 0 1 1 1 1 1 1 1 0 1 1 0 0 1

0 1 0 1 1 1 1 1 1 1 0 0 1 1 0 1

1 1 0 1 1 1 1 1 1 1 1 1 1 0 0 0

Sufficient conditions for a cycle through two disjoint edges in a tournament

In this section we restrict ourselves to the problem of the existence of a cycle through two disjoint edges in a tournament. As mentioned in Section 3 the tournament TI' shows that Theorem 3.1 is best possible in this case, too. Now we want to relax the condition on the number of internally disjoint paths from x; to y; in T - {xs-i,y3-i}, i = 1,2. We will also determine which connectivity will guaranty the existence of the desired paths. Here we already know from section 3, that there exist infinitely many 2-connected tournaments, which do not have such a cycle for suitably chosen edges y1

+

5 2 , y2

+ 51.

By Corollary 2.2, every 5-connected semicomplete digraph has a cycle through any two disjoint edges. The semicomplete digraph T in Section 3 shows that this is best possible in the case of semicomplete digraphs. As we shall see below, this is not the case for tournaments.

J. Bang- Jensen

30

Proposition 4.1. Let el = y1 22,e2 = y2 + 21 be disjoint edges in a tournament T. Suppose that there are four internally disjoint (xi,yi)-paths and three internally disjoint (234, ys-;)-paths in T for i = 1 or 2. Then T has a cycle through el and e2. --.)

Proof. Without loss of generality we may assume that i = 1. By the assumption, there are two internally disjoint (q,yl)-paths in T - ( 5 2 , y2}. Suppose that the local connectivity between 5 2 and y2 in T - { z l , y l } is 1. Let Qi,Q2,93 be internally disjoint (sz,yz)-paths in T. By our assumption, 21 E Q;,y1 E Qj,where i # j. Let x be the vertex immediately preceding y2 on &iand let y be the vertex immediately succeeding 2 2 on Qj. Then x # 21, y # y1, since T is a tournament and contains the edges e1,e2. We must have that 2 -+ y, since otherwise we get a contradiction to our assumption that the local connectivity between 2 2 and y2 is 1 in T - (21, yl}. Now we easily see that T has the desired cycle. Thus we may assume that there are two internally disjoint (22, yz)-paths in T - (51, yl}. If there are 3 internally disjoint ( z i , yi)-paths in T - ( 2 3 4 , y 3 4 } for i = 1 or 2, then the result follows from Theorem 3.1. Thus we may assume that this is not the case; Then for every set PI,.. ,P4 of internally disjoint (21,yl)-paths, 2 2 , y2 will belong t o different paths P;, Pj ,i, j E { 1,2,3,4}. Let u be the vertex immediately preceding y1 on P;, and let v be the vertex immediately succeeding 21 on Pj. Then u # 2 2 and v # 312, since T is a tournament and we know that y1 2 2 , ~-+~ 21. If v u then T has 3 internally disjoint (q,yl)-paths in T - {22,y2} contradicting our assumption above. Thus we may assume that u + v. Then Pi[22,u] U { u -+ v} U Pj[v,yz] and Pk are disjoint, where Ic E 0 {1,2,3,4} - {i,j]. Thus T has the desired cycle and we are done.

.

-+

--.)

The result in Proposition 4.1 is best possible in terms of the number of internally disjoint (zi,y;)-paths in T, i = 1,2. This is shown by the tournament T given with in Figure 2 (all edges not shown go from left to right.) The incidence matrix of T is shown below it. It is easy to see that T is 2-connected and has three internally disjoint (xi,yi)-paths for i = 1,2 and that T does not have a pair of disjoint (21, yl)-, (22, Yz)-Paths. Theorem 4.2. Let el = y1 -+ 2 2 , e2 = y2 + $1 be disjoint edges in a 3-connected tournament T. There exists a cycle in T through el and e2. That is, them exists a pair of disjoint (21, yl)-, (52, yz)-paths.

On the 2-Linkage Problem for Semicomplete Digraphs

31

Y2

Figure 2.

21 X

0 Y 1 91 1 a1 0 a2 1 a3 1 Y2 1 22 1 bl 1 b2 0 b3 1

0 0 1 0 1 1 0 0 1 0 0

1 0 1 0 1 1 0 0 1 0 0 0 1 0 1 1 1 1 1 0 0 1 0 0 0 1 0 0 0 0 0 0 0

0 0 1 0 0 0 1 0 0 0 0

0 0 0 0 1 0 1 0 0 0 0

1 0 1 1 0 0 0 0 0 0 0

1 0 1 1 1 1 1 0 0 1 1

0 1 0 1 1 1 1 1 0 0 0

1 1 1 1 1 1 1 0 1 0 0

1 1 1 1 1 1 1 0 1 1 0

We need some lemmas for the proof.

.

Lemma 4.3. Let X I , ..., x,, yl, .., ys, r,s 5 k be different vertices in a k-connected digraph, and let al, .. , a,, bl, .. , b, be positive integers such that

.

.

n

i=l

j=1

Then there exist lc internally disjoint paths from X = {XI,, . . ,xT} to Y = {yl, . ..,yB} such that precisely a; of the paths start in xi, and precisely bj of the paths end in Y j , for i = 1,2,. .., r and j = 1,2,. . ,s.

.

J. Bang-Jensen This lemma is a consequence of Menger's theorem [3].

L e m m a 4.4. Let el = y1 + 5 2 , e2 = y2 + x1 be disjoint edges in a 3-connected tournament. If there exists an (xi, g;)-path of length 3 in T {x3-;, y3-;} for i = 1 or 2, then T has a cycle through e1,e2. Proof. The proof is by contradiction. Suppose that T is a tournament satisfying the hypothesis of the lemma, but not the conclusion; i.e.,

T does not have a pair of disjoint (21, yl)-, (22, yz)-paths.

(*I

We derive a series of properties that T must have, and finally we show that these result in a contradiction.

For i = 1,2, there are two internally disjoint (x;,y;)-paths in T - { ~ 3 - i ,~ 3 - i ) .

(1)

Proof of (1): Suppose that the local connectivity between x; and y; is 1 in T - {x3-;, y3-i) for i = 1 or 2. Let PI,P 2 , P3 be internally disjoint (x;,y;)-paths in T. Then x3-; E P,, y3-; E P a , where T # s. Let y be the vertex immediately preceding y; on P,, and let x be the vertex immediately succeeding x; on P,. Then since T is a tournament and T contains the edges el, e2, we get that x # y3-;, y # x34. If y + x then P k and P,.[xg-;, y] U {y + z} U P,[x,y3-;] are disjoint, where k E {1,2,3} - { T , s } . This is a contradiction to (*). 1.e. we must have x + y. Now pk and x; + x -+ y + gi are internally disjoint (xi, y;)-paths in T - ( 2 3 4 , y3-;}, contradicting our assumption. There is no (xi, y;)-path of length less than 3 in T-{x3-;,

y3-;}.

(2)

Proof of (2): This follows directly from (1) and (*). Without loss of generality we can assume that T has an (xl,yl)-path of lenght 3 in T - ( 5 2 , yz}. Let x1 -+ x + y -+ y1 be such a path. There is no (xz,yz)-path in TI= T - {xl,x,y,yl}.

(3)

Proof of (3): This follows directly from (*).

Thus TI is not strong. Let TI,. .. ,T, be the strong components of TI. Then 312 E T; and x2 E Tj for some i , j with 1 5 i < j 5 s. Let

A = TI U .- - U Ti,B = Tj U . - .T8, C = Ti+l U - .. U Tj-1, i.e., A is the set of vertices u in TI, for which there is a (u,yz)-path in T', B is the set of vertices v in T', for which there is an (xz,v)-path in TI,

33

On the 2-Linkage Problem for Semicomplete Digraphs

and C is the set of the remaining vertices in T' (possibly C = 0). Then A dominates all of B U C and C dominates all of B. There are 3 internally disjoint (B, A)-paths in T, and there are two internally disjoint (x2,y2)-pths in T - {xl,yl} - C.

(4)

Proof of (4): By the 3-connectivity of T, there are 3 disjoint (B,A)-paths in T, and by (1),(3) and the definiton of C,there are two internally disjoint (z~,y2)-pathsin T - {xl,yl) - C. There exists a n (q,A)-edge and a (B,yl)-edge in T.

(5)

Proof of ( 5 ) : By directional duality it is enough to prove that there exists an (zI,A)-edge in T. Suppose that A dominates 21. Let U = ( 2 1 , ~ ) and V = {z,y,yl}. By Lemma 4.3 T has a system of internally disjoint (U,V)-paths Pl,P2,P3, such that exactly one of these, say PI starts in 21, the other two start in 2 2 , and exactly one of the paths ends in each of x,y and 31. If PI ends in y or y1 then we easily get, by (4), that T has a pair of disjoint ( q , y l ) - , (52, yz)-paths contradicting (*) (PI does not intersect A). Thus PI ends in x, and without loss of generality, we may assume that PI is simply the edge x1 + x. We can assume that P 2 is an (xz,y)-path and P3 is an (xz,yl)-path. Futhermore it is easy to see that P 2 and P3 lie entirely in T(B U {y,y1}). Suppose that y2 -+ x. Then we easily get from (4) that there exists a vertex a E A \ {yz} such that x -+ a and there exists a (y,yz)-path in P4 in T((A U {y}) \ {a}). Let b be the immediate predecessor of y1 on P3. Then b E B and b # z2, since T is a tournament and T contains the edge el. Thus x1 z a b + y1 and P 2 [ 2 2 , y] U P 4 are disjoint (21, y1)-, (22, yz)-paths contradicting (*). Thus x -+ y2. Let u be the immediate successor of 5 2 on P3. As before it is easy to see that u # y1 and u E B. We must have u x, since otherwise {x1 -+ x + u } U P3[u, y1] and P 2 [ 2 2 , y]P4 are disjoint (21, y1)-, (az,y2)-paths contradicting (*). Here P 4 denotes any (y, yz)-path in T(A U {y}). Let U = { 2 1 , 2 2 } and V = {x,y, y1} as before and, use lemma 4.3 to conclude that T has a system of internally disjoint (U,V)-paths Q1, Q2, Q3, such that &I,& start in $1, Q3 starts in 5 2 and exactly one of the paths ends in each of x, y and y1. We can assume that 93 ends in y1, since otherwise we easily obtain a contradiction to (*). We may assume that Q2 is an (zl,yl)-path and that Q1 is simply the edge x1 + x. Now u E Q2, since otherwise 5 2 + u -+ x y2 and Q 2 [ q , y ] U {y y1} are disjoint ( q , y l ) - , (zz,y2)-paths contradicting (*). Let 2) be the immediate successor of 2 2 on 93. Then 2, E B and v # y1 since T is a tournament and contains el. -+

--f

-+

--f

--f

--f

34

J . Bang-Jensen

We must have v + x, since otherwise {XI -+ x + v} U &s[v,y1]and U Q2[u, y ] U P4 are disjoint (21,y 1 ) - , (22, yz)-paths contradicting (*). Now & [ X I , y1] U { y -+ y1) and 2 2 + v + x y2 are disjoint (51,y ~ ) - , (22, yz)-paths. This contradiction to (*) completes the proof of ( 5 ) . {XZ + u }

-+

For every a E A, such that x1 -+ a we have: a separates y but not 2 from y2 in T(A U {x,y}). For every b E B , such that b + y1 we have: b separates 2 2 from x but not from y in T ( B u (2,3)).

(6)

Proof of ( 6 ) : This follows easily from (4), ( 5 ) , the fact that A dominates B, and ( t ) . For every w E C : w

-+

x1

and y1 + w.

(7)

Proof of (7): This follows easily from ( 5 ) , (6) and (*). For every a E A such that x1 -+ a we have: y b E B such that b -, y1 we have: b -+ x.

+ a.

For every

(8)

Proof of ( 8 ) : Suppose that 21 -+ a and a + y for some a E A. Then it follows from (6) and (4) that T - { x l , a , y , y ~ }has an (22, y2)-path contradicting (*). The other half of (8) follows by directional duality. dominates exactly one vertex a E A. exactly one vertex b E B .

21

y1

is dominated by

(9)

Proof of (9): By directional duality it is enough to prove the first assertion above. Let a E A with X I -+ a be arbitrarily chosen. Let

A1 = { u E A I there exists a ( u , yz)-path in T(A \ { a } ) } , A2 = A\(A1U{a}). Possibly A2 = 0. Then A1 dominates all of A2, there is an ( a ,AI)-edge and an (Az,a)-edge in T (if A2 # 0). If A2 # 0, then A2 dominates 21, for otherwise ( 5 ) , (6) and (8) imply that T has a pair of disjoint ( q , y ~ ) - , (22, yz)-paths contradicting (*). Suppose that 2 1 dominates a vertex a' E A ] . Then, by (6) and the fact that X I --+ a, a' + y , but by ( 8 ) and its proof we know that this contradicts (*). Hence A1 dominates 21, and (9) is proved. Let a E A, b E B be the uniquely determined vertices, such that 21 -+ a, b + y1. Let B1, B2 be defined analogously t o the definition of Al, A2 above. A2

dominates x, and y dominates

B2.

(10)

On the ZLinkage Problem for Semicomplete Digraphs

35

Proof of (10): Suppose that x -+ w for some w E A2. Then 21 x -+ w -+ b -+ y1 is a path in T. By (6), 2 2 can reach y in T minus that path. Since y -+ a , by ( 8 ) , and a can reach y2 in T(A1 U { a } ) by the definiton of A and A1, T has a pair of disjoint (xl,yl)-, (s2,yz)-paths contradicting (*). The last assertion of (10) follows by directional duality. -+

x

+

Y2, 2 2

+

Y.

(11)

Proof of (11): Suppose that y2 -+ x. By (4), (6) and the fact that a separates y from y2 in T(A U {y}), there exists a vertex u E A1, such that x --$ u and does not separate a from y2 in A. Thus it is easy to see from (6) and (8) that T has an (za,yz)-path, which is disjoint from the path x1 -+ x -+ u -+ b -+ y1, contradicting (*). The last part of (11) follows by directional duality. A1

= (921, B1 = i.2).

(12)

Proof of (12): Suppose A1 # {yz}. We must have that A1 \ {y2} dominates a (and thus a -+ y2), since otherwise (4) and (11) imply, that T has an (zz,ya)-path, which is disjoint from the path 21 -+ a w -+ y -+ y1, where a -+ w and w E A l , contradicting (*). (Note that by (6), A1 dominates y). Now A1 \ {yz} dominates a, 21,y. Thus by the 3-connectivity of T, there exists an edge 2 -+ u, where u E A1 \ {yz}. I.e., 21 -+ x --+ u -+ b -+ y1 and 2 2 -+ y -+ a y2 are disjoint paths by (8) and (ll),contradicting (*). The other part of (12) follows by directional duality. ---f

-+

Now we are ready to finish the proof of Lemma 4.4 by deriving a contradiction to (*) from (1)-(12). Since x,y1 dominate 2 2 and T - {y, b } is strong, we get from (12) that 2 2 -+ 21. From this and (7), (9) and the fact that T - {a,x} is strong we get that x1 must dominate a t least one vertex u in B2 ( b -+ 21 by (*)). Let B2 = Bi U B!, where

Bb = {w I There exists a (w,b)-path inT(B2 U { b } ) } . Then Bi dominates B{ and {q},since otherwise there exists an (x1,yl)path, which does not intersect the path 5 2 -+ y -+ a -+ 312. In particular u E B;. Now y, y1, b and Bi dominate B; by (9) and (10). Since T { 2 1 , 2 2 } is strong, there exists a vertex E B t such that w -+ 2. I.e., x1 -+ a -+ b -+ y1 and 2 2 -+ y -+ w -+ x -+ y2 are disjoint ( q , y l ) - , (22, yz)-paths, contradicting (*). This contradiction proves the lemma. 0

Proof of Theorem 4.2. Let 4,Pz,P3 be internally disjoint (21, y1)-paths in T, chosen such that they are minimal. By the proof of (1) above we

J . Bang- Jensen

36

easily see that we may assume, by Lemma 4.4, that PI and P2 lie in T {22,y2}. By Theorem 3.1 and the analogous argument showing that T has two internally disjoint (22, y2)-paths in T - (21, yl}, we may assume that P3 contains a t least one of 2 2 , ~ ~If. 22,y2 both lie on P3 then it is easy to show the existence of the desired cycle from the minimality of P3 and the fact that T is a tournament and contains el and e2. Thus we may assume without loss of generality that P3 contains 2 2 but not y2 (the proof in the opposite case is similar). We may assume, by Lemma 4.4, that the shortest (zi, yi)-path in T - {x3-;,y3-;} has length a t least 4. Let Q be an (22,yz)-path in T - {q,y1}. We may assume that Q intersects all the paths&, i = 1,2,3. Let a bethelast vertexofQ, whichlieson PlUPzUP3. Let j E {1,2,3} be chosen such that a E Pj. If j = 3 and a lies after 2 2 on P3 then it is clear that T has the desired cycle. Thus we may assume that this is not the case. Let y be the vertex immediately preceding y1 on P3 and let 2 be the vertex immediately succeeding 21 on Pj. Then y # 2 2 , since T is a tournament and contains the edge el. By Lemma 4.4, we may assume that y + 2. Let i E (1,2} \ { j } . Then the paths P; and P3[22, y] U {y + x} U Pj[x,a]U Q [ a ,yz] are disjoint. This proves Theorem 0 4.2. The tournament TI‘ in section 3 shows that Theorem 4.2 is best possible in terms of the connectivity. Remark. Actually Proposition 4.1 and Theorem 4.2 hold for a more general class of semicomplete digraphs than just tournaments. They hold for a semicomplete digraph D and special vertices 2 1 , 2 2 , y1, y2, satisfying that the only edge between 2 1 and y2 is y2 21 and the only edge between 2 2 and y1 is y1 2 2 . This can be seen by inspecting the proofs above, especially the places where we note that T is a tournament that contains the edges e1,e2. --+

--+

In a forthcoming paper with C. Thomassen we prove that the 2-linkage problem for semicomplete digraphs is in P,i.e., there exists a polynomial algorithm for the problem for semicomplete digraphs.

References [l] J. Bang-Jensen, “Edge-disjoint in- and out-branchings in tournaments and related path problems.” Submitted to J . Combin. Theory (B).

[2] J. A. Bondy and U. S. R. Murty, Graph theory with applications, Macmillan Press (1976).

On the 2-Linkage Problem for Semicomplete Digraphs

37

[3] G. A. Dirac, “Extensions of Menger’s theorem,” J . London Math. SOC. 38 (1963), 143-161. [4] S. Fortune, J. Hopcroft and J. Wyllie, “The directed subgraph homeomorphism problem,” Theoretical Computer Science 10 (1980), 111-121. [5] H. A. Jung, “Eine Verallgemeinerung des n-fachen Zusammenhangs fur Graphen,” Math. Ann. 187 (1970), 95-103. [6] P. D. Seymour, “Disjoint paths in graphs,” Discrete Math. 29 (1980), 293-309. [7] C. Thomassen, “2-linked graphs,” Europ. J . Combinatorics 1 (1980), 371-378. [8] C. Thomassen, “Connectivity in tournaments,” Graph Theory and Combinatorics, Academic Press (1984), 305-313. [9] C. Thomassen, “The 2-linkage problem for acyclic digraphs,” Discrete Math. 55 (1985), 73-87.

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Annals of Discrete Mathematics 41 (1989) 39-52 0 Elsevier Science Publishers B.V. (North-Holland)

The Fascination of Minimal Graphs R. Bodendiek Institute of Mathematics and its Didactics PH Kiel Kiel, FRG

K. Wagner WodanstraBe 57 5000 Cologne 91 FRG

Dedicated to the memory of G. A . Dirac The paper is a survey on properties of minimal graphs not embeddable in a given surface. In particular, the so-called spindle-surface 61 is

investigated,

1

Introduction

During the last 20 years many papers have been published which deal with new and difficult problems arising from the theorem of Kuratowski. As is well known the theorem of Kuratowski says briefly: A graph G is not planar, i.e., it is not embeddable in the plane, if, and only if, G contains a subdivision of K5 or K3,3.l In this formulation the theorem of Kuratowski is obviously an embedding-criterion for graphs in the plane. By means of the subdivision relation > 1 the theorem of Kuratowski can be formulated in a quite different way. In order to describe this formulation let G and N be two graphs. Then we can define: G >1 H if, and only if, G contains a subdivision of H . If I' is a non-empty set of graphs, and if > is a partial ordering on r, we call the set of all >-minimal graphs of I? the minimalbasis of I', denoted by M ( r , >). If especially I? is the set of all graphs not embeddable in a given surface 7 , then we also call the >-minimalbasis M(I',>) the minimalbasis of F. It will usually be denoted by M ( F ,>). Then obviously the theorem of Kuratowski is equivalent with the proposition: M(plane, > I ) = {A's, K3,3}. That means that the theorem of Kuratowski in this formulation is a theorem of the theory of minimalbases. 'In this paper, except where otherwise noted, all graphs mentioned are assumed to be finite and undirected.

39

R. Bodendiek and K. Wagner

40

About 50 years ago D. Konig, in [4], asked for the analogue of Kuratowski’s theorem for the torus. In the following we shall consider a simpler surface than the torus; the so-called spindle-surface Q1.

The spindle-surface

2

Generally, a spindle-surface B = B(G) is defined by means of a connected graph G.One obtains the spindle-surface G(G) from G by blowing up each edge k of G to a spindle B ( k ) ; i.e., to a surface G(k) with two peaks a and b being topologically equivalent with the sphere. The two peaks of E ( k ) are identical with the end points of k. The vertices of G (that are the peaks of all G ( k ) for each edge k of G) are called the singular points of G(G). If G is a graph consisting of only one vertex with one loop, and if we blow up this loop, the spindle-surface we obtain is always denoted by (71 can also be thought of as a surface arising from the torus by contracting one parallel of latitude to exactly one point (Figure 1).

Figure 1.

Y

2

Figure 2. Figure 2 shows a simple model of 81. It is a digon bordered by a doubleedge above and below. The vertices of the digon are equal to the singular point of 41. Finally, two points lying above the other on the borderline are identified. The surhce 5; is the only spindle-surface with exactly one singular point. 2&

can also be defined as a pseudosurface. See [6].

The Fascination of Minimal Graphs

41

Evidentially, on the basis of the definition of S1 every graph, being Furthermore, it follows embeddable in 91, is embeddable in the torus, from the rectangle model of the Klein bottle that 61 arises from the Klein bottle by contracting a meridian to exactly one point. Therefore, it is obvious that every graph, being embeddable in GI,is embeddable in Klein’s bottle, too. So we can say that with respect to embeddability, the spindlesurface 91 lies between the plane and the torus as well as between the plane and the Klein bottle. For that reason, 61 is a very interesting surface. Similar to the above formulation of Kuratowski’s theorem we have the following theorem for an arbitrary surface 3.

Theorem 1. A graph G is not ernbeddable in 3 ii and only ii it contains a subdivision of at least one graph of the >l-rninirnalbasis M ( F ,>I). The real problem here is to determine all graphs of M ( F ,> I ) explicitly. In order t o solve this problem, we want to study the minimal graphs with the aim to obtain some interesting necessary and sufficient conditions for a graph to be an element of M(3,>1). Let G be a member of M ( F ,>I). Then it necessarily follows that G is not embeddable in F,and that each vertex of G is incident with a t least three edges of G. This means: the minimal degree of G is at least 3. The first property of G follows directly from the definition of M(3,>1). The second property of G follows from the >I-minimality of G. The last conclusion is not trivial, for its validity is based on the property of 3, which is called preservation for embedding of 3. The concept of preservation of embedding of F is defined on the following way: Let G and H be two arbitrary graphs with G > 1 H . A surface F is said to preserve embedding with respect to >1 if, and only if, the embeddability of G in 3 implies the embeddability of H in F. Equivalently, if a subdivision of H can be embedded in T , then so can H . It is clear that all orientable and non-orientable surfaces and the spindlesurface 91 have got the property of preservation of embedding. It is also simple to prove that the spindle-surface G(K3)- K3 is the complete graph of order 3 - does not preserve embedding, provided that no edge of an embedded graph passes through a singular point of G(K3). In order to show this, we consider the two graphs H = K 5 and G = U(K5),where G is a subdivision of K 5 . 31f a graph G is embeddable in 61, we can assume without loss of generality that the singular point of 01 is a vertex of G . It follows immediately from this assumption that no edge of G passes through the singular point of GI.

R. Bodendiek and K. Wagner

42

We obtain G from H by adding two new vertices, x and y , and by replacing the old edge e = ( a ,b ) by three new edges el = ( a , z), e2 = ( 2 , y ) , and e3 = ( y , b ) . Then we see at once that although G is embeddable in G(113) with G > 1 H ,the graph H = 115 is not embeddable in G ( K 3 ) . Therefore, we only consider surfaces with the property of the preservation of embedding. In order to prove a further necessary condition for graphs of M ( F , >I), let G' be an arbitrary graph of M ( F , > l ) . Because of the >I-minimality of G', the graph G' - k is embeddable in F for each edge k of G'. It is not difficult to prove that these three above mentioned conditions are sufficient. In other words: It can be easily shown that a graph GI, fulfilling these three conditions, is an element of M ( F , > I ) . Summarizing, we have the following embedding-criterion for any surface F with the property of the preservation of embedding. Theorem 2. A graph G belongs to M ( F , > I ) if, and only if, the following three conditions hold: 1. G is not embeddable in T . 2. The minimal degme of G is at least 3 .

3. The graph G - k is embeddable in

F for each edge k of G .

The problematic nature of Kuratowski's theorem becomes evident from Theorems 1 and 2. According to Theorem 1, we can solve the embedding problem in the sense of Kuratowski with help of the minimalbasis. So we have to determine all graphs of the minimalbasis explicitly. As a result of Theorem 2, this problem seems not solvable without the help of a further embedding criterion. One can very well demonstrate this apparent dilemma investigating the examples of the plane and the spindle-surface GI,because the two following easy embedding cri terions hold. Theorem 3. A graph G is planar if, and only if, there exists at least one vertex a of G fulfilling the following condition: The graph G - a is embeddable in the plane, so that all neighbouring vertices of a in G are lying on the boundary of a face4 (region, or country) of G - a .

For example, it follows directly from this criterion and Theorem 2 that the two Kuratowski graphs K 5 and K 3 , 3 belong to the >I-minimalbasis 'It is well-known that a planar graph G divides the plane into connected components, which we call the regions, or faces, or countries of G.

The Fascination of Minimal Graphs

43

M(plane,>1). One considers the graph K5 - a = 1<4 for each vertex a of K5. It is obvious that the vertex a and all four vertices of 114 are adjacent. As the I i 4 only has a triangular embedding in the plane, the four vertices of 114 cannot lie on the boundary of one of the four faces of K 4 in which 114 divides the plane. Therefore, 115 is not planar. The proof of the non-planarity of K 3 , 3 is similar, so we can omit it. According t o Theorem 2, 115 and K ~ , are J minimal. If we cut the spindle-surface Q1 in the singular point of 61,then we find out directly that Q1 and the sphere are topologically equivalent. For that reason, we obtain the following analogous embedding criterion for GI.

Theorem 4. A non-planar graph G is embeddable in P1 if, and only if, there exists at least one vertex a of G fulfilling the following condition: The graph G - a is embeddable in the plane so that all neighbouring vertices of a in G are contained in the union of the boundaries of two faces of G - a . While Theorems 1 and 2 show that the embedding problem and finding a minimalbasis in Kuratowski’s theorem are equivalent, there arises a fascinating alternative from Theorems 3 and 4.

3

Applications t o the minimalbasis of the surface

It easily follows from Theorems 2 and 4 that the famous Petersen graph of Figure 3 belongs to the minimalbasis M(G1, >I).

P

h

Figure 3. can be shown in the following way: The non-embeddability of P in The two graphs P - a and P - b, depicted in Figure 4, obviously contain

44

R . Bodendiek and K . Wagner

P-b

P-a Figure 4.

a subdivision of K 3 , 3 each. Therefore, they are not planar. In accordance with Theorem 4, the graph P is not embeddable in GI. In order t o show the >I-minimality of P with help of Theorem 2, we have to prove for each edge Ic of P that the graph P - Ic is embeddable in 91. As P is symmetric, we only have to investigate the graph P - k with k = ( a ,b ) (Figure 5 ) .

b

-

R

Figure 5 . Since the graph (P - k) - c is planar, and since the three vertices of P, being joined by an edge to the vertex c, are contained in the union of the boundaries of two faces of (P- k) - c , the graph P - k is embeddable in 91, in accordance with Theorem 4. It follows from Theorem 2 that P is an element of M ( & , >I).

The Fascination of Minimal Graphs

45

Analogously, we can prove that the graphs5 G I = K5 U Ks, G2 = Kq* 1* K4, G3 = K3 * KZ* K3, G4 = 1 * K3,3, G5 = K3,5, G6 = K6,and the graphs G7,Gg,. . ., G12,depicted in Figures 6, 7, 8, and 9, belong to the minimalbasis M ( & , >I).

Figure 6. While according to Kuratowski's theorem, the >l-minimalbasis of the plane only consists of two graphs, the >l-minimalbasis of 61 has got a t least G2,.. ., G12,and the Petersen graph P. We these 12 minimal graphs G I , believe that the number of >l-minimal graphs of M(&,>I)is about 100. In 1975, Wang showed in [5] that the >I-minimalbasis M(proj. plane, > I ) consists of more than 100 minimal graphs. Recent investigations at Ohio State University tell us that the >I-minimalbasis M(torus, > I )has got more than 800 minimal graphs. For that reason, we are looking for a method to simplify the investigations. Toward that end, the authors have succeeded in reducing the large number of >I-minimal graphs of M(proj. plane, > I ) by a suitable refinement of the relation > I (See [l]and [a]). The reduced ' G I consists of two disjoint complete graphs of order 5 each; Ga and G3 consist of two complete graphs of order 5 being stitched together in a vertex, respectively in an edge; Gb consists of a K3,3 and one further vertex being joined by edges to all vertices of K3,3. Figures 6 and 7 show the graphs G T ,Gg, and G I Odrawn , in the model (circle with diametrical identification) of the projective plane.

46

R . Bodendiek and K . Wagner

G9

GlO

Figure 7.

3

b 7

4

Figure 8.

a

The Fascination of Minimal Graphs

a

A-

47

a

2

4

6 Gi2

Figure 9. number of graphs in the new minimalbasis is exactly 12. In order t o explain this new method, we first define three simple elementary relations, R2, R3, and Rq, on non-empty sets I' of graphs: Let G and H be two graphs of I'. Then &(G) = H means that the graph H arises from G by contracting an edge k of G;6 &(G) = H means that H arises from G be replacing a trihedral { a } * {el,e2,e3} of G by a triangle (e1,e2,e3,e1). Finally, &(G) = H means that H arises from G by replacing a double trihedral of G by a double triangle (Figure 10). Now we are able to-define the relation > j for i = 2, 3, 4 in the following way: Let G and H be arbitrary graphs. Then G > j H if, and only if, either G = H or there exists a sequence of graphs GI, GZ, . . , G, (n 2 2) with GI = G and G, = H so that G, = Rjv(Gv-l) with i, E (2,. . . ,i} for each Y = 2, . . . , n. We call the set of all >j-minimal graphs of a non-empty set I' of graphs the >i-minimalbasis of I' for each i = 2, 3, 4. This >i-minimalbasis is denoted by M(I',>i). In analogy to i = 1, the >j-minimalbasis is denoted by M ( 3 , >i),if 'I is the set of all graphs not embeddable in a given surface F. Since, according to the definition of the relation > j (j = 1, 2, 3, 4), G >;+I H always follows from G > i H for

.

~~

'We can assume that no vertex of G is a common neighbouring vertex of the two endpoints of k. So we avoid double edges in H, which could arise by applying R2 to G. We are similarly not allowed to apply Rs or RI to G, if double edges arise in H .

R. Bodendiek and K . Wagner

48

H-x t

Figure 10.

i = 1, 2, 3, the following inclusion chain holds for every surface:

For example, the >r-minimalbasis M(plane, > 4 ) = ( K 5 ) consists of only one graph; since R4(1c3,3) = K 5 , the equation M(plane, > 4 ) = ( K 5 ) is valid. Furthermore, the >4-minimalbasis M(proj. plane, > 4 ) consists of exactly 12 graphs as shown in [l]. Finally, the authors have explicitly determined 23 different >4-minimal graphs of M(torus, >4). We believe that the number of >d-minimal graphs of M(torus,>q) is less than 30. Another interesting property of the Petersen graph is worth mentioning. Since R i ( P ) = K6, and 1<6 E M(&,>i)for i = 1, 2, 3, 4, we have P E M(G1, >2) and P !$ M(Q1, >;) for i = 3, 4. Finally, it is not difficult

49

The Fascination of Minimal Graphs

.

t o prove that the >l-minimal graphs GI, G I , . . , G12 are also >l-minimal for i = 2, 3, 4. After these few remarks, we ask the question: How are we able to decide whether a graph G belongs t o M ( 3 , >;) for i = 1, 2, 3, 4? We get the answer by generalizing Theorems 1 and 2.

Theorem S (Embedding Criterion). A graph G is not ernbeddable in 3 if, and only if, there exists at least one graph H E M ( 3 , > j ) with G >; H for i = 1, 2, 3, 4.

It is very easy to formulate the corresponding criterion for minimal graphs : Theorem 6. A graph G is an element7 of M ( 3 , > j ) for i = 1, 2, 3, 4 iJ and only if, the following four conditions are fulfilled: 1. G is not embeddable in 3. 2. The minimal degree of G is as least 3.

3. The graph G - k is embeddable in F for each edge Ic of G. 4. Each graph Rj(G) is embeddable in 3 for j E (2,. with i = 2, 3, 4.

.., i }

Similarly to Theorem 1, we have to postulate that the surface 3 preserves embedding with respect to the relations R2, R3, and R4. It is obvious that the orientable and the non-orientable surfaces and spindle-surface Q1 have the property of preservation of embedding. The above mentioned inclusion chain induces the question of how we can determine the graphs of M ( F , >3) - M ( F , > 4 ) , of M ( 3 , > 2 ) - M ( 3 , >3), or of M ( 3 , > I ) - M ( 3 , > 2 ) with the help of the graphs of M ( F , >4). The following answer has been given in [2].

Theorem 7. Let 3 be a surface. For every graph G E M ( 3 , > ; ) M ( F , > i + l ) for i = 1, 2, 3, there exists a graph H E M ( F , > j + l ) with G = R;Y(H) ( n E N ) . For example, it follows for i = 2 that we can obtain every graph of M ( 3 , >2) - M ( 3 , >3) from a graph H E M ( F , >3) by replacing successively all triangles of H by trihedrals with a new vertex. Since this method must finish, it follows immediately that M ( 3 , > 2 ) must be finite, if M ( 3 , > 3 ) is finite. We can similarly conclude that the finiteness 'For i = 1, also see Theorem 2.

R. Bodendiek and I(. Wagner

50

of M ( F , >I)follows from the finiteness of M ( T ,> 4 ) . For that reason, we only have to consider the >d-minimalbasis M ( F , > 4 ) , if we want to solve the problem of minimalizing for a surface T . It is well-known that this minimalizing problem has been completely solved only for T equal to the plane and F equal to the projective plane. Finally, we consider the two graphs G' = Gl2 E M ( & , >4) (Figure 9) and GI' (Figure 11).

Figure 11. These two graphs contain the bipartite graph K 3 , 4 as subgraph with the two disjoint vertex sets V1 = {2,4,6} and Vz = {1,3,5,7}, so that each edge of the subgraph #3,4 has one point in Vl and the other in Vz. Furthermore, G' contains a subgraph Q', which consists of one edge with the end points a and b - the so-called center of G' - and two triples of edges in a or b. In the same way, GI' contains a subgraph Q" with the property that Q' and Q" are isomorphic. In [3], Q' and Q" are called Querstiiclie of G' or G" in relation to K3.4. The only difference between G' and G" is that the edge triple of G' has end-points 1, 2, 3, while the edge triple of G" has endpoints 1,2,4. One could call G' and G" twins although G' E M ( & , > 4 ) and G" 4 M(G1, > 4 ) . The last statement can be proved in the following way: According to Theorem 4, it follows immediately that G" is not embeddable in 91. Furthermore, we can easily see that the graph H arising from G" by deleting the three edges ( a , b ) , (1,2), and (1,4) is an element of A4(G1,>1) (Figure 12). Therefore, H is the fourteenth graph of A I ( G 1 , >I). The gra.ph arising from H by contracting the edge (1,6) of 11 is obviously equa.1 to

51

The Fascination of Minimal Graphs

Figure 12.

G5= #3,5

E M(G1, > 4 ) , so that the twin

GI‘does not belong to M ( Q 1 ,>4).

References [l] R. Bodendiek, H. Schumacher and K. Wagner, “Die Minimalbasis der Menge aller nicht in die projektive Ebene einbettbaren Graphen,” J. reine angewandte Mathematik 327 (1981), 119-142. [2] R Bodendiek, H. Schumacher and K. Wagner, “Uber Relationen auf Graphenmengen,” Abh. Math. Sem. Univ. Hamburg 51 (1981), 232-243.

[3] R. Bodendiek and I<. Wagner, “Zum Basisproblem der nicht in die projektive Ebene einbettbaren Graphs 11,” J. Combin. Theory 17 (1974), 249-265. [4] D. Konig, Theorie der endlichen und unendlichen Graphen, Leipzig (1936). [ 5 ] C. S. Wang, Embedding graphs in the projective plane, Dissertation,

Ohio State University (1975).

[GI A. T. White, Graphs, Groups and Surfaces, North-Holland, New York (1973); revised edition 1984.

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Annals of Discrete Mathematics41 (1989) 53-70 0 Elsevier Science Publishers B.V. (North-Holland)

Optimal Paths and Cycles in Weighted Graphs J. A. Bondy and G. Fan Department of Combinatorics and Optimization University of Waterloo Waterloo, Ontario, Canada

Dedicated t o the m e m o y of Gabriel Dirac A graph is called a weighted graph if each edge e is assigned a nonnegative number w ( e ) , called the weight of e . In this paper, some theorems on the existence of long paths and cycles in unweighted graphs are generalized t o weighted graphs and some related problems are proposed.

1

Introduction

Let G = (V,E ) be a simple graph (without loops or multiple edges); G is called a weighted graph if each edge e is assigned a nonnegative number w ( e ) , called the weight of e. For any subgraph H of G, V ( H )and E ( H ) denote the sets of vertices and edges of H , respectively. The weight of H is defined by

A cycle (path) is called an optimal cycle (path) if it is a cycle (path) of maximum weight. For each vertex v E V , N H ( V )denotes the set, and dH(v) the number, of vertices of H that are adjacent with v. We define the weighted degree of v in H by W(V,H)

=

c

w(vh).

h€NH(v)

When no confusion occurs, we will denote N G ( v ) , d c ( v ) , and w(v,G) by N ( v ) , d ( v ) , and w(v), respectively. An (2,y)-path is a path whose endvertices are 2 and y ; a z-path is one whose initial vertex is I. An internal vertez of a block of a graph G is a vertex which is not a cut vertex of G.

53

J . A. Bondyand G. Fan

54

An unweighted graph can be regarded as a weighted graph in which each edge e is assigned weight w(e) = 1. Thus, in an unweighted graph, W ( W ) = d(w) for every vertex w, and an optimal cycle is simply a longest cycle. The aim of this paper is to generalize the following three theorems on the existence of long paths and cycles in unweighted graphs to weighted graphs.

Theorem A (Dirac [2]). Let G be a graph and d an integer. If d(w) 2 d for every vertex v in G, then G contains (i) a path of length at least d , and (ii) if d 2 2, a cycle of length at least d 4- 1. Theorem B (Erdos a n d Gallai [3]). Let G be a 2-connected graph and d an integer. Let x and y be two distinct vertices of G . If d(v) 2 d for all v E V ( G )\ { x , y } , then G contains an ( 2 ,y)-path of length at least d . Theorem C (Dirac [2]). Let G be a 2-connected graph and d an integer. If d(w) 2 d for every vertex w in G, then G contains either a cycle of length at least 2d or a Hamilton cycle.

2

Weighted generalization of Theorem A

Theorem 1. Let G be a connected weighted graph on at least two wertices, z a specified vertex of G, and d a real number. Suppose that W ( W ) 2 d for every w E V ( G )\ ( 2 ) and w ( e ) > 0 for every e E E ( G ) . (i) Then G contains a z-path of weight at least d. Moreover, i f G contains no t-path of weight more than d , then each block B of G (a) is complete, (b) includes z , and (c) is weighted so that w ( w ) = (Y for all w E V ( B )\ { z } and w ( u v ) = p for all u,v E V ( B )\ { z } , where a t P(IV(B)I - 2) = d. (ii) If d(v) 2 2 for every v E V ( G )\ { t } , then G contains a cycle of weight more than d .

Proof. (i) By induction on IV(G)l. If IV(G)l = 2, the result is trivially true. Suppose now that IV(G)l 2 3. Let G' = G - z . If G' is disconnected, with components HI,H2, . , H,, then applying the induction hypothesis to the subgraph of G induced by V ( H i ) U ( 2 ) for each i, 1 5 i 5 rn, completes the proof. So we may suppose that G' is connected. Now let a = max{ w ( m ) : w E N ( t ) } , and choose z' E N ( z ) such that w ( t z ' ) = a. Set d' = d - a. Then w(w,G') 2 d' for all w E V(G'). By the induction

..

55

Optimal Paths and Cycles in Weighted Graphs

hypothesis, G' contains a a'-path P' of weight a t least d', and so G contains the a-path P = ta'P' of weight at least a d' = d. If G contains no z-path of weight more than d, then the maximum weight of a %'-path in G' is'exactly d'. Moreover, by the induction hypothesis, G' has the described structure. Let B' be a block of GI. Thus, B' is complete, includes z', and is weighted so that

+

w(wz') = a'

for all w E V ( B ' )\ { a ' } ,

w(uw) = p'

for all u, w E V ( B ' )\ { a ' } ,

and where a'+ p'(lV(B')I - 2) = d'.

If x E V ( B ' )\ { z ' } , then, using (l), w(x,G') = d'. Thus,

Since a > 0, z z E E(G). Moreover, by the choice of a', w ( z x ) = a. Thus,

w(wz) = a

for all w E V ( B ' ) .

It follows that any vertex of B' could have been selected as the vertex J'. This immediately implies that G' = B' and that a' = p'. The proof of (i) is completed by setting p = a'. (ii) Let P = woo1 . . .w, be a path in G of maximum weight starting a t 00 = a, and as long as possible subject t o these conditions. Then w, is adjacent only to vertices of P. Let

S = { i : Vizt,

€

E(G)} and j = minS.

Then for each i E S, wow1 ...wiw,wm-l wo = z. By the choice of P,

Summing this over all i E S, we obtain

Let

. . .wi+l

is another path starting at

J . A. Bondyand G. Fan

56

Then

w(C) 2

C w(v;vi+i) t w(v,vj)

2 d t w(vmvj) > d.

i€ s

Thus, G contains the cycle C of weight more than d.

0

Remark 1. It can be seen from the proof of Theorem l(ii) that, even if the condition that w ( e ) > 0 for every e E E ( G ) is dropped, the graph G contains (i) a z-path of weight at least d and (ii) a cycle of weight at least d. Theorem 2 below discusses the corresponding extremal graphs.

A subgraph B is said to be contracted if B is deleted and replaced by a new vertex b joined to every vertex v E V ( G - B ) for which G contains an edge uv for some u E V ( B ) . A pendant triangle is an endblock which is a triangle. Theorem 2. Let G be a connected weighted graph on at least two vertices, z a specified vertex of GI and d a real number. Suppose that w(v) 2 d > 0 f o r every v E V ( G )\ { z } . (i) Then G contains a z-path of weight at least d. Moreover, if G contains no z-path of weight more than d , then the graph G' obtained from G by recursively contracting blocks of weight zero until no such block remains has the following structure: each block B of G' ( a ) includes z , and (b) is weighted so that w(vz) = a for all v E V ( B )\ { z } with vz E E(G) and w(uv) = p for all u , v E V ( B )\ { z } with uv E E ( G ) , where a+@(IV(B)I-2) = d; moreover, (c) i f cr > 0 , then z is adjacent to every vertex of B - z , and i f /3 > 0 , then B - z is complete. (ii) I f d ( v ) 2 2 for every v E V(G)\{z}, then G contains a cycle of weight at least d. Moreover, i f G contains no cycle of weight more than d , then G has a pendant triangle with edge weights zero, zero and d .

Remark 2. In Theorem 2 ( i ) , the description of the graph G' in effect amounts to a complete characterization of the extremal graphs. On the other hand, Theorem 2(ii) provides no such characterization. Indeed, as the following example shows, the problem of characterizing the extremal graphs in Theorem 2(ii) is at least as hard as the problem of characterizing the graphs with Hamilton cycles. Let H be a 2-connected graph on n vertices. Assign a weight of to each edge of H ,where d > 0. Now append t o each vertex v of H an edge vv' of weight d , and attach a pendant triangle with edge weights zero, zero, and d to each vertex v', where the two edges incident with v' are the edges of weight zero. Call the resulting graph G.

5

Optimal Paths and Cycles in Weighted Graphs

57

Then G is a weighted graph on 4n vertices in which w(v) 2 d and d(o) 2 2 for every o E V ( G ) .Moreover, G contains a cycle of weight more than d if and only if H contains a Hamilton cycle.

Proof of Theorem 2. (i) The first assertion is Remark l(i). Thus we only need to prove the second assertion, and it suffices to prove this for graphs that contain no zero-weight block. Let G be such a graph. We use induction on lE(G)I. If (E(G)I = IV(G)l - 1, then G is a tree. Since G contains no zero-weight block, each edge of G is of positive weight. The result follows from Theorem l(i). Suppose now that IE(G)I 2 IV(G)l. If G contains an endblock B which does not include z, let b be the cut vertex of G contained in V ( B ) . From the first assertion, B contains a b-path PI of weight a t least d. Moreover, since G contains no zero-weight block and b # z, there is a (I,b)-path P2 of positive weight. Then PIU P2 is a z-path of weight more than d, a contradiction. This shows that every endblock of G includes z , or equivalently, every block of G includes z. Furthermore, we may suppose that G consists of a single block, for otherwise applying the induction hypothesis t o each block would complete the proof. If every edge of G is of positive weight, then the result follows from Theorem l(i). Suppose that zy E E(G)with w(zy) = 0. Let H = G - zy. We distinguish two cases: CASE1. H contains no z.ero-weight block. By the induction hypothesis, each block of H has the required structure. Case la. H is a block. If z # x, y, then H - z is not complete, and so by (c), /?= 0. Thus, w(xy) = 0 = p. If .z = z,then p(lV(B)I - 2) 2 w(y) 2 d. This implies, from (b), that w(zy) = 0 = a;the same conclusion holds if z = y. In either case, G = H xy has the required structure. Case l b . H has more than one block (in fact, by (a), exactly two blocks). In this case, we must have that w(e) = 0 for every e E E ( H - z), for otherwise an optimal z-path in G would be of weight more than d. This shows that G has the required structure (with LY = d and /3 = 0).

+

CASE2. H contains a zero-weight block. Since G consists of a single block, H is the union of blocks B1, B2, . . . , B,, m 2 2, where z E B1, y E B,, IV(B;)n V(Bi+1)1= 1, 1 5 i 5 m, and IV(B;)n V(Bj)l = 0 for j # i f 1. Without loss of generality, suppose that z f! V(B1 - b), where b is the cut vertex of H contained in B1. If B1 contains a b-path P1 of weight more than d, then extending PI to z yields a z-path in G of weight more than d. Thus, B1 contains no b-path of weight more than d. By the induction hypothesis, B1 has the properties (a), (b), and (c), which implies that B1

J . A. Bondy md G.Fan

58

has a (b,z)-path P of weight d. If there is an edge e of positive weight in some B;, 2 5 i’s m, then e and P lie on a cycle in G of weight more than d, and thus there is a r-path in G of weight more than d. Therefore, w(e) = 0 for all e E U Z a E ( B i ) . This implies that m = 2, y = z , and G = B1 U {zy,by}, where w(zy) = w(by) = 0. The above argument, with the edge zy replaced by the edge by, shows that B1 is a uniformly-weighted complete graph. Thus, G has the required structure (with a = 0 and

P = p7p-I. d (ii) The first assertion is Remark l(ii). We prove the second assertion by induction on IE(G)I. Since d ( v ) 2 2 for all v E V ( G )\ { z } , we have IE(G)I 2 IV(G)I. If IE(G)I = IV(G)I,then each component of G is a cycle and has weight a t least d , with equality only if it contains z. If G contains no cycle of weight more than d, it follows that G must consist of a single triangle with edge weights zero, zero, and d, where w ( z ) = 0. Thus the result holds for IE(G)I = (V(G)I.Suppose now that IE(G)I > IV(G)l. If G is not 2-connected, we apply the induction hypothesis t o a component of G or to an endblock B of G such that z is not an internal vertex of B . This will complete the proof. We may therefore suppose that G is 2-connected. Since w(v) 2 d > 0 for all v E V ( G )\ { z } and IE(G)I > IV(G)I, there are at least two edges of positive weight. If there is an edge e with w(e) 2 d, let e’ be any other edge with w(e’) > 0. By the 2-connectedness of G , there is a cycle containing both e and e’, which is thus of weight more than d. Therefore we may suppose that w(e)


for all e E E(G).

(2)

If w(e) > 0 for all e E E(G),then, by Theorem l(ii), G contains a cycle of weight more than d. Thus consider zy E E ( G ) with w(zy) = 0. If z # z , then d ( z ) 2 3, by (2) and the hypothesis that w(z) 2 d. The same conclusion holds for y. Hence, d ( v ) 2 2 for all v E V ( G- sy)\ { z } . By the induction hypothesis, G - zy contains a cycle of weight a t least d. However, G - zy does not contain a pendant triangle with weights zero, zero, and d, by (2). Thus, G - z y contains a cycle of weight more than d, and so does G. This completes the proof. 0

3

Weighted generalization of Theorem B

In this, and the next, section we need the following notation: Let C be a cycle in G with a fixed orientation. For any two vertices a and b on C, denote by C [ a ,b] the segment of C from a to b determined by this orientation.

Optimal Paths and Cycles in Weighted Graphs

59

Theorem 3. Let G be a 2-connected weighted graph and d a real number. Let x and y be distinct vertices of G. If w(v) 2 d for all w E V ( G )\ {x, y}, then G contains an (x,y)-path of weight at least d. Proof. Let IV(G)I = n. We use induction on n. If n = 3, let u be the third vertex other than x and y. Then the path xuy is of weight w(u) 2 d. Suppose now that n 2 4 and the theorem is true for all graphs on Ic vertices, 3 5 k 5 n - 1. Let H = G - y be the graph obtained by deleting y from G. We distinguish two cases:

CASE1. H is 2-connected. Choose y' E N(y) \ {x} (note that IN(y)l 2 2 by the 2-connectedness of G ) such that 4Y'Y) = m a 4 4VY) : v E "Y)

\ (41.

Then for all D E V ( H )\ {x},

w ( v , H ) = w(v,G)- W(VY)2 d - ~ ( y ' y ) . By the induction hypothesis, there is an (x,y')-path Q in H of weight at least d - ~ ( y ' y ) .Then P = Qy'y is an (5,y)-path in G of weight at least d.

CASE2. H is not 2-connected. Choose an endblock B of H such that x is not an internal vertex of B. Let b be the unique cut vertex of H contained in B and let B' be the subgraph of G induced by V ( B )U {y}. If yb E E(G),then B' is 2-connected and for all w E V(B')\ {y, b}, w(v,B') = w(v,G) 2 d.

By the induction hypothesis, there is a (y, b)-path P' in B' of weight a t least d. Extending P' in H from b to x, we obtain an (x,y)-path in G of weight a t least d. If yb f! E(G), we add yb to B' and set w(yb) = 0. Applying the above argument to the resulting graph, we obtain a (y,b)-path P' of weight at least d. If d > 0, then P' # yb, since w(yb) = 0; if d = 0, then we can choose P' such that P' # yb, since all we need is that w(P') 2 d. This shows that we still have P' C B'. Extending P' as before completes the proof of the theorem. 0

For later use, we need a slight variation, and direct consequence, of Theorem 3. Corollary 3.1. Let G be a nonsepamble weighted graph on at least two vertices and d a real number. Let x be a vertex of G. If w(v) 2 d for all w E V ( G )\ {x}, then G contains a path of weight at least d from x to any other vertex of G.

J. A. Bondy and G.Fan

60

4

Weighted generalization of Theorem C

In this section, we will prove the following theorem.

Theorem 4. Let G be a %connected weighted graph and d a real number. If w(v) 2 d for every vertex v in G, then either G contains a cycle of weight at least 2d or every optimal cycle in G is a Hamilton cycle. Before proving the theorem, we need the following lemma.

Lemma 1, Let C be an optimal cycle in a weighted gmph G. If there exists a path P in G - C , connecting vertices x and y , such that INc(x)l 2 1, INC(Y)l2 1, and "c(4 u NC(Y)I 2 2, then

Instead of proving Lemma 1, we are going to prove the following more general result, which is a weighted generalization of Lemma 2 in [l].

Theorem 5. Let C be an optimal cycle in a weighted graph G. Suppose that there exists a path P in G - C , connecting vertices x and y , such that INc(x)I 2 1, INc(Y)I 2 1, and INc(x) U Nc(Y)I 2 2. Define

X = Nc(z)\ N c ( Y ) , Y = Nc(y) \ N c ( x ) , and 2 = Nc(z)n Nc(y). Then

unless 121= 1 and either X = 0 or Y = 0, in which cases,

..

Proof, Let A = X U Y U Z and suppose that A = {a1,a2,. , a k } , where are in order around C. For each pair of vertices (a;,a;+l), we shall construct from C two new cycles by replacing the segment C[a;, ai+l] with two (ai,ai+l)-paths. These two paths are defined according to four cases:

a;

(i) a;,a;+l E 2. The two paths are a;xPya;+l

and

a;yPxai+l.

Optimal Paths and Cycles in Weigbted Graphs

61

(ii) ai E 2 and a;+l E X or Y.The two paths are

If a;+l E 2 and

a;

E X or Y ,the paths are defined in the same way.

(iii) a; E X and a;+l E Y or a; E Y and a;+l E X. The two paths are two copies of aixPyai+l or a;yPxa;+l. (iv) a;, a;+l E X or a;, a;+l E

Y.The two paths are two copies of

In each case, we have defined two paths t o replace the segment C[a;,a;+l] and hence formed two cycles (in (iii) and (iv) these two cycles are identical). Since there are k pairs of vertices (a;,a;+l) (i = 1, .. ., k), we obtain 2k cycles. In these cycles, every edge of C is traversed exactly 2k - 2 times; every edge from z or y to 2 is traversed twice, every edge from x t o X is traversed four times and, similarly, every edge from y to Y is traversed four times. Now suppose that the path P is traversed 1 times (we determine I later). Then the average weight of these 2k cycles is 1

-@(k 2k

- l)w(C) + 2

+

+

+

4 2)~t 2w(y, 2 ) 4w(z, X) 4w(y, Y ) !w(P)).

Since C is an optimal cycle, its weight is no less than this average weight. Thus ,

We now determine 1. If 121 2 2, then it is not difficult to see that 1 2 2121; if 121 = 1, X # 0, and Y # 0, then I 2 4 (at least twice from Case (ii) and twice from Case (iii)); if 121 = 0, then, noting that 1Nc(z)l 2 1 and INc(y)I 2 1, we see that X # 0 and Y # 0, and so that I 4 (from Case (iii)). Therefore, 1 2max{4,2121~ unless 121= 1 and either X = 0 or Y = 0. In the case 121= 1 and X = 0, we suppose, without loss of generality, that 2 = {al} and Y = ( ~ 2 , .. .,a k } . Since C is optimal,

J . A. Bondy and G. Fan

62

Similarly,

Also, for all i

#

1,k,

0

This completes the proof of Theorem 5.

Remark 3. It is clear that Lemma 1 is an immediate consequence of Theorem 5 . Proof of Theorem 4. Suppose that there exists a n optimal cycle C in G which is not a Hamilton cycle, and let H be a component of G - C. We consider two cases: CASE 1. H is nonseparable. Choose distinct vertices z and y in H (unless IV(H)I = 1, in which case necessarily 2 = y) such that (i) dC(2) 2 1, &(Y)

L 1, and

(ii) w(z,C) 2 w(y,C) 2 w(v,C) for all v E V ( H )\ {z,y}.

If INc(z)U Nc(y)I 1 2, then, by Lemma 1,

w ( C ) 1 2w(Y,c)

+ 2w ( P) ,

where P is an (z,y)-path in H . If IV(H)I = 1, then

w(P)= 0 2 d - ~ ( 9=)d - w(y,C).

63

Optimal Paths and Cycles in Weighted Graphs Otherwise, by the choice of x and y,

w(v, H ) = w(v) - w(v,C) 2 d - w(y,C)

for all v E V ( H )\ {z),

and so, by Corollary 3.1, we can choose P such that

In either case, ( 6 ) yields that

w(C) 2 2d. Thus we may suppose that N c ( z )U Nc(y) = { u } . Since G is 2-connected, there exists a vertex b E V ( C )\ { a } which is adjacent with some vertex J E V ( H )\ {z,y}. By the choice of x and y, for all v E V ( H ) .

w(w,H) = w(v) - w(v,C) 2 d - w(z,C)

By applying Corollary 3.1 to H , we have an (z,z)-path Q in H of weight

Thus ,

+

w(C)= w(C[a,b]) w(C[b,a])2 2(w(ax)

+ w(Q) + ~ ( b z )L) 2d.

This completes Case 1.

CASE2. H is separable. Let B1 and BZ be two distinct endblocks of H , and let b; be the unique cut vertex of H contained in Bi (i = 1, 2). For i = 1, 2, we choose xi E V ( B ; )\ { b i } such that (i) d c ( s ; ) 2 1, and (ii) w(x;,C) 2 w(v,C) for all v E V ( B ; )\ { b j } . It follows that

w(v, B;) = W ( V ) - w(v,C) 2 d - w(Zi,C)

for all

E V ( B ; )\ { b ; } .

For i = 1, 2, we apply Corollary 3.1 to B; and obtain a n (x;,b;)-path P; in B; of weight w(P;)2 d - w(z;,C). (7)

J . A. B o d y and G.Fan

64

If INc(zl)UNc(z2)1 2 2, then let P be an (zl,zz)-path in H of maximum weight. By (7),

w(P)1 w(P1)

+ w(P2) 1 d -

min(w(z1,C), w(z2,C)).

By Lemma 1,

w(C) 12min{w(zl,C), w(z2,C))

+ 2w(P) 2 2d.

If Nc(z1) = Nc(z2) = {ul}, then there exists a vertex u2 E V(C)\{ul} adjacent to some vertex b € V ( H ) . As (V(B1) \ { b ~ }n) (V(B2) \ {bz}) = 8, b cannot belong t o both V(B1) \ { b l } and V(B2) \ {bz}. We suppose, without loss of generality, that b @ V(B2) \ {b2}. Extending the path P 2 in H from b2 t o b, we obtain an (zz,b)-path P‘ in H of weight

w(P’)L w(P2).

(8)

Since C is optimal,

+ w(C[u2,u11) 1 2 ( w ( w 2 ) + w(P’>+ 4 u 2 b ) ) 1 2(w(u1z2) + w(P‘>).

w(C)= w(C[u1, 2121)

Now, using (8) and then ( 7 ) with i = 2, we have

+

+

w(C) 1 2 ( w ( ~ i z 2 ) w(P2)) 1 2d 2 ( ~ ( ~ 1 -~ 422 )2 , C ) ) = 2d. This completes the proof of Theorem 4.

5

0

Open questions

There are many other theorems on paths and cycles which have weighted analogues. We mention just two. One is Pbsa’s generalization [4] of Dirac’s theorem.

+

Theorem D. Let G be a 2-connected graph such that d ( u ) d ( v ) 2 k for every pair of nonadjacent vertices u and v . Then G contaim either a cycle of length at least k or a Hamilton cycle. This theorem suggests the following problem.

Problem. Let G be a 2-connected weighted graph with w ( u ) + w ( v ) 1k for every pair of nonadjacent vertices u and v . Is it true that either G contains a cycle of weight a t least k or every optimal cycle in G is a Hamilton cycle? Another candidate for generalization to weighted graphs is the following extremal theorem due to E r d k and Gallai [3].

65

Optimal Paths and Cycles in Weighted Graphs

Theorem E. Let G be a 2-edge-connected graph on n vertices and e edges. Then G contains a cycle of length at least

3.

We propose the following conjecture.

s.

Conjecture 1. Let G be a 2-edge-connected weighted graph on n vertices. Then G contains a cycle of weight at least 2w G Remark 4. Call a spanning tree T of a graph G a tritree if every fundamental cycle of T in G is a triangle, and a graph with a t least one tritree a trigraph. It is not hard to show that any cycle of a trigraph uses either zero or two edges of each tritree. Let G be a 2-edge-connected trigraph on n vertices, TI,Tz, .. , T,,,the tritrees of G, and w1,w2,, , w, nonnegative real numbers. To each edge e E E ( G ) ,assign the weight

.

..

w(e) =

c{

Then

wi : e E E ( T i ) } . m

i=l

Let C be a cycle of G. Then C uses a t most two edges of each Ti,and so

Thus, if Conjecture 1 is true, the graph G , with this weighting, is an extremal graph. The following is a weaker conjecture than Conjecture 1.

+.

Conjecture 2. Let G be a weighted graph on n vertices. Then G contains a path of weight at least 2w G Remark 5. To see that Conjecture 2 is implied by Conjecture 1, let G be a weighted graph on n vertices, where n 2 2 (the case n = 1 is trivial), and suppose that G is connected (otherwise, we can consider the nontrivial components of G). Let G’ be the weighted graph obtained from G by adding a new vertex y and joining y to every vertex of G by an edge of weight M, where M > w(G). Then G’ is 2-edge-connected. Let C be an optimal cycle of G’. Since M > w ( G ) and C is optimal, C must contain the vertex y. If Conjecture 1 is true, then

24G’) - 2 M G ) + n M ) - 2 4 G ) W(C)2 IV(G’)I - 1 n n Thus, C - y is a path in G of weight w(C) - 2 M 2

2M.

+. 2w G

66

J . A. Bondy and G. Fan

6

Proof of the conjectures for weighted complete graphs

Towards proofs of Conjectures 1 and 2, we settle the case in which G is a weighted complete graph. Before giving this result, we need some more definitions. A graph is called a vertex-weighted graph if each vertex of the graph is assigned a nonnegative number, called the weight of the vertex. If G is a graph on n vertices v1, v2, . , Vn, then G(wl,w2,. . . ,ton) denotes the vertex-weighted graph obtained by assigning to vertex D; the weight wi, 1 5 i 5 n. If w1 = 202 = - wn = t, then the notation is abbreviated to G(t). Suppose that G(w1, w2,. ,wn) is a vertex-weighted graph. Then the induced (edge-) weighted graph WG(w1, w2,. .,w,) is the weighted graph obtained by setting w(Vjvj) = wi w j for every edge v;vj. So WKn(t)is the uniformly-weighted complete graph on n vertices in which each edge is of weight 2t.

..

...

..

-+

.

Theorem 6. Let G be a weighted complete graph on n vertices where n 2 3. Then the maximum weight of a cycle of G is at least %.& n-l , with equality if and only if G = WK,(wl, w2,. . . ,Wn) for some w1, w2, . . . , w, such that C:,l W j =

s.

Remark 6. A complete graph K n is a trigraph whose tritrees are stars (one centered at each vertex), and the weighting of WKn( w1, w2,. . ,w,) is precisely the weighting described in Remark 4.

.

Proof of Theorem 6. If G = WKn(wl,w2,. . .,w,) and then, for any cycle C of G,

c:=lw; = A 9 n-1

7

s.

and equality holds when C is a Hamilton cycle of G. This shows that the maximum weight of a cycle of G is exactly 2w G Conversely, let G be any weighted complete graph on n vertices. Let C be the set of Hamilton cycles of G. Since G is complete, ICI = and each edge of G is contained in exactly ( n - 2 ) ! Hamilton cycles. Thus the average weight of a Hamilton cycle of G is

67

Optimal Paths and Cycles in Weighted Graphs It follows that the maximum weight of a cycle of G is a t least equality only if

m,with n-l

for all c E C.

w(C)= 2w(G) n-1

(9)

a.

Suppose, now, that the maximum weight of a cycle of G is exactly 2w G We first note that the edge weights of G satisfy the triangle inequality: for any x,y,z E V (G).

w(xy) i- w(xz) 2 w(yz)

(10)

Let P be a Hamilton path in G - x connecting y and z. Then

C = xyPzx

and

C'= yPzy

are cycles of G. Since C is a Hamilton cycle of G, we have, by (9), w(zy)

+ w(xz) + w(P) = w(C) =

n-1

2 w(C') = w(yz) + w ( P ) .

Thus (10) holds. Next, we observe that for any four vertices u, x, y, and z, w(xu)

+ w(yz) = w(xy) + w(uz).

(11)

Let P be a Hamilton path in G - {u,y} connecting x and z, and consider the two Hamilton cycles C = xuyzPx

and

C' = xyuzPx.

BY (9),

w(zu)

+ w(uy)t w(yz)+ w(P) = w(C) = w(C') = w(xy) t w(yu)

+ w(uz) + w(P),

and so (11) holds. We now assign, to each vertex x E V ( G ) ,the weight

w, = +(+Y)

t w(x.4 - W(YZ)),

where y and z are any two vertices other than z. By (lo), w, 2 0. To verify that these vertex weights are well-defined (that is, that w2 does not depend on the choice of y and z ) , it suffices to show that, for any u E V ( G )\ {z,y,z}, the assignment determined by u and z is the same as that determined by y and z , that is,

+

+

$(w(xu) w(xz) - w(uz))= i(w(xy) w(xz) - w(yz)).

J . A. Bondy and G . Fan

68

But this follows directly from (11). It remains to check that for any two distinct vertices x and y, Ws

+ Wy =

(12)

W(Z:y).

Let z be a third vertex. Then, by the definition, 'u)z

+

= +(ZY)

W(5k)

- w(y.z))

and

+ W ( Y 4 - 44), and so (12) holds. Therefore, G = W K n ( W 1 , . .,W n ) . Clearly, we have Wy

=

$(.)(YZ)

w2,.

CZ1Wi = 9, since, if C is a Hamilton cycle of G,

0

This completes the proof of Theorem 6.

e, .sj.

Corollary 6.1. Let G be a weighted complete graph on n vertices. Then the maximum weight of a path of G is at least 2w G with equality if and only if G = WKn(t),where t = Proof. If G = W K n ( t ) and t = ,=I,

then, for any path

P,

and equality holds when P is a Hamilton path of G. This shows that the maximum weight of a path of G is exactly 2w G Conversely, let G be a weighted graph on n vertices. As in Remark 5 , let G' be the weighted graph obtained from G by adding a new vertex y and joining y to every vertex of G by an edge of weight M , where M > w(G). By Theorem 6 and the argument used in Remark 5 , G contains a path of with equality only if G' = WKn+l(wo,w1,. . . ,wn), weight a t least where w is the weight of y. Thus, if G contains no path of weight more than then G = W K , ( t ) , where t = M - WO. Moreover, since a Hamilton path of G is of weight we have t = n-l as required. 0

+.

a,

A,

q,

e,,

Added in proof. The authors have recently proved Conjecture 1, and hence Conjecture 2. Moreover, they have shown that every extremal graph for Conjecture 1 is a weighted union of tritrees, as described in Remark 4,

Optimal Paths and Cycles in Weighted Graphs

69

and that the only extremal graphs for Conjecture 2 are the uniformlyweighted complete graphs and the graphs of weight zero. Conjecture 2 has also been verified (earlier, and by different methods) by A. M. Frieze, C. D. McDiarmid, and B. A. Reed.

Acknowledgement We gratefully acknowledge the support of this research by the Natural Sciences and Engineering Research Council of Canada.

References [l] J. A. Bondy, I. Ben-Arroyo Hartman and S. C. Locke, “A new proof of a theorem of Dirac,” Congressus Numerantiurn 32 (1981), 131-136. [2] G. A. Dirac, “Some theorems on abstract graphs,” Proc. London Math. SOC.(7) 2 (1952), 69-81. [3] P. Erdos and T. Gallai, “On maximal paths and circuits of graphs,” Acta Math. Acad. Sci. Hungar. 10 (1959), 337-356. [4] L. P h a , “On the circuits of finite graphs,” Magyar Tud. Akad. Mat. Kutat6 Int. Kozl. 8 (1963), 355-361.

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Annals of Discrete Mathematics 41 (1989) 7 1 -74 0 Elsevier Science Publishers B.V. (North-Holland)

A Note on Hamiltonian Graphs H. J. Broersma Depart men t of Applied Mat hematics Twente University of Technology Enschede, The Netherlands Dedicated to the memory of G. A . Dirac The closures of the graphs satisfying a sufficient condition by Dirac for the existence of a Hamilton cycle are determined. Infinitely many closures of odd order turn out to be noncomplete.

We use [2] for basic terminology and notations and consider only simple graphs. For a graph G we denote by SG( 5 i ) the number of vertices in G of degree at most i; SG(> i ) is defined analogously. P, will be the path on n vertices, C, the cycle on n vertices and kG the graph consisting of k disjoint copies of G. We first state some known results.

Proposition A. Let G be a graph. If V ( G ) contains a nonempty proper subset S such that w(G - S ) > IS[, then G is nonhamiltonian. Theorem B (Bondy and Chvdtal [l]). Let G be a graph with u 2 3 . If c ( G ) is complete, then G is hamittonian. Theorem C (Chvdtal [3]). Let G be a graph with u 2 3 such that, for each integer i with 1 5 i 5 f(u - l), either S G ( i) ~ 5 i - 1 or SG(> u - i) 2 i 1. Then G is hamittonian.

+

Let E denote the class of graphs G with v 2 5 such that either

G

Kj V

(Kv-2j

+j K , )

with

25j

k u - 1,

or u is odd and 71

H.J . Broersma

72

or

G c Ki(,-q v (21<3 t i ( u - ~ ) K I ) for u 2 7.

By Proposition A, every graph in E is nonhamiltonian.

Theorem D (Dirac [4]). Let G be a graph with u 2 3 and with the following properties: (a)

SG(5

1) = 0;

(b) if i is any integer with 2 or s c ( 2 u - i) 2 i;

5 i 5 iu - 1 , then either

s ~ ( i5) 5 i - 1

(4 S G ( 5 +(u - 1)) Ii(u t 5 ) ; (4 G f E * Then G is hamiltonian. Let B and C denote the classes of graphs satisfying the conditions of Theorem B and C, respectively. Furthermore, let D denote the class of graphs satisfying the conditions of Theorem D. All graphs in C have a complete closure [2, Theorem 4.51, in other words C C B . Obviously, C G D too. A natural question is whether D 2 B . Here we answer this question in the negative and determine D \ B . We use two lemmas. Lemma 1. Let G be a graph with closure c(G) and let k be an integer such that, for 1 5 i < k, S,(G)( 5 i) 5 i - 1.

Ifv is a vertex of c(G) with dc(G)(v)2 v - k, then d,(~)(v)= u - 1. Proof. Let v be a vertex of c(G) with d , . ( ~ ) ( v2) u - k. If d , ( q ( v ) = u - k’ < u - 1, then, since S , ( G ) ( ~ k’ - 1) 5 k’ - 2, one of the vertices that 0 is not adjacent to v has degree at least k’ in c(G), a contradiction.

A Note on Hamiltonian Graphs

73

Lemma 2. If c(G) E D , then, for each integer i with 1 5 i 5 S,(G)(<

3.

- 1,

i) 5 i - 1.

Proof. Suppose c(G) E D and that for some i with 1 5 i 5 i u - 1 we have that s , ( G ) ( ~i) 2 i. Let k = min{i I s,(G)(< i) 2 i} and K = {v E V ( c ( G ) )I d,(~)(v) 2 v - k}. Since c(G) E D, we have 1K1 2 k 2 2. By Lemma 1 all vertices of K have degree u - 1 in c(G). Since s,(G)(< k) 2 k and \ K \ 2 k, it follows that c(G)has exactly k vertices of degree u - 1 and a t least k vertices of degree k (and no vertices of degree at most k - 1). Hence c(G) Kk V (Kv-2k k K l ) , so that c(G) E E , a contradiction.

+

Define two classes G1 and G2 of graphs of odd order as follows:

r+3

r+3

r+3

r+3

r+3

k= 1

k=3

k=l

k=3

k=2

J

It is easily checked that G; E D \ B and that every graph in Gi is hamiltonian (i = 1,2). We now sharpen Theorem D using Theorems B and C.

Theorem 3. IfG E D , then c(G) is complete or c(G) E {Cs}U GI U G2. Proof. Let G E D. Except for Cs,which belongs to D and is isomorphic to its closure, all graphs in D with at most five vertices have complete -~ closures. Henceforth assume u 2 6. The fact that G E D implies that c(G) E D. As remarked above, c(G)is complete if c(G)E C.Now assume c(G) E D \ C and c(G)is not complete. We shall show that c(G) E GI U G2. Since c(G) $ C, there exists an integer T with 1 5 T 5 3(u - 1) such that s , ( G ) ( ~ r ) 2 r and s,(G)(> u - T ) 5 r . Lemma 2 now implies that T = fr(u - l ) ,or, equivalently, v = 2r + 1. The condition s , ( G ) ( ~u - T ) 5 T then amounts to saying that s,(G)(< r ) 2 T 1. Since c(G) E D we get that s , ( G ) ( ~r ) 5 T -t 3. Let K = {v E V ( c ( G ) )I d c ( G ) ( v ) 2 r -t 1). Applying Lemma 1 with k = r yields that all vertices of I( have degree u - 1 in c(G).

+

H . J . Broersma

74

Three cases may occur. CASE 1. s , ( G ) ( I T ) = T t 1 (IKI= T ) . Then the r 1 vertices of c(G)- K all have degree 0 in c(G) - K . It follows that c(G) K , V ( T l ) K 1 , so that c(G) E E, a contradiction.

+

+

CASE 2. ~ , ( q (T )<= T -t 2 (IK(= T - 1). Each of the T + 2 vertices of c(G)- I< has degree at most 1 in c(G)- K . Thus each component of c(G)-K is isomorphic to PIor P2. Since c(G) !$ E, it follows that c(G) E GI.

+

CASE 3. s,(G)(< T ) = T 3 (IKI = T - 2). Each of the T t 3 vertices of c(G)- K has degree at most 2 in c(G)- K . Thus each component of c(G)- K is a path or a cycle. Since c(G) E and sc(c)(< T - 1) 5 T - 2, it follows that c(G) E G2. It is worth noting that a good algorithm for deciding whether or not a graph is in G1 U Gz is easily constructed.

Acknowledgement I thank H. J. Veldman for his help in the preparation of this note and the referee for suggesting some substantial improvements.

References [l] J. A. Bondy and V. Chvital, “A method in graph theory,” Discrete Math. 15 (1976), 111-135.

[2] J. A. Bondy and U. S. R. Murty, Graph Theory with Applications, Macmillan, London and Elsevier, New York (1976). [3] V. Chvital, “On Hamilton’s ideals,” J . Cornbin. Theory ( B ) 12 (1972), 163-1 68.

[4]G. A. Dirac, “Note on Hamilton circuits and Hamilton paths,” Math. Ann. 206 (1973), 139-147.

Annals of Discrete Mathematics 41 (1989) 75-78 0 Elsevier Science Publishers B.V. (North-Holland)

Uniqueness of the Biggs-Smith Graph A. E. Brouwer Department of Mat hemat ics Technical University of Eindhoven Eindhoven, The Netherlands Dedicated to the memory of G. A . Dirac We give four descriptions of the distance regular graph with 102 vertices, diameter 7 and girth 9 found by Biggs and Smith, and we show that this graph is characterized by its intersection array.

A graph r is called distance regular (or metrically regular) with intersection array i ( r )= {bo, b l , . . . , b d - l ; c1,. . ,c d } when it is connected with diameter d and whenever s,y are two vertices at distance i (0 5 i 5 d ) , the vertex y has bi (respectively c i ) neighbours a t distance i 4- 1 (respectively i - 1) from x. (Here bd = co = 0. Clearly c1 = 1. Note that I’ is regular of valency bo.)

.

In [2], Biggs and Smith described a distance regular graph with 102 vertices, intersection array { 3 , 2 , 2 , 2 , 1 , 1 , 1 ; 1 , 1 , 1 , 1 , 1 , 1 , 3 } , diameter 7 and girth 9. Here we show that this graph is uniquely determined by its intersection array. Let us first describe the graph I?. We give three descriptions (and the uniqueness proof produces a fourth).

A. (This is the description given by B i g s and Smith [2].)

Figure 1. This pictures the following graph: take 17 mutually disjoint copies of the “letter H” (Figure l ) , so that we have the 6 . 1 7 = 102 vertices. The middle 75

A. E . Brouwer

76

vertices have valency 3 already; join the other vertices in four 17-gons, where the vertices labelled (171j) are in a 17-gon with vertices x o(j),. . . , 2 1( j6), and zjj) is joined to x j 2 j (indices modulo 17). Here is the vertex labelled (171j) in the ith copy of H.

XI' )

B. Take as set of vertices a conjugacy class of subgroups of PSL(2,17) isomorphic to Sym(4), and join two vertices if these subgroups intersect in a group isomorphic to the dihedral group of order 8.

C.Let C be the extended binary quadratic residue code of length 18 (for definitions, cf. Van Lint [3]; for an explicit description, see Berlekamp, Conway and Guy [l], Vol. 2, p. 435). It has weight enumerator 1

+ 102X6 + 153X8 + 153X1' + 102XI2 + X18.

Now I'2, the graph in which two vertices are adjacent whenever they have distance two in I', is (isomorphic to) the graph whose vertices are the 102 vectors of weight 6 and where two vectors are adjacent whenever they have Hamming distance 12. Note that I' can be recovered from I'2 - in fact, for unique triangle graph distance d r O 1 2 3 dh 0 8 12 8

4 6

5 6 10 8

7 6

Since the full automorphism group of C is PSL(2,17), the same holds for I'. Now let us prove the uniqueness. Theorem. There exists a unique (up to isomorphism) distance regular graph I' with intersection arragl i(r) = {3,2,2,2,1,1,1; 1,1,1,1,1,1,3}. This graph is distance tmnsitive.

Proof. Fix a vertex 00. The distance distribution diagram of I' around 00 is as in Figure 2. The notation of the figure means that, for example, there 0 3

1 39 2

1 69 2

p 22

1Yl

IYl

1Yl

398

Figure 2. are twelve vertices at distance 3 from 00, and each of them has one neighbour a t distance 2 and two neighbours at distance 4 - and no neighbours at distance 3.

Uniqueness of the Biggs-Smith Graph

77

Let rj(o0) be the set of vertices at distance j from 00 (0 5 j 2 7). Write x N y to denote that x and y are adjacent. Label each vertex z of where the xi are the six the graph with the ordered 6-tuple (d(x,x;)):'l, vertices in rz(w). Since the z i fall into three pairs at mutual distance 2, the ordered 6-tuple will be written as an ordered triple of ordered pairs. It is easy t o see what labels will occur: e.g., if x E rd(o0)then x has distances 3, 4 and 5 to the three neighbours of 00 and distances 2,4; 3,5; 5,6 t o the points 2;; only the ordering remains to be determined. We shall see that not all 3!23 = 48 orderings occur, but only half of them. This motivates the following description of a graph A: Take as vertex set the following collection of ordered triples of ordered pairs; we list representatives for the orbits of Sym(4), where Sym(4) acts by arbitrarily permuting the triple and interchanging the elements of an even number of pairs (like its action on the edge set of K 4 ) :

#

representative 1 (22,22,22) 3 (11,33,33) 6 (02,44,44) 12 (13,45,45) 24 (24,56,35) (34,67,46) 24 24 (45,66', 57) 8 (56,56,56)i (i = 1,2) Here the odd permutations of a triple in the last orbit change the index i. Join two 6-tuples whenever the corresponding 6 elements are ad6, i is adjacent to i t 1, and for i = 4,5,6 jacent, where for 0 5 i also i is adjacent t o i. The 6' behaves as 6 but serves to define the adjacency (45,66', 57) N (45,75,6'6) (and (45,66', 57) is not adjacent to (54,57,6'6), (54,75,66'), (45,57,66')). Concerning (56,56,56)1, it is adjacent t o the three cyclic shifts of (45,66', 57), while (56,56,56)2 is adjacent to the three cyclic shifts of (45,57,66'). This defines the graph A , regular of valency three. We claim that for an arbitrary graph l? with the above intersection array and an arbitrary vertex as 00 the labelling necessarily is as described. The 6' is defined by: if d(x, w ) = 6 and d(x,xi) = 6 then write 6' as i-coordinate of the label x whenever d(y,z;) = 7 for x y E I'5(oo). Now for any x with d ( x , w ) 2 3 and any y with y N x we see that if 2 has one of the allowed labels then y also has. Since r>3(oo)is connected, this shows that only the labels indicated occur. Given the label of a vertex x with 3 5 d(x, m) 5 7,

<

N

A. E . Brouwer

78

the labels of its three neighbours are uniquely determined. Consequently, if a vertex y had the same label as x then walking along a geodesic from z (respectively y) to 00 we would find distinct vertices x‘ and y’ with the same label, both adjacent to a vertex z , but this is impossible. Therefore, each of the labels indicated occurs precisely once. Finally we have to worry about I’,(oo). But since I’ has girth 9, the paths (45,66‘, 57) N

N

(35,56,24)

N

(34,67,46)

N

(46,67,34)

N

(24,56,35) (57,66’,45)

and (45,66’, 57) N

(13,45,45)

N

N

(34,67,46)

(24,35,56)

N

N

(24,56,35)

(34,46,67) N (45,57,66‘)

show that if (45,66‘, 57) (56,56,56)1 then (56,56,56)2 N (57,66’, 45) and (56,56,56)2 (45,57,66’). By symmetry this determines all adjacencies for vertices of I’7(00). This shows that any graph I’ with the above intersection array has the structure of A around any vertex, so there is a unique such graph, and it has a transitive group. But the description given above also shows that the stabilizer of a vertex 00 in r is isomorphic t o Sym(4) and acts transitively 0 on rj(o0) (0 5 j 5 7). So I’ is a distance transitive graph. N

N

Remarks. The graph r does not contain 10-cycles. Its spectrum is 3’, 218, 017, ( 1 9 , ) g ,Bj16, where the e j are the three roots of e3 + 3e2 - 3 = 0.

References [l] E. R. Berlekamp, J. H. Conway and R. K. Guy, Winning Ways, Academic Press, London (1982).

[2] N. L. Biggs and D. H. Smith, “On trivalent graphs,” Bull. London Math. SOC.3 (1971), 155-158. [3] J. H. van Lint, Introduction to Coding Theory, Springer Verlag, New York-Heidelberg-Berlin (1982).

Annals of Discrete Mathematics 41 (1989) 79-90 0 Elsevier Science Publishers B.V. (North-Holland)

Some Complete Bipartite Graph Numbers

- Tree

Ramsey

S. Burr* Computer Science Department City College (CUNY) New York, USA

P. Erdos Mat hematical Institute Hungarian Academy of Sciences Budapest, Hungary

R. J. Faudree, C . C. Rousseau and R. H. Schelpt Department of Mathematical Sciences Memphis State University Memphis, Tennessee, USA

Dedicated to the memory of G. A . Dirac We investigate T ( K , , , ~ , T for) a = 2 and a = 3, where T is an arbitrary tree of order n. For a = 2, this Ramsey number is completely determined by T ( K Z , ~ , Kwhere ~ , ~m ) = A ( T ) . For a = 3, we do not find such an “exact” result, but we do show that r ( K 3 , 3 , T ) 5 max{n [ ~ n ’ / r(K3,3, ~ ] , Kl,,,,)}. Except for the choice of c this result is best possible.

+

1

Introduction

Let T denote a tree of order n and maximum degree A ( T ) = m. The Rarnsey number T(K,,,,T) is the smallest integer p such that in every two-coloring of the edges of KP there is either a monochromatic or *Research supported by ONR under grant N000014-8s-K-00704. Research supported by NSF under grant DMS-8603717.

79

S. Burr, P. Erdos, R. J. Faudree, C. C. Rousseau and R. H.Schelp

80

else a monochromatic copy of T . By tradition, we shall let the colors be R (red) and B (blue) with the resulting edge-induced subgraphs denoted ( R ) and ( B ) , respectively. It is well-known that the computation of r ( K , , , , T ) is quite easy in some cases (e.g., for T = P,, a path of order n ) but difficult in general. In particular, the “star” case T = K1,,-1 is known to be complicated. In this paper, we show that for a = 2 the problem of computing r(K2,2,T ) ,i.e. r(C4,T),reduces to that of computing T(C4, K l , m ) . The following terminology will be used. An end-vertex is a vertex of degree one. An end-edge is an edge which is incident with an end-vertex. A suspended path in a graph G is a path ( z o , z ~ ,... ,x k ) in which z1,. . . ,x k - 1 have degree two in G.

2

C4-tree Ramsey numbers

Let T be a tree of order n and maximum degree A(T) = m. In this section, we shall prove that r(C4, T ) = max(4, n

+ 1, r(C4, I < I , ~ ) } .

Thus, r(C,, T ) is easily determined if f ( m )= r(C4, is known. In [3], Parsons proved that if q is prime power, then f ( q 2 ) = q2 q 1 and f ( q 2 1) = q2 q 2. A table of f ( m ) for small values of m is shown below.

+

+ +

m = f(m)=

1 2 3 4 4 6

+ +

4 5 7 8

6 7 9 11

8 12

9 13

10 14

We start with the following bit of graph-theoretic folklore, which however trivial, is the starting point for many results in extremal graph theory and Ramsey theory involving trees.

Basic Lemma. I f 6 ( G )2 n - 1 then G contains every tree of order n.

Lemma 1.1. I f F is any forest with q edges, then r ( C 4 , F ) 5 2(q + 1). Proof. For the case in which P is a tree, this follows easily by a double application of the Basic Lemma. For the general case, it follows by induction on the number of components. Except for some very special cases, the bound of Lemma 1.1 can be improved,

Some Complete Bipartite Graph - Tree Ramsey Numbers

81

Lemma 1.2. If F is a forest with q edges and two or more components, then r(C4, F ) 5 2q - 1 unless F 2 q K 2 or F 2 ( q - 2)Kz U K1,2. The next two results are well known. Lemma 1.3 is proved by Parsons in [3].

+ [m+ 1.

Lemma 1.3. For all m 2 2, T ( C ~ , K 5~ m ,~) Lemma 1.4. For all n 2 3, r(C4,P') = n

+ 1.

To prove Lemma 1.4, just consider a maximal length path in ( B ) . Its two end-vertices are joined to all of the remaining vertices in ( R ) . Now we prove the main result of this section. Theorem 1. If T is a tree of order n and maximum degree A(T) = m, then r(C4, T) = max(4, n 1,r(C4,K I , ~ ) } .

+

Proof. It is plain that p = max(4,n + l , ~ ( C 4 K l , ~is) }a lower bound for r(C4, T). Let ( R ,B ) be any two-coloring of the edges of Ii, in which ( R ) 2 C4. We claim that there is then an embedding of T into ( B ) . For n 5 6, the verification is straightforward. There are 13 trees to consider. These trees and the corresponding values of r(C4,T) are shown in Figure 1 overleaf. Our proof that the result holds for all n 2 7 will be by induction. Since the case in which T is a path has been settled, we may assume that m 2 3. To exhibit an embedding of T into ( B ) ,there are two basic strategies. (u)

Embed a maximum degme vertex first.

Let x be a maximum degree vertex of T and choose u ( x ) to be a maximum degree vertex in ( B ) . Thus, u ( z ) has degree k 2 m in ( B ) . Let N R and N B denote the neighborhoods of a(x)in (R) and ( B ) ,respectively. We wish to extend Q to an embedding of T into ( B ) . There are several cases in which such a strategy will be successful. (i) k 5 2(m - 1). Rewriting the inequality, we find k 2 2(k - m 1). From Lemma 1.1, we see that within the subgraph of ( B ) spanned by NB there is any desired forest with k - m edges. Let T' be a tree obtained from T by deleting (one a t a time) end-vertices other than those which are adjacent to z until there are precisely k 1 vertices left. Then F' = T' - z has k vertices and k - m edges. Thus, we are able to extend Q so that it provides an embedding of T' into ( B ) , and u maps V ( F ' ) onto N B . We now seek to

+

-+

82

S. Burr, P. Erdos, R. J . Faudree, C. C. Rousseau and R. H . Schelp r(C4,T ) = 4 : T(C4,T)= 5 : r(C4,T) = 6 :

+

Figure 1. r(C4,T ) for trees of order at most six. extend u to an embedding of T into ( B ) . Suppose that such an extension fails to exist. Then since p 2 n 1, there must exist a vertex y E V ( T ) , y # 2, such that a(y) is adjacent in (R) to two vertices of N R . In other words, (R) contains a C4, a contradiction. Henceforth we assume that k 2 2m - 1.

-+

(ii) T has a vertex of degree m 2 3 which is an "end-star." Suppose that x is adjacent to m - 1 vertices of degree 1. Then T - x consists of a tree T' and m - 1 isolated vertices. Choose o(z) to be a vertex of degree k 2 2m - 1 in ( B ) and attempt to extend the embedding as in case (i). In particular, let T" be a tree obtained from T' by successively deleting end-vertices (but keeping all vertices which are adjacent to z)until there are precisely k - m t 1 vertices left. Now we want to extend u so that it provides an embedding of T" into the subgraph of ( B ) spanned by N g . If this is possible, then o can be extended to an embedding of T into ( B ) . For, otherwise, there is a vertex other than a ( z )which is adjacent in (R) to two vertices of N R , so (R) contains a C4. Note that since T" is a subgraph

Some Complete Bipartite Graph - Tree Ramsey Numbers

83

of T , we have A(T") 5 m. Now for m = 3, the desired embedding of T" follows immediately from the induction hypothesis if k 2 r(C4,K1,3)= 6 . If k = 5 , then T" is the tree of order three and r(C4,T") = 4 < k . For m 2 4, use of the induction hypothesis shows that it suffices to verify that r(C4,K1,,) 5 2m - 1 . This is easily done; r(C4,1<1,4)= 7, r(Cq,K1,5)= 8 and m 1 5 2m - 1 for m 2 6 by an elementary calculation. Henceforth we assume that F = T - x has two or more nontrivial components.

+

+

(iii) F 2 mK2 or F 2 ( m - 2)Kz u K1,2. Consider first the case where F S mK2. Then r(C4,F) = 2m 1 for m 2 2 and we wish to prove that r(C4,T) = 2m 2 for m 2 3. If k = 2m - 1 , then the subgraph of ( B ) spanned by N g contains ( m - l)l12. The remaining vertex of N g is adjacent t o one of the two vertices in N R and this gives the desired graph. If k = 2m, then the subgraph of ( B ) spanned by N R U N B contains mK2 and again we find the desired graph. The case where F 2 ( m - 2)Kz U 1 1 1 ~is similar. In view of Lemma 1.2, we may now assume that r(C4,F) 5 2q - 1, when F is of size q .

+

+

(iv) 2m 2 n - 1 . Since F has n - m - 1 edges, then r(C,,F) 5 2(n - m - 1 ) - 1 5 2m - 1 5 k . In this case, the subgraph of ( B ) spanned by N B contains F so we have an embedding of T into ( B ) . Now suppose that strategy ( u ) fails. Then F = T - x has at least two non-trivial components and 2A(T) < n - 1. The first condition implies that T has (at least) three independent end-edges. This observation gives rise to the second strategy.

( b ) Complete the embedding of T using a matching. Let T' denote the tree obtained from T by deleting the three independent end-edges. Then T' is a tree of order n - 3 and maximum degree

A(T') 5 m < LPJ. We now claim that r(C4,T') = n - 2. This follows from the induction hypothesis if r(C4,I
+ [m+

84

S. Burr, P. Erdos, R. J. Faudree, C. C. Rousseau and R. H. ScheJp

constructed, there is a set of three vertices X = { q , x 2 , ~ 3 }for which a matching of X into Y in ( B ) would yield an embedding of T . Since the , vertex of X is adjacent subgraph of ( R ) spanned by Y contains a K I , ~every in ( B ) to at least one vertex of Y . Moreover, Hall’s condition cannot fail for any other subset of X since this would also imply a C4 in ( R ) . Hence, there is a matching of X into Y and so an embedding of T into ( B ) . Thus, in all cases either strategy ( u ) or strategy ( b ) is successful and the proof is complete. 0 The preceding theorem reduces the problem of computing T ( C T ~ ), t o lfl,m).Outside of special values of m (e.g., m = q2 one of computing ~(c4, or m = q2 l),little is known concerning exact values of f(m) for large m. The following result gives a modest improvement in our knowledge of the asymptotic behavior of this sequence of Ramsey numbers.

+

Theorem 2. Let pk denote the kth prime. If pk+l

- Pk

(*I

Pr

for all suficiently large k, then

~ O all T

suficiently large n.

Remark. At present, the best known value of a for which (*) holds is less than 11/20, but improvements in this value are being obtained rapidly. , , )determined to In any case, by Lemma 1.3 and Theorem 2, T ( C ~ , K ~is within 6 n g . Proof of Theorem 2. Let p be the smallest prime which exceeds ni. In [l]and [2], it is shown that there exists a graph Go of order N = p 2 t p t 1 which contains no C4 and in which the degree of each vertex is p or p t 1. 1 Set m = [nr - 6 n g J . We wish to establish the existence of a graph of order n m which contains no C4 and in which the degree of each vertex is at least m. To this end, we randomly delete d = N - ( n m ) vertices from Go to obtain a random graph G. A given vertex 5 which is of degree p in Go will be described as bad if it is not deleted and has degree < m in G. Let B, denote the event: “5 is bad.” Thus, the random graph G will belong t o B , if for some k > p - m, k vertices are deleted from the p vertices in the neighborhood of z and the remaining d - k deleted vertices do not include 2. It follows that

+

+

Some Complete Bipartite Graph - n e e Ramsey Numbers

where the sum extends over all k

85

> p - m. By elementary analysis,

where e = 2.71828.. .. By'our choice of p and m, all terms in the sum have k > p - m > 6 n 9 . Also, p < n i n 9 and N - d = n+ m > n. In the worst case, p (and therefore d ) are as large as possible while k and N - d are as small as possible. Even in this case, epd/k(N - d ) < O(1). It follows that n P ( B , ) + 0 as n + 00. Clearly, the same analysis holds if z has degree p 1. Consequently, there is a positive probability that all vertices are good when n is sufficiently large. Thus our random graph construction 0 is successful.

+

9+

+

K3,3-treeRamsey numbers

3

The following result shows that the Ramsey number r(K3,3,T) is determined by m = A ( T ) if m is quite close to n. Otherwise, our result yields only an upper bound.

Theorem 3. There exists a constant c such that for every tree of order n and maximum degree A ( T ) = m, 1'(1(3,3,T)L

ma+

+ rc.51

,r ( 1 ( 3 , 3 , f h , r n ) } .

To set the stage for the proof of this result, we first recall the simple argument by which one may obtain an upper bound for T(Ka,b,Kl,n). Namely, by the pigeonhole principle, if

~ , ~argument ). of Theorem 1 can be extended to then p > T ( K ~ , ~ , KThe show that there is a c such that r(K2,b,T) 6: max{ n c, T ( 1 ( 2 , b , Kl,,)} for all trees of order n and maximum degree m. Assembling these facts, we have the following result.

+

Lemma 3.1. Let T be an arbitrary tree of order n. Then, for all ciently large n, (i) (ii)

r ( ~ 2 , 3 , ~n)

<

+ 2n3,

< n +3ni,

T(K~,~,T)

(iii) r(K3,3,K I , ~<) n

+ 3n5. 2

SUB-

86

S. Burr, P. Erdos, R. J . Faudree,

C.C. Rousseau and R. H. ScheJp

Remark. The constants chosen here are rather generous, so the given inequalities hold for even quite small values of n. Proof of Theorem 3. The proof will be by induction. We shall argue that with c = 20 there is an integer no so that for all n > no, the truth of the theorem for all k < n implies its truth for n. Having done so, we need only adjust (if necessary) the value of c so the result holds for all n. In what follows there is alwaYs the tacit assumption that n is sufficiently large. Set p = max{n t [20ns1,~ ( 1 1 3 , 3K , l l m ) }and assume that (R, B ) is a two-coloring of the edges of K p in which ( R ) 2 1<3,3. There are four cases to consider. Use the first strategy of Theorem 1; if 2 is a maximum degree vertex in T , then make ~ ( 2 a) maximum degree vertex of ( B ) . Thus a(.) has degree k 2 m in ( B ) and, by Lemma 3.1, we may also assume that k > n - 3n3. Certainly, the number of vertices to which a ( z )is adjacent in ( R ) does not exceed 3ng t 20ni < 4n3. As in the proof of Theorem 1, let N R and N B denote the neighborhoods of C(Z) in (R) and ( B ) ,respectively. Describe a vertex in N B as powerless if it is adjacent to at least d = [lOnbl vertices of N R in ( R ) . If INRI = s and there are t powerless vertices, then

for, otherwise, there is a K3,3 in ( R ) with two powerless vertices and C(Z) on one side and three vertices from N R on the other. It follows that t < 6. This number is much less than then number of isolated vertices in F = T-x. The remaining vertices in N B will be called powerful. Note that there are a t least powerful vertices. This is more than enough to ensure that there is an embedding of the non-trivial components of F into the subgraph of ( B ) spanned by the powerful vertices. (Here we use the crude bound ~(K3,3 F, ) 5 3(q t 1) for a forest with q edges and note that our F has a t most 5 edges.) We now face the problem that there may not be enough vertices left over in N B to account for all the isolated vertices of F . To find an embedding of T into ( B ) we need to take appropriate “small” trees 2 (altogether involving at most 3n5 vertices) rooted at powerful vertices and find an embedding of an isomorphic tree where all vertices except the root reside in N R . Let ~ ( y denote ) a powerful vertex which is the root of such a tree. The existence of such an embedding is ensured by the fact that the vertex in question is powerful. As B U C ~ it , is adjacent in ( B ) to a t least

p

Some Complete Bipartite Graph - Tree Rarnsey Numbers

87

1

lOn3 vertices in excess of what may be needed to complete its tree. Let Y denote the set of vertices in N R t o which a ( y ) is adjacent in ( B ) . The subgraph of ( R ) spanned by Y cannot contain a K 2 , 3 , for then ( R ) would contain a K 3 , 3 . Suppose that within the subgraph of ( B )spanned by Y we need to find a forest F' with T vertices. Success is ensured since T < 3n8 and I Y> ~ r Ion: > T 2 T f r > T ( ~ 2 , 3 F'). ,

+

We now assume that A ( T ) <

+

q,

(ii) T has six independent end-edges. Since m = A ( T ) < %, we have

It follows that ( R )contains a K 2 , 4 . Let T' denote the tree obtained from T by deleting six independent end-edges. Then ~ ( K 3 , 3T') , 5 p - 6 , so we find a K 2 , 4 in ( R ) and a T' in ( B )which are vertex disjoint. Let Y denote the set of all vertices disjoint from the embedded T'. Thus lYl = p - n+ 6 2 14. In view of the construction of T', there is a set X = ( x l , x 2 , . . ,x6} for which a matching of X into Y in ( B ) would yield an embedding of T . Since the subgraph of ( R )spanned by Y contains I i 2 , 4 , every vertex of X is adjacent in ( B )to a t least two vertices of Y . Hence, failure of Hall's condition would yield a K,,, in ( R ) with T 2 3 and s 2 IYI - ( T - 1) > 3. Consequently, Hall's condition must be satisfied and there is an embedding of T into ( B ) .

.

(iii) T has a suspended path of length at least six. Let T' be a tree obtained from T by reducing the length of an appropriate suspended path by one. By induction, we find an embedding of T' into ( B ) . It is not difficult to show that if there are as many as eleven vertices external to the embedded T', then the absence of a K 3 , 3 in ( R ) implies that there is an embedding of T into ( B ) . In fact, there are more 1 than r20nal additional vertices, so this case causes no problem. (iv) The residue. By now, our tree is quite special. It has at most five vertices of degree three or more and it has no suspended path of length six. As usual, let x denote a maximum degree vertex of T . In view of the observations just made, we can be sure that x is adjacent to about 5 or more vertices of degree one. Furthermore, the forest F = T - z has at most four nontrivial components. Now we return to the first strategy and let a(x)by a maximum degree vertex in ( B ) .

88

S. Burr, P. Erdos, R. J . Faudree, C. C. Rousseau and R. H. Schelp

Note that almost all vertices of F are either isolated or have degree one. (The number of exceptions is limited by the fact that at most five vertices have degree three or more and the fact that the suspended paths which join these vertices are of length a t most five.) Suppose that ~ ( z has ) degree k in ( B ) . Delete end-vertices from F to obtain a forest F‘ of order k. Now delete all isolated vertices from F‘ t o obtain F“. We claim that there is an embedding of F“ into the subgraph of ( B ) induced by the set of powerful vertices in N B . This follows immediately from the induction hypothesis, since F“ is of order a t most k - 5 and has no vertex of degree > 4 , whereas there are at least k - 6 powerful vertices. Now we just need t o extend the embedding into N R to obtain an embedding of T into ( B ) . This only involves choosing the appropriate vertices within N R to play the role of end-vertices in T and (at the same time) possibly releasing some vertices within NB so they can play the role of end-vertices which are adjacent to 5. But there is no problem, because each powerful vertex is adjacent in ( R ) 1 1 t o at most lOnT vertices of N R and the whole graph contains 20nT vertices 0 in addition to the number needed for T . This completes the proof. Except for the value of the constant c, this result is best possible. Let p = 3 (mod 4) be prime and consider a tree T of order n = p3 - p + 2 which has exactly two vertices of degree greater than one. Of necessity, these two vertices are adjacent. We claim that ~ ( K 3 , 3T, ) 2 p3 1. To see this, we use the example of W. G. Brown [l]. This is a graph of order p3 which is regular of degree p2 - p and contains no 1<3,3. This graph has the property that for any two non-adjacent vertices, there are exactly p - 1 vertices t o which the two are commonly adjacent. It follows that the complement of this graph does not contain T since there are always p - 1 vertices which cannot be used in any attempted embedding. Now, if the degrees of the two “high degree” vertices of T are as balanced as possible, we ensure that 1 ~(K3,3 K, I , ~<)n 4- [ c n z l , at least when n is large. Thus, in this example the result ~ ( K 3 , 3T, ) 5 n 4- [ c n i ] cannot be improved except for the choice of c.

+

4

Open questions

There is still much more which could be known concerning the asymptotic behavior of f(n) = T ( C ~ K ,I , ~ )For . example, it is conjectured that for every constant c there are infinitely many n for which every graph of order n and minimum degree 2 f i - c contains a Cq. That is, infinitely often f ( m ) < m fic’. One of the authors (E.P.) offers $100 for a proof or

+

Some Complete Bipartite Graph - Tree Ramsey Numbers

89

+

+

disproof. Clearly, infinitely often f(n 1) = f ( n ) 1 and infinitely often f(n + 1) > f(n) 1. Is it true that f(n 1) = f ( n ) holds i.0. but that the density of these n is 0 and is it true that f(n 1) 5 f(n) 2 for all n?

+

+

+

+

References [l] W. G. Brown, “On graphs that do not contain a Thomsen graph,” Canad. Math. Buff. 9 (1966), 281-285. [2] P. Erdos, A. Renyi, and V. T. Sos, “On a problem of graph theory,”

Studia Sci. Math. Hungar. 1 (1966), 215-235.

[3]T. D.Parsons, “Ramsey graphs and block designs,” Trans. Am. Math. SOC.209 (1975), 33-44.

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Annals of Discrete Mathematics41 (1989) 91-110 Ekvier Science Publishers B.V. (North-Holland)

@

The Edge-Chromatic Class of Graphs with Maximum Degree at Least IVI - 3 A. G. Chetwynd Department of Mathematics , Cart me1 College University of Lancaster Lancaster, UK A. J. W. Hilton Department of Mathematics University of Reading Reading, UK Dedicated to the memory of G. A . Dirac We characterize the simple graphs G which satisfy A(G) 2 IV(G)l-3 and x’(G) = A(G), where x’(G) is the chromatic index of G and A(G) is the maximum degree of G.

1

Introduction

The graphs considered here are either simple graphs, with no loops or multiple edges, or are multigraphs, with more than one edge joining a pair of vertices, but no loops. An edge-colouring of a graph G is a map 4: E(G) C , where C is a set of colours and E(G) is the set of edges of G , such that no two incident edges receive the ~ a m colour. e The chromatic index x’(G) of G is the least value of IC( for which an edge-colouring of G exists. A well-known theorem of Vizing [7]states that, for a simple graph G , --+

+

A ( G ) I x’(G) I A ( G ) 1, where A ( G ) denotes the maximum degree of G . Graphs G for which A(G) = $(G) are said t o be Class 1, and otherwise they are Class 2 . The degree of a vertex v in G is denoted by dc(v),and the number of edges joining two vertices v and ‘u) is denoted by m ~ ( v , w ) .

91

A. G. Chetwynd and A. J . W. Hilton

92

We call a multigraph overfull if

An overfull graph is necessarily of odd order: since no colour class can consist of more than [iIV(G)IJ edges, an overfull graph must be Class 2. In [4], Chetwynd and Hilton made the following conjecture.

Conjecture. Let G be a simple gmph with A(G) > $lV(G)l. Then G is Class 2 i f and only if G contains an overfull subgmph H with A(H) = A(G). Of course, if G contains an overfull subgraph R with A(H) = A(G) then G is Class 2. The difficulty lies in proving the converse. The cases A(G) = IV(G)l - 1 for IV(G)l odd, and A(G) = IV(G)( - 2 for IV(G)( even, were proved by Plantholt ([5], [S]). The cases A(G) = IV(G)l- 2 for IV(G)l odd, and A(G) = IV(G)l- 3 for IV(G)l even, were proved by Chetwynd and Hilton ([l], [2]). The main object of this paper is to settle the problem when A(G) = IV(G)l - 3 for IV(G)I odd.

Theorem. Let G be a simple graph with A(G) = IV(G)I - 3. Then G is Class 2 if and only if G contains an overfull subgraph H with A(H) = A(G). We shall need the following lemma from [4].

Lemma 1. Let G be a simple graph of odd order, let A(G) = IV(G)l - 3, and let IE(G)I = A(G)~;~v(G)IJ. If G contains no owerful2 subgrnph H with A(H) = A(G), then G is Class 1. The non-trivial point in the proof of the theorem is the replacement of the equality (E(G)( = A(G)[+(v(G)(J in Lemma 1 by the inequality lW)I 5 l$W)IJ

-

2

S o m e useful results

We need the following further lemmas. The first is one of the results of Plantholt [5] mentioned above.

Lemma 2. Let G be a simple graph with A(G) = IV(G)l - 1. Then G is Class 2 if and only if G is overfull. Let Kin+l denote the multigraph obtained from KZn+l by doubling one edge. The following lemma is a corollary of [3, Lemmas 6 and 71.

The Edge-Chromatic Class of Graphs with Large Maximum Degree

93

Lemma 3. Let G be a multigraph which contains a vertex v* with which each multiple edge is incident. Let G have at most two vertices of maximum degree, and let G contain no subgraph on three vertices with more than t A(G) edges. Then G is Class 1 unless G = K2n+l for some n 2 2. Let K,',+, denote the graph obtained from K2n+l by removing one edge v*v', and doubling two other edges at v*. The following lemma is an immediate corollary of [4, Theorem 11.

Lemma 4. Let G be a multigraph which contains a vertex v* with which each multiple edge is incident. Let G have 3 vertices of maximum degpee, including v*, and let G contain no subgraph on three vertices with more than A(G) edges. Let A(G) 2 IV(G)l. Then G is Class 1 unless G = K,',+, for some n 2 2. Let T(G) denote the maximum number of edges in any submultigraph of G of order 3. We call two multiple edges independent if they have no common vertex, and a set of multiple edges is independent if each pair is independent.

Lemma 5. Let H be a loopless multigraph on four vertices vo, v1, 212 and 213. Let the underlying simple graph of H be K4. Let p 2 3, and let J be a simple graph on the vertices 0 4 , ..., vp. Let G be formed by joining each vertex of H to each vertex of J by a single edge. Then

either ( i ) x'(G) = max(A(G),r(G)) and, if r(G) > A(G) and V O , v1 and 0 2 ape vertices with altogether r(G) edges between them, then G can be edge-wloured with colours c1, ..., c+-J) with each colour c;, i E {A(G) 1,...,r(G)), used only on edges joining two of them, or (ii) x'(G) = A(G) 1, p is even, and either ( a ) IE(H)I = 6 and IE(G)I 2 $(J? 2), or ( b ) IE(H)l = 7 and J = Kp-3, or ( c ) dH(vi) = 4 ( 0 5 i 5 3 ) (so that IE(H)I = 8 ) , and IE(G)I 2 z1 ( P2 P t 21, or ( d ) p = 6, d ~ ( ~=i5 ) (0 5 i 5 3 ) ( S O that IE(H)I = lo), and IE(G)I = p t 8), or ( e ) p = 4 and IE(G)I 2 2A(R) 3.

+

+

+

+

h($ +

+

Proof. If IE(H)I = 6 then the result follows from Lemma 2. If IE(H)l = 7 then the result follows from Lemma 3. Suppose therefore that IE(H)l 1 8.

A . G.Chetwynd and A. J. W. Hilton

94

First, we prove the theorem in the case when either d(vi) = d(wj) (0 5 i 5 j 5 3) or (p = 4 and IE(G)I 2 2A(H) + 3 ) ; all the exceptional cases mentioned in (ii) are included here. If d H ( v i ) = 4 (0 5 i I 3), p is even, and IE(G)I >, i ( p 2 p + 2 ) , then

+

+

x‘(G) L klE(G)l/Pl = P + 2 = A(G) 1, and H contains one pair of independent double edges. Give one colour to a pair of independent single edges from this pair of independent double edges; the remaining edges of G can clearly be coloured with p 1 colours, and so G can be coloured with p 2 = A(G) 1 colours. Therefore, x‘(G) = A ( G ) 1. If d H ( v i ) = 4 (0 5 a’ 5 3), p i s even, and (E(G)I5 $ ( $ + p ) , then let F be a maximal independent set of single edges of G which contains one edge from each of the pair of independent multiple edges of H. Then G \ F is a simple graph with A(G\F) = p and IE(G\F)I 5 i ( p 2 + p ) - i p . To see this inequality, suppose that IE(G)I = $(p2 p) - y. Then a simple calculation - 7 p 8) - y. Furthermore, IE(G \ F)I 5 shows that (E(J)I = $($ p ) - y - 2. The inequality follows if y 2 $ p - 2, so suppose that y 5 $ p - 3. Then IE( J)I 2 3(p2 - 7 p t 8) - 21 p 3 = ( p - 4)( $ p - 2 ) - 1. Since p 1 6 is even, J can be edge-coloured with p- 4 colours, and so J has a colour class with at least i p - 2 edges in it. Therefore, IF1 2 $ p , and so IE(G \ F)I I p) - hp. Therefore, in each case the inequality holds, and so, by Lemma 2 , G \ F is Class 1. Therefore, G is Class 1, and so x’(G) = A(G). If d H ( v i ) = 5 (0 5 i 5 3) and p = 6 then IE(G)I 5 p 8). If 1 2 IE(G)I = S(P t p t 8) then

+

+

+

i($

+

+

+

+

+

i($+

i($ + +

x’(G) 1 [21E(G)l/pl = 9 = A(G)

+ 1.

Let F be a matching consisting of three single edges, including one from each of a pair of independent multiple edges. Then E(G\F) = +p+2) and, from above, x’(G \ F) = A(G \ F ) 1. Therefore, x‘(G) = A ( G ) 1.

i(p”

+

If either dH(vi) = 5 (0 5 i 5 3), p is even, and p or

dH(vi) = 5

(0

I a’ 5 3),

> 6,

p = 6, and IE(G)I 5 +($

+

(1)

+ p + 6),

then let F be a maximal matching which includes two single edges from a pair of independent multiple edges. Then

W I \ F)I 5

1

z(P

2

1 2 + P) + 4 - BP 5 z(P + P),

The Edge-Chromatic Class of Graphs with Large Maximum Degree

95

so, by the case above, G \ F is Class 1. Therefore, x’(G) = A(G). If d ~ ( v ;=) 6 (0 5

is 3)

(2) then let F be a maximal matching which includes two single edges from an independent pair of multiple edges. Then IE(G \ F)I 5 $(p2 p ) 3, so, as shown above, x‘(G \ F ) = A(G) \ F . Therefore, x’(G) = A(G). ) dg(vj) 6 (0 5 i < j 5 3), then If p is even, p > 6, and d ~ ( v i = H is the union of d ~ ( v 0pairs ) of independent edges. Select d ~ ( v 0-) 5 of these pairs, leaving at least one edge between each pair of vertices of H . Colour the selected edges with d ~ ( v 0-) 5 colours. Now, (1) above applies to the remaining edges, and so they can be coloured with 6 colours; G can be edge-coloured with dg(v0) 1 = A(G) colours. A similar argument applies when p = 6 and d ~ ( v ;=) d ~ ( v j1) 7 (0 5 i < j 5 3). Now, let us consider the case p = 4 and IE(G)I 2 2A(H) 3; this ) d ~ ( w j )(0 5 i < j 5 3), as in this includes the case p = 4 and d ~ ( v i= case we have IE(G)I = 2A(11) 4. We get and

p= 6

+ +

+

+

+

The graph H is formed either from A(H) - 3 pairs of independent single edges and a Kq, or from A ( H )- 4 pairs of independent single edges, a K4, and one further edge; in either case, G is formed from A(H)-3 independent sets of edges (each set containing 1 or 2 elements) and a K s ; it follows that G can be edge-coloured with A(H) 2 colours, so x’(G) = A(G) 1. If p is odd and d ~ ( v ;=) d ~ ( v j )(0 5 i < j 5 3), then H is the union of d ~ ( v 0pairs ) of independent single edges. Select d ~ ( v 0-) 3 of these pairs, leaving one edge between each pair of vertices of H . Colour these pairs with d s ( v 0 ) - 3 colours; the remaining edges can be coloured with p colours, so G can be edge-coloured with

+

+

colours. Therefore, x‘(G) = A(G). This proves the lemma when either d(v;) = d(vj) (0 5 i < j 5 3) or ( p = 4 and IE(G)I 2 2A(H) 3). From now on we may assume that I d ~ ( v ;-) d ~ ( v j ) l2 1 (0 5 i < j 5 3) and that, if p = 4 then IE(G)I 5 2A(G) 2. We may assume that d H ( U 0 ) 2 dH(U1) L dH(D2) L d H ( V 3 ) . Then dc(v0) = W). Let H contain 2 3 pairs PI, , , Pz+3 of independent single edges, where P,+1 U PS+2U P,+3 is the set of edges of K4. Let PI be extended to

+

+

+

..

A . G . Chetwynd and A. J . W. Hilton

96

a maximal matching F1 of G. Let

G * = { G \ (F 1uP 2u. . . U P z)

i f x 2 1, if x = 0.

Then A(G*) = A(G) - x and r(G*) = r(G) - z. Since d ~ ( w 0 )> d ~ ( v 3 ) , G* contains at least one multiple edge. Since (E(H)I 2 8, and since F1 is a maximal matching, if p # 4 then G* # I
+

since dG(V0)

and

= 2 t ( p - 3) t mc+o,

211)

t mc*(vo,212) t 1

+

7(G) = 2 t m c * ( ~ , ~ 2m) c * ( v o , ~ t i ) mc+(vo,v2); the formula above is obtained from these by subtraction. If r(G) > A(G), let F' (A(G) 5 i 5 r(G)) be distinct single edges of G* joining v1 and 212, and put Go = G* \ ( F ~ ( GU) *+ ~U FT(c)).Since T(G) > A(G), it follows that r(G*) > A(G*), and so m p ( v i , v j ) > p - 2 (0 5 i < j 5 3). However, m~0(211,212) = m c * ( v 1 , ~ 2-) T(G) A(G) = p - 2, ~ , - J o ( v o , v ~>) p - 2, and m~o(v0,212)> p - 2, and it follows that 7(G0) = A(Go) > max(dco(q),d,p(v2)), so A(Go) = r(Go) = dco(wo), and Go has exactly one vertex 00 of maximum degree. If T(G) 5 A(G) then put Go = G*. In both cases, dco(v0) = A(Go) 2 7(G0). We now show that Gois Class 1, except possibly when p = 4. First, suppose that Go has exactly one vertex v of maximum degree. Colour the edges joining two of 00, 211, and 212 with .(Go) 5 A(Go) colours. Now proceed to colour the remaining edges of Go with colours c,+1, , CA(GO)+,, using 0

+

...

97

The Edge-Chromatic Class of Graphs with Large Maximum Degree

Vizing's original argument [7]. Since Go has exactly one vertex of maximum degree and all multiple edges lie between two of vo, w 1 , and w 2 , this works provided the pivot vertex is always a member of V(Go) \ {WO, v l , Q}, and provided that an edge incident with vo is coloured last. Thus, Go is Class 1, as required. ) three ( v o , q , v 2 ) vertices of Now suppose that Go has two ( w o , ~ or maximum degree; this case only arises if T(G) 5 A(G). Let s = ~ G O ( W v2)~ , and let & + I , . . , Rr+a-l be s - 1 disjoint pairs of single edges, one edge of each pair joining 01 and 212, the other incident with vo but not v1 or v2; since ~ G o ( w o=) A(Go) 2 .(Go), it follows that s = m ~ o ( v 1 , v 2=) .(Go) - mGO('%,v1) - mGO(V0, W 2 ) 5 dGO(W0) - mGo (v0,v l ) - mGO(UO,v2) = p - 2, so the selection of & + I , ., Rr++1 is possible. If we put G' = Go \ (&+I U U Rr+d-l) then T(G') = .(Go) - s 1 and A(G') = A(GO) - s 1. We have ~ G o ( w ~ ,twmGO(w~9212) ~ ) p - 2 = dGo(v0) 2 dGO(W1) = mGO(WO,Wl)+1)+GO(vl,W2)+p-2, somGo(vO,~2) 2 mGO(v19~2). Extending this in a similar way, since d ~ o ( v 02) d ~ o ( v 12) ~ G O ( V ~we) , get , V~ ~G )o ( w O , W2~ mG0(2r1,212). ) Since there is at least similarly ~ G O ( V ~ 2 one multiple edge in Go, ~ G o ( w o , v2~ 2. ) Therefore,

.

..

+

+

+

A(G') = d ~ l ( v 0 ) = mGO(oO,vl)+mGO(oO,W2)+ { ( p - 2 ) - m G O ( W l , v 2 ) + 1 } 2 mGO(v0~0,vl)( p - 2 ) 1 2p+1.

+

+

Therefore, G' also has a t most three vertices of maximum degree; in fact, the vertices of maximum degree in Go are the vertices of maximum degree in G'. Furthermore, in G', each multiple edge is incident with WO. Suppose that Go, and thus G', has two vertices of maximum de ee. Since IE(H)I 2 8 and F1 was a maximal matching, we have G' # K2n+l

Y

for any n when p 2 5 . However, it is possible that G' = Kin+l when p = 4 (so that p 1 = 2n 1 = 5 ) and s = 1. In that case Go = G' = K i and x'(Go) = A(GO)+l;but then IE(G)I 2 2A(H)+3, a caae we have excluded here. So suppose now that G' # Kin+l for any n. Then, by Lemma 3, G' is Class 1, and so Go is Class 1. Now suppose that Go has three vertices of maximum degree; then so does G'. Again, since IE(H)I 2 8 and F1 was a maximal matching, if p 2 5 then G' # I(,*,+, for any n. However, it is possible that G' = Kin+, when p = 4 (so that p 1 = 2n 1 = 5 ) and s = 2. In that case, G consists of a K5 to which three further edges are added, so that there are 3 vertices of degree 6 and two of degree 4; but, as before, for this graph,

+

+

+

+

98

A: G. Chetwynd and A. J. W. Hilton

IE(G)I 2 2A(H) + 3, a case we have excluded here. Therefore, we may assume that G' # K;,+, for any n. But then, by Lemma 4, G' is Class 1 and so Go is Class 1. We now describe how to obtain the desired colouring of G. If G had 2 or 3 vertices of maximum degree, we colour the edges of &+I, . , Rz+s-l with colours c,+1, .. . ,C , + ~ - I , respectively. If Go # G* and T(G) > A(G), we colour the edges of FA(G)+1, . . ., F,(a)with colours cA(G)+1, ,c 7 ( ~ ) , respectively. If G # G* then we colour the edges of F1, P2, , P, with c 1 , .. .,c,, respectively. Finally, we colour the remaining edges with colours from ( c l ,, * c ~* (A *( G ) , ~ ( G ) )not } used so far. This completes the proof of 0 Lemma 5 .

..

. .. .. .

Lemma 6 . Let H be a loopless multigraph on two vertices, vo and v1, with IE(H)I 2 2. Let J be a simple graph on p - 1 > 0 vertices 212, .., vp. Let G be a loopless multigraph formed from J b y joining each vertex of H to each vertex of J by at least one edge. Then

.

either G is Class 1, or x'(G) = r(G) and there is a vertez v' E V ( J ) such that G has T(G)edges between 00, v1, and v', or G = K L l , where p is even, in which case x'(G) = A(G) t 1, or IE(H)I = 2 , p = 4, and $(G) = A(G) 1.

+

Proof. The result is clearly true if p = 2, so we shall assume that p 2 3. It is also clearly true if A(G) = 3, so we shall suppose A(G) > 3 and use induction on A(G). Let V ( H )= {WO, q}. If there is a vertex v* E {WO, v 1 ) with which each non-simple edge is incident, then G has a t most two vertices of maximum degree, one of which is v*, so, by Lemma 3, the result is true. Therefore, we shall suppose there is no such vertex v*. Since IE(H)I 2 2, the only way in which we could have x'(G) = T(G) is when there are three vertices including 00 and 01 with T(G) edges between them altogether. For the number of edges between vi, va, and Wb, where i E ( 0 , l ) and v,,vb E V ( J )is

and, if p 2 4, the number of edges between V a , V b , v , E V ( J )is at most 3 < p < A(G). Furthermore, unless p = 3, if x'(G) = T ( G ) then there is only one vertex v' such that VO, v1, and v' have T ( G )edges between them. For

The Edge-Chromatic Class of Graphs with Large Maximum Degree

99

Therefore, p 5 3 , and so p = 3. It follows that if x’(G) = r(G) then A(G) = r(G). We also observe that there is at most one vertex d1 E V ( J ) with d ~ ( v ” )= A(G). For suppose there were two such vertices, v” and vrN, then

Therefore, 21E(H)I 5 2, a contradiction. Also, if x’(G) = r(G) and there is a vertex v“ E V ( J )with d ~ ( v ” = ) A(G) then the vertices vo, v1, and v“ have r(G) edges between them. For if there were a vertex v, # v” such that W O , v1, and w, have r ( G ) edges between them, then

+ 21E(fi)I + mG(vO,v”>-k mG(v0,vz) + mG(vl,v”) + mG(vl,%) - dG(v0) + dG(v1) 5 2A(G) 5 dG(v”) + r(G) <

2(P- 3 )

5 (P- 2 ) + IE(H)I

+mG(~O,~”)-kmG(vO,v,)+mG(vl,v”)+ mG(vl,vz).

+

Therefore, p IE(H)I 5 4, and so p 5 2 , a contradiction. Let F be a maximal independent set of single edges, including one edge joining vo and 01 and, if there exists a vertex v“ E V ( J )with d ~ ( v ” )= A(G), one edge on v”. Clearly, A(G \ F ) = A(G) - 1 and r ( G \ F ) = r(G) - 1. Suppose IE(H)I > 2 and p # 4. Then

by induction since p # 4, and if p is odd then E((G $ ( p 2 - 3 p t 2 ) . Therefore,

\ F)\

x‘(G) - 1 5 max(A(G) - l , r ( G ) - 1) 5 x’(G) - 1.

(wo, q}) <

A. G.Chetwynd and A. J . W. Hilton

100

So x'(G) = max(A(G),.r(G)), as required. Now suppose that IE(H)I = 2 and that G contains no independent pair of non-simple edges. Then, by Lemma 4, we have x'(G \ F ) = max(A(G \ F ) , r(G \ F ) ) unless G \ F = Kin+,for some n 2 2. But, if n 2 3 then G \ F has at least two pairs of vertices which are not joined by an edge, so G \ F # I<;,+,. If n = 2 and G \ F = K; then p = 4 and x'(G) = A(G) 1. Next, suppose that IE(H)I = 2 and G contains an independent pair of non-simple edges. Let F* be a maximal independent set of single edges which includes one edge from each of an independent pair of non-simple edges. We may clearly assume that one of these edges is incident with v", if a vertex v" E V ( J ) with d ~ ( v " = ) A(G) exists, and if p > 3, that one of these edges is an edge between two of wo, w1, and wz, where there are .r(G) edges altogether between these vertices if x'(G) = T(G). Then, if p 6 {3,4}, we may argue as above when I E ( H ) )> 2, but with F' instead of F , and show that x'(G) = max(A(G),.r(G)), as required. If p = 3, the same argument works, except that there might be two vertices w,, instead of just one. Finally, suppose that p = 4 and that either IE(H)( = 2 and G has two independent non-simple edges, or that IE(H)I 2 3. Suppose also that x'(G) 2 A(G) 4-1. We must show that x'(G) = max(A(G) 4-~ , T ( G ) ) If. IE(H)I 2 3 then

+

x'(G) - 1 L x'(G \ F ) 5 max( A(G \ F ) 1,.r(G \ F ) ) 5 max(A(G), T(G) - 1)

+

(by induction)

5 X'(G) - 1, so x'(G) = max(A(G),.r(G) - 1). If IE(H)I = 2 then the same argument works, but with F* instead of F . Lemma 6 now follows by induction. 0

3

Proof of the theorem

We can now prove the theorem. The sufficiency is trivial. We shall prove the necessity. Let IV(G)I = 2n 1; if n = 1 then the necessity is trivially true, so we may assume that n 2 2. First, we add edges to G, forming a graph G* with as many edges as possible, subject t o G* being a simple graph, A(G*) = A(G) and G* also containing no overfull subgraph H * with A(H*) = A(G*).

+

The Edge-Chromatic Class of Graphs with Large Maximum Degree

101

Suppose first that G* contains a subgraph HI with A(H') = A(G*) which is maximally non-overfull, that is,

Then, since A(G*) = 2n - 2, either (i) IE(G*)I = A(G*) LiIV(G*)l] = (2n - 2)" (then H' = G*), or (ii) H' = Iizn and G* has an isolated vertex, or (iii) there are two vertices

01

and wz so that

(then H' = G* \ {w~,wz}).

If (i) holds then, by Lemma 1, G* is Class 1, and so G is Class 1. If (ii) holds then clearly G*, and therefore G, is Class 1. So suppose that (iii) holds. The graph G* \ { T J ~ , w ~is} a non-overfull graph of order 2n - 1 and maximum degree 2n - 2, and so, by Lemma 2, G* \ { q , WZ} is Class 1. Edge-colour G* \ {01,02} with 2n - 2 colours. Since IE(G* \ (211, WZ}) = ( n - 1)(2n - 2), it follows that each colour appears on n - 1 edges, and so is missing from exactly one vertex. Furthermore,

I

= (2n - 1)(2n - 2) - 21E(G* \

{2)1,02})1

= (2n - 1)(2n - 2) - (2n - 2)(2n - 2)

= 2n - 2,

+

so d ~ ( 0 1 ) d ~ ( ~ 52 )2n - 2. Therefore, the edges of G* between the sets {vl, v2} and V(G*) \ {v1,02} can be coloured, each edge receiving a different colour. If there are 2n - 2 edges between the sets { w ~ , w z }and V ( C )\ { W I , WZ} then 01 and 02 are non-adjacent, for otherwise

IE(G*)I = 1 + (2n - 2)

+ ( n - 1)(2n - 2) = n(2n - 2) + 1,

A. G . Chetwynd and A. J. W. Hilton

102

so G* would be overfull, a contradiction. If there are fewer than 2n-2 edges between the sets ( ~ 1 , 2 1 2 ) and V ( G * )\ { V I , ~ 2 then ) w102 E E(G*),and 211212 can be coloured with a colour not used so far on w1 and ~ 2 . Therefore, we may suppose that

I E(G*1I < NG)

14IW) IJ

(3)

',

and that

IEV* \ { V l , ' u 2 H I < W)[hlV(G)l - 11

(4)

for each pair ~ 1 , 1 ) 2E V ( G * ) . We may also assume that each pair of vertices of degree less than A(G) = 2 n - 2 is joined by an edge of G*. [Note, this is an important point in our proof here; it seems to be the point which prevents us proving Lemma 1 and the theorem together.] Let the vertices of G* of degree less than 2 n - 2 be ~ 1 . ,. . , I+,,and let the remaining vertices be vP+1,. . . , 02,+1. I f 0 5 p 5 1 then

IE(G*)I 2 $ ( 2 n

+ 1 - p ) ( 2 n - 2 ) 2 2n2 - 2n.

But, by (3), IE(G*)I < ( 2 n - 2 ) n , a contradiction. Similarly, if p = 2 then, by (3),

E(G*)= ${dG*(ZI1)

+ d G * ( V 2 ) + ( 2 n - 1 ) ( 2 n - 2 ) ) < ( 2 n - 2).

d ~ * ( v l 4) dp(V2)

< 2 n - 2. But on the other hand, by (4),

so that

${( 2 n -2)(2n- 1 )- dG* (.I) - d ~ (*V 2 ) ) 5 IE(G*\ { v i ,

V2))

I < (272- 2)(12-

1)

+

so that ~ G * ( v I ) dG'(W2) > 2 n - 2, a contradiction. Thus, p 2 3. Each vertex of {'uP+l,.. . ,02,+1} is joined to at least p - 2 of ( ~ 1 , .. . ,vP}, and so the total number of edges from {vP+l,. ,112,+1} to { q , . ,wP} is a t least (272 1 - p ) ( p - 2). Thus, the average degree of wl, . . . , w p is at least (P - 1) ( p - 2 ) ( 2 n 1 - p ) / p . Then

+

..

+

c

+

v€V(G*)

since

..

( ( 2 n - 2)

- d G * ( V ) ) 5 2 ( 2 n + 1 - 2p),

(5)

The Edge-Chromatic Class of Graphs with Large Maximum Degree

103

Therefore, IE(G*)I 2 3{(2n

+ 1)(2n - 2 ) - 2(2n + 1 - 2 p ) ) = 2n2 - 3n + 2 p - 2.

But, by (3), IE(G*)I < 2n2-2n. Therefore, 2n2-2n-1 2 2n2-3n+2p-2, so that p 5 i ( n 1 ) . Since p 2 3, we also have n 2 5 . Next, we describe how we adjoin to G* a further vertex wo and further edges on the vertices W O , .. ., wp in such a way that a regular Class 1 loopless multigraph Gt of degree 2n - 2 on vertices DO, . . , w ~ ~ +is lformed. After we have described how to construct G t , we shall show that it is Class 1. For 1 5 i 5 2n 1, let 6j = 2n - 2 - d~*(wi)and let 60 = 2n - 2. We may suppose that 60 2 61 2 ... 2 SP+1 = ... = 6zn+1 = 0 (otherwise, we relabel w1, ... , vp). We first note that

+

.

+

i=l

since P

+ 1)(2n - 2 ) - 21E(G*)I 2 (an + 1)(2n - 2 ) - 2{(2n - 2)"

6i = (2n i=l

- 1)

= 2n

> 60.

#[(c%,

To form Gt from G* we first add S i ) - 601 extra edges onto the vertices 81, . .. , wp one by one, at each stage selecting two vertices whose degree is least at that stage t o place the new edge on and keeping the degree of w1 least. After placing these extra edges on w1, . , , wp, if they all have a common vertex then that vertex will be w 1 , and all the vertices joined to w1 by an extra edge will have degrees differing from each other by at most one. , wpl for some pl E (2,. . ,p } . If We may suppose these vertices are w2, the extra edges do not all have a vertex in common then the degrees of all the vertices incident with an extra edge will differ from each other by at most one. In this case, we may suppose that these vertices are v 1 , . . . , w m for some pl E ( 2 , . . . , p } . Then the edges joining 00 t o ( ~ 1 , .. . ,wP} are inserted so as to make the regular loopless multigraph Gt of degree 2n - 2. By the construction we may assume

.

...

and

.

104

A. G. Chetwynd and A. J . W. Hilton

[It follows that r n c t ( ' u O , 'up) 2 1 since p 5 !j(nt 1) 5 2n - 2 , ;ts 5 5 3n.l From ( 5 ) , it follows that the number of edges placed on {q, . . ,vp} [ignoring the edges on W O ] is at most

.

+ 1 - 2p) - ( 2 n - 2 ) ) = n + 2 - 2p. From Gt, delete a vertex x E . ., ~ 2 ~ + 1 } Then . x was joined to 4{2(2n

{'up+l,.

at least p - 2 of 01, .. ., v,, had no multiple edges on it, and Gt \ {x} has two vertices of maximum degree 2 n - 2 , apart from WO. If possible, select 2 so that, in addition, in Gt \ {x}, at least one of these two vertices (f 'uo) of maximum degree is in {'up+l,. ,'uzn+l}. If this is not possible, then, again if possible, in addition select x so that there exists a vertex y E { w p + l , . ,w2,+1} such that y is adjacent to one of these two vertices (# vo) of maximum degree in Gt \ {z} and is non-adjacent to the other. The possibility that we cannot choose x to satisfy in addition either of these requirements is considered at the end. In the case when one of these two additional possible requirements for the selection of z can be satisfied, we show that Gt \ {x} can be edgecoloured with 2 n - 2 colours. When G is a multigraph and V' c V(G), let G(V') denote the submultigraph of G induced by V'. Let db(v) denote the degree of 'u in the multigraph Gt(w0,. . ,w,). First, we colour E(Gt(v0,. . . ,w,)) with 2 n - 2 colours. The method by which we do this is not straightforward, and we need to consider several cases.

..

..

.

CASE

I.

&u;)

t mct(wo,v;) 5 2 n - 2 for 1 5 i 5 p .

.

First, we colour the edges of G(v1,. . ,v,) with p colours. We then colour the further edges of Gt('u1,. ,' u p ) which are not in G(v1,. . ,wP) with some further colours. The number of such 'extra' edges is at most n 2 - 2p, so the total number of colours employed thus far is at most ( n 4- 2 - 2p) t p = n 2 - p 5 n - 1 5 2 n - 2. Finally, with wo as a pivot, we colour the 2 n - 2 edges on 'uo by Vizing's original argument in the case when there are multiple edges. This works provided that the numerical condition of this case is satisfied.

..

+

.

+

2. &v;) t rnGt(v0,w;) 2 2 n - 1 for gome i, 1 5 i 5 p . Since the number of 'extra' edges [i.e., edges of E(Gt(w1, ...,' u p ) ) E ( G * ( q , . , ' u p ) ) ] is a t most n + 2 - 2p, it follows that CASE

..

\

The Edge-Chromatic Class of Graphs with Large Maximum Degree

105

Therefore, for some i, 2n - 1 5 d!(v;) t mGt(vO,vi) 5 n

-+

- p t 1t 2mGt(vO, vi)

so that n 1 5 n 4- p - 2 5 2mGt(vo, w;). In other words, vo is joined to v; by at least t 1) edges. Since the degree of vo is 2n - 2, there are at most three such vertices v;. From our definition of pl, it follows that if there is more than one such vertex then pl 5 3 and mGt(vo,v;) 2 i ( n 1) for 1 5 i 5 pl. Further, if for 1 5 i < j 5 3 we have Gt({vl,.. .,v2,+l} \ {v;,vj)) = G*({vi,. . . ,vzn+i} \ (vi, vj}) then

i(n

+

Therefore, 4n plies (6).

- 6 2 2{mGt('UO,2);) t mGt(vO,Ifj) + mGt(vi,vj)}, which im-

CASE2(i). Imct(vo,v;) - mGt(vO,2)j)(5 1 for 15 i < j 5 pl. From the above remarks, pl 5 3, so G t ( { s , . ..,~ 2 ~ + 1\}{vi, vj}) = G*({Q, ...,~ 2 ~ + 1\ }{viyvj}) for 1 5 i < j 5 3, and so (6) is true. Let

Go(vo, ... ,vP) be the multigraph obtained from Gt( wo, . . . ,vP) by replacing each multiple edge joining one of {vo, v1, v2, v3} to one of ( ~ 4 , .. .,vP} by a single edge. Let A0 = A(GO(v0,. . , u p ) ) and TO = ~(G'(v0,. ,vp)). By Lemma 5 , except in some cases (specified in Lemma 5 ) , Go(vo, ...,vP) can be edge-coloured with max(A0, TO) colours, c1, ..., c ~ + , , ~ ) . If 70 > A0

.

..

A. G. Chetwynd and A. J . W. Hilton

106

..

then, since the number of edges of Gt(v1,. .., u p ) \ G*(v1,. ,vp) is at most n - p 1, vo is one of the three vertices between which there are TO edges; we may suppose that vl and v2 are the other two. By Lemma 5 , we can make the edge-colouring of Go(vo,. , u p ) have the property that colours CAo+l, . , cm appear just on the edges joining 01 and v2. In view of ( 6 ) , we can extend this edge-colouring to an edge-colouring of Gt(w0,. ,vP) with the 2n-2 colours c1, .. ,~ ~ - as 2 required. , If A0 2 TO, this is clearly also true. Now suppose that Go(vo,. . . ,vP) is one of the exceptional cases of Lemma 5 . Since (2n - 2) - p 2 (2n - 2) - +(n I) = +(3n - 5 ) 2 5 , in cases (ii)(a)-(ii)(d) of Lemma 5 , (2n - 2) - A0 2 2. In each of these cases, colour an edge joining 211 and 0 2 with CA,+1. From Lemma 5 , it follows that the remaining edges of Go(vo,. ,up) can be edge-coloured with c1, . ,C A ~ We . can then extend this to an edge-colouring of Gt(v0,. . ., u p ) with c1, , ~ ~ - as 2 , required. In case (ii)(e) of Lemma 5, we have p = 4. The only edges of Gt(v0,. , u p ) which are not in GO(v0,. ,vP) are all but one of the edges from vo to v4. Let G i = GO(v1,. . . ,vp), and let H,” = G;(vo,vl, w2,v3). Suppose there are y 1 edges from vg to 214. Then d p ( v 4 ) = 2n - 2 - y. Since IE(G;)I 2 2A(H,”) 3, it follows that for Go = Go(vl, . . . , v p ) , l d ~ o ( ~-; )dGo(vj)I 5 1 (0 5 i < j 5 3), SO d G i ( v i ) 2 2n - 2 - (Z 1) = 2n - 3 - y. Thus, there are at most y 1 edges from each of w l , v2, and v3 to ( ~ 5 , . .. ,v2,+1}, and at most one edge from v4 to ( ~ 5 , .. ,v2,+1}. There are no edges from 00 to ( ~ 5 , . . ,~ 2 ~ + 1 } Thus, .

+

..

..

..

.

+

..

..

. ..

..

..

+

+

+

.

+

.

4y+42 3~+42(2n+l-p)(p-2)=2(2~~-3)= 4n-6.

Therefore, y 2 n - 2 1 3. In this case, we colour one edge between the vertices v1, 02, and v3 with CAo+l and another with CA0+2, and then complete the edge-colouring of Gt(w0,. ,vp) in the same way as in cases (ii)(a)-(d) of Lemma 5, as described above.

..

CASE2(ii). lmGt(v0,vi) - mG+(vo,vj)l > 1 for some i , j E {I,. .. ,n}. All multiple edges are incident with vo or v1, so for 1 5 i < j 5 3 we haveGt({vi,...,vzn+1) \{v;,vj}) = G * ( { ~ I..., , ~ , + l } \ { v ; , q } ) , and so (4) holds. Since (4) holds, r(Gt) < A(G), and so, by Lemma 6, either

.. , u p ) ) = A(Gt(v0,. .. ,vp)), or

0

x‘(Gt(v0,.

0

Gt(vo,.

0

mGt(v0,vl) = 2, p = 4, and

.., u p ) = I
x’(Gt(vo,.

.. ,vP)) = A(Gt(v0,. .., u p ) )+ 1.

The Edge-Chromatic Class of Graphs with Large Maximum Degree

107

However, if Gt(v0,. ..,wP) = K i + , and p is even then vo is joined to ( 2 n - 2 ) - 1 vertices, so p = 2 n - 3. But p 5 + l ) , and it follows that n 5 2 , a contradiction. But suppose that mGt(vO,2)1) = 2 , p = 4, and x'(Gt('u0,. .. ,'u,)) = A(Gt(w0,. , u p ) ) 1. If there is just one vertex with which each multiple edge is incident then Gt = I<;. But then 2 n - 2 = d,g(vO) = 5, a contradiction. If there is no such vertex then, by the construction, 2 = mGt('u0,vl) 2 mGt(wO,vi) ( 2 I i I 4), so 2 n - 2 = dGt('uO) 5 7, and so n 5 4, a contradiction. Therefore, we are only left with the cme x'(Gt(w0,. . . ,vp)) = A(Gt(w0,. ..,wp)), and so Gt(w0,. .. ,vP) can be edge-coloured with C I , ,~ 2 ~ - 2as, required. This completes the colouring of E(Gt(v0,. . . ,wp)) with 2 n - 2 colours.

i(n

..

+

.. .

Next, we colour E ( G t \ ( 0 0 , . . . ,v,, z}) with 2 n - 2 colours. This can be done by Lemma 3 since Gt \ {WO, . . . ,vp, z} is a simple graph with a t most two vertices of degree 2 n - 2 , the remaining vertices having lower degree. } Finally, the edges joining Gt(w0,. . ,vP) to Gt({vp+l,...,~ 2 ~ + \1 {z}) are coloured. This can be done by Vizing's original argument if the pivot vertex is always in Gt \ (01,. . . ,vp,z} and if the final edge to be coloured is as follows. Let the two vertices of maximum degree in Gt \ {z} be b and c. If a t least one of these, say c, is in Gt \ (211,. . . ,'up, z}, and if b and c are adjacent then colour the edge bc last. If b and c are not joined then colour an edge on b next to last, and colour an edge on c last. If both b and c are in the set {q,. . ., ' u p } , but there is a vertex y E {wp+l,... ,v2,+1} \{z} joined to b but not to c, colour an edge on c next to last, and the edge yb last. Vizing's original argument will apply as at any stage during the construction of the fans there will always be a further colour available on the vertex at the other end of the pivot of the most recently adjoined edge; except at the final stage, such a further colour is then used to define the next stage of the fan. This yields the desired edge-colouring of Gt \ {z} with 2 n - 2 colours. In this edge-colouring, each colour is missing from exactly one vertex, and each vertex of degree 2 n - 3 has exactly one colour missing a t it. Therefore, this edge-colouring of Gt \ (z} with 2n - 2 colours can be extended to an edge-colouring of Gt with 2 n - 2 colours. Therefore, G*, and thus also G, is Class 1. It remains to consider the possibility that z cannot be chosen to satisfy either of the additional conditions. Since it is not possible to select 2 so that, in addition t o the other requirements, in Gt \ {z}, a t least one of the two vertices other than 00 of maximum degree is in {vp+l,.. . ,v2,+1}, it follows that the subgraph of Gt induced by { w u p + l , . , .,wzn+1} is complete. We already know that in Gt, each pair of vertices v;,vj (1 5 i < j 5 p )

.

A. G. Chetwynd and A. J . W. Hilton

108

are joined to each other by a t least one edge. Since A(Gt) = 2n - 2 , it follows that in Gt, each vertex of { v p + l , . .,~ 2 ~ + is 1 }joined to all but two of {vl, . ,vp}, and, since each vertex of ( ~ 1 , . . , u p } is joined to vo, each such vertex is not joined to at least three of { v p + l , . . . , ~ 2 ~ + 1 }Since . it is not possible to select x so that, in addition to the other requirements, there is a vertex y E {vp+l,. . .,~ 2 ~ + 1such } that y is adjacent to one of the two vertices of maximum degree in Gt \ {z} and non-adjacent to the other, it now follows that p is even and that, for some permutations ( ~ 1 , ,. , ,wp) of ( ~ 1 , ,vp) and ( w p + l , . , q n + 1 ) of ( v p + l , . . .,vqn+1), there are integers ro = 0, r1, ,rp12= 2 n t 1-p such that for 1 I i 5 i p , w2i-1 and w2i are both non-adjacent t o wp+r,-l+l, . . , wp+ri,but are both adjacent t o the rest of { w p + l , . . .,w2,,+1}, and that r; - r;-1 1 3. Let w o = vo. We may suppose moreover that rnct(wo,w4) 2 3. For if we could not make this assumption then it would be because rnGt(w0,w;)I 2 (1 5 i I p ) ; but that would imply

.

..

.. .

.

..

. ..

.

and so n 5 3, a contradiction. As p i s even and p 2 3, it follows that p 2 4. Let F1 be a 1-factor or near 1-factor avoiding, say, w* of Gt({vl,. . ,vP} \ {wl, w2, w3, w4, v1, 212, v3}), and let F2 be a minimal set of edges including w1wp+rl+l, w 2 ~ ~ + ~ ~ + and w3wp+3 such that each of { ~ 1 , ~ 2 , ~ 3 , v 1 , v 2 , ~ 3 \, w{wq} * } is in one edge of F2, and each edge of F2 joins a vertex of { v ~ , . ., u p } to a distinct v2n+l}. For each Y E {wl, w27w3, v1, v2, v37 w*}\ {w4}, vertex of {vp+l, let 4(y) be such that y4(y) E F2. Let F3 be a 1-factor of { w p + l , . , w ~ ~ + l } \ 4({'Wl,w2,'u)3,~1,~2,~3,~*}\{~4}) [as eachof{~l,...,wp}\{W4} is on an edge of F1 u F2, it follows that the number of vertices of { w p + l , . . ,,w ~ ~ + l } which are not on edges of F2 is even]. Let F = F1 U F2 U F3 U (wOw4); then F is a 1-factor of Gt. Notice that in Gt \ F , the vertices vo, 01, v2, and v3 are all joined to each other and that there is at least one multiple edge on vo [this observation will be of use later when we apply Lemma 5 or 6 to Gt \ F ] . We now proceed to show that (Gt \ F ) \ {wp+l} is edge-colourable with 2 n - 3 colours. First, we show that Gt(v0,. . . ,vp)\F is edge-colourable with 2n - 3 colours. Let Gtt denote the graph Gt \ F , and let dit(w;) denote the degree of wi in Gtt(v0,. ,vp). The argument is now just a slight variation on the earlier argument, and we just indicate the main points of it.

.

.

--

..

+

..

The argument in this case has no real difference to the earlier argument.

2 ,

The Edge-Chromatic Class of Graphs with Large Maximum Degree

CASE2'(i).

lmGt[vO,vi) - mGt(vO,vj)(

109

5 1 for 1 <_ i < j 5 p l .

Then pl 5 3 so (6') is true. Let Goo(vo,. . . ,vp) be the multigraph obtained from Gtt( vo, . . ,v,) by replacing each multiple edge joining one of {vo, v1, ~2,213)to one of ( ~ 4 , ... ,u p } by a single edge. Lemma 5 applies to Gooin this case since the construction of F was such that each pair of vo, v1, 02, and v3 are adjacent in Goo.The argument proceeds as in Case 2(i), with very minor alterations in detail.

.

> 1 f o r s o m e i , j E {I,...,m}. = w4 then mGtt(v0,2(1) 2 2. If 01 # w4 then, by the construction

CASE2'(ii).

ImGt(vO,vi)-mGt(vO,vj)l

If v1 of F , mGt(v0,2)1) = mGtt(v0,vl). In either case, m G t t ( v O , q ) 2 2, and Lemma 6 is applicable. All multiple edges are incident with 00 or v1, so Gt({vl,.*.,vPn+l} \ { v i , v j } ) = G*({vl,...,v2n+l} \ {v;,vj}), and therefore (6') holds. The argument now proceeds without much change from Case 2(ii); one case becomes slightly more complicated: if IE(H)[ = 2, p = 4, x'(Gtt(vo,.. .,v,)) = A(Gtt(v0 ,...,v,)) 1, and there is no one vertex with which each multiple edge is incident, then mGt(?&q) 5 3 and 2n - 2 = dGt(vo) 5 10, so n 5 6. This yields the contradiction that p 5 li(n 1)1 = 3.

+

+

Thus imitating the earlier argument, it follows that (Gt \ F ) \ {vp+l}is edge-colourable with 2n - 3 colours. Clearly, Gt(wg+2,. . .,w2,+1) \ F is edge-colourable with 2n - 3 colours. Finally, we colour the remaining edges of (Gt \ F ) \ {20,+1}; this can be done by applying Vizing's original argument rather carefully. There are now four vertices of maximum degree, namely 200, 201, w2, and 20,+2. The pivot vertices of the fans are always t o be in the set {wp+2, ,w2,+1}. The edge 20~+,.~+2w1 is coloured last but two, the edge wp+T1+~w2 is coloured last but one, and an edge on wp+2 is coloured last of all; Vizing's argument works because at each stage the pivot vertex is joined t o no vertices at which 2n - 2 colours have been used. In this edge-colouring of (Gt \ F ) \ {w,+l}, each colour is missing from exactly one vertex, and each vertex of degree 2n - 4 has exactly one colour missing. Therefore, the edge-colouring can be extended to an

...

110

A. G. Chetwynd and A. J . W. Hilton

edge-colouring of Gt \ F with 2n - 3 colours, and so Gt and therefore G’ and thus G can be edge-coloured with 2n - 2 colours. 0 This completes the proof of the theorem.

Acknowledgement The authors started to write this paper whilst they attended a workshop on latin squares at Simon Fraser University, organized by K. Heinrich. They would like to thank her for inviting them and also the Natural Sciences and Engineering Research of Canada and Simon Fraser University who sponsored the event, and the Open University, England, who provided additional support for the authors.

References [l] A. G. Chetwynd and A. J. W. Hilton, “Partial edge-colourings of complete graphs or of graphs which are nearly complete,” Gmph Theory and Combinatorics, Vol. in honour of P. Erdos’ 70-th birthday, Academic Press (1984), 81-98. [2] A. G. Chetwynd and A. J. W. Hilton, “The chromatic index of graphs of even order with many edges,” J . Graph Theory 8 (1984), 463-470. [3] A. G. Chetwynd and A. J. W. Hilton, “Critical star multigraphs,” Graphs and Combinatorics 2 (1986), 209-221.

[4]A. G . Chetwynd and A. J. W. Hilton, “Star multigraphs with three vertices of maximum degree,” Math. Proc. Cambridge Phil. SOC.100 (1986), 303-317. [5] M. Plantholt, “The chromatic index of graphs with a spanning star,” J . Graph Theory 6 (1981), 5-13. [6] M. Plantholt, “On the chromatic index of graphs with large maximum degree,” Discrete Math. 47 (1983), 91-96. [7] V. G. Vizing, “On an estimate of the chromatic class of a pgraph,” Diskret. Analiz 3 (1964), 25-30. [In russian.]

Annals of Discrete Mathematics 4 1 (1989) 111-1 16 0 Elsevier Science Publishers B.V. (North-Holland)

On some Aspects of my Work with Gabriel Dirac P. Erdos Mat hemat ical Inst it U t e Hungarian Academy of Sciences Budapest, Hungary

Dedicated to the memory of my friend and coworker G. A . Dirac Several results and problems from areas of discrete mathematics of joint interest to G. A. Dirac and the author are discussed.

I must have known Dirac when he was a child in the 1930’s, but I really became aware of his existence when I visited England for the first time after the war in February and March 1949. We met in London and he told me of his work on chromatic graphs. Dirac defined a k-chromatic graph t o be vertex critical if the omission of any vertex decreases the chromatic number and edge critical if the removal of any edge decreases the chromatic number. I immediately liked these concepts very much and in fact felt somewhat foolish that I did not think of these natural and obviously fruitful concepts before. At that time I was already very interested in extremal problems and asked Gabriel to prove that for every k an edge critical k-chromatic graph must have o(n2) edges. More precisely define ff’(n) to be the largest integer for which there is a G ( n ; f f ) ( n ) )(i.e. a graph on n vertices and f f ’ ( n ) edges) which is k-chromatic and edge critical. Estimate or determine ff’(n) as accurately as possible. Trivially f g ’ ( 2 n + 1) = fg’(2n 2 ) = 2n 1 and I expected that for k > 3, ff’(n) = o(n2). To my great surprise very soon Dirac showed [3]:

+

f t ’ ( 4 n t 2) 2 (2n t 112

+

+ 4n + 2 .

(1)

After I recovered from my surprise I immediately asked: is (1) best possible? This question is still open. Toft [16] proved that 111

P. Erdos

112

where fi")(n)is the largest integer for which there is a graph G ( n ;f$")(n)) which is k-chromatic and vertex critical. It seems quite likely that for every k24

but as far as I know (3) is still open, and it has not even been proved that the limits exist. I also asked what about f f ' ( n ) and ft)(n). I still hoped that perhaps fi"'(n) = o(n2). In 1970 Toft [14]proved that n

Toft's proof is based on the idea of Dirac for the case k 2 6. Simonovits and I proved that f i e ) ( n )5 $+n which was later improved t o f i " ' ( n ) 5 $. It would be very desirable to improve (4)or my result with Simonovits and t o determine limn+m f i e ' ( n ) / n 2 . By the way, the graph of Toft has many vertices of bounded degree. This lead me to ask: Is there an edge critical four chromatic graph of n vertices each vertex of which has degree > cln? I conjecture with some trepidation that such a graph does not exist. The strongest known result is due to Simonovits [lo] and to Toft [15]. They proved that there is a G ( n ) which is four chromatic and edge critical, each vertex of which has degree > cn1I3. Dirac's six chromatic edge critical graph is regular of degree 5 -t2. It is not impossible that there is a four critical regular graph of n vertices and degree cln, but this seems unlikely to me. As far as I know there is no example of a regular edge critical four chromatic graph of degree L for L 2 6, but I expect that such graphs exist for every 1. By the way all of the early examples of these critical graphs contained abnormally large bipartite graphs and also small odd circuits. At first I thought that this was not an accident and must be so, but I learned that Rijdl found counterexamples some time ago. He showed that there is a constant Cr so that there is a four chromatic edge critical graph on n vertices and c,n2 edges which contains no odd circuit of size 5 2r -t 1. Exact results are not known, of course, though no doubt it can be proved that cr tends t o 0 as T tends t o infinity, but probably showing cr < cr-1 will not be easy.

On some Aspects of my Work with Gabriel Dirac

113

I recently heard from Toft the following conjecture of Dirac: Is it true that for every k 2 4 there is a k-chromatic vertex critical graph which remains k-chromatic if anyone of its edges is omitted. If the answer as expected is yes then one could ask whether it is true that for every k 2 4 and T there is a vertex critical k-chromatic graph which remains k-chromatic if any T of its edges are omitted. Perhaps there is an f(n) so that for every k 2 4 there is a k-chromatic vertex critical graph on n vertices which remains k-chromatic if any f ( n ) of its edges are omitted. If so one could try to determine the largest such f(n). Recently Toft asked: Is there a four chromatic edge critical graph with n vertices and c1n2 edges which can be made bipartite only by the omission of c2n2 edges? The original example of Toft could be made bipartite by the omission of cn edges. Rod1 and Stiebitz [ll]constructed such a graph but it is quite possible that the largest such c1 for which this is possible will be less than &, and it might be of some interest to determine the dependence of c2 from c1. Dirac [5] and I independently noticed the following strengthening of Turin's classical theorem. Denote by f(n;k ( r ) ) the smallest integer for which every G(n;f(n;k ( r ) ) ) contains a complete graph k ( r ) on T vertices. We noticed that then our G(n; f(n;k ( r ) ) )already contains a k ( ~ 1 ) from which a t most one edge is missing. In fact we showed that there is a k ( r - 1) and c,n further vertices y1,. . .yt, t = crn, for which each of the y's is joined - The exact value of c, is known only for T = 3. to every vertex of our k ( ~ 1). Bollobbs and I conjectured and Edwards proved that every G(n;[n2/4] 1 ) contains an edge (21,22) and y's, each of them joined to both 2 1 and 22, and is best possible. It would be very nice if one could prove some analogue to our result for hypergraphs. The simplest problem would be: denote by f(n;rC(3)(4))the smallest integer for which every 3-uniform hypergraph G(3)(n;f( n;! ~ ( ~ ) ( 4 ) ) ) on n vertices and f(n;k(3)(4)) triples contains all four triples of some set of 4 elements. The determination of f ( n ; / ~ ( ~ ) is( na) classic ) unsolved problem of Turin. Is it true that our G(3)(n; f(n;k(3)(n)))contains 5 points 2 1 , z2, 23, y1, y2 and all the triples of the quadruples { q , z 2 , ~ 3 , y 1 }and { ~ 1 , ~ 2 , ~ 3 , ~ Unfortunately 2 } ? nothing is known about this. Our result led me to the following theorem. By the theorem of Koviri, V.T. S6s and /') a complete bipartite graph ~ ( T , T ) . Turin [9]every G(n;c ~ n ~ - ~ contains I showed that every G(n; c2n2-l/') contains a k ( ~ 1 , ~1 ) with at most one edge missing [8]. Simonovits and I have the following problem: Let f(n;H )be the smallest integer for which every G(n;f(n;H ) ) contains a subgraph isomorphic

+

+

+

+

P. Erdos

114

t o the graph H . Is it then true that every G ( n ; f ( n ; H )contains ) at least two subgraphs isomorphic to R? We could not decide this question for C4’s. H = C4. In fact we expect that every G(n;f(n;‘24)) contains Nearly thirty years ago I conjectured that every G ( n ; 3 n- 5 ) contains a topological complete pentagon, i.e. 5 vertices every two of which are joined by paths no two of which have any interior point in common. It is easy to see that if true this conjecture is best possible. I soon found out that Dirac [6] anticipated me; he made the same conjecture before me. The conjecture is still open. As far as I know the best result is due t o Thomassen [13] who proved that every G(n;4n - 10) contains a topological complete pentagon. Despite our many contacts Dirac and I only had one joint paper [7]. This paper was perhaps undeservedly neglected; one reason was that we have few easily quotable theorems there, and do not state any unsolved problems. We prove there, among other results, that if G ( n ; n 3) is planar then it contains two edge disjoint circuits. Finally, let me remind the reader of a nice conjecture of Dirac [2]. This was conjectured also independently and simultaneously by Motzkin. Let there be given n points 21,. ,z, in the plane not all on a line. Then for at least one zi there are - c distinct lines among the (z;,zj), In a weaker form this conjecture has recently been proved by J. Beck [l]and independently by Szemerbdi and Trotter [12]. Dirac gave several beautiful consequences of his theory of critical graphs. For example, he proved in [4] that any graph G on the orientable surface S ,, y 2 1, for which equality holds in Heawood’s inequality

+

..

must contain a complete graph on H ( 7 ) vertices. He also obtained the similar result for non-orientable surfaces. This was one of the most significant contributions to map-colour-theory since Heawood’s pioneering paper in 1890. (The case of the torus was first obtained by P. Ungbr). Dirac made many deep and significant contributions t o several other parts of graph theory than those mentioned above, among them paths and circuits, Menger’s theorem and connectivity, extremal results for contractions and subdivisions, and infinite graph theory. He seemed much influenced by the work of Konig. His own influence is now present everywhere in graph theory. Finally, I want to thank B. Toft for his help in writing this paper and for supplying some references.

On some Aspects of my Work with Gabriel Dirac

115

References [l] J. Beck, “On the lattice property of the plane and some problems of Dirac, Motzkin and Erdos,” Combinatorica 3 (1983), 281-297. [2] G. A. Dirac, “Collinearity properties of sets of points,” Quarterly J. Math. 2 (1951), 221-227. [3] G. A. Dirac, “A property of 4-chromatic graphs and some remarks on critical graphs,” J. London Math. SOC.27 (1952), 429-437. [4] G. A. Dirac, “Map colour theorems,” Canad. J. Math. 4 (1952), 480490. [5] G. A. Dirac, “Extensions of Turin’s theorem on graphs,” Acta Math. Acad. Sci. Hungar. 14 (1963), 417-422. [6] G. A. Dirac, ‘‘Homomorphism theorems for graphs,” Math. Ann. 163 (1964), 69-80. [7] G. A. Dirac and P. Erdijs, “On the maximal number of independent circuits of a graph,” Acta Math. Acad. Sci. Hungar. 14 (1963), 79-94. [8] P. Erdos, “On some extremal problems in graph theory,” Israel J . Math. 4 (19651, 113-116. [9] T. Koviri, V. T. S6s and P. Turbn, “On a problem of Zarankiewicz,” Coll. Math. 3 (1954), 50-57.

[lo] M. Simonovits, “On colour-critical graphs,’’ Studia Sci. Math. Hungar. 7 (1972), 67-81.

[ll] M. Stiebitz, Beitriige mr Theorie der fiirbungskritischen Graphen, Dissertation, Technische Hochschule Ilmenau (1985). [12] E. Szemerbdi and W. T. Trotter, ‘‘ Extremal problems in discrete geometry,” Combinatorica 3 (1983), 381-392. [13] C. Thomassen, “Some homomorphism properties of graphs,” Math. Nachr. 64 (1974), 119-133. [14] B. Toft, “On the maximal number of edges of critical k-chromatic graphs,” Studia Sci. Math. Hungar. 5 (1970), 461-470.

116

P. Erdos

[15] B. Toft, “Two theorems on critical 4-chromatic graphs,” Studia Sci. Math. Hungar. 7 (1972), 83-89. [16] B. Toft, “An investigation of colour-critical graphs with complements of low connectivity,” Annals of Discrete Math. 3 (1978), 279-287.

Annals of Discrete Mathematics 4 1 (1989) 117-130 0 Elsevier Science Publishers B.V. (North-Holland)

Bandwidth versus Bandsize P. Erdos Mat hemat ical Inst it U t e Hungarian Academy of Sciences Budapest, Hungary

P. Hell School of Computing Science Simon Fraser University Burnaby, B.C., Canada

P. Winkler Department of Mathematics and Computer Science Emory University Atlanta, Georgia, USA

Dedicated to the m e m o r y of G. A . Dirac The bandwidth (bandsize) of a graph G is the minimum, over all bijections p : V ( G ) + { 1,2,.. . , IV(G)l},of the greatest difference (respectively the number of distinct differences) Ip(v)-p(w)I for vw E E(G)* Weshow that a raph on n vertices with bandsize k has bandwidth e; between k and cnl-,, and that this is best possible. In the process we obtain best possible asymptotic bounds on the bandwidth of circulant graphs. The bandwidth and bandsize of random graphs are also compared, the former turning out to be n - c1 logn and the latter a t least n cz(logn)2.

1

Introduction

The problem of bandwidth minimization was motivated by the needs of matrix manipulations in structural engineering [l, 10, 191: there it is desirable to store the matrices in such a way that their non-zero entries are all as close to the main diagonal as possible. Simultaneous row/column permutations are applied to transform a given (large, sparse, symmetric) matrix to a form in which all non-zeros are in a narrow band of sub- and super-diagonals surrounding the main diagonal. Given a symmetric matrix M = (m,,j) (i, j = 1,.. . ,n ) , one may consider the graph G on n vertices in

117

118

P. Erdos, P. Hell and P. WinJder

which i is adjacent to j just if mi,j # 0. We now give the precise statement of the bandwidth minimization problem in terms of graphs; its correspondence to the above problem for symmetric matrices is self-evident (via the translation just given). Let G be a graph with n vertices. A numbering of G is a bijection p : V(G) -+ (1,. . . ,n}; the numbers Ip(u)- p(w)I for uw E E(G) are called the edge-diferences of the numbering p . The width of a numbering p is its largest edge-difference. The bandwidth of a graph G, bw(G), is the smallest width of any numbering of G, A matrix in which all non-zeros are in a narrow band is convenient for storage and computation. For some applications it may be enough to have all the non-zero entries concentrated in a small number of sub- and superdiagonals. (Such matrices also seem to arise in certain applications, e.g., in queueing network analysis of job line production models, cf. [8]Figures 1 and 2.) The corresponding graph-theoretic analogue is the following: The size of a numbering p is the number of distinct edge-differences of p; the bandsize of a graph G, bs(G), is the smallest size of any numbering of G. Of course, it follows from the definitions that bs(G) 5 bw(G). The actual motivation for the study of bandsize (as opposed to the possible application described above) originated from an investigation of spanning subtrees of the Ic-dimensional hypercube Q k . J. Malkevitch studied spanning subtrees of Q k , and derived a number of their properties; M. Rosenfeld observed that such trees must admit a numbering with k edgedifferences (in fact, with edge-differences 1,2,4,.. . ,2“-’),and was led t o ask if there was a bound to the number of edge-differences required by numberings of trees of maximum degree k. The notion of bandsize is formally introduced in [12],where it is shown that the bandsize of the complete 2. Since the maximum binary tree of height n, Tn,is between and degree in T, is three, this answers Rosenfeld’s question in the negative. Throughout the paper, we reserve the symbol “lg” for logarithms base 2, and “ln” for logarithms base e. The tree Tn has w = 2”+l- 1 vertices; thus its bandsize is roughly c lg w (for f < c < $). On the other hand, it can be shown that the bandwidth of Tnis as high as (cf. [4]for a tree similar to Tn).When the bandsize is so much smaller than the bandwidth, storing the matrix in the form we suggest, with few non-zero diagonals, would seem particularly attractive. In this paper we take up the comparison between bandwidth and bandsize. The largest bandwidth among all graphs of fixed bandsize is asymptot-

i

6,

9+

Bandwidth versus Bandsize

119

ically determined in the next section: a qraph with n vertices and bandsize k can have bandwidth as large as O(nl-r), but no more. Our method also yields best asymptotic bounds for the bandwidth of circulant graphs. In the last section we compare the bandwidth and bandsize of random graphs: it turns out that their values are quite close, n - c1 logn for bandwidth, and a t least n - c z ( 1 0 g n ) ~for bandsize. The bandwidth problem, i.e., the problem of deciding for a given graph G and integer k, whether there exists a numbering of G of width at most k, is well known to be NP-complete, even in the case of trees, [16, 91. On the other hand, if k is fixed, the problem of deciding if bw(G) 5 k can be solved in polynomial time, [17, 111. In contrast to this, the problem of deciding if bs(G) 5 k is NP-complete for every fixed k 2 2, [18]. A variety of other numbering problems have been studied recently, [6]. For instance the minsum (or optimal linear arrangement) problem [6] may be stated as follows: The sum of a numbering p is the sum of all its edgedifferences; the minsum of a graph G is the smallest sum of any numbering of G. The reader may find it amusing to note that the largest size of a numbering of G bears an obvious relation to graceful graphs; this notion, in some sense dual t o the notion of bandsize of a graph, may be called the gracesize of G, gs(G). Because of the famous graceful graph conjecture, it could be interesting to prove non-trivial lower bounds on the gracesize of trees. (In this terminology, the graceful graph conjecture asserts that the gracesize of any tree is equal to its number of edges.)

2 2.1

The extreiiial case General remarks

In this section, and the next, we prove the following theorem:

Theorem 1. Let k be fixed. A graph G with n vertices and bandsize k has bandwidth only O(nl-k). Moreover, this bound is best possible. Let I = { i 1 , i 2 , . . . ,il} be a set of integers, 0 < il < i 2 < . . . < il < n. The linear graph L n ( l ) (or L,(il,iz,.. .,il)) has the vertex set 2, = {0,1,. . . ,n - 1) and the edge-set { u v : Iu - E I } . The circulant graph C,(I) (or Cn(i1,i2,., . ,ir)) also has the vertex-set 2, and its edge-set is { u v : Iu - 01 = i (mod n ) for some i E I } . A graph is linear if it is isomorphic to some linear graph, and is a circulant if it is isomorphic to some circulant graph. Note that each L n ( l ) is a spanning subgraph of

VI

Cn(1).

P. Erdos, P. Hell and P. Winkler

120

Let G be a graph with n vertices. It follows from the definitions that the bandwidth of G is the smallest integer I such that G is isomorphic to a spanning subgraph of L,( 1,2,. . . , I ) (the isomophism taking a vertex numbered x to the vertex z - 1 of Ln( 1,2,. , . ,1 ) ) , and the bandsize of G is the smallest integer I such that G is isomophic to a spanning subgraph of some Ln( il ,i2, . . . ,il). Our proof of Theorem 1 (or rather of Theorem 2, below) depends heavily on the use of the natural “circular” extension of these notions: The circular bandwidth of the graph G (still with n vertices), cbw(G), is the smallest I such that G is isomorphic t o a spanning subgraph of C,(l, 2,. . . ,I); the circular bandsize of G, cbs(G), is the smallest I such that G is isomorphic to a spanning subgraph of some C,(il, i2, . . ,il). (Circular bandwidth shall play a central role in our proof; circular bandsize is introduced only for symmetry.) We shall also need the notion of “circular length”; formally the (circular) n-norm of a positive integer i, Ililln, is the unique lbl, - 5 6 _< F, such that i = a n 6. (Thus for i in Z,, the n-norm of i is the distance from 0 t o i in the graph Cn(l).) If we call the norm of a numbering the largest norm of its edge-differences, then the circular bandwidth of a graph is the smallest norm of its numberings. (If we call the normsize of a numbering the number of distinct norms of its edge-differences, then the circular bandsize of a graph.is the smallest normsize of its numberings.) The usefulness of circular bandwidth is due to the fact that

.

9

+

cbw(G) 5 bw(G) 5 2cbw(G). The first inequality follows directly from the definitions. To prove the second inequality it is enough to show that bw(C,( 1,2,. . . , l ) ) 5 21 :

+

Number all vertices x = O , l , . . . , - 1 by 2z 1, and all vertices y = [:I,, . . , n - 1 by 2(n - y). It is straightforward to verify that this is a numbering of Cn(1,2,. . . ,I) and that the edge-difference of any edge uv with w = u 1 (mod n ) is either 1 or 2; hence any edge-difference is at most 21. In the next subsection we shall prove the following result:

+

.

Theorem 2. bw(Cn(il,i2,. . ,ik)) I 4nl-k. Proof of Theorem 1 (from Theorem 2). A graph G with n vertices and bandsize k is (up to isomorphism) a spanning subgraph of L,(I) for

Bandwidth versus Bandsize

121

By Theorem som I with k elements; hence a spanning subgraph of Cn(I). 2, its bandwidth is only O ( n l - i ) . This bound is best possible, as for n = mk, the graph G = Ln(1,m,m2,...,mk-1 ) has bandwidth at least

(for k fixed and m + m). To see this, note that any two vertices of G can be joined by a path . m edges (at most m steps of type mk-', at most steps of at most of each of the types mk-2,.. . ,m, 1). Thus, whatever the numbering of G, the shortest path joining the vertices numbered 1 and n must have some edge-difference a t least

9

-n2- -1. v

m

2 k+l

mk-l

- 1. 0

Note that

e 4 is

maximized for k = Inn; hence

Corollary 1. The ratio of1bandwidth to bandsize for a graph on n vertices cannot ezceed (&) .4n1-

2.2

G.

The bandwidth of circulants

Here we prove Theorem 2; we restrict our attention to connected circulants (as the bandwidth of a graph is the maximum bandwidth of its components). In fact we shall show that (for any fixed k) .max bw(Cn(il,i2,. $1r...rlk

.., i k ) )

= @(nl-~).

. . ,i k ) )

= n(nl-i),

1

(2)

The lower bound, ma? bw(C,(il, i 2 , .

il

p...j$k

follows the same way it did for linear graphs. The remainder of this subsection contains the proof of cbw(Cn(i1,i2,. . .,ik)) 2 2n1-i,

(3)

which implies (2) and Theorem 2 because of (1). In other words, we seek an isomorphism of any Cn(i1,i 2 , . .. ,ik) onto a subgraph of cn(172,. . . , I ) with 1 5 2nl-k.

P. Erdos, P. Hell and P. Winkler

122

Lemma 1. Let n, k, and il,i2,,. . , i k be given positive integers. Then there exists an integer m with 0 < m < n such that Ilm ijll,, 5 n'-k for all j = 1 , 2 , . . ,k .

.

Proof. Note that the n-norm satisfies the triangle inequality. Let S ( n ,k) denote the torus [O,n)kwith entries taken modulo n. Let

-

A = { ( m i1,m. i2,.

..

,

m

a

ik)

E S(n,k) :0 5 m < n},

and for each m let 1 2

1

B, = { ( y l , y 2 , . . . , y k ) E S ( n , k ) :l l y j - m - i j l l n 5 - - . n l - r forall j } . The (k-dimensional Euclidean) volume of each B, is

Since the combined volume of the B,'s is nk (the volume of S ( n , k ) ) , and . . ,,zk) . in some since they are all closed sets, there is a point 2 = ( q , ~ BI, n B,u, m' # m". We then take m = Im'- m"l so that 0 < m < n and we have

for all i = ij, j = 1 , 2 , . . . ,k, as required. We are grateful to Mikl6s Simonovits for pointing out that Lemma 1 may also be derived from Dirichlet's theorem on simultaneous diophantine approximation, [2, p.1591. Because of its relation to [14], this may allow us to find the "multiplier" m figuring in Lemma 1 efficiently (cf. [14. p.5245251). Also note that Lemma 1 is sufficient to imply that

when n is prime. In fact, a long as the m from Lemma 1 is relatively prime to n, the mapping taking 2 to rn x mod n is a bijection 2, -+ 2, and hence an isomorpism of C,,(i1, i2, . . ,ik) onto a spanning subgraph of C n ( m i l , m i 2 , . . . ,m ik); and therefore onto a spanning subgraph of C,(l, 2 , . . . ,1) with I = Lnl-kJ.

-

.

-

Bandwidth versus Bandsize

123

If m and n are not relatively prime, then the above mapping is not a bijection. Nevertheless there always exists a bijection accomplishing our aims. This is not hard to see by arguing that the above mapping x + rn - x takes exactly g points of 2, onto each of the points g,2g,. .,n g = 0; thus it can be made bijective by local perturbations. To define such a bijection explicitely we can use the following facts:

.

Lemma 2. Let g = gcd(m,n) and d =

P.

(a) Each x E 2, can be uniquely written as and 0 5 r < d.

I

= dq

(b) Each x E 2, can be uniquely written as x = mu and 0 5 v < g.

+ r with 0 5 q < g

+ v with 0 5 u < d

+

Proof. Each x can be written as 2 = dq T with 0 5 T < d; since I E 2, and dg = 0 in Z,, q may be assumed to satisfy 0 5 q < g. The uniqueness in (a) follows from the fact that gd = n = lZ,l. Each I can also be written as x = mu v with 0 5 v < m; evidently u < d because md = * n 2 n. Since a m E g (mod n) for some a , v may be assumed to satisfy 0 5 v < g. 0 The uniqueness in (b) follows by the same argument as in (a).

-

+

Let F : 2, -+ 2, be defined as follows : if x = dq t r with 0 5 q < g and 0 5 T < d, then F ( I ) = m r q. According to Lemma 2, F is well. defined and a bijection. It remains to verify that F takes C,(il, i2,. . . ,ik) onto a spanning subgraph of Cn(l, 2 , . . . ,I) with 1 5 2n’-f;.

+

Lemma 3. If 111 - ~‘11,

= i, then [IF(.)

- F(x’)1In 5 Ilm - ill, t Ilgll,.

+

Proof. Let z = dq r , 0 5 q < g, 0 5 r < d , and x‘ = dq t T d(q f ) T’ where 0 5 T ’ < d. Case 1: 0 5 q t f < g. Then F ( x ) = m r q and F(x’) = mr‘ t q t f. Hence

+ +

+i =

+

I~F(I)

-

- F(z’)~], = l)m ( T - T ’ ) - f 1, = Ilm * ( d f - i) - f 1, = Ilm ‘ i f 1, L Ilm ’ ill, t Ilslln.

+

(Note that f 5 g because dg E 0 (mod n).) Case 2: 0 5 q t f - g < g. Then x’ = d(q f - g) T’ and F(d)= m . r ’ t q t f

+-

l l ~ ( x -) F(x’)lln =

=

+ - g. Hence Ilm - - r’) - f + glIn = Ilm. ( d - f - i) - f + 41, Ilm - i + f - glln L Ilm - ill, t 11g1In. (T

0

P. Erdos, P. Hell and P. Winkler

124

Lemma 4. Ifgcd(n,il,iz, ...,ik) = 1, and i f m < n has each llm.ij[l,,5 1 1 n1-F ( j = 1,2,. . . ,k ) , then g = gcd(m,n) 5 nl-b. Proof. If all Ilm. ijll,, = 0 then each m ‘ij is a common multiple of m and n, hence also a multiple of Therefore is a divisor of all ij as well as of n; according to our assumption = 1, contrary to g 5 m 5 It. O n the other hand, note that g divides m i j - a n for each j = 1 , 2 , , . . ,k, and every integer a. Assume that Ilm. ijll,, # 0; since llm. ijll,, is either r n . i j - a . n o r a . n - r n . i j for someintegera,gdivides JJrn.ij\l,,and hence 1 0 g 5 Ilm i j l l , , 5 nl-r as claimed.

a

y.

-

.

-

Proof of (3) (and thus of (2) and Theorems 2 and 1). Lemmas 1, 3 and 4 imply that the mapping F given above is an isomorphism of C,,(il, i2,. . ,ik) onto a spanning subgraph of C,,(l, 2,. . .,I), where 2 5 2n1-t, provided gcd(n,il, i 2 , . . .,ik) = 1. Since this condition is always satisfied for connected circulants C,( il, i 2 , . ,ik), (3) has been proved. 0

.

..

Remark. Let da(G) denote the number of vertices of G of distance s from a fixed vertex. We have studied, jointly with Martin Farber, the behaviour of d3(C,,(i1, i 2 , . . ,ik)) and d3(L,,(i1,iz,. . . ,ik)) for fixed k and believe that there exists a constant c depending only on k such that

.

d a ( ~ , ( i l , i 2 , . . . , i k )) 5 cn’-r

1

( 4)

for all n, s, and i1, i 2 , . . . ,ik. This would then offer another proof of Theorem 1: indeed any breadth first numbering (or so-called “level algorithm” in the terminology of [20])of L,,(il, i2,. . ,ik) would have width at most 2 c . n l - i . When k = 1 (4) is obvious (the constant c is 1 or 2 depending on the choice of the starting vertex); we were also able to prove (4) for k = 2 and k = 3. It turns out that to complete the proof of Theorem 1 this way, it would suffice to show that, for fixed k,

.

3

The random case

Our object in this section is t o compare the bandwidth and bandsize of random graphs. For 0 < p < 1 and n a positive integer, let G,,, be a graph with vertex set V,, = {1,2,. . . ,n} and edges defined randomly as follows: for each pair { i , j } of vertices with i # j the edge i j is included

Bandwidth versus Bandsize

125

(3

with probability p and excluded with probability q = 1 - p ; the choices are made independently. This construction has been studied extensively, especially by Erdos and Rdnyi [7]; see also BollobAs [3]. We wish t o compute bounds on bw(Gn,p) and bs(G,,,) for almost all G,,, when n is large. For simplicity we concentrate below on the case p = ?j; the results are easily generalized. Note that when p = the graph G,,,,which we denote below simply by G,, has the special property that every graph on the labelled set V, is equally likely to occur. The following theorem is proved in [13].

4

Theorem 3. With probability approaching 1 as n n - (2 t f i t o(1)). Ign

+ 00,

< bw(Gn) < n - (2 + &- o(1)) - Ign.

As we found, a number of recent papers studied the bandwidth of random graphs, [5, 13, 15, 20, 211. Theorem 3 is stronger than similar results in [20, 211, and weaker than the most general version of [13]. We have stated it in this way for simplicity, and also because this was the form of the result we had before discovering [13]; we had the constant 2 t 6in the lower bound, but not in the upper bound. (In [15] the authors study the average bandwidth of trees, which turns out to be between c l f i and czfilog n.)

Theorem 4. With probability approaching 1 as n bs(G,) 2 n Proof. Fix c

>

(A+

o(1))

00,

.

& and t = [c(lnn)21; we show that

Pr(G, has a numbering of size less than n

- t ) -P

0.

It suffices to show that Pr(G, has a numbering omitting lengths n - d l , . .. , n - d t ) O

tends to 0. For values of d l , dp, .. .,dt which are large relative to n, the constraints are severe and the probability that a given numbering satisfies them will be much less that n!. When the di’s are small, however, it becomes necessary to count only partial numberings, namely bijections from subgraphs of G, to the union of an initial and final segment of 1,2,., . ,n. To determine the

P. Erdos, P. Hell and P. Winkler

126

optimum size of these segments we compare the di’s to a certain exponential sequence. Let 0 < d l < d2 < ... < dt < n be a sequence of integers; set T = 1 where c > b > Note that for large n

-+ &

&.

In n

log, n =

In (1 -IIn n -

< hence dt dj < ~ j -


~ - ~ It.

di 2 T

~

i=l

2b2(ln n)3 2blnn - 1 c(lnn)2 - 1 5 t - 1,

follows that there is a least integer j such that for - ~ i < j and hence

i=l i- 2

Note that d j 2 j and that j -+ 00 as n -+ 00. We now consider partial numberings of G, into the set { 1 , 2 , . . . , d j } U {n-dj+l,n-dj+2,. . . , n } where edge-differences n - d 1 , n - d2,. . . , n - d j are avoided; since each difference n - di forbids exactly di edges within the domain of the partial numbering, altogether there are dl d2 ... d j forbidden pairs. For a given value of j there are fewer than nJ choices for d l d z , . . . ,d j and fewer than n2s partial numberings, where s = ~ j - l . Hence for fixed j the probability that G, has a partial numbering avoiding forbidden pairs which determine j is less than

+ + +

since 9 = ~

j - l is

forced like j to increase without bound as n -+ 00.

Bandwidth versus Bandsize

127

However, there are fewer than c(lnn)2 choices for j so Pr(bs(G,) < n - c(lnn)2) < c(lnn)2 n-’

--t

0,

and Theorem 4 is proved.

0

Corollary 2. With probability approaching 1 as n + 00,

n

- ~ ( l o g n 5) ~bs(Gn) 5 bw(G,) 2 n - c1 logn.

We have determined that the “correct” value of bandwidth of G, is n - clogn; in fact even the constant c is determined by the result of Kuang and McDiarmid, as stated in Theorem 3. We do not know at present whether the correct value of the bandsize is more like n - clogn or n c(1og n)2.

Acknowledgement We are grateful to Pauline van den Driessche, Martin Farber, Dale Olesky, Moshe Rosenfeld, and Mikl6s Simonovits for useful conversations concerning bandsize.

References [l] F. A. Akyuz and S. Utku, “An automatic node relabeling scheme for bandwidth minimization of stiffness matrices,” J. Am. Inst. Aeronautics and Astronautics 6 (1968), 728-730.

[2] T. M. Apostol, Modular functions and Dirichlet series in number theory, Springer Verlag, Berlin (1976). [3] B. Bollobtk, Random Graphs, Academic Press, London (1985). [4] P. 2. Chinn, J. Chvitalovi, A. K. Dewdney and N. E. Gibbs, “The bandwidth problem for graphs and matrices - a survey,” J. Graph Theory 6 (1982), 223-254. [5] P. Z. Chinn, F. R. K. Chung, P. Erdos and R. L. Graham, “On the bandwidth of a graph and its complement,’’ The Theory and Applications of Graphs (G. Chartrand and S. Alavi, eds.), Wiley and Sons (198l), 243-254.

128

P. Erdos, P. Hell and P. Winkler

[6] F. R. K. Chung, “On graph labellings,” Selected Topics in Graph Theory 111 (L. Beineke and R. Wilson, eds.). To appear. [7] P. Erdos and A. Rdnyi, “On the evolution of random graphs,” Publ. Math. Inst. Hungar. Acad. Sci. 5 (1960), 17-61. [8] R. E. Funderlic and R. J. Plemmons, “A combined direct-iterative method for certain M-matrix linear systems,” SIAM J. Alg. Disc. Meth. 5 (1984), 33-42. [9] M. R. Garey, R. L. Graham, D. S. Johnson and D. E. Knuth, “Complexity results for bandwidth minimization,” SIAM J . Appl. Math. 34 (1978), 477-495.

[lo] J. A. George and J. W. Liu, Computer solutions of large sparse positive definite systems, Prentice Hall (1981). [ll] E. M. Gurari and I. H. Sudborough, “Improved dynamic programming algorithms for bandwidth minimization and the mincut linear arrangement problem.” Manuscript (1984). [12] K. Heinrich and P. Hell, “On the problem of bandsize,” Graphs and Combinatorics 3 (1987), 279-284. [13] Y. Kuang and C. J. H. McDiarmid, “Probabilistic analysis of bandwidth problems,” Proc. British Combinatorial Conference 1985. To appear. [14] A. K. Lenstra, H. W. Lenstra Jr. and L. L O V ~ S Z“Factoring , polynomials with rational coefficients,” Math. Ann. 261 (1982), 515-534. [15] A. M. Odlyzko and H. S. Wilf, “Bandwidth and profiles of trees,” Proc. Fifth International Conf. Theory of Graphs, Kalamazoo, Michigan, July 1984. [16] C. H. Papadimitriou, “The NP-completeness of the bandwidth minimization problem,” Computing 16 (1976), 263-270. [17] J. B. Saxe, “Dynamic programming algorithms for recognizing small bandwidth graphs in polynomial time,” SIAM J . Alg. Disc. Meth. 1 (1980), 363-369. [18] I. H. Sudborough, “Complexity of bandsize,” A. M. S. Abstracts 7 (1986), 15 # 825-05-618.

Band width versus Bandsize

129

[ 191 R. P. Tewarson, “Row-column permutations of sparse matrices,” Computer J . 10 (1967), 300-305. [20] J. S. Turner, “Probabilistic analysis of bandwidth minimization algorithms,” Proc. 15-th Annual ACM Symposium on the Theory of Computing (1983), 467-478. [21] W. F. de la Vega, “On the bandwidth of random graphs,” Annals of Discrete Math. 17 (1983), 633-638.

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Annals of Discrete Mathematics 41 (1989) 131-140 0 Elsvier Science Publishers B.V. (North-Holland)

Circumference and Hamiltonism in Kl,3-Free Graphs E. Flandrin a n d I. Fournier

A. Germa

Laboratory of Informatics Research University of Paris-South Orsay, France

Informatics Department

ENST Paris, France

Dedicated to the memory of G. A . Dirac Let G be a 2-connected Kl,s-free graph of order n with d ( u ) + d ( v ) 2 h for any pair of non-adjacent vertices u and v. Then G has a cycle of length at least min(h 4, n ) , and if h 2 then G is hamiltonian. Furthermore, if G is a k-connected (k 2 2) Kl,s-free graph of order n such that the degree-sum of any k 1 independent vertices is a t least m, then G contains a cycle of length at least min( 4, n).

9 +

+

+

RQsumQ Soit G un graphe 2-connexe d’ordre n sans 1(1,3tel que pour toute paire de sommets non adjacents u et v on ait d(u) d ( v ) 2 h , alors G contient un cycle de longueur au moins min(h 4, n) et si h 2 alors G est hamiltonien. Par ailleurs, si G est un graphe k-connexe (k 2 2) d’ordre n sans K 1 , 3 tel que la somme des degrks de tout stable h k 1 Blkments est au moins m, alors G contient un cycle de longueur au moins min( 4, n ) .

+

+

9,

+

+

1

Introduction

In [2], M. Matthews and D. Sumner showed that if G is a 2-connected K ~ J free graph of order n, then G has a cycle of length at least min(26 4, n ) , where 6 denotes the minimal degree of a vertex in G, and if 6 2 then G is hamiltonian. These results are best possible if the minimum degree of G is $ - 1, as proved by the example of Figure 1. In this article, we generalize the above results by looking at the sum of two non-adjacent vertices instead of 6, and we prove the following two theorems .

+ F,

131

E. Flandrin, I. Fournier and A. Gerrna

132

Figure 1.

+

Theorem 1. Let G be a 2-connected K1,3-free graph of order n with d(u) d(v) 2 h for any pair of non-adjacent vertices u and v. Then G has a cycle of length at least min(h t 4,n). Theorem 2. Let G be a 2-connected K1,a-free graph of order n with d(u)+ d(v) 2 for any pair of non-adjacent vertices u and v. Then G is hamiltonian. Remark. The example of Figure 1 shows that these results are best possible if the minimum degree of G is - 1.

t

In [l],I. Fournier and P. Fraisse, answering a conjecture of J. A. Bondy, have shown that if G is a k-connected (k 2 2) graph of order n such that the degree-sum of any k t 1 independent vertices is at least m, then G contains a cycle of length at least min($,n). This result can be improved for Kl13-freegraph as follows.

Theorem 3. Let G be a k-connected (k 2 2) Kla-free graph of order n such that the degree-sum of any k 1 independent vertices is at least m. Then G contains a cycle of length at least min($k 4,n ) .

+

+

In this article, we shall use the two following results noted in [2] by

M. Matthews and D. Sumner.

Circumference and Hamiltonism in K1,3-Free Graphs

133

Lemma 0 . Let C = 2 1 1 ~ 2 . .. vpvl be a cycle of maximum length in a K1,3-

free graph. 1. If x $

2.

c is adjacent to Vk,

then Vk-1 must be adjacent to

Vk+l.

If x $ C and i f P and Q are two disjoint paths between C and x of length p and q, then between the extremities of P and Q o n C there are at least p

+ q + 2 edges of C.

Notations: 1. G will be a 2-connected K l ~ - f r e egraph (so G contains at least one cycle). 2. C will always denote a cycle of maximum length in G with an arbitrary orientation (for x c C , xf and x- are the following and preceding vertices of 2 on C according to that orientation) and R will be the subgraph induced by the vertices of G - C. Moreover, if a and b are two vertices of C , [ab]will denote the set of the vertices of C between a and b ( a and b included) according to the orientation of c. 3. A C.R-chain will be a chain with every vertex belonging t o R, except one of its extremities belonging to C.

4. For a vertex x in G, r ( z ) will be the neighbourhood of 2 in G, and if H is a set of vertices of G, I’H(x) will be the neighbourhood of x in H and d H ( x ) will be the cardinality of I?H(x). Moreover, if H is a set of vertices of C , I’&(x) will will denote the successors on C of the vertices of r H ( x ) . 5 . (ab) will always denote a “C-path,” (i.e., a chain of length at least 2, with every vertex in R, except the extremities a and b belonging to C ) , and we shall construct the vertices a and p as follows (Figure 2). 0

a is the first vertex of C following a+ not adjacent to a - .

0

p is the first vertex of C following b+ not adjacent to b-.

0

a and ,f3 are well defined because a- and b- are non-adjacent.

In our proofs, we shall need several times the following result.

134

E. Flandrin, I. Fournier and A. Germa

\-

ab chain

Figure 2.

Lemma 1. Let C , ab, a , and ICI - 4.

p be defined as above. T h e n d c ( a ) + d c ( P ) I

Proof. We first remark that a is before b- on C, because a- and b-- are not adjacent (and p is before a- on C). If a was adjacent to one of b, b+, b++, . . . ,p, then we could obtain a cycle larger than C, which is impossible. By hypothesis, a is not adjacent to a - , which implies, as G is a K1,3free graph, that a is not adjacent to a (for if it was, naming y the vertex of the C-path ( a b ) nearest to a , the subgraph induced by a, y, a - , a would be a #1,3, the edges ya- and ya being forbidden in G). Whence we have

Moreover, let u be a vertex of [ a t b - ] adjacent to a : u- is not adjacent to p; if it were, we could obtain a cycle larger than C, which is impossible. We deduce from the above remark that

Circumference and Hamiltonism in Kl,s-Free Graphs

135

Moreover, we have d[b,-](a)

=0

(5)

and

= 0. Summing the six above relations and using the fact that not edges of G, we obtain d[aa-](P)

and the proof of Lemma 1 is complete.

2

aa-

and pb- are

0

Proof of Theorem 1

Assume that G is not hamiltonian. Then Theorem 1 will be a corollary of the following Lemma. Lemma 2. Let C be a cycle of maximal length in G, a a vertex of C having at least one neighbour in R, P a maximal C.R-chain starting from a , and x the other extremity o f P not on C . Then d(x) 5 frlCl - 2. Proof. We know [l] that we can find a C-path ( a b ) which contains x and its neighbours in R. Let 1 be the length of (ab), and p the length of P. First we show that 1 5 fr1Cl- 2: a and b induce two paths on C, one of them of length a t least frlCl; from Lemma 0: frlCl 2 1 2. If 2 has no neighbours on C other than a , then b is not a neighbour of x, and so: 1 d ( x ) 5 1 - 1 5 -1CI - 3.

+

2

Assume now that x has a t least one neighbour on C other than a , and let u be such a vertex. Let ( a u ) be the C-path defined by P and the edge x u ; x is not adjacent to u+, u++ or u+++, otherwise we would have a cycle longer than C (Figure 3). Suppose that u is as near as possible to a. From Lemma 0 there must be at least p + 3 edges between a and u. Hence we have ICI 2 4 ( d c ( z ) - 2 ) + 2 ( p + 3), which implies dc(x) 5 a(lCl- 2p+ 2 ) and d(x) I $(lCl+ 2 p - a), since d ~ ( z=) d p - ( a ) ( x ) 5 p - 1,

E. Flandrin, I. Fournier and A, Gerrna

136

Figure 3. Here we have p -I- 1 = 1 5 31Cl - 2 (for b is a neighbour of s on C), so p 5 !jICl - 3, and 2 p 5 ICI - 6, whence d ( z ) 5 !jlCl - 2, which completes 0 the proof of Lemma 2. We can now prove Theorem 1.

Proof of Theorem 1. If G is not hamiltonian, we can find C, a , P , z and b as in Lemma 2 . Let cy and P be as in Notation 5. Two cases are possible. 1.

(or p) has a neighbour in R. We consider a maximal C.R-chain starting from a. If it has a common vertex with P , we can construct a cycle longer than C , which is impossible. Let I be the other extremity of that chain, z is not adjacent to s (else P would not be maximal). Moreover, Lemma 2 implies d ( z ) 5 41Cl - 2 and d ( z ) 5 31Cl - 2; so, from the hypothesis, we have h 5 d ( s ) + d ( z ) 5 ICI - 4 or ICI 2 h t 4.

cy

2 . Neither a nor /3 has a neighbour in R, which means d(a) = &(a) and d(P) = dc(P). But cy and ,f3 are not adjacent; so, from Lemma 1, we have h 5 d(a) d(P) = &(a) t &(/I) 5 ICI - 4 or ICI 2 h 4, which completes the proof of Theorem 1.

+

+

0

3

Proof of Theorem 2

Assume that G is not hamiltonian. By Theorem 1, we know that if C is a cycle of maximum length, we have:

2n - 5 2n+7 + 4 = -. ICI 2 h + 4 1 3 3

Circumference and Hamiltonism in Klt3-Free Graphs

137

Lemma 3, Let P and P’ be two maximal C.R-chains, x and x’ their extremities in R. Then x and x’ are adjacent.

Proof. Let p (respectively p’) be the length of P (respectively P’).

q.

a

If x has no neighbour on C other than a , then d ( x ) i: IR1 5

0

Otherwise, as in the proof of Lemma 2, we have d ( x ) 5 $(IC1+2p-2).

Using p

5 n - ICI

and ICI 2

y, we obtain

2n-ICI-2 4n-13 71-3 5 d(45 4 12 3 The same holds for x , whence d ( x ) + d(x’) 5 which implies that 0 x and x’ are adjacent.

-

y,

Lemma 4. With the hypothesis of Theorem 2, i f C is a cycle of maximum length and P a C.R-chain of maximum length, then V ( G )= V ( C )U V ( P ) . Moreover, if x is the extremity of P belonging to R, then d c ( x ) 2 2.

Proof. Let a be the extremity of P which is on C and y the vertex of P following a on P. 1. Let us prove that x has at least two neighbours on C ; otherwise we have d ( x ) 5 p 5 n - ICI + 1. The vertices a! and /3 being defined as in Notation 5 satisfy d(a) d ( P ) 5 ICl - 4 and therefore one of them, a! for example, satisfies d ( a ) 5 ;(lCl - 4);x is not adjacent to a, so we have 2n - 5 5 d ( x ) d ( a ) 5 n - ICI t 1 ICI - 4 3 whence ICI 5 in contradiction with ICI 2

+

+

9

+7

v.

2. Let b be a vertex of C adjacent to x and different from a ; let y be the extremity of a C.R-chain of maximal length starting from b, then if z and y are different they are adjacent. By considering successively the vertices of P we can prove that they are all extremities of a maximal chain and so the subgraph induced by P - a is a complete graph. 3. Let us suppose that there exists a vertex x which is not on P U C. Let Q be a chain between z and P U C. By 2., Q does not intersect P - { a } . Let Q’ be a maximal C.R-chain prolonging Q, and z’ its extremity in G - ( P U C ) ;x and z‘ cannot be adjacent so there is a contradiction with Lemma 3. This completes the proof of Lemma 4. 0

E. Flandrin, 1. Fournier and A. Germa

138

We can now give the proof of Theorem 2. Proof of Theorem 2. Let us consider a cycle C of maximum length and a C.R-chain P of maximum length p starting from a. Let us denote by x the other extremity of P . We know by Lemma 4 that V(G)= V ( C )U V ( P ) , and so we have n = ICI p . Since, from Lemma 4 , x has at least two neighbours on C, let us denote by b a vertex of C adjacent to x and different from a.

+

0

a being defined as in Notation 5 , we know that a is not adjacent to a , a- nor to itself.

is not adjacent to any vertex of P

0

(Y

0

a is not adjacent to b, b+, b++.

0

- {a}.

Let d be a neighbour of x on C which precedes a and is as near as possible to a. There are at least p 2 vertices of C on the interval [d+,a-] and a is not adjacent to d, d+, . . . , d+(P+l), and all these vertices are different from a- .

+

We deduce from the above remarks that: d(a) = &(a) L ICI - 3 - 3 ( d c ( x ) - 2 ) - ( p + 2 ) n -p - 3 - 3(d(x)-p 1) 6 - p - 2 =n -p-3-3d(x)+3p-3+6 -p-2 = n p - 3d(x) - 2.

+ +

+

As a 'and 5 are not adjacent, we deduce from the hypothesis and the previous inequality that:

so Similarly we have:

Adding these two inequalities and using Lemma 1 (with ICI = n - p ) , we obtain n - p - 3 5 d ( a ) t d(P) 5 n - p - 4, a contradiction. This completes the proof of Theorem 2.

Circumference and Hamiltonism in K ~ JFree - Graphs

4

139

Proof of Theorem 3

Assume that G is not hamiltonian. First we establish the following lemma.

Lemma 5. Let C be a cycle of maximal length in G, a a vertex of C having at least one neighbour in R , P a maximal C.R-chain starting &om a and x the other extremity of P ; let Q be the first vertex of C following a+ nonadjacent to a - . Then we have d ( x ) &(a) 5 ICI - 4.

+

Proof. 1) Assume first that I has no neighbours on C - { a } . Let (ab) be a C-path containing x and its neighbours in R and let /3 be defined as in Notation 5 . We have seen in the proof of Lemma 1 that Q is not adjacent t o a , b, b+. Moreover, Q is not adjacent to itself, nor t o /3+q for 0 5 q 5 d ~ ( x (Figure ) 4 (a)), so &(a) 5 ICI - 4 - ( d ~ ( x ) 1). Since d c ( x ) 5 1 we have:

+

2) We suppose now that x has at least one neighbour on C different from a. Let b be the neighbour of x as near as possible to a before a. There are at least d ~ ( x )5 vertices of C on the interval [ba]and a is not adjacent to b , b+, . . . , b+(dR(”)+2), a-, a (Figure 4 (b)).

+

Figure 4. Moreover, a is not adjacent to itself, and if v is a neighbour of z different from a and b, then Q is not adjacent to v+. All this implies &(a) 5 ICI - ( d ~ ( xt) 5 ) - (&(I) - 2) - 1 whence d ( x ) &(a) 5 ICI - 4. 0

+

Theorem 3 results from Lemma 5 exactly as in [l],so we omit the end of the proof.

E. Flandrin, I. Fournier and A. Gerrna

140

References [l] I. Fournier and P. Fraisse, “On a conjecture of Bondy,” J. Cornbin. Theory (B) 30 (1985), 17-26.

[a] M. Matthews and D.

Sumner, “Longest paths and cycles in I T ~ ~ - f r e e graphs,” J. Graph Theory Q (1985), 269-277.

Annals of Discrete Mathematics 41 (1989)141-170

0 Elsevier Science Publishers B.V. (North-Holland)

The Prism of a 2-Connected, Planar, Cubic Graph is Hamiltonian (a Proof Independent of the Four Colour Theorem) H. Fleischner Institute of Information Processing Austrian Academy of Sciences Vienna, Austria To the memory of my friend Gabdr who taught me a lot (and not just maths) It is an open question whether the graphs of simple d-polytopes contain hamiltonian cycles for arbitrary d 4. Using the Four Colour Theorem, Rosenfeld and Barnette showed that the prism over any simple 3-polytope (which is a simple 4-polytope) is hamiltonian, [ll]. Later, Rosenfeld and Goodey were able to give a proof avoiding the use of the Four Colour Theorem provided the 3-polytope is cyclically .l-line-connected, [8]. Here we present a proof that the prism over any planar, 2-connected, 3-regular graph is hamiltonian, without using the Four Colour Theorem. This establishes a nontrivial necessary condition for the validity of the Four Colour Theorem.

>

1

Introduction and statement of the main theorem

In [9,p. 3651, Victor Klee states that it is unknown whether every simple d-polytope contains a hamiltonian cycle for d > 3 (a d-polytope is simple if its graph is regular of degree d). However, Barnette conjectures that every simple 4-polytope is hamiltonian, [ll](note that the reference given there does not contain this conjecture). For d = 3 several nonhamiltonian examples are known, namely those corresponding (by Steinitz’ Theorem) t o planar, 3-regular, 3-connected graphs, and which are therefore counterexamples t o Tait’s famous conjecture. Also, for every d > 3, there exist nonhamiltonian d-polytopes which are not simple.

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However, in 1973 Rosenfeld and Barnette showed that the prism over any simple 3-polytope P is hamiltonian provided the facets of P are 4colourable, [ll,Theorem 31. In fact, they prove a slightly more general result, namely: If G is a planar, 2-connected, 3-regular graph and if G is 3-line-colourable, then P ( G ) - the prism over G - is hamiltonian. As the authors point out, if one can find a planar, 2-connected, 3-regular G with nonhamiltonian P(G), then the Four Colour Conjecture is false. Hence it is desirable to find a proof of the above result without using the validity of the Four Colour Theorem. This is partly achieved in [8] where - in graph theoretical terms - it is proved that any planar, 3-regular, cyclically 4connected graph G has a hamiltonian P (G) . The proof of this result uses the existence of a Tutte cycle C in such graph (i.e., a cycle C such that every line is incident to a point of C), and - under the given hypothesis - the existence of C is guaranteed by Tutte’s famous Bridge Theorem (from which also follows the existence of a hamiltonian cycle in any planar 4-connected graph). However, this as well as the result quoted first do not preclude the existence of a nonhamiltonian P(G)for some planar, 2-connected, 3-regular G if the Four Colour Theorem is not assumed. For, while such a 3-linecolourable graph G has a hamiltonian P(G),it is conceivable that based on such a G which is not 3-line-colourable, and which does have a hamiltonian P(G),one may construct a planar, 2-connected, 3-regular G’ which is also 3line colourable but for which P(G’)is not hamiltonian. On the other hand, if one considers the Four Colour Theorem an established fact, then it is still of interest to ask for the practical relevance of that theorem in a proof of [ll, Theorem 31. In fact, the following result (which is a slight generalization of [11, Theorem 31) uses comparatively little effort for its proof if one uses the Four Colour Theorem, while elaborate induction techniques are needed here to produce a proof avoiding the use of the Four Colour Theorem.

MAIN THEOREM. For every planar, 2-connected, 3-regular graph G, the prism over G is hamiltonian. Thus, our main theorem is a partial solution to Barnette’s conjecture (since P ( G ) is a simple 4-polytope if G is 3-connected), as well as a nontrivial necessary condition for the Four Colour Theorem.

2

Preliminary discussion and results

For concepts not defined in this paper, see the quoted references, especially [9] and [lo]. However, contrary to [lo] we permit a graph to have multiple lines, and a 3-regular graph is shortly called a cubic graph.

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Definition 2.1. 1. A graph S is called a BEPS-graph iff it is connected and bipartite, and if it can be written in the form S = E U P,where a)

E is an Euler graph,

b) P is a set of totally disjoint paths (path-forest), c)

E and P are line-disjoint.

2. For a graph G we call S C G a BEPS-graph of G (G contains a BEPS-graph S, respectively), if S is a BEPS-graph spanning G.

3. If G is cubic we write, for a BEPS-graph of G , S = K U P , for, in this case, li = E is a set of totally disjoint (even) cycles. Moreover, for some BEPS-graph S = K U P of G and some v E V ( G ) incident to the lines e , f , g , we call S an (e,f;g)-BEPS-graph of G iff a) e , f E E ( W ,

b) g

6W).

Definition 2.2. For a graph G, the prism over G is the Cartesian product G x K 2 , shortly P(G). For our purposes, we define P ( G ) equivalently in a more descriptive way. Take G and a copy G’ of G with v E V ( G ) corresponding to v‘ E V(G’) and define

V ( P ( G ) )= V ( G )U V(G’), E ( P ( G ) )=: E(G)U E(G’) U {[v,v’] I v E V ( G ) ,v’ E V(G’)}. We call a line of the form [v,v’] a pillar of P(G). We note in passing that a BEPS-graph S of a cubic graph G is precisely what Goodey and Rosenfeld call a graph of type S. However, we choose the above notation because a BEPS-graph is a bipartite EPS-graph ([1,2,5,6,7]). In fact, as will be discussed in Section 4 of this paper, in dealing with hamiltonian cycles in P (G ), BEPS-graphs fulfill a function almost analogous to that of EPS-graph with respect t o hamiltonian cycles in squares of graphs.

Remark 2.3. If S is a BEPS-graph of a cubic G , each component of G - S := G - ( E ( S )U {v E V ( G )I degs v = degG v}) is a path or a cycle. As we shall see below, if such G contains a BEPS-graph a t all, then it contains one such that G - S is a path-forest.

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Generalizing the method of constructing a hamiltonian cycle in P ( G ) developed in [ll],Goodey and Rosenfeld [8, Lemma 3.11 showed, in effect, that if a cubic graph G contains a BEPS-graph, then P ( G ) is hamiltonian. What went unnoticed and is expressed in Theorem 2.4, is that the converse is true as well.

Theorem 2.4. For a cubic graph G, P ( G ) is hamiltonian if and only if G contains a BEPS-graph.

Proof. Because of [8, Lemma 3.11, it suffices to show that the existence of a hamiltonian cycle in P ( G ) implies the existence of a BEPS-graph in G. Let C be a hamiltonian cycle of P(G). We define a graph H with

V ( H )= V ( G )and E ( H ) consisting of “bottom” and “top” lines as follows: If C passes through [z,Y]E E ( G ) , then [ z , Y ]E~ E ( H ) ; and if C passes through [d,y’] E E(G’), then [z,ylt E E ( H ) . Pillars passed by C are disregarded in H . That is, H appears as a projection of G’ onto G with precisely those lines passed by C retained in H and distinguished by b and t depending on whether they originally belonged to G or G’. Thus, C is transformed into an Eulerian trail T of H and, obviously, degHv E (2.4) for every v E V ( G )= V ( H ) . The following properties of H now follow easily from the fact that C is a hamiltonian cycle of P(G) and that G is cubic: a) for every w E V ( H ) ,there are as many “bottom” lines incident to v as there are such “top” lines; b) if, for some w E V ( H ) ,degH w = 4, then there is some w E V ( H )such that {[V,W]b,[v,W]t} c E ( H ) ; c) the graph P C G which is induced by the lines [z,y] E E(G) for which { [ z , Y ][z, ~ ,y]t} c E ( H ) , is acyclic; hence, since degH v 5 4 for every w E V ( H ) ,P is a path-forest. Since degH 5 4 for every w E V ( G ) and because of b ) it follows for K O := H - {[Z,y]b,[Z,y]tI [z,y] E E(P)}that degK, v E {0,2}; hence, every component of KO is an isolated point or, by a), an even cycle (since “bottom” and “top” lines alternate in such components). Clearly, K O corresponds in G to a set Ii’ of totally disjoint even cycles. By definition, P and K are line-disjoint. Moreover, P U K is connected and spans G since T is an Eulerian trail of H . Now let P‘ C P contain as few lines as possible such that S = P‘U li is connected and spans G . Then the lines of P‘ are the bridges of S, and hence

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- since P' is also a path-forest he-disjoint with K - S is a BEPS-graph 0 of G. We note that by methods analogous to those developed in the proof of Theorem 2.4, and [ 5 ] , one can show that if P(G) has a hamiltonian cycle C for some arbitrary graph G, then C induces an EPS-graph of G (for a more detailed discussion, see Section 4 of this paper). Before we continue to deal with BEPS-graphs of cubic graphs, we want to exhibit the graph of Figure 1 which is not bipartite and has a hamiltonian prism, while none of its spanning subgraphs has this property.

b

t

t

b

2

15

t

b

lb 10 1'

14

Figure 1. First we want to point out that in the proof of Theorem 2.4, the few times that we used the hypothesis of G being cubic, were in fact decisive. For, in general, a point of degree 4 in H does not imply that this point

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must lie on P. Hence, in general, C can be of the form with ( 2 , ~ n ) {u,w} = 0. Thus, in general, we cannot conclude as in the proof of Theorem 2.4, that two “bottom” (“top”) lines are not adjacent in the same cycle of K o , K respectively. A discussion of Figure 1 (which we omit for brevity’s sake) shows that G = S = E U P does not have a bipartite E . We note that this graph has exactly one point with degree exceeding 3, and that it has lines in P which are not bridges in S = G. In fact, for every hamiltonian cycle C of P ( G ) and the EPS-graph S = E U P induced by it, we must have {[1,2],[14,15]} C P; from this follows automatically E = K 1 U K 2 U K 3 . We conclude S = E U P = G for every hamiltonian cycle C in P(G) with E and P uniquely determined. As one can see easily, for G in Figure 1, P(G)has exactly four hamiltonian cycles,

Remark 2.5. The graph G of Figure 1 is not the smallest example having the properties mentioned above. There are graphs resembling G but having fewer endpoints. However, we chose G because we wanted to emphasize the line labeling with b and t (which stands for “bottom” and “top” line, respectively). Also note that G can be turned into a graph H 3 G with similar properties such that 5 is the only vertex of degree different from 3; for details of such a construction see Section 4. Now we turn again to general considerations on BEPS-graphs of cubic graphs. We look at B = G - S; since G is cubic and S spans G, each component of F is a path or a cycle. Clearly, Ir‘ U P is spanning; but it need not be connected, no matter what the shape of P in S = lir U P is. Trivially, if 7 contains a cycle, then K U 7 cannot be connected. However, since we ultimately want to show that every planar, 2-connected7 cubic graph contains a BEPS-graph, we can try to find a BEPS-graph 5’ in G such that P “behaves” almost like P. This is established in the next theorem and will be useful in the proof of the main result.

Theorem 2.6. Suppose every 2-connected, planar, cubic graph G contains a BEPS-graph. Then G contains such a BEPS-graph S = IC U P such that every component of 7 = G - S has a point i n common with Ii and is, consequently, a path-forest; and the bridges of S coincide with the lines of P . Proof. We form a new graph G* from G by replacing every line [a,b] E E(G) with four lines [a,all, [ b l ,b ] ,[a1,b1]1, [ul,6112 where a1 # 61 and

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a 1 , b l f V ( G ) (in other words, we subdivide each line of G with two points and add an extra, line joining these two points). Clearly, G* is a 2connected, planar, cubic graph. Hence, by hypothesis G* contains a BEPSgraph S* = I<* U P*. We can assume, w.l.o.g., S* chosen in such a way that as many lines of G* as possible belong to K*,and that, subject to this restriction, as few lines as possible belong t o P*.By this we obviously have that either [ a , a l ] , [ b l , b ] E E ( K * ) ,or the digon 2 = al,[al,b1]1,bl,[bl,a1]2 is a component of K*;moreover, the lines of P* are the bridges of S*. Let us now determine the graph S C G induced by S* C G*. We proceed as follows: For every [a,b] E E ( G ) and the corresponding lines [ a ,all, [ b l ,b] we define [a,61 E

E ( K ) iff [a,4,[bl, bl E E(h'*),

E(P)iff [ a , a l l , [ b l , b l E E ( P * ) , [a,b] E E ( F )iff {[a,al],[bl,b]} g' E ( K * )u E(P*). [a,bl E

This definition obviously yields a partition of E ( G ) into three classes Ii',P,P. We first show that S = Ii U P is a BEPS-graph. We note the fact that max{degK v , degp v, degpv} = 2 for every v E V ( G )since S* is a BEPS-graph of G*. Obviously, a component C of K corresponds t o a component C* of I<*, and 1(C)= $ ( C * ) by definition of G* and K. Hence, K is a set of totally disjoint even cycles since Ii* is such a set of cycles. It follows from the connectedness of S* and A(P) 5 2 that no component of P is a cycle. Hence, P is a path-forest. By definition, K and P are line-disjoint while A(F) 5 2 guarantees that S = I< U P spans G. Also, the definition of P and the connectedness of S* ensure the connectedness of S. Hence, S is a BEPS-graph of G; and since the lines of P* are the bridges of S*, the construction of S implies that the bridges of S are the lines of P. Turning finally to it follows from A(F) 5 2 that each component PI of P is a path or a cycle. Again, the connectedness of S* and A ( P * ) 5 2 imply that, PI has a point in common with K if PI is a path. Suppose, therefore, that S* was chosen such that subject to the original conditions, P has as few cycles as possible, and suppose PI = C F is a cycle,

is'=

~ o , [ v o , ~ 1 ] , v ..., 1 , [ ~ k - l , ~ k ] , ~ k , [ ~ k , ~ O1] 1. ,~O,~

Since S spans G, the lines [ t i ,vi] E E ( G ) not belonging to lie in P. Define

c,i = 0, ...,k,

P = K ,P / = P U [ v o , v l ] - [ t l , v l ] , F = P U [ t 1 , v 1 ] - [ v 0 , v 1 ]

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and let F ( z , t l ) c F be the path with z E V ( K )if tl $ V ( K ) ,P ( z , t l ) = 0 if z = t i E V ( K ) . Hence

is a path-component of P' with z E V ( K ) (z = tl possibly). Moreover, S' = K ' u P' is a BEPS-graph of G with IE(K')I = IE(K)I, IE(P')I = IE(P)I, and the same equations hold with respect to the BEPS-graphs of G* corresponding to S and S'. Obviously, the lines of P' are the bridges of S' since this property holds with respect to P and S. However P' has fewer cyclic components than 7 has. Therefore, by the choice of S*, P is a path-forest, and - as has been shown above - every component of P has 0 a point in common with K . This finishes the proof of the theorem.

3

The existence theorem implying the main theorem

By Theorem 2.4, the question whether a cubic graph has a hamiltonian prism is being decided by answering the question whether that graph has a BEPS-graph S = K U P. Because of Theorem 2.6 it is reasonable t o look for those S such that every component of B = G - S has an endpoint in common with K . In fact, we shall prove the following theorem.

Existence Theorem for BEPS-Graphs (Theorem 3.1). Let G be a 2-connected, planar, cubic graph; let z E V(G) be arbitrarily chosen, and let e , f , g denote the lines incident with z. Then G contains an ( e , f ; g ) BEPS-graph S = K U P such that, for P = G - S, 1. every component of 'B has at least one end in K ,

2. the component of 7 containing g has both ends in K . Before dealing with the actual proof ot Theorem 3.1, some discussion on the proof's structure and on certain steps of the proof is appropriate. The proof is by induction and long and tedious. The reason for this obstacle lies in the possible appearance of triangles. However, we shall use triangles in several reduction steps in order to get in G what we want.

Remark 3.2. Suppose the cubic graph G has a triangle A and a BEPSgraph S = K U P as stated in Theorem 3.1. Suppose further that E(A) n E ( K ) = 0. Then S-V(A) is a BEPS-graph of G-V(A) with the properties 1. and 2 . quoted in the theonern. This is clear if one observes that no two

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of the three lines e l , e2, e3 incident with A (but not in A) can belong t o the same component of P,B respectively. This follows from property 1. stated in the theorem. Hence, each of the points of A is an endpoint of a component of P or P; henceforth the component of P containing g does not contain any e;, i = 1 , 2 , 3 (see property 2. of the theorem).

Remark 3.3. Suppose G contains a path

P6

of length six

with [213,v5]E E(G). Let G’ be the cubic graph homeomorphic to the graph G-{[vi,v2],[v&, us]} where vi, v& are not in P6 but lie in the neighbourhood of v2,v6 respectively. Suppose G’ has a BEPS-graph S’ = K’ U P‘ inducing a BEPS-graph S = K U P in G. Let us see in which way S’ can induce S. Let us denote by A = (v3,~ 4 , ~the s ) triangle induced by 213, v4, v5. We only deal with the case K‘ n V(A) # 0. Then exactly two of the lines [vl, v3], [v5, v7], [v:, v4] lie in K’ (v: is defined the same way as vi, v; above). If E ( K ’ ) 3 {[q,v3], [v3, v5], [v5, v7]} then it is easy t o see that E ( K ) 3 {[vl,v2], [v2,v3],[v3,v5],[v5, v6],[v6, v7]}; i.e., K contains exactly two lines more than K’does. If E ( K ’ ) 5) {[q, v3], [v3,v4], [v4, v5],.[v5,v7]}, then Ii‘ contains the six lines passed by Ps; i.e., K contains, again, two lines more than I<’ does. If, however, &(I{‘) 3 {[v:, ~ 4 1[v5, , 0711, then 3 {[v4, ~ 3 1[v3,v5]} , if and only if K‘ 3 { [ v 4 , ~ 5 ] } , and K 3 {[v4,v5]} if and only if K’ 5) {[v4,

~ 1 [v3,v51). ,

Formulating it differently, we can say that in the first two cases, 214 E V ( K ) if and only if v4 E V(K’),while in the last two cases, v3 E V ( # ) if and only if 213 4 V(K’). We argue analogously in the other cases where K’ passes through two or three points of A . As we shall see in the actual proof of Theorem 3.1, there are also other ways by which points of A will be “absorbed” by Ii‘ (even if K’does not “absorb” points of A ) . Since such situations will happen in the proof of Theorem 3.1, we shall say that K is obtained from K’ by A-absorption of A (or shortly, Ii‘ absorbs A ) rather than use a more formalistic approach. This will shorten the proof without losing clarity.

Remark 3.4. Let us consider what happens locally to P , P respectively if S is induced by S‘, and if K’ absorbs A . We use the notation of Remark 3.3. If Ii’ absorbs all points of A and K as well, then [vi,v4] belongs to P if and only if this line belongs to P’ (P‘). If, however, K‘ absorbs

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all points of A while K absorbs w3, w5 but not 214, say, then [2)3,2)4] E P if and only if [v4, v5] E P (regardless of where [wi,wq] lies). In fact, in this case we can choose arbitrarily [w3,w~q]or [2)4, 051 to lie in P (Prespectively). In particular, property 2. of Theorem 3.1, will not be destroyed by this what we call P-extension, 15-esctension, respectively (depending on whether [wi,w4] lies in P’ or p’>. Similarly, if K and K’both absorb exactly two points of A, say v3, v5, then in 2)4 the “shape” of P , P respectively will be the same as in P’,P’ respectively. If, however, K’absorbs 2)3,w5 but not w4, while K absorbs all points of A, then both P and P will lose a line incident to w4 in comparison with P‘ and P’. On the other hand, P does contain [v3,v!j] (which does not lie in P’ U P’) in this case. Again, property 2. of the theorem will hold in S if it holds in S’. Since it is obvious from this discussion how, in the case of a A-absorption, P, P respectively, deviates from P‘, P‘ respectively, locally, we shall not explicitly note the changes of the corresponding path-forests in the points of a triangle A if this A is absorbed by K unless this appears necessary with regard t o properties 1. and 2. of the theorem. In other words, for the sake of relative shortness, we shall use formalized arguments only when necessary and ask the reader to use some paper and pencil to check certain details. It also follows from property 1. of the theorem that we need not worry whether each line of P is a bridge of S (the theorem does not deal with this question). For, if e E E ( P ) is not a bridge of S,and S satisfies the theorem, then 5’1= K U (P- e) and = P U e satisfy the theorem as well. We now turn to the actual proof of Theorem 3.1. Proof of the existence theorem for BEPS-graphs. Suppose the theorem is false. Then there is among all planar 2-connected, cubic graphs at least one for which the theorem fails, and which contains as few points as possible. Since all such graphs with up to 6 points satisfy the theorem, a smallest counterexample, call it G, must contain a t least 8 points. Together with G we consider G* obtained from G by contracting all triangles of G not containing z. Thus, also G* is planar 2-connected and cubic. G* = G is possible, but if G*contains a triangle A $ { z } , then IV(G*)I< IV(G)l. We divide the proof into 8 steps which we want to list first and then show, one by one, the validity of each of these steps. In what follows, U F ( S )(U;(z)) denotes the set of faces of G (G*) which have z as a boundary point. Moreover, in the proof to follow, we shall often add paths to a given set K of even cycles to obtain K U P ; by the above we can always modify P

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so as t o create no odd cycles while keeping K U P a BEPS-graph. We shall not elaborate on the instances. The steps are: Step 1. X(G) = K ( G )= X(G*)= rc(G*)= 3. Step 2. G* - & ( z ) does not contain any n-gon F with n 5 4. Step 3. G' - &(z) contains two neighbouring faces F I ,F2 where F1 is a pentagon, and F2 is a pentagon or a hexagon. Step 4. A pentagon of G* - Ug(z)corresponds in G either to a pentagon or to a 10-gon. Step 5. A hexagon of G* - U ; ( J ) corresponds in G to a 7-gon. Step 6. G* - U g ( z ) does not contain two neighbouring pentagons. Step 7. G - UF(Z)does not contain a pentagon neighbouring a 7-gon corresponding in G* t o a hexagon. Step 8. In all reductions employed in Steps 4-7 one can assume that the smaller graphs are 2-connected.

Proof of Step 1. Suppose X(G) = 2. Then G contains lines el = [ a , ~ e2 ], =

[b,d] such that G - { e 1 , e 2 } has exactly two components G1,Gz. W.1.o.g. a , b E V ( G l ) ,c , d E V(G2),and IE(G1) n { e , f,g}l 2 2. We form G: from G;, i = 1,2, by adding to Gi a new line which we denote by h = [a,b] for i = 1 and m2 = [c,d] for i = 2. Let ml, m3 denote the other lines incident with x E { c , d}. Since G: has fewer points than G , i = 1,2, and is obviously planar, 2-connected and cubic, Theorem 3.1 can be applied to G:. In particular G/, has an ( e ,f ;g)-BEPS-graph 5'; = It; U Pi satisfying properties 1. and 2. of the theorem. If {e1,e2} n { e , f,g} # 0 , then e , or f , or g stands for h. We distinguish three cases and assume that the BEPS-graphs considered in G2 automatically fulfil properties l., 2. of the theorem. case 1): h E E ( K [ ) . Then we take an ( m l ,m2; mS)-BEPS-graph Sg = K; UP; in Gi and define S = I t U P by K = ( K i UK; - { h ,m z } ) U {e1,e2} and P = Pi U Pi. (By deleting h,m2 in the corresponding cycles of It[, It; respectively, we obtain two paths which become a cycle of K by adding e 1 , e z ) . It is easy to see that S is a BEPS-graph having properties 1. and 2. regardIess of whether h E { e , f } or h @ { e , f}.

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case 2): h 6 E(P{). Then {el, e2} fl {e, f,g} = 0 must hold anyway. Now we take an (ml, m3; mz)-BEPS-graph Si = I<; U Pi of Gi. For

K = K;

U K;,

P = (Pi- h ) U Pi U {el,e2},

S = K U P is a BEPS-graph of G as required: For m2 is not a line of Si; therefore, adding the lines el, e2 in forming P guarantees the connectedness of S (even if S; - h is disconnected). Moreover, g E E(G:) implies that - in the worst case - each of the two components of - m2 (where Pt/C is the component containing mp and has property 2.) have still an end in K . Thus, properties 1. and 2. are fulfilled by S. case 3): h E E(P,'). Suppose, w.l.o.g., g = el if g E {e1,e2}. We choose again an (ml,rn3;m2)-BEPS-graph 5'; of Gi, and define K as in case 2). However, in order to secure the connectedness of S, we define

P = P; u Pi u {ez} unless the component P'

c

containing h either is of the form

,..., t

s ,..., a , h , b

with s E V ( K i ) ,t 4 V ( K ; )and , therefore cannot contain g, or else g lies between b and t in the above F.In this case we form

P = P{ u P; u {el} and assume ml,m3 incident with d. Now it is clear that in both cases S = KUP is as required (in particular, for g E F,one end of the component of P containing g lies in I<;, the other in I<;). Since X(G) = K ( G )for cubic graphs anyway, and since the contraction of a triangle does not alter the connectivity of a cubic graph (though, possibly, the cyclic connectivity), the proof of Step 1 is finished. From now on we can always consider G as a plane graph because the 3-connectedness of G as such determines already the face boundaries regardless of how G is embedded in the plane.

Proof of Step 2. Because of Step 1, G* does not contain a digon. Suppose first, G* - Us(%)does contain a triangle A* = ( s , t , u). By construction of G*, A* corresponds to a 5-, or 7-, or 9-gon C in G having chords which form triangles together with parts of C. Form G" from G by contracting C to a point o. Since {e,f,g} fl E(A*)= 8 follows from A* $! U;(z), therefore, an (e, f ;g)-BEPS-graph S** = K**U P** of G" as required exists.

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If v E K,**C K**, then I<;* can be extended to an even cycle K1 in G by A-absorptions of all points of those triangles having a line in common } P = P**, S = K U P is with C. Clearly, for K = ( K * * - { K , * * } ) U { K 1 and an (e, f ;g)-BEPS-graph as required (for the sake of simplicity we consider lines of G*, G** as lines of G). If, however, v 4 V(K**), then we find an even cycle K1 containing all but one point of C and containing only lines of C and one of its chords. Thus, for K = K** U {Kl}and P = P**U { h } where h E E ( C ) - E ( K 1 ) , S = K U P is as required. So we have to consider the possibility of G* - U $ ( z ) containing a 4-gon 0 = ( ~ 1 , 2 0 2 ,w3, w4). 0 corresponds to a 4-, or 6-, or 8-, or lo-, or 12-gon C in G. Denote in G by v1, v2,2)3,2)4 in a cyclic order those points of C which are adjacent to points v;, vi, vi, v: respectively, not in C (these points v;, v:, i = 1,2,3,4, exist because of Step 1). Form G' = (G - V ( C ) )U {z,[z,v:] I i = 1,2,3,4} where

2

$ V(G). G' is connected and bridgeless. Thus either G1 = (G' - ). u {[v:,41, [v;,41}

or G2 = (G' - ). u {[v:,41,

[4,41}

is a 2-connected, cubic, planar graph (this is a consequence of [3, Theorem 11). W.1.o.g. GI is 2-connected. Denote shortly e l = [vi,v:], e2 = [vi,41, and let S1 = K1 U PI be an (e,f;g)-BEPS-graph of G1 with properties 1. and 2. Clearly, since 0 is in G* - U;(z) and G1 is 2-connected, a t most one of the points 01, v4 can be incident with one of the lines e, f,g; the same holds for 0 2 and v3. That is, e, f , g correspond to three different lines of GI;thus, if [v:,vl] = g, say, the (e,f; g)-BEPS-graph S'1 has to be read as (e, f;el)-BEPS-graph. As we shall see from the ensuing discussion, such case will not hamper the construction of a required BEPS-graph of G; hence we speak of an (e, f ;g)-BEPS-graph of G1 in any case. Again we distinguish three cases. case 1): {e1,e2} c E(K1). Then we extend the cycle I<11 of 1<1 with el E K11 by absorbing in G ~ 1 , ~and 4 all other points of C which lie on triangles containing v1 or v4 (if such triangles exist). Thus, K11 has been extended by an even number of points. Performing the same type of extension for Ir'l2 with e2 E K 1 2 C K 1 ( K 1 1 = K 1 2 might hold), we obtain K from K1 by absorbing all points of C. For P = PIwe obtain S = K U P as required.

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case 2): 0 # {e1,e2} n E ( K 1 ) # {e1,e2}. W.1.o.g. el E E ( K 1 ) . Let K11 be KO of G such that V ( C ) c V(lG-,).For the component 5 2 of PI containing e2 we see that To = Ti2 - (e2) U {[$A,~ 2 1[v$, , vs]) with 2'12 = P12, T12 = 52 respectively, has exactly two components each of which has an endpoint in K O .This suffices for S = K U P with as in case 1). Clearly, K11 can be extended to a cycle

(m (m

K = Kl - {K11) u {KO} and

P = PI (if

T12

-

= S Z )P, = PI - {P12) U To (if

2'12

= PI^),

respectively, to be a BEPS-graph as required. case 3): el,e2 4 E(K1). Then let Ii' = K 1 U { C } , and by a two-fold application of transforming a path into two paths as we did in case 2), we obtain P if { e l , e2) n # 0. If, however, e l , e2 E then contains a component T with {e1,e2} n T # 0 and such that one of the components of (T- {el,ez}) U {[vI,v;] I i = 1,2,3,4) does not contain g but has an end in K 1 , For this component and the corresponding [v;,v;] contained in it, let [v:,v;] belong to P but let [v:,vi] belong to P. For the rest into P (p> see above. Clearly, S = K U P is as of transforming PI required in any case.

K,

z- (m

Proof of Step 3. Denote U$(z) = (F!,F2,F3},v = IV(c")l, e = lE(G*)l, f = lF(G*)l,let a: denote the number of faces of length i of G* different from F1, F2, F3. Finally, let L = E l , 4(Fi). Then we have

+ 54; + 6aA + vah v17 L + 56; + 6 ~ ; + 7( f - 3 L + 7f - 21 - ( 2 4 +

3v = 2e = L

2 =

U;

- u&)

Ui).

From this we obtain

- 7f + 21 2 L - ( 2 4 + UA). By applying the Euler Polyhedron Formula 7v - 7e + 7f 3v

= 14 and

3v = 2e, the last inequality becomes 0

--2 + 7 2 L - ( 2 4 + UL).

(*I

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Suppose now that outside U F ( z ) no pentagon borders a pentagon or hexagon. Then we get as an estimate for 2t

and the more

+2 4 5

40;

2)

- 1;

substituting in (*) we then have

+7>L+-

1 2

which implies !(F;) = 2, i = 1,2,3. This obvious contradiction (v > 2 since IV(G)l 8) finishes the proof of Step 3.

>

Proof of Step 4. Suppose the claim to be false. Denote a pentagon of G* - U;(z) which does not correspond to a pentagon or 10-gon in G, by F5 = (vl,v2,v3,v4,v5) and by v[ the point adjacent to D;, 1 5 i 5 5, but not in F5. For simplicity's sake we consider a point D; obtained in G' by contracting a triangle A of G as a point of A c G onto which this A has been contracted; the same should hold for any other point of G*. (We shall make use of this interpretation of a contraction of a triangle from now on until the end of the proof of Theorem 3.1; thus V(G') c V ( G ) ) .For such v; denote by c; and d; the points of G such that A; = (vi,c;, d;) is a triangle of G. Suppose the notation chosen such that the 21;'s are indexed according to the counterclockwise order of the points of F5 and such that c;1'ies o n a shortest path from v;-1 to 0; (we set v;+l = v1 for i = 5 ) . Because of the indirect procedure of the proof we have to assume that F5 corresponds in G to a C,, or Cg,or C1l, or c13 with t(C2i+1) = 2i 1 for 3 5 i 5 6. W.l.o.g., the cases to be considered in G are: case 1): Exactly 01 lies on a triangle A l . case 2): Exactly 212 and 215 lie on triangles A2,As respectively. case 3): Exactly v3 and v4 lie on triangles A3, A4 respectively. case 4): Exactly v2, v3,vg lie on triangles A2, A3, A5 respectively. case 5): Exactly ~ 2 ~ ,3 , 2 1 4lie on triangles A2, A3, A4 respectively. case 6): Exactly 172, v3, v4,05 lie on triangles Az, A3, A4, A5 respectively. In all six cases we perform the same reduction: We delete all lines except vl, v2,vs and add the lines and points of the corresponding [q, v2], (v2,vs], q], [v;, The graph thus obtained is planar and cubic, and has fewer points than G. Call the new graph G' and denote A' = ( q , v ~ , v 5 )C G'. We now assume that G' is 2-connected in all cases (to

+

41.

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show that such an assumption is no loss of generality is the task of Step 8). Note that e, f,g correspond to three lines of G’,and only one of e,f,g could have been transformed into [v;, 41; if this is the case, then we retain in G’ the corresponding label for [vk,41. Thus we can choose in G’an [e, f ;g]BEPS-graph S’ = K’ U P‘ satisfying properties 1. and 2. of the theorem. Next we define Mo = {[v;,41,[v:,~ iI i ]= 1,2,5) and no = IE(h”) n Mol.

Note that M On E(S’)# 8 in any case. A) no = 0. Since in each of the cases 1)4)the cycle Czi+l taken together with its chords belonging to triangles contains an even cycle C2i, we can define K = Ii‘ u {Czi}.

In all cases 1)4)except case 3) we can assume w.1.o.g. that the corresponding Cz; contains both 213 and v4, while in case 3) we assume w.1.o.g. that CScontains all points of Cg except v4. As for obtaining P from P’ and P from P’, since K absorbs every triangle whose points lie in it follows from Remark 3.3 how one has to proceed around the points v1,2)2,2)5. As for [vi,v:] denote by T’the component of P’, P’ respectively containing this line. Then 2’’ - {[v;, 41) U {[v;, ~ 3 1[v:, , 041) consists of exactly two components one of which is being enlarged by a P or F-extension in case 3). Thus we obtain two paths of P, P respectively, having at v3, A4 respectively an end in K. This guarantees the preservation of properties l., 2. in S = K U P which is obviously a BEPS-graph of G. B) no = 1. This implies MOn E(Ii’) = [v;, 41. Denote by Ii; the cycle in K’ containing this line, and consider C2i (see A)) as well. It is easy to see how one obtains an even cycle K1 containing V(1i;) U V(C2i) by looking at ( K i - {[v;, 04’1)) U { [vi,v3], [v:, vq]} and Cai except in case 3). In this case K1 will contain v4 $! V(C2i) but not c4 E V(C2i). As for obtaining P and P from P‘, respectively, see Remark 3.3 and A). Again, for K = (K’ - { K i } ) )U {Kl}, S = K U P is as required. C) no = 2. Then [v;, vi] $! E(If’). In the two cases where [v:, v1] E it40 n E(K’) we proceed analogously to what we did in B). However, depending on whether IV(K‘) n V(A’)l is 3 or 2, one has to pick up all points of C2i+l or all but one of these points. Similarly, if [vi,v1] 4 Mo n E ( K ’ ) , and if Ki C K’is the cycle with E ( K ; ) n Mo # 0, then If; can be enlarged to an

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even cycle K1 in the cases 2)-6) by absorbing all points of C2i+l- 01 or all but one of these points (depending on whether v1 does not or does lie in K ; ) . In case 1)however, ~ 3 , 1 7 4I$ K1 and K1 contains as many points of A1 as Ki does with respect to A' (consequently, v2,v5 E V(K1)). In all but this last of the cases, P and P are obtained form P' and P' analogously to how this has been done in A) by using Remark 3.4 (note: in the cases 2)-6), A1 does not exist; i.e., for [vi,v1] 4 E ( K ' ) P - and F-extensions are possible without violating property l., while this, in general, would not be so if case 1) had been handled like cases 2)-6)). To finally handle case 1) for [wi,v13 4 E(K'), take T' as in A) and form

T = T' - {[vi, 4]}U {[vi, ~ 3 1[v3, , Q], [ v 4 , 4 ]

(n

which is a path. Moreover, [v2, v3] and [vq, vg] form two paths which belong to P if T belongs to P ( P ) . Since in this case K1 absorbs Al, a P - and P-extension is applicable at v1 if necessary. Summing up we can say that also for no = 2, S = K U P is as required (with K as defined in B) and P as just explained). D) no = 3. Then [vi, v:] E E ( K ' ) anyway. In case 1) we obtain K from K' by replacing [vi,v;] with [v:, v3], [v3,v4], [v4,v:] and absorbing A1 (depending on how K' absorbs A' we will have IlV(1C) nV(A1)l- IV(K') n V(A')ll equal to 0 or 1 which does not matter since A2 and A5 do not exist in this case). Moreover, by letting [v2, v3], [v4,v5] belong to p,it is clear how P (F)is obtained from P' (F). For the cases 2)-6) we first observe that A, and A j exist for at least one i E {2,3} and at least one j E {4,5}, while A1 does not exist. Denote K i , K i E K' such that [vi,vi] E E ( K { ) , [v:,vi] E E(K4) for two i in {1,2,5}. Form in G

-

(Ki U K; - E(A') - {[&

411) U {[vk,v31, [vi, v4lh

this is a forest consisting of two disjoint paths PI,P2 (regardless of whether Ki = K; or K: # K i ) . The endpoints of PI U P2 are v3,vq and v,-,v,, r E {1,2}, s E {1,5}, s # r . Clearly, C2i+l taken together with its chords

contains pairs of disjoint paths from vr to 03 and v, to v4. Among all possible choices far these pairs, choose P(v,-,v3) and P(v,,v4) such that

PI u P2 U P(vr, ~ 3 U) P(vs, ~

4 )

is either an even cycle or consists of two disjoint even cycles. This is always possible because that expression is a cycle or consists of two disjoint cycles anyway and P(v,, v3) absorbs A,, and P(ws,04) absorbs Aj (see above).

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Thus we obtain K from K' with all vi's except possibly v1 contained in K. Since A1 does not exist in cases 2)-6) it is clear how a P- and F-extension yields P , P respectively from PI- E(A'),P'- E(A') respectively, such that S = I( U P is as required. Since no > 3 is impossible, the proof of Step 4 is finished. Proof of Step 5. As in the proof of Step 4 we proceed indirectly. Thus we assume G* - U ~ ( Jto) contain a hexagon F6 which corresponds in G to a cycle c6, or c 1 0 , or c 1 2 , or c 1 4 , , or CIS, or CIS, with t ( c 2 ; ) = 2i. For F6, let v;,vi,c;,d;,A;, 1 5 i 5 6, be defined the same way as has been done for F5 in Step 4. Since we shall use two different reductions we define two classes of cases depending on the position of the chords of the C 2 i ' ~ . First we state the 12 nonisomorphic possible cases. case 1): None of the v; lies on a triangle of G. case 2): v1 and v6 lie on triangles of G. case 3): 0 2 and v5 lie on triangles of G. case 4): v2 and 06 lie on triangles of G. case 5): 01, v2 and vf3 lie on triangles of G. case 6): 01, v 2 and 05 lie on triangles of G. case 7): v2,v4 and 2)6 lie on triangles of G. case 8): v l , Q, v5 and v6 lie on triangles of G. case 9): 212,214, v5 and 06 lie on triangles of G. case 10): v1, v 2 , v4 and v 5 lie on triangles of G. case 11): All v; except v3 lie on triangles of G. case 12): All v; lie on triangles of G. Class I contains the cases l), 7) and 9)-12); the other cwes form class 11. In any case we form G" = G - V(C2;) fm the corresponding C 2 i and define G' to be for the cases of class I

for the cases of class I1

G' = G"U {[v:,o~],[v:,v~],[v~,v:]}. As in the proof of Step 4 we assume in all cases G' t o be 2-connected (the proof of Step 8 will show this assumption not to be a loss of generality). Clearly G' is planar and cubic and has fewer points than G; so we can choose a BEPS-graph S' = N' U P' of G' according to Theorem 3.1. For

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159

the given e , f,g the same consideration as in the proof of Step 4 applies; thus S' can be called an ( e , f;g)-BEPS-graph. Analogously to what we did in the proof of Step 4, define

for the respective classes I and 11, and let us first deal with what can be handled more or less simultaneously for both classes. A) no = 0. Then in all 12 cases

K = K' u { Cz;} for the corresponding C2j, 3 5 i 5 9, i # 4, is a set of disjoint even cycles. Replacing in P' and P' the corresponding elements of MO with the corresponding pairs of elements of {[vi,vj] 1 1 5 j 5 6}, and adding the chords of C2j t o F,we obtain P, P respectively, such that S = K U P is a BEPS-graph as required unless MO C E(P'). For this latter case, S as just constructed, has all properties of a required BEPS-graph of G except that S is not connected; one of the two components of S is C2;. In particular, S satisfies property 2. of Theorem 3.1. N o matter how many elements of MO are contained in the component of P' containing g, it follows from the construction of K that P contains yet another component T with both ends in K ,with one endline of T being [vh,vk], where 1 5 k 5 6, while g $ E(T). Throwing [vb,vk] from into P, we obtain what we want. B) no = 1. For all 12 cases, this situation can be handled analogously to what we did in the proof of Step 4 for no = 1, as long as for class I1 MOn E ( K ' ) # [4,2): Note ].that since C2j is an even cycle, the extension of the cycle Ki c K' with K: 3 MOn E(K') to an even cycle K1 of G will absorb all points of C2i. As for P, B respectively, we proceed analogously as before. Thus for K = (K' - { K i } )U K l , S = K U P is as required. If, however, one of the cases of class I1 applies, and MonE(K') = [vi,41, then we extend Ki (defined as above) to an even cycle .Kl of G by passing through all points of C2; except v3,wq. Note that A3,A4 do not exist for the cases of class 11. Thus, if we replace (in P',P' respectively) [vi,v:] with [v$,o3],[v3,v~q],[v4,v~] and let [22,2)3],[v4,y5](with 2 2 = d2 if A2 exists, y5 = c5 if A5 exists, and 2 2 = v2, y5 = v5 otherwise) belong to P ( P ) if [v$,vA] E E(P') (E(P')), and if we proceed at the other points of C2j in constructing P and P from P' and P' as we did in A), then for this P and K defined as above S = K U P is as required.

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C) no = 3. In order to handle class I and class I1 simultaneously, we define for the cases of class 11:

S" = (S'- {[VL

4],[4,411) u { [ 4 , 4 1 9 [4, 411.

Thus S" resembles locally S' of the cases of class I. However, S" might not be connected anymore, and I<" (which is the actual part of S' changed in the definition of S") might contain exactly two odd cycles. But by noting that for the cases of class I1 A, and A, exists for at least one r E {1,2} and a t least one s E { 5 , 6 } one obtains a set of disjoint even cycles, K ,from K' by either absorbing the entire set V(C2;) or - in the cases of class I1 where h"' has odd components - by absorbing all of V(C2;) except exactly one point of each of Ar,A8. As for obtaining P (P)from P' (P'),we define

P = P' for the class I cases and P = P' u ( P ( v t ,K ) 1 t = 3,4} u P(u1,?46)for the class II cases, where P ( q , K) is a path in C2i from wt to K starting with the line [vt, q - 1 1 , [wt,yt+l] respectively, depending on whether t = 3 or t = 4 (for these two lines, see B)); and P(u1, us) is a path in C2i from u1 E V(K) i l V(A1) to ug E V ( K )i lv(&) such that E(P(ul,u6)) n E ( K ) = 8 (i.e., depending on how the A-absorption is performed in A,,& respectively, u1 = v1 or u1 = c1 or u6 = 2)6 or U 6 = de respectively). Now S = K U P is as required in all 12 cases. D) no = 2. Now we handle classes I and I1 differently. Class I. a) Case 1): W.1.o.g. M On E ( K ' ) = { [vi, 41, [vh, wi]}. Then by

we obtain the necessary I<. By defining for E WT'),

I;.;.[

T' c P' or T' c P' with

and letting [v1,v6],[vq,v5]belong to P (P)if T' belongs to P (P),one obtains the necessary P. Clearly, S = K U P is as required. b) Other cases of class I: We observe that regardless of the shape of Mo n E ( K ' ) , if we take K i , K ; E K' with K{nMo # 8,i = 1,2, possibly Ki = A ';, then ( V ( K i )U V(K4)U V(C2i)) either has a hamiltonian cycle Kl (which is even) or is spanned by two disjoint cycles C1,C2 each of which contains

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V(A;) for a t least one i. Thus, by a proper A-absorption C;can be turned into an even cycle K ; of G. Thus,

K = (K' - { K i , K i } )u {K;I i = 1, i = 1,2 respectively} is a set of disjoint even cycles of G. Moreover, it is clear from all the previous constructions and P-, F-extensions, how P (F)can be obtained from I" (p')such that S = K U P is as required. Class 11. If MonE ( K ' ) = { [vi, 41, [vi, vi]}, then we proceed analogously t o what we did in b) above. For the other two choices of M o n E ( K ' ) we can extend the cycles Ki,K i (see b) above) in such a way t o even cycles K1, K2 disjoint or identical of G that K1 U K2 IIV(C2;).It is clear how P is obtained from P' (P') such that S = K U P is as required. Summing up, we have shown that for all possible choices of no, G contains a BEPS-graph according to Theorem 3.1 if G' does. This finishes the proof of Step 5.

(9

Proof of Step 6. Suppose G* - V $ ( z ) does contain two neighbouring pentagons F5f, F,**. Denote their points in counterclockwise ordering by v1, v2,. ,v8 with E(F,* n Fz*) = [vd, 2181. Because of Step 4 either F; U F,'* c G or A; = (c;, v;,d;) exists for 1 5 i 5 8. We use the same notation as above. Thus we either have an 8-gon or a 24-gon, i.e., an even cycle C anyway, as a subgraph of the graph H induced by v1,. . . ,v8, respectively by Ug,V(Ai). Define for MO = {[v;, 1.741, [oh,41, [vk,vk]}

..

G' = (G - V ( C ) )U Mo and suppose it is 2-connected (the proof of Step 8 will justify this Assumption). Since it is planar and cubic anyway, we can choose an (e,fg)-BEPSgraph S' = K' U P' of G'. (As for - exactly - one of the lines e,f,g becoming an element of Mo, see what we said before in the proofs of Step 4,Step 5 respectively). Define no as before. A) no = 0. Then we form K = K' U {C} and replace in every component T' of P',P' respectively, with E(T') n MO # 0 the corresponding element of MO with the appropriate two lines of the form [v:,~;];the elements of E ( H ) - E(C)are added to P'. Thus we also obtain P, P respectively, such that S = KUP is as required provided not every line [v:, v;],i = 1,2,3,5,6,7 belongs to p. If, however, [vi,v;] E E ( n for i = 1,2,3,5,6,7, then we conclude as we did in part A) of the proof of Step 5 , that P contains a

H . Fleischner

162

component T with g 4 E ( n , and T has both ends in K ;and one of these ends is a certain 0;. By forming for this i

Po= Pu{[v;,v;]},

~=P-{[v:,v;]}

we see that SO = K U PO is as required. B) no = 1. Then V ( C ) can be absorbed by extending the cycle Ki E K' with E ( K i ) n Mo # 0 to a cycle K1 of G. This is obvious regardless of which line of Mo belongs to Ki. Since Ki and C are even, so is K1. Thus we can define K = (I<' - { K i } )u { K l } . By the same replacement procedure employed for T'in A) and by adding E ( H ) \ E ( K ) to P', we obtain P and H,as in A). Clearly, S = K U P is as required. C) no = 3. Then, similar to B), K' can be transformed into K c G by absorbing all of V ( C ) : One simply picks up the points of C by following the route indicated by the elements of Mo. That is, in H U {[v:,oi] I i = 1,2,3,5,6,7} exist three disjoint paths P(vi,vi), P(vh, v;), P(vL, v;) covering V ( H )and each containing an even number of points of H. Thus, by replacing in E(K') the elements of Mo with these three paths, we obtain K from K'. Hence, for

P = P',

P = p'u

( E ( H )- E ( K ) )

S = K U P is as required. D) no = 2. Here we have to distinguish cases according to the size of C. 1) t ( C )= 8. Replace the elements of Mo with the three paths defined in C), call them shortly Pl,P2,P3, and let E ( H ) - U;=!=,E(P;) belong to P (H) if n M O # 0 (P'n MO # 0). Thus, S' has been appropriately extended to S c G. 2) l ( C ) = 24. Let {[v:,v;.],[v~,v:]} = Mo n E ( K ' ) . Then either for the pairs ( i , ~and ) ( j , s ) or the pairs ( i , s ) and ( j , ~exist ) two disjoint paths P I ,P2 with V(P1)u V(P2)- {v;,v;,v;,v:} = V(H). Clearly, each of Pl,P2 absorbs at least two A;'s, 1 5 i 5 8. Thus, by replacing the elements of MOn E ( K ' ) with PI and P2 we obtain I(, a set of disjoint cycles in G. If K contains an odd component, then it contains two - call them Kl,l<2 - such that P; c I<;, i = 1,2. This follows from the evenness of l ( C ) . If K has odd components, replace P; with Ptr obtained from P; by altering the A-absorption in exactly one triangle of

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H , i = 1, 2, and obtain this way a set of disjoint even cycles I<-. Let

KO = K or K O = I<-, depending on which one has even cycles only. For

[v;, wi] E MO- E(K‘) we can assume w.1.o.g. that v k , v l E V(K0). After having replaced [v;, vi] with [v;, vk], [vi,vq] in the corresponding component of P‘,P’ respectively, it is an easy task to split up the lines of E ( H ) - E(K0) into classes T‘,F to finally obtain P and P with T‘ C P, F c H,such that S = KOU P is as required. This finishes the proof of Step 6. Proof of Step 7. Since by Step 6, G’ - & ( z ) does not contain two neighbouring pentagons, it must contain, by Step 3, a pentagon F: neighbouring a hexagon F:. It follows from Steps 4 and 5 that Fz corresponds to a pentagon F5 in G while F,*corresponds in G to a cycle c 8 with V(C8) 3 V(A) for a triangle A . Denote counterclockwise the points of F,+UF,*by v1,. . . ,vg such that [w2,07] = E(F: n F;) = E(F5 n CS).We use the same notation for the objects involved as in previous steps. W.1.o.g. A = A5 or A = A,. Form the graph (regardless of A = A5 or A = As), G’ = G - V(c8) u {[vl,

081,

[vi, 4 1 7 [vi,v:]}

and let MO = {[vi,41, [vi,wg], [v;,vi] I i = 1,8,9}. As before, let S’ = K‘ U P‘ be a BEPS-graph in accordance with the theorem (for we assume G‘ to be 2-connected and leave it t o the next step to show that this is not a loss of generality). With the experience gathered before, the reader is in a position to handle easily the cases no = 0 as well as 72.0 = 1 with [vi,w;] E E(K’) and no = 2 with E(K’)nE(A’) # 0 (note for the first two cases P’nE(A’) # 0). Thus we have to consider A) no = 1 and [v&,vk]E E(K’). Since E(A‘) n E(K’) = 0 in this case, it follows that S“ = S’ - V(A’) is a BEPS-graph of G” = G’ - V(A’) satisfying Theorem 3.1. Now extend I<” t o I< in G by absorbing all of A and v3,vq, and respectively. Clearly, K is a set of disjoint even cycles. Define PO = [v3,v2],[ U 2 , vl], Lvl, 2191, [v9, b 8 , v71 and P = (P”- [w:, v;]) U Po U A ( w ~w:),

v5,v6

%I,

where A(vi, vg) = 0 if [v:, wk] $! E(P’) and A(vi, v:) = [w:, [vi,vg] E E(P,’) C E(P’), otherwise.

vq], [v:, 051

with

H. Fleischner

164

Clearly, S = I< U P is as required (we need not detail the construction

of

F).

B)no = 2 with ,??(A‘) n E ( K ‘ ) = 0. Here, all points of c8 are absorbed in the transition from K’t o K with [wQ,w3],[w3,w2],[wz,2171, [w7,x ] being used in replacing [w;, w;] with a path of G (z = ds if A = A6, and x = otherwise). Define PO and P similar to PO and P in part A). Now, S = K U P is as required. C) 71.0 = 3. This implies

E ( W n W‘) # 0 # E(I<’)n

m,41,[4,4} # {[4,41,[4,.hI}.

Let K; E I<’ denote the cycle for which E ( K ; ) n E(A‘) # 8, and let K; E K’denote the cycle which contains the third element of M O in I<’. 1) I<; # I<;. a) [w:, wi] E E ( K i ) . Then one constructs an even cycle KO c G with

E(Ko) 3 ( E ( K i )- E ( A ’ ) )u (E(K2) - [w:, w~])u {[w:,~],

[~~,~5],[~4,~3],[~3,~2],[~2,~i],[~s,~7],[v7,x1[x*, YI)

where x = d g , x* = cg, y = w 5 for A = A,, and x = 2)g = x* and y = d5 for A = As; and the character of the A-absorption at A depends on that of the A-absorption at A’. Now it is easy to construct P from P‘ (7) once

(9

K = I<’- { K ; , I < ; } u { K O }

is given. Thus, S = K U P is as required. b) [w;, w;] E E ( K ; ) . As for the replacement of [v$,w;] with a path in G we proceed as in part A) with the A-absorption absorbing all of A. Thus we obtain K2 from I<;. Similarly, K 1 is obtained from I<; by containing either at most two lines of F5 (and w2,07 !$ V(K1))or at least three lines of F5 (and w2,07 E V(K1))depending on the A-absorption of A’ by K;.Hence, with

Ii =(K’-{K;,K~})U{K1,Ii2} defined and using the experience previously acquired in constructing P from P’, it is easy to see that S = Ii U P is as required. 2) I{; = I<; =: I<;. a) [w:, wh] E ,??(I<;). Now replace the lines of A’nKt, with the corresponding subset of { [wl,wg], [ v ~2),8]}, and replace [w:, wi] with

[v:,

~ 1 P(%, , w5), [w5,wh1

The Prism of a 2-Connected, Planar, Cubic Graph is Harniltonian

165

where for the path P(v*,v5) we have

the value depending on the A-absorption of A' by KA. Thus we obtain from I<; the even cycle KO C G which absorbs 2 or 3 points of A. It is trivial t o find for Ir' = (K' - {KA})u { K O } the corresponding P,H respectively, such that S = K U P is as required. b) [v$,vi] E E(K6). To obtain KO C G from K6 we perform the same replacement procedure employed in case 1) b) and form K as in 2) a). Again, P, H can be obtained easily from P', P' respectively. D) no = 4. Denote by I<:, i = 1,2,3, the elements of K' such that E ( K ; )n E(A') # 0, [vi,v5] E E ( K ; ) , [v;, 41 E E(KA). 1) I<; # I<; # I<; # K;. Then we obtain from K i and I<; one even cycle KO C G such that KO 3 v(c8)- (~2,217). K1 is obtained from Ir'; the same way as in case C). 1) b). For K = (I<'- { K ; , K ; , K ; } ) u{ K o , K 1 } we have locally around 2 1 2 , the ~ ~ same behaviour of K as in case C) 1) b). Hence it is easy again to find P and P such that S = K U P is as required. 2) K ; = I<; # I<;. Proceeding as in case 1) will yield an S as required (with KO= K l ) . 3) I<; = I<; # I<;. Again we obtain one even cycle K O from K i and I<; = I<$ by using iv$, 2131,[v37212]7 [212,111],[%3, w],[v7,2],[v6,'41 (2 = d6 for A = As, 2 = 2)6 otherwise) in order to "hook" K ; - E(A') onto I<; - [vi,41; and the corresponding path of G from v: to vi replaces [v:,vi]. Clearly, the type of A-absorption performed at A depends on the A-absorption of A' by I<;. Define K = (I<'- {I<;,K;})u {KO}.

To find P and from P',P' is as easy as in case C) 1) a). 4) I<; = I<; = I<; =: I<&. In this case the replacement procedure is the simplest one: [vi,vk] is re, is replaced placed by a path P(vi,vk) c G with v3 @ V ( P ( v i , v i ) ) [wk,vA] by a path P(u4,v;) c G with [v2,07]E E(P(w4,v;)), and E(A') n E(Kh) is replaced with the corresponding subset of [vl, 7191, [vg,vg]. Thus we obtain an even KO c G where the type of A-absorption of A by KO depends on the A-absorption of A' by K ; . The remainder of this case is trivial.

H. Fleischner

166 Thus, the proof of Step 7 is finished.

Proof of Step 8. If in any case of the reductions performed in the Steps 4-7 the resulting planar, cubic graph G' is not two-connected, then it is clear that G' is connected and every bridge of G' must lie on one of those newly created faces F' of G' for which the inequality IE(F') n Mol 2 2 holds. Otherwise, G has already a bridge or has connectivity 2, a contradiction to Step 1. Hence we can conclude from the types of reductions performed that G contains a line-cut C consisting of three elements two of which lie on F5,Fs, F,*uF,**- E(F,*nF,**),F{ UF; - E ( F{ nF;) respectively. Therefore, IC n {e,f,g}I 5 1. Denote the two components of G - C by G1,Ga such that z (the point incident with e , f , g ) lies in Gz. If follows from Steps 1 and 2 that G1 and G2 have more than one point each. Now consider among all line-cuts C having exactly three elements, one for which IV(G,)l is as small as possible but IV(G1)l > 1 and therefore, and since X(G) = 3, IC n { e , f , g } l 5 1. We claim that G1 must contain at least one of the configurations considered in Steps 4-7. First we note that C is a line-cut in G* as well. Thus, if we let

such that w.1.o.g.

21, y1, tl

E GI, and introduce a new point

21, to

form

then every face of H1 not containing z1 in its boundary, is a face of G as well. In particular, this is true for triangles of HI, thus H;* is obtained from HI analogously to how G* is obtained from G. Moreover, Hi obviously is, by Steps 1 and 2, a planar, cubic, 3-connected graph having no 2-, 3- or 4-gons outside U ~ ( z 1 )Applying . Step 3 to H1 with z1 in place of z , we find in H1 one of the subgraphs considered in Steps 4-7 which are corresponding subgraphs of GI and therefore G as well. Performing in such subgraphs the reduction procedures employed in the proofs of these steps, the resulting graphs G' must be 2-connected. Otherwise, we would find a line-cut Co having exactly three elements such that - as can be seen easily - the corresponding GY satisfies GY 5 GI. (Clearly, Con C # 8 may hold; this implies, however, ICon CI = 1.) This contradicts the choice of C. Summing up what we have proved in Steps 1-8 we can say: A smallest counterexample G to Theorem 3.1 must be 3-connected (Step l ) , G*-U$(z)

The Prism of a 2-Connected, Planar, Cubic Graph is Hamiltonian

167

cannot have a face boundary of size 2, 3 or 4 (Step 2), and therefore - by Step 3 - must contain at least one of the subgraphs considered in Steps 4-7. By Step 8, we can choose these subgraphs such that the reduction of G to G' in each case yields a two-connected G'. By Steps 4-7, a BEPS-graphs S' of G' satisfying properties l., 2. of Theorem 3.1 yields such BEPS-graph S of G. By the choice of G, S' exists while S supposedly does not. This contradiction proves Theorem 3.1 true.

4

Final remarks

When the author first dealt with finding EPS-graphs in connected bridgeless graphs G, [l],and in subsequent papers as well, a proof by induction was employed in which the reduction step used a whole cycle C whose lines were deleted. Then in G - C block-chains were considered. In view of Step 1 of the proof of Theorem 3.1 respectively with regard to properties l., 2. of that theorem, one might very well ask whether it is possible to develop a similar procedure for finding BEPS-graphs. Of course, the cycle C with E ( C ) to be deleted must be even. This procedure was, in fact, tried out by the author in a first attempt t o prove Theorem 3.1 in a somewhat weaker form. However, as it turned out one cannot prescribe an arbitrary even cycle to belong to I<. Although the example to this fact is rather trivial, it took the author weeks to find it. Take the K 4 embedded in the plane and replace a point of the boundary of the outer face with a triangle. In this graph replace the two inner points with triangles. Now subdivide every line e of this new graph with two new points ve, We and add the new lines [v,, we](i.e., into every line a digon is inserted). Denote the graph obtained by G. As the initial cycle C take the boundary of the outer face of G which is a 12-gon. It is easy to see that for this C, there is no BEPS-graph S = K U P of G with C C K. However, although that first attempt to find BEPS-graphs had failed, an interesting result remained untouched by this failure. As a side product, so to say, it was shown that in any planar, 2-connected, cubic graph G, there is an even cycle G having a prescribed line e as a chord provided e does not belong to a line-cut of size two. This result is contained in [4]. But the connection between BEPS-graphs in the context of hamiltonian cycles in prisms of cubic graphs and EPS-graphs in the context of hamiltonian cycles in squares of graphs, is not all that superficial as the above remarks might indicate. In fact, Theorem 2.4 (a cubic G has a hamiltonian P ( G ) iff G contains a BEPS-graph) has its analogue in the squares of graphs, namely: the total graph T ( G ) is hamiltonian iff G contains an

H . Ffeischner

168

EPS-graph (see [6, Theorem 11). In fact, if we want to prove Theorem 2.4 such that the BEPS-graph S = K U P obtained from C has the additional property that the lines of P are the bridges of S, we can achieve this by choosing C such that it uses as many pillars of P ( G ) as possible. This corresponds t o the choice of a hamiltonian cycle C in G2 using as many lines of G as possible (see [5]) which led to [6, Theorem 13. Moreover, in both cases the choice of C implies a covering walk W in S such that each line of P is passed exactly twice, namely once in each direction. This would also be true in the case of hamiltonian cycles in P(G) for arbitrary G. Plus, exactly the lines of P are passed by W more than once in each case. However, these choices of C permit only local conclusions, since - in all cases - a “good way” of checking whether these C are optimal (with respect to the prescribed lines passed) would be tantamount to a good algorithm for checking if G itself is hamiltonian or not. It should be noted, however, that the dissimilarities between the constructions of C in G2, P ( G ) respectively, are the basic reason why S has to be a bipartite EPS-graph in the case of C c P(G)for cubic G. We note in passing that EPS-graphs played a central role in all the referenced papers on squares of graphs as well as many other papers on squares of graphs. However, the reason why the proof techniques for Theorem 3.1 differ essentially from those for finding EPS-graphs (see, e.g. [1,2,6]) has been demonstrated by the graph exhibited a t the beginning of this section. On the other hand, the extraordinary length of the proof of Theorem 3.1 lies in the possible appearance of triangles. This marks a qualitative difference to finding a 1-factorization or a hamiltonian cycle in cubic graphs. For,if G has a triangle A,and if the cubic graph G1 obtained from G by contracting A to a point v, has a 1-factorization or even a hamiltonian cycle, then G has, trivially, the respective property. Such a conclusion, however, cannot be drawn regarding the property of having a BEPS-graph of the type stated in Theorem 3.1. For suppose none of the three lines ei,ei ,e$ incident t o v in GI belong to K I C SI.Suppose ei, e&lie in PI c S1, and therefore, e$ E E(PI) holds. If one extends S1 to a BEPS-graph S of G, then the points of A lie, in general, in the same component of P c S. Hence, a line of A will belong to B with the ends of this line lying in P. A similar argument can be made if ei, ei lie in We have shown - in essence - in Theorem 3.1 that for any planar, cubic, 2-connected graph G, P(G)is hamiltonian. The proof did not deal with 3-line-colourability whatsoever but the Euler Polyhedron Formula was employed. Thus, for the proof of Theorem 3.1 the planarity of G was

r,

The Prism of a 2-Connected, Planar, Cubic Graph is Hamiltonian

169

essential. Nevertheless one might ask if the hypothesis of planarity can be deleted. The Petersen graph indicates that a negative answer to this question might require a nontrivial effort to get it: For, the Petersen graph has a cycle of length 8. But even if one stays with planarity but drops the hypothesis of 3regularity, the problem of finding new classes of graphs G having a hamiltonian P(G)might be rather difficult to tackle. This is indicated by the graph of Figure 1. So what if we try to determine all planar cubic graphs having a hamiltonian prism? Well, let us leave our hands off that question, for it is tantamount to ask for the planar, cubic graphs which are hamiltonian or a t least 3-linecolourable. To see this let G be any cubic graph. Form GA from G by replacing each point v of G with a triangle A. Subdivide every line e of GA with two new points ae,be; denote this graph by H * . Now take two copies Ta,, Tb, of that graph TOwhich has - except for one endpoint vo - points of degree 3 only such that TO- 00 is homeomorphic to the tetrahedron K q . Let w.1.o.g. the endpoint of Ta, (Tb,) be denoted by a, ( b e ) . Identify for each e E E(GA)the point a, ( b e ) of H* with the point a, ( b e ) of Ta, (The). The graph H thus obtained is planar and cubic. It is easy to see that any BEPSgraph of H induces a hamiltonian cycle of G, and vice versa. Similarly, if we restrict the above subdivision and identification procedure (of the e’s, ae’s,be’s) to the new triangles of GA,then the analogous statement can be made regarding 3-line-colourability of G: any BEPS-graph of H induces a 3-line-colouring of G, and conversely. Of course, the above constructions and statements on hamiltonicity and 3-line-colourability of G remain true if G is just cubic, but not planar (compare this with V. Chvital’s note [13] on squares of graphs). Thus we have very similar reductions of both the hamiltonian cycle problem and the 3-line-colouring problem of cubic graphs t o the hamiltonian cycle problem for prisms of cubic graphs (with the reduction from 3-line-colourability giving, not surprisingly, the smaller instance). This indicates that determining whether an arbitrary cubic (or planar and cubic) graph G has a hamiltonian P ( G ) is an NP-complete problem.

Acknowledgement The author wishes to express his thanks to his Ph.D. student E. Wenger who wrote up in German a first version of the proof of Theorem 3.1 based on a series of lectures by the author at the University of Vienna. Mr. E. Wenger’s valuable comments helped t o formulate the proof as presented in this paper.

H. Fleischn er

170

References [l] H. Fleischner, “On spanning subgraphs of a connected bridgeless graph and their applications to DT-graphs,” J. Combin. Theory 16 (1974), 17-28.

[2] H. Fleischner, “Hamiltonsche totale Graphen von ebenen Graphen,” Math. Nachr. 68 (1975), 83-91. [3] H. Fleischner, “Eine gemeinsame Basis fur die Theorie der Eulerschen Graphen und den Satz von Petersen,” Monatsh. f Math. 81 (1976), 267-278. [4] H. Fleischner, “Even cycles with prescribed chords in planar cubic graphs,” Discrete Math. 44 (1983). [5] H. Fleischner and A. M. Hobbs, “A necessary condition for the square of a graph to be hamiltonian,” J . Combin. Theory 19 (1975), 97-118. [6] H. Fleischner and A. M. Hobbs, “Hamiltonian total graphs,” Math. Nachr. 68 (1975), 59-82. [7] H. Fleischner and A. M. Hobbs, “Hamiltonian cycles in squares of vertex-unicyclic graphs,” Canad. Math. Bull. 19 (1976), 169-172. [8] P. R. Goodey and M. Rosenfeld, “Hamiltonian circuits in prisms over certain simple 3-polytopes,” Discrete Math. 21 (1978), 229-235. [9] B. Griinbaum;Convez Polytopes, John Wiley & Sons Ltd. (1967).

[lo] F. Harary, Graph Theory, Addison Wesley, Mass. (1969). [ll] M. Rosenfeld, D. Barnette, “Hamiltonian circuits in certain prisms,” Discrete Math. 5 (1973), 389-394. [12] W. T. Tutte, “A theorem on planar graphs,” Trans. Am. Math. SOC.8 2 (1956), 99-116.

[131 P. Underground, “Note on graphs with hamiltonian squares,” Discrete Math. 2 1 (1978), 323.

Annals of Discrete Mathematics4 1 (1989) 171-178 0 Elsevier Science Publishers B.V. (North-Holland)

A Note Concerning some Conjectures on Cyclically 4-Edge Connected 3-Regular Graphs H. Fleischner Institute of Information Processing Austrian Academy of Sciences Vienna, Austria

B. Jackson Department of Mathematics Goldsmiths’ College, University of London London, UK

Dedicated to the memory of G. A. Dirac Five conjectures about 4-edge connected graphs are considered. It is shown that three of them are equivalent.

1

Introduction

All graphs considered are finite and without loops but may contain multiple edges. Given a graph G and subsets S1 and S2 of V ( G )we shall use G[S1] to denote the subgraph of G induced by S1 and e c ( S 1 , S ~ )to denote the number of edges of G joining vertices of S1 to vertices of S2. Let P( G ) = {(Sl,S2)l(S1, S2) is a partition of V ( G ) } , Pe(G) = {(Sl,SZ)E P(G)(G[SI] and G[S2] contain edges of G}, and P,(G) = {(Sl,5’2) E P(G))G[S1] and G[S2] contain cycles of G}. Put

171

H. Fleischner and B. Jackson

172

Thus X ( I i 4 ) = 3 and A e ( K 4 ) = 4 = X c ( I i d ) . For k an integer, we shall say that G is k-edge connected if X(G) 3 k , that G is essentially k-edge connected if Xe(G)2 k, and that G is cyclically k-edge connected if X,(G) 2 k . Clearly X,(G) 2 Xe(G) 2 X(G). We shall use the following elementary results:

Lemma 1. If each vertez of G has degree at least $k 4- 1 then G is essentially k-edge connected i f and only i f G is cyclically k-edge connected. Corollary 1. A 3-regular graph is essentially 4-edge connected if and only if it is cyclically 4-edge connected. Our purpose is to consider the following conjectures.

Conjecture 1 (Bondy [l]). There ezists a constant c, 0 < c < 1, such that every cyclically 4-edge connected, 3-regular graph G has a cycle of length at least c IV(G)l. Conjecture 2 (Jaeger [5]). Every cyclically 4-edge connected 3-regular graph G has a cycle C such that G - V ( C ) is acyclic. Conjecture 3. Every cyclically 4-edge connected 3-regular graph G has a cycle C such that G - V(C)is an independent set of vertices. Conjecture 4. Every essentially 4-edge connected graph G has an Eulerian subgraph H such that G - V ( H ) is an independent set of vertices. Conjecture 5 (Thomassen tonian.

[el). Every 4-connected line graph is hamil-

We shall show that: (a) The truth of Conjecture 2 would imply the truth of Conjecture 1 with 1 c= 5 .

(b) Conjectures 3, 4 and 5 are equivalent.

Some Conjectures on Cyclically 4-Edge Connected 3-Regular Graphs 173

Remark 1. Jaeger [5] has pointed out that the truth of Conjecture 2 would also imply the result of Seymour [7] that every bridgeless graph has a nowhere zero 6-flow. Remark 2. A modified version of Conjecture 3, with the added hypothesis that G is not 3-edge colourable, was made by Fleischner in [2]. Remark 3. Some evidence in favour of Conjecture 4 may be deduced from the result that every 4-edge connected graph has a spanning Eulerian subgraph. This result was pointed out to us by F. Jaeger [4]and C. Thomassen [8]. Its proof follows easily from the result of Kundu [6] that every 4-edge connected graph has two edge-disjoint spanning trees.

2

Proof of (a)

Proof. Let G be a 3-regular cyclically 4-edge connected graph and suppose G has a cycle C such that G - V ( C )is acyclic. Let IV(C)l = m. Then the number of edges in G - V ( C ) is at most IV(G)l - m - 1. Since G is 3-regular, it follows that the number of edges from V ( G )\ V ( C )to V ( C ) is at least

3(IV(G)I - m ) - 2(IV(G)I - m - 1) = IV(G)l- m + 2. On the other hand, the number of edges from V ( C )to V ( G )\ V ( C )is at most IV(C)l = m. Thus m 1 IV(G)l- M 2 and m 2 iIV(G)( 1.

+

3

+

Proof of (b)

We shall use the following operation. Let G be a graph, t a vertex of G of degree t , and u1, ~ 2 , .. . ,ut an ordering of the neighbours of z , allowing repetition of neighbours in the case of multiple edges. Let H be the graph . by obtained from the disjoint union of G - z and the cycle ~ 1 . ~ 2 .. ztz1 joining t; to ui for all i , 1 5 i 5 t. Then H is said to be an inflation of G at z and denoted by G(z;211,212,. . . ,u t ) . If every vertex of G has degree at least three then we may obtain a 3-regular graph J by successively taking inflations at each vertex of degree greater than three in G. We shall call such a graph J a 3-regular inflation of G. Abusing notation so that for each inflation G(z;~1,212,. . . ,ut)we give the new edge z;u; the same label as the old edge zu; we may consider E ( G ) c E ( J ) .

H. Fleischner and B. Jackson

174

Lemma 2. Let G be an essentially 4-edge connected graph and z be a vertex of G. Suppose d ( z ) = t 2 4 and d(w) 3 for all w E V ( G )\ { z } .

>

Then some inflation of G at

t

is essentially 4-edge connected.

Proof. Choose an arbitrary inflation H = G(z;u1,u2,. . . ,ut) of G at z and suppose H is not essentially 4-edge connected. Then there exists a partition (M1,M2) E P e ( H ) such that eH(Ml,M2) 5 3. Since d ( z ) 4, Mi # {z1,z2,. . . , z t } for i E {1,2}. Furthermore, since Xe(G) 2 4, ( ~ 1 ~ 2 2. ., ,.z t } $ Mi for i E {1,2}. Since eH(Ml,M2) 5 3 we may deduce that

>

(i) N I = { z r , z r + l , * - * , z s c } M I and N2 = {zs+1,zs+2,...,zr-1} c M2 for some r , s , 1 5 T , s 5 t , here subscripts are t o be read modulo t. (ii) G - z is either disconnected or else contains a bridge. We now construct an essentially 4-edge connected inflation of G at z. Consider G I = G - z and let G2 be the graph obtained by deleting all bridges from G I . Let F be the forest obtained from G1 by contracting each component of G2 into a single vertex. Then E ( F ) is the set of bridges of G I . Consider the following two cases.

(4JV) # 8. Let B1 and B,+1 be two components of G2 corresponding to end-vertices of F . Let {B2,B3,.. . , B p } and (B1, Bp+l,. . ,B q } be the sets of components of G2 corresponding to isolated vertices and end-vertices of F , respectively. Since Xe(G) 4 and since d(v) 3 for all v E V ( G ) , we may choose three neighbours zl;, vui+q and v;+zq of t in B; for 2 5 i 5 p and two neighbours vj and v j t q of z in Bj for j = 1 and p + 1 5 j 5 q , where we allow repetition of neighbours in the case of multiple edges. Label +2,. the remaining neighbours of z arbitrarily as ~ 1 + 2 ~ , v , + ~ ~ + l , v ~ + .2.~,q. Let EI = G ( ~ ; v i , v z , ..,wt). . If Xe(G) < 4, then there exists a partition (M1,M2) E Pe(H) as in the first paragraph of the proof. By (ii), we may choose components B; and Bj of G2 such that V ( B ; ) C M I and V(Bj) C M2. If (.zj,zj+q} n M I # 8 and { z i , ~ i +n~ M2 } # 0 then

.

>

>

~ H ( M M2) I,

>

eH(Bi, { z i , .z;+q} n M2) eH(N1,N2) 2 1+1+2=4

Hence we may assume that {zi,zi+q} n M2 # 8. Thus

{tj,zj+q}

c

~ H ( J I I , M1~e )~ ( B i , { z i , z i + n~ M } 2)

+ eH(Bj, M2.

{zj,zj+q}

n M I )t

Using (i) it follows that

+ e ~ ( N l , N 2>) 1t 2 = 3.

Some Conjectures on Cyclically 4-Edge Connected 3-Regular Graphs 175 Hence e ~ ( M 1 , M 2 = ) 3 and equality must hold throughout the above inequality. Since Bi could be chosen to be any component of G2 contained in MI and eH(M1 \ NI,M2) = e ~ ( B i , M 2 )= 1, it follows that B; is the only component of G2 contained in M I . Furthermore, since eH(B;,M2 \ { z ; , zi+,}) = 0 we deduce that B; is a component corresponding to an isolated vertex of F. We next consider Bj. If Bj corresponds t o an isolated vertex of F then, as eH(Bj,MI) = e ~ ( i t 4 2\ N2,Ml) = 0 , we must have zj,zj+,,zj+29 E M2. Thus, by (i), at least two vertices of z;, z;+,, z;+pq must belong to M2. This contradicts the fact that eH(Bi,M2) = 1. Since Bj could be chosen to be any component of G2 contained in M2, and since B; is the only component of G2 contained in M I , we deduce that i = p = 2 and that B1, BJ, . ,B, are contained in M2. As eH(B1,MI) = e ~ ( M \2 N 2 , M l ) = 0 , it follows that (z1,zzq) c M2. BY (i), {zz,zq+2} c M2. Thus eH(B2,it42) L 2. This contradicts the fact that e ~ ( B z , M 2= ) 1 and completes the proof of case (a).

..

(PI E(F)= 0.

Let B1,B2, ...,B , be the components of G2. Choosing neighbours w;,w;+,, w;+2, of z in Bi for 1 5 i 5 p , we obtain a contradiction in the 0 same way as in case (a). Corollary 2. Let G be an essentially 4-edge connected graph of minimum degree at least three. Then some 3-regular inflation of G is essentially 4edge connected. Proof. Immediate.

0

In proving statement (b) we also need a result of Harary and NashWilliams [3].

Theorem 1. The line graph of a graph G is hamiltonian i f and only i f G has an Eulerian subgraph H such that G - V ( H ) is an independent set of vertices. Proof of statement (b). The statement (b) now follows from (l),( 2 ) and ( 3 ) below. (1) Conjecture 3 + Conjecture 4. Suppose Conjecture 3 is true. Let G be an essentially 4-edge connected graph. Let V1

= {w E V(G)ld(w)5 2,w has exactly one neighbour in G}.

V2 = {w E V ( G ) l d ( u )= 2,w has two distinct neighbours in G } . Since X,(G) 2 4, it follows that

H . Fleischner and B. Jackson

176

(iii) if v E V1 then the neighbour of v in G has degree a t least four in G - V1. (iv) if v E Vz then both neighbours of v have degree at least four in G-

V1.

Let G1 be the graph obtained from G - V1 by suppressing all vertices of degree two. Then GI is essentially 4-edge connected and has minimum degree at least three. By Corollary 2, G1 has as essentially $-edge connected, 3-regular inflation J. By Corollary 1, J is cyclically 4-edge connected. Since Conjecture 3 is assumed to be true, J has a cycle C such that J - V ( C ) is an independent set of vertices. Thus C must contain at least one vertex from each “inflated vertex” of GI. By considering E(G1) c E ( J ) , we see that E ( C ) n E(G1) induces an Eulerian subgraph H of G1 which contains all vertices of degree at least four in G I , and is such that GI - V ( H ) is an independent set of vertices. Using (iii) and (iv), we deduce that the subgraph of G corresponding to H is an Eulerian subgraph of G satisfying the conclusion of Conjecture 4. (2) Conjecture 4 + Conjecture 5 . Suppose Conjecture 4 is true. Let G be a graph such that the line graph L(G) is 4-connected. If L(G) is complete then L(G) is clearly hamiltonian. Hence assume L(G) is not complete. It follows that P,(G) # 0 and since L(G) is 4-connected, X,(G) 2 4. Using the validity of Conjecture 4 and Theorem 1, we deduce that L(G) is hamil tonian.

+

(3) Conjecture 5 Conjecture 3. Suppose Conjecture 5 is true. Let G be a cyclically 4-edge connected, 3-regular graph. Then L(G) is 4-connected. Using the validity of Conjecture 5 and Theorem 1, we deduce that G has an Eulerian subgraph H such that G - V ( H ) is and independent set of vertices. Since G is 3-regular it follows that H is the required cycle of G. 0

References J. A. Bondy. Personal communication.

H. Fleischner, “Cycle decompositions, 2-coverings, removable cycles, and the four-colour disease,” Progress in Graph Theory (J. A. Bondy and U. S. R. Murty, eds.), Academic Press (1984), 233-246. F. Harary and C. St. J. A. Nash-Williams, “On eulerian and hamiltonian graphs and line graphs,” Canad. Math. Bull. 8 (1965), 701-710.

Some Conjectures on Cyclically 4-Edge Connected 3-Regular Graphs 177 [4] F. Jaeger, “Flows and generalized colouring theorem in graphs,” J. Combin. Theory ( B ) 26 (1979), 205-216. [5] F. Jaeger. Personal communication. [6] S. Kundu, “Bounds on the number of disjoint spanning trees,” J. Combin. Theory (B) 17 (1974), 199-203. [7] P. Seymour, “On nowhere zero 6-flows,” J . Combin. Theory ( B ) 30 (1981), 130-135. [8] C. Thomassen, “Reflections on graph theory,” J. Graph Theory 10 (1986), 309-324.

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Annals of Discrete Mathematics 4 1 (1989) 179-194 0 Elsevier Science Publishers B.V. (North-Holland)

On Connectivity Properties of Eulerian Digraphs A. Frank* Department of Computer Science Eotvos University Budapest, Hungary

Dedicated to the memory of G. A . Dirac Directed counterparts of theorems of Rothschild and Whinston and of Lovbz concerning Eulerian graphs are proved. As a consequence, a polynomial time algorithm is presented to solve the integral t w e commodity flow problem in directed graphs in the case where all capacities are ‘Eulerian’.

1

Introduction

Connectivity results in graph theory start with Menger’s theorem which has (at least) four versions, according to whether we are interested in the maximum number of edge-disjoint or node-disjoint paths from s to t in a directed or an undirected graph. The max-flow-min-cut (MFMC) theorem can be considered as an extension of the directed edge-version of Menger’s theorem. T. C. Hu [4] proved a ma-flow-min-cut theorem for two-commodity flows in undirected graphs. Unfortunately, in this case the maximum flow is not necessarily integer-valued. Rothschild and Whinston [141 found an integervalued version of Hu’s theorem for Eulerian graphs. Two other interesting connectivity results concerning Eulerian graphs are due to L, Lovbsz [7]. The purpose of the present note is to prove directed counterparts of these results. We also describe a polynomial time algorithm for integral two-commodity flows in directed graphs when the capacities are Eulerian (that is, at every node the sums of out-going and in-going capacities are equal). *Supported by a grant from the Alexander von Humboldt Stiftung and by the Institute of Econometrics and Operations Research of the University of Bonn.

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180

The proofs make use of the operation of splitting off incident edges at a node. This powerful technique was used in [14] and in [7]. It also played a basic role in connectivity results of Mader ([8] and [9]) concerning non-Eulerian graphs. In an undirected graph G = ( V , E ) ,splitting 08two edges wu and ut means replacing wu and ut by a new edge wt. Similarly, in a directed graph, splitting 08two edges wu and ut is an operation that replaces wu and ut by a new edge wt. If w = t , we leave out the resulting loop wt. Throughout, we work with connected loopless graphs (directed or undi:= V - X. Note that A c B rected) on node set V. For X C V, put means that A C_ B, but A # B . For s , t E V , a subset X C V is called an s h e t if s E X and t $ X. We do not distinguish between an element w and the one-element set {w}. For a graph G = ( V , E ) , d c ( X , Y )denotes the number of edges with one end in X - Y and one end in Y - X. (When it is not ambiguous, we leave out the subscript G). We use d c ( X ) for d ~ ( X , m For . two graphs G = (V, E) and H = (V, F), G H denotes a graph with node set V and edge set EUF. In a directed graph G, e c ( X ) (respectively S G ( X ) )denotes the number of edges entering (leaving) X. For a function c: E -, R ,

+

The connectivity c ( X , Y )between two disjoint subsets X and Y of nodes of an undirected graph is the maximum number of edge-disjoint paths connecting X and Y. In a digraph, the di-connectivity d c ( X , Y ) from X to Y is the maximum number of edge-disjoint directed paths from X to Y . An undirected graph G = (V,E) is Eulerian if d ~ ( w )is even for every w E V. A digraph G = ( V, E) is Eulerian if eocw) = 6 ~ ( w )for every w E V. A graph G is called acyclic if it does not contain directed circuits.

2

Edge-disjoint paths

The edge-disjoint paths problem is as follows: Given an undirected graph G = (V, E) and k pairs of nodes, find Ic edge-disjoint paths in G connecting the corresponding terminals. In general, the problem is NP-complete [6],

181

On Connectivity Properties of Eulerian Digraphs

but for fixed k it is polynomially solvable [13]. For arbitrary k, some important special cases are well characterized ([3], [12], [lo], [ll],and [IS]). See also Schrijver’s survey paper [15]. Sometimes the following reformulation is useful. Mark each terminal pair by an edge. The graph H = (V,F) of marker edges is called a demand graph, while we call the original graph G a supply graph. The edge-disjoint paths problem is: In G H , find k edge-disjoint circuits such that each of them contains one demand edge. A natural necessary condition for the solvability is the

+

Cut Criterion.

d c ( X ) 2 djy(X)

for every X

c V.

If H consists of k parallel edges, then the cut criterion is sufficient (: undirected-edge Menger). The following result of Rothschild and Whinston is a strengthening of Hu’s two-commodity flow theorem [4].

Theorem 2.1 ([14]). If H consists of two sets of purallel edges and G + H is Eulerian, then the cut criterion is necessary and suficient for the solvability of the edge-disjoint paths problem. The directed edge-disjoint paths problem can be defined analogously. Here we formulate only the corresponding second version: Let G = (V, E ) and H = (V,F) be directed graphs. Find IF[ edge-disjoint directed circuits in G H , so that every circuit contains one edge of H . Such a circuit will be called a good circuit. Again, the following criterion is obviously necessary.

+

Directed Cut Criterion.

e c ( X ) 2 6jy(X)

for every X

c V.

If H consists of parallel edges of the same direction, the cut criterion is sufficient; this is exactly the directed-edge version of Menger’s theorem. If H consists of two oppositely directed edges (i.e., if H is a two-edge circuit), the directed edge-disjoint paths problem is NP-complete [2], and the directed cut criterion is not sufficient. Even if G H is Eulerian, the directed counterpart of theorem 2.1 does not hold in general. This can be seen from the graph of Figure 1. Here the solid lines denote the edges of H ; G H is Eulerian, and the directed cut criterion holds, but there is no solution. However, we have the following theorem.

+

+

Theorem 2.2. Suppose that G + H is an Eulerian digraph, H consists of two sets of parallel edges, and H is acyclic. Then the directed cut criterion is necessary and suficient for the solvability of the directed edge-disjoint paths problem.

182

A . Frank

Figure 1.

Proof. Let G + H be a minimal counter-example. Assume that H consists of kl edges from tl to s1 and k2 edges from t2 to s2. Let T = {si,s2,ti, t z } be the set of terminal nodes. By Menger's theorem, k1 > 0 and k2 > 0. Call a pair {vu,u t } of edges feasible if splitting them off does not destroy the directed cut criterion. Since G t H is a minimal counter-example, there cannot be feasible pairs { v u , u t } for u E V - T . Call a set X tight if e c ( X ) = S H ( X ) .

CLAIM1. If X is tight, then so is

r,

PROOF. Since G + H is Eulerian,

ec(x)t e d x ) = M

X )tMX),

whence the claim follows.

CLAIM2. If X and Y are tight sets, then d H ( X , Y ) 2 d G ( X , Y ) , and if equality holds, then both X n Y and X U Y are tight. PROOF.Using (1) and the directed cut criterion, we immediately get

ec(xn Y )- eH(X n Y )t e c ( X u Y )- eH(X u Y )t d c ( X , Y )- ~ H ( XY ,) = eG(x)- e d x )t m ( Y )- e d y ) =o+o, from which the claim follows.

CLAIM 3. If X and Y are tight, then d H ( X , y ) 2 d G ( X , j p ) , and if equality holds, then both X - Y and Y - X are tight. PROOF. Immediate from Claims 1 and 2.

On Connectivity Properties of Eulerian Digraphs

183

CLAIM4. For i = 1, 2, sit; is not an edge in G .

PROOF.Otherwise, a demand edge t;s; and sit; form a good circuit C. After deleting the two edges of C, the hypotheses of the theorem continues to hold, so the resulting digraph contains IF1 - 1edge-disjoint good circuits. These circuits, together with C, constitute IF1 edge-disjoint good circuits in G H , a contradiction.

+

CLAIM5 . T c V . PROOF.Let T = V. Since Icl > 0 and sltl is not in G, there are two edges G such that s1 # z # y # s1. Since every pair is infeasible, there is a tight set A for which z E A and s1,y $! A. Since ~ G ( A>) 0 and A is tight, e H ( A ) > 0. Similarly, ~ H ( A>) 0. Hence, tlsl leaves A, and t2s2 enters A. By Claim 4, tl # x . So we must have y = t2 and z = s2, contradicting Claim 4.

s1z and x y in

Since G is connected, there is a node u E V - T such that us E E for some s E T. The next lemma immediately implies the Theorem. 0 Lemma. There exists an edge wu in E such that {wu,us} is feasible.

Proof. Assume that s = tl (the other possibilities are analogous). If there is no tight sE-set, then any vu edge will do. Assume first that there is exactly one maximal tight sn-set, denoted by X. We claim that there is an edge wu E E such that v !$ X . Otherwise, we have eG(x

+ u ) < e G ( x ) = b H ( X ) = 6 H ( x + u),

contradicting the directed cut criterion. Note that the pair {wu,us}is feasible. Next, suppose that there are two maximal tight sE-sets, denoted by X and Y . Since their union is not tight, by Claims 1 and 2, d ~ ( x , Y>) 0 and ~ H ( XF) , > 0. Since tl E X n Y, among the two edges of H only t2s2 can contribute to d H ( X , Y ) . Therefore, we cannot have three maximal tight sE-sets, X I , X2, X,, since otherwise S2t2 would connect Xi - X j and X j - X i , 1 5 i < j 5 3, which is impossible. Assume that t2 E X - Y and s2 E Y - X. We claim that there is an edge wu in E such that v 4 X U Y. Otherwise, ~ G ( X=) 0 (since is tight), and

x

contradicting the directed cut criterion. Now the edge pair {wu,us} is feasible. 0

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184

Notice that Theorem 2.2 easily implies Theorem 2.1. Indeed, a theorem by Ford and Fulkerson [l]says that the undirected edges of a mixed graph can be oriented in such a way that the resulting digraph is Eulerian if and only if every node has an even number of incident edges (directed or undirected), and every cut contains at least as many undirected edges as is the difference between the numbers of entering and leaving directed edges in this cut. Apply this result to the mixed graph obtained from G t H by directing the edges of H from tl to s1 and from t 2 t o s2. By the undirected cut criterion, the necessary and sufficient condition above is satisfied, so there is an Eulerian orientation of G H . The undirected cut criterion also implies the directed cut criterion for this orientation. Thus, Theorem 2.1 follows from Theorem 2.2. By a star we mean a digraph in which either all the edges enter the same node or all the edges leave the same node.

+

+

Theorem 2.3. If G H is an Eulerian digraph, and H is the union of two stars, then the directed cut criterion is necessary and sufficient for the solvability of the directed edge-disjoint paths problem. Proof. We derive this result from Theorem 2.2 by using an elementary construction. Assume that H is the union of two stars, HI and H z . If every edge of Hi leaves ti, adjoin a new node s; t o V . For every edge t ; x of IT;, add a new edge six to G, and in H i , replace tix by tisi. Analogously, if, in Hi, every edge enters s i , adjoin a new node t; t o V . For every edge xs; of H i , add a new edge xt; t o G, and in H i , replace xs; by tisi. In the resulting problem, the directed cut criterion continues to hold, so we can apply Theorem 2.2. 0 A kind of converse to Theorem 2.3 is also true. A digraph H = ( V , F ) is called suitable if, in G H , the directed cut criterion is sufficient for the directed edge-disjoint paths problem for every digraph G for which G+H is Eulerian. Theorem 2.3 says that the union of two stars is suitable.

+

Theorem 2.4, If H is suitable, then H is the union of two stars. Proof. First, observe that a subgraph H’ = (V,F’) of a suitable H is also suitable. Indeed, assume that G’ H’is Eulerian for a certain G’, and the directed cut criterion holds, but there are not IF’] edge-disjoint good circuits in G’ H‘. Then the same statement is true for G t H , where G arises from G’by adding the edges in F - F’ oppositely directed. In Figure 2, we list some digraphs (along with G), which are not suitable. The edges of H are drawn by solid lines.

+

+

On Connectivity Properties of Eulerian Digraphs

185

Figure 2. It is an easy exercise to see that if a digraph H does not contain any of these graphs as a subgraph, then H is the disjoint union of two stars. 0 Theorems 2.3 and 2.4 can be considered as a counterpart t o a theorem by Papernov [12], saying that in the undirected case, a graph H is suitable if and only if it is the union of two (undirected) stars, or it arises from K (complete 4-gon) or C (circuit of five edges) by adding parallel edges. See also [17]. On the other hand, we cannot be overjoyed with our characterization since there is another natural necessary condition for the edge-disjoint paths problem:

Covering Criterion. For any subset F’ E F of the demand edges, the good circuits, using an element of F’, cannot be covered by less than IF’I edges of G. This criterion immediately implies the cut criterion, but not the other way around. The graph G + H in Figure 1 satisfies the cut criterion, but not the covering criterion. Note that in the undirected case, the (corresponding) covering criterion is equivalent to the cut criterion [18]. It is tempting to try to find classes for the directed case, where the covering criterion is sufficient, while the cut criterion is not. I conjectured that the covering criterion is sufficient if the digraph G + H is Eulerian and planar. (By a theorem of Seymour [18], if G + H is undirected, Eulerian and planar, then the cut criterion is sufficient.) However, C. Hurkens (Tilburg University) found the counter-example of Figure 3. (Relying on the above theorem of Seymour, one can easily show that if every face of G -t H is a directed circuit, then the directed cut criterion is sufficient.)

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A. Frank

Figure 3.

3

Two-commodity flows

The proof of Theorem 2.2 suggests an algorithm to find the required good circuits. Actually, the theorem itself gives rise to an algorithm, since it implies that if the directed cut criterion holds, then either the number of edges in G from s; to t ; is at least k; (i = 1, 2), or there is a feasible pair of edges in G. In order to find feasible pairs, we have to be able t o test a pair of edges for feasibility. This can be done if the directed cut criterion can be checked efficiently. By Menger’s theorem, the directed cut criterion is satisfied (for G and H in Theorem 2.2) if and only if, in G, there are k1 edge-disjoint paths from s1 to t l , and there are IC2 edge-disjoint paths from s2 to t 2 , and there are kl+ k2 edge-disjoint paths from ( ~ 1 , s ~ ) t o { t l , t 2 } . Therefore, the directed cut criterion can be checked by three max-flow-min-cut computations. Begin the algorithm with checking the directed cut criterion. Second, repeat the following procedure as long as there is a node u E V for which eG(u) > 0 and ~ G ( u )> 0: Choose an edge us E E . One by one, check every pair {vu,us}, vu E E, whether it is feasible. By Theorem 2.2, we can be sure that at least one of these pairs is feasible. Once a feasible pair is found, split it off. When the splitting phase terminates, the resulting digraph contains ki edges from s i to ti (i = 1, 2). Each of these edges corresponds t o a path of G from which the edge has arisen during the splitting phase. The third part of the algorithm consists of reconstructing these kl k2 edge-disjoint paths.

+

On Connectivity Properties of Eulerian Digraphs

187

Let us consider the following directed two-commodity flow problem. We are given a digraph G = ( V , E ) endowed with a non-negative integral capacity function c: E -+ 2. Two (ordered) pairs of terminals sj, ti (i = 1,2) are specified along with a prescribed flow value k;. The problem is to find two flows x1 and $2 such that xi 2 0 is a flow from s; to ti of value ki, and

for every e E E . We solve this problem under the assumption that ec(w) = 6,(w)

if w

# s;,t;

and ec(si)

+ ki = 6,(si)

and ec(ti) = b,(t;)

+ Ici

(i = 1, 2).

Theoretically, the problem goes back to the uncapacitated case by replacing each edge e by c(e) parallel edges. This reduction gives rise t o an algorithm that is not polynomial, since the required number of splitting operations is proportional to the maximal capacity. This difficulty can be overcome if we split off a feasible pair {wu,us} more than once at a time. How many times can a pair be split off, without violating the cut criterion? If M denotes this number, then

M = min(c(vu),c(us),m) where

m = min(e,(X)

- ~ H ( X:)u 6 X and

v , s E X).

Like the way we tested the directed cut criterion, m can be computed by three max-flow-min-cut computations. By a weighted splitting we mean an operation that, given two edges wu and us, reduces c(vu) and c(us) by M and introduces a new supply edge from w to s of capacity M. We say that a weighted splitting operation is critical if M = m. Now, the weighted algorithm is the same as the unweighted one, except that each time we perform weighted splittings. That is, the algorithm begins with checking the directed cut criterion. The second part consists of iterating the following procedure for u = w;, i = 1, 2, ..., IVI: Perform a weighted splitting for every pair (wu,us}. The third part builds the two required flows 21 and 22 by iterating the next procedure: Assume that xl(e) and q ( e ) have already been computed for an edge e = ws having

A. Frank

188

arisen by splitting off wu and us. Increase zi(vu) and si(us) by s i ( e ) for i = 1, 2 , and reduce z ; ( e ) to 0. Although this algorithm is a natural extension of the uncapacitated algorithm, proving that the number of weighted splittings is bounded by a polynomial in IVl and IEI requires some more work. First, observe that during the weighted splitting phase, once a subset of nodes becomes tight, it remains tight throughout. Consequently, once a pair of edges becomes infeasible, it never again becomes feasible. Lemma. In the course of the weighted splitting phase at most 0 ( l V l 2 ) weighted splittings are critical. Proof. Let T = {sl,s 2 , t l , t2) be the set of terminal nodes. Let .f denote the family of tight sets. For 2 c T, let .fz = { X : X E .f and X n T = 2). By Claim 2 (in the proof of Theorem 2.2), Fz is a ring family. (A family of subsets is called a ring family if it is closed under union and intersection.) Thus, F is the union of at most 16 ring families. It is well-known that for every ring family R C 2" for which O,V E R, there is a unique (transitive) digraph D1 = (V,El) such that

R = { X : 61(X)= 0 and X C V }, where &(X) is the number of edges leaving X in D1. (Namely, put El := { uv : there is no uv-set in R}.) Consequently, for a sequence R1 c R2 C -.. C 'Rk of ring families, we have Ic 5 lVI2. Thus, if F1 c . f 2 C ..- C 3 1 is a sequence of families of tight sets during the weighted splitting phase, then I 5 161VI2. Since a critical splitting operation strictly increases the family of tight sets, the number of these operations is a t 0 most 161VI2. Since at a node u, at most

non-critical weighted splittings can occur, the total number of weighted splittings is a t most IVl * (El2 161VI2.

+

4

Connectivity in Eulerian digraphs

In this section we prove directed counterparts of two theorems of Lovdsz [7]. Let G = (V,E) be a directed graph. Throughout this section, is used t o denote the underlying undirected graph of G. Let A = (211, 212,. ,o k } C V

..

On Connectivity Properties of Eulerian Digraphs

189

be a specified subset of nodes. A path connecting two distinct elements of A is called an A-path. Let d; = dc(v;,A - v;) for i = 1, 2, .. ., k.

Theorem 4.1. Let o x be an edge of an Eulerian digraph G such that v 4 A . There exists an edge yv such that splitting o g y v and v x does not reduce dc(vi,A - v;) for i = 1, 2, ..., k .

Proof. Call a set X i-critical for some i = 1, 2 , 6 ( X ) = dc(vi,A - ~ i ) .

.. .,k if X n A = {w;}

and

CLAIM.If X and Y are i-critical, then so are X n Y and X U Y , and d ( X , Y ) = 0 . If X is i-critical, and Z is j-critical (i # j ) , then X - Z is i-critical, and Z X is j-critical, and d ( X , Z ) = 0 .

-

PROOF.Functions d and c below concern c, functions 6 and dc concern G. Since G is Eulerion, d ( X ) = 2 6 ( X ) and c ( X ,Y ) = 2dc(X, Y ) for disjoint By (2), we have sets X and Y . Thus, it suffices to work with

e.

2c( v;,A - v;)

+

= d(X) d(Y) =d(XnY)+d(XuY)+2d(X,Y) 2 ~ c ( v ;A, - v i ) 2d(X,Y )

+

whence the first statement follows. Similarly,

+

C(vj,A - v;) C(vj,A - vj) = d( X) d( 2) = d ( X - 2)t d(Z - X ) t 2 d ( X , Z ) 2 c(vi,A - v;) c ( v j , A - vj) 2 d ( X , Z )

+

+

+

which implies the second statement, and thus proves the claim. The edges yv and wz can be split off without reducing di if and only if there is no i-critical set X , for which either v E X and z,y 4 X , or v # X and z,y E X . Thus, if there is no critical set M with I{v,x} n MI = 1, then for any edge gv, the pair {yv,vx} can be split off. If there is such set, we have two cases to consider.

CASE1. There exists a critical vT-set M. Let M be minimal, and suppose that M is i-critical for some i = 1,

2,

... , k. There is an edge yv such that y E M , for otherwise

dc(v;,A- v;) 5 6(M - V ) < 6 ( M ) = dC(vi, A - v i ) ,

A. Frank

190

a contradiction. We claim that yw and wz can be split off. Indeed, there is no i-critical zr-set C,since then d(C, M ) > 0 contradicts the claim. There is no i-critical wg-set C, for otherwise C n M is i-critical, contradicting the minimality of M . Similarly, there is no j-critical y2i-set C, since then M - C would be i-critical, contradicting the minimality of M , and there is no j-critical vr-set, since d(M,-e) > 0.

CASE2. There is no critical wr-set, but there is a critical xZi-set M . Let M be maximal, and suppose that M is i-critical for some i = 1, 2, . . ., k. Then there exists an edge yw such that y $ M . We claim that yw and wz can be split off. Indeed, by assumption there is no critical wT-set. Similarly, there cannot be a critical @-set C, since C is i-critical; then C U M is also i-critical, contradicting the maximality of M , and if C is 0 j-critical ( j # i), then d(M,iT) > 0, contradicting the claim. The following theorem is immediately implied by Theorem 4.1.

Theorem 4.2. In an Eulerian digraph G = (V,E), the maximum number t of puirwise edge-disjoint A-paths is k i=l

.

where V,,V,, . ., Vk are disjoint subsets of V and V;. n A = {w;} (i = 1, ., , k). Furthermom, t = C d;, and if M ; denotes the minimal set for which Mi n A = {w;} and 6(M;)= d;, then the sets M; (i = 1, 2, . .. , k) are pairwise disjoint, and they form an optimal solution to (3).

2,

.

Remark. Lov&z’s theorem on A-paths of an undirected Eulerian graph [7] easily follows if we apply Theorem 4.2 to any Eulerian orientation of G. On the other hand, the converse derivation is not difficult either. Let G = (V,E ) be an Eulerian digraph, and let denote the underlying undirected Eulerian graph. We show that there are t directed edge-disjoint A-paths in G, provided that there are t edge-disjoint A-paths in E. For a directed A-path or circuit P , we say that an internal node w of P is in-bad (on P ) if both edges of P incident to w enter u; w is said to be out-bad if the two incident edges on P leave w. Let b(P) denote the number of bad nodes of P . Let P be a decomposition of E into undirected A-paths 5,Pz, , Pt and some circuits such that

z

...

x ( b ( P ) :P E

P)

On Connectivity Properties of Eulerian Digraphs

191

is minimal. We claim that every path P is a directed path in G (although not necessarily a simple path). Suppose, indirectly, that a path P E P contains an in-bad node v (say). Since G is Eulerian, v must be an out-bad node of another member Q of P. At least one of the two possible switches of P and Q at v gives rise to another partition P' of E into t (possibly not simple) A-paths and some circuits for which C ( b ( P ) : P E P') is smaller, a contradiction. See Figure 4.

Figure 4. Switch of P and Q at v.

Remark. From an algorithmic point of view, the edge-disjoint A-paths problem and its capacitated version (when to every edge a non-negative capacity is assigned such that ec(v) = &(v) for every v E V) can be solved quite analogously to the two-commodity flow problem analysed in Section 3. We leave out the details. Theorem 4.3. Let G = ( V , E ) be an Eulerian digraph and vx E E . There exists an edge yv E E such that splitting 08yv and vx does not reduce dc(s,t ) for any s , t E V - v. Proof. Call a set C critical with respect to a pair of nodes s,t (# v) if C is a tT-set and d c ( s , t ) = e(C). If C is critical for s,t , then C is critical for t , s . The set C is said to be critical with respect to s , t in if I { s , t } n CI = 1 and c ( s , t ) = d(C). Obviously, C is critical in G if and only if C is critical in 5. The next lemma concerns 5.

c

Lemma. I f X and Y are critical, then either (i) X n Y and X U Y are critical and d ( X , Y ) = 0 , or (ii) X

- Y and Y - x are critical and d(x,T)= 0.

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Proof of Lemma. Let X be critical for q , ~ and , let Y be critical for y l , y2. We have three possibilities t o consider. (a) XnY separates oneof the pairs { q , z 2 } , {yl, yz}, and XUY separates the other. (A set A is said to separate 2 and y if ( A n (2,y}( = 1.) Then

d(X)

+ d(Y)= d ( X n Y )+ d ( X u Y )+ 2d(X,Y)

+

2 d ( X ) t d(Y) 2d(X,Y) from which (i) follows. (b) X - Y separates one of the pairs { q , x 2 } , {yl,yz}, and Y - X separates the other. Applying (a) to the sets X and p,we obtain (ii). (c) X separates { y l , y ~ } ,and Y separates { t 1 , ~ 2 } ,and one of the pairs {XI, ZZ}, {y1,y2} is separated by both X n Y and X U Y . Now

Suppose that the pair (y1, y2} is separated by both X Then we have 4x1 = d(Y)= C(Y1, Y2) and

+

n Y and XU Y .

+ + 2 C(Yl,Y2) + C(Y1, Y2) + 24x9 Y )

d ( X ) d(Y)= d(X n Y ) d ( X u Y ) 2d(X, Y )

+

+

= d ( X ) d(Y) 2d(X, Y ) from which (i) follows. 0

This completes the proof of the lemma.

The lemma implies the corresponding assertion for G. Therefore, if X and Y are critical in G, and the edge w t enters both, then X U Y is also critical. Hence, there is a unique maximal critical tv-set M (if there is one at all). If no such M exists, any edge yw can be split off. If we have such an M , there is an edge yw with y 4 M . Indeed, if M is critical for s, and t and the required edge yw did not exist, then

a contradiction. By the construction of M , the edges yw and split off without reducing dc(s,t) for any s , t E V - w.

212

can be 0

On Connectivity Properties of Eulerian Digraphs

193

Remark. W. Mader proved [9] that, given a not necessarily Eulerian digraph G = (V, E) and a node v such that dc(s,t ) 2 k for every s, t E V - v and e(v) = 6(v), a pair of edges yv,vz can be split off such that the connectivity from any s € V - v to any t € V - v continues t o be at least k. Remark. B. Jackaon [5] also proved Theorems 4.1 and 4.3.

References [l] L. R. Ford and D. R. Fulkerson, Flows in Networks, Princeton Univ. Press, Princeton, N. J. (1962).

[2] S. Fortune, J. Hopcroft and J. Wyllie, “The directed subgraph homeomorphism problem,” Theoretical Computer Science 10 (1980), 111121. [3] A. Frank, “Edge-disjoint paths in planar graphs,” J. Cornbin. Theory (B) 39 (1985), 164-178. [4] T. C. Hu, “Multicommodity network flows,” Operations Res. 11 (1963), 344-360. [5] B. Jackson, “Some remarks on arc-connectivity, vertex splitting and

orientation in digraphs.” Preprint (1985). [6] R. M. Karp, “On the computational complexity of combinatorial problems,” Networks 5 (1975), 45-68. [7] L. LovBsz, “On some connectivity properties of Eulerian graphs,” Acta Math. Acad. Sci. Hungar. 28 (1976), 129-138. [8] W. Mader, “A reduction method for edge connectivity in graphs,” Annals of Discrete Math. 3 (1978), 145-164. [9] W. Mader, “Konstruktion aller n-fach kantenzusammenhangenden Digraphen,” Europ. J . Combinatorics 3 (1982), 63-67.

[lo] H. Okamura, “Multicommodity flows in graphs,” Discrete Applied Math. 6 (1983), 55-62. (Proc. London Math. SOC.(3) 42 (1981), 178192).

[ll] H. Okamura and P. D. Seymour, “Multicommodity flows in planar graphs,” J . Combin. Theory ( B ) 31 (1981), 75-81.

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[12] B. A. Papernov, “Feasibility of multicommodity flows,” Studies in Discrete Optimization (A. A. Friedman, ed.), Isdat. “Nauka,” Moskow (1976), 230-261. (In Russian). [13] N. Robertson and P. D. Seymour, “Graph minors - a survey,” Surveys in Comb~naton’cs(I. Anderson, ed.), London Math. SOC.Lecture Notes 103 (1985), 154-171. [14] B. Rothschild and A. Whinston, “Feasibility of two-commodity network flows,” Operations Res. 14 (1966), 1121-1129.

[ 151 A. Schrijver, “Min-max results in combinatorial optimization,” Mathematical Progmmming - The State of the Art (A. Bachem, M. Grotschel and B. Korte, eds.), Springer Verlag (1982), 439-500. [16] P. D. Seymour, “Disjoint paths in graphs,” Discrete Math. 29 (1980), 293-309. [17] P. D. Seymour, “Four-terminus flows,” Networks 10 (1980), 79-86. [18] P. D. Seymour, “On odd cuts and plane multicommodity flows,” Proc. London Math. SOC.(3) 42 (1981), 178-192.

Annals of Discrete Mathematics 41 (1989) 195-210 0 Elsevier Science PublishersB.V. (North-Holland)

Some Problems and Results on Infinite Graphs R. Halin Mathematical Seminar University of Hamburg Hamburg, FRG

Dedicated to the memory of G. A . Dirac The structure of infinite graphs is investigated. Especially locally finite and so-called bounded graphs are examined, as well as infinite maximal planar graphs.

1

Introduction

G.A. Dirac was among that minority of graph theorists who have a genuine interest in infinite graphs. I had several discussions with him on this topic, especially when I enjoyed an invitation by him t o h h u s in 1977. Some of the material presented in this paper was suggested by these conversations. With gratitude and respect I dedicate this report on some of my current research to his memory. In Section 3 the concept of maximal planar graph is discussed. There are several equivalent statements in the finite case which however split up highly if carried over to infinite graphs. It becomes evident that only the concept based on Kuratowski’s theorem and its extension to countable graphs (see Dirac and Schuster [2]; it is attributed to P. Erdos by these authors) leads to a sensible definition of maximal planar graphs in the infinite case. Further in Section 3 some results of [5] are reported and extended. In Section 4 the structure of countable maximal planar graphs is investigated especially with respect to infinite paths. Every 3-connected planar graph has a locally finite spanning subtree. Hence every infinite maximal planar graph contains a 1,oo-path. We characterize those maximal planar . it is shown that an infigraphs which do not contain a 2 , ~ - p a t h Further nite maximal planar graph has a VAP-free representation in the plane iff it 195

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has only one end (see below for definitions); then it has a t most 2 vertices of infinite degree. In Section 5 we discuss a possible classification of infinite graphs by reducing “modulo finiteness”; the concept seems especially useful for locally finite graphs because they can be represented by trees. In Section 6 we introduce “bounded graphs”, a class of countable graphs generalizing the locally finite graphs. A countable graph is called bounded if to every function f : V(G)-+ 1zv there is a sequence 71,72,73,. of natural numbers which, for each 1,oo-path U C G, majorizes the sequence of f-values of the vertices in U apart from a finite number of exceptions. The problem of characterizing the bounded graphs by forbidden subgraphs is solved for the class of trees.

..

2

Terminology and notation

We follow the standard terminology and notation in graph theory; some additional remarks are necessary. All graphs in this paper are undirected and do not contain loops or multiple edges; they are allowed to be infinite. A graph is looally finite if all its vertices have finite degree. A one-way infinite path is briefly denoted as a 1,w-path, or simply as a PI,^. Similarly, a 2, w-path, or a is a two-wuy infinite path. 1,oo-paths U , U’ in a graph G are separated by a finite T C G if there are infinite subpaths of U and U‘ which belong t o different components of G - T . 1,ca-paths U,U’ in G are called equivalent if they cannot be separated by a finite T ; we then write U -G U’. N G is an equivalence relation; any equivalence class with respect to N G is called an end of G. U NG U’ is equivalent with the existence of a third 1,ca-path which meets both U and U‘ infinitely often [3]. A graph G is muzimal in a class I’ of graphs (or with respect to a graph theoretical property defining I’), if G E I’ and any graph arising by adding a new edge (Lee,which joins two non-adjacent vertices of G) to G no longer belongs to I’. If @ is a set of finite graphs, I?(@) denotes the class of all graphs (finite or infinite) not containing a subdivision of any F E @. The class of maximal graphs in I?(@) is denoted by I?(@). Using Zorn’s lemma one easily proves that each G E I?(@) can be extended, by adding new edges, to an element of f(@)[5, (3.1)) The elements of f(@)have a simplicia1 decomposition with finite or countable prime members (see [5]; [8], Chap. 10; or [13] for more details).

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By G * H we denote the disjoint union of graphs G and H , in which, in addition, every possible edge is drawn between V ( G )and V ( H ) . In G * n, n a cardinal, n is understood as the edgeless graph with n vertices. A graph G is called planar if i t can be represented (drawn) in the plane. Sometimes we identify an (abstract) planar graph with one of its representations in the plane, If we use the attribute ‘%lane” we refer to a special representation (figure) in the plane. A VAP-free representation of a graph in the plane is a representation such that the vertex set (as point set in the Euclidian plane) does not have an accumulation point.

3

On planar embeddings and maximality of countable graphs

If G is a finite graph with a t least 3 vertices, then the following statements are equivalent:

(a) G has a maximal plane representation $J.This means: It is not possible t o draw an edge (= Jordan arc) between two non-adjacent vertices without meeting +(G) in an inner point of this arc. (b) G has a plane representation $ which triangulates the plane. This means: .RR2 - $(G) has, as its connected components, regions (= open and connected point sets) whose boundaries are triangles of G. (c) G is a maximal planar graph (i.e., maximal in the class of planar graphs). (d) G is maximal in

I’(K5, K 3 , 3 ) .

This equivalence is no longer valid at all in the case that G is countable. Of course each embedding of a maximal planar graph must be a maximal plane representation, hence (c) ==+ (a). But for instance PI,^ has a maximal plane representation, hence (a) does not imply (c). For let PI be an edge [u,b],represented as a vertical line segment in the plane. Assume the finite path P, starting in u to be represented in the plane by vertical and horizontal line segments. Then continue P, to a path P,+l by adding a path in its end vertex # a such that this continuation is represented by horizontal or vertical line segments and each point of P, has on both sides (left and right) points of this continuation in distance < 2-,. (This can be done by tracing P, “very closely” on both sides). The

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union of the P, gives a representation of in which no additional edge can be drawn. are e.g. such that all points of the Other “foolish” embeddings of plane are accumulation points of the vertex set. (Let D be a countable dense subset of R2 such that each horizontal and vertical line contains only finitely many points of D ;then by traversing D using horizontal and vertical line segments we find a straight line representation of PI,^ such that V ( P l , , ) 2 D.)Also, we can draw PI,, in such a way that a given finite or countable number of connected components of R2- PI,^ arises. (c) does not imply (b), for the “iterated cube” CJ is a maximal planar graph without a triangle. (Q is defined as follows: Let Q1 be a rectangle in the plane. Assume Q n to be constructed as a finite plane graph in which each region is bounded by a rectangle. Then in each of these rectangular regions a further rectangle is drawn, the vertices of which are connected to the vertices of the boundary by 4 additional edges such that a cube graph arises. The union of the Q n is Q.) Of course every triangular embedding gives a maximal plane representation, so (b) implies (a). But (b) does not imply (c). For instance let a straight line representation of K 4 be given with vertices a , 6 , c , d , vertex a inside the (bounded) triangle b, c, d. Let a sequence of points pi # a , b on the segment be given which converges to a , and draw all edges [pi,c], bi,d ] . Then the resulting configuration is a plane graph which triangulates the plane, but this graph is not maximally planar, because an edge [a,6]can be added. By the Erdos-Dirac-Schuster extension [2] of Kuratowski’s theorem to countable graphs we see that (c) and (d) are equivalent, and along this line only it makes sense to investigate the maximality of countable planar graphs. We call the elements of I ’ ( K s , K 3 , 3 ) formally planar. Among the (maximal) formally planar graphs those of cardinality 5 No are the (maximal) planar graphs. In [ 5 ] , Satz 4 the structure of maximal formally planar graphs is described. We call an edge (or a K z ) in a graph G of type 1 if it is contained in a non-separating triangle of G; otherwise it is of type 2. Then we have:

Proposition 3.1. Every maximal formally planar graph G has a decomposition of the form

G = UGx X
where u is an ordinal, the Gx are induced subgraphs of G such that

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for every r , 0 < r < u, and where each Gx either is a K B or a n at most countable 3-connected maximal planar graph, further each S, is of type 2 in (at least) one of the graphs UX<, Gx, G,. Vice versa: Every graph which has such a decomposition is a maximal formally planar graph. For what follows it is important to mention ([5], Satz 5):

Proposition 3.2. If an element o f l ? ( K s , K 3 , 3 ) has an edge of type 2 then it contains a P Z , ~ . In the iterated cube Q every edge is of type 2.

- Let

Q1 be equal to

Q. Assume Q n to be defined, and attach, along every I i 2 of Qn, a copy of Q, to obtain graph Qn+l. Put Qoo = U;=,Qn. Then Qoo is a maximal planar graph in which every edge is of type 2 and every K 2 separates Qoo into infinitely many parts. It is interesting to consider our problem of maximality also in relation to the characterizations of planar graphs by Whitney and Mac Lane. By their theorems, for a finite graph G, (a)-(d) are also equivalent with the statemen ts (e) G is maximal with respect to the property of having a dual; ( f ) G is maximal with respect to the property of having a 2-basis. The notions of “2-basis” and “dual” have been extended to infinite graphs in the deep investigations of C. Thomassen (see [14], [15]). So (e), ( f ) make sense also for infinite (especially countable) graphs. In the case that G is locally finite and 2-connected, the graphs in (e) and (f) have been characterized by C. Thomassen as the strong and the weak (respectively) triangulations of the plane; see [14], [15] for these theorems and the required definitions. Weak and strong triangulations are maximally planar graphs, but not vice versa, even if one restricts oneself to locally finite graphs. Namely in [14],Theorem 7.4, Thomassen shows that a 2-connected graph has a 2-basis if and only if it has a VAP-free plane representation. For finite G trivially (a)-(f) are also equivalent with

(g) G is maximal with respect to the property of having a VAP-free representation in the plane.

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For countable and 2-connected G, (f) and (g) are equivalent by Thomassen’s theorem. For general G however one has to take into account that G has a 2-basis if and only if each block of G has a 2-basis, whereas the analogous statements is not true for VAP-free representability. Examples show that none of (f) and (g) implies the other for countable graphs in general. The graph P Z *, 2~fulfills (g), but is not maximally planar. (Mind that P Z ,* 112 ~ is planar too.) If in this graph P2,00 is replaced by the strip between two P2,00’swhich is triangulated by additional vertices and edges, we get a graph of the same kind; also if we paste together several graphs isomorphic to P Z ,* 2~ along vertices of infinite degree. (Hence graphs of this kind need not to be 2-connected.) Other graphs of this kind are obtained by pasting together copies of PI,^ * K 2 along K 2 ’ s or along vertices (of infinite degree) in an appropriate way. It would be interesting to obtain a characterization of all graphs satisfying (g) without being maximally planar.

4

Infinite paths and ends in maximal planar graphs

We begin with the following result: Theorem 4.1. If G is infinite and n-connected (1 5 n < oo), then either G contains a subdivision of K,,, where /A = IV(G)l if IV(G)l is regular or p is any cardinal less than IV(G)l i f IV(G)l is singular, or G (is countable and) has a locally finite spanning subtree.

Proof. If G is uncountable, the assertion follows from Theorems 9.1 and 9.4 in [7]. Let G be countable. If a finite T C G exists such that G - T has infinitely many components Gi, then G contains a subdivision of K N ~ (as ,, is seen by choosing a vertex w; in each Gj and n paths from wi to T having pairwisely only vi in common). Otherwise G - T has only finitely many components for each finite T C G, and then by Satz 7 in [9]the existence 0 of a locally finite spanning subtree follows.

As corollaries we can state: Theorem 4.2. If G is a 3-connected countable planar graph, then G has a locally finite spanning subtree. Theorem 4.3. Every maximally planar infinite graph contains a 1,oopath.

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Proof. If G is 3-connected, the assertion follows from Theorem 4.2 and a well-known theorem of D. Konig. If G is separated by 2 vertices, then by 0 Propositions 3.1 and 3.2 a P Z ,is~contained in G. Theorem 4.2 gives rise to the following problem. D. Barnette [l]has shown that every finite 3-connected planar graph has a spanning subtree in which every vertex has degree at most 3. Can this result be extended to countable planar 3-connected graphs? Other related questions are of interest, for instance: Has every 4-connected planar graph a locally finite 2-connected spanning subgraph? Next we characterize those maximal planar graphs which have a 1 , ~ path, but not a 2, w-path.

Theorem 4.4, Let G be a countable graph. G is a maximal planar graph without a 2, oo-path i f and only i f G has the form 00

G = UGn n= 0

where the G, are finite planar triangular graphs, (2)

is a triangle contained in Gn-1 ( n = 1,2,3,. . .), all the triangles Sn have a common edge [a,b], and no Sn separates any of the graphs G;.

Proof. If G has the form described in the theorem and U c G is a 1,mpath, then U must meet all the vertices # a , b in Sn for sufficiently large n; hence there cannot be two disjoint infinite paths in G. Since, for each n, G1 U . .. U Gn is a finite planar triangulation, G is maximally planar. Assume, on the other hand, G to be a maximally planar graph without a 2,oo-path. G must be 3-connected by Proposition 3.2 and has a 1,wpath P by Theorem 4.3. By Satz 3 in [4] there exists a finite F such that G - F has the form Ho U H1 U H Z U . . where each Hi is without a 1,wpath and, for i 2 1,Hi n ( H I U . ., U Hi-1) is a single vertex ai which is in Hi-1 - ( H I U . . U Hj-2). Hence the a; are articulations of G lying on P for i sufficiently large. Assume F and the H ; chosen in such a way that IV(F)I is minimal. If Hi (i 2 1) is not connected, then all components of H i which do not contain ai or ai+l are transferred to Ho. Further, if there is a vertex w in F from which there are not edges to infinitely many H;, then there is

.

.

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k such that v is not adjacent to any vertex of H k + l U H k + 2 U . . .; then we replace Ho by the subgraph induced by w and Ho U . . . U H k , to obtain a contradiction to the minimality of F . Hence, each v E V ( F ) sends edges to infinitely many Hi. IV(F)I 2 3 would imply that G contains a subdivision of K 3 , 3 (mind that, for sufficiently large i, each Hi contains an a;,a;+l-path C P and is connected). Hence I V ( F ) 5 2. Since F together with each a; separates G, F must consist of exactly two elements a,b. a,b together with a; span a separating triangle S; for each i, by [5], Satz 6. If G; denotes the subgraph induced by H ; and S;, we find that the G; and the Si fulfill equations (l), (2) in the theorem. Each Gi must be maximally planar; for otherwise G itself would not be maximally planar. By Theorem 4.3 each G;must be finite, hence a finite triangular graph. No S, separates a Gi; otherwise Gi together with G,-1 or G, would infer a subdivision of K 3 , 3 . Hence G has 0 the form described in the theorem.

a

Suppose G to be a maximal planar graph with a given planar representation and C a circuit in G. Then, if the interior and the exterior of C both contain infinitely many vertices of G , there are 1 , ~ - p a t h sC G in both these regions1, and it is clear that C separates the corresponding ends. In this way we see that Q has uncountably many ends. One may ask which end numbers a (locally finite) maximally planar prime (i.e., 4connected) graph can have. We consider the following procedure: In the interior of a rectangle R draw one rectangle R' or two disjoint rectangles R', R" such that each of R', R" lies in the exterior of the other, and then draw edges between the vertices of R and R' (and R", resp.) such that the area between R and R' (R, R', and R", resp.) is triangulated and no separating triangle arises. It is easy to see that, by iterating this procedure in an appropriate, way, we get the following:

Theorem 4.5. For any n E N U {N0,2'0} there is a locally finite 4connected maximally planar graph which has exactly n ends. If G is a plane graph and p an end of G, then p is called unbounded if there is a 1,oo-path in p whose vertex set is not bounded in lR2. In general there may be many unbounded ends in G. Assume G to be maximally planar. If G is separated by K 2 , then G may have more than one unbounded end. But suppose, on the other hand, G to be 3-connected. Let p # p' be 'Consider for instance C together with the interior part of G . Then, by adding a vertex in the exterior and connecting it with all the vertices of C by edges, we obtain a maximally planar graph; Theorem 4.3 yields a 1, oo-path in the interior of C.

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ends of G and T a finite set of vertices separating p, p' in G. If IT1 is chosen minimal, by Satz 6 of 151 T induces a circuit C. G - T has components H # H' which contain elements of p and p', respectively. Each vertex of T is adjacent t o vertices of both H and HI, by the minimality of T . Therefore one of the ends p, p' must lie in the interior, the other in the exterior of C. Hence we have:

Proposition 4.6. I n every plane representation of a maximally planar infinite 3-connected graph there is at most one unbounded end. Now we can prove:

Theorem 4.7. A maximally planar graph G has a VAP-free representation in the plane i f and only if it has exactly one end. I n this case G must be 3-connected; it can have at most 2 vertices of infinite degree, and these vertices must be adjacent to each other.

Proof. Assume first G to be not 3-connected. Then by Proposition 3.1 and Satz 5 of [ 5 ] there are in G an edge [a,b], a P2,cr,c G - { a , b} and internally disjoints a , b-paths . ..P-2,P-1, Po,P I ,P2,. . . such that P Z ,meets ~ all the Pi, in the order indicated by their indexing. It is clear that then G has no VAP-free representation and possesses at least two ends. Therefore in what follows we suppose G t o be 3-connected. If G has at least two ends, then by Proposition 4.6 in any plane representation of G there is a 1,oo-path of G whose vertex set is bounded, i.e. has an accumulation point; so G has no VAP-free representation. If, vice versa, G does not admit a VAP-free representation, then, by the characterization of the graphs with this property by Halin and Schmidt (see [6], p. 212) it follows that in any plane representation of G there must be a circuit C such that its interior as well as its exterior contains infinitely many vertices of G. Therefore G has a decomposition of the form G = G' U G", G' n G" = C , where GI, G" both are 2-connected and infinite; by the argumentation following the proof of Theorem 4.4 there are 1,oo-paths in GI, G", which belong to different ends of G. If a , b are non-adjacent vertices in G, then, by the maximality of G, any smallest set of vertices separating a , b must induce a circuit C. Then in any plane embedding of G one of the vertices a,b belongs t o the interior, the other t o the exterior of C. If G has a VAP-free representation, not both of the vertices a , b can have infinite degree. Hence the vertices of infinite degree induce a complete graph. It is clear that the latter cannot have order 2 4.

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Assume T = { t l ,t2,t 3 ) is the set of vertices of infinite degree. Then G - T is locally finite and has a t most two components, exactly one of which is infinite; we denote the latter by H . Let U be a 1,oo-path of H ; mind that H has one single end. By Satz 5 in [3] for each i = 1 , 2 , 3 there is an infinite sequence ( t i j ) j E mof neighbours of t; in H such that there lead finite paths P,j c H from t i j t o U ;without loss of generality we can assume the t ; j , P;j to be chosen in such a way that any two of the Pij are disjoint ( i = 1 , 2 , 3 ; j = 1 , 2 , . . .). We see that a subdivision of K 3 , 3 is in G (with t l , t 2 , t 3 in one of its colour classes). Hence G cannot have more than 2 vertices of infinite degree. (Of course 2 vertices of infinite degree are possible, as for instance K 2 * PI,^ shows.) 0

5

On a classification of locally finite graphs

In his studies on infinite graphs Dirac basically asked: What makes a graph infinite? And he answered this question by certain infinite substructures which must be present in an infinite graph with more or less specified properties. In what follows we try to quest for the very essence of the infinity of an infinite graph from another point of view ,namely by searching for those structural properties of a graph which cannot be destroyed by any finite reductions of the graph, which, so to speak, are preserved if the graph is reduced “modulo finiteness”. One possible way to give these intuitive ideas a precise meaning is indicated in what follows. Let G be an infinite graph. Let 7 be a partition of V(G)into finite non-empty subsets such that each T E 7 induces a connected subgraph of G. Let G I 7 denote the graph which arises by contracting each T E 7 onto a single vertex VT and letting V T , vs, for distinct T , S E 7, be adjacent if and only if there exists an edge in G connecting vertices of T and S. We may say that G and G’ have the same infinity structure if there are 7 and 7‘(of the kind considered above) such that G / 7 and G’/7‘ are isomorphic. This relation between graphs is reflexive and symmetric, but not transitive. In order to get an equivalence relation we form the transitive closure of the relation in question; we denote this equivalence relation by w W . Each of the corresponding equivalence classes will be called an infinity type of graphs. If G is an infinite graph, we get all members (up to isomorphisms) of the infinity type of G in the following way: Replace each vertex w of G by a connected finite graph F,, (such that, for w # v’, F,, n F,,I is empty) and draw edges (at least one) between F,, and F,,I iff w and v‘ are adjacent in G;

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then, if the arising graph is denoted by G, form some G / 7 , as indicated. We call a structural property P of graphs an infinite property if the following holds: If one of the members of an infinity type has P, then P is shared by all its members. To study infinity properties it is sufficient to consider an infinite graph G, a 7 as above, and to find out which structural features are preserved by the canonical mapping IC of G onto G / 7 and its inverse I C - ~ We . shall see that infinite degree, number of components, and end structure are infinite properties. (5.1). If T E 7, then VT has infinite degree d in G I 7 i f and only i f there is a vertex v in T which has infinite degree d in G,

Hence: (5.2).

Locally finiteness is an infinite property.

(5.3). The number of connected components is an infinite property; thus also connectedness is an infinite property.

If U is a 1,oo-path in G, then K ( U ) is an 1,oo-path in G / 7 ; and, vice versa, if there is a 1,oo-path 0 in G / 7 , then there is a t least one 1,oo-path U in G with K ( U ) = 0.

(5.4).

However, not necessarily every 2, oo-path in G is mapped onto a 2,mpath in G / 7 , namely if the two "halfs" of the 2, m-path belong to the same end of G. But the structure of ends is preserved: (5.5).

If U , U' are 1,m-paths in G which are separated by a finite F

C

V ( G ) then K ( F )separates I C ( U ) IC(U') , in G / 7 ; and this statement can be reversed as follows: If 0,0'are 1,oo-paths in G / 7 which are separated by a finite E , then there are 1, oo-paths U 2 K - ' ( O), U' I C - ' ( ~which ' ) are separated in G by I C - ~ ( ~ ) . Thus, free ends of G correspond to free ends of G / 7 ; the orders of ends corresponding under K in G and G I 7 are equal. (Here the term "order of an end" is used in the sense of Jung [lo].)

(5.6).

In order to get a classification of infinite graphs one would like to choose a characteristical, especially simple member out of each infinity type. Here one naturally thinks of trees. But not all infinity types have a tree among

its members, as for instance the graphs we have:

* 1 and P Z ,* 1~show. However

R. Halin

206

Proposition 5.7. Every locally finite infinite connected graph G has a (countable) tree in its infinity type. Proof. Let A0 be a finite connected non-empty induced subgraph of G. G - A0 has finitely many components C1,. . . ,C,; let Fi be the finite set of vertices in Ci which are adjacent to some vertex of Ao; and then let Ai be a finite connected induced subgraph of Ci containing Fi (i = 1,.. . ,r ) . Ci- Ai has finitely many components Cil, . ,Ciri ( i = 1,... ,r ) ; in each Cij choose a finite connected induced subgraph Aij containing all the vertices of Cij from which there leads an edge into Ai. Then delete Aij in Cij, consider the finitely many components, and so on. One sees that, by the 0 A , . ., a partition 7 of V ( G )is defined such that G / 7 is a tree.

..

In order to get a classification of locally finite connected graphs it is therefore necessary to obtain a complete set of invariants for the locally finite countable trees. A locally finite tree is called a full ramification if it is a subdivision of a tree in which each vertex has degree 2 3. A countable tree T has uncountably many ends if and only if it contains a full ramification, and then there exists a maximal full ramification which contains all full ramifications of T as subgraphs (Kaluza [ll]).It is not difficult to show that all full ramifications belong to the same infinity type; it is formed by those locally finite connected infinite graphs which do not possess a free end [3]. The locally finite trees without a full ramification are those in which each end has an order in the sense of Jung [lo]. A locally finite tree can either be built up by a transfinite process indicated by this concept of order, or it arises from a full ramification by attaching “branches”, consisting of trees of the first kind (i.e., in which each end has an order). The existence or non-existence of a full ramification and the orders of the ends in the attached trees form a set of invariants in our classification which however is far from being complete. It seems to be rather difficult to get such a complete set of invariants.

6

Bounded graphs: A generalization of locally Anite graphs

The 1,oo-paths in locally finite graphs have the following property:

Proposition 6.1. Let G be a countable locally finite graph with the vertices v1, ~2,213,.. . . Assume f to be a function which associates to every vi a natural number f ( v ; ) . Then there exists a sequence y1,72, y3,. . . of

Some Problems and Results on Infinite Graphs

207

natural numbers such that for each 1,oo-path U in G, say with the vertices ao, a l , a2, . . . and the edges [a;,a;+l],the following holds: There is an no = n o ( U ) such that for all i 2 no we have f(u;) < yi. Proof. Let T$ be the set of vertices of G which, from some vj with j 5 i, can be reached by a path with edge number 5 i (i = 1,2,. . .). By the locally finiteness of G each V;: is finite. Put

yi = 1

+ maz(f ( v ) I v E T$)

for i = 1,2,. . . . Now let U as in the theorem be given. a0 appears as some V N . Then for i 2 N each a; belongs to V,; hence we have f ( a ; ) < y; for all i2N. 0

If especially f ( v ) is chosen as the degree of v in G and H is an arbitrary graph which contains a 1,oo-path with vertices ao, a l , a2,. . . and edges [ai,a;+l] such that each a; has degree 2 y;, then H cannot be isomorphic to a subgraph of G. So it follows that in the class of locally finite countable graphs there is no universal element (result of de Bruijn; see Rado [12]). The property considered in Proposition 6.1 can be used to define a new class of (countable) graphs which lie between the locally finite graphs and the general countable graphs. We call a countable graph G bounded, if to every function f : v ( G ) --+ Llr there is a sequence (-y;);€m of natural numbers such that for every 1,oo-path U C G with the vertices ao, a l , . . . and edges [a;,u;+l] there exists an no = no( f , U ) with f(a;) < y; for i 2 no. We shall say that (y;);€m majorizes f . By Proposition 6.1 all locally finite graphs are bounded. The countable complete graph and the No-regular tree T, are not bounded. It is not difficult to show: (6.2). If G is bounded, then every countable subgraph and every subdivision of G is bounded.

(6.3). If G is not bounded, then every subdivision of G is not bounded too. Let (y;);€m, ( 6 i ) i E ~be two sequences of natural numbers. We say that (6i) majorizes (y;), if there is an no E Llr such that 6, > y, for all n 2 no.

If (yji)iEm, j = 1,2,. . ., are countably many sequences of natural T numbers which manumbers, then there is a sequence ( C ? ; ) ; ~of~ natural jorizes them all. For let 6i = 1 maz(yjil j = 1,. . ,i). (6.4).

+

From (6.4) we conclude:

.

208

R. Halin

If G is countable, F c G is finite and the components of G - F are all bounded, then G itself is bounded.

(6.5).

Apart from the subdivisions of T, we get two other classes of unbounded graphs as follows: Let Bi be the union of countably many internally disjoint finite paths connecting two vertices ai # bi ( i = 1 , 2 , . . .), and for i # j let B; n Bj be empty. Further add disjoint finite bt. , ai+1paths Pi (for i = 1 , 2 , . . .) which have nothing in common with the Bi apart from their end vertices. Then the union of the Bi and P;forms a connected countable graph which is not bounded; we call such a graph a bundle graph. Further let P be a 1, oo-path with vertices q ,02,03,.. . (in their natural order). For every natural number k add a 1, oo-path PI,with initial vertex w 2 k such that the PI, are disjoint and have only ZI2k in common with P;for each k draw all possible edges from W ~ I , - to ~ PI,, We get a graph F which is obviously not bounded. Any subdivision of F will be called a fan graph. We conjecture that a graph is not bounded if and only if it does not contain (aa subgraph) a subdivision of T,, a bundle graph, or a fan graph. We can prove this conjecture only for trees.

Theorem 6.6. Let G be a countable tree. Then G is bounded i f and only if G does not contain a subdivision of the No-regular tree T,. Proof. By (6.2) we have only to prove the if-part. Therefore suppose G to be not bounded. Choose a root T in G and let < denote the order induced by r on V ( G ) . Call an edge e = [z, y] essential if z < y and the branch B, induced by y (i.e. the subtree consisting of all vertices 2 y) is not bounded. In this case also all the edges of the r,y-path c G are essential, and we find that all essential edges form a tree TO,more precisely a rooted subtree (with root r) of G. The non-essential edges form finitely or countably many subtrees T; ( i = 1,.. . ,n or i E nV) such that each Ti has exactly one vertex in common with To and G is the edge disjoint union of To and these Ti. Each Ti ( i 2 1) is bounded. If TOwere locally finite, we choose some function f : V ( G )--+ LV. Let fi denote the restriction o f f onto Ti. By definition of boundedness and by Proposition 6.1, we find, for each i, a sequence (yij)jEmwhich majorizes fi. According to (6.4) the sequences (7ij)jEm can be majorized simultaneously by a sequence ( & ) ; E N . Since each 1,oo-path c G ends in some Tj we find that ( & ) ; E N majorizes f , and G itself would be bounded. Therefore we conclude that TOcontains a vertex t o of infinite degree. The neighbours > to of to in TOdetermine infinitely many branches which themselves are not bounded. By the same argumentation in each of these

Some Problems and Results on Infinite Graphs

209

branches we find a vertex of infinite degree (with respect to To), and by repeating this procedure we get a subdivision of T , in G. 0

Acknowledgement The author wishes to express his thanks to Carsten Thomassen for his careful reading of the manuscript and several helpful comments.

References [l] D. Barnette, “Trees in polyhedral graphs,” Canad. J. Math. 18 (1966), 731-736. [2] G. A. Dirac and S. Schuster, “A theorem of Kuratowski,” Nederl, Akad. Wetensch. Proc. Ser. A 57 (1954), 343-348.

[3] R. Halin, “Uber unendliche Wege in Graphen,” Math. Ann. 157 (1964), 125-137. [4] R. Halin, “Charakterisierung der Graphen ohne unendliche Wege,” Arch. der Math. 16 (1965), 227-231. [5] R. Halin, “Ein Zerlegungssatz fur unendliche Graphen und seine Anwendung auf Homomorphiebasen,” Math. Nachr. 33 (1967), 91-105.

[6] R. Halin, “Some topics concerning infinite graphs,’’ P r o b l h e s combinatoires et thdorie des graphes, CNRS, Paris (1978), 211-213. [7] R. Halin, “Simplicia1 decompositions of infinite graphs,” Advances in graph theory (B. BollobAs, ed.), Annals of Discrete Math. 3 (1978), 93-109.

[8] R. Halin, Graphentheorie I, IT, Wiss. Buchges., Darmstadt (1980/81). [9] R. Halin and H. A. Jung, “Uber Minimalstrukturen von Graphen, insbesondere von n-fach zusammenhangenden Graphen,” Math. Ann. 152 (1963), 75-94. [lo] H. A. Jung, “Wurzelbaume und unendliche Wege in Graphen,” Math. Nachr. 41 (1969), 1-22.

[ll] T. Kaluza, “Struktur- und Machtigkeitsuntersuchungen an gewissen unendlichen Graphen mit einigen Anwendungen auf lineare Punktmengen,” Math. Ann. 122 (1950), 235-258.

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R. Halin

[12] R. Rado, “Universal graphs and universal functions,” Acta Arithm. 9 (1964), 331-340. [13] C. Thomassen, “Infinite graphs,” Selected topics in graph theory vol. 2 (L. W. Beineke and R. J. Wilson, eds.), Academic Press (1983). [14] C. Thomassen; “Planarity and duality of finite and infinite graphs,” J. Combin. Theory ( B ) 29 (1980), 244-271. [15] C. Thomassen, “Planarity and duality of graphs,” Congressus Numerantium 2 8 (1980), 5-14.

Annals of Discrete Mathematics 41 (1989) 211 -220 0 Elsevier Science Publishers B.V. (North-Holland)

On a Problem Concerning Longest Circuits in Polyhedral Graphs J. Harant a n d H. Walther Section of Mat hematics Institute of Technology of Ilmenau Ilmenau, GDR Dedicated to G. A . Dirac We prove that the shortness exponent of r(q,w ) is strictly less than 1 for any q 2 136 and any w , where I'(q,w) is the class of planar, 3-regular, 3-connected graphs for which each longest circuit has a bridge with at least w touch points, and for which the boundary of each face has at most q edges. In 1950, G. A. Dirac [l]was the first to find a sufficient condition for the existence of a hamiltonian circuit (this is a circuit containing all vertices) in not necessarily planar graphs by proving that a graph having n vertices possesses a hamiltonian circuit if no vertex has a degree less than 3. In 1931, H. Whitney [7] had already shown for planar graphs that 4connected triangulations possess a hamiltonian circuit. W. T. Tutte [5] generalized this theorem in 1956 showing that each planar 4-connected graph has a hamiltonian circuit. With the first example of a non-hamiltonian planar 3-connected 3regular graph presented in 1946, W. T. Tutte [4] initiated the development of the construction of non-hamiltonian polyhedral graphs, with the number of vertices of a longest circuit being kept as small as possible. Consider the class r of all 3-regular polyhedral graphs, that is, the class of the planar 3-connected graphs in which each vertex has degree 3. Let C be a circuit of a graph G E r. If one removes from G the vertices of C and all edges incident with them, the resulting graph G - C disintegrates into connected components H I , H 2 , . . , Hk. If is a hamiltonian circuit of G, then G - C is obviously empty, but in all other cases, however, holds k > 0. Let C' be the set of those vertices of C having at least one neighbour in H i . Since each vertex has the degree 3 in G, a vertex

.

211

c

212

J . Harant and H. Walther

from C' is incident with exactly one edge which does not belong to C, that is, this vertex is adjacent to one vertex from H i . In addition, there exists for any vertex 2 E C a t most one index i such that x is adjacent to one vertex of H i . If one adds to H i the vertices of Ci and also all edges whose one end-vertex belongs to Ci and the other one t o H ; , then the resulting graph Bi is called a bridge of G over C . Calling also an edge whose two end-vertices belong to C, but not the edge itself, a bridge, then we see that t o each vertex of C there exists exactly one bridge H in which this vertex lies. A vertex belonging to both C and B; is called a touch point of the bridge Bi. Let a bridge over a path be defined accordingly. Let I'(z0) be the class of graphs G in I' in which there exists for any longest circuit C of G a bridge B over C with at least w touch points. In accordance with [2], we understand by the shortness exponent a(A) of a class A of graphs G

o(A) := lim inf 1% c(G) GEL\ logn(G) with c(G) being the length of a longest circuit of G and n(G) the number of vertices of G. In [6] it has been shown that:

Theorem 1.

a(I'(2o))

< 1 for any integer w > 0 .

Let f be any elementary face of a graph G E number of the bordering edges o f f . Let

I' and let q ( f ) be the

and rqc I' the class of all graphs G with q(G) 5 q ( q being an integer) as well as r(q,W) := rqn I'(z0). We shall prove:

Theorem 2. o(I'(q,w)) < 1 for any q 2 136 and for any

W.

Before proving this theorem by a suitable construction, let us explain in a less formal way what it says: It is possible to find 3-regular polyhedral graphs G with all elementary faces having a limited number of bordering edges (that is, there exist at most 136-gons), in which to any longest circuit C there exists a bridge B with any large number w of touch points, and in which the ratio of the length of a longest circuit to the total number of vertices is as small as desired.

On a Problem Concerning Longest Circuits in Polyhedral Graphs

213

Proof of Theorem 2. We shall carry out a series of constructions. Let M = M ( u , b, c; d ) be a planar 3-connected graph of degree 3 with the following properties: (i) There exists in M a vertex P that is incident with an ( u a ( b i- 1)-gon, and a ( c 1)-gon.

+

+ 1)-gon,

(ii) Any other elementary face of A4 possesses at most d bordering edges.

Figure 1. Let the structure U = U ( u , b , c ; d ) be emerging from M ( a , b , c ; d ) by deleting the vertex P and maintaining the three edges originally incident with P as half-edges. Thus, the structure T of Figure 1 is a T(3,5,4;5) and T' of Figure 2 a T'(3,3,3; 4), etc. 1. (Cf. [4]) The structure T(3,5,4; 5) of Figure 1 has the property that a longest path connecting the two half-edges 2 and y and going through T (in short: an (2,3)-path through T ) omits exactly one black vertex, whereas an (z,r)-path or a (y, 2)-path through T contains all vertices.

J . Harant and H. Walther

214

Figure 2.

2. The structure V results from T by replacing each black vertex by a structure T" (Figure 2), whereas the structure W results by replacing the vertex A of T by a structure T' and each other black vertex of T by a structure T". It can immediately be seen: V = V(5,11,8; 12), W = W(5,11,8; 12), and an (z,y)-path through V omits exactly one structure T" (that is, one black and ten white vertices), whereas an (5,z)-path or a (y, %)-pathcontains all vertices. An (z,y)-path through W omits exactly the structure T' which has been substituted into the black vertex A (that is, it omits one black and six white vertices), an ($,%)-path or a (y,z)-path through W contains all vertices. Consider in W a bridge B containing the half-edge z over an (2,%)-pathN . Then, B contains the structure T' substituted into the vertex A of T, and the two white vertices A1 and A2 incident with A in T are the touch points of B over N in W . 3. If, according to Figure 3, two copies V1 and V2 of V and a structure W are put together to form a TI, then we have:1'2 = T1(10,22,16; 24), and an (zl,yl)-path of TI omits exactly one black and six white vertices (namely just the stucture T' of W which has been substituted into the vertex A of T). A (yl,zl)-path or a (zl,zl)-path of TI omits exactly one black and ten white vertices (namely just one structure T" of V1 or Vz). Further, we have: The bridge containing z1 over an (21, yl)-path has two touch points, namely A1 and A2 of W.

On a Problem Concerning Longest Circuits in Polyhedral Graphs

215

Figure 3. 4. Let Tj+l result ccordi g to Figure 4 (TI h s been constructed in 3) by putting together 10 copies of Ti (referred to as ql, . .. , T;”) and 19 further vertices. It is easy to see that T; = T;(4,8,8; 136), i = 2, 3, ... . By checking all possibilities one can easily convince oneself of the fact that an (zj+l, Yj+l)-path through Tj+l takes a course as indicated by Figure 4. The bridge containing z;+l over an (z;+l, yi+l)-path N has 2 (twice the number of touch points of the bridge containing z; over an (sj,yj)-path in T,) touch points, namely G, H , and the touch points of the bridge containing z; over an (z;,yj)-path through Tf and T:. Let b; be the number of the touch points of a bridge containing zj over an (zj, y;)-path through T;. Then, because bi+l = 2 2bj and bl = 2, it follows that bj = 2’+l - 2. It can easily be seen that an (xi+l, zi+l)-path or a (yj+l, zj+l)-path through Ti+l also passes through all 10 structures Ti,but contains 4 less vertices than an (zi+l,yj+l)-path through T;+1. 5 . If one puts together 3 structures Ti, according to Figure 5 , to form a graph Fj, then the following holds: The course of a longest circuit through F; is as indicated in Figure 5. The bridge containing Q over a longest circuit has 3(2’+l - 2) touch points. There are at most 136-gons in F;. Let Q be a white vertex.

+

+

J. Harant and If. Walther

216

Figure 4.

Figure 5.

On a Problem Concerning Longest Circuits in Polyhedral Graphs

217

Now let w be a given integer. In what follows we choose i such that 3(2’+’ - 2) 2 w holds, with i being minimal, and put G1 := Fi. Let Sj and Wj be the numbers of black and white vertices, respectively, of Gj (for j > 1 these graphs still have to be constructed), and let wj and sj be the numbers of white and black vertices, respectively, in a longest circuit I<j of Gj.

Figure 6.

6. If, according to Figure 6, three structures W (cf. 3) are put together to form a structure U , then we have: U = U(19,19,19; 15), and any ( z , 2.)path through U omits exactly one structure T‘, that is, one black and six white vertices. Designate with S, and W, the number of black and white vertices, respectively, of U , and with s, and w, the number of black and white vertices, respectively, of a (z,r)-path through U . Then it holds: S, = 27, W , = 3(8 10 + 1 ~6 + 6) = 276, s, = 26, and w, = 270. 7. Let G1 be constructed according t o 5. Let Gj+l result by replacing each black vertex of Gj by a structure U = U(19,19,19; 15). Since in G1 there occur at most 136-gons, and since each black vertex is adjacent only to white vertices, and since black vertices lie only in 6-gons (as they all have been “encircled” by a T‘ or T”), at most 24-gons will be created when substituting structures U in black vertices, and thus Gj ( j= 1, 2, . . .) contains a t most 136-gons.

-

J. Harant and H. Walther

218

Let us now calculate Sj, Wj, sj, w j ( j= 1, 2,

. . .):

A circuit containing in G j a black vertex X can contain (after substituting a structure U into each black vertex) at most 26 of the 27 black vertices of U when passing through Gj+l, that is: sj+l = 26sj = 26jsl. Consider a longest circuit Kj in Gj. It contains sj black and w j white vertices. When generating Gj+l, I<jcan be “blown up” t o form a longest circuit Kj+1 of Gj+l, and Kj+l then contains the white vertices originally lying in Kj (namely w j vertices), and exactly wusj new white vertices of the structures U passed through by I<j+lof Gj+l (that is, there are exactly s j such structures U):

Thus, the length

C(Gj+l)

of a longest circuit of Gj+l becomes:

and the number n(Gj+l) of vertices of Gj+l becomes:

The numbers 4, Wl, 81, w1 being fixed (cf. the generation of G1 in 5), it can easily be seen that to any E > 0 a j’= j ‘ ( E ) can be found such that logc(Gj) log26
+E

holds for all j 2 j‘.

As a longest circuit of G2 had a bridge with at least w touch points, all white, this is clearly true for all Gj, j 2 1. So a ( r ( q ,w)) < 1 for any w, and q 2 136, and Theorem 2 has been proved. 0

On a Problem Concerning Longest Circuits in Polyhedral Graphs

219

The following problem remuins open (cf. [3,8]): Let P * P 2 + + (we assume p l < pz < ... < p , = q ) be the class of the 3-regular polyhedral graphs for which it holds: To any elementary face f there exists a nylmber p; ( i E {1,2,. . ,r } ) such that f forms a pi-gon. The least number r bas to be found such that there exist numbers pl, a,. . , pr with @P1rM*-+*pr n I’(p,., w))< 1. We conjecture that r = 2.

.

.

References [l] G. A. Dirac, “Some theorems on abstract graphs,” Proc. London Math. SOC.2 (1952), 69-81. [2] B. Griinbaum and H. Walther, “Shortness exponents of families of graphs,” J. Corilbin. Theory 14 (1973), 364-385.

[3] P. J. Owens, “Shortness parameters of families of regular planar graphs with two or three types of faces,’’ Discrete Math. 30 (1982), 199-209. [4] W. T. Tutte, “On Hamiltonian circuits,” J. London Math. SOC.21 (1946), 98-101. [5] W. T. Tutte, “A theorem on planar graphs,” Trans. Am. Math. SOC. 82 (1956), 99-116. [6] H. Walther and H.-J. VoB, h e r Kreise in Gruphen, VEB Deutscher Verlag der Wigsenschaften (1974), 58-62. [7] H. Whitney, “A Theorem on graphs,” Ann. of Math. 32 (1931), 378390. [8] J. Zaks, “Non-Hamiltonian simple 3-polytypes having just two types of faces,” Discrete Math. 20 (1980), 87-101.

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Annals of Discrete Mathematics41 (1989) 221 -228 0 Elsevier Science Publishers B.V. (North-Holland)

Interpolation Theorems for the Independence and Domination Numbers of Spanning Trees S.

F. Harary* Department of Mathematics

Schuster

Department of Mathematics Carleton College Northfield, Minnesota, USA

University of Michigan

Ann Arbor, Michigan, USA

Dedicated to the m e m o r y of Gabriel Dirac For each invariant i = i(T)of a tree T , we say that i interpolates over (the set S of all spanning trees of) a given graph G if the following statement holds: If n and m are the maximum and minimum values (respectively) of i(T) over all T E S, then for any k, m < k < n, there is a tree T E S such that i(T) = k. In previous work, it was shown that the diameter and the number of endnodes of a tree are interpolating invariants over each 2-connected graph and each connected graph (respectively). We now prove that the node and edge independence numbers and the domination number are interpolating invariants over all connected graphs, but the independent domination number is not,

1

Introduction

In earlier work, the current authors established interpolation theorems for two graphical parameters of the spanning trees of a given graph. Specifically, the following were proved in [2]and [3], respectively.

Theorem A. If G is a 2-connected graph having two spanning trees of diameter m and n with m < n, then for every integer k with m < k < n, G contains a spanning tree of diameter k. *Present address: Department of Mathematical Sciences, New Mexico State University, Las Cruces, NM 88003 USA.

221

222

F. Harary and S. Schuster

Theorem B. If G is a connected graph having two spanning trees with m and n endnodes, m < n, then for every k such that m < k < n, G contains a spanning tree with exactly k endnodes. Our present purpose is to extend this work by proving interpolation theorems for the node independence number Po, the edge independence number P1, and the domination number a00 of the spanning trees of a given connected graph G. In addition, we show that no such interpolation theorem can hold for the independent domination number aho. We deem it particularly appropriate to have this work appear in a volume dedicated to our friend, Gabriel Dirac, for the question of whether such interpolation theorems held was first raised as we were traveling from the railway station in Sonderborg, Denmark t o the conference held at Sandbjerg Castle. Further, the solution was essentially achieved at that marvelous gathering of scholars who were colleagues of Dirac. The terminology and notation we use will be standard, [l].In particular, the node independence number of a graph G,denoted by Po(G),is the maximum cardinality of an independent set of nodes. The edge independence number ,&(G), is defined similarly. The domination number aoo(G) is the smallest number of nodes that cover V ( G ) ,while the independent domination number a&(G) is the minimum cardinality of an independent set of nodes that cover V ( G ) .

2

The edge-exchange procedure

A well-known method of transforming one spanning tree TOof a graph into another spanning tree T* of G is a basic tool in our proofs. We therefore begin with a description of this transformation. Let e be an edge of T* that is not an edge of TO.Then TO e contains a cycle which, of necessity, possesses an edge f which is not present in T*. For the first stage of this transformation, we define

+

and note that TI is a spanning tree of G having one more edge in common with T' than does TO. If TI = T*, the transformation is complete; otherwise, we repeat this edge-exchange procedure, defining a sequence of spanning trees T I ,T2,,.. until obtaining T'. If G has order p , the spanning tree T* is obtained from TOin a t most p - 1 stages. If p - 1 stages are required, then the two original spanning trees must have been edge-disjoint.

Interpolation Theorems for Spanning Trees

223

The node independence number

3

We now apply the edge-exchange procedure to prove that the node independence number ,L?o is an interpolating invariant over any graph G. Lemma 1 . Let 01, v2 be two nonadjacent nodes of a graph G and let G+ = G 0 1 ~ 2 . Then Po(G) - 1 I Po(@) I Po(G), (1)

+

i.e., Po(G+) = Po(G) or Po(G+)= Po(G) - 1. Proof. Every set S of nodes independent in G+ is also independent in G. Hence, Po(@) L Po(G). In order t o prove the second inequality, we let M be a largest independent set of nodes in G and consider two cases, depending on whether the given nodes 211 and 212 are members of M . If at least one of v1,v2 is not in M , then joining them leaves M as an independent set in G+. Hence, Po(G+) = Po(G) in this first case. However, if v1 and v2 are both in M , then M - (01) is an independent 0 set of nodes in G+.Thus, ,&(G+) 1 Po(G) - 1 in the second case. We note that in the last case in the above proof, both possible values for Po(G+) may occur. In Figure 1, where the darkened vertices are those in M , the first graph G1 illustrates po(G+) = Po(G) while G2 satisfies Po(G+) = Po(G) - 1.

P Figure 1. The two cases of the inequalities (1).

Corollary 1. For each edge e of a graph H ,

i.e., Po(H

- e ) = P o ( H ) or ,&(H - e ) = Po(H) + 1.

F. Harary and S. Schuster

224

Proof. Since every independent set of nodes in H is also independent in H - e , we have Po(H) 5 Po(H - e ) . The remaining inequality follows 0 immediately by applying Lemma 1 with G = H - e and G+ = H . Theorem 1. Let T and T* be any two spanning trees of a given graph G with Po(T) = m and Po(T*) = n where m < n. Then for every integer k with m < k < n, there exists a spanning tree TI such that Po(T') = k.

Proof. We employ the edge-exchange procedure to transform T into T* via a sequence of spanning trees T = TO,T I , .. . ,T, = T'. Since each stage of the procedure is reached from the previous one by the addition and deletion of a single edge, Lemma 1and Corollary 1imply that Po(T;-I)-l 5 Po(Ti) 5 Po(T;-l) + 1 for i = 1 , 2 , , , . , T . Hence, the sequence of integers 0 Po(T;) contains (m,m 1,m 2,. . . ,n) as a subsequence.

+

4

+

The edge independence number

Having shown that PO is an interpolating invariant over any graph G, it is easier to demonstrate the same conclusion for the edge independence number PI. Lemma 2. Let vl, 212 be two nonadjacent nodes of a graph G and let G+ = G vlv2. Then

+

i.e., PI(G+) = &(G) or Pl(G+) = Pl(G)

+ 1.

Proof. Since every independent set of edges in G is also an independent set in G+, we have &(G) I Pl(G+). In order to prove the remaining inequality, we let N be a largest independent set of edges in G+. In the event that 0 1 0 2 f N , it is clear that PI(G) = Pl(G+). We therefore examine the situation in which q w 2 E N by considering N1 = N - { ~ 1 ~ 2 ) . If N1 is alargest independent set of edges in G, then P1(G) = Pl(G+)-l. On the other hand, if N1 is not a largest independent set in G, then let N2 be such a set. Of course, N2 is also an independent set of edges in G+, which implies IN11<

IN21

I IN.

We conclude that IN21 = I N [ , from which we see that

P1(G)

= P1(G+).

0

In terpofation Theorems for Spanning Trees

225

Corollary 2. For each edge e of a graph H , P l ( H ) - 1 5 PdH - e ) I Pl(H),

(4)

i.e., & ( H - e ) = &(a) or P l ( H - e ) = & ( H ) - 1. Proof. Since every independent set of edges in H - e is also an independent set in H , we have p l ( H - e ) 5 & ( I f ) . The remaining inequality follows directly from Lemma 2 by letting H - e = G and H = G+. Theorem 2. Let T and T* be any two spanning trees of a given graph G. If P1(T) = m and P1(T*)= n where m < n, then for every integer k with m < k < n, there exists a spanning tree T' such that P1(T') = k, Proof. The proof is precisely the same as that for Theorem 1. We employ the edge-exchange procedure and observe that Lemma 2 and Corollary 2 imply that Pl(Ti-I) - 1 I Pl(Ti) 5 P l ( T t - I ) 1. Hence, the sequence of edge-independence numbers contains the subse0 quence (m,m 1,m 2,.. . ,n).

+

+

5

+

The domination number

By definition the domination number aoo(G) is the minimum number of nodes in a set S C V = V ( G )such that each node in V - S is adjacent to some node in S. It is next established that a00 is an interpolating invariant over any graph G.

Lemma 3. Let 0 1 , v2 be two nonadjacent nodes of a graph G and let G+ = G + 01212. Then

aoo(G)- 1 5 aoo(G+)I aoo(G),

(5)

i.e., aoo(G+)= aoo(G) or aoo(G+)= aoo(G)- 1. Proof. Any set of nodes that covers V ( G ) also covers V(G+). Hence aoo(G+)I aoo(G). Let M be a smallest cover of V(G+). There are three posibilities. If v1,v2 $! M , then M is also a smallest cover for V ( G )so that aoo(G+) = aoo(G). If 211,v2 E M , then again M is a smallest cover for V ( G )as well as for V(G+).Finally, suppose 01 E M and v2 $! M . Then it may be that 02 is not covered in G by any node in M . Thus M U (212) does cover V ( G ) , 0 S O aoo(G) I ~ j o ( G + ) 1.

+

F. Harary and S. Schuster

226

In Figure 2 both possibilities for the inequalities ( 5 ) are exhibited by showing two graphs: Gf = GI t v1v2 and G t = G2 ~ 1 ~ here 2 ; aoo(G1) = aoo(Gf) whilst aoo(G2) = aoo(G$) 1 .

+

+

Figure 2. The two cases of the inequalities ( 5 ) .

Theorem 3. Let T and T* be two spanning trees of a graph G . If aoo(T) = m and aoo(T*) = n, where m < n, then for every integer k with m < k < n, there exists a spanning tree TI such that aoo(T') = k . The proof is omitted as it is precisely the same as for the previous interpolation theorems. The fact that no such interpolation theorem holds for the node independent domination number aho is clear from the example in Figure 3 showing a graph G and two spanning trees T and T * , where ado(T)= 2 and abO(T*) = 4, while no other spanning trees exist.

Interpolation Theorems for Spanning Trees

227

Figure 3. A graph to illustrate aho.

References F. Harary, Graph Theory, Addison-Wesley, Reading (1969).

F. Harary, R. J. Mokken and M. Plantholt, “Interpolation theorems for diameters of spanning trees,” IEEE Trans. Circuits and Systems 7 (1983), 429-432. S. Schuster, “Interpolation theorem for the number of end-vertices of spanning trees,” J . Graph Theory 7 (1983), 203-208.

This Page Intentionally Left Blank

Annals of Discrete Mathematics 4 1 (1989) 229-254 0 Elsevier Science Publishers B.V. (NorthRolland)

Ein zum Vierfarbensatz Aquivalenter Satz der Panisochromie H. Heesch Institute of Mathematics University of Hanover Hanover, FRG

Dedicated to the memory of G. A . Dirac A motivation for a partition of the four colour theorem into 11 cases is presented. Of these 11 cases we can exclude 9. The remaining pair of cases can be formulated as a property of certain planar 4colourings. The property is called panisochromaticity. As far as I know there has been no studies of this concept for its own sake - at least not with respect to those aspects of interest for the four colour problem. The theme panisochromaticity also applies to graph colouring in general, e.g., to non-planar or infinite graphs. In the final part of this paper we exhibit a question on panisochromaticity equivalent to the four colour problem.

1 1.1

Triangulationen Tv und Figuren Fv Triangulationen T,

Betrachtet werden die Triangulationen T,der Kugel mit 21 Ecken, e = 3w - 6 Kanten, f = 2v - 4 Dreiecksflachen, 21

-e

+f =2

(Euler),

(1)

mit der Einschrankung, daB jedes Dreieck von Tv eine der f Dreiecksflachen ist. Hieraus folgt, da8 der Grad g einer Ecke V mindestens 4 ist und da8, mit alleiniger Ausnahme des Tetraeders, v = 4,

v 2 6. 229

(2)

H. Neesch

230

Die Liste der Anzahlen des T, fur die 7 Eckenzahlen 6 5 v

5 12 ist

Eckenanzahlv 6 7 8 9 10 11 12 1 1 2 4 10 25 87 Anzahl T, Bekanntlich folgt aus der Vierfarbbarkeit der oben definierten T,, fur alle v die Vierfarbbarkeit aller planaren Graphen.

1.2

Figuren F,,

AuDer den T, werden Graphen betrachtet, die durch Loschen einer beliebigen Kante aus einer T,, erzeugt gedacht werden konnen; sie werden Figuren F, genannt. Im Unterschied zu den randlosen Tv sind die F, berandete Graphen mit dem Randlcreis C4 (Bild 1).

A

c Bild 1. Die eine “Seite” einer Figur

Fvist mittels

inneren Ecken trianguliert. Zu einer T,, gehoren durch Loschen jeweils einer Kante 3v-6 Figuren F,,; diese sind nicht notwendig alle voneinander verschieden. In den Randkreis einer Figur F, laflt sich eine der beiden Diagonalen ( A , C ) oder ( B , D ) einsetzen. Wird eine von ihnen in F,, eingesetzt, so entsteht

T! im Falle ( A , C ) , T; im Falle ( B , D ) .

(5)

231

Ein zum Vierfarbensatz Aquivalenter Satz der Panisochromie

Zur Menge der Figuren F, wird hinzugerechnet die einzig mogliche Figur mit der Eckenzahl w = 5 , F5, vi = 1, das V4-Rad. Es wird hinzugerechnet, obwohl es keine T5 gibt, aus der F5 durch Loschen einer Kante hervorgeht. Die Liste der Anzahlen der F, fur die 8 kleinsten Eckenzahlen w, 5 5 w 5 12, lautet Eckenanzahl Anzahl der F,

5 6 7 1 1 2

8 6

9 10 11 12 18 68 282 1309

(6)

Die zu den vier ersten dieser Eckenanzahlen, 5 5 w 5 8, gehorige Figurenliste lautet (Bild 2):

1001

4003

2001

3001

4001

4002

4004

4005

Bild 2. Liste der 10 Figuren F, fur 5

3002

4006

5 w 5 8.

H . Heesch

232

Bild 3. Als Beispiel fur die Zusammengehorigkeit einer T,, mit ihren 3u - 6 Figuren F,, werde die Ti2), Bild 3, gewahlt. Je 4 ihrer 8 Ecken haben den Grad 4 bzw. 5. Die 3v - 6 = 18 Kanten der Ti2)bilden unter der Symmetriegruppe von Ti2), die die Ordnung 8 besitzt, 4 Klassen aquivalenter Kanten. Und zwar gibt es hierbei 4 aqu. Kanten mit dem Eckengradpaar 5,5,

2

”

8

”

4

”

9,

,9

97

9,

9)

9,

7,

77

n

9,

9,

,,

474, 495, 4,5.

Entsprechend gibt es aus Ti2)nach Loschen jeweils einer Kante im ganzen 4 topologisch verschiedene Figuren F8, je in denselben Anzahlen 4, 2, 8, 4;in der hierzu gehorigen Reihenfolge sind es in Bild 2 die vier F8 mit den Namen 4002,4004,4001, 4003.

2

Farbungen, Randfarbungsklassen

Gefarbt werden die Ecken mit hochstens 4 Farben, wobei die beiden Ecken einer Kante verschiedene Farben tragen. Zwei Farbungen werden genau dann als dieselbe gezahlt, wenn sie sich hochstens durch eine Permutation der Farben unterscheiden. Nach dieser Festsetzung gibt es die Anzahlen nT,, oder nF, aller Farbungen einer T,, oder einer F,,. In jeder Farbung von F tragt der stets mitgefarbte Randkreis Cq, Bild 1, eine seiner 4 moglichen Randfarbungen a

a

b

a

1

b

b

C

2

a d

b

a

3

a

d

b

C

4

b

(7)

Ein zum Vierfarbensatz Aquivalenter Satz der Panisochromie

233

Daher bilden die n F Elemente der Farbungsmenge @ ( F )einer Figur F die 4 Klassen von Farbungen gleicher Randfarbung aus (7). Die Farbungsanzahlen in den einzelnen Klassen seien

Dann gilt

auch wird geschrieben

Die Anzahl der von null verschiedenen unter den 4 Zahlen nl bis wird r(Fv) genannt; fur jede Figur F, beliebiger Eckenzahl v ist

n4

Fur 6 Figurenbeispiele aus Bild 2 werde r angegeben in Bild 4.

3

Gedankengang der weiteren Arbeit

Der folgende Satz 3.1 stellt eine Motivation dar fur eine Aufgliederung der Behauptung des Vierfarbensatzes in 11 Faille. Von diesen 11 Fdlen werden 9 in Abschnitt 4 und 5 ausgeschaltet; es bleibt allein ein Fallpaar ubrig fur eine noch nicht auszuschlieaende Moglichkeit von Funfchromatizitat eines planaren Graphen, wie in Abschnitt 6 formuliert wird. Diese Moglichkeit wird in Abschnitt 7 als eine innerhalb der Vierfiirbbarlceit zu formulierende besondere Eigenschaft bestimmter planarer Graphenfarbungen aufgewiesen; diese Eigenschaft wird Panisochromie genannt. Meines Wissens ist Panisochromie - wenigstens in dieser fur das Vierfarbenproblem interessierenden Akzentuierung - noch nicht als Gebiet der Graphenfarbung fur sich behandelt worden. Das Thema “Panisochromie” besteht fur alle (d.h. auch fur nichtplanare wie auch fur unendliche) Graphenfarbungen. In Abschnitt 8 wird diejenige Teilfrage aus dem Gebiete der Panisochromie der endlichen planaren Graphen aufgezeigt , die dem Vierfarbenproblem aquivalent ist. Damit ist das Ziel dieser Arbeit erreicht.

H. Heesch

234

F5 : 1001

= 3

Bild 4. Sat2 3.1. Gilt f6r eine Figur F, mit v

25

so sand die beiden Triangulationen TvI und

T;, aus deren jeder, wie in Ab-

schnitt 1 dargelegt, F, durch Loschen einer Kante hervorgeht, vierfarbbur. Beweis. Man kann sich T,I zusammengesetzt denken aus der Figur F,, die

Ein zum Vierfarbensatz Aquivalenter Satz der Panisochromie

235

von ihrem Randkreis C4 berandet ist, und aus einem weiteren Randkreis C4, der mittels lediglich der Diagonalen ( A ,C) trianguliert ist; T,I entsteht, indem man diese beiden Randkreise indentifiziert. Fur den diagonalisierten C4 lautet (10) (np.*)= (O,l,O,l), fur Fv gilt nach Voraussetzung (12). Da, wie bei jeder vierelementigen Menge, so auch bei (7), jede wenigstens dreielementige Teilmenge mit einer beliebigen zweielementigen Teilmenge wenigstens ein gemeinsames Element hat, folgt Satz 3.1 fur T,.I Unter Vertauschung der Diagonale ( A ,C) mit der Diagonale ( B ,0)bei hierzu 0 gehorigem ( n F , ) = (O,l, 1,O) folgt Satz 3.1 auch fur T;. Unmittelbar folgt aus Satz 3.1: Satz 3.2. E i n Beweis dafur, daj’ bei beliebiger Eckenanaahl v 2 5 fur jede Figur F,, die Ungleichung (12) gilt, ist ein Beweis des Vierfarbensatzes.

Anders gesagt: Satz 3.3. Das Gelten von (12) f u r sumtliche Figuren F,, jeder beliebigen

Eckenanzahl v 2 5 ist eine hinreichende Bedingung f u r das Gelten des Vierfarbensatzes. DaB hierbei zumindest die Bedingung 2 5 nicht fehlen kann, wird deutlich aus der Existenz der beiden schon im Beweis von Satz 3.1 verwendeten Figuren F4, den C4-Diagonalisierungen, fur deren jede r(F4) = 2 gilt. Bezieht man auch den durch Identifikation eines Gegeneckenpaares, A = C oder aber B = D, entarteten C4 (das ist: je ein Kantenpaar) mit v = 3, in die Betrachtung ein - zu den beiden Kantenpaaren gehoren (n*) = ( l , O , l , O ) und (n,) = ( l , O , O , l ) , beide ebenfalls mit r = 2, - so enthalt die Liste von Figuren mit r 5 2 bereits 4 Verwirklichungsbeispiele. Wie in Bild 4 aufgefuhrt, gilt (12) ausnahmslos fur die samtlichen Figuren zu allen Triangulationen T, fur die kleinstmoglichen v-Werte 5 , 6, 7. Fortan werde ein Gelten von (12) fur alle F, zu allen T,, 5 5 u, 5 v - 1, angenommen, wahrend fur eine bestimmte zu einer Triangulation Tv gehorige Figur F,, die Ungleichung (12) als nicht geltend angenommen werde. Fur ein solches F,, ist damit als geltend angenommen:

Diese Ungleichung hat, wegen (7), genau die folgenden 11 Verwirklichungs-

H. Neesch

236 moglichkeiten Fall 1 bis 11:

(l)

Fall 1 bis 6 : r(F,,) = 2 :

Aus (7) sind auf = 6 Weisen zwei Rand farbungen ausw ahlbar . Fall 7 bis 10: r(F,,) = 1 : ist auf jeweils eine von 4 Weisen realisierbar. Fall 11: T(F,,) = 0 : ist auf eine Weise realisierbar.

Hierbei ordnen sich zunachst die 6 Faille T = 2 zu 3 Paaren gleichartiger Falle. Die Verschiedenheit dieser 3 Fallpaare macht verschiedene Methoden fur ihre Behandlung erforderlich. Fur das erste Fallpaar, Fall 1 und 2, zu dem sich auch der Fall 11, T = 0, als hinzuzufugen erweist, wird in Abschnitt 4 bewiesen, daB es uberhaupt keine zugehorige Figur F,, gibt. Fur das zweite Fallpaar, Fall 3 und 4, zu dem sich auch die vier F d l e 7 bis 10, T = 1, als hinzuzunehmen erweisen, wird in Abschnitt 5 gezeigt, daB es keine Figur F,, gibt, in der sich einer dieser 2 t 6 = 8 Falle verwirklicht. So bleibt von den 11 Verwirklichungsmoglichkeiten (14) allein das Fallpaar 5 und 6, fur T ( F,,) = 2, ubrig, wobei der eine Fall symmetrisch zum andern ist, weshalb auch von nur noch einem iibriggebliebenen Fall gesprochen wird,

4

Die drei Falle Nr. 1, 2 und 11 von (14)

Unter der Induktionsvoraussetzung,daB fur alle F,, 5 5 w 5 n, T ( F,) 2 3 ist, wird jetzt die Existenz einer Figur F,, mit v = n t 1angenommen, fur die T(F,,) = 2 in der Weise verwirklicht ist, daB die folgenden zwei Gleichungen und die zwei Ungleichungen gelten: n a = n a = 0; bab

bad

n a > -1, b,d

Dies bedeutet: In samtlichen Vierfarbungen von F,, tragen die beiden Randgegenecken A und C des Randkreises C, von F,, verschiedene Farben. In dieser F,, gehe jetzt der Randkreis C4 = ( A , B , C, 0)durch Identifikation A 3 C in das Kantenpaar ( B , A 1 C , D ) iiber; F,, ist hierbei in die Triangulation Tv-l ubergegangen. Laut Induktionsvoraussetzung ist Tv-l vierfarbbar. Es liege daher jetzt T,,-1 mit einer seiner nT, Vierfarbungen gefarbt vor. Nun werde die Identifikation A = C wieder aufgehoben, wobei alle Ecken ihre Farbe behalten und die jetzt wieder verschiedenen Ecken A und C beide die Farbe behalten, die sie bei der Identifikation hatten. Jetzt

Ein zum Vierfarbensatz Aquivalenter Satz der Panisochromie

237

liegt eine Vierfarbung der ursprunglichen F,, vor, fur die wenigstens eine der beiden Ungleichungen n u > -1 bab

oder

n u 21 bad

gilt. Das steht im Widerspruch zu dem Gleichungspaar in (15). Hierdurch ist bewiesen, dafl es keine Figur F,, gibt, die die Bedingungen (15) erfullt. Aus Symmetriegrunden ist hiermit auch der zweite der 11 Falle (14) erledigt, dessen Definition sich aus (15) durch Platztausch von n a mit bad

n

bcb

ergibt.

Da beim Aufbau des Widerspruchs nur die zwei Gleichungen in (15), aber keine der zwei Ungleichungen benutzt wurden, ist zugleich bewiesen, dafl es auch fiir den Fall 11 von (14) mit seiner Bedingung r(F,,) = 0 keine Figur gibt, die diesen Fall 11 verwirklicht. Hieraus folgt, wie beilaufig bemerkt sei, 5 5 w 5 v - 1, r(F,,,) 2 3 gilt, sind auch noch alle Figuren F,, vierfarbbar. Satz 4.1. Unter der Vomussetzung, dajl fur alle Figuren F,,

5

Die Falle Nr. 3, 4 sowie 7 bis 10 aus den 11 Fallen (14)

Hier wird mit derjenigen Verwirklichung von r ( F ) = 2 begonnen, bei der n u >1, bad

n u > -1; b,b

dies sind die Definitionsbedingungen fur den Fall 3 aus den 11 Fallen von (14). Die Einsicht, dai3 keine Figur F,, existiert, die (17) genugt, wird durch einen Blick auf die einfachsten Eigenschaften von Kempeketten in F,,Farbungen gewonnen: Kempeketten sind maximal zusammenhangende zweifarbene Teilgraphen in einem mit hochstens 4 Farben gefarbten Graph. Mit den Kempeketten eines Farbpaares werden meist zugleich auch die Kempeketten des komplementaren Farbpaares betrachtet. Ein Paar komplementarer Paare der 4 Farben wird als Farbpaarwahl bezeichnet. Es gibt die 3 Farbpaarwahlen ab, cd; ac, bd; ad, bc.

(18)

H. Heesch

238

dz

Fur eine Figur F, mit Randkreis Cq = existiert bei jeder Farbung wenigstens eine Farbpaarwahl derart, dafl das eine der 2 Randgegeneckenpaare (2.B. A , C) nur mit Farben des einen Farbpaares - sagen wir: a , c, - das andere ( B , D ) nur mit Farben des komplementaren Farbpaares b,d - gefarbt ist. In (7), z.B., trifft das bei jeder der 4 Farbungen 1, 2, 3, 4 fur die Farbpaarwahl ac, bd zu. Wegen der (in der Definition geforderten) Maximalitat des Zusammenhangs einer jeden Kempekette liegen daher fur dieses Farbpaar immer die beiden Randecken genau eines der beiden Gegeneckenpaare in einer Kempekette. Diese Kempekette trennt zugleich immer die beiden verschiedenen komplementaren Kempeketten voneinander, deren jede genau eine der beiden ubrigen 4 Randecken enthalt. Vertauscht man in einer dieser beiden letzteren Kempeketten deren beide Farben miteinander, so wird dabei (bekanntlich) die Farbungsvorschrift (siehe Abschnitt 2, Anfangssatz) nirgends verletzt. Man erhalt dadurch also eine andere Farbung der F,, bei der auch genau eine der 4 Randecken ihre Farbe in die andere ihres Kempekettenfarbpaares gewechselt hat. Iterierung dieser Operation fuhrt auf die Ausgangsfarbung einschlieDlich Ausgangs-Randfarbung zuruck. Das heifit: Den F,-Farbungen ist eine Puarstruktur eigen, an der auch die 4 Randfarbungen partizipieren. Fur die Randfarbungen lauten die moglichen Paare (Randfarbungsnumerierung siehe (7)): 1 - 3 oder 1 - 4 oder 2 - 3 oder 2 - 4. Weil, wie gesagt, bei der Umfarbungsoperation nur immer genau eine Randecke eines Gegeneckenpaares die Farbe zu wechseln vermag, gibt es kein Paar 1 - 2 und kein Paar 3 - 4, das durch die besprochene Operation hatte entstehen konnen, Das Gesagte ist darum nunmehr folgendermaflen zu formulieren: S a t z 5.1. Bei jeder Figur F,, far bu ng

5

5 v, tritt mit einer Farbung der Rand-

1 wenigstens auch eine Furbung mit Randfarbung 3 oder 77 n Y? 79 2) n 2 3 ?’ n 79 97 97 n 77 3 1 ” 99 Y9 n # n 4 1 ” )>

4 auj 4 ”, 2 ”, 2 ”,

Eine einfache Folgerung aus diesem Satz lautet: Satz 5.2. Es gibt keine Figur

F,, deren samtliche Furbungen dieselbe

Randfiirbung haben. Hierdurch ist fur die 4 F a l e Nr. 7 bis 10 der 11 Falle (14) - dies sind genau die samtlichen 4 Falle mit r(F,) = 1 - bewiesen, dafi es keine Figur F, gibt, durch die einer von ihnen verwirklicht wurde.

Ein zum Vierfarbensatz Aquivalenter Satz der Panisochromie

239

Ferner ist der Fall Nr. 3, mit T(F,) = 2, der in (17) definiert ist, infolge von Satz 5.1 ohne mogliche Verwirklichung. Ebenso ist auch der aus (17) durch Vertauschung der ersten beiden Randfarbungen mit den letzten beiden Randfarbungen hervorgehende Fall Nr. 4, na >1, bab

n a > -1 , bed

laut Satz 5.1 ohne Verwirklichung durch eine Figur F,.

6

Beschreibende Formulierung des letzten der 11 Falle (14)

Wie schon im Schluflsatz von Abschnitt 3 gesagt, bleibt jetzt fur F, bei Ansetzen von (13) allein der Fall Nr. 5 (mit seinem symmetrischen Gegenstuck Nr. 6) von den samtlichen 11 Verwirklichungsmoglichkeiten (14) zu untersuchen. Definiert ist Fall 5 durch die folgenden 4 Bedingungen - 2 Gleichungen und 2 Ungleichungen -

Fall 6 durch na >1, bab

n u >1. beb

Unter der in Abschnitt 3 vor Ungleichung (13) ausgesprochenen Induktionsvoraussetzung, daB fur alle Figuren F ! , 5 5 w 5 v - 1, die Ungleichung (12) gilt, wird jetzt das Vorliegen einer Figur F, angenommen, fur die (13) in der Form der 4 Bedingungen (20) gilt. Diese Bedingungen besagen: Fur F, ist Satz 4.1 in folgender Weise erfullt : Zunachst: Wegen der 2 Gleichungen in (20) tragen die Randkreisecken A und C des Randkreises Cq = .“d) in allen Vierfarbungen von F, dieselbe Farbe. Alsdann gilt wegen der 2 Ungleichungen in (20): Es gibt wenigstens eine Vierfarbung von F,, in der auch B und D gleichgefarbt sind, und es gibt ferner ebenfalls wenigstens eine weitere Vierfarbung von F,, in der B und D verschiedengefarbt sind. Wird jetzt aus F, durch Hinzufugen der Diagonale (B,D) die Triangulation T, E T,- erzeugt, so tragen ebenfalls in siimtlichen - wegen der

H. Heesch

240

letzten Ungleichung in (20) sicher existierenden - Vierfarbungen der T,,die Ecken A und C dieselbe Farbe. Wird indessen, anstelle von ( B , D ) , die andere Diagonale, ( A , C ) , in den Randkreis von F, eingezogen, so liegt TvI vor. Jetzt ist es - wegen der beiden Gleichungen in (20) - nicht moglich, auch fur die Kante ( A ,C) die Farbungsvorschrift (siehe Anfangssatz des Abschnittes 2) zu erfullen, daB auch deren beide Ecken A und C verschiedene Farben tragen. Dies heiat aber: Ti ist nicht vierfarbbar, T,,I ist funfchromatisch. Damit steht man vor einer neuartigen Situation: Erste Erfahrungen mit ihr vermitteln den deutlichen Eindruck, dafl eine Erledigung auch dieses letzten Falles von (14) in einem ahnlichen Rahmen wie fur die 9 in Abschnitt 4 und 5 abgehandelten Falle nicht erkennbar ist. Zwar lassen sich mit geringen Mitteln bemerkenswerte Eigenschaften solcher T, beweisen, z.B. dafl eine funfchromatische T,, keine Ecke des Grades 4 enthalt. Die Hauptfrage indessen ist die Frage nach der Existenz von Figuren F,,, die (20) genugen. Nun gibt es eine Algorithmik, in der diese Hauptfrage direkt und unkompliziert formulierbar ist; sie gehort indessen einem m.W. bisher wenig entwickelten Gebiet an. Wie schon ervahnt, handelt es sich dabei um die Panisochromie; von ihr wird darum in Abschnitt 7 ein Abrifl gegeben. Dieser Abrifi kann kurz sein, weil der (unbewiesene) Satz aus dem Gebiete der Panisochromie, dessen Beweis das in Abschnitt 3 bis 5 Gebrachte zu einem Beweis des Vierfarbensatzes erganzt, in seiner Formulierung einem noch recht einfachen Teilgebiet der Panisochromie zugehort. Diese Formulierung geschieht in Abschnitt 8, womit das Ziel dieser Arbeit erreicht ist.

7 7.1

Kurzer Abrifi der Panisochromie Allgemeine Einfuhrung

In der Graphenfarbungstheorie sind meist die Eclcen die Trager der Farben (fur planare Graphen durch Dualisierung die “Lander”). Die Farbungstheorie hat die Frage zum Gegenstand, wie in verschiedenen Graphen und Graphenklassen die “Farbungsforderung” verwirklicht wird. Diese Farbungsforderung lautet: Die beiden Ecken einer jeden Kante tragen verschiedene Farben. Die Objektklasse, in deren samtlichen Elementen die Farbungsforderung zu verwirklichen ist, ist also die Klasse der adjazenten Eckenpaare eines Graphen. Hat man dabei die planare Problematik - sei es fur endliche Graphen

Ein zum Vierfarbensatz Aquivalenter Satz der Panisochromie

241

(Kugel) oder unendliche (euklidische Ebene) - im Auge, so ist man besonders an den Ergebnissen bei mazimal planaren Graphen, d.i. bei Triangulationen T interessiert. Ist, im Kugelfall, die Eckenzahl v , so gilt die Eulersche Polyedergleichung (1) mit 2e = 3 f ;

(22)

hieraus ergibt sich die Anzahl der [Kanten oder der] adjazenten Eckenpaare einer jeden T, zu (siehe Textzeile 2 auf Seite 1) e = 3v - 6. Diese 3v - 6 Eckenpaare sind also, wie schon gesagt, der genaue Objektbereich, auf dem die Farbungsforderung operiert. Nun gibt es in der T, im ganzen (g) = &(v - 1 ) Eckenpaare; das heifit: es gibt in einer T, 1

(i)- ( 3 -~6) = -(v - 3 ) ( -~ 4) = ( " T ~ ) 2 nichtadjazente Eckenpaare. Direkt sind diese nicht betroffen durch die Farbungsforderung. Ihr gegenuber ist daher als allgemeines Verhalten eines beliebigen dieser ('i3) nichtadjazenten Eckenpaare anzunehmen, daB in den , der Tv die beiden Ecken dieses (im allgemeinen zahlreichen) n ~ Farbungen Paares bei einigen dieselbe Farbe, bei den ubrigen verschiedene Farben tragen. Die Beobachtung zeigt nun, daB es Graphen T, gibt, fur die die gewohnliche Farbungsforderung, in allen Farbungen verschiedene Farben zu tragen, uber die adjazenten Eckenpaare hinaus auch fur weitere Eckenpaare erfiillt ist. Jedes solche Eckenpaar wird panheterochromes Eclcenpaar zur Farbenanzahl x (hier meist x = 4 ) genannt (vgl. das bekannte Beispiel des Ikosaedergraphen, 7.2, Bsp. 3). Ferner zeigt die Erfahrung, dafi fur nichtadjazente Eckenpaare in Graphen auch das (fur adjazente Eckenpaare ausgeschlossene) Kontrare vorkommt, namlich Gleichgefarbtheit der beiden Ecken in allen Farbungen, das ist: deren Panisochromie. Dabei ist dies Vorkommen nicht auf Paare beschrankt, es zeigt sich auch bei Eckentripeln, -quadrupeln, . . ., -nl-tupeln. In jedem solchen Falle sagt man, der Graph enthalte eine nl-elementige Menge panisochromer Ecken mit n1 2 2 . Ja, unter diesen gibt es Graphen, in denen je eine solche Menge zu zwei oder drei der x Farben, bis, hochstens, zu den samtlichen x Farben vorliegt (Beispiel siehe 7.2, Bsp. 6). In einem Graph mogen V,W ein panisochromes Eckenpaar sein, und es gebe in dem Kreis der mit W adjazenten Ecken wenigstens eine Ecke X mit einem Abstand 2 2 von V . Als Folge der Panisochromie von V,W ist das Eckenpaar V , X panheterochrom. Jede in ahnlicher Weise als Folge

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vorkommender Panisochromie auftretende Panheterochromie werde dependente Panheterochromie genannt. Hingegen heiBt jede in einem Graph ohne ein panisochromes Eckenpaar vorkommende Panheterochromie independent. Fur independente Panheterochromie ist der unmittelbar vorher erwahnte Ikosaedergraph ein Beispiel. Siehe auch die Weiterfiihrung dieser Unterscheidungen 7.3.4!

7.2

Beispiele

Beispiel 1. Die ungeraden Doppelpyramiden (als Panisochromie-Besp.): Der Aquator werde durch 2n - 1 Ecken in 2n - 1 Kanten geteilt; jede Ecke wird mit dem Nord- und dem Sudpol durch je eine weitere Kante verbunden. Der Aquatorkreis besitzt 82n-1 = (22("-1)- 1)/3 Eckenfarbungen mittels 3 Farben. Durch Hinzufugen einer vierten Farbe fur jede der beiden Polecken hat man die samtlichen ebenso vielen 4-Farbungen der ungeraden Doppelpyramiden; die beiden Polecken bilden die zweielementige panisochrome Eckenmenge.

Beispiel 2. Nachdem die 7-seitige Doppelpyramide Tg,l schon unter Beispiel 1 fie1 und eine T9,2 erst unter Beispiel 4 erwahnt werde, mogen hier die beiden ubrigen der 4 moglichen Tg besprochen werden: T9,3 entsteht aus F7 in Bild 4, Zeile 3, und aus FG in Bild 4, Zeile 5, durch Identifizierung der beiden Randkreise. T9,3 hat 5 Ecken des Grades 4, d.i. n4 = 5; n5 = 126 = 2. Die beiden Ecken A , C des Grades 5 bilden ein panisochromes Paar.

Bild 5. T9,4

T9,4 entsteht (Bild 5) durch Einsetzen je einer Ecke des Grades 4 in jede der 3 rechteckigen Seitenflachen eines dreiseitigen Prismas. Es ist 72.4 = 3 n5 = 6;die Anzahl Farbungen ist nrs,, = 2. Die Ecken des Grades 4 bilden eine dreielementige panisochrome Eckenmenge.

Beispiel 3. Der Ikosaedergraph mit der Anzahl 10 seiner Farbungen hat kein panisochromes Eckenpaar. Uber das bereits in Abschnitt 7.1 uber

Ein zum Vierfarbensatz Aquivafenter Satz der Panisochromie

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ihn Gesagte hinaus werde weiter vermerkt: In allen 10 Farbungen sind simultan die Ecken eines jeden seiner 6 Eckenpaare des Abstandes 3 verschiedengefarbt. Bei x = 4 Farben gibt es (?$ = 6 verschiedene Paare ungleicher Farben. In jeder Ikosaederfarbung kommt jedes dieser 6 Paare, ah, ac, a d , bc, bd, cd in genau einem der 6 genannten panheterochromen Eckenpaare des Abstandes 3 vor. 6 ist also die Maximalanzahl denkbarer panheterochromer Eckenpaare in irgendwelchen 4-gefarbten Graphen. Fur panheterochrome Eckentripel = 4. Meines Wissens ist uber hierzu ist die Maximalanzahl nur (t) = gehorige Verwirklichungen nichts bekannt. Der Ikosaedergraph, eine der 87 Triangulationen 7'12 (siehe Anzahlliste nach G1. (3)) ist nicht die kleinste (das ist v-minimale) Kugeltriangulation : trianmit independenter Panheterochromie. Diese ist vielmehr T ~ JMan guliere einen Aquatorkreis c 6 einmal (Sudhalbkugel) mittels einer Ecke v 6 und ein zweites Ma1 (Nordhalbkugel) mittels einer VsV5-Kante. Es gibt bei nT,,, = 6 Farbungen kein panisochromes Eckenpaar. Wohl aber ist jedes der 3 diametralen Eckenpaare des Cs panheterochrom und also independent panheterochrom.

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Beispiel 4. Hier wird eine bestimmte Ti5 betrachtet (Bild 6) mit n ~ , ,= 74. Das (einzige) panisochrome Eckenpaar in Ti5 hat Abstand 3. A

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Bild 6. Bild 7 und Bild 8 zeigen die 36 Farbungen der Ti5 zur Randfarbung 1 = bfb. In Bild 9 und Bild 10 sind die 19 Farbungen der Ti5 zur Randfarbung 4 = b:b gezeichnet. Diese stehen mittels der Symmetrie der Ti5 in ersichtlicher bijektiver Zuordnung zu den ebenfalls 19 Farbungen zur Randfarbung 3 = @, im ganzen gibt es also nT,, = 36+2.19 = 36+38 = 74 Farbungen der 7'15.

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Bild 10. In allen bisherigen Beispielen zur Panisochromie gab es (wenigstens) ein panisochromes Eckenpaar des Abstandes 2. Ein solches gibt es nicht in der T15. Deren einziges panisochromes Eckenpaar ist F, J mit dem Abstand 3.

Beispiel 5. Unter den (laut Liste nach G1. (3)) 25 Triangulationen 2'1 findet sich eine, T11,9,(Bild ll), in der 2 Paare panisochromer Ecken so vorkomrnen, daf3 je eine Ecke des einen Paares mit je einer des anderen Paares adjazent ist. Es liegt also Panisochromie eines Kantenpaares vor. Es ist n ~ = 10. ~ Die ~ beiden , ~ panisochromen Eckenpaare sind F , I und G, J ; das panisochrome Kantenpaar ist ( F ,G), (I,J ) . In jeder Farbung der T11,gtragt das Eckenpaar F , G dasselbe Farbpaar - und zwar ausschliealich in derselben Reihenfolge - wie auch das Eckenpaar I,J. Die Symmetrie der ist eine Vierergruppe: Spiegelung an der Ebene des Funfecks ( A , B , C, D ,E ) und an der d a m orthogonalen Ebene durch A , H , IC und drei Kantenmittelpunkte, Drehung urn K urn die Schnittgeraden der beiden Spiegelebenen. Die erstgenannte Spiegelung permutiert die beiden panisochromen Kanten, die letztgenannte permutiert innerhalb jeder der beiden Kanten die Ecken.

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E Bild 11. T11,g

Beispiel 6 . Obwohl fur die Fragen in dieser Arbeit die in Unterabschnitt 1.1 fur Triangulationen Tv getroffene Einschrankung sachgerecht ist, gibt es Operationen wie z.B. das Kontrahieren von Kanten, die zum Erkennen von Eigenschaften panisochromer Graphen dienlich sind, bei deren Ausfiihrung indessen jene Einschrankung durchbrochen wird, indem sich als Kontraktat eine “gewohnliche” Triangulation ergibt mit Ecken vom Grad 3 sowie mit beidseits (noch mittels innerer Ecken) triangulierten Dreiecken. Ein in dieser Weise begriindetes Oszillieren zwischen den beiden T,-Begriffen werde (ohne weiter erwahnt zu werden) bei derartigen Betrachtungen akzeptiert, darunter auch bei diesem Beispiel 6, auf das in 7.1 hingewiesen wurde, da es die dortige Bemerkung illustrieren moge: Werden 2 Exemplare des Tetraedergraphs 1<4 durch Identifizieren je eines ihrer Dreiecke zusammengesetzt, so entsteht eine dreiseitige Doppelpyramide, eine T5. Aufsetzen eines dritten K 4 auf ein Dreieck von T5 ergibt eine TSusf. Alle so entstehenden Graphen mogen “reine K4-Derivate” heiDen. Jeder von ihnen hat nur eine Farbung. Jetzt werden nur diejenigen reinen K4-Derivate betrachtet, in denen wenigstens ein K q vorkommt, bei dem auf jeder seiner 4 Dreieckflachen ein weiterer K 4 aufgesetzt ist; dies garantiert, daD jede der 4 Farben (in der einzigen Farbung des reinen K4-Derivats)

Ein zum Vierfarbensatz Aquivalenter Satz der Panisochromie

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wenigstens zweimal vorkommt. Aufgrund der Definitionen in 7.1 enthat jede hierzugehorige T,die Maximalanzahl x = 4 von panisochromen Eckenmengen. Fur jede dieser Triangulationen T,, gilt ferner, daa jede ihrer Ecken Element in einer der 4 panisochromen Eckenmengen ist; diese Graphen enthalten also uberhaupt keine Ecke auj7erhalb einer panisochromen Eckenmenge.

7.3 Einige m.W. offene Fragen der Panisochromie 7.3.1. In 7.2, Beispiel 5, ist der Graph T11,9 angegeben, bei dem auf3er der bisher allein betrachteten Klasse von Panisochromietragern, den nichtadjazenten Eckenpaaren, auch ein Kantenpaar als Trager von Panisochromie vorkommt. Angesichts dieses Vorkommnisses bei der T11,gist die Frage naheliegend: Gibt es ferner auch Triangulationen Tv - im Sinne von 1.1 mit einem Paar von Panisodreiecken, gar auch von Paniso-dreieckspaaren bei einer gemeinsamen Kante der beiden Dreiecke? Und welche Teilgraphen sind die groatmoglichen Trager von Panisochromie in einer T,,? 7.3.2. Als Beispiel 4 in 7.2 ist der (vielleicht eckengeringste) Graph 2’15 angegeben, in dem - ohne daa ein panisochromes Eckenpaar vom Abstand 2 vorkommt - ein panisochromes Eckenpaar vom Abstand 3 vorkommt. Gibt es in Triangulationen T, einen maximalen Wert fur den kleinsten Abstand der Ecken in jeder der vielleicht mehreren panisochromen Eckenmengen, und, wenn ja, welcher ist es? 7.3.3. Als Beispiel 3 in 7.2 ist der Ikosaedergraph mit der Maximalzahl )(; = = 6 je zweielementiger independenter panheterochromer Eckenmengen gegeben. Weil darin diese Maximalzahl 6 verwirklicht ist, tritt der Ikosaedergraph in eine “Negationsanalogie” zu den im Beispiel 6 in 7.2 besprochenen speziellen reinen 11‘4-Derivaten: In diesen letzteren ist jede Ecke Element in einer der in ihrer Maximalzahl 4 vorkommenden panisochromen Eckenmengen; und die Menge dieser lid-Derivate ist unendlich. Im Ikosaedergraphen ist jede Ecke Element in genau einer der gleichfalls in ihrer Maximalzahl6 vorkommenden panheterochromen Eckenpaare. Andere Graphen dieser Eigenschaft (oder auch geringmodifizierter weiterer Eigenschaften) sind m.W. nicht bekannt.

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7.3.4. Am Schlua von 7.1 war dependente von independenter Panheterochromie unterschieden worden. In Weiterfuhrung des dort Gesagten ist denkbar, daa in einer T, zugleich vorkommen:

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b) wenigstens ein panheterochromes Eckenpaar, wobei keine einzige Ecke eines der vielleicht mehreren panheterochromen Eckenpaare zugleich Ecke auch nur eines einzigen der panisochromen Eckenpaare ist. Auch diese Eckenpaare sind independent panheterochrom. Ob es Kugeltriangulationen mit zugleich paniso- und im soeben erlauterten Sinne independent panheterochromen Eckenmengen gibt, ist mir nicht bekannt.

7.3.5. In einer Triangulation T, komme ein panisochromes Eckenpaar A , D vom Abstand 2 (Bild 12) vor, und zwar in folgender Weise: AuBerhalb von Bild 12 sei jeder Weg in T,, der A und D verbindet, von einer Lange 2 3; die mittlere Ecke P des einzigen Weges ( A , P , D ) zwischen A und D der Lange 2 habe nach jeder der beiden Wegseiten wenigstens 2 weitere von ihr ausgehende Kanten.

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D Bild 12. Diese Situation habe ich noch in keiner der mir bekannten Triangulationen mit einem panisochromen Eckenpaar A , D des Abstandes 2 fur eine mittlere Ecke P eines Weges ( A ,P,0)verwirklicht gefunden. In jedem dieser Vorkommen gibt es fur jedes derartige P wenigstens eine ihrer beiden Seiten des Weges ( A , P , D ) ,nach der genau nur eine weitere Kante ( P , Q ) ausgeht. Hierbei folgt sofort, dafi die andere Ecke Q der Kante (P,Q)die Existenz eines weiteren Weges ( A ,Q,D), gleichfalls der Lange 2, zwischen demselben panisochromen Eckenpaar A , D garantiert. Diese Erfahrung ist auch so auszusprechen: In allen mir bekannten Triangulationen T, mi t einem panisochromen Eckenpaar des Abstandes 2 gibt es (wenigstens) eine Kante, deren Gegeneckenpaar es ist.

Ein zum Vierfarbensatz Aquivalenter Satz der Panisochromie

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Ein offenes Problem ist daher zunachst die folgende

Frage 7.3.5.1. Gibt es zu jedem panisochromen Eckenpaar des Abstandes 2 in einer T,, wenigstens ein Dreieckspaar mit gemeinsamer Kante, dessen Gegeneckenpaar es ist? Moglicherweise ist es einfacher, eine Antwort (oder Teilantwort) auf folgendene Verfeinerung dieser Frage zu finden: Frage 7.3.5.2. Gibt es in der Menge der mehreren panisochromen Eckenpaare des Abstandes 2 immer wenigstens eines, das Gegeneckenpaar einer Kante ist? 7.3.6. Eine insbesondere im Hinblick auf Abschnitt 8 wichtige offene Frage der Panisochromie ist die folgende: Es ist m. W. kein einziges Vorkommen eines panisochromen Eckenpaares bekannt in Triangulationen T,, ohne Ecken vom Grad 4,d.h. fur Triangulationen T,, mit w4 = 0. Praziser: Der bisher vorliegende (noch recht kleine) Atlas von Triangulationen T,, mit wenigstens einem panisochromen Eckenpaar enthdt ausschlieBlich solche T,, die 114 2 3 haben. Dadurch ist es nahegelegt, als Vermutung zu formulieren: V e r m u t u n g 7.3.6. Jede T,, mit wenigstens einem panisochromem Eckenpaar enthdt mindestens 2 Ecken des Grades 4. Eine Begrundung, warum hier gegenuber der vorher mitgeteilten Erfahrung 04 2 3 wieder auf 2 zuriickgegangen wird, wurde den Rahmen dieser Note sprengen. Im Hinblick auf die vorangehenden Fragen in 7.3.5 sei bemerkt, dafl die soeben a l s stets existierend vermuteten Ecken des Grades 4 durchaus nicht auch schon in einem panisochromen Eckenpaar vorkommen mussen. Zwar ist dies, wie 7.2, Beispiel 2, Bild 5 , zeigt, moglich; aber 2.B. in Bild 11 gehort keine der 4 Ecken des Grades 4 in der T11,g einem panisochromem Eckenpaar an.

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Vierfarbensatz aquivalent einem Satz der Panisochromie

Wie im Anfangs- und Schluflsatz von Abschnitt 3 und im Anfang von 6 gesagt, ist der Vierfarbensatz bewiesen, wenn zu den Ergebnissen in 2 bis 5 noch der Beweis hinzukommt, dafl keine Figur F,, existiert, die den 4

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Randbedingungen (20) genugt. Diese Situation soll jetzt weitergefuhrt werden, indem die Nichterfillbarkeit von (20) durch eine Figur F,,als ein Satz aus dem Gebiet der Panisochromie formuliert wird. Hierzu sind die Bedingungen (20) in ihren Modellierungen vor den Blick zu riicken: 2 1, besagt: Identifiziert man in 1. Die erste Ungleichung in (20),n

F,, die Randecken B I D (siehe Bild l),so entsteht die (bereits wegen der Induktionsvoraussetzung (siehe 4, Eingangssatz)) vierfarbbare T,,-1. Die zweite Gleichung in (20) besagt: Die T,,-l hat das panisochrome Eckenpaar A , C. 2. Die zweite Ungleichung in (20) besagt: Wird in einer beliebigen, (20) genugenden Figur F,, die Randkreisdiagonale ( B ,0)eingefiigt, so ist eine vierfarbbare Triangulation T; entstanden, die wegen der ersten Gleichung in (20) ebenfalls in A , C ein panisochromes Eckenpaar hat. Und zwar liegt ein panisochromes Eckenpaar des Abstandes 2 als Gegeneckenpaar der Kante ( B , D ) vor. (Bemerkung: Im Vergleich zu der weniger elementaren Struktur der Beispiele 7 , 4 oder 5 , sowie auch der in 7.3 geschilderten offenen Fragen (von deren keiner hier jetzt eine Antwort gebraucht wird) wird man die Situation der beiden soeben vor den Blick geruckten Triangulationen T,,-1 und T; mit dem panisochromen Eckenpaar A,C in jeder der beiden als eine recht einfache Situation aus dem Gebiete der Panisochromie bezeichnen konnen.) Nur geht j a TU,1 aus T; durch Kontraktion der Kante ( B , D ) hervor; und T; geht durch Wiederprotrahieren der Kante ( B , D ) mittels Aufschlitzens des Kantenpaares ( A , B = D,C) und Einsetzens der Kante ( B ,0)hervor. 8.1. Wer sich mit Farbungen von Triangulationen befaBt, in denen einer Kante ein panisochromes Eckenpaar gegenuberliegt, und wer dabei - ohne jeden Gedanken an das Vierfarbenproblem - auch das Kontrahieren dieser Kante in sein Spielen einbezieht, wird leicht gewahr, daB mit der Kante stets auch die Panisochromie des (dabei in seinen topologischen Daten mitveranderten) Eckenpaares verschwindet. Und so mag er wohl bald formulieren die Vermutung 8.1. In einer Triangulation T, liege einer Kante ( B , D ) das Eckenpaar A , C gegeniiber. Durch die Kontraktion von (B,D) zur Ecke B = D gehe T,, in T,-1 uber. A , C sei in T,, ein panisochromes Eckenpaar; dann ist A , C kein panisochromes Eckenpaar in T,-1. In Vermutung 8.1 wird nicht vorausgesetzt daB jedes Dreieck von T,, oder von TU-1eine der Dreiecksflachen ist. Aber es wird vorausgesetzt daB

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T,- ( B , D ) ein F, ist. Ein Beweis, der aus dieser Vermutung ein Theorem macht, beweist zugleich die Nichtexistenz von Figuren F,,die den Bedingungen (20) geniigen. Damit erganzt ein Beweis der Vermutung zugleich das in Abschnitt 2 bis 5 Gebrachte zu einem Beweis des Vierfarbensatzes. 8.2. Wer indessen vom Vierfarbensatz ausgeht und dann die durch Loschen

einer Kante aus den Triangulationen T,, v 2 5 , entstehenden Figuren F, daraufhin befragt, ob unter ihren Farbungen wohl immer wenigstens 3 mit paarweise verschiedenen Randfarbungen vorkommen, mi t anderen Worten, ob wohl(l2) universe11fur sie gilt, der findet nach Erledigung aller ubrigen Moglichkeiten z.B. durch die in Abschnitt 2 bis 5 aufgezeichneten Begrundungen, dai3 als gesamter Restfall nur genau eine Begriindung dafiir noch aussteht, dafi keine Figur F, existiert, die die Verneinung von (12) in Gestalt der Bedingungen (20) erfullt. In Zusammenfassung der Uberlegungen von 8.1 und 8.2 gilt daher Satz 8.2. Die Vermutung 8.1 ist Equivalent dem Vierfarbensatz.

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Annals of Discrete Mathematics 41 (1989) 255 -266

0 Elsevier Science Publishers B.V. (North-Holland)

The Existence Problem for Graph Homomorphisms P. Hell School of Computing Science Simon Fraser University Burnaby, B.C., Canada

J. Neietiril Department of Applied Mathematics Charles University Prague, Czechoslovakia Dedicated t o the memory of G. A. Dirac Graph homomorphisms have been investigated since the nineteen sixties [l, 2, 4, 7-10, 12-14, 16-20], and have enjoyed renewed interest lately, because of their relation to grammar interpretations [16]. (A different notion of graph homomorphism was investigated by Dirac, Wagner, and others, [3, 211.) A homomorphism G -+ H is a mapping of the vertex set of G to the vertex set of H such that adjacent vertices have adjacent images. Because a homomorphism c: G -+ I<, is just an n-coloring of G, a homomorphism G + H is also called an H-coloring of G. The following H-coloring problem has been the object of recent interest: Instance : A graph G. Question : Is it possible to H-color the graph G? Several authom have studied the complexity of the H-coloring problem for various (families of) fixed graphs H [l, 2, 17, 181. Since there is an easy H-colorability test when H is bipartite, and since all other examples of the H-colorability problem that were treated (complete graphs, odd cycles, complements of odd cycles, etc., [l,17, IS]) turned out to be NF-complete, the natural conjecture, formulated in several sources [17, 181 (including David Johnson's "P-completeness column [12]) asserts that the H-coloring problem is NP-complete for any non-bipartite graph H. We have proved this conjecture, and will publish a full proof elsewhere [ll]. Here, we outline the ideas behind our proof.

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Introduction

Let G and H be graphs. A homomorphism f: G + H is a mapping f of V(G) to V(H) such that f(g) and f(g’) are adjacent vertices of H whenever g and g‘ are adjacent vertices of G. Since a homomorphism c: G I(, is just an n-coloring of G, the term H-coloring of G has been employed to describe a homomorphism G -+ H. Homomorphisms and H-colorings have been studied in various contexts [ l , 2, 4, 7-10, 12-14,16-201; in particular for their relation to grammars and interpretations [16]. Here, we study the H-coloring problem, i.e., the decision problem: “Is a given graph G H-colorable?” Clearly, each H-coloring problem is in the class NP. It is easy to see that if H is a bipartite graph then G is H-colorable if and only if G is 2-colorable. For some non-bipartite graphs H, the H-coloring problem is NP-complete. Obviously, this is the case of K’,-coloring; moreover, Czk+l-coloring is NP-complete according to [17, 181, where several other NP-completeness results of this type were obtained for special classes of graphs. (Also see [l,2, 7, 12-14,191.) We proved that the H-coloring problem is NP-complete for any non-bipartite graph H. This was conjectured in [17]; cf. also [18] and [12]. Our proof is interesting not so much for the reductions we use, which are similar t o those previously used, but rather for the intricate interplay of the various graphs, some quite complex, which must be employed in these reductions. These complications may help to explain why the problem had previously resisted solution. Thus, we have the following theorem. -+

Theorem 1. If H is bipartite then the H-coloring problem is in P . If H is not bipartite then the H-coloring problem is NP-complete.

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A. The indicator construction Let I be a fixed graph, and let i and j be distinct vertices of I such that some automorphism of I maps i to j and j to i. The indicator construction (with respect t o ( I , i , j ) ) transforms a given graph H into the graph H* defined t o have the same vertex set as H and to have as the edge set all pairs hh’ for which there is a homomorphism of I to H taking i to h and j t o h’. Because of our assumption on I, the edges of H will be undirected.

Lemma 1. If the H*-coloring problem is NP-complete then H -coloring problem.

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Before introducing the next construction we need to review the following concepts [9, lo]: If H is a subgraph of H' then a retraction of H' to H is a homomorphism T : H' -P H such that r ( h ) = h for all vertices h of H. A graph is a cow (or minimal graph [3]) if it does not admit a retraction to a proper subgraph; equivalently, H is a core if it does not admit a homomorphism to a proper subgraph. It is easy to see [lo, 31 that every graph H' contains a unique (up to isomorphism) subgraph H which is a core and admits a retraction T : H' + H ; we call H the core of H'. Note that if H is a core of H' then there are homomorphisms H 4 H' (the inclusion) and H' + H (a retraction); thus, G is H'-colorable if and only if it is H-colorable. This allows us to restrict our attention to cores H. (The core of a bipartite graph H is K2; the core of a graph H with loops is one loop. Now, it follows that in both cases testing for H-colorability is easy. In particular, this proves the first half of Theorem 1.)

B. The sub-indicator construction

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. ..

. ..

Lemma 2. Let H be a core. If the fi-coloring problem is NP-complete then so is the H -coloring problem.

C. The edge-sub-indicator construction Let J be a fixed graph with a specified edge j j ' and t specified vertices kl, k2, . . . , kt, such that some automorphism of J keeps each vertex ki fixed while exchanging the vertices j and j ' . The edge-sub-indicator construction transforms a given core H with t specified vertices h l , h2, , ht into its subgraph determined by those edges hh' of H which are the images of the edge jj' under retractions of W (defined as in B) to H. Note that are again undirected. because of our assumption on J , the edges of

...

a

Lemma 3. Let H be a core. If the I?-coloring problem is NP-complete then so is the H -coloring problem.

P. Hell and J . Nesetfjl

258

Special cases of the first two constructions have been used in [17, 181. The third construction is somewhat more cumbersome, but is crucial for our proof. To prove the NP-completeness of the H-coloring problem for a particular non-bipartite H, we may appeal to an indicator construction, and reduce the problem to proving the NP-completeness of the H*-coloring problem; we shall always choose I, i , and j in such a way that H*, in addition to being undirected, has no loops, contains all the edges of H, and at least one more edge. Or, we may appeal to a sub-indicator construction, and reduce the problem to proving the NP-completeness of the g-coloring problem; we shall always choose J, j , and the ki's so that k is still non-bipartite, but has fewer vertices than H. (We cannot use the edge-sub-indicator construction by itself because it reduces the number of edges and thus counteracts the effect of the indicator construction. However, we shall only use it when it can be immediately followed by a sub-indicator construction.) Hence, let H be a non-bipartite graph for which the H-coloring problem is not NP-complete and such that the H'-coloring problem is NP-complete for any non-bipartite H' 1. with fewer vertices than H, or 2. with the same number of vertices as H, but with more edges. Clearly, if Theorem 1 does not hold then such an H must exist. Moreover, since each K,-coloring problem is NP-complete, H has n > 3 vertices and m < edges. We shall proceed to derive a number of structural properties of the graph H, which will eventually imply that it cannot exist, thereby proving the theorem. It follows from our earlier remarks that H is a core.

(3

3

The structure of triangles

Our first goal is to prove that each edge of H belongs to a unique triangle. We do this in a sequence of steps: ( A l ) H contains a triangle.

(A2) H contains no K 4 . (A3) Each vertex of H belongs to a triangle. (A4) Any two vertices of H have a common neighbor.

(A5) There is no homomorphism S + H . (Figure 1.) (A6) H contains no K,' . (Figure 1.) (A7) Each edge of H belongs to a unique triangle.

259

The Existen ce Problem for Graph Homomorphisms

S

Figure 1. This now follows from (A4) and (A6). In particular, the neighborhood of any vertex of H induces a perfect matching. Our next objective is to investigate the interconnections among the triangles of H .

(A8) In H , any triangle abc and edge cc' (c' a subgraph .'2 (Figure 2.)

# a, b,

c) are contained in

3

U Figure 2.

(A9) For any homomorphism U are adjacent in H .

4

H (Figure 2), the images of i and j

(A10) In H , any two triangles, abc and ab'c', are contained in a subgraph P (= K3 x K 3 ) . (Figure 3.)

P. Hell and J . Negetfil

260

U

b

C

P Figure 3.

4

The structure of squares

From now on, we base our considerations on a fixed vertex r , chosen to be a vertex of maximum degree in H. By (A7), the neighborhood of T consists of k (say) disjoint edges ala;, u2uk, .. , ,a&. Let R denote the subgraph of H induced by the remaining vertices V ( H )- { r ,a l , a ; , . ,ak,a i } ; according t o (A4), each vertex z of R is adjacent to some ai. By (A7), each edge uv of R belongs t o a triangle uvw; if w = ai, we label the edge uv by aj. (Of course, the whole triangle uvw could belong to R, in which case none of the edges uv, uw, and vw would be labelled.) If v in R is adjacent to some a; then the edge a;v lies in a triangle ajvw where w is also in R; hence, v is incident with an edge labelled ai. Note that (A6) implies that each edge obtains a t most one label, and that two edges of the same label cannot intersect or have two of their endpoints adjacent. We shall state this briefly as follows.

..

(Bl) Two edges of the same label cannot be incident or adjacent.

A similar proof to that of (Bl) shows that no vertex can belong to both an edge labelled ai and an edge labelled a:. For any i # j, we can apply (A10) to the two triangles TU;U: and raja: t o conclude that there is in R a four-cycle with edges consecutively labelled ai, a j , a:, and a;. (Figure 4.) Such a four-cycle will be called a square; there may, of course, be fourcycles in H (or even in R) which are not “squares.” There are a t least (:) squares, and they may intersect. Their structure is analyzed in this section, and it leads to a proof of Theorem 1. (B2) The squares are edge-disjoint.

The Existence Problem for Graph Homomorphisms

26 1

T

Figure 4.

(B3) H is2k-regular. It follows from the proof of (B3) that each labelled edge belongs to a square (thus to a unique square), and that H contains exactly one square labelled a;aja:ag for each i # j .

(B4) Each vertez of R belongs to a square. (B5) The subgraph RL of R formed by the labelled edges is bipartite. (B6) Each component of RL is complete bipartite. We conclude from ( B l ) and (B6) that each label aj (or a:) occurs at most once in a component of RL. It also follows from (B6) and (A7) that an unlabelled edge of R joins two vertices of diferent components of RL. Note that the unique triangle containing an unlabelled edge of R has both other edges unlabelled (and in R).

(B7) Suppose x y z is a triangle in R with all three edges unlubelled, x is incident with an edge labelled air yv is any edge of R such that v is incident with an edge hbelleda:, and z is incident with an edge labelled a j . Then v is also incident with an edge labelled a j . If uv is a n unlabelled edge of R, and if u is incident with an edge labelled ai while v is incident with an edge labelled a:, we mark the edge uv by the index i. Note that each edge obtains at most one mark.

( B 8 ) Every unlubelled edge of RL is marked by exactly one index i. (B9) Each vertex of R belongs to at least two squares.

P. Hell and J. NeSetNl

262

(B10) H does not exist. The proofs of (Al)-(AlO) and (Bl)-(BlO) involve a number of indicators, sub-indicators, and edge-sub-indicators, some of which are illustrated in Figure 5.

I kl

A, J

J

I

I Figure 5.

Remark. The situation is less clear for directed graphs. Even a conjecture anticipating which H-coloring problems are polynomial and which are NPcomplete does not suggest itself. Only a few results are known [2,17]. There are some simple digraphs H (paths, cycles, transitive tournaments, etc.) for which polynomial H-coloring algorithms exist [2, 171. Typically, they make

The Existence Problem for Graph Homomorphisms

263

use of results of the following type: there is a homomorphism D + H if and only if there is no homomorphism H’ + D (for some fixed digraph H’, depending on H) [2, 7, 13, 191. (These results may be viewed as proving that H-colorability is in NP n co-NP, and are, in some sense, prototype results of this type; this line of study is pursued in [13,19].) There are also a few classes of digraphs H with NP-complete H-colorability problems [17]. We note in passing that many NP-complete H-coloring problems may be produced by using the construction (of G*)from the proof of Lemma 1, with a suitable choice of the indicator I. Specifically, let (I,i, j ) be a digraph indicator such that for graphs G and H, there is a homomorphism G -+ H if and only if there is a homomorphism of digraphs G*+ H*. Such indicators are called ‘strongly rigid’ in the terminology of [8]; they can be constructed to satisfy many additional properties - assuring for example that H* is an acyclic, or even balanced, digraph. (A digraph is acyclic if it has no directed cycles; it is balanced if it has the same number of forward and backward arcs on any cycle.) In any event, if D = H* for such an I and a non-bipartite H then the D-coloring problem is also NP-complete. Thus, there are balanced (and hence also acyclic) digraphs H for which the H-coloring problem is NP-complete. Acyclic digraphs H with NPcomplete H-coloring problems were also constructed by S. Burr, and by W. Gutjahr and E. Welzl (personal communications). Finally, it should be mentioned that any digraph H such that the ‘symmetric part’ H s of H (all pairs uv of vertices for which both uv and vu are arcs of H) is a non-bipartite graph also results in an NP-complete H-coloring problem. This is an easy corollary of Theorem 1 and the observation that a graph G admits a homomorphism G --+ H s if and only if it (viewed as a digraph) admits a homomorphism G -+ H.

Acknowledgements We are grateful to David Kirkpatrick and Emo Welzl’for many inspiring conversations and helpful ideas.

P. Hell and J . NeSetfil

264

References [l] M. Albertson, P. Catlin and L. Gibbons, “Homomorphisms of 3-chromatic graphs 11,” Congressus Numerantium 47 (1985), 19-28. [2] G. Bloom and S. Burr, “On unavoidable digraphs in orientations of graphs,” J . Graph Theory 11 (1987), 453-462. [3] G. A. Dirac, “Homomorphism theorems for graphs,” Math. Ann. 153 (1964), 69-80. [4] W. D. Fellner, “On minimal graphs,” Theoretical Computer Science 17 (1982), 103-110. [5] M. R. Garey, D. S. Johnson and H. C. So, “An application of graph coloring to printed circuit testing,” IEEE Trans. Circuits and Systems 23 (1976), 591-598. [6] M. R. Garey and D. S. Johnson, Computers and Intractability, Freeman (1979). [7] R. Haggkvist, P. Hell, D. J. Miller and V. Neumann-Lara, “On multiplicative graphs, and the product conjecture,” Combinatorica 8 (1988), 71-81. [8] 2. Hedrlin and A. Pultr, “Symmetric relations (undirected graphs) with given semigroups,” Monatsh. f. Math 69 (1965), 318-322. [9] P. Hell, “Absolute planar retracts and the Four-color conjecture,” J. Combin. Theory (B) 17 (1974), 5-10.

[lo] P. Hell and J. NeSet81, “Cohomomorphisms of graphs and hypergraphs,” Math. Nachr. 87 (1979), 53-61. [ll] P. Hell and J. NeSetiil, “On the complexity of H-coloring,” Simon Fraser University, School of Computing Science Technical Report T R 86-4. To appear in J . Combin. Theory (E). [12] D. S. Johnson, “The NP-completeness column: An ongoing guide,” J. of Algorithms 3 (1982), 89-99.

[131 P. KomArek, “Some new good characterizations of directed graphs,” Cas. pro Pest. Mat. 51 (1984), 348-354.

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[14] L. A. Levin, “Universal sequential search problems,” Problems o f Information Transmission 9 (1973), 265-266. [15] F. T. Leighton, “A graph coloring algorithm for large scheduling problems,” J . Res. Nat. B. Standards 84 (1979), 489-496. [16] H. A. Maurer, A. Salomaa and D. Wood, “Colorings and interpretations: a connection between graphs and grammar forms,” Discrete Applied Math. 3 (1981), 119-135. [17] H. A. Maurer, J. H. Sudborough and E. Welzl, “On the complexity of the general coloring problem,’’ Information and Control 51 (1981), 123-145. [18] J. Neietfil, “Representations of graphs by means of products and their complexity,” MFCS (1982), 94-102. [19] J. Neietfil and A. Pultr, “On classes of relations and graphs determined by sub-objects and factor-objects,” Discrete Math. 2 2 (1978), 287-300. [20] G. Sabidussi, “Graph derivatives,” Math. 2. 76 (1961), 385-401. [21] K. Wagner, “Beweis einer Abschwachung der Hadwiger-Vermutung,” Math. Ann. 153 (1964), 139-141.

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Annals of Discrete Mathematics 41 (1989) 267-280 0 Elsevier Science Publishers B.V. (North-Holland)

On Edge-Colorings of Cubic Graphs and a Formula of Roger Penrose F. Jaeger * LS D IMAG Grenoble, France

Dedicated to the memory of G. A . Dirac We study a formula, due to Roger Penrose, which gives the number T(G) of edge-3-colorings of a cubic connected plane graph G in terms of certain families of cycles of G. We present an equivalent formula which gives T(G)in terms of the set of embeddings of G on orientable surfaces. Then, using the theory of bicycles and leftright paths of Rosenstiehl and Shank, we obtain another equivalent formula, which refers only to algebraic properties of G. Finally, we prove the new formula for all cubic connected graphs, thus generalizing Penrose’s formula to non-planar graphs.

1

Introduction

The physicist Roger Penrose presented in 1969 a paper, entitled “Applications of negative dimensional tensors” [l],where he obtained a number of remarkable formulas for the number of edge-3-colorings of cubic connected plane graphs. As the title of the paper suggests, Penrose’s methods of proof are quite original, and they appear to be very powerful. We shall not discuss these methods here. T h e purpose of the present work is t o study and generalize one of Penrose’s formulas. In Section 2 we present this formula and give a n interesting interpretation in terms of the set of embeddings of a cubic connected plane graph on arbitrary orientable surfaces. We also *Full address: Laboratoire de Structures Discrktes, Institut IMAG, BP 68, 38402 St. Martin d’Hbres Cedex, France.

267

F. Jaeger

268

exhibit a counter-example to a natural extension of the formula to nonplanar graphs. In Section 3 we show how the formula can be expressed in purely algebraic terms by using the theory of bicycles and left-right paths of Rosenstiehl[2] and Shank [3]. Finally, in Section 4 we prove the validity of the algebraic version of the formula for arbitrary cubic connected graphs. The definitions not given here will be found in [4] or [5], and we consider only finite undirected graphs, which may have loops or multiple edges.

Presentation of the formula

2 2.1

Statement of the formula

Let G = (V,E ) be a connected cubic plane graph; T ( G )denotes the number of edge-colorings of G with 3 colors (edge-3-coZorings, for short). As is well known, the statement that T ( G )# 0 whenever G is bridgeless is equivalent t o the Four Color Theorem, and this is a rather powerful motivation for the study of non-trivial expressions for T(G). Let Y be a subset of V . Following [l],we shall associate to Y a set the additional amount of of cycles of G, which we shall denote by C(Y); formalization introduced here will be helpful in the sequel. Consider the algorithm A described below, which defines a cycle ( e l ,v1,. . . ,ek, W k ) of G (ei E E , vi E V , i = 1, ..., Ic): (u)

Select an edge el of G and choose a travel direction on e l .

( 6 ) Traveling on e; with the specified direction leads to the vertex

0;.

( c ) Perform a right turn at if vi belongs to Y , and a left turn otherwise; this leads on the edge ei+l with a specified travel direction.

( d ) If ei+l is equal t o el and the travel direction assigned to ei+l in step ( c ) is the same as the travel direction assigned to el in step ( a ) , stop. The resulting cycle is ( e l , q ,. . , e , , v ; ) .

.

( e ) Otherwise go to step ( 6 ) with i replaced by i

+ 1.

We define C ( Y )as the set of cycles of G which can be obtained from algorithm A , with the convention that two cycles corresponding to the same directed cyclic sequence of edges and vertices are identical. It is then easy to check that for each edge of G with a specified direction, there exists exactly one cycle of C(Y)which takes this edge in this direction.

On Edge-Colorings of Cubic Graphs

269

Example 1. If Y = V then C(Y)is the set of face-boundaries of G taken in the clockwise direction. Example 2. Let G be the graph of Figure 1 and assume that Y = {v}. Then C ( Y )consists of 3 cycles: (f,w,g, w,f,w, e, v), ( 9 ,w)and ( e , w).

Figure 1.

Example 3. Let G be the graph of Figure 2 and assume that Y = {v}. Then C(Y)consists of the unique cycle ( e , w,g, v , f,20, e , w , g , w,f,w).

Figure 2.

We may now state the formula we want to study ([l],p. 240):

For instance, in Example 2, IC(Y)I = 3 for all Y V and this immediately gives T ( G )= 0; in Example 3, IC(Y)I = 1 if IYI = 1, IC(Y)I = 3 otherwise, and both sides of ( Fl ) equal 6.

2.2

Interpretation in terms of embeddings on orientable surfaces

A theorem of Edmonds [6] allows t o represent any 2-cell embedding (ernbedding, for short) of a connected graph G on an orientable surface in terms

F. Jaeger

2 70

of rotations. For this purpose, each edge of G is considered as a pair of oppositely directed edges. For a vertex w of G, a rotation ofv is a cyclic permutation nu on the set of directed edges incident to v and directed away from w. A rotation system o f G is a family ( ru,w E V ) , where rVis a rotation of v. To every rotation system of G corresponds an embedding of G on some orientable surface; the face-boundaries of this embedding can be determined using the algorithm A introduced in the previous section, with the only difference that in step (c) the turn a t w; is performed in such a way that ei+l (directed away from wi) is the image under A, of ei (also directed away from w;). Conversely, every embedding of G on some orientable surface corresponds in this way to a unique rotation system. For a detailed treatment of this classical “permutation technique,” the reader can refer to [7]. Now, if G is a cubic connected plane graph, we may associate to each subset Y of V a rotation system n(Y)= ( A ~ w, E V ) such that A, geometrically corresponds to a counter-clockwise rotation around w if w E Y , and to a clockwise rotation otherwise. Let I ( Y ) be the embedding of G corresponding to n ( Y ) . It is easy to check that the set of face-boundaries of I(Y)is precisely C(Y).Let g(Y)be the genus of the embedding I ( Y ) (that is, the genus of the associated orientable surface). Then, by Euler’s formula:

Then Penrose’s formula ( F l ) becomes

or, equivalently

(- 1)Iy141-s(y).

T ( G )= YGV

Example 4. Let G be the complete graph on 4 vertices embedded in the plane. This graph has precisely two embeddings of genus zero (corresponding to the opposite rotation systems A(V) and n(B)), and each contributes 4 t o the right-hand side of (F2). By Euler’s formula, the other embeddings must be of genus 1, and their total contribution to the right-hand side of (F2) is -2. Thus, both sides of (F2) equal 6.

Remark. Since the correspondence I between subsets of V and orientable embeddings is one-to-one, (F2) can be viewed as a summation over the set of

On Edge-Colorings of Cubic Graphs

271

orientable embeddings of G (where embeddings corresponding t o opposite rotation systems are considered distinct).

2.3

Failure of a geometric generalization

Let G = ( V , E )be an arbitrary cubic connected graph and e = ( e,, o E V ) V , let A(Y,e) = ( A,, v E V ) be the be a rotation system of G. For Y rotation system of G such that A, = (e,)-' if o E Y , x , = e, otherwise. Let I(Y,e) be the embedding of G corresponding to n(Y,e), C(Y,e) be the set of face-boundaries of this embedding, and let g(Y,e) be its genus. Let

Thus, Penrose's formulas (Fl) and (F2) assert that T ( G ) = Q ( G , e ) whenever e defines a planar embedding of G. Note that the set of rotation systems of G is { n(Y,e) I Y G V }, with in particular e = n(8,e). Furthermore, it is easy to see that for all Y G V , V , A ( Y , A ( Z , ~=) A(Y ) Z , e ) (where denotes the symmetric Z difference of sets). Hence, we also have C(Y,n(Z,e)) = C(Y 2 , ~and ) g(y,~ ( 2 el), = g(Y 2,el. Consider now the rotation system e' = ~ ( 2 e)., Then

-+

+

"+"

+

Q(G,e')= YCV

YCV

and thus, setting Y' = Y

+ 2: (-1)Iy'+Zl4'-g(Y',Q)

Q(G,e') =

= (-l)lZ1&(G,e).

Y'GV

Hence, for any two rotation systems e and e', IQ(G,e)I = IQ(G,e')I, and we shall denote this number by Q(G). Now we may adapt an argument used by Penrose in [l]for the study of another formula to show that Q(G)is not in general equal to T ( G ) .For this purpose, consider the two drawings in Figure 3 of the complete bipartite graph K 3 , 3 in the plane.

F. Jaeger

272

X

X

t

t

Figure 3. Note that the two drawings define the same graph G on the vertexset {x,y, z, t, u , v}. Let e (respectively @’) be the rotation system of G which geometrically corresponds to the clockwise rotation around each vertex in the plane drawing of Figure 3(a) (respectively 3 ( b ) ) . Then obviously Q(G,e) = Q(G,e’) because the two drawings are essentially identical (this idea could be made more precise with the notion of isomorphism of maps on orientable surfaces). On the other hand, e and e’ coincide on exactly 3 vertices x,z , u, that is, e’ = n({y, t , v}, e ) ,and hence Q(G,e ) = -Q(G, e’). It follows that Q(G,e) = Q ( G ,e‘) = 0 and hence Q ( G )= 0 # T( G) . In the next section we present an algebraic expression of Penrose’s formula (Fl).

3 3.1

An equivalent algebraic form Spaces and graphs

Let X be a finite set; P ( X ) denotes the set of subsets of X . For A , B in P ( X ) , we denote by A t B the symmetric difference of A and B . For A in P ( X ) and a in GF(2) we define a A by: a A = 0 if Q = 0, a A = A if a = 1. Then P ( X ) together with the two operations defined above is a vector space over GF(2). We shall call any subspace of P ( X ) a space on X. For instance, a family ( Ai, i E I ) of subsets of X generates a space on X which we denote by ( Ai, i E I ) . In particular, for A C X , ( { a } , a E A )

On Edge-Colorings of Cubic Graphs

273

is a space on X which we identify with P ( A ) . We define the scalar product A B E GF(2) of two subsets A , B of X as equal to 0 if [ An BI is even and equal to 1 otherwise. If F is a space on X , then the set { A E P ( X ) I VB E F,A . B = 0 ) is alsoa spaceon X which is denoted by F’. For instance (X)’ is the space of subsets of even cardinality of X . It is known that (F’)’ = F,and 3 and F1 are said to be orthogonal spucas on X . Let G = ( V , E ) be a graph. The boundary of an edge e with ends v, v’ is d ( e ) = {v} (0’) (so that d ( e ) = 8 if e is a loop). For F E the boundary of F is d ( F ) = C e E F a ( e ) . Thus, 0 is a linear mapping from P ( E ) to P ( V ) . The kernel of 8, denoted by C(G),is the cycle space of G. For S C V , the set of edges of G with exactly one end in S is called the cocycle o f S and is denoted by w(S). Since w ( S ) = ~,,,o({v}), w is a linear mapping from P ( V ) to P(E). The image of o,denoted by K(G), is the cocycle space of G. It is easy to show that C(G) and K(G) are orthogonal spaces on E . The space C(G)n K(G) is called the bicycle space of G and is denoted by B(G). Its elements are called bicycles of G (see [2], [3]).

+

3.2

s

Bicycles and left-right paths

Let G = ( V , E ) be a connected plane graph. Consider algorithm A’ below which defines a cycle ( e l ,v1, . . . ,ek, vk) of G (ej E E , vi E V , i = 1, . .. ,k):

( a ’ ) Select an edge el of G, choose a travel direction on el and a turning behavior for el which can be R (right) or L (left). (6’) Traveling on ej with the specified direction leads to the vertex w;.

v; (that is, enter the rightmost edge) if current behavior is R, and a left turn otherwise; this leads on edge ei+l with a specified travel direction; change the turning havior (that is, the behavior for ej+l is chosen as different from behavior for e i ) .

( c ’ ) Perform a right turn at

the the bethe

(d’) If ei+l is equal to e l , and the travel direction and turning behavior assigned to ei+l in step (c‘) are the same as the corresponding parameters assigned to el in step (a’), stop. The resulting cycle is

( e l 3 v1 (el)

* * * 3

ei, vi).

Otherwise go to step (b’) with i replaced by i

+ 1.

F. Jaeger

274

The cycles of G produced by algorithm A’ are called the left-right paths of G. They were first studied by Rosenstiehl[2] and Shank [3]. Let us adopt the convention that two left-right paths corresponding to the same directed cyclic sequence of edges and vertices, or to two opposite such sequences, are identical. Let then L R ( G ) = {PI,. , ,P p } be the set of left-right paths of G. It is easy to check that each edge of G appears twice in the elements of L R ( G ) ; to be more precise, it appears either twice in one of the Pi’s and not in the others, or once in two different Pi’s and not in the others. In fact, it can be shown that the Pi’s are the face-boundaries of an embedding of G on a (not necessarily orientable) surface. Let us call the set of edges which appear exactly once in P; the core of Pi. It is clear that the core of P; belongs to C(G). By considering the geometrical dual G* of G, it is easy to see that, similarly, the core of Pi belongs to K(G) (this is because LR(G*) can be identified with L R ( G ) while C(G*)can be identified with K(G)). Hence, the core of Pi is a bicycle of G. The following stronger result is proved in [2] and [3]: The cores of any T - 1 of the Pi’s form a basis of the bicycle space B(G). We shall retain the following corollary:

.

Proposition 1. Let G be a connected plane graph with exactly r left-right paths. Then dim B(G) = T - 1. We now use this result to derive a new expression for Penrose’s formula (Fl).

3.3

An algebraic formula equivalent to ( F l )

Let G = ( V , E ) be a connected plane cubic graph. For F 2 E , we denote by G : F the graph obtained from G by subdividing each edge of F , that is by replacing each edge of F by a path of length 2.

Proposition 2. For all subsets Y of V , IC(Y)I = ILR(G:E - w(Y))I. Proof. Let Y be a subset of V . It is clear that the graph G : E - w(Y) is bipartite. More precisely, we may color the vertices of this graph with two colors R and L in such a way that each edge has one end of each color, with the vertices of degree three corresponding in G to the vertices of Y (respectively V - Y) colored R (respectively L ) . Consider now a left-right p a t h P = ( e l , q , ..., e k , v k ) ofG:E-w(Y). Wenote that in theexecution of algorithm A‘ of Section 3.2 which produces P , either for all i = 1, . . . , k, the turning behavior a t vi is equal to its color, or for all i = 1, . .., k,

275

On Edge-Colorings of Cubic Graphs

the turning behavior a t wi is distinct from its color. Moreover, if the sequence ( e l ,~ 1 , .. . ,e k , w k ) is in one of these situations, the opposite directed cyclic sequence ( e l ,vk,. . .,e2, w1) is in the other situation. We recall our convention that two opposite sequences correspond to the same left-right path. Hence, we may identify the set L R ( G : E - w ( Y ) ) with the set of face-boundaries of the embedding associated to the rotation system which geometrically corresponds t o a counter-clockwise plane rotation a t vertices colored R and to a clockwise rotation at vertices colored L . Finally, it is immediate, using the straightforward correspondence between this embedding of G : E - w ( Y ) and the embedding I ( Y ) of G defined in Section 2.2, that LR(G : E - w ( Y ) ) is in one-to-one correspondence with the set C ( Y ) 0 of face-boundaries of I ( Y ) . This completes the proof. Now, using Propositions 1 and 2, the formula ( F l ) becomes (-1)

T ( G )= (3)i'"'

IY 121+dim &3( G:E-w( Y))

YGV

Since G is cubic, and the graph obtained from G by identifying the vertices of V - Y into a single vertex has an even number of vertices of odd degree, IYI = Iw(Y)I (mod 2) for all Y V . Thus, the formula (Fl) becomes 1 31VI-1

T ( G )= ( 5 )

C(-1)1

w(Y)l2dimB(G:E- w ( Y ) )

YCV

Finally, we observe that for K E K(G) there exists exactly two subsets of V whose cocycle equals K. Hence, the formula ( F l ) is equivalent to the following 1 ;Ivl-2

T ( G )= ( 2 )

c

(-1)

IKl2dim B(G:E-K)

(F3)

KEK(G)

In the next section we prove that this formula holds for all cubic connected graphs G.

4

The main result

Theorem. Let G z (V,E ) be a cubic connected graph, and let T ( G ) be the number of edge-3-colorings of G. Then

W )= ( 12 )W l - 2

c

(-1)

K€IC(G)

IK12dim &3(G:E-A')

F. Jaeger

276

For a fixed K in K ( G ) ,let us first evaluate dimB(G : E - K) in terms of spaces on E . Denote by E' the edge-set of G : E - K, and let q5 be the unique linear mapping from P ( E ) to P(E')such that: for e in K , 4({e)) consists of the edge of G : E - K corresponding to e; and for e in E - K , 4({e)) consists of the pair of edges in series in G : E - K corresponding to e. It is clear that the restriction of 4 to C(G) is an isomorphism from C(G) t o C ( G : E - K ) . Let BK(G)= + - l ( B ( G : E - I i ) ) . Thus,BK(G) is aspace on E isomorphic to B(G : E - K) and we may write

Now, let us present a convenient description of BK(G). We may write successively for a subset X of E

,XE f?jh'(G)iff $(X)E C(G : E - Ii)n K(G : E - I<) (by the definition of Bjh'(G)) iff

X E C(G) and 4 ( X ) E K(G : E - K ) (because C(G :E - I<) = 4(C(G)) )

iff X E C(G) and, for all Y in C(G), 4 ( X ) # ( Y ) = 0 (again because C(G :E - I<) = c$(C(G)), together with the orthogonality of C(G: E - IC) and K(G:E - K ) ) iff X E C(G) and, for all Y in C(G), IX n Y n KI = 0 (mod 2) (because, for all X and Y in C( G ) , n & ( Y )n 4 ( -~I<)! = 2 1 n~Y n ( E and I#(X) n 4 ( Y )n #(#)I = IX n Y n 11') )

I+(x)

K)I

iff X E C(G) and X n Ii E K(G) (by the orthogonality of C(G) and K(G) ).

To conclude BK(G)= { C E C(G) I C n Ir'E K ( G ) } .

(3)

On Edge-Colorings of Cubic Graphs

277

Note that a space of dimension d over GF(2) has cardinality 2d. Hence, by ( 2 )

KEK(G)

CEC(G) CnK€IC(G)

For every C in C(G),let Dc(G) = { K E K(G) I C n K E K(G)}. Then (4) becomes

S(G) =

c c

(-l)lKl.

(5)

CEC(G)KEVc(G)

Let us evaluate (for a fixed C in C(G))the sum

c

(-1)IKl.

KEZ)c(G)

Clearly, Dc(G) is (a)

Dc(G) Then

space on E . Two cases may occur:

(E)', that is, all elements of Dc(G) have even cardinality.

(-1)lK1 = IDc(G)I = 2di"vc(G). KEWG)

( b ) 3Ko E Dc(G)with ( K o l odd. Then I< E Dc(G) is of even cardinality iff Ii KOE Dc(G) is of odd cardinality and hence

+

c

(-1)IKl = 0.

KEVc(G)

Then ( 5 ) may be rewritten as

S(G) =

2hZ)C(G).

(6)

CEC(G) 'DC(G)E(E)

Now, we observe that an equivalent definition for Dc(G) is as follows

Dc(G) = { X E E I C n X E K(G),( E - C)n X E K ( G )}. This is in turn equivalent to

Dc(G) = ( K ( G )n P(C))@ (K(G)n P(E - C))

(7)

F. Jaeger

278

where the symbol @ denotes the direct sum of spaces. By the orthogonality of C(G) and K(G), K(G) n P(C)2 ( E ) * . Hence, Dc(G) C (E)* iff K(G) n P ( E - C) C (E)'-, that is, iff all cocycles of G disjoint from C are of even cardinality. This is equivalent to say that the graph obtained from G by contracting the edges of C is eulerian. Now, since G is a cubic graph, the element C of C(G)has this property iff it is an even 2-factor, that is a 2-factor, all components of which are even cycles. Let us denote by F(G)the set of even 2-factors of G. It then follows from (6) and (7) that S(G) =

C

2~IC(C)nP(C)+dimfC(G)nP(E-C)

(8)

C€3(G)

We now observe that if C E F(G), the space K(G) n P(C)is isomorphic t o the cocycle space of the graph obtained from G by contracting the edges of the perfect matching E - C. Since this graph is connected on 31Vl vertices, dimK(G) n P(C)= klVl- 1. Similarly, the space K(G) n P(E - C) is isomorphic to the cocycle space of the graph obtained from G by contracting the edges of C.Then, if p ( C ) denotes the number of connected components of C, dimK(G)nP(E-C) = p(C) - 1. Hence, (8) is equivalent to

2 ;IVI -l+P(C) -1

S(G)= C€3(G)

that is CE3(G)

Now let f: E -t { 1,2,3} be an edge-3-coloring of G. Then f-l({ 1,2}) is an even 2-factor of G which we denote by a ( f ) . Moreover, for every C in F(G), there exists exactly 2P(c) edge-3-colorings f: E + { 1,2,3} such that a(f)= C. Hence

The result now follows from (9) and (10).

0

On Edge-Colorings of Cubic Graphs

279

References [11 R. Penrose, “Applications of negative dimensional tensors,” Combinatorial Mathematics and its Applications, Proceedings of the Conference held in Oxford in 1969,Academic Press, London (1971), 221-244. [2]P. Rosenstiehl, “Bicycles et diagonales des graphes planaires,” Cahiers du C.E.R.O. 17(2-3-4) (1975), 365-383. [3] H.Shank, “The theory of left-right paths,” Combinatorial Mathematics 111,Lecture Notes in Math. 452, Springer-Verlag, Berlin (1975),42-54. [4]C. Berge, Graphes et Hypergraphes, Dunod, Paris (1974). [5]J. A. Bondy and U. S. R. Murty, Graph Theory with Applications, MacMillan, London (1976). [6]J. Edmonds, “A combinatorial representation for polyhedral surfaces,” Notices Am. Math. SOC. 7 (1960),646. [7]A. T. White, Graphs, Groups and Surfaces, North-Holland, Amsterdam, London (1973).

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Annals of Discrete Mathematics 4 1 (1989) 281-298 0 Elsevier Science Publishers B.V.

(North-Holland)

Longest &Paths in Regular Graphs H. A. Jung Department of Mathematics Technical University of Berlin

Berlin, FRG Dedicated to the memory of G. A . Dirac Some results about longest paths between given vertices in regular graphs are given. One result is that if a graph G is regular of valence d, 3-connected and has at most 3d - 1 vertices, then either G is a certain given graph, or G has a hamiltonian path joining any pair of vertices, except if G is bipartite and the pair of vertices belong to the same vertex class of G.

1

Introduction

It a well-known result of G. _-.Dirac [2], that a graph with IGl 2 3 vertices and minimum valence dmin has a hamiltonian circuit if [GI 5 2dmin. Moreover, ICI 2 min(lGI,2dmi,) for a longest circuit C in G if G is 2connected. A 2-connected graph G, which is regular of valence d, has a hamiltonian circuit if [GI 5 3d, as was shown by B. Jackson [ 5 ] . In [4] (Corollary 1.1), G. Fan proved ICI 2 min(lGI,3d) for a longest circuit C in G if G is 3connected and regular of valence d. In this paper analogous results for longest ab-paths, i.e., longest paths joining given vertices a and b, are obtained. A consequence of Theorem 1 below is the following. Corollary 1. Let G be regular of valence d and 3-connected. Further let e be an edge of G. If [GI 5 3d - 1, there is a hamiltonian circuit containing e. If G is 4-connected, there is a circuit C containing e such that ICI = [GI or ICI 2 3d - 2.

If G is regular of valence d and bipartite, where the edges join the sets V, and V,, then clearly Ihl = IVZl and ]PI < IGl for any ab-path P 281

H. A . J u n g

282

in G, whenever a, b E V1 or a, b E V2. We say in this case, that G has the hamiltonian property, if for a longest ab-path P in G one has IPI = (GI whenever a E V1 and b E V2, and IPI = [GI - 1 whenever a,b E VI or a, b E V2. A non-bipartite graph G is said to have the hamiltonian property if G is hamiltonian connected, i.e., ]PI = (GIfor any longest ab-path in G . Let G(d) be the graph, which is obtained from a complete graph with d vertices 21,z2,.. . ,zd by adding vertices q,22,.. . ,zd, y1, y2,. . .,yd-1 and joining each x; to z; and to each yj. The graph G ( d ) is d-connected and has 3d - 1 vertices, while a longest xi, xk-path in G ( d ) has 3d - 2 vertices. For G(3) see Figure 1.

Figure 1.

Theorem 1. Let G be regular of valence d . 1. I f G is 3-connected, [GI 5 3d ha rn iltonia n p r o p rty.

- 1 and G # G ( d ) , then G has the

2. If G is 4-connected, then G has the hamiltonian property or IPI 2 3d - 2 for any longest ab-path P in G.

A path P in a graph G is called dominating, if each edge of G is incident with a vertex of P . Theorem 2. Let G be regular of valence d . 1. If G is 4-connected and IGI is dominating.

5 4d - 5, then each longest ab-path in G

2. If G is 5-connected, then each longest ab-path P in G is dominating or IPI 2 4d - 7.

A by-product of the proofs is the following result.

Longest ab-Paths in Regular Graphs

283

Corollary 2. Let G be regular of valence d and 5-connected. If [GI 2 6d - 13, then /PI2 4d - 7 for any longest ab-path. Let GI, G2,. . . ,Gk be disjoint complete graphs on d vertices each, where 1 1 , ~ 2 , .. . , z k be k additional vertices. It is easy to distribute kd edges, each with one endvertex in { 2 1 , 2 2 , . . . ,zk} and the other in GI, G2,. . . ,Gk, so that a k-connected d-regular graph Gk,d arises. Clearly IGk,dl = kd k and lPl = (k - l)d k for any longest 2122-path in Gk,d; also a longest 2122-path is not dominating. For the proof of the above results, we need a result from [6] about longest ab-paths in arbitrary graphs. For other results on longest ab-paths see [l],[3], and [7].

k 5 d, and let

+

+

2 Non-trivial components In this section a 2-connected graph G and two distinct vertices a, b are fixed. If K , L are subgraphs of G or subsets of V ( G )let N K ( L ) denote the set of vertices of K ,which in G are adjacent to some vertex of L , further dK(v) = I N K ( ~ ) Ifor vertices v in G and e(K;L) = C V E L d ~ ( v )Thus . e(I<;L) is the number of edges joining K and L where the edges with both end vertices in K and L are counted twice. For KIn K 2 = 0 clearly 2)’ if e(K1 U K2;L) = e(K1; L ) e(K2; L). Set d(v) = dG(v) and w E NG(v’). Paths in G are considered as subgraphs of G. For vertices v and v‘ on a path P let P[v,w’] denote the subpath of P with terminal vertices 21and 0’. P ( v ,v’)(P[v,v‘), P(w,v’]) arises from P [ v ,v‘] by deletion of v and v’ (resp. v‘, v). A subgraph H of G is called a stronghold in G if H is complete and N G - H ( H ) = N G - H ( ~for ) all v E V ( H ) .

+

Proposition 2.1. Let P be a longest ab-path in G and H some component of G - P, which is not a stronghold in G. There is a vertex v in H such that 1. [GI 2 3d(v) if G is 3-connected,

2. [PI2 3d(v) - 2 if G is 4-connected,

3. IGI 2 4d(v) - 4 if G is 4-connected and IHI 2 3, and 4. [PI2 4d(v)

-7

if [HI 2 3 and G is 5-connected.

H . A. Jung

284

Proof. See Theorem 3 and Theorem 4 in [6].

0

In the following two lemmas P is a longest ab-path and H a component of G- P. Label N p ( H ) = { z 1 , z 2 , . ,z,} according to the order on P from a to b. Abbreviate IHI = h,Pi = P(zi,zi+l) for 1 _< i < 8, Po = P [ u , z ~ ) and P, = P(z8,b] (possibly Po = 0 or Pa = 0). For z on P let z+ and zdenote respectively the successor and the predecessor of z on P.

..

Lemma 2.2. Let the vertices z on P ( z j , . ~ j + l ]and z‘ on P(zk, z k + l ] , where 1 5 j < k 5 s , be joined by a path Q in G - H such that V ( Q )n V ( P ) = { z , z ’ } . Then (P(zj,z)l IP(zk,z’)( = 191- 2 E with E 2 1. Moreooer, E 2 h if H is hamiltonian connected and N H ( Z j , z k ) = V ( H ) .

+

+

Proof. Let L = L[zj,zk]b e a path in G, such that V ( L ) - { z j , z k } E V ( H ) and ILI 2 3. If N ~ ( z j , z k = ) V ( H ) and H is hamiltonian connected one can choose L , such that ILI = h+2. Then P [ u ) z ~L], ,P[Q,21, Q and P[z’,b] 0 form an ab-path P‘. From IP’l 5 [PIthe claim follows.

+

+

Lemma 2.3. Let H be a stronghold in G, lPjl 5 h 1 and IPkI 5 h 1 where 1 5 j < k 5 s - 1. There is at most one path Q joining a vertex on Pj to a vertea: on Pk such that IV(Q) n V(P)I = 2 . Moreover, IQI = 2 for such a path Q . Proof. Let Q = Q[z,z’] be a path in G according to the hypothesis. Then IP(zj,z)l IP(zk,z‘)I 2 h -i-191- 2 by Lemma 2.2. By symmetry, IP(z,zj+r)l IP(z‘,zk+i)l 2 h I&[ - 2. Since lPjl 5 h 1 and IPkI 5 h+ 1combination of the twoinequalities yields 2h 2 2h+2(1&1-2), hence I&[ = 2. Now let e = [v,w’] be a second edge, such that w E V ( P j ) and v’ E v ( P k ) . Without loss of generality assume v # z, say v E V ( P ( z j , z ) ) .The equalities IP(Zj,z)l IP(ak,z‘)l = h and IP(zj,v)l IP(zk,v’)l = h yield v’ E V ( P ( z ’ z, k + l ) ) . Consider the ab-path P’ defined by

+

+

+

+

+

+

For the remainder of this section, we assume that G is regular of valence

d.

Longest ab-Paths in Regular Graphs

285

Lemma 2.4. Let P be a longest ab-path in G and H a component of G - P , such that [HI 2 3. 1. If G is 4-connected, then [GI 2 4d - 4. 2. If G is 5-connected, then [PI 2 4d

- 7.

.

Proof. let N p ( H ) = ( Z I , Z ~ , .. ,z8} and P; (0 5 i 5 s) be defined as above. By Proposition 2.1 it remains to consider the case, in which H is a stronghold in G. Then /Pi[ 2 1H1 for 1 5 i < s. Abbreviate [Ell = h, N p ( H ) = 2 and 6 = /Polt IPS[ C:ii(IP;l- h). Clearly d = s t h - 1. Since IPI = (s - l)(h 1) t 1 6 = 4(s h - 1) ( s - 5)(h- 3 ) - 11 6 and [GI 2 4(s h - 1) t (s - 4)(h - 3 ) - 8 6 the claims follow if b 2 4. Let an integer i with 1 5 i < s and lP;l = h be given. The path P i , obtained from P by substitution of P; with a hamiltonian path of H , is a longest ab-path. By Proposition 2.1 we may assume that the component H ; of G - Pi, which contains Pi, is a stronghold in G. We then have V ( H ; )= V ( P i ) ,since a vertex in V ( H i )- V ( P ; )would be in V ( G )- V ( P ) and adjacent to subsequent vertices of P;,contrary to the maximality of P. Assume that some vertex on Pi is joined to some I E V ( P j ) ,where 0 5 j 5 s and j # i. If 1 5 j < s then 'z E N p ( z ) yields IP(zj,z)I 2 h by Lemma 2.2; symmetrically IP(z,zj+l)l 2 h, hence lPjl 2 2h 1 2 h + 4. If j = s then also IP(zj,z)l 2 h, hence IPS[2 h+ 1 2 4. Similarly [Pol 2 4 if j = 0. Assuming 6 5 3, we have seen that each vertex of Pi,where lP;l = h, is adjacent t o each z E N , ( H ) . If there exist at most two integers j , such that 1 5 j < s and lPjl > h, then s(s h - 1) = e ( 2 ; G )2 ( s - 2)hs 4 and hence 0 2 (s - 3 ) ( h - 1) - 2 t a contradiction. Therefore there are 3 integers j , such that 1 5 j < s and lPjl > h. Then 6 = 3, and for such = 0. an integer j we must have lPjl = h 1, further lP0l = ]Pal If s 2 5 then s(s h - 1) 2 e ( 2 ; G ) 2 (s - 3)hs 6 and hence 0 2 (s - 4)(h - 1) - 3 !, a contradiction. Thus 2. is proved. Assume [GI 5 4d - 5 and s = 4. Since [PI h 2 4d - 8 t 6 = 4d - 5 , we have also V ( G ) = V ( H ) U V ( P ) . If lPjl = lPkl = h 1, there is, by Lemma 2.3, a t most one edge, which joins Pj and Pk. A vertex on Pj, which is not incident with such an edge, has at least s - 1 neighbours in 2. Therefore e(Pj;2 ) 2 ( h l)(s - 1) - 2. A count of e ( 2 ; G ) yields 0 s(s h - 1) 2 (s - 3)hs 3(h l)(s - 1) - 6, a contradiction.

+

+

+

+

+

+

+

+

+

+

+

4, +

+

+

+

+

+

+

+

+ +

Lemma 2.5. Let P be a longest ab-path in G and H a component of G- P , such that [HI 2 2. If G is 3-connected, then \GI 2 3d. If G is 4-connected, then IPI 2 3d - 2.

H . A. Jung

286

Proof. In view of Propositon 2.1 we may assume that H is a stronghold in G. Continuing the notation of the preceding proof, we further may assume that H ; is a stronghold in G , whenever 1 5 i < s and lP;l = h. Also d = s h - 1 and (PI = (s - l)(h t 1) t 1t 6. Since (s - l ) ( h 1) 1 = 3(s + h - 1) (s - 4)(h-2) - 5 and [GI 2 3(s+ h - 1) (s - 3)(h - 2) - 3 + 6, it remains to consider the case in which 6 5 2. If there is at most one integer j , such that 1 5 j < s and lPjl > h, then, a8 in the preceding proof, (s h - 1)s = e ( 2 ; G )2 (s - 1)hs 2, hence 0 2 (s - 2)(h - 1) - 1 f , a contradiction. Thus lPjl = lPkl = h t 1 for two integers 1 5 j < k < s, further lPol = IPS[= 0 and [Pi1 = h for the remaining indices. First we treat the case in which s 2 4. Since (s t h - 1)s = e ( 2 ;G ) 2 (s - 2)sh 4 implies 0 2 (s - 3)(h - 1) - 2 t t, we have s = 4, h = 2 and e ( 2 ;G ) = (s - 2)sh 4. For a component H' of G - P with H' # H therefore holds N z ( H ' ) = 0 hence N p ( H ' ) C V ( P ; )for some i, by Lemma 2.3. Then h 1 2 IP;I 2 2(Np(H')I - 1 2 5 , a contradiction. Thus V ( G )= V ( P )U V ( H ) . By Lemma 2.3 there is at most one edge joining Pj and Pk, Thus e(Pj;2 ) 2 ( h t l ) ( s - 1) - 1 > 2, a contradiction. Let s = 3 and [GI 5 3d - 1. Then H is the only component of G - P since IPI IHI 2 3d - 3 6 = 3d - 1. By Lemma 2.3 we have e ( 2 ;P;) 2 (h+l)(s-1)-1 for i = 1,2. Hence s(s+h-l) 2 ( ~ - 2 ) h s t 2 ( h t l ) ( s - 1 ) - 2 , a contradiction. 0

+-

+

+

+

+

+

+

+ +

+

+

+

+

Theorem 2.6. For each longest ab-path P' in G , let each component of G - PI have at most two vertices. Let P be a longest ab-path in G. If G - P has a non-trivial component, then IPI 2 4d - 7 and IG) 2 4d - 4.

For the proof of Theorem 2.6 pick a component H of G - P, such that IHI = 2, say V ( H ) = {c,c'}. We continue the notation N p ( H ) = { . q , z ~ , . , . , z , } = 2 and P; (1 I i < s), PO = P[a,zl) and P, = P(zs,b]. Further set t = INp(c) n Np(c')l. Clearly s = 2d - t - 2.

Lemma 2.7. If z: E N p(v) or ;2 E Np(v) for some j and for some o E V ( G )- V ( P )- V ( H ) ,then [PI2 4d - 7.

Proof. Assume z; E N p ( v ) for some v E V ( G )- V ( P )- V ( H ) and [PI 5 4d - 8. Let H' be the component of G - P containing o. Abbreviate ~i= Ip(zi,zi+lII - ~ I N P ( H ' )n V(p(z;,t.;+l])l IN,(H') n {z;+l}I,for 1 i i < 8, EO = IP[a,zl]l- 21NP(H1) n V(P[a,zl])l IN,(H') n {z1}1 and E s = IP(zs,b]- 21qJ(H') n V(P(kd,b])l. If H' has no neighbour on P(z;,z,+l],where i > 0, then we have that E ; = IP(zi, z;+l]l. Let HI have a neighbour on P(z;,%;+I] where i # 0, s , j .

+ +

Longest ab-Paths in Regular Graphs

287

Then z+,z++ 6 N,(H‘) by Lemma 2.2 hence ~i 2 2, moreover E; 2 3 if N ~ ( z i=) V ( H ) . Similarly E j 2 O , E O 5 0 and 2 0 (respectively E j 2 -1 i f j = s). ThereforeCboe; 2 2s+t-6. SinceC;=,E; = IPI-21Np(H’)I+ INz(H‘)I and s = 2d - 2 - t, we obtain 2INp(H’)I - INz(H’)I - t 5 2, that is INP(H’)l t INP-Z(H’)I 5 t t 2. (*I If INp(H’)I 2 d then INp(H’)I 5 t 1 by (*). In that case d = INp(H’)I = t + 1, hence H is a stronghold in G, and INp-z(H’)I = 1. If H is a stronghold in G and INp-z(H’)I = 1, then INp(H’)I = INz(H’)I 1 5 121= t. Therefore INp(H’)I 5 d - 1. We deduce that IH’I = 2 and H‘ is a stronghold in G. Since t j E N p ( H ) and z; E N p ( H ’ ) we can interchange the roles of H and H‘ and conclude that also H is a stronghold in G. Notice that CboE; 5 2 s t t - 5 since otherwise d-l+INp-z(H’)I 5 t = d - 1 , contrary to Np-z(H’) # 0. In particular j < s. Also INp-z(H’)I 5 2 by (*I. Now let P(z+,z;+l]n N p ( H ’ ) # 0. If i # j then, by Lemma 2.2, the first 4 vertices on P; are not in Np(H’). If the fifth vertex of P(z;,zi+l] is in Np(H’) then the ab-path P’,constructed in the proof of Lemma 2.2, has [PIvertices, and G - P’ would have a component of a t least 4 vertices. Thus i # j implies E ; 2 5, a contradiction to C ~i 5 2s t - 5. If i = j then t++,zr++ 6 Np(H’), hence ~j 2 1. Therefore N z ( H ’ ) C (21,zj+l} with N z ( H ‘ ) 2 (11) or C E; 2 2s t - 5. If C E ; 2 2s t - 5, then INp-z(H’)( 5 1. Thus in any case d - 1 = INp(H’)I 5 3. Since a vertex in N z ( H ’ ) is joined to all vertices in V ( H U H‘), we must have N z ( H ’ ) = 0. But INp-z(H’)I 5 2 hence d 5 3, 0 contrary to INc(zT)l 2 4. The proof of the lemma is complete.

+

+

+

+

+

Our next aim is to estimate e ( 3 ;P - 2 ) where 3 = {z: : 1 5 i < s} U{Z,+~ : 1 5 i < s}. Abbreviate u; = ti+,w ; = ztql and = { u i , w;} for 1 5 z < s. Since P is maximal, clearly {ui : 1 5 i < s} and {w;: 1 5 i < s} are independent sets, hence e ( 3 ;P - 2 ) = e ( 3 ;S ) where S = V ( P )- 2 { u i : 1 5 i < s, u; = w;} and 7 = U:z;$. Let t 2 be the number of i, such that 1 5 i < s and lPil 2 2. If H is no stronghold, t 2 2 t + l , since then, for each v E V ( H )there is an i, such that N H ( z ; ) = {v}, and N ~ ( z ; + 1#) {v} or i = s; hence lP;l2 2 or i = s. L e m m a 2.8. Let z,z’ E V ( P j ) , z N uk and t’ N W I ,where 0 5 j 5 s. If z’ = z+ or z’ = z-, then j 5 1 < k, I < k 5 j , or k 5 j 5 1 . If I N ~ ( t h , t l + 1 ) I= 2 and z’ = z++ or I’ = t - - , then j 5 1 < k, 1 < k 5 j, or k 5 j 2 1.

H. A. Jung

288

Proof. Let L = L[Zk,z[+I]be a path in G with L(Zk,z[+1)C_ V ( H )and ILI = 4if ( N H ( z zi+1)( ~ , = 2. First assume k 5 1. If j < k and 2' = z+ (resp. z' = z++), then P[a,21,P[Uk,W I ] P[z', , Z k ] , L and P[zl+l,b] would define an ab-path P' with IP'I > IPI. If j < k and z' = z- (resp. z' = z--), similarly a contradiction is obtained. The case, in which k 5 1 < j , is symmetric. Thus it remains to consider the case in which 1 < j < k . If 2' = z+ (resp. z' = z++), then P[a,wl],P[z',z k ] , L , P[z[+1,21,P[Uk,b] define an ab-path P' with (P'I > [PI,a contradiction. If z' = z- (resp. z' = z - - ) , similarly a contradiction is obtained. 0 The proofs of the following two lemmas are very similar to the preceding one and are omitted.

Lemma 2.9. Let Z,Z' E V ( P j ) , z uk and z' ' 1 ~ 1 where k # 1 a n d 0 5 j 5 s. I f . z ' = z+, then j < 1 < k, 1 < k 5 j , or k 5 j < 1 . If z ' = z++ and INH(zk,zi)l = 2, then j < 1 < k , 1 < k 5 j , or lc 5 j < 1. N

N

Lemma 2.10. Let z,z' E V ( P j ) , z wk and z' wl where k # I and O L j i s . Ifz'=z-,thenkil< j , l < j < k , o r j < k < l . Ifz'=z-and INH(Zk+i,Zi+i)1= 2, then k < 1 < j , 1 < j 5 k , or j 5 k < 1. N

N

Proof of Theorem 2.6. In the following i, j , k , l always mean elements of { 1,2, ...,s - 1 ) if occurring as indices. We first seek estimates for e ( 3 ; Pj) where lPjl 2 2. Set E i j = e(ui;Pj) e(w;;Pj)- lPjl + 2 unless lP;l = 1 and 2e(ui,Pj) - lPjl < -2, in which j 0. Further, ~j = Cfi,'Eij. For i # j lemma 2.8 yields case set ~ i = e(ui;Pj) e(wi;Pj) 5 lPjl - 1, since W i $ N p ( w j ) and Z- 4 N P ( W ~ ) whenever z E Npj(ui). Therefore E i j 5 1 whenever i # j . Moreover if ~ ;= j 1 and i # j then

+

+

z- E Np(w;)U { z j } whenever2

E V ( P j )- NP(u;).

(1)

+

Further note e(u;;P,) e(w;;P,) 5 lP,l for m = 0 and m = s. Let j be fixed where lPjl 2 2. Call j of type ( a ) if ~ j 2j 4. Then lPjl 2 4. If s = 2 then d 5 3 hence (PI 2 4d - 6 and [GI 2 4d - 4. If i # j and z+ E N p j ( u j ) , then z 6 Np(ui) by lemma 2.9. Hence e(uj;Pj) e(u;;Pj) 5 lPjl and, by symmetry, e(wj;Pj) e(w;;Pj) _< lPjl (i # j ) . Thus E j j e(ui;Pj) e(wi;Pj) - lPjl 2 5 4 and consequently ~ i 5j 0 if i # j . Notice that t2 2 2 implies E j j E i j 5 4 for some i with ~ l # i wi and i # j . For j of type (a)we therefore have the claim or E j 5 4. Call j of type ( p ) if & k j = 1 for some k such that = 1. Since subsequent vertices on Pj cannot be both adjacent to uk, we have that

+

+

+

+

+

+

Longest ab-Paths in Regular Graphs

289

the vertices on Pj,beginning with u; and ending with w i , are alternately in N p ( u k ) . In particular INH(zk,zj)l = 1 and NH(Lk+l,Zj+l)l = 1. By Lemma 2.9 vertex u j is not adjacent to z+ whenever z E Np,(uk). Hence 2 e ( ~ j ; P j )5 lPjl - 1. By symmetry, 2e(wj;Pj) 5 lPjl - 1 hence E j j 5 1. Assume ~ ; =j 1 for some i, where i # k , i # j. Since uk = wk E N p ( u f ) we ) Lemma 2.8 and Lemma 2.10, hence uf ui by (1). have wi 4 N ~ ( u jby We obtain / N ~ ( z zj)l i , = 1. For j of type ( p ) we have seen that ~j 5 d - t. There are at most t 2 - t - 1 integers j of type (p). We call j of type (7) if lPjl 2 2, ~ , = j 1 for some i # j and j is not of type (a)or (0). If ~ j =j 1 and i # j then by (1) either u; uf hence I N ~ ( z ; , z j )= l 1 or w; u j . If wi N u j let v; be the last vertex on Pj such that V(P[uj,v;]) Np(w;). Let j be of type (7) and assume J N ~ ( . z j=) l 1. If I N ~ ( z i ) = l 1 whent2 - t 3 since ~ j j 3. Let ever ~ i =j 1 and i # j , then clearly ~j ( N H ( z ~=) ~2 , i # j and ~ ;=j 1. Then ui 4 Np(uf). Further vf = wj, since otherwise of+ N u; by (1) and I N ~ ( z ; , z i + l ) l = 1 by Lemma 2.8. In particular IN~(z;+1,zj+l)l = 1 by Lemma 2.10. Also e ( w j ; P j ) 5 1 by Lemma 2.10 hence ~ j 5j 2. For any such i let ~ ( ibe) the last integer such that I N H ( Z i + l , * * * 7 y i ) )I = 1. For any other i such that ~ ; =j 1 and i # j set cp(i) = i. If tp is injective, then ~j 5 t 2 - t 3. Now let p(k) = tp(1) where 1 < k. Then I N ~ ( z k )= l 1 since otherwise cp(1) < k < cp(k). Therefore cp(lc) = k and INH(zk,zk+l)l = 2. Since INH(zl+1,zj+l)l = 1 we have I N ~ ( z k + l , z j + l ) l= 2. Further uk N W j , since otherwise wk wT by (1) and hence I N ~ ( z j + l , z k + l ) l= 1, a contradiction. Let z be the first vertex on Pj such that V(P[z,wj]) E N p ( u k ) . Then ul = z since otherwise z-- N Wk by ( l) , hence [ N H ( Lzk+l)l ~ , = 1 by Lemma 2.8, a contradiction. Also V ( P [ u j , w r ] ) E Np(wl) as shown above. If tp(i) = i for all i # 1,j such that ~ ; =j 1, then ~j 5 t 2 - t 3. Let tp(i) # i for some i # 1. Since u j N w; and N uk we have that j < i < k , i < k < j , or k < j < i by Lemma 2.8. In particular tp(i) # j . Set q = tp(i) and assume cp(q) = q. Then ul uq and u j N wi hence j < 1 < q, 1 < q < j or q < j < 1 by Lemma 2.8. If j < 1 < q, then j < k < i. If 1 < q < j, then k < i < j. If q < j < 1, then i < j < k. Therefore tp(q) is not defined and cp(i) # tp(m), for all m # i , 1 5 m < s. We have shown that j is not in the image of cp and k , l is the only pair of distinct integers such that cp(k) = cp(2). Therefore in fact E j t 2 - t 3 if j is of type (7) and INH(zj)l = 1. In the case where lNH(Zj+l)[ = 1 and j is of type (7) we have ~j 5 t 2 - t 3 by symmetry. Now let j be of type (7) and NH(zj) = N H ( z ~ +=~ V) ( H ) . If ~ i=j 1 and i # j then W j N Uj by (1) since ui f Np(uf). Also vi # w; since otherwise w; N wJ: and N

N

N

+

<

<

+

N

+

ut

N

<

+

+

H . A. Jung

290

INH(Zj+l,Zi+l)l = 1. Therefore vt+ u; by (l),hence I N ~ ( z ; , z i + l )= l 1. Thus E j 5 t2-t+3 also in this case. If lPjl 2 2 and j is not of type (a),(p) or (7)then E j 5 3 by definition. NOW e ( u i ; P j ) e ( w ; ; P j ) 5 lPjl - 2 -+ E i j which yields 2 e ( z P j ) 5 (-S;((IPjI-2)+2~ij(equality if lP;l 2 2). Thus 2e(TF;Pj) 5 [ S I ( I P j l - 2 ) + 2 ~ j . Let E be the sum of all E j , for which 1 5 j < s and lPjl 2 2. Then N

+

2 e ( S ; ~5)

191(lsl- 2t2) + 2

~ .

(2)

Introducing e ( S ; S ) = dlSl - e ( T ; Z ) , which we may assume in view of Lemma 2.7, and e ( 3 ; 2 ) 5 ds - 2d 2 , we obtain

+

With the abbreviations t* = t 2 - t - 1 and a = 10d - 13 have Is1 = s - 1 t 2 = 2d - 2 t* and hence

+

+

-+ t 2 - 3 ) + - 2~ 5 ISl(lSlCase 1: t 2 2 t -+ 1 and t 2 - t -+ 3 2 d - t. t*(4d - 2t

(Y

2t

+ 3t2 - t we

- t 2 + 2).

(4)

Then E 5 ta(tz-t+3) = tzt*+4t2, and ( 4 ) attains the form t*(4d-2t-t23 ) Q - 8t2 5 Isl(lSl- 2t - t z 2). From t 2 5 s - 1 = 2d - t - 3 we deduce 4 d - 2 t - t 2 - 3 2 2 d - t anda-8t2 2 4t+2.Therefore ISI-2t-t2+2 > 0, which yields [PI> 4d-7, since IPI = 2s-l+ISI-t2 = 4d-5+ISI-2t-t2.

+

+

+

Case 2: t 2 2 t + 1 and t 2 - t 3 5 d - t - 1. T h e n & < ( t z - t - l ) ( d - t ) + ( t + l ) ( t z - t + 3 ) = t * ( d + 1 ) + 4 t + 4 , a n d (4) attains the form t*(2d - 2t t 2 - 5 ) (Y - 8t - 8 5 Isl(lSl - 2t - t 2 2). S i n c e d 2 t + 5 a n d a - 8 t - 8 = 10d-21-9t+3t2 2 10d-18-6t 2 4 d - 6 we again have IS1 - 2t - t 2 2 > 0 , and hence [PI> 4d - 7.

+

-+ +

+

Case 3: t 2 5 t . Then H is a stronghold in G, t 2 = t - 1 = d - 2 and 1 31= 2d case E j 5 4 for all j (unless s = 2). Now (4) attains the form

6d - 12 - 2~ 5 (2d - 4 ) ( l S l - 3d Let i

Eij

+

#

-+ 6 ) .

- 4. In this (5)

j . We have shown that ~ i -tj E j j 5 4 and E i j 5 0. Assuming E j j = 4 we obtain e(ui; Pj) e ( u j ;Pj) = lPjl. For E N p j ( u i ) - { ~ j }

-+

+

we have v+, v++ $ N p ( u j ) by Lemma 2.9. On the other hand e(ui;Pj) e(uj;Pj) 2 l P ' l yields Wj E N p ( u ; ) and v+ E N p ( u j ) whenever v E

Longest ab-Paths in Regular Graphs

29 1

V(Pj) - N p ( u ; ) - {wj}. For the first vertex z on Pj in N p ( u i ) we therefore have V(P[z, wj]) N p ( u i ) and V(P(Uj, z]) C Np(Uj). By symmetry wi u j hence z # u: and z # u:+ by Lemma 2.8. Now u: @ Np(w;) since otherwise (without loss of generality assume that z; lies on P [ a , z j ] ) P[a,zi], c, c', P[zi+l, uj], P [ z - , u:], P[w;, ui] and P [ z ,b] would define an abpath P' with IP'I 2 ]PI 2. Again by symmetry w; @ N p ( u ; ) that is z = wj. So if we cannot find any i # j (for fixed j ) such that ~ ;+j ~ j 5j 3 , then u j is adjacent t o all w; and all D E V(Pj) - {uj}, a contradiction since t 2 = d - 2, lPjl 2 4 and zj u j . Thus in fact ~j 5 3 for all j (1 5 j < 8 ) . NOWE5 3t2 = 3d-6 and ( 5 ) yield ISI-3d+6 2 0, that is IPI = d-l+ISI 2 4d - 7.

+

N

It remains to consider the subcase in which IS/ = 3d+6 and G- P = H . Then ~j = 3 for all j and e(y;2 ) = ds-2d+2, that is e(S-3; 2 ) = 0. Notice that lPjl = 2 implies ~j 5 ~ j =j 2. Therefore lPjl 2 3 for all j . In view of 4 d - 7 = IPI 2 4 ( t - 1)+1we may deduce Po = P, = 0 and lPjl = 3 for all j. Also u j E N(wj) since otherwise ~j 5 ~ j5j 1. From u;, wi $! Np(uf) for i # j a n d e ( S - g ; Z ) = O w e o b t a i n e ( u f ; S - S ) 2 d - 2 , t h a t i s u f NU: for some i. This means a contradiction, since we can construct an ab-path P', such that IP'I = IPI + 2 using the edges [uf,uf],[uj,wj] and [ui,wi]. 0 The proof of Theorem 2.6 is complete. Notice that Theorem 2 is a consequence of Lemma 2.4 and Theorem 2.6.

Proof of Corollary 2. In view of Lemma 2.4 and Theorem 2.6 we may assume that G - P is independent for any longest ab-path P in G. Pick a longest ab-path P in G and set R = V(G - P). Case 1. There exist vertices c,c' E R and z E V(P) such that c z and c' z+. In order to apply Lemma 2.2 we set H = { c} and N p ( c ) = { z1, z2, .. . ,zd}. Let a = zj. By Lemma 2.1 there is no i # j such that E N p ( c ' ) ; also there exists at most one i such that zf E Np(c'). Consider a vertex z; E N p ( c ' ) n N p ( c ) where i # 1 and i # j 1. Let D; be the first vertex on P(z;-l,zi] in Np(c'). By Lemma 2.2 we have D; # z z l and v; .# ; :2 Moreover w; # zL;+ since otherwise the path P' constructed in Lemma 2.2 is a longest ab-path and P(zi-1, vi) is contained in a component of G - P',contrary to assumption. Hence lP(ai-1, v;)l 2 3. We obtain IPI 2 2 l N p ( c ) u Np(c')l - 3 2(INp(c) n Np(c')I - 2 ) , that is IPI 2 4d - 7 . N

N

+

+

H. A. Jung

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Case 2. Any two elements of N p ( R ) are not subsequent on P. Then ]PI2 2INp(R)I-1 and e ( R ;P ) = dlRl 5 (d-2)INp(R)1+2. Assuming IPI 5 4d- 8 we obtain INp(R)I 5 2d- 4 hence dlRl 5 ( d - 2)(2d- 4) t 2 = 0 d(2d - 6 ) - (2d - 10) and [GI 5 (4d - 8 ) i- (2d - 6 ) .

3

The “hopping” case

In this section distinct vertices a and b in the graph G are fixed. In view of Lemma 2.5 for the proof of Theorem 1 it remains to consider the case, in which G - P is independent for each longest ab-path P. Since the following variation of Woodall’s “hopping lemma” may be useful in related investigations we give it a more general setting than we actually need. Let P be an ab-path in G with vertices w1, w2,. . . ,wn, and let c be an isolated vertex of G - P. Assume that there is no ab-path P‘, such that V(P’) = V ( P )U {c}. Set recursively YO = 0, Xj = N p ( { c } U 5 - 1 ) . . and Y2 and Y j = {w; : wi-1 E Xj and wi+l E Xj}. Clearly Y1 N p ( c ) = X1 c x 2 E ..

s .

..

.

Lemma 3.1. Let a positive integer j and disjoint paths a = al, a2, . . ,Uk and b = b l , b 2 , . . . ,bl in G - c be given, such that U k , b l E Xj, further q-1 {al, a29 * * * yak, bl 9b2r * * * 3 bl) and

c

(Hj)

if j‘ < j , K < k and a, E Y j t , then a,-1, a,+l E Xjt; if j‘ < j, X < I and bx E yjf, then bx-1, bx+l E Xjt.

Then there is an ab-path with vertex set {al, a2,. . . , a h , bl, b2,.

. . ,bl, c}.

Proof. By induction on j . For j = 1 the claim follows from a h , bl E X1 = N p ( c ) . Assume, that the claim holds for j - 1 where j 2 2. Notice, that subsequent vertices wt, w t + l on P cannot be both in Xj-1, since otherwise the lemma could be applied t o the paths w1, v2,. . . ,wt and wn, w , + ~ , . . . ,wt+l to obtain an ab-path with vertex set V ( P ) U{c}. Also Xj-1 n yj-1 = 0 since wt E Yj-1 implies wt-1, wt+l E Xj-1. Since ak E N p ( w ) for some w E {c} u Yj-1 we cannot have a k E Yj-2. For otherwise a k E Yj-2 n XI or w E Yj-l f l Xj-1. By symmetry, bl $ Yj-2. If a k , bl E Xj-1 the claim follows by induction hypothesis. Case 1: ak $! Xj-1 and bl E Xj-1. There exists v E Np(ak) n 5-1. Clearly w $! Yj-2. If w = a, then a,-1, a,+l E Xj-1 by ( H j ) . Application of the induction hypothesis to the paths a l , az, ,a,, a h , ak-1,. ,a,+1 and bl, b2,. ,bl yields the claim.

...

..

..

293

Longest ab-Paths in Regular Graphs

In the remaining subcase we have w = b, for some s < 1. Since b,-1 E Xj-1 by ( H j ) , application of the induction hypothesis t o the paths al, a2,. . .,a k , b,, b,+l,. . .,bl and b l , b2,. . . ,b,-l yields the claim. Case 2: ak E xj-1 and bl $! xj-1. That case is symmetric to Case 1. Case 3: ak 6 Xj-1 and bl $! Xj-1. There exist wh E N p ( a k ) nYj-1 and w; E Np(bl) nYj-1. Clearly wh, w; $! q - 2 . If w h = bl then the paths w1, w2,. . ,wh-1 and w,, ~ ~ - 1. ., ,.w h satisfy wh-1 E Xj-1,wh E Xj and ( H j ) . By Case 2 there would exist a path with vertex set V(P)U { c } . Thus w h # bl and symmetrically Wi # a k . Therefore oh and w; are inner vertices of a l , a2,. ..,a k or b l , b2, ... ,bl, and their neighbours on those paths are members of Xj-1. If oh = a, and w; = b,, the paths al, a2,. . ,a,, a k , a k - 1 , . . . ,a,+l and b1,b2,. . . , b S, b l , bl - l , . . . ,b,+l satisfy (Hj-1). If w; = a, and w h = b,, the paths a l , a2, .. . ,a,, bl, bl-1,. .. ,b,+l and b l , b2,. . . ,b,, a k , a k - 1 , . . . ,a,+l satisfy (Bj-1). If w h = a, and w; = a,, where r < s, then the paths a l , a2, . . . ,a,, a k , Uk-1, . . . ,a,+l and b l , b2,. . .,bi, a,, as-l, . . ,a,+l satisfy (Bj-l). If wh = a, and w; = a,, where s 5 r , the paths a l , a2,. . . ,a,-l and b l , b2,. . .,bl, a,, a,+l,. ..,a,, U k , U k - 1 , . . . ,a,+l satisfy (Bj-1). In each of these subcases the claim follows by induction hypothesis. The subcases 0 where oh = b, and V i = b, are symmetric.

.

.

.

we obtain that two vertices Setting X = CZ,Xj and Y = C;,Yj of X are not subsequent on P. Consequently two vertices on Y are not subsequent on P and X n Y = 0. These assertions follow by the argument at the outset of the foregoing proof. Corollary 3.2. If there is no ab-path P’, such that IP’I = /PIand G - P’ has less components than G - P , then NG(Y) G V ( P ) .

Proof. If y E Y is adjacent to d E V(G) - V ( P )and y = w; then paths w1, w 2 , . . . ,wi-1 and wn, ~ ~ - 1. ., ,.w i + l obey the hypothesis of Lemma 3.1 with the minimum j , such that y E q, Hence there is an ab-path P’ with vertex set V(P’) = V ( P )- {y} U { c } . Clearly G - P’ has less components than G - P. 0 For the remainder of this section we assume that G - P is independent for any longest ab-path P. Clearly in this case Lemma 3.1 and Corollary 3.2 apply to any longest ab-path P and any vertex c E V ( G )- V ( P ) .Let a longest ab-path P in G and a vertex c E V(G - P)be fixed, and let X , Y

H.A. Jung

294

be constructed as described a t the beginning of this section. We further assume from now on that G is regular of valence d. Let PI, Pz,. . . ,Pt be all components of P - X of the form P;= P ( z ,2') where x,z' E X and IPiI 2 2 (1 5 i 5 t ) . Clearly t = 1x1 - IY1 - 1. Define ui,wi for 1 5 i 5 t by P; = P[u;,w;].If a $! X let PO = P[a,wo]be the component of G - P containing a. Similarly define Pt+l = P[ut+l,b] if b $! X . Abbreviate U = { u ; : 1 5 i 5 t t l}, W = {w;: 0 5 i 5 t } , S = V ( P )- X - Y and R = V(G)- V ( P )- {c}. Lemma 3.3. The sets U U Y and W u Y are independent. Furthermore, e ( v , U ) 5 1 and e ( v , W ) 5 1 for all v E R.

Proof. Assume that there is a path Q = Q[u,v],such that U , T I E U U Y and IV(Q) n V(P)I = 2. Let u lie on P[a,v). If u E Y then IQI = 2 by Corollary 3.2, hence w E X by definition, a contradiction. If U , T I E U - Y the paths P [ u ,u-] and P[b,T I ] U Q U P [ u ,T I - ] obey the hypothesis of Lemma 3.1 (with j chosen, so that u-,w- E Xj).There would be an ab-path P', 0 such that V ( P ' )2 V ( P )U V ( Q )U {c}, a contradiction. Since we have that e ( X ;S) = d l X l - e ( X ;Y U {c}) and e ( X ; Y U {c}) = (IYIt 1)d we get

- e ( X ;X

) - e ( X ;R)

Proof of Theorem 1. For the rest of the proof of Theorem 1 we consider 4 cases. Case 1: t = 0. Then e ( X ; X ) = e ( X ; R ) = 0 hence R = S = 0. Therefore G is bipartite, where the edges have one end vertex in X and the other in Y U { c } . Further ]PI= [GI - 1 and a , b E X , hence the claim. Case 2: e(v,X ) 5 1 for some TI E R. By Lemma 3.3 there are a t most two pairs of subsequent vertices on

P with elements in N p ( v ) U X . Therefore IP( 2 2lX U Np(w)I - 3 = 21x1 + 2d - 2 d x ( v ) - 3 2 4d - 5. By assumption G is 3-connected, hence IPI 2 3d - 2 and [GI 2 2 t [PI2 3d. Before we turn to Cases 3 and 4 we prove a lemma on the distribution of edges joining U U W and S.

Longest ab-Paths in Regular Graphs

295

Lemma 3.4. Let k , l be distinct elements of (0, 1,. . . ,t e(Uk;fi)

+ 1). Then

+ e ( w k ; q )5 181if 1 5 k 5 t;

(7)

the inequality is strict i f 1 5 l 5 t.

+

e(%; 8) 4- e ( w ;8 )5 if 1 5 k 5: t 1 and 1 5 1 5 t + 1; ( 8 ) the inequality is strict if 1 5 t and U k 4 N p ( w l ) .

+

e(wk;Pl) e ( q ;P I) 5 141if 0 5 k 5 t and 0 5 1 5 t; the inequality is strict if 0 5 l and W k $ Np(ul).

(9)

Proof. Let z N U k , where t E V(P1).If < k and 2- E Np(Wk), then the paths P[U,2-1 U [t-,Wk]U P[Wk,U k ] U [ U k , t]U P[., U i ] and P[b,w,'] would obey the hypothesis of Lemma 3.1 (with j chosen, so that u;, w,' E X j ) ; there would be an ab-path PI, such that V ( P ' ) = V ( P )U {c}. Similarly z- 4 Np(Wk) if k < 1. Notice that t # ul (if # 0) and wl $ N(Wk) (if # t 1) by Lemma 3.3. Therefore Iv(Pl)- NP(Wk)l 2 INfi(Uk)l with strict inequality if 1 5 1 5 t. Hence (7). For the proof of (8) assume t N U k , where z E V ( 4 )- {wI}.Then z+ $ N p ( u l ) . Otherwise we would obtain a contradiction, as in the proof of (7). Thus IV(Pl) - Np(ul)I 2 INp,(Uk)( with strict inequality if l 5 t and W l $ NP(Uk). 0 (9) is symmetric to (8).

+

Case 3: t 2 2. S e t S ; = { u i , w i } f o r 1 5 i 5 t a n d S = z U ... u s . L e t 1 5 l S t and pick i # 1. Then e(E;Pl) t e ( 5 9)5 2191 by (8) and (9). Combining that inequality with inequalities (7) for all k. # i, 1, we obtain e ( 3 ;4 ) 5 t p 1 - t 2. Similarly combination of (7) for all k yields e(T;Pl)5 t1Pl1 ( 1 = 0 and 1 = t 1). Hence

+

+

In view of Case 2 we may assume e ( X ; R ) 2 214. Since e ( 3 ; R ) 5 21RI by Lemma 3.3, we have e ( X ; R ) 2 e ( 3 ; R ) and e(S;G) = 2dt 5 tlSl - t ( t - 2) e ( 3 ; X ) e ( X ; R ) . From (6) and e ( 3 ; X ) 5 e ( S ; X ) we infer 2dt 5 tlSl - t ( t - 2) td - e ( X ; X ) , hence d 5 (5'1 - t 2. Since IPI 2 21x1 + IS1 - t - 1 and 2 1 x 1 1 = d , it remains to consider the subcase, in which X = X I , that is N G ( Y )= X for all y E Y . Case 3.1: e ( z U Pl)= 2191 for some i # 1 in {1,2,. . . ,t}.

+

+

5,

+

1x1

+

H. A. J u n g

296

Then u; N wl and w; ul by (8) and (9). We shall show that u;N 201 and wi N ul imply Y = 0. To this end we exploit the symmetry of the condition and assume i < I for definiteness. Since P is a longest ab-path . and ui N wl,vertex w: is not adjacent to vertices y E Y on P[u;,w ~ ]Since wi ul,vertex uf is not adjacent t o y E Y on P[a,wi]U P[ui,b]. But Nc(y) = X for y E Y , hence Y = 0. From u; wl and w;N u1 also follows lP;l 2 3 and lPil 2 3. Therefore IPI23t+3=3d. N

N

N

Case 3.2: e ( 5 U %,PI) 5 21Pll

- 1 for all i # 1 in { 1 , 2 , . . . ,t}.

The estimates a t the beginning of Case 3 now yield e ( 3 ; S ) 5 tlSl IS1 - t - 1 2 3d - 2. This proves 2. in Theorem 1. In the remaining subcases we have R = 0 and equality in all used inequalities. In particular e ( X ; X ) = 0 and e ( 3 ; X ) = e(S; X ) hence a,b E

t(t - 1) and d 5 IS1 - t + 1, hence IPI 2 21x1

+

X. For any pair i # I in {1,2, ...,t } moreover e ( z U G 4 ) = 21PiI - 1, which yields u; wl or w; ul by (8) and (9). Case 3.2.1: u1 N wt.Since P is a longest ab-path, vertices u:, wr are not adjacent to any w E U U W U Y on P(u1,wt). In view of e ( X ; S )= e ( X ; y ) and e ( X ; X ) = 0 we obtain N ~ ( u ~ ,Ew{c,u1,wt} ~) U Y , hence d 5 3 lYl= 2 1x1 - t = d - t + 2 . we deduce From t = 2, IS1 = d t 1 and e(S - s ; G ) = e(S Ns(w) = S - v for all v E S - 3.If u t # w1, then u1 N w2 and u r N 212, a contradiction to the maximality of P . Therefore u: = w1 and, by symmetry, w; = 212. We infer IS1 = 4,hence d = 3 and Y = 0. Now u2 $ Np(w1), hence N G ( U 2 ) = (212, u t } , a contradiction. Case 3.2.2: 201 N ut. Determine k < I with wk N u1 and 1 - k as small as possible. Since P is a longest ab-path, vertices u r , wl are not adjacent to c V(P[wk,u,]). any w E Y U U U W on P[a,wr,)UP(u1,b].Thus Np(w~,ur) If k 1 = 1, then NG(W;) = {c,wk,u~}U Y, hence t = 2 and N S ( V )= S-{v} for all w E S-r,as in Case 3.2.1. On the other hand wi $ N p ( w l ) , hence w; = uk. By symmetry 141 = 2 and therefore IS1 = 4 = d 1. Further Y = 0 , V ( P ) = { a , u k , w k , wl,ul,wl,b},Np(uk) = { a , w k , b } and Np(w1) = {b,ui,a}. From this it readily follows, that G is isomorphic to G(3) (where the pair a,b corresponds to the pair a’,b’ in Figure 1). Let k t 1 < 1. Then wk+l $ N p ( q ) , hence uk+l N wl. By the maxihas no neighbour w € Y U U U W on P(uk+l,wl). mality of P vertex From N Y ( U ~ +=~Ny(wk+) ) = Y we deduce Y G V ( P [ w t , By symmetry Y G V ( P [ w L l , u ; ] ) ,hence Y = 0. In particular ui+l = wt and N

+

+

N

3;s)

+

+

297

Longest ab-Paths in Regular Graphs N p ( w l ) = {wk,uk+l) (since e ( w i ;S to d = INp(c)l 2 t -t 1 2 4. This settles Case 3.

s> = 0). Therefore d = 3, contrary

Case 4. t = 1.

If IS1 > d then [PI 2 21x1 - 1 + IS1 - 1 2 3d - 1. Thus assume IS[5 d, further e ( v ; X ) 2 2 for all v E R (in view of Case 2). Then e ( X ;S) 5 d - e ( X ;X ) - 2 1 4 by (6).

Case 4.1: a $ X and b E X . By(9)wehavee(wo,wl;Ej) 5 IEjIforI=Oandl,thatise(wO,wl;S) 5 IS[.Since e(w0,wl;R) I IR( by Lemma 3.3 and e ( w 0 , w l ; X ) < e ( S ; X ) , we obtain 2d = e(w0,wl; G ) = e(wo,wl; S ) e(w0, wl; X ) e(w0,wl; R ) < IS1 d - IRI, contrary to assumption. The subcase, in which a E X and b $ X , is symmetric.

+

+

+

Case 4.2: a $ X and b $ X . From (8) we obtain e(u1,uz;Pl U P2) 5 1 4 U P2l. If z N u1, where z E V(Po),then z+ 212 by the argument of Lemma 3.4. Hence e ( q , u 2 ;Po)[I lP0l 1 and further e(u1, 212; S ) 5 IS1 1. By symmetry e(w0,wl;S) I IS1 and therefore e(U U W ; S ) I 21SI t 2. Now 4d = e(U U W ; G ) = e(U u W ; S ) e(U u W ; X ) e(U u W ; R )I 215’1 2 d - 21RI 21RI. But IS1 5 d by assumption, hence d I 2, giving a contradiction.

+

+

+

+

+

+ +

+

Case 4.3: a E X and b E X . For v E R clearly IS\ 2 2ds(v) - 1, hence d 2 2ds(v) - 1 and dx(v) = d - ds(w) 2 If IRI 2 2, then (6) yields e ( X ; S ) 5 d - e ( X ; R ) 5 d - 2 9 = 1, a contradiction. Next assume R = {v}. If IS( 2 d, then [PI2 21x1 (SI - 2 2 3d - 2 and and [GI 2 3d, and we are done. If IS1 I d - 1, then ds(v) 5 e ( X ;S) 2 d - dx(v) I On the other hand e ( X ;S) 2 ISl(d - 15’1 1)d ds(w) 2 2(d - 1) - 5d > 2, a contradiction. Finally assume R = 0. If 15’1 I d - 1, then e ( X ;S) 2 ISl(d - 15’1+ 1) 2 2(d - 1) contrary to (6). Therefore 15’1 = d by assumption. Now e ( X ;S) 2 IS1 and e ( X ;S) 5 d-e(X; X ) by ( 6 ) . Hence e ( X ;S ) = d, and S is complete. Further e ( X ; X ) = 0. If 1x1 > 1x11,then [PI2 21x1 15’1 - 2 2 3d. If X = X I , then G - X has d - 1 one-element components and one d-element component. Since each one-element component is adjacent to each x E X , the d edges issuing from S, end in distinct vertices of X. Hence G is isomorphic t o the graph G(d). The proof of Theorem 1 is complete. 0

9.

+

c.

+

+

298

H. A. Jung

Acknowledgement The author wishes to express his gratitude to the referee, who found two errors in the first version, for his helpful remarks.

References [l] J. A. Bondy and B. Jackson, “Long paths between specified vertices in a block.” To appear. [2]G.A. Dirac, “Some theorems on abstract graphs,” Proc. London Math. SOC.(3) 2 (1952), 69-81. [3]H. Enomoto, “Long paths and large cycles in finite graphs,” J. Graph Theory 8 (1984), 287-301.

[4]G.Fan, “Longest cycles in regular graphs,” J . Combin. Theory (B) 30 (1985), 325-345. [5] B. Jackson, “Hamilton cycles in regular 2-connected graphs,” J. Combin. Theory (B)20 (1980), 27-46.

[6] H. A. Jung, “Longest paths joining given vertices in a graph,” Abh. Math. Semin. Univ. Hamb. 56 (1986), 127-137.

[7] S. C. Locke, “A generalization of Dirac’s Theorem,” Combinatorica 5 (1985), 149-159.

Annals of Discrete Mathematics 4 1 (1989) 299-306 0 Elsevier Science Publishers B.V. (North-Holland)

On Independent Circuits in Finite Graphs and a Conjecture of Erdos and P6sa. P. Justesen Nprrresundby Gymnasium 9400 Npjrresundby Denmark

Dedicated to the memory of G. A . Dirac We state and prove some results about the existence of many mutually disjoint circuits in a graph. The results were obtained during thesis work supervised by G. A . Dirac, and most of them have been announced by him several years ago. In some cases, however, proofs have never before appeared.

1

Introduction

In this article is presented refinements of some theorems, which provide sufficient conditions on either the valencies or the number of edges t o ensure a certain number of mutually disjoint circuits in finite simple graphs. The results have been announced by G. A. Dirac a t the 5. British Combinatorial Conference in 1975 in Aberdeen, and the theorems without proofs can be found in the proceedings from that conference. Until now the publishing of the proofs has been limited to a few copies of my thesis. In 1978, in his book “Extremal Graph Theory” [l]Bollobds is thus restricted t o quote my result and present the proof of the theorem of Erdos and P6sa on the number of edges required in a graph with n vertices to ensure k independent circuits. This theorem only deals with n 24k, whereas Erdos and P6sa conjectured that the theorem with a few modifications can be extended to n 2 3k. The proof of this conjecture shall be the main purpose of this article. The results were obtained in connection with my work on my master’s thesis in 1975 at the University of Aarhus. The work was strongly encouraged by Gabriel Dirac, who was my supervisor, and by Roland Haggkvist, who at the same time stayed at our institute as a research student.

>

299

300

P. Justesen

2 Notations In this paper I shall only deal with simple finite graphs, i.e., graphs with no multiple edges and no loops. For a graph G , [GI will denote the number of vertices, e ( G ) the number of edges and v ( X ) the valency of the vertex

X. Two or more subgraphs of a graph are said to be independent if they are mutually disjoint. ( n ) denotes the complete graph with n vertices. ( p , q ) denotes a complete bipartite graph with p and q vertices in the two classes. ( ( p ) , ~ is) the graph obtained from a ( p , q ) through adding all possible edges between vertices of the class with p vertices. O k will denote a graph consisting of k independent circuits. First we shall recapitulate some theorems. In [4], Erdos and Pbsa mention that they proved the following theorem.

Theorem 1. Let G be a graph with n 2 6 vertices. If e ( G ) 2 3n - 6 then G 2 O2 or G is isomorphic to a ( ( 3 ) , n - 3 ) . A proof due to H. J. Voss can be found in [5]. In [2], CorrAdi and Hajnal proved the following.

Theorem 2. Let G be a graph and k denote a natural number. If [GI 2 3k and the minimal valency in G is 2 2k then G 2 Ok. Theorem 2 was proved in a more complicated form giving upper bodnds on the lengths of the circuits. The proof was very long and technical. Corrbdi and Hajnal themselves mention that they tried to carry out an easier proof giving no bounds on these lengths. In my thesis I gave a somewhat less complicated proof of the following generalization:

Theorem 3. If G is a graph with 2 3k vertices, k 2 1, satisfying that

for any two nonadjacent vertices X and Y , then G _> Ok.

I do not find that the generalization itself justifies an inclusion of the proof, which is still pretty long. A copy can be required through direct application to me.

Independent Circuits in Graphs and a Conjecture of Erdos and Pdsa 301 Define, to shorten the notation,

f ( n ,k ) = ( 2 k - 1)(n - k ) , which is the number of edges in a ( ( 2 k - l ) , n - 2k

+ 1) (Figure 1 ( u ) ) .

<3k-1>

Figure 1. Erdos and P6sa proved in [4] that any graph with [GI = n 2 24k, k 2 2 and with f ( n , k ) edges will contain an O k or be isomorphic t o a

+

( ( 2 k - l ) , n - 2k 1). Their theorem cannot simply be extended to n 2 3k as for sufficiently large k, there will be graphs of the following types (see Figure 1 ( b ) ) with more than f ( n , k ) edges and obviously without an Ok: The graphs consist of a (3k - 1) and one or more vertices of valency 1. The number of edges g(n,k) in these graphs is 1 g(n,k ) = -(3k - 1)(3k - 2 ) t n - 3k t 1. 2

Erdos and P6sa also conjectured that any graph with [GI = n 2 3k, k 2 2 , and more than m a x ( g ( n ,k ) , f ( n ,k ) } edges will contain an Ok.In my thesis I proved:

Theorem 4. Let G be a n y graph with n vertices and k 2 2 an integer. If n 2 3k and G contains at least max{g(n, k ) l , f ( n ,k)} edges then G contains an O k or G is isomorphic with a ( ( 2 k - l),n - 2k -t 1).

+

Before the proof it will be convenient with the following lemma.

P. Justesen

302

Lemma 1. Let G be a graph with n verices and e edges, and k 2 2 an integer. Assume n 2 3k and e 2 f ( n ,k ) . If G contains a triangle C with the following properties:

a) C contains a vertex of valency 5 2k

- 1 in G,

b) G - C is a ( ( 2 k - 3 ) , n - 2 k ) ,

then G 2 O k or G is isomorphic to a ( ( 2 k - l ) , n - 2k

+ 1).

Proof of Lemma 1. Let C = C(Ql,Q2,&3)denote a triangle with the properties a) and b), say with w(Q1) 5 2k - 1. By b) we get 3

C v ( Q ; ) = e ( G ) - e ( G - C )t 3 i= 1

2 f ( n , k ) - f(n - 3,k - 1) t 3 2 2nt2k-3. Hence, by a), v(Q1) = 2k - 1, v ( Q 2 ) = n - 1 and v ( Q 3 ) = n - 1, and by b) it follows that G - Q1 % ( ( 2 k - l ) , n - 2 k ) .

If Q1 if joined to 2k - 1 vertices of valency n - 1 in G then alternative 0 2 holds. Otherwise it is not difficult to find an O kin G.

Proof of Theorem 4. First some observations. For k 2 2 and n 2 3k we have g ( n , k ) t 1 2 f ( n ,k ) if and only if 13k-4 1 (1) n i r + m with equality occuring simultaneously. Note that for k = 2 and k = 3 equality sign (only) is consistent with n 1 3k (viz. n = 3 k ) . 1 g ( n , k ) - g(n - 1 , k ) = f(n,k)- f(n-1,k) = 2k-1

(2)

g ( n , k ) - g(n - 3,k - 1 ) = 9k - 9 f(n,lc)- f ( n - 3,lc - 1 ) = 2n t 2k - 6

(3)

And now to the proof, which will be by induction on k . For k = 2 , f ( n , k ) = 3 n - 6 and the theorem therefore follows from Theorem 1. Assume next k >_ 3 and that the theorem holds for k - 1. The induction step in k will be performed by induction on the number of vertices.

Independent Circuits in Graphs and a Conjecture of Erdos and P6sa 303 Let G denote a graph with n vertices satisfying the conditions for k and let XI denote a vertex of minimal valency in G. Put " ( X I )= T . a) Assume first n = 3 k . If T 2 2k then it follows from Theorem 2 that G 2 O k , so assume henceforth T 5 2k - 1. Also T 2 2 since e(G) 2 g ( 3 k , k ) 1. At least two of the neighbours of XI,say X2 and X3, must be adjacent for otherwise

+

3k - 1

Then

3k- 1

3

C "(Xi)5 2k - 1+ 2 ( n - 1)= 8k - 3 . i=l

Define G' = G - X I

- XZ- X3.

Then

e(G') 2 g ( 3 k , k ) t 1 - ( 8 k - 6 ) L g ( 3 k , k ) t 1 - ( 9 k - 9) i.e., e(G') 2 g ( 3 k - 3 , k - 1) t 1. Hence by (1) and the induction hypothesis in k , either G' 2 Ok-l or G' is isomophic to a ( ( 2 k - 3 ) , n - 2 k ) , and hence by Lemma 1 the theorem holds for n = 3 k . b) Assume next that n > 3k and that the theorem holds for n - 1 vertices. As before we only need to consider the case " ( X I )= T 5 2k - 1. If T I: 1 then, by (2),

and hence by the induction hypothesis in n , G - X I _> Ok. So 2 5 T I 2k - 1 remains to be considered. We shall treat the alternatives: X is contained or not contained in a triangle. Assume first that XI is not in any triangle in G. The graph G' obtained through contracting XI with one of its neighbours satisfies IG'l = n - 1 and e(G') = e(G) - 1. Hence by ( 2 ) and the induction hypothesis in n, G' _> O k (G' has more than f ( n - 1,k) edges), and hence also G 1 O k ,as it is easy to see that to every O kin G' there corresponds an O k in G. Assume next that X1 is contained in a triangle C(X1, X2, X3). Then 3

c v ( X i ) 5 2 k - 1+ 2 ( n - 1) = 2n + 2 k - 3. i=l

P. Justesen

304 Define G" = G - XI- X2 - X 3 . w e get

e(G") t 2n + 2k - 6 2 e(G) 2 f(n, k) and hence by (3), Furthermore,

e(G") 2 f ( n - 3, k - 1). e(G") 2 g(n - 3 , k - 1) t 1.

Proof of (6): Assume on the contrary that

g(n - 3, k - 1) t 1 > e(G"). Then, by (5) and (l), n<3+

13(k - 1) - 4 4

1 + 2(k - 1) - 2

- 13k - 5 4

+-2 k1- 4 '

but on the other hand, by (4)and the assumptions,

g(n,k)

- g(n

- 3, k - 1) < e(G)- e(G") I 2n + 2k - 6,

which gives

n> Hence

14k - 6 4

+-

14k-6 13k - 5 1 < (7) 4 2k-4' 4 which contradicts k 2 3, as (7) implies k(k - 3) < 0. So (6) holds. Then by the induction hypothesis in k and Lemma 1, it follows that the 0 theorem holds for G, which completes the proof of Theorem 4. Finally I should like to mention another result. Dirac proved in [3] a theorem similar to that of Corrbdi and Hajnal. He showed for k 2 1 that any graph with at least 3k vertices and minimal valency 2 i ( n + k ) contains k independent triangles. With the help of Theorem 3 I could generalize this to:

Theorem 5. Let G be a graph with 2 3k vertices, k 2 1. If w ( X ) w(Y) 2 n t k for any two nonadjacent vertices X and Y in G , then G contains k independent triangles.

+

As the proof only involves minor changes of Dirac's proof I shall not include it here.

Independent Circuits in Graphs and a Conjecture of Erdos and P&a 305

References [l] B. Bollobk, Eztrernal Graph Theory, Academic Press, London (1978). [2] K. Corrbdi and A. Hajnal, “On the maximal number of independent circuits in a graph,” Acta Math. Acad. Sci. Hungax. 14 (1963), 423439. [3] G. A. Dirac, “On the maximal number of independent triangles in graphs,” Abh. Math. Semin. Univ. Hamb. 26 (1963), 78-82. [4] P. Erdos and L. Pbsa, “On the maximal number of disjoint circuits of a graph,’’ Publ. Math. Debrecen 0 (1962), 3-12. [5] H. Walther and H.-J. Voss, Uber Krezse in Graphen, VEB Deutscher Verlag der Wissenschaften, Berlin (1974).

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Annals of Discrete Mathematics 41 (1989) 307-310 0 Elsevier Science Publishers B.V. (North-Holland)

Contractions to Complete Graphs L. K. Jorgensen Department of Mathematics a nd Computer Science Institute of Electronic Systems, Aalborg University Aalborg, Denmark Dedicated to the memory of G. A . Dirac We survey some extremal problems concerning contraction to complete graphs, including two new theorems of the author. We also show an application to a conjecture of Las Vergnas and Meyniel. In this paper we only consider finite simple graphs. Edge-contraction is defined in such a way that the contraction of an edge in a triangle does not create parallel edges. For any two graphs G and H , we say that G is contractible to H if H can be obtained from a subgraph of G by successively contracting edges. For other definitions see Bollobb [l]. We consider the following problem: determine the maximum number of edges in a graph G with n vertices such that G is not contractible to the complete graph K,, for integers n 2 p 2 2. Denote this maximum by fp(n). By considering the graph K P - 2 En-p+2we see that

+

Dirac [3] proved that equality holds for p 5 5 , and Mader [9] proved that equality holds for p 5 7. As noticed by Mader, the complete 5-partite graph with two vertices in each vertex class, K5(2), shows that this equality does not hold for p = 8 and n = 10. The author [5] has proved that the only graphs with n vertices and more than 6n-21 edges which are not contractible to 1i-8 are the Ks(2)-cockades. A K5(2)-cockade is a graph obtained by identifying K5’s in copies of X5(2) (as in the definition of MP;-cockades below). A K5(2)-cockade has n = 5m vertices, m 2 2, and 6n - 20 edges. Therefore 6n - 21 unless 5 divides n, f8(.) = 6n - 20 if 5 divides n.

{

307

L. K. Jerrgensen

308

The author also characterized the extremal graphs to Mader’s theorem for p 5 7. To state this result we need the following recursive definition of MPi-cockades, where i is a non-negative integer: 1. If H is a maximal planar graph then H

+ K; (H if i

= 0) is an

MPi-cockade.

2. Let G1 and G2 be disjoint MPi-cockades, let 2 1 , . . . ,zg+; be the vertices of a K3+; in G I ,and let y ~, ,. ,y3+i be the vertices of a K3+i in G2. Then the graph obtained from GIU G2 by identifying x j and y j for j = 1,.. . , 3 t i is an MP;-cockade.

.

Theorem [4].

a) If G is a graph with n 2 7 vertices and f 7 ( n ) = 5n - 15 edges, and G is not contractible t o K 7 , then either G is an MP2-cockade or G is the complete 4-partite graph K2,2,2,3. b) If G is a graph with n 2 6 vertices and fe(n) = 4n - 10 edges, and G is not contractible t o IC6, then G is an MP1-cockade. c) If G is a graph with n 2 5 vertices and f5(n) = 3n - 6 edges, and G is not contrnctible to K 5 , then G is an MPo-cockade. Parts b) and c) follow from part a). Part c) also follows from Wagner’s characterization [ll] of the graphs not contractible to K 5 . Using Euler’s polyhedron theorem and the fact that a graph contractible to K 5 is nonplanar, it is easy to se that the MPi-cockades have the correct number of edges and are not contractible to KS+;. To see that IC2,2,2,3 is not contractible to K 7 , note that if a graph with nine vertices is contracted to K7, then at most two edges are contracted and so the graph contains a K 5 . A similar argument shows that K 5 ( 2 ) is not contractible to K 8 . The argument can be generalized to prove that some (Turin-) graphs are not contractible to a complete graph of a given order, but these graphs do not have considerably more edges than K p - 2 En-p+2. An upper bound on f p ( n ) for all values of p was first given by Mader [8]. He proved that there is a constant cp for each value of p 2 2, such that f p ( n ) 5 cpn, and if cp is the infimum of all such constants (for a fixed p ) , then c p 5 [2cp-11. In [9], Mader proved that cp 5 8p10g2(p). A lower bound on c p was found by de la Vega and by Thomason. In [2] and [lo] the following bound, for large p, was proved using random graphs:

+

CP

1$Pdrn.

Contractions to Complete Graphs

309

Kostochka [6] proved that this is the correct order of magnitude of c,, i. e., that for some constant c. Thomason [lo] proved that c, 5 2 . 6 8 p d m for large p . We shall now apply the extremd results to the following conjecture of Las Vergnas and Meyniel [7]: Conjecture. If a graph G contains p disjoint vertex-sets Vl, . . . ,V, such that G[K U Vj] is connected for all i and j, 1 5 i < j 5 p , then G is contractible t o K,.

In an attempt to prove this conjecture by induction (it clearly holds if IGI = p ) , we may assume that every vertex of G belongs to one of the sets Vl,. .. ,V,, and that G[V,] has no edges, 1 5 i 5 p , for otherwise we can contract an edge in G[K]. We may also assume that G[K U V,] is a tree, 1 I i < j 5 p . Thus the number of edges in G is e(G) = E e ( G [ K u V , ] ) i<j

It follows from the theorems of Mader ( p 5 7) and the author ( p = 8), that if p I:8 then G is contractible to K,. So the conjecture of Las Vergnas and Meyniel holds for p 5 8. It remains an intriguing open question for p 2 9.

References [l] B. Bollobas, Eztrernal Graph Theory, Academic Press, London-New York-San Francisco (1978).

[2] W. F. de la Vega, “On the maximum density of graphs which have no subcontraction to K’,”Discrete Math. 46 (1983), 109-110. [3] G. A. Dirac, “Homomorphism theorems for graphs,’’ Math. Ann. 153 (1964), 69-80.

310

L. K.Jtwgensen

[4]L. K. Jprrgensen, “Extremal graphs for contractions to I<,,’’ Ars Combinatoria 25C (1988), 133-148. [5]L. K. Jprrgensen, “Contractions to KB.”Preprint, Institute of Electronic Systems, Aalborg University (1988). [6]A. V. Kostochka, “Lower bound of the Hadwiger number of graphs by their average degree,” Combinatorica 4 (1984), 307-316. [7]M. Las Vergnas and H. Meyniel, “Kempe classes and the Hadwiger conjecture,” J. Combin. Theory ( B ) 31 (1981), 95-104. [8] W. Mader, “Homomorphieeigenschaften und mittlere Kantendichte von Graphen,” Math. Ann. 174 (1967), 265-268. [9]W.Mader, “Homomorphiesatze fur Graphen,” Math. Ann. 178 (1968), 154-168.

[lo] A. Thomason, “An extremal function for contractions of graphs,” Math. Proc. Cambridge Phil. SOC.95 (1984), 261-265. [ll] K. Wagner, “Uber eine Eigenschaft der ebenen Komplexe,” Math. Ann. 114 (1937), 570-590.

Annals of Discrete Mathematics41 (1989) 311-324 Elsevier Science Publishers B.V. (North-Holland)

Triangulated Graphs with Marked Vertices H.-G. Leimer* Hoechst AG Clinical Research Frankfurt, FRG

Dedicated to the memory of G. A . Dirac We investigate strongly decomposable graphs, a special class of triangulated graphs with marked vertices. In statistics, graphs with marked vertices represent models for multivariate data, where some variables are qualitative and some quantitative. Strongly decomposable graphs correspond to models, which admit simple interpretation and solution to the estimation problems. We show that a graph G is strongly decomposable if and only if a suitably defined modification of G is triangulated. This result allows an easy generalization of characterizations, properties and algorithms for triangulated graphs to strongly decomposable graphs.

1

Notation and preliminaries

We shall only consider graphs G = ( V , E ) , which are simple (i.e., without loops and multiple edges) and undirected. The vertex set V is assumed t o be finite and an edge is denoted by {v,w} E E with w,w E V,v # w. We say that v and w are adjacent or neighbours if {w,w} E E . Adj(v) := {w E V : {v,w} E E } denotes the adjacency set of a vertex v E V and if W V , Adj(W) := (UVEwAdj(v))\ W and G ( W ) := ( W , E ( W ) )is the subgraph of G induced by W , where E ( W ) := { {TI,w} E E : w, w E W } . We shall consider graphs with two types of vertices. A V is the set of vertices, we shall call marked, and I' := V \ A contains the unmarked vertices. The definitions and results below will typically be non-symmetric in I' and A. 'Full address: Hoechst AG, Clinical Research, P.O.Box 80 03 20, D 6230 Frankfurt 80, FRG.

311

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312

A set W c V is complete if any two different vertices in W are adjacent. Especially W is complete if IWl 5 1. A graph G is called complete if its vertex set is complete. A c2ique is a maximal (w.r.t. “c”)complete subset of the vertex set V . By a path between two vertices v,w E V we mean a sequence of distinct vertices v1,. . . ,w k E V with {v~,vk} = {v,w} and {vi, v;+i} E E for i E [l,k- 11, where [n, rn] := { z E 2 : n 5 I 5 m} for n,m E 2. If furthermore {v1,vk} E E , then the sequence v1,. . ,vk, 01 is called a cycle of length k. Every other edge among these vertices, i.e., {vi, vj} E E with i , j E [l,k] and 1 < li - jl < k - 1, is called a chord of the cycle. A graph is called triangulated if every cycle of length at least four has a chord. Triangulated graphs are also called rigid circuit graphs or chordal graphs. If a, b are distinct vertices with {a, b} $? E , then a set C E V is called an a, b sepumtor if every path between a and b contains a vertex in C. And C is called a minimal a, b separator if C, but no proper subset of C,is an a, b separator. If A, B and C are pairwise disjoint sets of vertices, then we say that C separates A and B if C is an a,b separator for all a E A and b E B. The triple (A,C,B) is a decompositionof G intoG(AUC) and G(BUC) if V = A U B U C,A # 8 f’ B , C separates A and B and C is complete. The decomposition ( A ,C, B ) is strong if at least one of the three conditions C C A, A C I?, B I? holds. This definition will be motivated in Section 2. If A is complete, then every decomposition is strong. Henceforth we assume that A and B are arranged in such a way that C E A or B E I?. A graph G is called (strongly) decomposable if G can successively be reduced to complete subgraphs, i.e., G is (strongly) decomposable if G is either complete or if there exists a (strong) decomposition (A,C,B) of G into (strongly) decomposable subgraphs G(A U C) and G ( B u C). This recursive definition makes sense since [A U CI < IVI and IB LJ CI < IVI. Several other characterizations of (strongly) decomposable graphs will be given in Theorems 1, 2 and 2’ below one being the well known result, that a graph is decomposable if and only if it is triangulated, see e.g. Dirac [2] or Lauritzen, Speed and Vijayan [4]. Note that, in fact, Lauritzen and Wermuth [5]defined strongly decomposable graphs through the equivalent property iv)’ in Theorem 2’. Both the notion of strong decomposition above as well as Theorem 1 below has led to a considerable simplification of the theory. We close this section by mentioning some obvious implications for decomposable and strongly decomposable graphs.

.

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313

Remark 1. i) G (strongly) decomposable, W G V =+ G ( W ) (strongly) decomposable. ii) G strongly decomposable =+ G decomposable. iii) G decomposable, A complete or V = A + G strongly decomposable.

2

Introduction

Suppose that a subset A of the vertex set V of a graph G = (V,E ) has been marked and that the vertices in I' := V \ A are unmarked. One possibility to characterize the strong decomposability of G is the existence of a so-called reducible ordering such that the vertices in A are ordered lower than those in I?. Expressed in another commonly used terminology this is the existence of a monotone transitive, zero fill-in or perfect elimination ordering for the vertices such that the elements of A are ordered highest. Strongly decomposable graphs were first considered by Lauritzen and Wermuth [ 5 , 6 ] in the following statistical context: let X = (XI,. . .,Xn) be a random vector, where some of the components are discrete (qualitative) and the others are continuous (quantitative) variables, e.g. in a given investigation we might have X1 = sex (male, female), X2 = type of work (skilled, unskilled), X3 = weight and X4 = systolic blood pressure. The aim is to describe (in)dependences among these variables. If every X; is associated with the vertex i of a graph G = (V,E ) with vertex set V = (1,. ,n}, edge set E and A := {i E V : X; is discrete}, then the mixed graphical model M ( G ) for X is defined by the following two conditions:

..

i) The conditional distribution of (Xc I c E I') given ( X , Gaussian, and

Id

E A) is

ii) X; and Xj are statistically independent given (Xr I 1 E V \ { i , j } ) for all { i , j } 6 E . Condition i) is a distributional assumption, while the independence conditions in ii) are directly related to the graph G = (V,E ) , Graph theoretical terminology and results are an important tool for the investigation of these models and conversely these models raise new graph-theoretical problems. This is illustrated by some examples:

I) If A , B , C E V are pairwise disjoint sets of vertices such that C separates A and B in G, then the two groups of variables (X, I a E A )

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I b E B ) are conditionally independent given the variables ( X c I c E C) under the model M ( G ) (see Corollary 4.4 in [ 5 ] ) . Hence, it is very simple t o find from the graph groups of variables that are conditionally independent under any model M ( G ) .

and ( X ,

Assume the model M ( G ) for X = (XuI 2) E V) and W C V. It is important to know when M ( G ) implies the model M ( G ( W ) ) for Xw = ( X , I w E W ) ,since then statistical problems concerning these variables can be investigated in the smaller-dimensional model. If C := Adj(V \ W ) is complete then the conditional independences, which define M ( G ( W ) ) ,are implied by M ( G ) : Assume s,t E W,s# t, with{s, t } f! E . We have to show that X , is independent of X t given ( X , I w E W \ { s , t } ) . C complete implies that not both of s and t are in C and so W \ {s, t } separates {s} and { t } in G. Using I) we get the desired result.

To guarantee that the distributional assumption corresponding to i) for M ( G ( W ) )is implied by M ( G ) , we require that C 5 A or B := V \ W C I' (see Propositions 3.1 and 3.2 in [S]). In that case ( A , C , B ) is a strong decomposition of G into G ( A U C) and G ( B U C). 111) If ( A , C , B ) is a strong decomposition of G , then the estimators for the parameters in M ( G ) can be derived in a trivial fashion from the model M(G(AU C)) and the model corresponding to G ( B U C) (Proposition 9.1 in IS]). If G is strongly decomposable, then M ( G ) can successively be decomposed in that way into simple models corresponding t o complete subgraphs. In the same way as triangulated graphs, strongly decomposable graphs can be characterized by the existence of certain orderings of the vertices or cliques (see Theorem 2 below). Such an ordering explicitly describes a decomposition sequence for M ( G ) . Further statistical applications of these orderings are given in Section 4. Strongly decomposable hypergraphs can be investigated with the same methods as used here for graphs. For details we refer to Leimer [7]. The main tool to investigate strongly decomposable graphs is Theorem 1, where we show that a graph G is strongly decomposable if and only if a suitably defined modification G, of G is decomposable. Several characterizations of decomposable graphs are summarized in Theorem 2 and the corresponding characterizations for the strong case are given in Theorem 2'. In Section 5 we prove Theorem 2' and some useful relations between orderings of the cliques and vertices of a graph G and its modification G,.

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Tarjan and Yannakakis [ll]described a simple linear-time algorithm to test decomposability of graphs. Using Theorem 1 we generalize this algorithm to test strong decomposability (Section 6). The generalized algorithm contains two previously proposed algorithms for the strong case, SMCS and MMCS, (see Leimer [7]) as special cases.

3

Characterization of strong decomposability by the decomposability of a modified graph

With every graph G = (V,E ) ,V = I’ U A, we associate a graph Gm, Gm := (Vm,Em) := (V U {m},E U { { d , m } : d E A}), where m is some vertex not contained in V. We shall show in Theorem 1 that strong decomposability of G and decomposability of G, are equivalent. Lemma 1. i ) Let ( A ,C , B ) be a decomposition of Gm. We can assume m E A U C without loss of generality. If A # { m } then

is a strong decomposition o f G . ii) Let (A, C, B ) be a strong decomposition of G; then ii.1) (A U {m},C,B) is a decomposition of G, if B ii.2) ( A , C U { m } , B ) is a decomposition of Gm if C

E r and E A.

Proof. i): (A’,C’,B’)is a strong decomposition of G, since B’ m E A and C’ A if m E C,respectively. ii) is obvious.

c

c

I’ if 0

If in Lemma 1 i) (A, C, B) is a decomposition of G, with A = { m } then A E C is complete which implies that every decomposition of G is strong. Theorem 1. G is strongly decomposable if and only if Gm is decomposable.

Proof. The equivalence follows from Lemma 1 by induction on IVI. We 0 omit the details of the proof. The graph Gm was derived from G by adding one vertex to the graph. Instead, we can modify G by adding several vertices, so that again the modified graph is decomposable if and only if G is strongly decomposable

H.- G. Leimer

3 16

(see Theorem 1‘). We shall use this in Section 6 to derive an algorithm to test strong decomposability of graphs. For every k E N ,we define: Gm,k := ( V U { m l , * * * , m k } ,

E

u { {mi, m j } , { m i , d } : i7.i E [1,k],i

+ .i,d

EA

1)

where ml, . . .,mk are k different vertices not contained in V. Identifying m and ml we get G, = G,J.

Theorem 1‘. For every k E N , G is strong2y decomposable if and only if Gm,k is decomposable. Theorem 1‘ can be proved in exactly the same way as Theorem 1.

4

Further characterizations of (strongly) decomposable graphs

In this section we state a number of well-known characterizations of decomposable graphs (Theorem 2) and the corresponding characterizations of strongly decomposable graphs (Theorem 2’). In the next section we shall show how Theorem 2‘ can easily be derived from Theorem 2 using Theorem 1. First we define orderings for the cliques and vertices of a graph, which will characterize (strong) decomposability. If C1, ,CT is a sequence of sets, typically the cliques of a graph, then for all t E [1,T], Rt := Ct n U (;: Cq)and St := Ct \ Rt. Note that the St’s are disjoint and R1 = 0 by definition.

.. .

Definition 1. The sequence Cl,.. . ,CT is said to be D-ordered if for all t E [2, TI there is a p < t with Rt 5 C,. If C1,. .. ,CT are subsets of V = I’ U A the sequence is said to be SD-ordered if furthermore for every t E [2,T] Rt c A or St c r. The letters (S)D stand for “(strong) decomposition”, since an (S)Dordering of the cliques of a graph defines a sequence of (strong) decompositions. It is also said that C = {Cl,. , , C T } has the running intersection property if the elements of C can be D-ordered (see Beeri et al. [l]). It will be convenient to write an ordering of the vertices of a graph as a bijection a : [ x , x t IVl - 11 -, V with z E 2,typically z = 0 or x = 1. a-’(w) is called the number of a vertex E V. For any fixed ordering a we define the monotone adjacency set of a vertex v E V as MAdj(w) :=

.

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317

Adj(v) n {w E V : a-l(w) 5 a-'(v)}. Note that some authors define MAdj( v) with the inequality reversed. Following Lauritzen and Wermuth [5] we define:

Definition 2. An ordering a of the vertices of a graph G = (V,E) is reducible, if MAdj(v) is complete for all v E V. a is strongly reducible if furthermore MAdj(v) C A for v E A. The reverse ordering of a reducible ordering is also called a monotone transitive (Rose [9]), perfect elimination (Rose, Tarjan and Lueker [lo]) or zero fill-in ordering (Tarjan and Yannakakis [111). Strongly reducible orderings of the vertices of a graph G are very important in the statistical context, since for such an ordering a the mixed graphical model M ( G ) is equivalent to the so-called recursive graphical model M ( G ,a),which takes the order structure of the variables into account (see Proposition 5.2 and Section 7 in Lauritzen and Wermuth [5]). Similarly M ( G ) is equivalent to a so-called graphical chain model, which assumes a partial ordering of U ST, where S1,. . . ,ST the variables given by the partition V = S1 U corresponds t o an SD-ordering of the cliques of G (Proposition 9.1 in [5]). We show in the next lemma how (S)D-orderings of cliques and (strongly) reducible orderings of vertices are related. If C1, ,CT is an (S)D-ordering of the cliques of a graph, then an ordering a : [l,lVl] + V is said t o be (S)D-generated by C1,..., CT if:

.. .

.. .

or in the strong case, respectively,

Lemma 2. Every (S)D-generated ordering of the vertices is (strongly) reducible. Proof. The proof of the lemma follows from the following fact: if v E St and the ordering of the vertices is (S)D-generated by C1, .., ,CT then 0 MAdj(v) E Ct. Note that the converse of Lemma 2 is not true. (Strongly) reducible orderings, which are (S)D-generated, are not only useful for the proof of the following theorems. In a separate paper we have described an algorithm for the (strong) decomposition of an arbitrary graph, which depends on the existence of such orderings (Leimer [8]).

H.- G. Leirner

3 18

Theorem 2. The following conditions for a graph G = (V,E ) are equivalent: i) G is decomposable; ii) for every clique C ' of G there exists a D-ordering Ci,. . ,CTof the cliques of G with C" = C1; iii) there exists a D-ordering of the cliques of G; iv) there exists a reducible ordering of the vertices of G; v) G i s triangulated; vi) if v, w E V , v # w and {v, w} 4 E , and C is a minimal v, w separator, then C is complete; vii) if v,w E V , v # w and (v, w) !$ E , then there exists a complete v, w separator. The equivalences in Theorem 2 are more or less standard. The equivalence of iv) - vii) is proved aa Theorem 1 and 2 in Rose 191 and some parts of it also earlier by Dirac [2] and Fulkerson and Gross [3]. Dirac [2] also showed that a graph satisfying v) can successively be decomposed into v). i) + ii) can be proved by complete graphs and vice versa, i.e., i) induction on IVI using Lemma 3 of Section 5 (see also Lauritzen, Speed and iv) is proved by Lemma 2. Vijayan [4]). ii) =+ iii) is obvious, while iii) Not all known characterizations are listed in Theorem 2. Further characterizations can be found e.g. in Lauritzen et al. [4], Theorem 2, or in Beeri et d.[l],who gave twelve characterizations for H := (V,{Cl,. .. ,CT}) t o be an acyclic hypergraph, one of them being condition iii) of Theorem 2. The following Theorem 2' generalizes Theorem 2 to the strong case.

.

Theorem 2'. The following conditions for a graph G = (V,E ) , V = I'u A , are equivalent: i)' G is strongly decomposable; ii)' for every A0 C A, with A0 complete, there exists an SD-ordering C1,. .. ,CTof the cliques of G with A0 E C1 and i f A is complete with A C C*for some clique C" of G, then we can choose C1 = C*; iii)' there is an SD-ordering of the cliques of G; iv)' there exists a strongly reducible ordering of the vertices of G; v)' G is triangulated, and for every path d l , c1,. ,cn, d2 with n 2 1, d1, d2 E A and c1,. . ,c,, E I ', we have { d l , dz} E E; vi)' if v, w E V , v # w and {v, w} $! E , and C is a minimal v ,w separator, then C is complete and C C A if v,w E A; vii)' if v, w E V , v # w and { v ,w} $ E , then there exists a complete v ,w separator and ifv, w E A , then there exists a complete v, w separator C C A.

.

..

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319

An algorithm t o generate a strongly reducible ordering a of the vertices of a strongly decomposable graph will be given in Section 6. In Section 5 we shall indicate a proof of Theorem 2' using Theorems 1 and 2. Note that conversely Theorem 2 is just Theorem 2' for I' := V or A := V . Given any strongly reducible ordering a the following reordering rule generates a strongly reducible ordering a*,which order the vertices in A before those in I". Let a be an ordering of the vertices in V = I' U A, S := lAl, n := IVI, {q,.. . ,v6} := A, { v ~ + ~. .,,vn} . := I' with a-l(v1)< . . . < a-l(v6) and a-'(ub+l) < ... < a-l(vn). Then define a* : [1,n] -+ V by a*(i):= vi for all i.

Proposition 1. If a is strongly reducible then a* is strongly reducible. Proof. Proposition 1 follows directly from the definition of a strongly re0 ducible ordering. Since every reducible ordering a with a - l ( d ) < a-'(c) for all d E A and c E I' is strongly reducible, we get that:

Proposition 2. G is strongly decomposable if and only if there is a reducible ordering a of the vertices with a-l(d) < a-'(c) for all d € A and c E r. In the statistical context it is often convenient to use a reducible ordering with the above property.

5

Equivalent properties of G and G,

Before proving Theorem 2' from Theorem 2 we need some lemmas. Let G, G,, Gm,k,etc. be defined as in the previous sections. The following lemma shows how an (S)D-ordering of a sequence of sets can be maintained, if a set is deleted, which is contained in another set of the sequence. Let C1,.. . ,CT be an (S)D-ordered sequence of sets with Ct C C, for some t # p . Assume that p is minimal with that property for t fixed. Lemma 3. i) C1,. . . ,Ct-l, Ct+l,. . .,CT is (S)D-ordered if p

< t;

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320

Proof. The proof of Lemma 3 follows from the definition of an (S)Dordering, since for p > t , R, = C, n (UC : ), = Ct by the minimality of p . 0

.

Assume C1,. . ,CTis a D-ordering of the cliques of G, with m E Ci and Ci := Ct \ { m } for t E [l,T].

Lemma 4. The sequence Ci,. ., ,Ch is SD-ordered and the maximal sets ur.r.t. “c” in this sequence are the cliques of G. Proof. The sequence C; ,., , ,Ck is obviously D-ordered. Fix t E [2, TI and (7;). We have to show that R A or S := Ci \ R I’. R := Cl tl Define D := C,n A, then D U { m } is complete w.r.t. G m and therefore D u { m } C C, for some p , p minimal. We must have p 5 t because otherwise m E C1,D c Ct, D U { m } & C, implies D U { m } c R9, which contradicts the minimality of p . If p = t then m E Ct,hence R C, A. Ifp < t then D 2 R and therefore S c I’. The remaining part of the assertion follows 0 directly from the definition of Gm.

c

c

Applying Lemmas 3 and 4 together one can construct an SD-ordering of the cliques of G given a D-ordering C1,.. ,CT of the cliques of G m with m E C1. Similarly one can construct a D-ordering of the cliques of Gm, given an SD-ordering of the cliques of G. We omit the details of that. The next lemma is used only partly in this section, but we shall make full use of it in Section 6. For k E N fixed, let

.

a,,k

: [-(k

- I), lvl]

+

..

v u {ml,. . . ,mk}

be an ordering of VU { m l , . ,mk} with a,,k([l, IVl]) = V and let restriction of a,,k to [I,IVI].

L e m m a 5. f o r G.

a m , & is

Q

be the

reducible for Gm,k if and only if a is strongly reducible

Proof. The lemma follows directly from the definition of (Ym,k, Gm,k and 0 of a (strongly) reducible ordering. We are now ready to prove that for a graph G the properties i)’ to vii)’ of Theorem 2’ are equivalent.

Proof of Theorem 2’. Applying Theorem 2 to Gm and using Theorem 1 it is sufficient to show that ii) =+ ii)’ + iii)‘ =+ iv)’ + iv) and vi) =+ vi)’ =+ vii)’ I=$ v)’ + v), where the numbers without a dash correspond to properties of Gm i ~ 6described in Theorem 2. Using the previous lemmas,

Triangulated Graphs with Marked Vertices

32 1

the proofs of these implications are easy. Therefore we only give details for ii) + ii)'.

ii) + ii)' : Fix A0 C A, A, complete w.r.t. G. Then A0 U { m } is complete w.r.t. Gm and there exists a D-ordering C1,. . . ,CTof the cliques of Gm with A0 U { m } C C1. It follows then from Lemmas 3 and 4 that there is an SD-ordering Ci,. . . ,C& of the cliques of G with A0 C C;. To prove the second part of ii)' assume that C* is a clique of G with A E C*. Gm is decomposable by Theorem 2 since it satisfies ii), hence G = Gm(V) is decomposable. There exists a D-ordering C1,. . ,CTof the ' = C1 by Theorem 2. This sequence is also SD-ordered cliques of G with C since A C C1.

.

A simple linear-time algorithm to test strong decomposability

6

Tarjan and Yannakakis [ll]described a simple linear-time algorithm t o test whether a graph is decomposable and to derive a reducible ordering of the vertices if the graph is decomposable. This algorithm can be generalized to the strong case using Theorem 1' and Lemma 5 , and we show that the generalized algorithm contains the two previously described algorithms for the strong case, SMCS and MMCS, (see Leimer [7]) as special cases. The proposed algorithm to test strong decomposability of a graph G = ( V , E ) , V = I' u A, is essentially equivalent to applying the algorithm of Tarjan and Yannakakis to Gm,k for any k E N . The algorithm for the strong case consists of two steps: 0

0

STEP 1: For any k E N , generate an ordering cr of the vertices of G by the procedure MCS k described below.

+

STEP 2: Test whether

a!

is strongly reducible for G.

It follows from Theorem 2' that G is strongly decomposable if STEP 2 is successful, i.e., if cr is strongly reducible. We shall show the converse in Theorem 3. Hence a graph is strongly decomposable if and only if STEP 2 is successful. Let G = ( V , E ) be a graph with V = I' U A and n := IVl. For k E N U (0) we define a procedure, which we call MCS i-12, to generate an ordering of the vertices of G. In every step of the procedure let n(v) denote the number of previously numbered vertices adjacent to a so far unnumbered vertex v E V , N ( v ) := n(v) t k if v E A and N ( v ) := n ( v ) otherwise. Then MCS k is defined as follows:

+

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322 0

Number the vertices from 1 to n in increasing order according t o the following rule: as the next vertex to be numbered select an unnumbered vertex v for which N ( v ) is maximal. Break ties arbitrarily.

+

If the vertices v1,. , . ,vn of V can be chosen in that order by MCS k, then every ordering a : [z 1,z -+ n] + V ( z E 2) with a(z i) = o; for i E [1,n]is said t o be MCS + k generated. For k = 0, MCS + k always chooses that vertex v with the maximal cardinality of the set of previously numbered neighbours as the next vertex to be numbered, i.e., MCS t 0 is the rnazirnurn canlinality search (MCS) procedure described by Tarjan and Yannakakis [ l l ] ,(apart from a reverse labelling of the vertices). Hence we shall use the terms MCS and MCS + 0 synonymously. Expressed in our terminology, Tarjan and Yannakakis showed:

+

-+

0

If G is decomposable then every MCS generated ordering is reducible

for G.

The corresponding result for the strong case is given in the following theorem.

Theorem 3. If G is strongly decomposable, k E N and MCS -+ k, then a is strongly reducible for G .

CY

is generated b y

Proof. Let G, k and a : [ l , n ]+ V be as in Theorem 3 and Gm,kwith vertex set V,,k := VU(m1,. .. ,mk}as in Section 3. Let a,,k : [-(k - l ) , n ] -+ Vm,k be an extension of a to an ordering of Vm,k, i.e., such that a is the restriction of a,,& to [ l , n ] . Then cr,,k is MCS generated w.r.t. Gm,k since {ml,. .. ,mk}is complete. But Gm,k is decomposable by Theorem 1’ and therefore a,,k is reducible w.r.t. Gm,k by the above mentioned result of Tarjan and Yannakakis. Theorem 3 now follows directly from Lemma 5. 0 Tarjan and Yannakakis [ l l ]showed how MCS and the test for reducibility can be implemented t o run in O(lVl+lEl) time. In STEP 1 , MCS+k can be implemented exactly as MCS except for a change in the initial values, and it requires only one additional inspection of every {v, w} E E to test strong reducibility in STEP 2 instead of reducibility. Hence the complexity of the above described algorithm for the strong case is also O(lVl [ E l ) . For k 3 lAl, MCS+k numbers the vertices in A first. To prove this, one that the first i selected vertices can show by induction over i , i E [ l ,la[], are in A if k 2 lAl. It follows from this result that the MCS t k procedures are equivalent for k 2 lAl. Furthermore, it can easily be seen that they

+

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are also equivalent t o the scxalled modified maximum cardinality search (MMCS) procedure described in Leimer [7]. There is another procedure, called strong maximum cardinality search (SMCS), to generate an ordering of the vertices in the strong case. SMCS was also described in Leimer [7]. For a graph G = (V,E ) , V = I' U A it can be defined as follows: 0

SMCS for G is defined as MCS but breaking ties always in favour of an element of A.

SMCS is equivalent t o MCS + 1 if ties are broken in favour of an element of I' (contrary t o SMCS !) in the MCS 1 procedure. To prove this, one can again show by induction on i , i E [I,IV(],that the assertion is true for the first i selected vertices. SMCS has the advantage that it is a special case of MCS and that an SMCS-generated ordering of the vertices is also SD-generated by the cliques of G (see Section 4 for the definition of an SD-generated ordering and Theorem 4.3.1 in Leimer [7] for a proof). On the other hand, generating an ordering by MCS k, k 2 lAl, has the advantage that the vertices in A are numbered as 1,. .,[A[. Then the ordering is strongly reducible if and only if it is reducible, which simplifies STEP 2. Note that we can also combine MCS k with the reordering rule of Proposition 1, Section 4, so that A gets the numbers 1,., . ,lAl even for small k E N . But the orderings derived in that way are in general not the same for all k E N . Example 4.2.2 in Leimer [7] shows that an ordering might be generated by MCS 1 (even by SMCS) with renumbering according to Proposition 1, but that it cannot be generated by MCS k for k 2 [A[. By a little extension of that example one can also show that the converse is not true either. Summarizing, we can say that there are several different ways to generate an ordering of the vertices in STEP 1, which in general are not equivalent. We might use MCS k for any k 5 IAI, especially SMCS, and we can apply the reordering rule or not. The choice will depend on the application.

+

+ .

+

+

+

+-

7

Acknowledgements

This paper was completed while the author was with the University of Mainz (FRG) and was visiting Aalborg University (Denmark). Especially he wants to thank Andries Brouwer and Steffen L. Lauritzen for stimulating discussions which initiated this paper.

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H . - G . Leimer

References [l] C. Beeri, R. Fagin, D. Maier and M. Yannakakis, “On the desirability of acyclic database schemes,” J . Assoc. Cornput. Mach. 30 (1983), 479-5 13. [2] G. A. Dirac, “On rigid circuit graphs,” Abh. Math. Sem. Univ. Hamb. 25 (1961), 71-76. [3] D. R. Fulkerson and 0. A. Gross, “Incidence matrices and interval graphs,” Pacific J . Math. 15 (1965), 835-855. [4] S. L. Lauritzen, T. P. Speed and K. Vijayan, “Decomposable graphs and hypergraphs,” J. Austral. Math. SOC.( A ) 36 (1984), 12-29. [5] S. L. Lauritzen and N. Wermuth, “Mixed interaction models,” R e . report R 84-8, Institute of Electronic Systems, Aalborg University (1984). [6] S. L. Lauritzen and N. Wermuth, “Graphical models for associations between variables, some of which are qualitative and some quantitative,” Res. report R 87-10, Institute of Electronic Systems, Aalborg University (1987). [7] H.-G. Leimer, “Strongly decomposable graphs and hypergraphs,” Res. report 85-1, Department of Mathematics, University of Mainz (1985). [8] H.-G. Leimer, “Optimal decomposition by complete separators,” Res. report 86-3, Department of Mathematics, University of Mainz (1986). [9] D. J. Rose, “Triangulated graphs and the elimination process,” J. Math. Anal. Appl. 3 2 (1970), 597-609. [lo] D. J. Rose, R. E. Tarjan and G. S. Lueker, “Algorithmic aspects of vertex elimination on graphs,” SIAM J. Cornput. 5 (1976), 266-283. [ll] R. E. Tarjan and M. Yannakakis, “Simple linear-time algorithms t o test chordality of graphs, test acyclicity of hypergraphs, and selectively reduce acyclic hypergraphs,” SIAM J . Cornput. 13 (1984), 566-579.

Annals of Discrete Mathem'atics41 (1989) 325-332 0 Elsevier Science Publishers B.V. (North-Holland)

On a Problem upon Configurations Contained in Graphs with Given Chromatic Number F. Lescure Department of Mathematics University of Pierre and Marie Curie Paris, France

H. Meyniel Department of Combinatorics Center of Social Mathematics Paris, France

Dedicated to the memory of G. A . Dirac We consider the following problem: Do simple graphs with chromatic number y contain the following configuration: 7 vertices pairwise linked by a system of paths pairwise edge disjoint? We prove this is the case for y = 5 and 7 = 6.

1

Introduction, definitions and notations

Definitions and notations are classical, see [l]. Graphs considered here are undirected, finite, without loops and simple, except for one edge or two which can be double. An immersion of K k + l denoted by k k + l is constituted by a set of k 1 vertices xi, 1 5 i 5 k 1, pairwise linked by elementary paths pij with pij not containing xp, t # i, j and the pij being pairwise edge disjoint. The xi, 1 5 i 5 k + 1 are the basic vertices of k k + l . A. Bouchet [2], as a consequence of results on isotropic systems, proved that every four-regular graph contains an immersion of Ks. Of course, one may think that this can be generalized to k-regular graphs, k any integer (as it is obviously true for CC = 2, and also for k = 3 in view of a well known theorem by Dirac which asserts that if a graph has minimum degree at least 3, it contains a homeomorphic copy of I<*).

+

+

325

F. Lescure and H. Meyniel

326

Professor Mader suggested us to study the following problem: What happens if one supposes that the minimal degree is at least k? This has led us to give a “graphical” proof of a statement stronger than the theorem by Bouchet, namely that, if a graph G has minimum degree a t least 4, then it contains a k!j. As proven also in this paper, if the minimum degree of G is at least 5, then G contains a k ~and , F. Lescure [5]has proven the next case, with minimum degree at least 6 (but the proof is rather long). P. Seymour’ found counterexamples to the natural problem:

Problem 1. Is it true that if a graph has minimum degme at least k it contains a Rk+l ?

The counterexample by P. Seymour is the nine-regular graph constituted by K l z in which one deletes the edges of four triangles. It can be extended to higher degrees by adding vertices linked to every other vertex. Hence Problem 1 is still open for minimal degree 7 or 8. But no counterexample has yet been found for our Problem 2:

Problem 2. Is it true that if a graph G has chromatic number y then it contains a K-,? As a consequence of the results of this paper, this is indeed true for y = 5 and y = 6, and F. Lescure has also proved it for y = 7.

Note that Problem 2 is related to a conjecture by Hajbs, namely that if a graph has chromatic number 7, it contains a homeomorph of I
2

The results

In Section A we establish the proof of the case k = 4 for Problem 1.

2.1

Section A

In fact, in order to prove the theorem announced above, we need to prove a stronger statement, namely the following theorem. ‘This counterexample has been communicated to us by N. Robertson whom we are pleased to thank.

Configurations Contained in Graphs with Given Chromatic Number

327

Theorem 2.1. Let G = (V,E ) be a simple graph with at least two vertices and all degrees 2 4 except maybe for one edge which can be double and a vertex which can have degree less than 4, the double edge and the vertex with degree less than 4 being adjacent i f they both exist simultaneously. Then G contains a k5. Proof. The smallest graph satisfying the assumptions is K5 itself. Our proof goes by induction on the number of vertices or on the number of edges and we consider a certain number of cases: in each case we derive from G a graph G' with strictly less vertices or strictly less edges, which satisfies the induction hypothesis. In order t o make the proof easier to read, we shall give only the construction of G'. In each case, the reader is invited to check that G' satisfies the induction hypothesis and that any 8 5 found in G' can be transformed into a 8 5 in G. Case 1: G contains a vertex x of degree dG(x) 5 3. Case 1.1: d c ( x ) E {O,l}: delete x. Case 1.2: d ~ ( z=) 2. Subcase 1.2.1: & ( z ) = 2 and I'G(x) = {y}, i.e., there is one double edge adjacent to x: delete x. Subcase 1.2.2: edge ( Y , 4

dG(s) =

2 and r G ( x ) = { y , z } , y

# I: delete x

and add one

Case 1.3: & ( x ) = 3. Subcase 1.3.1: ~ G ( z =) 3 and r G ( x ) = { y , z } , i.e., there is one double edge adjacent to x: delete x and add one edge ( y , z). Subcase 1.3.2: dG(x) = 3 and r G ( 2 ) = { y , z , t } , one of the vertices y , z , t (let us say y ) has degree 2 5: delete the edge ( x , y ) . Subcase 1.3.3: d ~ ( z=) 3 and r G ( X ) = { y , z , t } , d G ( y ) = d ~ ( z=) dG(t) = 4 and one of the edges (y, z ) , ( y , t ) ,( z ,t ) is missing: add one of the missing edges and delete vertex x. Subcase 1.3.4: If we are not in one of the preceding cases, the subgraph induced by x U r G ( X ) and the neighbours of z U r G ( X ) must take one of the shapes of Figure 1 (there can exist other edges only between u,v , w). In situation (a): delete x u r G ( X ) . In situation (b): contract x U r G ( X ) into one vertex x' (which has u , v, w as neighbours).

F. Lescure and H.Meyniel

328

Figure 1. In situation (c): contract z U r G ( 5 ) into one vertex x', obtaining a double edge (z',u ) and a single edge (z',v). Case 2: Now we consider a graph with

all degrees 2 4. Let us suppose

there is some double edge ( z , y ) . Case 2.1: One of the vertices

z,y has degree 2 5: delete one edge

(2,

y)

((z,y) becomes a simple edge).

= 4. Let {z1,22,y} be the neighbours of z and {yl, y2, z} the neighbours of y. If one of the (not necessarily distinct) vertices z1,22, y1, y2 has degree 2 5, delete one of the corresponding edges ( z , z 1 ) , ( z , z z ) , ( y , y 1 ) or (Y,Y2). Case 2.3: Now we have Case 2.2: d ~ ( z )= d&)

dG(z) = dG(Y) = dC(z1) = d G ( z 2 ) = dG(y1) = dG(y2) = 4, and let us suppose one of the edges ( q , z 2 ) , (z1,y) or (zz,y) is missing (let us say ( 2 1 , y), for example, is missing): add an edge (z1,y) and delete the edge (z,z1) and one of the edges (2, y). (Similarly if (21,z2) or ( 5 2 , y) is missing). (The same thing can be done if we consider the edges (z,yl), (x,y2) or (y1, y2) and the vertex y). Case 2.4: If we are not in one of the preceding cases, the subgraph induced by X , y and r G ( Z ) U r G ( Y ) has the shape of Figure 2. But we notice immediately (as dc(z1) = d ~ ( z 2 )= 4) that it is sufficient to consider a graph G' obtained from G by contraction of { z , y , z 1 , z ~ }into one vertex u.

Configurations Contained in Graphs with Given Chromatic Nurn ber

329

Figure 2. Case 3: Now

dG(x) 2 4 for each z E V and there is no double edge.

Case 3.1: dG(x) 2 5 for each z: delete any edge. Case 3.2: Consider one vertex x with degree 4 ( r G ( x ) = ( x 1 , x 2 , x 3 , ~ 4 } ) and suppose there is one edge in the subgraph induced by the neighbourhood of x which is missing (let us say the edge ( q , x 2 ) is missing): add an edge (z1,22) and delete edges ( % , X I ) , ( 2 , ~ ) . Case 3.3: Now d G ( z ) = 4 and r G ( X ) is a complete graph: we have indeed found the required configuration. 0 2.2

Section B

We now prove the case k = 5 of our Problem 1. As in the case k = 4 we derive a graph G' with strictly less vertices or edges but the demonstration is a bit more complicated and we shall give eventually some indications on the way we get a k6 in G from a k6 in G'. As in the case k = 4 we need to prove a stronger statement, namely our theorem 2.

Theorem 2.2. Let G be a simple graph with at least two vertices and at most one or two double edges and at most one vertex x of degree dG(x) 5 4, with the two following additional conditions: (a) the vertex of degree 5 4, i f it exists, is adjacent to the double edge(s) if it (or they) exists, (b) i f there are two double edges, they are adjacent to a vertex y of degree d G ( y ) 5 6. Then G contains a k6. Proof. The smallest graph satisfying the assumption is K6 itself. We now proceed by induction. Let x be the vertex with degree dG(x) 5 4 if it exists.

F. Lescure and H. Meyniel

330

We can indeed suppose that r G ( 2 ) is a complete graph and that for any x' E r G ( 2 ) , d ~ ( d =) 5. (If any edge ( u , v ) with u,v E r G ( 5 ) is missing delete edges (x,u ) ,( 5 , ~and ) add one edge ( u , v). If 5' with 8' E I'G(z) has degree 2 6, delete edge (x,~').) Case 1: d ~ ( z E) {0,1,2}, The demonstration is the same as in the case

k = 4.

Case 2: d ~ ( z=) 3.

Case 2.1: ~ G ( z = ) 3 and

rG(Z)

= {y, z } . Delete x and add one edge (y, 2).

Case 2.2: d ~ ( x )= 3 and r G ( Z ) = {y,z,t}. Contract vertices s,y into one vertex u ( u is joined by double edges to z and t and has degree 6 in GI). Subcase 2.2.1: Suppose a 2 6 in G' has not u for basic vertex; then it is easy t o derive a 2 6 in G, remarking that if a path uses successively the edges ( z , u ) and ( u , t ) , we can eventually replace them by edges ( x , z ) and (x,t). Subcase 2.2.2: If u is a basic vertex in a 2 6 of G', we can take y as a basic vertex for the required 2 6 in G unless the path uses the two double edges. In this case, we can use t or z as a basic vertex for the 2 6 in G. Case 3:

d ~ ( 2= )

4.

Case 3.1: d ~ ( x )= 4 and r G ( Z ) = {y,z}. As an edge (y,z) exists and d ~ ( y= ) & ( z ) = 5 , contract (2,y, z } into one vertex. Case 3.2: d ~ ( x )= 4 and r G ( Z ) = {y,z,t}, and let us suppose the edge is double, then contract {x,y} into one vertex.

(2,y)

Case 3.3: ~ G ( z )= 4 and r G ( Z ) = { y , z , t , u } . Consider the set N of neighbours of {y, z, t, u} in G - {x,y, z, t , u } . Subcase 3.3.1: IN1 = 1. Delete vertices x,y, z , t , u. Subcase 3.3.2: IN1 = 2. Let N = {v,w} and let v be adjacent to vertices {y, z, t} and 'w to u (for example). Delete vertices 5,y, z, t , u and add one edge (v, w). Subcase 3.3.3: IN1 = 2 and v, 20 are adjacent t o two vertices among vertices { z , y,t,u}, or IN1 = 3, or IN1 = 4. Contract z,y,z,t,u into one vertex.

G has all degrees 2 5 and two double edges. Let y be the vertex adjacent to the double edges (d&) 5 6). As in the beginning of our proof we can suppose that all vertices adjacent to y have degree 5 and that r G ( 9 ) is a complete graph. Then the cocycle related t o y U r G ( Y ) is of size 2 if d ~ ( y =) 6 and of size 4 if d ~ ( y =) 5 . If this cocycle is of size 2, Case 4: Now

Configurations Contained in Graphs with Given Chromatic Number

331

contract y U rG(y) into one vertex; if it is of size 4 the Cases are similar to Cases 3.3.2 and 3.3.3.

G has all degrees 2 5 and one double edge (x,y). Case 5.1: One of the vertices adjacent to z or y, say x, (let it be z‘) has degree 2 6, delete the edge ( x ’ , ~ ) . Case 5.2: All vertices adjacent to x and y have degree 5 and one edge is missing in rG(Z); let it be ( u , v). Delete edges (z, u ) , (x,v) and add an edge Case 5: Now

(u,

4.

Case 5.3: All vertices adjacent to x,y have degree 5 and rc(x)is a complete graph. The cocycle related to the set of vertices z U rG(Z) is then of size 3. Subcase 5.3.1: The three edges of this cocycle are adjacent, delete SurG(Z). Subcase 5.3.2: In the other case, contract z U r G ( $ ) into one vertex.

G is a simple graph with all degrees 2 5. Case 6.1: All degrees are 2 6: delete any edge. Case 6.2: One edge is missing between two vertices ( u ,v) adjacent to some vertex x with degree 5: delete edges (x,u),(x,v)and add one edge (u,v). Case 6.3: Let x be a vertex with degree 5 and rG(5) be the complete graph, Case 6:

then

5

U rG(5) generates the required configuration.

0

Acknowledgement We are pleased to thank Professors Bouchet and Mader for helpful conversations on this subject.

References [l] C. Berge, Graphes, Gauthier-Villars, Paris (1983). [2] A. Bouchet, “Isotropic Systems” To appear. [3] P. A. Catlin, “Hajos’ graph colouring conjecture, variations and counterexamples,” J . Combin. Theory (B) 26 (1979), 268-274. [4] P. Duchet and H. Meyniel, “Must a p-chromatic graph contain a pscheme,” Progress in Graph Theory (A. Bondy and U. S. R. Murty, eds.), Academic Press (1984), 525-527. [5] F. Lescure, Thtse J t m e cycle, Universitd Paris VI (1985). To appear.

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Annals of Discrete Mathematics 4 1 (1989) 333-340 0 Elsevier Science Publishers B.V. (North-Holland)

On Disjoint Paths in Graphs W. Mader Institute of Mathematics University of Hanover Hanover, FRG

Dedicated to the memory of G. A . Dirac We survey some results on the maximum number of edge-disjoint paths in a graph, where a pair of upper bounds is assigned to each vertex, one for the number of paths ending in it, the other for the number of paths passing through it. As an application, we prove a conjecture due to M. Hager.

All graphs and multigraphs considered in this paper are supposed to be finite. Multigraphs may contain multiple edges, but no loops, whereas graphs have only simple edges. We consider a multigraph G in conjunction with two functions e and p from the vertex set V ( G ) to the non-negative integers N , and we denote such a ‘valued’ multigraph by Gg. In [l], T. Gallai posed the problem to determine the maximum number p(Gg) of edgedisjoint paths in a valued multigraph G: such that every vertex z is an endpoint of at most e(z) paths and an intermediate point of a t most p ( z ) paths. In this problem, and hence in this paper, a path always has length a t least 1 and different endpoints. The analogous problem, which arises by admitting also ‘closed paths’, was solved by T. Gallai in [l] (cf. also [7]). In [6],we determined p(G:) as the minimum of certain values of the partitions of V ( G ) , and we specialized this result by various choices of the functions e and p . We shall survey these results here and apply them to prove a conjecture of M. Hager [2]. Let us first fix some notation. The edge set of the graph G is denoted by E(G) and its set of components by L ( G ) . A C E L(G) is called trivial, if (CI = 1. The edge joining the vertices z and y in a graph is denoted by [z,y]. Let d ( z ; G ) be the degree of the vertex z in G. For disjoint subsets C1, , ,Ck of V ( G ) , K(CI,. . ,Ck;G) denotes the number of edges , Ck in G. For X & V ( G ) , let joining any two different classes C1, B ( X ; G )be the set of those vertices of X ,which are joined by an edge to

..

.

.. .

W. Mader

334

a t least one vertex of V ( G )- X , and let G ( X ) be the subgraph spanned by X in G . For x E V ( G ) , N ( x ; G ) denotes the set of neighbours of 2 in G. An x , y - p a t h in G is a path with (different) endpoints x and y, and for X , Y V ( G ) ,an X,Y-path is an x,y-path with x E X and y E Y . The domain of a function f is denoted by Dj. Let G be a graph and A V ( G ) . If we choose the functions e , p such that e ( x ) 2 d (x;G )and p(x) = 0 for all x E A and e ( x ) = 0 and p ( x ) = 1 for all x E V ( G )- A, then p(GZ) becomes the maximum number p ( A ;G) of openly disjoint A-paths in G, that means paths with both endpoints, but no intermediate point in A, which are disjoint on V ( G )- A . The determination of p ( A ; G )was the key for the solution of Gallai’s general problem. Obviously, we may confine ourselves to an independent A E V ( G ) , that means a set of vertices A such that E ( G ( A ) )= 8.

c

Theorem 1 [5]. For independent A

c V(G),

. ..

where the minimum is taken over all partitions (Co, C1, ,C,) of V (G)- A such that G - [CO U U E l E(G(Ci))]does not contain an A-path. The first proof of Theorem 1 was given in [5] by applying Menger’s Graph Theorem. (On the other hand, Theorem 1 easily implies Menger’s theorem.) In [3], L. Loviisz deduced a somewhat generaliaed version of Theorem 1 from his polymatroid matching theorem. He considered disjoint vertex sets A1,. . . ,Ak C V ( G ) ,and he determined the maximum number P(A1,... ,Ak; G ) of disjoint paths P in G, with each path P being an A;,Ajpath for any i # j. It is no problem to get one of the numbers p ( A ; G ) and P(A1,. ,Ak; G ) from the other. Given A = {al,. . . ,ah} E V ( G ) ,we split every vertex ai E A into an independent set Ai of d(ai;G) vertices of degree 1 to obtain a graph, which we denote by Then p ( A ; G ) = P(A1,. ,Ak; Vice versa, given disjoint vertex sets A l , . , Ak in G , define F by adding k new vertices { a l , ,a k } =: A to G and joining ai to all the vertices of Ai for i = 1, ... , k . Then P(A1,. ,Ak;G ) = p ( A ; It is also easy to get a formula for P(A1,. ,Ak; G ) , if the sets A1, . . . , AI, are not necessarily disjoint. We will derive such a formula only for k = 2.

..

..

z.

m.

Theorem 2. For A1,Az

.. . ..

c V(G),

..

..

q.

335

On Disjoint Paths in Graphs

where the minimum is taken over all partitions (Co,C1,. .. ,C,)of V ( G ) such that ICn(A1 uA2)l 2 2, but C n A 1 = 0 or C n A 2 = 0 holds for every non-trivial C E L(G - [COU UyZl E(G(Ci))]). Proof. Let A := {u1,a2} U { a , : z E A1 n A * } be a set of 2 elements such that A n V ( G )= 8 and define the graph G by

+ IAl n A21

V ( c ):= V ( G )U A and

E ( c ) := E(G)U { [ a l , x ] : x E A1 - A2 } U{[aa,~]:x€Az-A1} U{[a,,x]:xEA1nA2). Then p(Al,A2;G) = p ( A ; g , and by Theorem 1 there is a partition (CO, C1,. .. ,Cn) of V ( G ) such that G - [COU U z 1 E(G(Ci))]does not contain an A-path and

holds. Choose such a partition (Co,C1,. possible. 0 bviously,

. . ,Cn)so that

n is as small as

B(Ci;G - Co) = B(C;; G - Co) U (Ci n (A1 U A2)). Let C be a non-trivial component of G - [COU U b l E(G(C;))].We may assume C n Ci = 0 for i = 1, . . . , k, and C n C; # 0 for i = Ic 1, . . . , n. As C is non-trivial, we have k I n - 2. Of course,

+

ci := c1, . . ., ci := c k , ci+1:= c k + l u is a partition of V ( G )- COsuch that - [COU U!:

* * *

u cn

E(G(C,'))]does not

contain an A-path. Let us suppose IC n (A1 U A2)l I 1. Then

W. Mader

336 and hence n

C

I B ( C i + l ; E - C O )5~ l B ( C ; ; E - Co)I - ( n - k i=k+l

- 1).

But this contradicts the choice of n. Hence IC fl (A1 U A2)1 2 2, and C n A1 = 0 or C n A2 = 0, as G - [COU UyZl E(G(Ci))] does not contain an Al, Az-path. As on the other hand n

F(Ai,Az;G) I lcol t C [ $ l B ( C i ; G

- Co) U (Ci n (A1 U A2))Il

i=l

.

for every (Co,C1,. , ,C,) satisfying the condition of Theorem 2, the proof 0 of Theorem 2 is complete.

.

If A1 = A2 in Theorem 2, the partitions (CO, C1,. . ,Cn) considered there have the property that all components of G - [Co U lJy.l E(G(Ci))] are trivial; hence, for i = 1, , . , n , Ci consists of components of G - Co. So Theorem 2 implies immediately the generalization of Tutte's famous l-factor theorem (i.e., the case A1 = A2 = V(G)) given by Gallai in [l]:

.

c

For independent A V(G), we consider the sets T C V(G) - A with the property that G - T does not contain an A-path; we call such a vertex set T A-separating and define .r(A;G) := min{ IT/ : T A-separating}. Of course, we have always p(A; G) 5 T(A; G), and Menger's theorem is the equality p(A; G) = T ( A ;G) for /A[ = 2. But obvious examples show that equality does not hold in general. For instance, let G arise from a complete graph K 2 n + l by adding an independent set A of 2n 1 vertices and joining these vertices by 2n 1 independent edges to K 2 n + l . Then p(A; G) = n = '.r(a;G). It was conjectured by T. Gallai in [l]that always p(A;G) 2 TTT(A;G). This conjecture is easily derived from Theorem 1.

+

+

Corollary 1 [6]. For every independent vertea: set A in a graph G,

p(A; G ) 2 ~ T ( AG). ;

If we define

c

T(A1,A2; G) := min{ IT1 : T V(G) such that G - T does not contain an A,,A,-path},

we get a similar corollary from Theorem 2.

337

On Disjoint Paths in Graphs Corollary 2 . For d l vertex sets A1, A2 in F(A,,A2;G)

Q

graph G,

1 3%41,A2;G).

Proof. By Theorem 2, there is a partition (CO, C1,. . . ,Cn) of V(G) such that G - [CO U UZ1 E(G(Ci))] does not contain an Al,Az-path and

holds, where Bj := B(Cj;G - Co) U (Ci n (A1 U A2)) for i = 1, . . ., n. P ut B: := Bi,if lBjl is even, and Bi := Bj - {b;} for any bj E Bi, if IB;(is odd. Then every A1, A2-path in G contains a vertex of T := COU UZ1 Bi, and so we get n

'f(A1,A2;G) 5 IT1 5 2lcol

+ CIBII = 2@41,A2;G).

0

i=l

We shall point out now that Corollary 2 immediately implies Conjecture 1 of the paper [2]. First we need a concept introduced by L. Montejano and V. Neumann-Lara in [8]. For vertices 5 # y in a graph G, they defined pzn(z7 y; G) := max{ IS1 : S system of openly disjoint x , y-paths of length a t least n } and TZn(5,Y;G) := min{ IT1 : T V(G) - {x,Y} such that G - T has no z,y-path of length a t least n } , and they proved in [8] for all n 2 2,

For n = 3, M. Hager [2] improved this to

and he conjectured p23(5,?/;G)

1 +Tz3(5,Y;G).

This conjecture is immediately implied by Corollary 2 (and vice versa, too), because P23(5,Y; G) = P("; G),N(y; G); G - (5, Y})

W. Mader

338 and

713(5,9;G)= T ( N x ; G ) , N ( y ; G ) ; G- b , Y } ) for non-adjacent vertices x # y. Let us now return to Gallai's general problem, which was solved in [6] by applying Theorem 1 to a suitably modified graph. Let GZ be a valued multigraph, where e and p are functions from V ( G ) to N . As usual, we e(x) for X V ( G ) . The set of all (ordered) write e ( X ) instead of EXEX partitions (C,,,, Cm+l,. ,Cm+k) of V ( G )for any integers m 5 0 5 m k is denoted by P(G). Let P = (C,,,, ,Cn) E P(G). We call a function f : Df-+ { m ,.. .,-1) P-admissible, if

..

+

.. .

n

n

i=l

i= 1

U~ ( c -i u:=~ ; ~ Cj) E ~f E Uci and f ( x ) = f ( y ) for all adjacent vertices x and y, which belong to different ones of the classes C1, . . . ,Cne For i > 0 and P-admissible f,

+ + + ~ ( c iUjj =1m C j ; G ) - C

vfp(Ci):= e(C;) e(Ci n Df) p(Ci n Dj)

~ ( 5~ f, ( x )G). ;

XECinDf

Furthermore,

The following theorem solves the above mentioned problem of T. Gallai.

Theorem 3 [6]. For every multigraph G valued by functions e , p from V ( G ) to N ,

If we choose the function p such that p(x) = 0 for all x E V ( G ) ,then p(Gg) becomes the maximum number of edges in a (spanning) subgraph H

of G satisfying the condition d ( s ; H ) 5 e(x) for all x E V ( G ) . Hence, Theorem 3 yields Tutte's criterion for the existence of an e-factor [9](cf. Theorem 4 below), as the existence of an e-factor is equivalent to p(G$')2 i e ( V ( G ) ) .If we choose a set of two vertices A = { a l , a 2 } and the functions e and p as in the paragraph before Theorem 1, then p(GZ) = p( { a l , up};G).

Disjoint Paths in Graphs

339

So, Theorem 3 is a common generalization of Menger’s theorem and of Tutte’s theorem. Let us turn now to an analogue of Corollary 1. We call a path A C G (with different endpoints) admissible in GZ, if e ( z ) > 0 for both the endpoints z of A , and p(x) > 0 for all the intermediate vertices z of A. A triple (T,S, El) with T U S C V(G) and E’ E E(G) is called GZ-separating, if every admissible path in Gf has an endpoint in T , an intermediate vertex in S, or an edge in El. (For instance, if we take p as in the first sentence of the preceding paragraph, then (T,0,E’) is Gf-separating, iff (G - E’) - T does not contain adjacent vertices z and y with e ( z ) > 0 and e ( y ) > 0. If, in addition, e ( z ) = 1 for all z E V(G), then (T,0,0)is Gz-separating, iff T is a vertex cover, that means E(G - T ) = 0.) The inequality p(GZ) 5 r(GZ) := min{ e ( T )

+ p(S) + IE‘I : (T,S, El) Gf-separating}

is again obvious, but equality does not hold in general. But Theorem 3 gives the best lower bound for p(Gf).

Corollary 3 [6]. p(G{) 2 ~T(GZ). Specializing the functions e and p in different ways, one gets various interesting results (cf. [6]), but we will confine ourselves to one further case. Let A be a subset of V(G), and the functions e , p may satisfy the condition e ( z ) = 0 for all x E V(G) - A, p ( z ) = 0 for all x E A, and p ( z ) large enough, say p(z) 2 IE(G)I, for all x E V(G) - A. Then X(A; G+) := p(GE) depends only on the restriction elA of e to A and is the maximum number of edge-disjoint A-paths such that every x E A is contained in not more than e ( z ) of these paths. The following theorem is easily derived from Theorem 3.

Theorem 4 [el. Given a rnultigraph G, a subset A e: A --+ N , then X(A;G,) =

V(G), and a function

/

min

(e(C0) +tc(Cl,...,Ck;G)

(co,ci,...,Ck)E’P(A;G)

where P(A;G):=

{ (CO,Cl,...,Ck):

Co,(71,. . .,Ck disjoint subsets of V(G) (k E N arbitrary) with Co A and ICi n A1 = 1 for all i = 1, .. . , k}.

W. Mader

3 40

Taking A = V(G) in Theorem 4, we have lCll = ... = Ickl = 1, and we get immediately Tutte’s general e-factor theorem. If we choose the function e so that e(x) is large enough for all x E A, then X(A;G,) is the maximum number of edge-disjoint A-paths. Then the minimum in Theorem 4 is attained by a (Co,Cl,. . . ,Ck)E P(A; G) satisfying Co = 8 and C n A = 0 for all C E L(G-U;=,C;);hence k = 1Al. We get, therefore, the main result of [4]: The maximum number of edge-disjoint A-paths is equal to min

{ ~(ci,. . . ,CI,;G) t Cc,=qc-Utlc,)L+(C, U L c i ; G ) J : k = IAJ and C1,.. . , CI, disjoint subsets of V(G) satisfying IC; n A! = 1 for dl i = 1, ... , k

}.

References [11 T. Gallai, “Maximum-Minimum-Satze und verallgemeinerte Faktoren von Graphen,” Acta Math. Acad. Sci. Hungar. 12 (1961), 131-173. [2] M. Hager, “A Mengerian Theorem for paths of length at least three,” J. Graph Theory 10 (1986), 533-540.

[3] L. Lovisz, “Matroid matching and some applications,” J . Combin.Theory (B) 28 (1980), 208-236. [4] W. Mader, “Uber die Maximalzahl kantendisjunkter A-Wege,” Arch. der Math. 30 (1978), 325-336. [5] W. Mader, “Uber die Maximalzahl kreuzungsfreier H-Wege,” Arch. der Math. 31 (1978), 387-402. [6] W. Mader, “Uber ein graphentheoretisches Problem von T. Gallai,” Arch. der Math. 33 (1979), 239-257.

[7] W. Mader, “Uber ein graphentheoretisches Ergebnis von T. Gallai,” Acta Math. Acad. Sci. Hungar. 35 (1980), 205-215. [8] L. Montejano and V. Neumann-Lara, “A variation of Menger’s theorem for long paths,” J. Combin. Theory (B) 36 (1984), 213-217.

[9]W. T. Tutte, “The factors of graphs,” Canad. J. Math, 4 (1952), 3 14-328.

Annals of Discrete Mathematics41 (1989) 341-346 0 Elsevier Science Publishers B.V. (North-Holland)

Conjecture de Hadwiger: Un Graphe k-Chromatique Contraction-Critique n'est pas k-Regulier J. Mayer UniversitC Pa u l ValCry Montpellier, France De'die' & la me'moire de G. A . Dirac In the problem of contraction-critical graphs with chromatic number E , it is known that the degree of every vertex is 2 k. If every vertex is a Vk (i.e., is a vertex of degree k), the graph is chromatically reducible, and so it is not contraction-critical. There are two cases of reducibility: lo) 3 0 U r(V(z0)) contains two paths of length 2; 2 O ) r ( V ( z 0 ) ) consists of two disjoint complete subgraphs for every 20 = V k : then a cycle of v k can be reduced.

1

Introduction et notations: but de l'htude

Nous supposons connues les notions de base concernant les graphes. Les notations utilisdes seront, sauf indications contraires, celles de Berge [l]. Les graphes considdrds sont non orientes et simples (sans boucles ni arEtes multiples). Les definitions des contractions, de l'homeomorphie (appel6e aussi sous-contraction [3]) sont donndes dans [2]. Notations: I' est le graphe considere, suppos6 k-chromatique contraction-critique et non isomorphe au graphe complet de k sommets (IL'k). v;: sommet de degre i dans I'. I'(V(zo)), configurntion de voisinage de 50: sous-graphe de I' induit par les sommets voisins de $0. Si 20 est un v;,on notera I'(V(z0)) = I'i. P(G): nombre de stabilite (nombre maximum de sommets inddpendants) du graphe G. (Un stable est un ensemble de sommets ind6pendants.) La pr6sente etude a pour objet de montrer que, si I'est r6gulier de degre k, il ne peut Qtre k-chromatique contraction-critique.

J . Mayer

342

PropriMs des graphes contraction-critiques

2

Un graphe I' k-chromatique contraction-critique non isomorphe 8. ( I c 2 6) possbde les propribtds suivantes:

KI,

(2.1). Le degre' de tout sommet de I' est 2 k (Dirac). (2.2).

I' 2 Kk,l (Dirac);

d'ou, pour tout sommet

2 0 , I'(V(z0))

2 Kk-2.

(2.3). Pour tout sommet vi, I'i ne contient pas de stable de i - Ic sommets (Dirac). Donc, en particulier, P(I'k) = 2.

+3

Structure de r k

3

est de stabilitb 2 d'aprhs (2.3). La proposition suivante concerne les graphes de ce type: I'k

(3.1). Si un graphe G de p sommets ( p 2 6 ) est de stabilite' 2, il ve'rijie l'une des trois conditions suivantes: a)

G 2 Kp-2 ;

b) G contient cinq sommets distincts dont deux forment un stable et les trois autres induisent une chahe de longueur 2; c) G est constitue' de deux cliques disjointes K,., K , ( p - 3 2 r 2 s 2 3). Preuve. Par hypothbse, G n'est pas complet. Puisque P ( G ) = 2,si G n'est pas connexe, il est constitud de deux cliques disjointes: si l'une d'entre elles est d'ordre 2 p - 2, G v6rifie (a); sinon, c'est le cas (c). Suppoaons donc G connexe et comportant un stable { a , b}; si G - { a , b} est une clique, G vbrifie (a); si G - { a , b } est connexe et # K P - 2 , il contient un stable { c , d } et une chaine de longueur 2 2 joignant c et .d: c'est le cas (b); si { a , 6 } dbconnecte G et G 2 K P - 2 , il existe un stable { c , d } , puisque G - { a , b } a deux compoaantes connexes; ou bien on peut choisir { c , d } tel que a et b soient joints par une chaine de longueur 2 2 disjointe de { c , d } , ou bien G est isomorphe 8. une subdivision de K p - 2 , une ar6te 6tant remplacde par une chaine P ( a ,c, 6, d ) ; dans ce dernier cas, puisque G a au moins six sommets, il existe deux sommets e, f tels que P(a,e, d) et {c,f} soient respectivement 0 la chaine et le stable cherchbs: c'est b nouveau le cas (b). (3.2). Si I'k contient un stable { a , b } et une chaine induite II(c,e,d), e e'tant de degre' k dans I?, I' n'est pas contraction-critique.

Un Graphe k-Chromatique Contraction-Critique n'est pas k-Mgulier 343 Preuve. Soit xo un Ok de I' et soit I'k sa configuration de voisinage. Dans l'hypothbse de l'Qnonc6, xo U I'k contient deux chaines II(a,XO, b) et l I ( c , e , d ) . Contractons chacune de ces chaines en un sommet: nous obtenons un graphe r', (k'- 1)-colorable puisque, par hypothbse, I' est kchromatique contraction-critique. Une (k - 1)-coloration de I" induit une ( k - 1)-coloration de I' - x o oii les couleurs de a et de c sont rQpQtkesrespectivement en b et en d; la coloration de e ne pose aucun problkme, puisque e est de degrk k - 1 dans I' - xo et que la couleur de c se retrouve en d . I1 s'ensuit que I'k n'emploie que k - 2 couleurs; donc la (k - 1)-coloration de I' - xo s'dtend k I', contrairement B l'hypothhe selon laquelle I' est 0 k-chromatique. D'aprBs (2.2), (3.1) et (3.2), si I' est k-rkgulier, toutes les configurations sont constituQes de deux cliques disjointes IT,,K, ( r s = k; k - 3 2 r 2 s 3 3). La section suivante sera consacrCe B la reduction de ce cas.

+

I'k

4

Cycles reductibles

Supposons que, dans I', (1) toute arbte (vk,vk) fasse partie d'un triangle de 'Uk, et que (2) toute configuration I'k soit constituhe de deux cliques disjointes K r , K 8 ( r 2 s 2 3). Ces deux conditions sont satisfaites si I' est k-rCgulier (voir section prQcCdente). Sous les conditions (1) et (2), un cycle C,, sans diagonales, de m sommets de degr6 k est rhductible. Nous montrerons d'abord que: (4.1). I' contient un stable A , de m sommets, dont chacun est joint a un sommet dinkrent dans C , (ou, si m = 4, duns un cycle C' de m i m e

longueur). Nous envisagerons successivement A) m = 4;B) m

2 5.

Cas A) (Tous les indices, sauf l'indice 4, s'entendent modulo 4). Soit

dans I' un cycle C4 = (c1,c2, c3, cq) sans diagonales, dont tous les sommets sont de degre' k. Il existe un stable A4 = {a1,a2,a3,a4} tel que chaque ar6te ( a ; , c ; ) appartkenne ci I', ou un cycle Ci et un stable A& avec la m i m e proprie'te'.

Preuve. Appelons Id') la clique maximale de I' contenant l'ar6te (c;,c;+l). On peut supposer, sans perte de gQnQralit6,que I<(') et I d 3 ) sont d'ordre ~ + 1K,( 2 )et I d 4 ) Qtantd'ordre s+l ( r 2 s). Deux cliques I d i - ' ) et K ( ' ) - c ; Qtantdisjointes, la recherche de A4 se rambne B celle de deux stables { a l , a3} et {a2,a4} (a; E I{(;)). Si le stable { a 1 , ~ 3 n'existe } pas, tous les sommets

J . Mayer

344

- c3 - c4 sont joints par des arttes: la - c1 - c2 et ceux de de contraction des arBtes (c1, c4) et (c2, c3) fait apparaitre une clique Kzr 2 Kk,contrairement B l’hypothkse (I’ est contraction-critique). Si le stable {a2,a4} n’existe pas, tous les sommets de 1d2)-c2-cg et ceux de 1 d 4 ) - c 4 c1 sont joints par des arttes: il existe en particulier, d’aprks la condition (1)’ une artte (ci, c:) joignant deux sommets de degr6 k (ca E c& E JJ existe alors un cycle Ci = ( c l ,c ~ , c ;c&). , c; et c&font partie d’une clique maximale de I’ de r + 1 sommets: d4signons-la par (4 La stable A: est constitu6 de deux sommets non joints appartenant aux cliques K ( l ) et (4 (ils existent n4cessairement, sinon r pourrait Ptre contract6 en K 2 r 2 K k ) , de c4 et d’un sommet quelconque E I d 2 ) - c2 - c3 - ch. Cas B) (Tous les indices, sauf m , s’entendent modulo m ) . Soit C ,

..

=

( c ~ , c z , . , c m ) ( m 1 5 ) un cycle de vk de longueur minimum dans I’, si l’on excepte les cycles K3. L? existe un stable Am de m sommets ai tel que les arites (aj, c ; ) ( i = 1’2,. .. ,m ) appurtiennent ci I?.

Preuve. Chaque artte (ci, ci+l) fait partie d’un triangle de v k ( c i , Ci+l,ai) (condition (1) ci-dessus). Les ai constituent le stable A , cherchd, puisque, s’il existait une artte (ajyaj), celle-ci complbterait deux cycles de v k dont l’un au moins serait plus court que C,: celui-ci ne serait donc pas de longueur minimum dans I’. 0

Si dans I’ chaque configuration r k est constitute de deux cliques disjointes, un cycle Cm sans diagonakes, dont chaque sommet est joint par une arite a un sommet distinct appurtenant ci un stable A , de m sommets, constitue une configuration chromatiquement rtductible.

(4.2).

Preuve. Le sous-graphe C,UA, peut iitre contract6 en un sommet unique. Cette contraction produit un graphe I“ (k - 1)-colorable, puisque c’est un contract4 de I’, et dont toute (k- 1)-coloration induit une (k - 1)-coloration de l7 - C m oh tous les aj ont la mtme couleur (disons la couleur 1). Cette (k - 1)-coloration est directement extensible B I’ de la manibre suivante: on colorie c1 avec l’une des k - 2 couleurs (autres que 1)’ qui ne figure pas dans son voisinage. Notons que c1 a k voisins dont deux (a1 et a,) portent la couleur 1 et dont deux autres (cp et c,) ne sont pas colori6s. On a donc le choix entre deux couleurs, soit Q et P. Si l’une des deux couleurs, disons P , figure dans le voisinage de c2, on coloriera c1 avec l’autre couleur (a). Puis on coloriera c2 avec une couleur diff6rente de Q et de /?. On pourra ensuite colorier de proche en proche tous les ci, jusqu’g c,. Si ce dernier sommet nkcessite l’emploi de la couleur a,on recoloriera c1 avec la couleur

P.

Un Graphe k-Chromatique Contraction-Critique n’est pas k-Rbgulier 345 Si les couleurs a! et p manquent toutes deux dans le voisinage de c2, c’est par c2 que l’on commencera le coloriage de Cm, donnant ensuite B c3 une couleur diffbrente de a! et de p. Ce processus n’est en ddfaut que si les couleurs a et /3 manquent dans le voisinage de tous les ci. En ce cas, si m est pair, C, est colorable avec ces deux couleurs. D’autre part, soit Cm de longueur impaire: alors toutes les cliques auxquelles appartiennent les , seront du mbme ordre t 1; elles comprendront, arbtes successives de C outre les sommets non colorids cj et c;+l, un sommet E A, portant la couleur 1 et - 2 sommets de couleurs diffdrentes de 1, a!, p. Si une couleur 7 manque dans deux cliques consdcutives K j - 1 et K j , le sommet c j est colorable avec cette couleur et le reste du cycle Cm est colorable avec les couleurs a!, p. Sinon, m Qtant impair, la couleur 7 est prQsentedans deux cliques conshcutive$ Kit-1 et K j t , et il existe une couleur 6 absente de ces deux cliques, avec laquelle on peut colorier cjt, le reste de C, Qtantcolorable avec a! et p. r est donc (k - 1)-colorable, contrairement B l’hypothkse. 0

4

5

Comme l’absence d’un cycle C, ( r n > 3) impliquerait l’existence dans I’ de sommets de degrd < k, ce qui contredirait (2.1), on a montrd qu’un graphe rQgulierde degrQk ne peut btre k-chromatique contraction-critique.

Remerciements Nous avons plaisir b remercier les rdviseurs de l’article ainsi que le professeur L. D. Andersen pour leur examen soigneux et leurs utiles remarques.

RQfQrences [l] C. Berge, Graphes et hypergraphes, Dunod, Paris (1973) (28me Qd.). [2] G. A. Dirac, “On the structure of 5- and 6-chromatic abstract graphs,” J . reine angewlandte Mathematik 214-215 (1964), 43-52.

[3] 0. Ore, The four-CoZorproblem, Acad. Press, London-New York (1967).

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Annals of Discrete Mathematics 41 (1989)347-354 0 Elsevier Science Publishers B.V. (North-Holland)

A Theorem on Matchings in the Plane M. D. Plummer* Department of Mathematics Vanderbilt University Nashville, Tennessee, USA

Dedicated to the memory of Gabriel Dirac It is shown that any connected planar graph G with at least 8 vertices and having a perfect matching contains a set S of three independent lines with the property that S is not contained in any perfect matching of G .

1

Introduction and terminology

Let G be a graph with IV(G)l = p points and IE(G)I = q lines. A matching in G is any set of lines in E(G) no two of which are adjacent. Matching M in G is said to be a perfect matching, or p.m., if every point of G is covered by a line of M. Let G be any graph with a perfect matching and suppose positive integer n 5 (p - 2 ) / 2 . Then G is n-extendable if every matching in G containing n lines is a subset of a p.m. The concept of n-extendability gradually evolved from the study of elementary bipartite graphs (which are l-extendable) (Hetyei 1964, [3],Lovbz and Plummer 1977, [S]), and then of arbitrary l-extendable (or “matchingcovered”) graphs by LovLz in 1983 [5]. The study of n-extendability for arbitrary n was begun by the author in 1980 [lo]. In this paper we are concerned with matchings in planar graphs. When we speak of an imbedding of planar graph G in the plane, we mean a topological imbedding in the usual sense (see White [ll]) and would remind the reader that such an imbedding is necessarily 2-cell. (See Youngs [12].) ‘Work partially supported by ONR Contract derbilt University Research Council

# N00014-85-K-0488and by the Van-

M.D.Plummer

348

If we wish to refer to a planar graph G together with an imbedding of G in the plane, we shall speak of the plane graph G. The main result of this paper is to show that no planar graph is 3extendable. Throughout this paper, we will assume that all graphs are connected, that mindeg(G) 2 3 and that mindeg*(G) 2 3, where mindeg*(G) denotes the size of a smallest face in an imbedding of G. For any additional terminology, we refer the reader to Harary [2], to Bondy and Murty [l]or to Lov$sz and Plummer [7].

2

Some planar considerations

One of our main tools will be the secalled theory of Euler contributions initiated by Lebesgue in 1940 [4] and further developed by Ore in 1967 [8] and by Ore and Plummer in 1969 [9]. Let v be any point in a plane graph G. Define the Euler contribution of v, Q(v), by Q(v) = 1 -

deg v

deg v 2

i=l

where the sum runs over the face angles at point v and z; denotes the size of the ith face a t v. We shall require several simple lemmas. We include the proofs for the sake of completeness. The first is essentially due to Lebesgue [4]. L e m m a 2.1.

If G is a connected plane graph, then C, Q(v) = 2.

Proof. Let p = IV(G)I, q = IE(G)I and planar imbedding of G. Then deg v V

V

by Euler's classical formula.

T

be the number of faces in any

deg v

i=l 0

L e m m a 2.2. Let G be a connected plane gmph with mindeg*(G) 2 3 . Then for all v E V(G), Q(v) 5 1 - degv/6.

Proof. Since z; 2 3 for all i, we have Q(v) 5 1 - degv/2 t degv/3 and 0 the result follows.

A Theorem on Matchings in the Plane

349

It follows from Lemma 2.1 that there must exist a point v in any plane graph G with @(v) > 0. Let us agree t o call any such point v E V(G) a control point (since such a point will be seen t o “control”, or limit, the degree of matching extendability in G). It is well-known, of course, that any planar graph has points v with degv 5 5. We would like t o emphasize, however, that Lemma 2.2 tells us that we must have control points with degree 3 , 4 or 5 . Moreover, for any control point v, we have the inequality deg v

ci1 > i d e g v -

1.

i=l

Since we are assuming that each x; 2 3, inequality (1) yields the following three diophantine inequalities: degv = 3: degv = 5:

x:=lf > $ - 1 = z 1

1 ;ET

> 5 - 1 = T3

We shall see in the next section that we shall need solutions to these inequalities only in the degv = 4 and degv = 5 cases. The solutions for these two inequalities are listed below: deg v = 4:

deg v = 5:

( 3 , 3 , 3 ,x ) (3,3,4,4 (393,594 (3,4,4,x> (3,3,3,3,z)

x x x x x

.

= 3,4,. , = 4, ...,11 = 5,6,7 = 4,5 = 3,4,5

(Note that for the sake of conciseness, we list each solution in monotone non-decreasing order, although other cyclic orderings of faces of these sizes about a point are certainly possible and must be considered. See Ore and Plummer [ 9 ] . )

3

The main result

We shall need two basic results about n-extendable graphs. The proofs may be found in Plummer [lo]. Theorem 3.1. If n extendable.

2 2 and G is n-extendable, then G is also (n - 1)-

M . D. Plurnrner

350

Theorem 3.2. I f n 2 1 and G is n-extendable, then G is (n+l)-connected. Of course, since no planar graph can be 6-connected, this immediately tells us that no planar graph is 5-extendable. However, we now show that this result can be sharpened.

Theorem 3.3. No planar graph is 3-extendable.

Proof. Suppose G is a 3-extendable plane graph. Then by Theorem 3.2, graph G is 4-connected and hence mindegv 2 4. But then by the results of Section 2, graph G must contain a control point v of degree four or five. The possible facial configurations about point v are listed in Section 2 and we proceed to treat each. (Note that since our graphs are, in particular, 3connected, the mod 2 line sum of the face boundaries of the faces at control point v is always a cycle.)

Figure 1.

Figure 2.

( 3 , 3 , 3 , 2 ) . In this case we must have the configuration of Figure 1 and we see that { e , f } cannot be extended to a perfect matching. Hence G is

351

A Theorem on Matchings in the Plane

not 2-extendable. But then G is not 3-extendable by Theorem 3.1 and we have a contradiction. (3,3,4, z). Here x 2 4 and we must have either the configuration of Figure 2a or 2b. In the former, {e,f,g} does not extend to a perfect matching. In the latter, {e,f} does not extend and again G is not 3extendable by Theorem 3.1. So in either case we get a contradiction. (3,3,5,2). Here x 2 5 and we must have one of the configurations of Figure 3a and 3b. In the former, {e, f,g} does not extend and in the latter { e , f } does not extend. As before, we have a contradiction.

f e

Figure 4.

M. D. Plurnrner

352

(3,4,4,z). Here z 2 4 and we must have one of the configurations of Figure 4a or 4b. In both, the matchings {e,f,g} do not extend, a contradiction. (3,3,3,3, z). Here z 5 3 and we have the configuration of Figure 5. Let us label the neighbors of v in clockwise order as u1, 212, u3, 214 and us.

Figure 5. Suppose there is a point w $ {uz, 113, ~ 4 , 1 4 5 ,v}, but w is adjacent to u1. Then {ulw,~ 2 ~u4u5) 3 , is a matching of size three which cannot extend to a perfect matching, a contradiction. So the neighborhood of u1, N(u1) C_ {u2,113,u4,u5,v}. We know that {u2,v,u5} N(ul), but since G is 4connected, we have that degu1 2 4, and so u1 is adjacent t o a t least one of u3 and u4. Suppose u1 is adjacent to u3. Then degu2 = 3, a contradiction. By symmetry, a similar contradiction is reached if u1 is adjacent to u4. 0

4

Concluding remarks

In the decomposition theory of graphs with perfect matchings (see Lovkz and Plummer [7]), two important classes of “building blocks” are (1) 1extendable bipartite graphs and (2) bicriticd graphs. A graph G is bicm’ticd if G - u - v has a perfect matching for all choices of distinct points u and v. There is a nice relationship among 2-extendable graphs, l-extendable bipartite graphs and bicritical graphs, In particular, we have the following result. For the proof, see Plummer [lo].

A Theorem on Matchings in the Plane

353

Theorem 4.1. If graph G is 2-eztendable, then G is either (a) bipartite or (b) bicritical. (Note that no bicritical graph can be bipartite, so the two classes in the conclusion of the preceding theorem are disjoint.) Bicritical graphs - especially those which are 3-connected - are still not completely understood. Thus in light of Theorem 4.1 the study of graphs which are n-extendable, for n 2 2, may help us to better understand the structure of $connected bicritical graphs, as well as being of interest in its own right. The present paper is concerned with the planar case. Although we now know that no planar graph is 3-extendable, there are many such graphs which are 2-extendable. The dodecahedron, the icosahedron and the cube are but three familiar examples. We shall present a more detailed study of 2-extendable planar graphs in a subsequent paper.

Figure 6. Let us conclude by noting that there there do exist 3-extendable graphs which can be imbedded on the surface of the torus. The Cartesian products of two even cycles C2, x C,,, (m,n 2 2) are such graphs. See Figure 6 for an imbedding of C4 x C+

354

M. D. Plummer

References [l] J. A. Bondy and U. S. R. Murty, Graph Theory with Applications, American Elsevier, New York (1976).

[2]F. Harary, Graph Theory, Addison-Wesley, Reading, Mass. (1969). [3]G. Hetyei, “Rectangular configurations which can be covered by 2 x 1 rectangles,” P h i Tan. F6isk. Kozl. 8 (1964), 351-367.

[4]H. Lebeague, “Quelques consbquences simples de la formule d’Euler,” J . de Math. 9 (1940),27-43. [5]L. Lovkz, “Ear-decompositions of matching-covered graphs,” Combinatorica 2 (1983), 395-407. [S]L. Lovkz and M. D. Plummer, “On minimal elementary bipartite graphs,” J. Combin. Theory ( B ) 23 (1977), 127-138. [7]L. Lovkz and M. D. Plummer, Matching Theory, Annals of Discrete Math. 29, North-Holland, Amsterdam (1986). [8]0 . Ore, The Four-Color Problem, Academic Press, New York (1967), 54-61. [9] 0.Ore and M. D. Plummer, “Cyclic colorations of plane graphs,” Recent Progress in Cornbinatorics, (W. T. Tutte, ed.), Academic Press, New York (1969), 287-293. [lo] M. D.Plummer, “On n-extendable graphs,” Discrete Math. 31 (1980), 201-210. [ll] A. T.White, Graphs, Groups and Surfaces, North-Holland/American Elsevier, Amsterdam (1973).

[12]J. W. T. Youngs, “Minimal imbeddings and the genus of a graph,” J. Math. Mech. (1963),303-315.

Annals of Discrete Mathematics 41 (1989) 355-362 0 Elsevier Science Publishers B.V. (North-Holland)

Removing Monotone Cycles from Orientations 0. R. L. Pretzel Department of Mathematics Imperial College of Technology London, UK

Dedicated t o the memory of G. A . Dirac We prove the following generalization of a result of K. M. Mosesian: If a graph G of girth at least 2k has an orientation in which every cycle is either monotone or has at least k edges oriented each way, then G has an orientation in which every cycle has at least k edges oriented each way. In [3]K. M. Mosesian proved the following remarkable theorem. If a finite graph G has girth 2 4 and a n orientation in which every cycle is monotone or has a t least two edges directed each way, then G has such an orientation without monotone cycles, that is G can be oriented as the (Hasse-)diagram of a partially ordered set. Unfortunately Mosesian’s paper has remained untranslated and not easily available. The purpose of this paper is to prove a generalization of Mosesian’s Theorem in which 2 is replaced by an arbitrary k. This requires a generalization of the standard result that edge-critical strongly connected graphs have vertices of degree 2. Our proof of this generalization, Lemma 2, also yields, we believe, a new proof of the original result. The same is also true of our proof of our main theorem, but there the differences are not so striking. In [l] M.Aigner and G . Prins introduced the concept of a k-good orientation, generalizing that of a diagram orientation. An orientation is k-good if every cycle has at least L edges oriented each way. Clearly, a graph with a k-good orientation must have girth 3 2k, but this condition is not sufficient. Indeed, NeSetfil and Rod1 [4]have proved that deciding whether a graph has a 2-good orientation is NP-complete. Our theorem can be stated as follows. Let G be a graph of girth 2 2k. If G has an orientation in which every cycle is either k-good or monotone, then G has a k-good orientation.

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The reason we are interested in this result is that in previous work we have only been able to derive information on reorientations which do not change the number of edges oriented each way in any cycle. In the survey by Pretzel [6] finding more general reorientation operations with good properties, was given as one of the major open problems of the area. Our present result represents a small step in this direction. It is possible that a stronger generalization of Mosesian’s result is true and we shall discuss this at the end of the paper’. We adhere to the terminology of our previous papers, but repeat the definitions with a few new ones for the convenience of the reader. Our graphs will be finite and we will in principle permit loops and multiple edges, but note that requiring girth 2 2 prohibits loops and requiring girth 2 3 prohibits multiple edges. A walk W from a to 6 in a graph G is a sequence of vertices v; and edges ej : a = voelvl.. envn = b, such that the endpoints of e i are vi-1 and vi. The inwerse walk -W from b to a is obtained by reversing the sequence. The length of W is the number of edges n . If a = b the walk is called a circuit, We do not distinguish cyclic shifts of circuits but we do distinguish inverse circuits. A walk is simple if the vertices v1, ... ,on (and the edges e l , . , ,en) are distinct. The underlying graphs of simple walks and circuits are called paths and cycles. Given an orientation R of the graph G, the edges of any walk W fall into two classes, the forwad edges W+ (those edges e; that are directed towards vi by R ) and the backwards edges W - . Clearly W - = -W+ so any universal statement about W+ also holds for W - . W is a forward walk if W - = 8. The underlying path or cycle of a forward walk or circuit is called monotone. R is acyclic if there are no monotone cycles and strong if every edge lies on a monotone cycle (a strong orientation of a connected graph makes it into a strongly connected digraph). With this terminology we can make the definition of “k-good” precise. R is k-goal if IC+I 2 k for every simple circuit C. In order to state our generalization of Mosesian’s Theorem we define R to be k-fair if IC+(
.

.

Theorem 1 . Let G be a finite graph of girth _> 2k. If G has a k-fair orientation it has a k-good orientation. The proof of the theorem follows from three lemmas, the first of which shows that edges of G that do not lie on monotone cycles of a k-fair orientation R do not present obstacles to making R k-good. ‘See note added in proof.

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Lemma 1. Let R be a k-fair orientation of the graph G and define A as the set of edges of R-monotone cycles and B as the edges not in A. If a simple circuit contains an edge of B , then IC+ n BI 2 k. Proof. Colour the edges in A red and those in B blue. Then R is strong on the subgraph induced by the red edges and to every red edge e directed from the vertex z to the vertex y by R there exists a forward walk P, from y to z all of whose edges are red. We replace all the red edges of C+ by the corresponding walks -Pe to obtain a possibly non-simple circuit K . The blue edges of K are precisely those of C and K traverses no blue edge twice. Furthermore K + consists of precisely the blue edges of C+.So it will suffice t o show IK+l 2 k. K contains a blue edge f traversed by K , say, from a to b. Then K contains a simple walk Q from b to a , disjoint from f. Thus the circuit L = af b Q a is simple and L+ E K+. Now as L contains a blue edge L is not backwards monotone. Hence L+ # 0. As R 0 is k-fair it follows that IL+I 2 k and our claim is proved. Corollary 1. Under the hypotheses of the lemma, if S is a k-good orientation of the subgraph induced by A, then the orientation which agrees with S on A and R on B is a k-good orientation of G. For the other two lemmas we need some definitions for paths in a graph

G. The vertices of a path other than its endpoints are called internal. Two paths are disjoint if no internal vertex of one lies on the other. A path is unramified if its internal vertices have degree 2 in G. We will admit a cycle with a distinguished vertex v as an unramified path if all its vertices except possibly v have degree 2 in G.

Lemma 2. Let G be a graph with a strong k-fair orientation R. If G is not a single cycle it contains two disjoint monotone unramified paths of length 2 k. Proof. The proof is by induction on the number of edges in G. Assume G is not a single cycle. Let C be a monotone cycle of G and let G be obtained from G by contracting C to a single vertex C and deleting all the edges of C. We first verify that the induced orientation fi of G is k-fair (it is clearly still strong) . Suppose some simple circuit I? of G has 0 < [I?+[< k. The edges of l? cannot form a circuit in G, so must be a vertex of I? and the edges of l? form a simple walk K in G from a vertex z of C to a vertex y # z of C. This walk contains no other vertices of C. Now C contains a forward walk

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W from x to y. So the simple circuit L = zKy - W z has L+ = K + = E+ violating the assumption that R was k-fair. We may assume that the lemma is true for G. Case 1. G consists of a single cycle I?, Then is necessarily monotone and has length 2 k. K corresponds in G to a forward walk K from a vertex x of C to a vertex y on C (which may be equal t o z). Again K contains no other vertices of C. If x = y we can take C and K as our two paths. If x # y, C contains a forward walk W from x t o y and the circuit L = xKy - Wx is simple with K = L+ and W = L - . As R is k-fair it follows that IKI 2 k and IWl 2 k. So we can take K and W as our two paths. Case 2. G has two disjoint monotone unramified paths p and Q. If is not an internal vertex of either P or Q then their edges form the required paths P and Q. As p and Q are disjoint C can be internal to only one of them, say P, and then 6 is not a vertex of Q so Q corresponds to a monotone unramified path Q of G with no vertex in C. As is an internal vertex of the unramified path it follows that d ~ ( c=) 2. Hence dG(v) = 2lcl 2.

c

c

+

VEC

Then if P corresponds t o a path P in G with a unique vertex x in C, then x has degree 4 and all the other vertices of C have degree 2. We can thus choose C as our second unramified path. For C is monotone by assumption, of length 2 k by k-fairness, disjoint from Q and unramified with x as distinguished vertex. Otherwise P corresponds to two unramified forward walks PI from a t o 5 E C and P z from y E C to b, where x # y. Thus x and y have degree 3 in G and all other vertices of C have degree 2. Since p is unramified and & is strong there is a monotone cycle in G containing an edge of p and that cycle must contain the whole of P . Let the cycle be aP'bwa. Then corresponds in G to a forward walk from b to a containing no other vertices of C U PIU P2. Let U be the forward walk from y t o x on C. The simple circuit L : aP1x - UyPzbWa has L- = U = 0. Thus by k-fairness IUI 2 k. Now U and Q are two disjoint monotone unramified paths of length 2 Ic and the lemma is proved. CI Before proceeding t o the next lemma we observe that if a circuit C does not contain a simple circuit, then C is built up out of retraced walks W - W and thus IC+I = 1C-l. Hence if all simple circuits have 1 0-1 2 k the same holds for arbitrary circuits of length 2 2k.

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Lemma 3. Let G be a graph of girth 2 2k and P and unramified path of length k in G. Let H be obtained from G by deleting the edges and the internal vertices of P. Then any k-good orientation R of H can be extended to a k-good orientation of G. Proof. We shall show that we can find a vertex v on P such that it is possible to extend R to a k-good orientation of G by directing all the edges of P towards v. Let the endpoints of P be x and y . If they lie in different components of H , then no edge of P lies on any cycle of G and we can choose v = y. Otherwise let U and W be two (not necessarily distinct) walks from x to y in H . As G has girth 3 2k both U and W have length 2 k. Thus the circuit U - W which may not be simple has length 2 2k and as R is k-good it follows that I(U - W ) + )2 k or IU+I IW-1 2 k. Let f = minIU+I and b = min 117-1 over all walks from 2 to y in H . Then f b 2 k. Choose v as the point on P such that d ( v , y ) = k - f if f < k and choose v = y if kI f. If a simple circuit of G contains an edge of P it contains the whole of P so it is of the form k(U - P ) where U is a walk from x t o y in H . Consider the circuit h ' = U - P. Then

+

+

and

As R was assumed to be k-good on H it follows that IC+I 2 k for all 0 simple circuits C and our orientation is k-good on G. Before proving our main theorem we digress with a corollary of Lemma 3 that is perhaps of some interest. Corollary 2. Let G be a graph and H I and H2 subgraphs such that G = U Hz and H1 n H2 consists of two vertices 2 and y such that the distance from x to y is 1 k. Assume that both HI and Hz have k-good orientations. Then any k-good orientation R of HI can be extended to a k-good orientation of G.

HI

Proof. Let P be an unramified path of length k from x to y disjoint from G. The given k-good orientation of H I can be extended to H I UP in such a way

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that all the edges of P are directed towards some vertex w of P. Similarly an arbitrary k-good orientation S' of H2 can be extended to H2 U P. Now by a theorem of Mosesian [2] (see also Pretzel [5], where a proof is given) it is possible to modify the orientation of H2 U P by "pushing down" to produce an orientation S such that if there is any path from w E H2 U P t o w , then there is a forward path from w to w. This does not disturb the property of k-goodness. But then it follows from k-goodness that any path of length 5 k ending in w is directed towards w (this easy fact is also proved in Pretzel [5]). Thus the orientation of H1 U P and H2 U P agree on P. Now let T be the orientation of G agreeing with R on H1 and S on H2. We must show that any simple circuit K with edges in both H1 and H Z has lK+l 2 k. As K is simple it must contain both z and y and be the union of a walk U from, say, z to y in H1 and a walk W from y to z in H 2 . Now K+ = U + U W - and IU+I 2 k - d(w,y) while IW-1 2 k - d(w,z) because the circuits U - P and P W lie in H1 U P and Ha U P respectively. But 0 now IK+I 2 2k - d(v,y) - d(w,s) = k.

+

Proof Theorem 1. The proof is by induction on the number of edges of G. If this is < 2k, then G has no cycles and the theorem is empty. If G has exactly 2k edges then either G has no cycles or it consists of a single cycle and the theorem is clearly true. Now assume G has more than 2k edges and the theorem is true for all graphs with fewer edges than G. Let R be a k-fair orientation of G. We distinguish two cases. Case 1. G contains an unramified path P of length k. Let H be obtained from G by deleting the edges and internal vertices of P. Then R reduces t o a k-fair orientation of H. By the induction hypothesis R has a k-good orientation S and by Lemma 3, S can be extended to G. Case 2. G contains no unramified path of length k. Let A be the subgraph of G induced by the edges of monotone cycles of R. By Lemma 2, A has unramified paths of length k. Thus A has fewer edges than G. By the induction hypothesis there is a k-good orientation S of A and by the corollary to Lemma 1 the orientation that agrees with S on A and R on the other edges is k-good. (7

There is a stronger possible generalization of Mosesian's theorem that we have been unable t o decide. So we phrase it as a conjecture. Conjecture 1. Let G be a graph of girth 2 2k. If G has an orientation in which no simple circuit C has IC+I = k - l, then G has a k-good orientation2. 2See note added in proof.

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For k = 2 this agrees with our theorem. The only evidence we have for the validity of the conjecture for k > 2 is the smallest graph we know that has a 2-good orientation but no %good orientation and yet girth 6. This graph is illustrated in Figure 1. The properties stated above can easily be verified by following the proof that Ms has no 2-good orientation given in Pretzel [5]. An exhaustive computer search by A. E. Brouwer has confirmed that every orientation of this graph has a simple circuit C with IC+( = 2, as the conjecture requires.

Figure. 1.

Added in proof. Conjecture 1 was disproved by a counterexample found by Dale Youngs of Imperial College in January 1988.

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References [l] M. Aigner and G . Prins, “lc-Orientable graphs.’’ Unpublished preprint, Free University Berlin (1980).

[2] K. M. Mosesian, “Some theorems on strongly basable graphs,” Alad. Nauk. Armian. SSR. Dokl. 64 (1972), 241-245. [In Russian.] [3] K. M. Mosesian, “Basable on strongly basable graphs,’’ Akad. Nauk. Armian. SSR. Dokl. 66 (1972), 83-86. [In Russian.] [4] J. Neietfil and V. Rijdl, “Complexity of diagrams,” Order 3 (1987), 321-330. [5] 0. Pretzel, “On graphs that can be oriented as diagrams of ordered sets,” Order 2 (1985), 25-40.

[6] 0. Pretzel, “Orientations and reorientations of graphs,” Proceedings of the Joint A.M.,!?.-I.M.S. Summer Reseamh Conference on Combinatorics and Ordered Sets (I. Rival, ed.), A.M.S., Providence, R.I., Contemporary Mathematics 57 (1986), 103-125.

Annals of Discrete Mathematics 41 (1989) 363-370

0 Elsevier Science Publishers B.V. (North-Holland)

Disentangling Pairings in Trees G. Sabidussi Department of Mathematics and Statistics University of Montreal Montreal, Quebec, Canada

In Memory of Gabriel Andrew Dirac A disentangling pairing of a set A of vertices of a tree T is a set of non-degenerate edge-disjoint paths whose endpoints form the set A , no two of the paths having a common endpoint. We characterise those sets of generators of an infinite tree T which have a disentangling pairing covering T in terms of the degrees of the vertices of T . This

characterisation is used to obtain two results about decompositions of eulerian trees into paths and circuits.

Disentangling pairings are perhaps the only useful graph-theoretical concept having fallen into oblivion. They were introduced by Dirac in [l]in the course of an investigation on cycle decompositions, where, as a glance at that paper will show, their usefulness and pertinence is immediately apparent. It seems, however, that Dirac remained alone in this endeavour. This note is intended as an effort to bring this elegant concept back to life. Let G be a (finite or infinite) graph. Given a set A of vertices of G a disentangling pairing (d.p.) for A in G is a set A of non-degenerate paths of G whose set of endpoints is A , any two distinct paths, P,Q E A having a t most one vertex in common, the common vertex (if any) being an endpoint of at most one of P and Q . Clearly the paths of a d.p. are edge-disjoint. The main question one can ask about disentangling pairings is whether they exist. In order that a set A have a d.p. it is of course necessary that (A1 be even or infinite (Dirac restricted his definition to the case where A is finite, allowing G t o be infinite). We shall consider the question of the existence of d.p.’s in the simple case where G is a tree. It was shown in [3] that if A is an arbitrary set of vertices in a tree, then A itself, or A minus any one of its elements, has a d.p. Here we sharpen this result under the additional hypothesis that the disentangling pairings cover G.

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Let A c V ( G ) . A d.p. A for A in G is said to cover G or to be a covering d.p. if and only if GA = G, where GA := U{P : P E A}. It will be useful t o define K ( G ) := { x E V ( G ) : d ( x ; G ) = i (mods)}, i = 0,1,where d ( z ;G ) is the degree of x in G. When no confusion is likely we shall write V; for K ( G ) . The following is immediate from the above definition.

Lemma 1. If A has a covering d.p. in G and x is a vertex of finite degree of G , then d(x; G ) is odd if and only if x E A ; in other words, A 3 V1 and V ( G )- A 2 Vo.

A set A with this property will be called pure. Observe that V1 is pure and that if G is locally finite, V1 is the only pure set. In general purity is not sufficient for A to have a covering d.p. It is the main result of this note that in the case of a tree, purity does indeed imply the existence of a covering d.p., provided A is what we shall call a set of generators of T . Given a tree T and a,b E V ( T ) ,Tab will denote the unique path in T joining a and b. For A C V ( T ) ,TA := U a , b E A T a b is the subtree of T generated by A (i.e., the smallest subtree containing A). A is a set of generators of T if TA = T . Two vertices a , b E V ( T )are A-neighbours in T if and only if a, b E A , a # b, and T a b contains no vertex of A beside its endpoints. Given a set A of generators of T and any B C A we define the A-closure of B to be := V ( T B )n A. In a tree T , a disentangling pairing for A is simply a set A of nondegenerate edge-disjoint paths whose set of endpoints is A , no two distinct paths in A having a common endpoint. If A generates T , then clearly a n.a.s.c. for A to cover T is that TA be connected. Occasionally it is convenient to represent A by its associated pairing function p : A -, A : for any a E A , p(a) is the other endpoint of the unique path in A one endpoint of which is a. p is characterised by the properties that p(a) # a = p2(a), and Ta,p(a), Tb,p(b) are edge-disjoint if b E A - {a,p(a)}, a E A.

Theorem 1. Let A be a set of generators of T . Then A has a covering disentangling pairing i f and only i f A is pure. We only have to prove sufficiency. In order to do this we shall make use of Nash-Williams’s cycle decomposition theorem [2],Theorem 3, in the following equivalent form: A graph G has a cycle decomposition if and only i f every finite subgraph X of G can be embedded in a finite eulerian subgraph Y of G. We need:

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Lemma 2. Let A be a pure set of generators of T . Then for every finite subtree F of T there is a finite subset B of A such that B = = V ~ ( T B ) and TB 3 F .

Proof. Consider the set U := (Vo(F)f l A) U ( & ( F ) - A). Since A is pure, each x E U is incident with an edge e, of T\ F . Let FObe the tree obtained from F by adding the edges e,,z E U . Then for any x E V(F0) with d ( x ;Fo)> 1 we have that d ( x ;Fo)is odd ($ x E A. Hence if all pendant vertices of FO belong to A the proof is complete. In the opposite case, let 5 1 , . . . ,z, be the pendant vertices of Fo not in A. Since A generates T there is a pair of A-neighbours a, b such that Tab contains the unique edge of FO incident with 5 1 . Clearly, in F1 := FoUTab the number of odd vertices not in A has decreased by at least one (in fact, Vl(F1)-A C Vl(F0)- (AU { X I } ) ) . Let i 2 be the least subscript such that zil E Vl(F1)- A , and repeat the procedure to obtain Fz, etc. After r 5 n steps all vertices in VI(F0)- A 0 will have been used up, and B := Vl(F,) is the desired set. Proof of Theorem 1. Let A be a pure set of generators of T . Form a new graph G by joining a new vertex z t o all vertices in A, i.e., V ( G )= V ( T )U { z } , where z 4 V ( T ) ,and E(G) = E ( T )U { [ a , z ]: a E A} . Claim: G satisfies the hypothesis of Nash-Williams’s theorem. For a given finite X c G consider F , the subtree of T generated by V ( X )n V ( T ) . Let TB ZI F be the corresponding subtree of T given by Lemma 2, and let Y be the induced subgraph of G on V ( T B )U { z } . From the properties of B and the fact that d ( z ;Y ) = IBI = IVI(TB)I is even, it follows that Y is eulerian. Trivially Y 3 X . By Nash-Williams’s theorem G has a cycle decomposition 27. Then A = 0 {C - z : C E D} is a covering d.p. for A in T . An immediate consequence of Theorem 1 is:

Corollary 1. Let T be a locally finite tree. If V1 generates T , then it is the only set of generators having a covering d.p.; if Vl does not generate T , then no set of generators has a covering d.p. In particular, V ( T )has a covering d.p. if and only i f every vertex of T is odd. Corollary 2. Let A be a pure set of generators of T , B = B C A such that (i,) d ( z ; T ) < 00 for every x E V ( T B )n V(T\TB), and (ii) B has a covering d.p. A B in TB. Then A B can be extended to a covering d.p. for A in T . Proof. Let Ti, i E I , be the components of T\TB, x ; the unique vertex in V ( Z )n ~ ( T B Ai ) , := A n V ( Z ) ,i E I . By (i),

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+

d ( x i ;Ti) d ( x ; ;T B )= d(xi, T ) < 00. Suppose that 5; E A. Then d(xi;T;)must be even, for otherwise both d(z;,Ti)and d(xi;T)are odd so that by Lemma 1 (with G = TB),xi $?! B , i.e., x; f A since B = B . Hence we have that xi f A; and since d(s;T;)= d ( x ; T )for any x E V ( q ) x, # x;, it follows that A; is pure in Ti, i E I . Claim: A; generates T;, i E I . Let e E E ( z ) . In T there is a pair a , b of A-neighbours such that e E E(T,b) and at least one of a, b does not belong to B , say a f B . Then a $?! V ( T B )because B = B,hence a E Ai. If also b E A;, then e E E ( T A ~ )If. b $ Ai, then x; E V(T,b). Let el be the edge of Tazi = Tabn T; incident with x;. Since x; $?! A; (i.e. d(z;;Ti) even or infinite) there is an edge e2 E E(T;)- { e l } incident with xi. Let c be an A-neighbour of b in T for which e2 E E(Tb,). Then e E E(Tac)and a,c E Ai. By Theorem 1, Ai has a covering d.p. A; in T;, i E 1. Then clearly 0 A := AB U UiErAi is a covering d.p. for A in T extending AB.

-

Theorem 2. Let A be a set of generators of an eulerian tree T . Then A has two complementary disentangling pairings A,,A,, i.e., A0 n A, = 8 and A0 U A1 is a decomposition of T . Proof. Form a new tree S by attaching two pendant vertices at each a E A . Formally, put B = A x {0,1} and let

V ( S )= V ( T )U B , E ( S ) = E ( T )U { [ a , ( a , i ) ]: a E A , i = 0 , l ) . It is clear that B is a pure set of generators of S. Hence by Theorem 1,

B has a covering d.p. A in S. Assume for the moment that A satisfies p(a,i) # ( a , 1 - i) for all a E A, i E (0, l},

(1)

where p is the pairing of B associated with A. Define q; : A -+ A, i = 0,1, by p(a,i) = (q;(a),

*

1,

a

E A.

Then qo, q1 are the desired complementary d.p.'s for A in T . It remains to show that there always is a covering d.p. of B which satisfies (1). Given an arbitrary bijection f : B' + B', where B' c B , put

D ( f ) := { a E A : f(a,i)# ( a , l - i) for some ( a , i ) E B'}.

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Let p be an arbitrary covering d.p. of B in S. Index D ( p ) x {0,1} by some initial segment of the ordinals, D ( p ) x ( 0 , l ) = {w, : cy < a},say. We shall define an increasing sequence of sets B, c B such that U,. Put

+

and define p , : B,

3

B, by

and p , = p p otherwise (Figure 1).

0

Figure 1. Theorem 2 can be reformulated in terms of decompositions of T into edge-disjoint circuits (circuit = 2-way infinite path).

Theorem 3. For every set A of generators of an eulerian tree T there exists a circuit decomposition C of T such that every C E C is generated b y A n V(C).

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Proof. By Theorem 2, A has a pair of complementary d.p.'s PO, pl. Take any a E A and define a sequence a,, n E Z, inductively by a0 = a, and for n 2 0: an+1 = Pl-i(an) if n I i (mod 2), i E {O,l}. a-(n+l) = P i ( a - n ) Then C, := UnEZTa,a,+l is a circuit and is generated by A n V(C,) = 0 ( a , : n E Z}. {C, : a E A } is the desired circuit decomposition of T . Not every circuit decomposition of T need have this property. In the tree of Figure 2 let A be the set of all vertices not belonging to C. Then {C} U { C, I n E Z } is a circuit decomposition one of whose members, C , contains no generators at all.

C Figure 2. We conclude with the observation that Theorems 1, 2 and 3 are equivalent. The preceding proofs show that Theorem 1 =+ Theorem 2 + Theorem 3.

Proposition. Theorem 3 implies Theorem 1. Proof. Let A be a pure set of generators of T. Take a family of pairwise ~ , disjoint from T , and form disjoint rays (1-way infinite paths) ( R a ) a ~each a new tree T' by identifying each a E A with the origin of the corresponding

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ray R,. T’ is eulerian and is generated by A’ := V ( T ’ ) - V ( T ) .By Theorem 3, T‘ has a circuit decomposition C such that for each C E C, A’ n V ( C ) generates C.This implies that C n T is finite, i.e., a path whose endpoints are in A . Since each a E A is the vertex of attachment of exactly one R,, A := {C n T : C E C} is a d.p. for A in T , and since UC = T‘ it follows that A covers T . 0

Acknowledgement The author wishes t o thank the referee for pointing out some minor errors in the original version. Support from the Natural Sciences and Engineering Research Council of Canada, Grant A-7315, is gratefully acknowledged.

References [l] G. A. Dirac, “Note on the structure of graphs,” Canad. Math. Bull. 5 (1962), 221-227.

[2] C. St. J. A. Nash-Williams, “Decomposition of graphs into closed and endless chains,” Proc. London Math. SOC.(3) 10 (1960), 221-237. [3] G. Sabidussi, “Infinite Euler graphs,” Canad. J. Math. 16 (1964), 821838.

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Annals of Discrete Mathematics 4 1 (1989) 371-396 0 Elsevier Science Publishers B.V. (North-Holland)

Colour-Cri tical Graphs with Vertices of Low Valency H. Sachs a n d M. Stiebitz Section of Mathematics Institute of Technology of Ilmenau

Ilmenau, GDR Dedicated to the memory of Gabriel A . Dirac All graphs considered are finite, undirected, and simple. A graph G is called Ic-critical if x(G) = k and x(G’) < Ic for every proper subgraph G‘ of G ( x ( G )is the chromatic number of G ) . A low (-valency) vertex of a &-critical graph is a vertex which has valency Ic - 1. This paper contains, in a condensed form, all results known about critical graphs having low vertices.

1

Introduction

The chromatic number seems to be the most popular graphical invariant, its popularity being mainly due to the renowned Four Colour Conjecture which assumes that all planar graphs are 4-colourable. Nevertheless, very little is known about the chromatic number. Its computation is difficult, and there is no good characterization of k-chromatic graphs (i.e., graphs with chromatic number k) except for the trivial cases k = 1,2. G.A. Dirac [2]-[lo] attacked this problem and about 1950 he suggested an approach by investigating what he called k-critical graphs, namely, those graphs which are k-chromatic and are inclusion-minimal with respect to this property. His investigations resulted in some of his well-known theorems on graphs in general. The observation that a k-critical graph has minimum valency no smaller than k - 1 led T. Gallai [ll]to the following concepts. Any k-critical graph G can be divided into two parts (possibly empty): the subgraph of G induced by the vertices of valency k - 1, called the low-vertex subgraph, and the remaining part, called the high-vertex subgraph of G. Generalizing

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Brooks’ theorem [l](see Proposition 3), T. Gallai [ll],[12] obtained a simple and constructive characterization of the set of all low-vertex subgraphs of critical graphs (see Theorems 2 and 3). In June 1980, G.A. Dirac and the first author (who was then a visitor of Aarhus University) discussed the question how to characterize the set of all high-vertex subgraphs of critical graphs with non-empty low-vertex subgraph. Later the authors jointly developed a general procedure (called the construction K ) for construction k-critical graphs which in particular enabled them to show that, for k 2 4, every proper subgraph of any kcritical graph is the high-vertex subgraph of some k-critical graph [18]. The construction K is described in Section 3. In Section 4 a modified proof of the original embedding theorem is given which also yields an improved result concerning embeddings of graphs in critical graphs as subgraphs (see Theorem 9). In his doctoral thesis [21], the second author showed that as long as the high-vertex subgraph of a k-critical graph G is (k - 2)-colourable ( k 2 4), G can be obtained by means of the construction K (provided the low-vertex subgraph is connected). Further, he found a good characterization of the set of all critical graphs with complete high-vertex subgraphs. These results [20] (see also [21])are contained in Section 5.1. As the proof of Theorem 12 given in [20] is not quite correct, in Section 5.2 a corrected version of the original proof is given. Problems connected with the low vertices tend to be much more easily manageable than questions concerning high vertices. In a way, the whole theory can be split into an (almost hopeless) hard part concerning high-vertex problems and another part, not quite as hard, where - by the presence of low vertices - some characterization theorems can be proved in a “constructive” manner.

2 2.1

Some basic definitions and theorems Notation

Most of the concepts used in this paper can be found in [15 pp. 528-5401. All graphs considered are finite, undirected and have neither loops nor multiple edges. A gmph G = (V, E) consists of a finite set V = V(G) of elements called the vertices of G and a finite set E = E(G) of 2-subsets of V called the edges of G. If e = {P, Q } E E(G) then P and Q (the end vertices of e ) are called adjacent in G. For P E V(G), N(P : G ) denotes the set of all

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vertices of G which are adjacent to P. The valency of a vertex P with respect to G is vaZ(P : G) = IN(P : G)I. E(G). The subgruph o f G induced by X, Let X C V(G) and F denoted by G[X], is defined by

V(G[X]) = X, E(G[X]) = {ele E E(G) A e C X}; further,

G - X := G[V(G) - XI, G - F := (V(G), E(G) - F).

A k-colouring of G is a mapping c of V(G) into the (colour-) set {1,2,.. .,k} (k 2 1) such that c ( P ) # c ( Q ) for any two adjacent vertices P,Q E V(G). Let Ck(G) denote the set of all k-colourings of G. A graph which admits a k-colouring is called k-coloumble. The chromatic number x(G) of a non-empty graph G is the smallest integer k for which G is k-colourable. If x(G) = k then G is called kchromatic. The empty graph is O-chromatic. A graph G is called colour-critical or, briefly, critical if x(G') < x(G) for every proper subgraph G' of G; it is called edge-critical if X(G-{e}) < x(G) for every edge e of G, and it is called k-(edge-)critical if it is (edge-)critical and k-chromatic. Obviously, a connected graph G is k-critical if and only if it is k-edgecritical. Removing the isolated vertices from a k-edge-critical graph (k 2 2) results in a k-critical graph. 2.2

S o m e basic properties of critical graphs

The classes of all k-colourable, k-chromatic, or k-critical graphs are denoted by Col(k), Chr(k) and Cri(k), respectively. Note that for k > 1, Chr(k) = Col(k) - Col(k - l), Cri(k)

c Chr(k) c Col(k) c Col(k + 1).

Proposition 1. (Immediate). Every k-chromatic gmph has a k-critical subgraph.

A vertex of G whose removal increases the number of components (connected parts) of G is called a separating vertex. Proposition 2 . (Immediate). IfG is k-critical then G is connected, has no separuting vertex, and vaZ(P : G) 2 k - 1 for every vertex P E V(G). The vertices of a k-critical graph G whose valencies are equal to k - 1 are called the low vertices, and the others are called the high vertices, of G;

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the subgraphs of G induced by the set of the low vertices or the set of the high vertices are called the low-vertez subgraph L(G) and the high-vertez subgraph H(G), respectively. Let K, and C, denote the complete graph and the circuit on n vertices, respectively. The problem of determining all critical graphs without high vertices is settled by: Proposition 3. (An immediate consequence of Brooks’ theorem [l]).A k-critical graph G has no high vertices if and only if either G 2 Kk (k 2 1) or k = 3 and G E C29+1 ( q 2 1).

The graphs mentioned in Proposition 3 act as the “modules” of the low-vertex subgraphs (see Section 2.4), therefore, we shall call them bricks: B E K k is a k-brick, B 2 C29+1is a 3-brick. If B is a k-brick then we shall say that B has size k and write s ( B ) = k. For k < 4 the k-bricks are the only k-critical graphs: therefore, in what follows we shall assume k 2 4; for k 5 3 the theory is almost trivial whereas for k = 4 all the main difficulties are already present. Proposition 4. (Immediate; see [3]). If G is k-critical and K is a complete graph contained in G as a proper subgraph then G - V ( K ) is connected.

2.3

Two fundamental construction principles

The join G = G1 v G2 of the disjoint graphs GI, G2 is defined by

V(G) = ~ ( G IU) V(G2), E(G) = E(Gd u E(G2) u {{Pl,P2)IPl E V(Gl),P2 E V(G2)). 0bviously, X(G1 v G2) = x(G1) t X(G2). Proposition 6. (The Dirac construction, see [ll]).If G1 E Cri(k1) and G2 E Cri(k2) then GI v G 2 E Cri(b1 t k 2 ) .

Starting from long circuits of odd length (which are 3-critical) and isolated vertices, for each k 1 6 arbitrarily many k-critical graphs without low vertices can be constructed just by repeatedly forming joins. T. Gallai [ll] was the first t o show, by an ingenious construction, that such graphs exist also for k = 4,5. G.A. Dirac [9] and G. Haj6s [14] invented a construction, later generalized by T. Gallai [ll, Satz (2.12)] and G.A. Dirac (see Proposition 7),

Cofour-Critical Graphs with Vertices of Low Valency

I

Qi

Qi

375

Q2

e Q2

Figure 1. which proved a powerful tool in the theory of b-critical graphs. The following proposition describes this construction in its simplest form.

Proposition 6. (The Dirac-Haj6s construction). Let G 1 and G2 be disjoint k-critical graphs ( b 2 3 ) . For i = 1 , 2 , let Pi,Qi c V ( G ; ) ,{ P ; , Q ; }E E(Gj)- Delete the edges el = {Pi,Ql}, e2 = (Pz,Qz}, identify PI with P2 (thus creating a new vertex PO),and connect Q1 with Q2 b y an additional edge e (see Figure 1). Then the resulting graph G := [ G 1 , P l , Q 1 ; G2, P2, Q2] is k-critical.

# PI,P2

of G1 or G2 retains its valency in G = [Gi,P1,Q1;Gz,P2,&2], i.e., if P E V ( G ; )and P # P; then we have val(P : G ) = val(P : G ; ) ;further, if k 2 4 then vaZ(P0 : G) 2 2 ( k - 2 ) 2 k: this means that POis a high vertex of G . An extended version of Proposition 6 was already given in [18, $61; unfortunately, there is a line missing, therefore we repeat it here. Note that every vertex P

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Proposition 7. (G. A. Dirac; before 1980). Let G1,Gz be disjoint kcritical graphs (k 2 3). For i = 1,2, let H ; be a complete graph on h vertices, contained as a subgraph in Gi, where 1 5 h 5 k - 2; let V(H;) = { P: ,pi2, . ,P,"}, Suppose that GI, G2 satisfy the following conditions.

..

( A ) There are vertices

and edges

(B) h 5 2, or h > 2 and for j = 2 , 3 , . ..,h either {P..,Q1} E E(G1) or {Pi,Q2} E E(G2) (or both). Apply the following construction: Delete the edges el, e 2 , Identifi P{ with Pj,j = 1 , 2 , . ,h, Replace double edges by single ones, And Connect Q1 with Q2 by an additional edge.

..

Then the resulting graph

The proof is left to the reader. The interesting question whether in Proposition 7 the hypothesis (B) may be dropped - aa Dirac conjectured (or knew) - is, as far as the authors could find out, still open (possibly, Dirac himself had a proof). However, it is not difficult to prove that, if the hypothesis (B) is dropped in Proposition 7, then the resulting graph G = [GI,HI,QI;G2, H2, Qz] is still k-chromatic and every edge e of G which does not connect a pair of vertices obtained by the identification process satisfies x(G - {e}) < x ( G ) (i.e., e is a secalled critical edge). Clearly, the Dirac-HajBs construction described in Proposition 6 can be applied to any pair of disjoint graphs GI, G2 having edges. It is particularly interesting to note that, in a sense, the converse of Proposition 6 is also true:

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P r o p o s i t i o n 8. Let GI and G2 be disjoint graphs, each having at least one edge, and let { ~ , Q I }E E(G1) and {Pz,Q2) E E(G2). Then, for k 1 4, the graph G = [Gl,P1,Ql;G2,P2,Q2] is k-critical if and only if both G1 and G2 are k-critical. (Not true for k = 3.) The proof, as an interesting exercise, is left to the reader. In this connection, a fundamental result of G. H a j b (before 1959), shedding particular light on the Dirac-Haj6s construction and its significance for the theory of criticality, should be mentioned: Let G and H be disjoint graphs and let (Op. 1) and (Op. 2) be the following operations. (Op. 1) Identify two non-adjacent vertices of G; (Op. 2) Apply a Dirac-Hajb construction (see Proposition 6) to G and H. T h e o r e m 1. (G. Haji, [14]). Any k-critical graph ( k 2 1 ) can be obtained from disjoint copies of the complete graph on k vertices by (repeatedly) applying the operations (Op. 1) and (Op. 2). 2.4

The fundamental theorems of T. Gallai

A maximal subgraph B of a graph G such that any two edges of B are contained in a circuit of G is called a block of G. The set of all blocks of G is denoted by B(G). Obviously, a vertex of G is a separating vertex if and only if it is contained in more than one block of G. An end-block of G is a block which contains at most one separating vertex of G. Two blocks which have a vertex in common (they cannot have more than one vertex in common) are called adjacent. Definition 2.1. A connected graph all of whose blocks are bricks is called a Gallai tree; a Gallai forest is a graph all of whose components are Gallai trees. A k-Gallai-tree (-forest) is a Gallai tree (forest) all of whose vertices have valencies 5 k - 1. Figure 2 shows an example of an 8-Gallai-tree. The following theorems are fundamental.

Theorem 2. (T. Gallai [ l l ] ) . If the graph G is k-critica then L(G) is a k - Gallaai-forest (possibly empty). T h e o r e m 3. (T. Gallai [ll]). Let k 14 and let L' be a k-Gallai-forest which does not contain a complete graph on k vertices as a component (L' may be empty). Then there is a k-critical graph G with L ( G ) 2' L'.

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n

B6

Figure 2. 2.5

The theorem of D. Greenwell and L. Lovdsz

D. Greenwell and L. LovQsz [13] characterized the class of all proper subgraphs of k-critical graphs. We need some preparation. Let G be an arbitrary graph and e = {P,Q} E E(G). The graph G/e is obtained from G - {e} by identifying P and Q (the edge e is “contracted” to a single vertex R , see Figure 3). Put, for r 2 3, Con(r) := { H I H E Col(r) A H / e E Col(r) for every edge e E E(H)}. Note that Col(r - 1) c Con(r) c Col(r). Now let G be a k-critical graph and e = {P,Q} an arbitrary edge of G. Then, by definition, G - { e } is (k - 1)-colourable. Since G is not (k - 1)colourable, we conclude that c ( P ) = c ( Q ) for every (k - 1)-colouring c of G - {e}, hence X(G/e) 5 k - 1. This implies that if H is a proper subgraph of a k-critical graph then H E Con(k - 1). That the converse of this statement is also true was proved by D. Greenwell and L. Lovisz in 1974:

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Figure 3.

Theorem 4. (D. Greenwell and L. Lov&z [13]). A graph H is a proper subgraph of some k-critical graph if and only if H E Con(k - 1). Note that this characterization of the set of all proper subgraphs of b-critical graphs is not of such a simple form as Gallai's characterization of the class of all low-vertex subgraphs since all graphs H and H/e have to be tested for (k - 1)-colourability (this problem is NP-complete). But note that testing for the membership in Con(k - 1) requires only testing for colourability whereas testing for the membership in Cri(lc) requires testing also for non-colourability.

3

A general procedure for constructing k-critical graphs

In this section, a procedure is described which, starting from a suitable Gallai tree T and a vertex set N, disjoint from V ( T ) , enables many kcritical graphs to be constructed (see [17], [18] and [20]). We need some preparation. Definition 3.1 Let T be a Gallai tree. A subset B1 of B(T) is called a matching of T if every vertex of T is contained in a t most one block of B1. A matching B1 of T is called perfect if every vertex of T is contained in exactly one block of B1.

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Let T be a Gallai tree which has a perfect matching B1. Note that B1 is uniquely determined, therefore we shall write B1 = Bi(T). For the Gallai tree T of Figure 2, &(T) = {BI, Bq,B5,B6, B8,Blo} is its unique perfect matching.

Definition 3.2 For k 2 4 and q 2 1 let Tk(Q) denote the set of all Ic-Gallai trees T having a perfect matching B1(T) such that (1) for each B E B1(T), s ( B ) 2 k - q, (2) for each B E B(T) - B1(T), s ( B ) 5 q

+ 1. .

Note that ICk and Kk-1 belong to all sets Tk(Q) (q = 1,2,. .).

3.1

The construction K (see [lS])

Let q 2 1 and Ic 2 4 be given integers, put

T

= Ic - q and N, = {1,2,.

. . ,q } .

(Kl)Denote by Uk(Q)the set of all pairs (T,u) satisfying (A) T E Tk(q), (B) u is a mapping of B(T) into the power set of N, such that: (Bl) for each B E B1(T), Iu(B)I = s ( B ) - T ; (B2) for each B E B(T) - B1(T), lu(B)I = s ( B ) - 1; (B3) for any two adjacent blocks B', B" E B(T), u(B')nu(B''>=

0.

(K2)For (T,u) E Uk(q)and P E V(T)put v(P) = U u(B), where B runs through all blocks of T containing P. (K3) For (T, u) E Uk(Q),define the graph H = H(T, u,q ) by

V ( H ) = N,; E(H) = { { i , j } l i # j and there is a B E B1(T) such that { i , j } N, - u(B)} '

(K4)For a given pair (T,u) E Uk(q) and an arbitrary graph H with V ( H ) = N, and V ( T )n N, = 0, define the graph G = G(T,u, q; H ) by

V ( G )= V(T)U N,; E(G) = E(T) U E ( H ) U {{P,i}lPE V ( T ) ,i E N, - v ( P ) } . (K5)Put G(T,u, q ) := G(T,u, q; H(T, u,q)). The following theorem was proved in [18].

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Theorem 5. Let (T,u) E Uk(q) and let H E Con(k - 1) be an arbitrary graph with V ( H ) = N, containing H ( T , u,q ) as a subgraph. Let G stand for G(T,u,q; H ) . Then (a) val(P : G ) = b - 1 for all vertices P E V ( T ) ;

(c) x(G - {e}) = k

- 1 for all edges e E ( E ( G )- E(H)) U E ( H ( T ,u,q));

(d) G(T,u,q ) is k-edge-critical.

3.2

The construction K’ (see [20])

The construction K has the disadvantage that Gallai trees without a perfect matching are excluded. In what follows we shall describe a possible extension of K that can be applied to arbitrary Gallai trees.

(K‘l) Denote by U; the set of all pairs ( T ,u) satisfying (A’) T is a k-Gallai-tree;

(B‘) u is a mapping of B(T)into the power set of Nk-1 such that (Bl‘) for each B E B(T),Iu(B)I = s ( B ) - 1; (B2’) for any twoadjacent blocks B’, B“ E B(T),u(B’)nu(B’’) =

0.

(K’2) For ( T ,u) E U; and P E V(T)put v(P) = through all blocks of T containing P.

U u(B), where B

runs

(K’3) For a given pair ( T ,u) E Ui define the graph G = G(T,u) by

V ( G ) = V ( T )U Nk-1 ( V ( T )n Nk-1 = S), E(G) = E(T)U {{i,j}li # j and i, j E Nk-1) U{{P,i}lP E V ( T ) , iE Nk-1 - v ( P ) } . The following theorem was proved in [20] (see also [21]). Theorem 6. Let ( T , u ) E Ug-,. If T does not have a perfect matching then G = G(T,u) is a k-critical graph where every vertex of T is a low vertex of G.

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3.3

Some remarks

It should be remarked that, despite of their lengthy descriptions, both constructions are, in fact, quite simple and in all their details straightforward. The constructions K and K’ which, seemingly, have little to do with the classical methods of constructing k-critical graphs are, in fact, a very general form of the (extended) Dirac-HajCls construction (see Proposition 7) where many special k-critical graphs simultaneously act together in creating a new b-(edge-) critical graph. A detailed discussion of the constructions can be found in [18] and [21].

Remark 1. Let (T,u) E Uh(q) and let I denote the set of all isolated vertices of G(T,u,q). Then

I=

n

u(B)cN,,

BEBi(T)

and provided H ( T , u, q ) E Con(b - 1) the graph G(T,u, q ) - I is k-critical. If T is not a complete graph on k vertices, then I is a proper subset of N, and G(T,u,q) - I is a graph of the form G(T,u’,q’) where q’ = q - 111 and u’(B) = u(B) - I for every B E B(T).

Remark 2. Let T be a k-Gallai-tree having a perfect matching. Then the following statements hold: (a) there is an integer qo such that T E Tk(q) for every q 2

40, and

(b) there is an integer qo(T,k) such that every graph G(T,u,q) with (T,u) E Uk(q) and q 2 qo(T,k) has isolated vertices.

Remark 3. For a given integer k 2 4 and a given k-Gallai-tree T the set of k-critical graphs obtainable by means of the construction K or K’ is finite.

4

Embedding given graphs in critical graphs

Theorem 7. Let k 2 4 and d 2 1 be given integers. Then for every non-empty graph H E Con(k - 1) there is a k-critical graph G satisfying (1) H(G) % H ; ( 2 ) L(G) is a non-empty graph which has the same number of components as

H(G);

( 3 ) v a l ( s : G) 2 d for every s E V ( H ( G ) ) .

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Theorem 7 was proved by the authors in 1980. The proof given in [18] is constructive in the following sense: there is a polynomial algorithm which, given a graph H E Con(k - l), constructs a k-critical graph G such that H ( G ) E H . We shall give here the proof in a slightly modified form, which yields an improved embedding result (see Theorem 9). First we shall prove the following lemma.

Lemma 1. (See [IS]). Let k 2 4 be an integer and let m 2 3 be an odd integer satisfying m(k - 3) 2 4. Further, let H E Con(k - 1) be a connected graph with IV(H)I > 1. Then there is a pair (T,u) E Uk(q) satisfying

(4 H ( T , u,q>2 H ; (b) vaZ(z : G(T,u,q ) ) 2 m(k - 3) for every

(4 IT)I= m(k - 2)

2

E N,;

IE(H)I. Proof. Let V(H) = N , , q > 1. Further, let L denote the line graph of H (i.e., V(L) = E(H), E(L) = {{e,f}le,f E E(H) A [ e nfl = 1)). L being the union of q complete graphs such that every vertex of L is contained in exactly two of these graphs, it is easy to see that L contains a spanning tree D (i.e. D C L is a tree with V(D)= V ( L ) )such that every vertex of D has valency 5 4 in D. Part I: Construction of a Gallai tree T . Corresponding t o each e E E(H) = V(D),let Fk(e,m ) denote a special Gallai tree defined as follows: Fk(e,m ) consists of m blocks BF, . . . , BZ which are (k- 2)-bricks and one block Be S C,,,(recall that m 2 3 is odd), where every vertex of Be is contained in exactly one of B r , . . . ,I?:; clearly, Fk(e,m) E T k ( q ) and BI(Fk(e,m))= {BF,.. . ,.I? Gallai :}. trees assigned to distinct edges of H are assumed to be disjoint. Let us call a vertex of a Gallai tree which is not a separating vertex, a boundary vertez. Clearly, Fk(e,m) has m(k - 2) vertices and m(k - 3) boundary vertices , Next, the Gallai trees Fk(e,m),e E E(H), are interconnected by additional 2-bricks (i.e. edges) B(e,f) according to the following rules. *

(i) A 2-brick B(e,f) between Fk(e,m) and Fk(f,m)is introduced if and only if e and f, considered as vertices of L, are adjacent in the spanning tree D of L . (ii) B(e,f)connects a boundary vertex of Fk(e,m) with a boundary vertex of Fk(f,m). (iii) The bricks B(e, f) are pairwise disjoint.

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Because of v d ( e : 0)5 4 5 m ( k - 3 ) for each e E V ( D )= E(H), these conditions can always be satisfied. The graph T resulting from this construction is obviously a Gallai tree. Clearly, the set UeB1(Fk(e,m)) is the perfect matching Bl(T) of T and T E Tk(q). Further, IV(T)I = m(k - 2) * IE(H)I.

Part 11: Construction of a mapping u. Put r = k - q. (a) For each (k - 2)-brick Bf E Bl(Fk(e,m)) C Bl(T), e E E(H), put u(BF) = N, - e. Then we have Iu(BF)l = q - 2 = (k - 2) - T = s(B%)- r ; and (see (K3))e E E(H(T, u, q)). (b) Foreach3-brickBe E B(Fk(e,m))-B1(F,(e,m)) 2 B(T)-Bl(T) put u(Be) = e. (c) For a 2-brick B (e,f) (which, necessarily, belongs to B(T) - B1(T)) put u(B(e,f)) = e n f . Now it is easy to check that (T, u) E Uk(q) where, due t o the definition of u, H(T,u,q) 2 H . If P is a boundary vertex of Fk(e,m) as well as of T then, in G(T,u, a), P is adjacent to both end vertices of e 2 N,. If P is a boundary vertex of Fk(e,m) but not of T then P is contained in exactly one 2-brick B(e,f) for some f E E(H) = V ( L ) = V ( D ) . If e = {x,y} and e n f = {x} then, in G(T,u, q), P is adjacent to y but not t o x. But x is an end vertex of f # e in H ( T , u,q). Therefore, we have v d ( z : G(T, u,q)) 2 m(k - 3) for every x E N,. Thus Lemma 1 is proved.

Proof of Theorem 7. Lemma 1 establishes Theorem 7 for connected graphs H E Con(k - 1) (for H 2 K1 see Theorem 11). In order to finish the proof of Theorem 7 we shall show by induction over the number c of components of H that Theorem 7 holds for disconnected graphs, too. Let H E Con(k - 1) have precisely c > 1 components, say H I , .. . ,H,. Denote the subgraph of H consisting of the components H I , .. . ,H,-1 by H1 and put H , = H 2 ; then, clearly, H1,H2E Con(k - 1). By the induction hypothesis, there are k-critical graphs GI, Gz satisfying H(Gi) ?! H i ; L(G;) is a non-empty graph which has the same number of components as H(G;) and v d ( x : Gi) 2 d for every x E V ( H ( G i ) ) (i = 1,2). We may assume that G2 has a low vertex P2 which is adjacent to no high vertex. For H 2 li K1 this follows from Theorem 11, and for graphs

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385

H 2 that have more than one vertex it follows from the proof of Lemma 1 (any vertex P of T with val(P : T ) = k - 1 may be taken for Pz). Let PI be a high vertex of GI and Q1 a low vertex of GI adjacent to PI,and let 9 2 be any vertex of G2 adjacent to Pz. Now apply the Dirac-Haj6s 91 and G2, P2,92 (see Proposition 6): clearly, this construction t o GI,PI, , 91; G2, P2,9 2 1 with H ( G ) 2 H ; results in a k-critical graph G = [ G I PI, further, L(G) has just as many components as H(G) and val(z : G ) 2 d for every 2 E V ( H ( G ) ) . This proves Theorem 7.

0

I f G is a k-critical graph which has low vertices, then H(G) is a proper subgraph of G and therefore, H(G) E Con(k - 1) (as remarked a t the beginning of Section 2.5). Theorem 8. Let k 2 4. For any non-empty graph H the following three statements are equivalent: (1) EI E Con(k- 1). (2) H is the high-vertex subgraph of some k-critical graph which has low

ve rtices. ( 3 ) H is a proper subgraph of some k-critical graph. Note that Theorem 8 in particular implies Theorem 4. Let us mention the following consequence of Lemma 1. Theorem 9. For every k 2 4 there is a constant c = C k such that any connected graph H E Con(lc - 1) with at least one edge is contained as an induced subgraph in a k-critical graph G with IV(G)l 5 clE(H)I 5 c l V ( H )l2*

By a result of B. Toft [26] it is known that for every k 2 4 there exists a constant it& such that any graph H E Col(k - 2) is contained as a subgraph (not necessarily induced) in a k-critical graph of a t most 21V(H)I Mk vertices. Whether or not this remains true if Col ( k - 2) is replaced by Con(lc - 1) is an unsolved problem. The k-critical graphs G constructed in the proof of Theorem 7 have the property that their low-vertex subgraphs have precisely as many components as their high-vertex subgraphs (see statement (2) of Theorem 7). This is in accordance with an old conjecture of T. Gallai, namely, that if G is critical and L(G) # 0 then H ( G ) cannot have more components than L(G). In 1981, Gallai’s conjecture was confirmed by the second author who proved the following theorem.

+

H. Sachs and M . Stiebitz

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Theorem 10. ([19]). I f G is a colour-critical graph which has low vertices then its high-vertex subgmph has no more components than its low-vertex subgraph.

Characterization of k-critical graphs whose high-vertex subgraph is (k - 2)-colourable, or is a complete graph

5

5.1

The main results

In 1963 T. G d a i [ll]characterized the set of all k-critical graphs having exactly one high vertex; this result is implicitly contained also in a paper of G.A. Dirac [9] (1957).

Theorem 11. (G.A. Dirac [9]; T. Gallai [ll]). (1) Let k 2 4 and let G be a k-critical graph which has exactly one high vertex. Then L ( G ) E Tk(1).

( 2 ) Let k 2 4, let T E Tk(1) and assume that T has at least two blocks; connect every non-separating vertex of T with an additional vertex Q . Then the resulting graph G is k-critical, L(G) = T , and H ( G ) consists of the single vertex Q. The construction described in (2) of Theorem 11 is a special case of the construction K: this is just the way to obtain G(T,u, 1) for ( T ,u) E Uk(1). The second author [20] proved that a considerably larger class of critical graphs can be characterized by means of the constructions K and K’ (see Theorems 13, 14 and 15 below). Before formulating these results we need the definition of an important concept and a more general theorem.

Definition 5.1. Let G be a graph. The set X called a k-set if (1) for all P E X , val(P : G ) 5 k

V ( G ) with X # 0 is

- 1;

( 2 ) G [ X ]is connected;

(3) G - X E Col(k

- 1).

Note that if G E Cri(k) and L ( G ) is non-empty and connected then V ( L ( G ) )is a k-set of G.

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387

Theorem 12. ([20]). Let k 2 4 and G E Chr(k). Further, let X C V(G) be a k-set of G, and suppose that H = G - X is ( k - 2)-colourable and V ( H ) = N, ( q 2 1). Then the following statements hold. (a) For all P E X , val(P : G ) = k

- 1.

(b) T = G [ X ]is a k-Gallai-tree which has a perfect matching. (c) There is a mapping u such that ( T ,u ) E Uk(q) and G = G(T,u,q ; H ) with H ( T , u , q )C H . Theorems 5 and 12 immediately imply Theorem 13. Theorem 13. (See [20]). Let G E Cri(k) for k 2 4 and suppose that L ( G ) is connected and H(G) E Col(k - 2 ) . Put IV(H(G))I= q. Then the following statements hold. (1) T = L ( G ) is a k-Gallai-tree which has a perfect matching. (2) There is a mapping u such that ( T ,u) E Ulc(q) and G

G(T,u,q).

Making use of Proposition 4,we obtain: Theorem 14. ([20]). Let G E Cri(k) for k 2 4 and suppose that H ( G ) is a complete graph on q vertices (1 5 q 5 k - 2). Then T = L ( G ) E Tk(q) and there is a mapping u such that ( T ,u) E Uk(q) and G G ( T ,u,q).

For the proof of the next theorem, the reader is referred to [20]. Theorem 15. ([20]). Let G E Cri(k) for k 2 4 and suppose that H ( G ) is a complete graph on k - 1 vertices. Then T = L ( G ) is a k-Gallai-tree and there is a mapping u such that ( T ,u) E U; and G 2 G(T,u).

5.2

Proof of Theorem 12

Let us start with some definitions and two auxiliary propositions. Let G' be a subgraph of G. The colouring c of G is called an extension of the colouring c' of G' (briefly: c' c ) if, for all P E V(G'), c( P ) = c'( P ) . A subset C' of Ck(G') is called extendable with respect to Ck(G) (briefly: C' + C k ( G ) )if, for every c' E C', there is a c E Ck(G) such that c' C c. Instead of {c} -+ c k ( G ) we shall briefly write c Ck(G). --f

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Lemma 2. (See [19]). Let G be a graph and let X G. Then the following statements hold.

(2)

If there is a vertez P +

E X with val(P : G)

s V(G) be a k-set of

5 k - 1 then Ck-l(G - X)

Ck-l(G).

( 3 ) Let P E X be a vertez which is in G adjacent to 91,Q2 E V(G - X)

(91

# 92). I f c

E Ck-l(G

- X) satisfies c(Q1) = c(Q2) then c

+

c k - 1(G)*

(4) For all P E X, Ck-I(G - X) -+ Ck-i(G - {P}). (5) For all e E E(G) - E(G - X), Ck-l(G

- X) + Ck-i(G - {e}).

Lemma 3. (See [19]). Let G be a graph with x(G) > k k-set and let X C V(G) be a k-set of G. Then

- 1 which has a

(2) for all P E X, v a l ( P : G) = k - 1;

(3) G[X] is contained as a subgraph in every k-critical subgmph of G; (4) G[X] is a k-Gallai-tree.

We now come to the proof of Theorem 12. In what follows let G denote a k-chromatic graph ( k 2 4) and let X E V(G) be a k-set of G where H = G - X is a ( k - 2)-colourable graph with V(H) = N, ( q 2 1). Put G[X] = T . We shall use the following notation: for B E B ( T ) and P E V(B) put

N(P) := N ( P :G) n N,; M(P : B ) := N(P : G) - V(B). From Lemmas 2 and 3 we obtain as an immediate consequence:

Proposition 9, (A.l) T is a k-Gallai-tree; (A.2)v d ( P : G) = k - 1 for all P E X; (A.3) Ck-l(H) + Ck-l(G - V(B)) for all B E B ( T ) . Next we shall prove the important Proposition 10.

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Proposition 10. Let B E B(T). Then ( B . l ) IM(P : B)I = k - s ( B ) for all P E V ( B ) ; (B.2) c ( M ( P : B ) ) = c(M(Q : B ) ) for all P,Q E V ( B ) and for all c E Ck-l(G - V ( B ) ) .

+

Proof. By (A.2), val(P : G) = k - 1 = val(P : B ) I M ( P : B)I = s ( B )- 1+ IM(P : B)I, therefore, IM(P : B)I = k - s ( B ) , i.e., ( B . l ) holds. Proof of (B.2): Assume that there is a c E Ck-l(G - V ( B ) )such that c ( M ( P : B ) ) # c ( M ( Q : B ) ) for some pair P,Q E V ( B ) . Since B is connected we may assume that P and Q are adjacent in B . Let i E c ( M ( P : B ) ) - c(M(Q : B ) ) and put Y := V ( B )- { Q } . Clearly, there is a ( k - 1)-colouring c‘ of G - Y with c & c’ and c’(Q) = i. The graph B - { Q } being connected, Y is a k-set of G and, by Lemma 2 (3), 0 c‘ + Ck-l(G) - contradicting x(G) = k. Proposition 11. H [ N ( P ) ] is a complete graph for all P E X. Proof. Assume that there are two non-adjacent vertices in H [ N ( P ) ] .Because of x(H) k - 2, there is a ( k - 1)-colouring c of H with ~ ( x=) c(y). Now Lemma 2 (3) implies x(G) k - 1, contradicting the assumption. 0

<

<

Proposition 12. Let B E B(T) and let P E V ( B ) be a non-separating vertex o f T . Then N(Q) C N ( P ) = M ( P : B ) for all vertices Q E V ( B ) . Proof. Obviously, M ( P : B ) = N(P). Now suppose that there is a vertex Q E V(B) such that N ( Q ) N(P). Let 5 E N ( Q ) - N(P). Because of x ( H ) 2 k - 2, there is a ( k - 1)-colouring c‘ of H such that x is the only vertex of H with ~ ’ ( x=) k - 1. By (A.3), c’ can be extended to a ( k - 1)-colouring c of G-V(B). Then weobtain c ( M ( P : B ) ) = c ( N ( P : B ) ) c c ( V ( H ) - { x } ) E : {1,2, ..., k - 2 ) a n d k - l E c ( N ( Q ) ) c c ( M ( Q : 0 B ) ) ,hence c ( M ( P : B ) ) # c ( M ( Q : B ) ) - contradicting (B.2). Combining Propositions 10, 11 and 12, we immediately obtain:

Proposition 13. Let B be an end block of T and let Q E V ( B ) denote the only separating vertex of B (if there is not such a vertex, then T = B and for Q an arbitrary vertex of B may be taken). Then there is a vertex set M ( B ) V(H)= N, such that ( C . l ) N(P) = M(P : B ) = M ( B ) for all P E V ( B )- { Q } ; (C.2) N ( Q ) c M ( B ) ; (C.3) IM(B)I = k - s ( B ) 5 Q; (C.4) s ( B ) 2 k - q = r ; (C.6) H [ M ( B ) ]is a complete graph.

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Next we shall prove: Proposition 14. Let B , B' E B ( T ) be two adjacent blocks of T . an end block then every vertex of B' is a sepamta'ng vertex of T .

If B is

Proof. Suppose that, on the contrary, there is a vertex P' E V(B') which is a non-separating vertex of T . Let P be the vertex of T belonging to both B and B'. Then P # P' and N(P') = M ( P ' : B ) 5 V(H). Because of x ( H ) 5 k - 2 there is a c E Ck-l(G - V(B')) with c(V(H)) E {1,2,. .. , k - 2) and therefore, c(M(P' : B')) C {1,2,. . ,k - 2). Since B is an end block, by Proposition 13 the graph G' = H[M(B)] ( B - { P } ) is a subgraph of G - V(B') with x(G') = k - s ( B )t s ( B )- 1 = k - 1. Therefore k - 1 E c(V(G')) and thus k- 1 E c(V(B - { P } ) ) . The fact that B is a brick implies that H[M(B)] v B is a k-critical graph and we conclude that k - 1 E c ( M ( P : B')). Thus c(M(P' : B')) # c ( M ( P : B')) - contradicting (B.2). 0

.

v

Proposition 14 immediately implies: Proposition 15. The end blocks of T are pairwise disjoint. Proposition 16. Let B1,Bz E B ( T ) (B1 # B2) be end blocks o f T and let B' E B ( T ) be a block with V(B') n V ( B i ) = { P i } (i = 1,2). Further, suppose that for i = 1,2 the vertex P; is contained in no block of T other than B' and Bi. Then (D.1) M(B1) - N(P1) = M(B2) - N(P2); (D.2) IM(B1) - N(P1)I = IM(B2) - N(P2)J = s(B') - 1 5 q. Proof. Suppose that, contrary to ( D . l ) , there is a vertex x satisfying x E M(B2) - N(P2) and x 4 M(B1) - N(P1). Because of x ( H ) 5 k 2, there is a (k - 1)-colouring c of G - V ( B ' ) with C ( X ) = k - 1 and c(V(H) - { x } ) 5 {1,2,. . . ,k - 2). According to (C.l), all vertices Q of B2 - {PZ}are adjacent to x. The vertex P2 is not adjacent t o x and M(P2 : B') = N(P2 : B2) U N(P2), therefore, h - 1 4 c(M(P2 : B')). Now we shall show that k - 1 E c(M(P1 : B')). If x is adjacent to PI then, clearly, k - 1 E c(M(P1 : B')) (note that C ( X ) = k - 1). If x is not adjacent to PI our conclusions run as follows. From x 4 M(B1) - N(P1) weobtainx $ M(B1), thusc(M(B1)) G c(V(H-{x})) E {1,2, ...,k-2}. Because of (C.1) and (C.5), the graph H[M(B1)] (B1 - {PI}) is a (k - 1)-chromatic subgraph of G - V(B'), therefore, k- 1 E c(M(P1 : B')). Thus c(M(P1 : B')) # c(M(P2 : B')) - contradicting (B.2). This contradiction proves (D.1).

v

Colour-Critical Graphs with Vertices of Low Valency

39 1

Proof of (D.2). From (A.l) and Proposition 13 we obtain

thus

This proves Proposition 16.

0

We are now ready to prove Theorem 12 by induction over (B(T)I.

Proof of Theorem 12. For IB(T)I = 1, Theorem 12 is an immediate consequence of Proposition 13. Assume IB(T)I > 1. Then, by Proposition 14 and 15, there is at least one block in B(T), say B', with IV(B')I = t which has the following properties: The vertices of B' can be so labelled that V(B') = {PI,&,.. ., P t } and for i = 1 , 2 , . . ,t - 1, the vertex Pj is contained in exactly one block other than B', say Pi E V(Bi), Bj # B', and all the B; are end blocks. Now put

.

(1) M ( B ' ) := M(B;) - N ( P ~ (i ) = i , 2 , . .. , t - 1; see (DJ)), (2) T' := T - V(B1) - V(B2) - * * . - V(Bt-I), (3)

G1

:= G[V(T') U NqJ,

{

(4) F := {Pt, i}li E M(B')}, and

(5) G' := (V(Gl),E(Gi)U F).

Claim I : x(G') = k . Proof:Clearly, V(T')is a k-set of G' and therefore, all we have to show is that x(G') 2 k . Suppose that, on the contrary, there is a (k- 1)-colouring c' of G'. Note that c'(Pt) 4 c(M(B')). The colouring c' can be extended to a (k - 1)colouring c of G in the following way. - {Pi} (i = 1 , 2 , . .. ,t - 1) are coloured: First, the vertices P E V(B,) this can be done in such a way that, during the colouring procedure, each vertex of Bi - { P i } (note that Bi - {Pi} is a connected graph) not yet coloured is adjacent t o no more than k - 2 vertices already coloured, thus,

H . Sachs and M. Stiebitz

392

in this step, & - 1 colours suffice; P being adjacent to all vertices of M ( B ’ ) (see (C,l)), the colour of P is not contained in c’(M(B’)). of B‘ - { P t } are coloured: the Eventually, the vertices PI,P2,. s(B‘) - 1colours of c‘(M(B‘))can be used. Note that, for i = 1,2,. . . ,t - 1, N(Pi) and M(B’) are two disjoint subsets of M(Bi), and H [ M ( B ; ) ]is a complete graph (see (C.2), (C.5), (D.l), and (D.2)). Thus we have proved x(G) 5 k- 1- contradicting the general hypothesis that G E Chr(k). This proves Claim I.

..

Evidently, G’ and X‘:= V(T‘)satisfy the assumptions of Theorem 12. Thus, by virtue of the induction hypothesis, there is a mapping u’ satisfying (T‘, u’) E uk(q)and G’= G(T’,u’,q) with H(T’,u‘, q ) C H . In particular, T’E Tk(Q).If B1(T‘) is the perfect matching of T‘, then B1(T) = BI(T’) U {B1,B2,,. .,B+l} is the perfect matching of T and, by ((3.5) and (D.2), T E Tk(q). Now let u be the following mapping: (i) u(B) = u’(B) for B E

B(T‘),

(ii) u(B’) = M(B’), and (iii) u(B;) = N,

- M(B;) (i = 1,2,. . . ,t - 1).

Claim 11: (T, u) E Uk(q). Proof: Because of (C.3) and (D.2), (T,u) satisfies (Bl) and (B2) (see (K.2), Section 3.1), therefore, all we have to show is that u(B’)nu(B2) = 0 for any pair of adjacent blocks B’, B2 E B(T). It is sufficient to consider the case where B1 = B’ and B2 is any block of T’ containing Pt. Now, in G‘ the vertex Pt is adjacent to all vertices from M(B’) = u(B’) and from the definition of G‘ = G(T’,u‘,q) it follows that u(B’) n u(B2) = M(B’)n u ’ ( B ~ = ) 8. This proves Claim 11. From Proposition 13 and 16 we now conclude that G = G(T,u,q ) and H ( T , u, Q) G H . 0 Theorem 12 is thus proved.

6

Concluding remarks

Let us conclude our investigations with some remarks concerning the close relationship between computational hardness and general theoretical difficulty.

Colour-Critical Graphs with Vertices of Low Valency

393

1. Let an integer k 2 4 and a k-critical graph G be given. It is not too difficult to show that it can be decided in polynomial time whether or not G is obtainable by means of one of the constructions K or K’, i.e., whether or not there is an integer q and a pair (T,u) E Uk(q), or a pair (T,u) E Ug, such that G 2 G(T,u,q), or G E G(T,u), respectively. 2. Also, making use of Theorems 14 and 15 and the above remark, it can be decided in polynomial time whether or not a given graph is a critical graph which has a complete graph as its high-vertex subgraph. 3. However, given k 2 5, no algorithm is known which would allow to decide in polynomial time whether or not a given graph G has the following properties. (i) G E Cri(k), (ii) L(G) is non-empty and connected, (iii) H ( G ) E Col(k - 2). (Note that, for k 2 5 , testing for (k - 2)-colourability is NP-complete.) 4. Given an integer k 2 4 and a k-Gallai-tree T , there are infinitely many pairwise non-isomophic k-critical graphs whose low-vertex subgraph is isomorphic to T (this follows immediately from Theorem 3 and Proposition 6). However, Remark 3, in connection with Theorem 13, implies that among these graphs there are only finitely many whose high-vertex subgraphs is (k - 2)-colourable (which can all be obtained by means of the construction K). 5. From a certain point of view, the main procedures described above may be considered variations of the Dirac-Hajb construction (see also [18 $61). However, as is clear from Haj6s’ theorem (Theorem l ) , there is one other important operation needed for the construction of all k-critical graphs, namely, the identification of non-adjacent vertices. Thus, in order t o master the particular difficulties connected with k-critical graphs G satisfying H ( G ) E Chr(k - 1) or L(G) = 0 (if these difficulties - which have not been attacked in this paper - can be mastered at all), the effect of identifying non-adjacent vertices of a graph on its criticality properties should be studied. In addition to the references given overleaf, some more references concerning the general theory of colour-critical graphs can be found in [17].

Added in proof. In September 1987 the authors submitted a paper to Annals of Discrete Mathematics entitled “On constructive methods in the

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H. Sachs and M. Stiebitz

theory of colour-critical graphs,” which contains a list of about 60 references.

References [l] R. L. Brooks, “On colouring the nodes of a network,” Math. Proc. Cambridge Phil. SOC. 37 (1941), 194-197. [2] G. A. Dirac, “Some theorems on abstract graphs,” Proc. London Math. SOC.(3) 2 (1952), 69-81. [3] G. A. Dirac, “A property of 4-chromatic graphs and some remarks on critical graphs,” J. London Math. SOC.27 (1952), 85-92. [4] G. A. Dirac, “The structure of k-chromatic graphs,” Fund. Math. 40 (1953), 42-55. [5] G. A. Dirac, “Theorems related to the four colour conjecture,” J. London Math. SOC.29 (1954), 143-149. [6] G. A. Dirac, “Circuits in critical graphs,” Monatsh. f. Math. 59 (1955), 178-187.

[7] G. A. Dirac, “Map colour theorems related t o the Heawood colour formula,” J. London Math. SOC.3 1 (1956), 460-471. [8] G. A. Dirac, “Map colour theorems related to the Heawood colour formula (II),” J. London Math. SOC. 32 (1957), 436-455.

191 G. A. Dirac, “A theorem of R. L. Brooks and a conjecture of H. Hadwiger,” Proc. London Math. SOC.(3) 7 (1957), 161-195.

[lo] G. A. Dirac, “On the structure of 5- and 6-chromatic abstract graphs,” J . reine angewandte Mathematik 214-215 (1964), 43-52. [ll] T. Gallai, “Kritische Graphen I,” Magyar Tud. Akad. Mat. Kutat6 Int. Kozl. 8 (1963), 165-192. [12] T. Gallai, “Kritische Graphen 11,” Magyar Tud. Akad. Mat. Kutatd Int. Kozl. 8 (1963), 373-395. [13] D. Greenwell and L. LovAsz, “Applications of product colouring,” Acta Math. Acad. Sci. Hungar. 25 (1974), 335-340.

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[14] G. H a j b , “Uber eine Konstruktion nicht n-farbbarer Graphen,” Wiss. Zeitschrift Martin-Lu ther- Univ. Halle-Wit ten berg, Math .-Nat . X/1 (1961), 113-114. (See also: G. Ringel, Furbungspmbleme auf FZZchen und Graphen, Berlin (1959), 27-30.) [15] L. Lovisz, Cornbinatorial Problems and Exercises, Akad. Kiadb, Budapest (1979), Problems 9.16-24 (pp. 60-61). [16] V. Muller, “On colorable critical and uniquely colorable critical graphs,” Recent Advances in Graph Theory, Proc. 2nd Czechoslovak. Sympos., Prague (1974), 385-386. [17] H. Sachs, “On colour critical graphs,” Fundamentals of Computation Theory, FCT ’85 Cottbus, GDR, Sept. 1985, Lecture Notes in Cornpu ter Science 199, Springer-Verlag, Berlin-Heidelberg-New YorkTokyo (1985), 390-401. [18] H. Sachs and M. Stiebitz, “Construction of colour-critical graphs with given major-vertex subgraph,” Combinatorial Mathematics, Annals of Discrete Math. 17 (1983), 581-598. [19] M. Stiebitz, “Proof of a conjecture of T. Gallai concerning connectivity properties of colour-critical graphs,” Combinatorica 2 (1982), 315-323. [20] M. Stiebitz, “Colour-critical graphs with complete major-vertex subgraph,” Graphs, Hypergraphs and Applications, Proc. Conf. Graph Theory, Eyba, Oct. 1984, Teubner-Texte zur Mathematik 73, Teubner Leipzig (1985), 169-181. [21] M. Stiebitz, BeitrZge zur Theorie der fiirbungskritischen Graphen, Dissertationsschrift (B), Technische Hochschule Ilmenau (1985). [22] M. Stiebitz, “Subgraphs of colour-critical graphs.” Submitted to Combinat orica. [23] B. Toft, “On the maximal number of edges of critical k-chromatic graphs,” Studia Sci. Math. Hungar. 5 (1970), 461-470. [24] B. Toft, Some contributions to the theory of colour-critical gruphs, Doct. Diss. Univ. of London (1971); Var. Publ. Ser. 14, Aarhus Universi ty. [25] B. Toft, “Two theorems on critical 4-chromatic graphs,” Studia Sci. Math. Hungar. 7 (1972), 83-89.

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[26] B. Toft, “Color-critical graphs and hypergraphs,” J . Combin. Theory (B)16 (1974), 145-161. [27] B. Toft, “On critical subgraphs of colour-critical graphs,” Discrete Math. 7 (1974), 377-392. [28] B. Toft, “Some problems and results related to subgraphs of colour critical graphs,” Graphen in For.qchung und Unterricht (ed. R. Bodendiek et d.),B. F’ranzbecker Verlag, Salzdetfurth (1985), 178-186.

Annals of Discrete Mathematics41 (1989)397-400 0 Elsevier Science Publishers B.V. (North-Holland)

About the Chromatic Number of l-Embeddable Graphs H. Schumacher Institute of Mathematics and its Didactics PH Kiel Kiel, FRG Dedicated to the memory of G. A . Dirac Following G. Ringel, for any surface S we let xl(S) denote the maximum chromatic number of a graph which can be drawn on S in such a way that each edge intersects at most one other edge. Ringel showed that for the torus SI and the Klein bottle S,l, xl(S1) = x(S,l) = 9. In this paper it is proved that every graph attaining this maximum for one of these two surfaces must contain ICg. This is applied to a result about maps on the same two surfaces, concerned with colourings of vertices and countries at the same time.

Let S be a closed surface and let G be a graph embeddable into S. If G is represented on S such that each edge is not crossed over by any other edge, G divides S into faces (countries), edges (borders) and vertices. We call such a representation of G on S a map M on S , if G is a graph without loops and if every country is a topological representation of the plane. In this case we say G is the border-graph of M . Two countries are said t o be adjacent if they have got a least one border in common. Two vertices are adjacent if they have at least one edge joining them. Now let M be a map on a closed surface S. We look for the smallest number a ( M ) of colours which are needed t o colour M in such a way that every country and every vertex of M contains exactly one of the colours, every two adjacent countries and adjacent vertices are coloured differently and the colour of every country is different from the colour of every vertex on the border of the country concerned. % ( M )is a new chromatic number of the map M . The maximum 2 ( M ) of all maps M on S is said to be a new chromatic number of the surface S, shortly named 2(S). These new chromatic numbers 2 ( M ) and SO) of maps on the sphere SOhave been introduced by G. Ringel in [4].

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On the other hand, G. Ringel considers in [4]and in [6] graphs being 1embeddable into a closed surface S. A graph G is said to be 1-embeddable into S if and only if it can be drawn on S in such a way that every edge intersects at most one other edge of G. In [6],G. Ringel writes x l ( S ) for the maximum x(G) of all graphs G being 1-embeddable into S and shows in case of the torus S1 and Klein's bottle S; that xl(S1) = x,(Sz) = 9. It can easily be shown that

for every closed surface S. Since M is a map on S, we choose one point in the interior of every country as a new vertex and call it the capital of the country. Then we connect every capital with all the vertices lying on the border of the country concerned by new edges. Those edges can be drawn in such a way that they do not intersect any edges of the border-graph G of M . Furthermore, two capitals are connected by a new edge if and only if their countries are adjacent. Such an edge can be drawn in such a way that there is an intersection with exactly one edge of the common border. The adjacency on the vertex-set of the obtained graph G * ( M )obviously delivers the chromatic number x ( G * ( M ) )= k ( M ) . Because of its construction G * ( M )is a graph being 1-embeddable into S, and we obtain (1). Figure 1 shows a map M and the respective graph G * ( M ) on the torus whereas the double-edges of G ( M ) are extinguished in G * ( M ) .

M Figure 1. Now the question must be considered whether we can use the sign of equality in case of (1) in a way similar to the usual chromatic numbers

About the Chromatic Number of 1-Embeddable Graphs

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of maps and graphs on the same surface. In case of the sphere So Ringel proved in [4]that 6 I ?(So) I Xl(S0) 57, and Borodin showed in [3] that

In case of the projective plane S; it is proved in [7] that

In case of the torus 5’1 and Klein’s bottle Sz, however, we can show

g(S1) = 8 and 7

I k(Sz) I 8.

(3)

(4)

Hence equality does not hold in (1) for the torus and for Klein’s bottle.

Theorem 1. If G is a graph which is 1-embeddable into the torus or into Klein’s bottle and if G does not contain a complete graph Iig, then x ( G ) 5 8. Proof. Suppose that G is a graph with x ( G ) = 9 which is 1-embeddable into S1 or 1-embeddable into S; and which does not contain a complete graph I i g . Then we can find such a graph G with a minimal number of vertices a. In this case every vertex v of G has got a degree 7 ( v , G )2 8. If this is not the case, then there exists a vertex v of G with 7 ( v , G ) 5 7 and with the property that G - v can be coloured with 8 colours because of the minimality of a. Because 7 ( v , G )5 7 we obtain x ( G ) I 8, and this - i)lKl in [7] is a contradiction to x ( G ) = 9. It follows from 0 2 &(8 that 7 ( v , G )= 8 for every vertex v of G. But this is a contradiction to Brook’s theorem [8],since there must exist 0 a vertex v of G with r(v,G) 2 x ( G ) . So Theorem 1 is proved.

Theorem 2. If M is a map on the torus S1 or IClein’s bottle S;, then k(M) 5 8 . Proof. Suppose that there exists a map M with ? ( M ) = 9 on a closed surface S of Euler characteristic 0. From Theorem 1 we know that G * ( M ) then contains a complete graph K9. A subgraph Iig of G * ( M )contains at least 9 pairs of intersecting edges, shortly called pairs of diagonals. If 119 contains less than 9 pairs of diagonals, we obtain a graph with 9 vertices and at least 28 edges being embedded into S by deleting exactly one diagonal

H. Schurnacher

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of every pair of diagonals. The fact that a1 5 3ao for each graph with a0 vertices and a1 edges which can be embedded into S (without intersection) [5], gives us the contradiction required. So let Kg be a subgraph of G * ( M ) with at least 9 pairs of diagonals. Since one of the diagonals of each pair is incident to two capitals and the other one is incident t o vertices of the border-graph G of M , at least 5 vertices of I i g must belong to the border-graph G and at least 5 vertices of K g must belong to the set of capitals. Since the vertex-set of G and the set of capitals are disjunctive, this would imply that the K g has got at least 0 10 vertices, and we obtain the contradiction required. The proofs for (3) and (4) follow from (2), Theorem 2 and the map M on the torus with G * ( M )= K g which is shown in Figure 1.

References [l] R. Bodendiek, H. Schumacher and K. Wagner, “Bemerkungen zu einem Sechsfarbenproblem von G. Ringel,” Abh. Math. Sem. Univ. Hamb. 53

(1983), 41-52. [2] R. Bodendiek, H. Schumacher and K. Wagner, “Uber l-optimale Graphen,” Math. Nachr. 117 (1984), 1-17.

[3] 0.V. Borodin, ‘‘Regenie zadat Ringelja o versinnogranevoj raskraske ploskich grafov i o raskraske I-planarnych grafov,” Metody Diskretnogo analixa v izuc‘enii realizacii fogiteskich funkcij. 41 (1984), 12-26.

[4]G. Ringel, “Ein Sechsfarbenproblem auf der Kugel,” Abh. Math. Sem. Univ. Hamb. 20 (1965), 107-117. [5] G. Ringel, “Map Color Theorem,” Grzlndlagen der rnathernatischen Wissenschaften, Band 200, Springer-Verlag, Heidelberg (1974). [6] G. Ringel, “A Nine Color Theorem for the Torus and the Klein Bottle,” The Theory of Applications of Graphs (Gary Chartrand, ed.), Wiley (1981). [7] H. Schumacher, “Ein 7-Farbensatz l-einbettbarer Graphen auf der projektiven Ebene,” Abh. Math. Sem. Univ. Hamb. 54 (1984), 5-14. [8] K. Wagner, Graphentheorie, B I Hochschultaschenbucher, Mannheim (1970).

Annalsof Discrete Mathematics41 (1989) 401-416 0 Elsevier Science Publishers B.V. (North-Holland)

Problems and Conjectures in the Combinatorial Theory of Ordered Sets W. T. Trotter* Department of Mathematics Arizona State University Tempe, Arizona, USA

Dedicated to the memory of G. A . Dirac 49 questions from the theory of ordered sets are stated, and the present knowledge about each is surveyed.

1

Introduction

The purpose of this article is to survey a number of combinatorial problems and conjectures for ordered sets. This area of combinatorial mathematics is relatively new and is experiencing rapid growth. Accordingly, it is impossible to claim that the article is exhaustive in scope. Rather, we concentrate on problems which fall under the heading of “extremal problems”. Furthermore, we have chosen problems which exhibit the single most attractive feature of combinatorial mathematics in that the problems can be understood by nonspecialists. Throughout the paper, we consider an ordered set as a pair ( X , P ) where X is a set and the partial order P is a reflexive, antisymmetric and transitive relation on X .

2

Dimension problems

The dimension of a n ordered set ( X ,P ) , denoted dim(X, P), is the least t for which P is the intersection o f t linear orders on X. Our first conjecture is one of the best known problems in dimension theory. *Research supported in part by the National Science Foundation

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W. T.Trotter

402 Conjecture 1. If (X, P ) is any ordered set with distinct points 5, y E X so that dim(X, P ) 5 1 -k dim(X

1x12 3, then there exist

- {z, y } , P(X - {z, y } ) ) .

This conjecture arises from attempts to provide a simple inductive proof of the following basic inequality due to T. Hiraguchi [13]: dim(X,P) 5 lXl/2 when 1x1 2 4. We believe the first published reference to the conjecture is [3], but the problem is so natural that one cannot be certain. See Kelly’s note [16] for additional background information on this problem. Conjecture 2. For every pair m, n of positive integers, there exist ordered sets (X,P) and (Y, Q ) so that dim((X,P) x (Y,Q)) = max{dim(X,P),dim(Y,Q)}. This problem comes from the following elementary inequalities governing Cartesian products of ordered sets: max{dim( X, P ) ,dim( Y,Q)} 5 dim( (X, P ) x (Y, Q)) dim((X,P) x (Y,Q))5 dim(X, P)

(1)

+ dim(Y,Q)

(2) In many cases, the second inequality is tight. For example, K. Baker [2] showed that this occurs whenever the two ordered sets are nontrivial and have greatest and least elements. However, very little is known about the accuracy of the first inequality. In [34], Trotter showed that for each n 2 3, there exists an n-dimensional, ordered set (X, P ) with the property that dim((X, P ) x (X, P ) ) = 2n - 2. This ordered set is the standard example, i.e., the family of all l-element and (n - 1)-element subsets of an n-element set ordered by inclusion. Perhaps it is true that for each n 2 1, there exists an ordered set (X, P ) so that dim(X, P ) = n and dim((X, P ) x (X, P))= n. The standard example figures in another intriguing dimension theory problem. Problem 3. For each n 2 1, determine the least integer f(n) so that if (X, P ) is any ordered set in which each point is comparable with at most n other points, then dim(X, P ) 5 f (n).

Z. Fiiredi and J. Kahn [lo] have supplied a clever argument to show that there exists an absolute constant c so that f (n) 5 c n (log n)2. On the other hand, the standard example provides the trivial lower bound f (n) 2 n + 1. Although no better bound is known, it appears likely that f(n)/n tends t o infinity.

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An ordered set (Y,Q)is called a homeomorph of ( X , P ) if the diagram for (Y,Q) is obtained from the diagram of ( X , P ) by inserting points on edges.

Problem 4. Is it true that if (Y,Q)is a homeomorph of ( X , P ) , then dim(X, P ) 5 dim(Y, Q) 5 2 dim(X, P ) ? There are many analogies between dimension for ordered sets and chromatic number of graphs. Here is a problem which comes from attempts t o see how far this analogy extends. In order for a graph to have large chromatic number, it must contain a relatively small subgraph which has relatively large chromatic number. To be somewhat more precise, Kierstead, Szemeredi and Trotter [21]proved that if G is a graph on n vertices, and G has no subgraph H whose chromatic number exceeds c and whose radius in G is a t most 2kn11k, then the chromatic number of G is at most k(c - 1) 1. The special case of this theorem when c = 2 was conjectured by P. E r d k .

+

Problem 5. What is the maximum value f(n,rn,Ic) of the dimension of an ordered set on n points in which no subodered set on m points has dimension exceeding k ? A special case of this problem may be of particular interest.

Problem 6. What is the maximum value f ( n ,k ) of the dimension of an ordered set on n points which does not contain a standard example of dimension k ? A solution to the preceding problem may shed some light on the following problem which is due to B. Sands.

Problem 7. What is the least integer f ( n ) for which there exists a lattice of dimension n having f ( n ) points? Of course, the distributive lattice 2n consisting of all subsets of an nelement set ordered by inclusion has dimension n so f ( n )5 2n. But in [12], B. Ganter, P. Nevermann, K. Reuter and J. Stahl observed that the lattice of subspaces of the projective plane of order 3 has 28 points and dimension a t least 5. More generally, they showed that f ( n ) < cn for every c > 1. This result follows from their observation that if lIn denotes the lattice of all partitions of an n-element set, then there exist absolute constants c1, c2 so that c1n2 < dim(II,) < c2n2, Perhaps there is an exact answer for the dimension of the lattice of partitions of a set. At least it should be possible

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t o determine the limit of dim(II,)/n2. As far as I know, the dimension of the lattice of partitions of an integer has not been investigated. These same authors also show that the lattice of subspaces of an ndimensional linear vector space over GF(2) also provides a subexponential bound on f ( n ) . For this lattice, they show that the dimension is a t least 2 " / n and a t most 2". It should be possible to obtain a good estimate for the dimension of this lattice. Recently, Z. Fiiredi and J. Kahn [ll]observed that a much better bound on the minimum size of an n-dimensional lattice can be derived from the lattice obtained by adding a zero and a one to the ordered set formed by the points and the lines of a finite projective plane. They observed that if the order of the projective plane is n, then the dimension of the lattice is a t least En where E is an absolute constant. This shows that f(n) < cn2. Probably the exponent is right in this estimate. The following problem is due to J. Ginsburg. Define the cutset number of ( X , P ) as the least nonnegative integer n so that for each x E X, there exists a subset S of at most n points with each point of S incomparable t o z so that every maximal chain contains a point from S U {x}.

,Problem 8 . Does there exist a function f(n) so that dim(X, P ) 5 f(n) whenever the cutset number of (X, P ) is n ?

3

Problems on linear extensions

Let E ( P ) denote the set of all linear extensions of a partial order P on a finite set X. For a distinct pair x,y E X, define PROB(x < y) as the ratio of the number of extensions in E ( P ) which place x before y divided by the total number of extensions in E ( P ) . The following beautiful conjecture was made by M. Fredman [9]: Conjecture 0 . I f ( X , P) is not a chain, there exist distinct points x, y E X so that 1/3 2 PROB(z < y) 5 2/3. Linial [25] proved that the conjecture holds if the width of (X, P ) is a t most two. Aigner [l]classified the width two ordered sets for which the inequality is tight. However, the best result to date for arbitrary width is due t o Kahn and Saks [15]. They have shown that there exists x , y so that 3/11 5 PROB(x < y) 5 8/11. They also make the following conjecture.

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Conjecture 10. For every E > 0 , there exists an integer n so that if (X, P ) is any ordemd set with width(;", P ) 2 n, then there exist distinct p i n t s X , Y E X SO that 1/2 - E 5 PROB(2 < y ) 5 1/2 E .

+

Additional conjectures of this general flavor are given in Saks' note [31]. There are a number of interesting problems concerning the cardinality of E ( P ) . The following problem is relayed by J. Kahn.

Problem 11. Does there exist an eficient algorithm which accepts an ordered set (X, P ) as input and which outputs a number m satisfying that m 5 IE(P)I 5 m2n, where n = [ X I ? Problem 12. For integers n,k with 0 5 k 5 (y), find the greatest integer f ( n , k ) for which there exists an ordered set ( X , P ) with n = 1x1, IE(P)I = f ( n , k ) , and with ezactly k of the (Y) pairs o f X comparable in P. Some information is known about the extremal ordered sets. It is relatively easy to establish that if ( X , P ) is an ordered set with n points and k comparable pairs and IE(P)I = f ( n , k ) , then there exists a function g from X to the set R of real numbers so that z < y in P if and only if g ( z ) 1 < g ( y ) . Such ordered sets are called semiorders (also unit interval orders). Here is a variant of the preceding problem posed by I. Rival.

+

Problem 13. For integers n,k with 0 5 k 5 n2/4, find the greatest integer g ( n , k ) for which there ezists an ordered set ( X , P ) with 1x1 = n, IE(P)I = g(n,k), and with k edges in the diagram of (X, P ) . It may also be of interest to find those ordered sets with a specified number of comparable pairs or a specified number of edges in the diagram for which the number of linear extensions is minimum. Let ( X , P ) be an ordered set. For each L E E ( P ) , let jum p(l; P ) count the number of pairs of elements which are consecutive in L and incomparable in P . Then define the jump number of (X, P ) , denoted jump(X, P ) , by jump(X, P ) = min{jump(L; P ) : L E E ( P ) } . Call ( X ,P) jump critical if jump(X - {z}, P ( X - {z})) < jump(X,P) for every z E X.

Problem 14. For each integer k 2 1, find the least integer f(k) so that i f jump(X,P) = k and ( X , P ) is jump critical, then 1x1I f ( k ) . El-Zahar and Schmerl [GI show that f ( k ) 5 (k

+ l)!.

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406

Perhaps this inequality is far from best possible. The best known lower bound comes from examples constructed by El-Zahar and Rival [5] which yield f(k) 2 ck where 3 < c < 4. Let Jn denote the ordered set consisting of the linear sum of n two element antichains, i.e., points from distinct antichains are comparable. Then let An denote an n-element antichain. J. Schmerl (321 has proved that an ordered set with large jump number either contains a large Jn or a large A,.

Problem 15. For positive integers m and It, find the least integer f (m, n ) so that i f ( X , P ) is any ordewd set with jump(X, P) 2 f(rn,n), then either ( X , P ) contains J , or A,.

4

Interval orders and generalizations

Let C be a collection of distinct closed intervals of the real line. Define a partial order P on C by A < B in P if and only if a < b on the real line for every a E A and every b E B . An ordered set isomorphic to one obtained by this definition is called an interval order. Let f(n) be the least integer so that if ( X , P ) is an interval order and length(X,P) = n, then dim(X,P) 5 f(n). Rabinovitch [29] proved the existence of f(n) and Bogart, Rabinovitch and Trotter (41 showed that there exists a constant c > 2 so that loglogn 5 f ( n ) < log, n. The upper bound has been improved by Rod1 and Trotter [30] to f(n) < (loglogn)Cfor some c. Probably f(n)< (loglogn)'+" and perhaps f(n) < cloglogn. However, there has been little work done on the dual problem.

Problem 18. Find the least f ( n ) so that if ( X , P ) is an interval order of width n, then dim(X, P) 5 f(n). Call an ordered set ( X , P ) dimension critical if, for every dim(X - { z } , P ( X - {z})) < dim(X,P).

2

E X,

Problem 17. How many dimension critical interval orders are there on n points? Problem 18. How many dimension critical interval orders of dimension n are there? Let I ( n ) denote the interval order determined by the set of nondegenerate closed intervals with integer endpoints from {1,2,0 ,n}. 0

Problems and Conjectures in Combinatorid Theory of Ordered Sets

407

Problem 19. Does there exist a function f ( n ) so that if ( X , P ) is an interval order and dim(X, P ) > f ( n ) , then ( X ,P ) contains I ( n ) ? The interval count of an interval order is the minimum number of different length intermis required in a representation. P. Fishburn [7] d'iscusses many intriguing problems for interval count and we mention here two of them.

Conjecture 20. If ( X ,P ) is an interval order and X contains 3n points, then the interval count of ( X ,P ) does not exceed n.

+

The interval order which is defined by {[-2i - 1,2i 11 : 0 5 i 5 n} U {[-2i - 1, -2i] : 1 5 i 5 n } U {[2i,2i + 11; 1 5 i 5 n } shows that this conjecture if true is best possible.

Problem 21. Does there exist a constant c so that the removal of a point from an interval order decreases the interval count by at most c? Consider an angle in the Euclidean plane as determining an infinite region bounded by two incident rays. An ordered set isomorphic to a family of regions formed by angles in the plane and ordered by set inclusion is called an angle order. Fishburn and Trotter [S] proved that every interval order is an angle order and that every ordered set with dimension at most four is an angle order. They also proved that there exists an ordered set of dimension 7 which is not an angle order.

Problem 22. Find the least integer t for which there exists an ordered set of dimension t which is not an angle order. Problem 23. If ( X , P ) is an angle order, is the ordered set formed by adding a zero to ( X ,P ) also an angle order? Call an ordered set a circle order if it is isomorphic to a family of circular regions in the plane ordered by inclusion. Problem 24. Is every three dimensional ordered set a circle order? Problem 25. Find the least integer t for which there exists an ordened set of dimension t which is not a circle order.

5

Ramsey fheoretic problems

Let ( X , P ) and (Y,&)be ordered sets. We write (Y,Q) --$ ( X , P ) when every partitioning of the points of Y into 2 classes yield a subset X' C Y

W. T. Trotter

408

so that (X’,Q(X’)) is isomorphic to ( X , P ) and all points of X’ belong to the same class.

Problem 26. For each n 2 2, find the least integer f ( n )so that if ( X ,P) is an ordeed set of width at most n, then there’excistsan ordered set (Y,Q) of width at most f ( n ) so that (Y,Q) --t ( X , P ) . It is trivial to see that 2n - 1 5 f ( n ) 5 n2. In [23],H . Kierstead and W. Trotter show that f ( n ) 2 2n. The argument for this bound seems t o leave room for improvement, and I suspect that f ( n ) > & n 2for some constant E

.

Problem 27. For each n 2 1, let g(n) be the least positive integer so that if ( X , P ) is any ordered set with 1x1 = n, then there exists an ordered set (Y,Q) with lYl = g(n) so that (Y,Q) ( X , P ) . --+

Again, we start with the trivial bounds 2n- 1 5 g(n) 5 n2. But in this 1. The reader may naturally ask why we do not pose the analogous problems for dimension and length. The answer is that these problems have been completely solved by Nebetfil and Rod1 [26]. They proved that if dim(X,P) = n , then there exists (Y,Q) with dim(Y,Q) = n so that (Y,Q) 4 ( X , P ) . Furthermore, if length(X, P) = n , there exists (Y,Q) with length(Y, Q)= 2n - 1 so that (Y,Q) -+ ( X , P ) . Clearly, these results are best possible. It follows immediately from Ramsey’s theorem that if (X, P) is an intervalorder, then there exists an interval order (Y,Q) so that (Y, Q) + ( X , P ) .

case, we can at least improve [23]to n2/45 g(n) 5 n2 - n

+

Problem 28. If (X, P ) is an interval order on n points, what is the least f ( n ) so that there exists an interval order (Y,Q) so that (Y,Q) + ( X , P ) and the cardinality of Y is at most f ( n ) ? Problem 29. If ( X , P ) is an interval order on n points and has length (respectively width, dimension) k, what is the least f ( n ,k) so that there exists an interval order (Y,Q) so that (Y,Q) -+ ( X , P ) and the length (respectively width, dimension) of (Y, Q ) is at most f ( n ,k) ? In particular, does f ( n ,k) depend on n ? On the other hand, it is relatively easy to see that there exists an angle order ( X , P ) for which there is no angle order (Y,Q) so that (Y,Q) --$

(X, P). Problem 30. If ( X , P ) is a circle order, does there exist a circle order (Y,Q) so that (Y,Q) -+ ( X , P ) ?

Problems and Conjectures in Combinatorial Theory of Ordered Sets

409

Define the angle dimension of (X, P ) , denoted Adim(X, P), as the least t for which there exists t binary relations A1, A2, *. At on X so that P = A1 n A2 n . . . n A , and (X,Ai) is an angle order for i = 1,2,.-.,t. a ,

Problem 31. Find the least f ( n , t ) so that if (XI = n, Adim(X,P) = t , and z E X, then Adim(X- {z}, P ( X - (2))) 2 t - f (n,t). In particular, does there exist a constant c so that f ( n ,t ) 5 c for all n, t ? Problem 32. If

1x1 = n, what is the maximum value of Adim(X, P ) ?

Problem 33. Does there ezist a function f ( n ) so that i f Adim(X, P ) = n then there ezists (Y,Q ) with Adim(Y, Q ) 5 f ( n ) so that (Y,Q ) + (X, P)? Of course, we chn also define the circle dimension of an ordered set in the obvious fashion and rephrase the last three problems in terms of this parameter, Needless to say, the concept of dimension is one that admits an endless number of variations. Some of these have been explored, and perhaps more work on this theme is worthwhile.

6

Problems from recursive combinatorics

Consider the following game theoretic setting involving two players A and B , a class C of ordered sets and an integer t. It is further assumed that if (X, P) E C, then X = {1,2, ,m } where m is some positive integer. The players alternate moves with Player A constructing an ordered set from C and Player B partitioning this ordered set into chains. At Round i , where i 5 m, Player A provides the binary relation determined by the restriction of P to {1,2, - - ,i}. After receiving this information, Player B makes an irrevocable assignment of i to one of t sets C1, C2, - ,Ct each of which is a chain in ( X , P) . The game terminates and A is the winner if at some step, say Round i with i I. m , Player B has no admissible move, i.e., for each j = 1,2, ,t , the assignment of i to Cj produces a set which is no longer a chain. If Player B is able t o make admissible assignments t o chains at each Round i for i = 1,2,. -.,m, then Player B is the winner. We say that the class C can be recursively partitioned into t chains if there exists a strategy for Player B which will enable him to defeat Player A regardless of the strategy followed by Player A.

--

-

-.

--

-

Problem 34. What is the least f ( n ) for which the class of finite ordered sets of width n can be recursively partitioned into f ( n ) chains?

W.T.Trotter

410

It is not a t all clear that f ( n ) exists. In a beautiful paper [19]H. Kierstead showed that f ( n ) 5 (5n - 1)/4, The best lower bound is due to Szemerddi (see [17]), f ( n ) 1 It is of particular interest to decide whether f ( n ) is polynomial or exponential. This problem is likely to be difficult since even the precise value of f(2) is not known. It is either 5 or 6. Kierstead and Trotter [22]have shown that the class I ( n ) of interval orders of width n can be recursively partitioned into 3n - 2 chains. This result is best possible. In an obvious manner, we can speak of recursively partitioning a class of ordered sets into antichains, recursively coloring a class of graphs, etc. The antichain partitioning problem has been completely solved. J. Schmerl (see [18])proved that the class L ( n ) of ordered sets of length n can be recursively partitioned into antichains and Szemerbdi [18]has shown that the result is best possible.

(nzl).

(7')

Problem 35. Does there exist a function f ( n ) so that the collection of all compambility graphs in which the maximum size of an independent set is n can be partitioned into f ( n ) complete subgmphs? Some nice results have been obtained for recursive dimension. H. Kierstead, G. McNulty and W. Trotter [20]showed that the family of width three ordered sets does not have finite recursive dimension. The difficulty comes from the class of height one ordered sets which are called crowns. These ordered sets contain k maximal elements a1 ,a2,' . ,U k and k minimal elements b l , b2, *. bk with ai > bi and ai > bi+l (cyclically) with k 1 3. Crowns are three dimensional ordered sets and they are dimension critical. They play a key role in a number of combinatorial problems for ordered sets.

-

a ,

Problem 36. What is the least f ( n ) so that the class of width n crown-free ordered sets has recursive dimension at most f ( n ) ? A double exponential upper bound and an exponential lower bound for f ( n ) is established in [20]. The proof techniques strongly support the following conjecture. Conjecture 37. For every k 2 3 and every n 5 k , the class of ordered sets of width n which do not contain a crown on 2m points for any m with 3 5 m 5 k does not have finite recursive dimension. We refer the reader t o Kierstead's survey article [18]for additional problems from recursive combinatorics.

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411

7 Planarity An ordered set is planar if it is possible t o draw a diagram for it in the plane without edge crossings.

Problem 38. Which ordered sets are planar? Develop a good algorithm for testing an ordered set for planarity.

It makes good sense to extend this problem to the class of oriented graphs which do not contain directed cycles. Just as is done with ordered sets, we can use diagrams for acyclic oriented graphs in which the orientation of edges is indicated by the orientation in the plane. In this way, an acyclic oriented graph can be nonplanar when the underlying undirected graph is planar. Problem 39. Which acyclic oriented graphs are planar? Provide an efficient algorithm which tests planarity. Also characterize planarity for acyclic oriented graphs by providing a list of forbidden subgmphs. Some examplw of forbidden subgraphs are given in [35]. One of the original motivations for studying planarity for ordered sets comes from dimension theory. A planar ordered set with greatest and least elements has dimension at most two. Trotter and Moore [36] showed that a planar ordered set with a least (or greatest) element has dimension at most three. However, D. Kelly [17]constructed planar ordered sets of arbitrary dimension.

Problem 40. Do there exist dimension critical planar ordered sets of arbitrary dimension? I have just learned from W. Schnyder of the following beautiful result he has recently proved. With a graph G = (V, E ) , we associate an ordered set ( X ,P ) defined by X = V u E with x < e in P if and only if the vertex x is an endpoint of the edge e. Schnyder proved [33] that the graph G is planar if and only if the dimension of the associated ordered set is a t most three.

Problem 41. Does there ezist a function f ( n ) so that i f G is a gmph of genus n, then the dimension of the associated ordered set is at most f ( n ) ? Problem 42. Does there exist a function g(n) so that i f ( X , P ) is the ordered set associated with the graph G, and dim(X,P) = n, then the chromatic number of G is at most g(n)?

W.T. Trotter

412

The preceding problem is related in spirit to the famous problem involving the existence of graphs with bounded maximum clique size and arbitrarily large chromatic number. The reverse problem is easier since it is easy to show that there exists an absolute constant c so that if (X, P) is the ordered set associated with graph G, and the chromatic number of G is n, then dim(X,P) < cloglogn. Apart from the value of the constant c, this result is best possible. This is shown by the complete graph on n vertices.

8

Miscellaneous problems

We have not included in this paper any Sperner theory problems only because in most cases, the formulation of these problems requires additional background material. However, we must comment that there are a large number of important problems in this area and we refer the reader to Griggs’ survey article [13]for a sampling, For similar reasons, we have not discussed problems from combinatorial lattice theory. There are a tremendous number of challenging and easily accessible problems dealing with families of subsets of a set. Here is one of my favorites. It comes from P. Frankl.

Problem 43. Prove that there exists an absolute constant E so that whenever F is a family of sets closed under unions, i.e., A U B E F for every A , B E F , then there is an element which belongs to at least E I F ~of the sets in F . Actually, Frank1 conjectures that the value E = 1/2 works. This value is best possible as the family of all subsets of a set shows. Here is another posed by D. Kleitman.

Problem 44. How many colors are required to assign colors to the subsets of { 1,2, ,n) so that for each i = 1,2,-..;n, the family Fi of subsets assigned color i is completely union free, i.e., if A l , A2, * ,Ak, Ak+l E F;, then Al U A2 U ... U A&# &+I?

--

-

+

9

It is easy to see that n 1 colors suffice since we can assign a set A its cardinality as a color. An easy inductive argument shows that the number of colors required is a t least n/2. Call a graph G a cover graph if there is a drawing of G in the plane which yields the diagram of an ordered set. It is obvious that a cover graph has no triangles. 0. Pretzel [28]has constructed a graph of girth six which

Problems and Conjectures in Combinatorial Theory of Ordered Sets 413 is not a cover graph. J. NeGetfil and V. Rod1 [27] have shown the existence of graphs with arbitrarily large girth which are not cover graphs.

Problem 45. Give an explicit construction for graphs of arbitrarily large girth which are not cover graphs. The following problem comes from P. Erd&.

Problem 46. Let f ( n ) denote the minimum value of the independence number among all cover graphs on n vertices. A n old construction of ErdOs shows f ( n ) = o(n). Is it true that f ( n ) 2 nl-&? J. Negetiil asks the following question.

Problem 47. If C1 is a cover graph, does there exist a cover graph C2 so that i f the edges of C2 are partioned into two classes, then there exists an induced copy of CI so that all the edges in this copy belong to the same class? There are a number of difficult problems involving the existence of hamiltonian cycles in graphs, in particular the well known conjecture of L. Lovasz for vertex transitive graphs. Here is a special case for ordered sets. I am not sure who first posed the problem. I first heard about it from I. Dejter who credits it to P. Erdos, but perhaps someone else should be credited. D. Kelly has posed the problem at conferences in Banff and Oberwolfach, and so has I. Have1 in Prague.

Problem 48. Consider the bipartite graph formed by the vertices in the middle two levels in the diagram for the ordered set consisting of all subsets of a set of 2n + 1 elements ordered by inclusion. Is this graph hamiltonian? We close this paper with a problem of Linial, Saks and Schor

[as]:

Problem 49. What is the largest integer f ( n ,d ) so that every ordered set on n points contains a d-dimensional subordered set on f ( n ,d ) points? These three authors have shown that there exists a constant c so that every ordered set on n points contains a subordered set on c f i points in which all maximal chains have the same length. Of course the existence of c is easy t o establish, but they have also shown that the result is best possible except for the precise value of the constant c. This problem is representative of the type of problem which deserves more attention in the future.

W.T.Trotter

414

Acknowledgements The author would like to express his appreciation to the University of Paris/Orsay and to the organizers of the Dirac Conference i Sgnderborg, Denmark and the Conference on Graphs and Ordered Sets in Arcata, California in the summer of 1985. The author would also like to acknowledge the assistance of many friends and colleagues in the formulation of the problems and conjectures formulated in this manuscript, especially P. Erdos, P. Fishburn, Z. Fiiredi, J. Griggs, J. Kahn, H. Kierstead, G. McNulty, J. NeBetfil, M. Paoli, I. Rival, V. Rodl, M. Saks, J. Schmerl, J. Walker and D. West.

References [l] M. Aigner, “A note on merging,” Order 2 (1985), 257-264. [2]

K.Baker, “Dimension, join independence and breadth in partially ordered sets.” Unpublished notes.

[3] K. Bogart and W. Trotter, “On the complexity of posets,” Discrete Math. 16 (1976), 71-82. [4] K. Bogart, I. Rabinovitch and W. Trotter, “A bound on the dimension of interval orders,” J . Combin. Theory (A) 21 (1976), 319-328. [5]

M. El-Zahar and I. Rival, “Examples of jump-critical ordered sets,” SIAM J . Alg. Discr. Meth. (1985). To appear.

[6] M. El-Zahar and J. Schmerl, “On the size of jump-critical ordered sets,” Order 1 (1984), 3-6. [7]

P. Fishburn, Interval orders and interval graphs, Wiley, New York (1985).

[8] P. Fishburn and W. Trotter, “Angle orders,” Order 1 (1985), 333-344.

[9] M. Fredman, “How good is the information theory bound in sorting ,” Theoretical Computer Science 1 (1979), 355-361. [lo] Z. Fiiredi and J. Kahn, “On the dimension of partially ordered sets,” Order. To appear.

[ll] Z. Fiiredi and J. Kahn. Personal Communication.

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[12] B. Ganter, P. Nevermann, K. Reuter and J. Stahl, “How small can a lattice of dimension n be?” To appear. [13] J. Griggs, “The Sperner property,” Annals of Discrete Math. (1984), 397-408.

23

[14] T. Hiraguchi, “On the dimension of partially ordered sets,” Sci. Rep. Kanazawa U. 1 (1951), 77-94. [15] J. Kahn and M. Saks, “Balancing poset extensions,” Order 1 (1984), 113-126. [16] D. Kelly, “Removable pairs in dimension theory,” Order 1(1984), 217218. [17] D. Kelly, “On the dimension of partially ordered sets,” Discrete Math. 35 (1981), 135-156. [18] H. Kierstead, “Recursive ordered sets.” To appear. [19] H. Kierstead, “An effective version of Dilworth’s theorem,” Trans. Am. Math. SOC.268 (1981), 63-77. [20] H. Kierstead, G. McNulty and W. Trotter, “A Theory of recursive dimension for ordered sets,” Order 1 (1984), 67-82. [21] H. Kierstead, E. Szemerhdi and W. Trotter, “On coloring graphs with locally small chromatic number,” Com binatorica 4 (1984), 183-185. [22] H. Kierstead and W. Trotter, “An extremal problem in recursive combinatorics,” Congressus Numerantium 33, 145-153. [23] H. Kierstead and W. Trotter, “A Ramsey theoretic problem for ordered sets,” Discrete Math. To appear. [24] N. Linial, “The information theoretic bound is good for sorting,” SIAM J . Comput. 13 (1984), 795-801. [25] N. Linial, M. Saks and P. Schor, “Largest induced suborders satisfying the chain condition,” Order 2 (1985), 265-268. [26] J. NeSetfil and V. Rodl, “Combinatorid partitions of finite posets and lattices,” Algebra Universalis 19 (1984), 106-119. [27] J. Negetfil and V. Rodl, “On a Probabilistic Graph-Theoretic Method,” Proc. Am. Mat. SOC.72 (1978), 417-421.

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[28] 0.Pretzel, “A Non-Covering Graph of Girth Six,” Discrete Math. To appear. [29] I. Rabinovitch, “An Upper Bound on the Dimension of Intervals Orders,” J . Combin. Theory (A) (1978), 68-71. [30] V. Radl and W. Trotter, “An Improved bound on the Dimension of Interval Orders.” To appear. [31] M. Saks, “Balancing Extensions of Ordered Sets,” Order 2 (1985), 327-330.

[32] J. Schmerl, “Posets with Small Width and Large Jump Number,’’ Order. To appear. [33] W. Schnyder. Personal Communication.

[34] W. Trotter, “The Dimension of the Cartesian Product of Partial Orders,” Discrete Math. 5 3 (1985), 255-263. [35] W. Trotter, “Order Preserving Embeddings of Aographs,” Springer Verlag Lecture Notes in Math. 642 (1978), 572-579. [36] W.Trotter and J. Moore, “The Dimension of Planar Posets,” J . Combin. Theory (B) 2 2 (1977), 54-67.

Annals of Discrete Mathematics 41 (1989) 417-420 0 Elsevier Science Publishers B.V. (North-Holland)

A Proof of Kuratowski’s Theorem H. Tverberg Mathematical Institute University of Bergen Bergen, Norway

Dedicated to the memory of G. A . Dirac A new, short proof of the difficult half of Kuratowski’s theorem is presented, 1. This classical theorem, first published by Kuratowski in 1930 ([3]) has been proved many times. The first relatively simple proof was given in 1954 by Dirac and Schuster [l],and many other proofs have been found (cf. Thomassen’s recent paper [4]).See also a discussion of its history by Kennedy, Quintas and Syslo [2]. We state the theorem as follows:

Theorem. A graph is planar if and only if it contains no subdivision of KE; or K 3 , 3 . The purpose of the present paper is to present a new proof of the “if” part of the Kuratowski theorem. The proof is carried out by studying an imagined counterexample having a minimal number of vertices. 2. Let G be a non-planar graph containing no subdivision of K5 or K3,3 and

having as few vertices as possible. We can clearly assume G to be simple, connected and without cutpoints. After adding edges we may also assume that G contains a subgraph H , which is either a subdivided K; (K5 minus one edge) or a subdivided Kq3, and that G is edge-minimal under that assumption. We now decompdse G into H and the bridges relative to R. Say that two edges not in H are equivalent if they have a common vertex outside H . This generates an equivalence relation on these edges, and the subgraph formed by the edges in one equivalence class is a bridge. Let first We shall see that H can be assumed t o be a subdivided H be a K ; (undivided) and B a bridge (which must exist, as H is planar, and meets at least two vertices, P and Q,of H , as G is without cutpoints). 417

418

H. Tverberg

P and Q are connected by a path in B, and so cannot be those two vertices of H which are non-adjacent (as G contains no subdivided Ks). With P and Q being adjacent in H, simplicity forces the existence of a third vertex in the path. Thus, upon replacing the edge PQ by the path PQ,we may assume H to be a properly subdivided I<:, The reader will now finish this argument by verifying that each of the two subdivided K c ’ s on 6 vertices contains a unique K i 3 . From here on H is a subdivided K i 3 . In view of the equivalence between spherical and planar embedding, we shall attempt embedding on the boundary of a tetrahedron T with vertices P,Q,R, S. We start by embedding H in the l-skeleton of T,with H having, besides P,Q,R, S,also at least the vertices h(P t Q)and !j(Rt S). Consider a bridge B of G. It cannot contain both of the vertices +(Pt Q ) and !j(R S),as G would then contain a subdivided K 3 , 3 . But each of i ( P + Q ) and i ( R + S ) must be at least 3-valent in G, by vertex-minimality, and so there must be another bridge B1. The edge-minimality of G then shows that the graph H U B is planar. We conclude that the vertices of attachment of B to H are either contained in an edge of T (then B is called a small bridge) or in the union of two or three of the boundary edges of a boundary triangle of T (then B is large). We now assume that H has been chosen so that there is a large bridge Bo (if possible), and so that Bo has a maximal number of edges. The problem of embedding only the large bridges splits into four completely independent subproblems, one for each face of T. If it wits unsolvable for, say, the face PQS,it would still be unsolvable if some bridge containing i ( R S) were removed, and edge-minimality would again be contradicted. We embed the large bridges and note that the problem of embedding the small ones also splits into six subproblems, one for each edge. Thus minimality shows that the small bridges are all attached to one edge, which cannot, for instance, be PQ,as again nonembeddability would persist on removal of all bridges containing i ( R t S). We may assume the troublesome edge t o be PS. It is now clear that there must be some (in fact at least two) large bridges. Consider Bo, as defined above. Bo cannot influence the embeddability of the small bridges. (For, if it did, there would be vertices X,Y, 2 along PS, with X and Z on some small bridge and Y ,between X and 2, on Bo. Replacing the path XZ of H by a path in the small bridge, from X to 2,we change H and add at least two edges to Bo, which conflicts with the choice of H.) Thus after removal of Bo, we would still have a counterexample, contradicting at least the edge-minimality of G.

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References [l] G. A. Dirac and S. Schuster, “A theorem of Kuratowski,” Indagationes Math. 16 (1954), 343-348.

[2] J. W. Kennedy, L. V. Quintas and M. M. Sys€o,“The theorem on planar graphs,” Historia Math. 12 (1985), 356-368. [3] K. Kuratowski, “Sur le problkme des courbes gauches en topologie,” Fund. Math. 15 (1930), 271-283. [4] C. Thomassen, “Kuratowski’s theorem,” J . Graph Theory 5 (1981), 225-241.

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Annals of Discrete Mathematics41 (1989) 421-436 0 Elsevier Science Publishers B.V. (North-Holland)

Finite and Infinite Graphs whose Spanning Trees are Pairwise Isomorphic P. D. Vestergaard Department of Mathematics and Computer Science Institute of Electronic Systems, Aalborg University Aalborg, Denmark

Dedicated t o the memory of G. A . Dirac A theorem characterizing graphs with only one isomorphism class of spanning trees was given in [2], [3], [5], [6], and [14]. Here, two new proofs of the theorem are given, one of which has the merit that it uses only the property that an isomorphism from one tree to another maps a path of length d to a path of length d . The other proof is by induction, and it is interesting because it rapidly shows that it is sufficient to examine the unicyclic graphs with the property that every second vertex on the cycle has valency two. Two new analogues of the theorem are stated: one with labelled vertices (given in the induction proof) and one for oriented graphs. A counterexample shows that the characterization does not hold for infinite graphs, but a conjecture is stated.

1

History

In 1970, Klaus Wagner ([13], p. 50) posed the problem of characterizing the connected graphs in which any two spanning trees are isomorphic. In 1971, Bohdan Zelinka as the first published a solution obtained by considering invariants of a tree. The article ([14])is in Czech with an ultra-short summary in English. Zelinka refers to the problem as a conjecture by JiFi SedlAEek. In 1973, two different solutions appeared by Fischer and Friess. Fischer considered isomorphisms from one tree to another ([2]), and Friess considered invariants of a tree ([3]).

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Bert L. Hartnell solved the problem by an elegant method of labelling vertices ([5], [6])and also gave a solution t o the problem of graphs with two spanning trees up to isomorphism. The case of oriented graphs is treated in this article and in [8]. The problem of characterizing graphs with one or two isomorphism classes of spanning unicyclic graphs has been considered in [9] and [lo]. A survey is given in [ll]. As a further generalization, a characterization has very recently been given in [l] of the graphs having one isomorphism class of connected spanning subgraphs H with E ( H ) - V ( H )t 1 = r , r E N O .

2

Definitions and notation

Unicyclic graph

A graph G is called unicyclic if it is connected and contains exactly one circuit. A tree T with a distinguished vertex v E V ( T ) is called a rooted tree ( T ,v). By abuse of language, we may omit v and just write the rooted tree T , when it is clear which root v E V ( T ) ,we intend. Assume k 2 3; for i = 1, 2, . . . , k, let (Ti,vi) be a rooted tree. Any unicyclic graph G is of the form G = C(T1,T2,. , .,Tk); i.e., G is a graph consisting of k disjoint rooted trees (T;,vi) and the edges e; = (vi,vi+l), i = 1, 2, . . . , k. Indices are reduced modulo k, so vk+1 = v1 (see Figure 1). Two rooted trees, ( T , v ) and (T',v'), are said to be root-isomorphic, denoted T & T', if T and T' are isomorphic under an isomorphism taking v to v'. The height of a rooted tree (Ti,v;) is h(T;)= max,,,v(Z) d(vi,v). C e n t r a l element

For vertices u,v in a graph, d ( u , v ) denotes the length of a shortest path with end-vertices u and v. A central vertex of a tree T is a vertex q with d ( q , v ) = min,€V(T)[maxv€V(T)4%41. the property that rna-&J€V(T) It is well-known that any tree has either just one central vertex or two central vertices joined by an edge ([4], p. 35, and [7], p. 63). The central element of a tree denotes the unique central vertex or the unique edge joining the two central vertices. If the central element of T is q = ( u , v ) and s E V ( T )then we define d ( q , s ) = min{d(u,s),d(v,z)}.

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Figure 1. The central element of a tree T is identical t o the central element of any longest path in T. The central element as well as the intersection of all longest paths of a tree are isomorphism invariants.

Labelled vertices Let G be the unicyclic graph G = C(T!, T2,.. . ,Tk). Let each vertex of G be labelled with an n-tuple. Two labelled graphs, GI and G", are said to be label-true isomorphic, denoted GI N t G", if they are isomorphic and corresponding vertices have the same label. If two rooted trees, (Ti,v;) and (Ti, vi), are root-isomorphic as well as label-true isomorphic then we write Ti zt Tj.

Deletion of leaves For a tree T (unicyclic graph G) and a set of vertices X C V ( T )(a set of vertices X C V ( G ) ) ,let T ( X ) (respectively G ( X ) ) denote the minimal subtree of T (the minimal unicyclic subgraph of G) containing X , and let XI denote the end-vertices of T ( X ) (respectively G ( X ) ) ,

X' = { z E

x I dT(X)(4= 1

(dG(X)(X)= 1)

1.

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The graph D x T (the graph DxG) obtained by deleting leaves on X from T (respectively G ) is produced in the following way: for each x' E X ' , delete S - x' from T (respectively G ) , where S satisfies (i) S is a subtree of T (respectively G ) rooted a t x', (ii) V ( S )n T ( X )= {z'} (V(S)n G ( X ) = {x'}), and (iii) S is maximal with respect to (i) and (ii). Oriented edges Let G' and GI' be oriented graphs. We write GI 3 G" if there is an isomorphism 4 between G' and G" preserving both incidence and orientation. Furthermore, if 4 takes a distinguished vertex v' E V ( G ' )to a distinguished vertex v" E V(G") then we write (G', v') 2 (G",v"), or just G' 5 GI' when it is clear which distinguished vertices are intended. A circuit C = {v1,v2,. ..,oh} is said to have monotone orientation if its edges are oriented ( v x ) , i = 1, 2, . , , k; C is said to have alternate c-orientation if ( v x ) and ( v i , v i + ~ )implies (vi+l, vi+2) and ( ~ i + l , ~ i + p ) , respectively.

.

-

____).

Each of the next three sections will present characterizations of graphs with only one spanning tree modulo different isomorphisms. First, we present the classical theorem together with a new proof. Another new proof will appear as a special case from the theorem in Section 4.

3

The classical version

Theorem (Classical version). Let G be Q connected graph. ( u ) Any two spanning trees of G are isomorphic

is a tree, or ( c ) for some k 2 3, G = C(T1,T2,. . . ,Tk) with Ti & T;+2 f o r i = 1, 2, ..., k. Note. For odd k,

( c ) implies

T; & Ti+l for i = 1, 2, . . . , k.

To prove ( a ) + ( c ) for a unicyclic graph G = C(T1,T2,. . . ,Tk), we state a lemma about the circuit C = { q , v2,. . . ,vk}.

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Lemma. Let G = C(T1,T2,. . . ,Tk) satisfy ( a ) . If II is a longest path in G then either (i) II contains all edges of C but one, or

(ii) II contains all edges of C but two. If E ( C ) \ E ( l I ) = ( e , e'} then e and el have a common vertex v E V(C) and the central element q of G - e (as well as q' of G - el) is on C diametrically opposite to v. Proof of Lemma. The edge-set E ( C ) \ E(II) is all of C or a path. If IE(C)\E(IE)I 2 2 then let e; and e;+l be two adjacent edges of E(C)\E(II). The central element q of II is also the central element of the trees G - ei and G-ei+l, respectively, because II is a longest path of G-e;+s, 6 = 0 , l . By hypothesis ( a ) , there is an isomorphism q5 from G - ei onto G - e;+l; +(q) = q, because q5 maps the central element q of G - ei onto the central element q of G - ei+l. Therefore, the two sets of numbers,

{ dG-ei(Q, V ) I V E V(G) I and { dG-ei+l (Q,v)

I

E V(G)1,

are identical. The tree T;+l is rooted at vi+l E V ( C ) , the common neighbour of e; and e;+l. For all v in V(G)\V(Ti+l),we have dG-ei(q, v) = dG-eitl (q, v), and for some constant a, we have for all v in V(Ti+l)that dG-ei(q,v) = dG-ei+l(q,v) a. Therefore a = 0, and in particular dG-ei(q,vi+l) = dG-e,+l ( q , v;+l), which implies that q is either on C diametrically opposite to v;+l or q is in a tree Tj attached to C diametrically opposite to v;+l. If E(C)\E(rI) contained another pair of neighbouring edges, they would point out a different position for q. Hence, IE(C)\E(II)l 2 2 and q belongs 0 to C. The lemma is proven.

+

Proof of Theorem. It is immediate that ( b ) or ( c ) implies ( a ) . Conversely, [l]and [2] show by almost identical arguments that if G is not a tree then ( a ) implies that G is unicyclic. We shall therefore restrict ourselves to showing that for a unicyclic graph G = C(T1,T2,. . . , T k ) , ( a ) implies (c).

CASE 1. Suppose any longest path of G contains k - 1 edges from C. A longest path of G - e; is also a longest path of G and hence must contain k - 1 edges from C. PROPOSITION 1. All the trees TI,T2,. . . , Tk have the same height. PROOFO F PROPOSITION 1. If not, two neighbouring trees Ti and T;+1 will have different heights, say h(T;) > h(Ti+1). A longest path of G - e;

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by hypothesis contains E ( C ) - e; and has length h ( Z ) k - 1 h(Ti+i). A path starting in T; and ending in T;+2 is shorter; therefore, we have h(Ti+2) 5 h(T;+l). A longest path of G - e;+l contains E ( C ) - ei+l and has length h ( Z + l )t k - 1 h(Ti+2) < h(T;)t k - 1t h(Ti+l),which is a 0 contradiction. Proposition 1 is proven.

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CASE1 (CONTINUED). Until the end of the proof of the Theorem, i will be an arbitrary but fixed number, 1 5 i 5 k. Case l a . Suppose k is odd. Proposition 1 implies that wi is the central element of the tree G - e i + F . The intersection of all longest paths in G - e i + U will contain E ( C ) - e i + y . Because central element, intersection of longest paths and distance are isomorphism invariants, the isomorphism 4, which by hypothesis exists from G - e . k - 1 to G - e i + l + q , $+-2will take wi to v;+l and E(C)- e i + y to E ( C ) - e i + l + y . This implies 4(Z) = Z+l, and we have T; & T;+1. Together with the corresponding result for i 1, we have T; & Ti+2, which is (c). Case I b . Suppose k is even; then ei is the central element of G - e;+g, and the intersection of all longest paths of G - ei+i contains E ( C ) - ei+;. The isomorphism q5 from G - ei+i to G - ei+*+4 takes ej to e;+l. If +(wi) = w;+l and $(wi+l) = w;+2 then we deduce T; & Ti+l and Ti+l 4 Ti+2; if 4(w;) = w;+2 and 4(w;+l) = w i + l then we deduce Ti & Ti+2. Hence, for k even we have the claim of ( c ) : Ti &Ti+2. i CASE2 . Suppose G contains a longest path 11 with exactly k - 2 edges from C.

+

PROPOSITION 2 . The number of trees, k , is even, and all trees Tj with index j of one parity (say odd) have the same height h, and all trees Tj with index of the .other parity (even) have height less than h.

PROOF OF PROPOSITION 2. Suppose, if necessary by change of indices, that e1,e2 E E ( C ) \ E(II). We shall show that G also has longest path containing E ( C ) \ {e3, e4). By (ii) in the lemma, the central element of 11 is on C diametrically opposite to w 2 , and that implies h(T1) = h(T3) = h. Furthermore, h(T2) 5 h - 1 and h(T4) 5 h - 1, because otherwise G - el and G - e3, respectively, would have too long a path. If, in G - e4, any longest path must contain E(C)- e4 then h(T5) 2 h, because h(T4) 5 h - 1. On the other hand, h(T5) > h would produce too long a path beginning in T3 and ending in T5. If G - e4 has a longest path containing only k - 2 edges of C then the two edges missing are either {e4,e5} or {e3,e4}. In the first case,

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h(T6) = h(T4) 5 h - 1 and the path is too short; therefore { e 3 , e 4 } is the missing pair of edges, and we get h(T5) = h(T3) = h. This can be continued around C. If k were odd, we would get h(Tk) = h, and G - e k would have too long a path, so k is even, and h(Tj) = h for all odd j, and h(Tj) < h 0 for all even j . Proposition 2 is proven.

CASE2 (CONTINUED). Suppose, if necessary by change of indices, that G has a longest path not containing e l and e2; then v2+k is central element z for G - el as well as for G-e2. The isomorphism 4 from G- el t o G-e2 will preserve the two vertices belonging to the intersection of all longest paths and each having distance - 1 to the central element; i.e., 4 ( ( 0 1 , 0 3 } ) = {01,03}. As v1 has different valencies in G - el and G - e2, 4 ( ~=) 0 1 is impossible. It follows that TI & T3. By Proposition 2 , one sees that G will have a longest path not containing the next pair of edges e 3 , e 4 , so v4+; will be central element, and T3 & T5. By continuing around C ,we

4

get Tj & Tj+z for all odd j . The other class of isomorphisms, Tj & Tj+2 for all even j , is for k G 0 (mod 4) obtained by noting that in G- el the rooted tree attached at the central vertex to the intersection of all longest paths is ( T 2 + fv, 2 + t ) ,while the corresponding rooted tree in G - e3 is (T4+g,v4+f). This argument is repeated around C. Otherwise, k = 2 (mod 4), and we obtain TI+$ & T3+k by considering the isomorphism 4: G - el --t G - e2 that q = w 2 + f is the central element of G - e l and also of G - e 2 . In G - el as well as in G - e2, 01 and 03 are characterized as the two vertices belonging to the intersection of all longest paths and having distance - 1 to q. The two vertices are (01) distinguishable from one another by their valencies, &'-el (03) > and dG-e2(2)1) > d ~ - ~ ~ ( vIn3 )G. - e l , we find V l + k at distance 1 from q on ?i the qv3-section of the intersection of all longest paths. The corresponding vertex in G - e2 is v 3 + k . Using 4 ( 0 3 ) = 01, ~ ( w I )= 03, 4(q) = q , we 5 get 4("1+$ = 03+g and 4(T,+;) = T3+4' Also this argument is repeated around C,and Tj & Tj+2 is now shown for all even j. This proves the 0 theorem.

4

The labelled version

Theorem (Labelled version). IfG-e; then Ti 6it Ti+2 for i = 1, 2, . . . , k.

N~

G-ej f o r a l l i , j , 1 5 i , j 5 k ,

Proof. Let us first show by induction on IV(G)l that it is sufficient to

P.D.Vestergaard

428

prove the theorem for graphs G of type (*): k is even, and ~ G ( w ;=) 2 for all indices i of one parity (say even), and ~ G ( w ;= ) d, d > 2, for all indices i of the other parity (odd). This part of the proof is due t o Lars Dprvling Andersen. The theorem is obviously true if G is a circuit. Suppose G is not a circuit, and suppose the theorem holds for graphs with fewer than IV(G)l vertices. Let €(G) denote the set of end-vertices of G:

€(G) = { TI E V ( G )I ~ G ( v =) }. For all i , j , we have that G - e; N~ G - e j implies that G - ej - € ( G - ei) ~t G - e j - €(G - e j ) .

CASEA. Suppose &(a;) > 2, i = 1, 2, . .. , k; then G - ei - € ( G - e i ) = G - E(G) - e; for all i, so that H = G - €(G) has fewer vertices than G; furthermore, H is unicyclic, H = C(T1 - € ( G ) ,T2 - €(G), . . . ,Tk - €(G)), H - e; N H - e j for all i , j , and, by the induction hypothesis, we obtain T; - €(G) k t Ti+2 - €(G) for all i. The remaining part to show T, G t Ti+2 is done by sticking labelled vertices correctly back on T; and T;+2: Each vertex v of the graph H is furnished with an extra ( n -t 1)-st coordinate in its label telling both how many end-vertices of G have been deleted at w and also what their n-tuple labels were. Using the ( n t 1)-st coordinate of the label of each vertex of T; - €(G), we can reconstruct Ti with n-coordinate labels, and for a l l i = 1 , 2 , ..., k,weobtainfrornT;-€(G)ktT;+2-€(G) thatTicStT;+2. CASEB. Suppose there exists an i with d ~ ( v ; = ) 2. For each i, 1 5 i 5 b , we count the number of edges of the line graph L(G)of G minus the number of edges of the line graph L ( G - ej): IE(L(G))I- I E ( L ( G - e i ) ) l equals the valency in L(G) of the vertex corresponding to e; E E ( G ) ,and this number is d ~ ( v ;t) dG(W;+l) - 2. For i = 1,2, . . . , k, this number is a constant d , so on the circuit every second vertex has valency 2, and every second vertex has valency d. If d = 2 then G is a circuit, which we have supposed is not the case; hence, d > 2 and G is of type (*). To finish the proof for graphs of type (*), we apply deletion of leaves.

PROPOSITION 3. For a graph G of type (*) satisfying the hypothesis of the theorem we have with X = { v E V ( G )I dc(w)# l , d } ; l X # 0 for all odd i, Either (Bl) V ( T ; )r or (B2) V ( T ; )n X = 0 for d l odd i.

PROOFOF PROPOSITION 3. Suppose, if necessary by change of indices, that V(T1)n X # 0 and V(T3)n X = 0. Let Xi = { z E V ( G - e;) I

Graphs whose Spanning Trees are Pairwise Isomorphic

429

# l , d }, i = 1, 2. In G - e;, delete leaves on X i , i = 1, 2. But the two resulting graphs do not have the same number of vertices in 0 contradiction to G - el N G - e 2 , so Proposition 3 is proven. dG-e,(Z)

CASEB (CONTINUED). Using the notation

we reason as follows. If ( B l ) holds for G, let H = DxG. Extend n-tuple labels of 21 E V ( H ) with an ( n + 1)-st coordinate describing both structure and labels of what was possibly deleted at 21. The labelled graph H has IV(H)I < IV(G)l by ( B l ) , and H = DxG is unicyclic and satisfies the induction hypothesis H - e; czt H - e j for all i , j , 1 5 i , j 5 k , because H - e; = Dx,(G - e ; ) for all i, 1 5 i 5 k . We conclude that the tree rooted at 21; in H is &t-isomorphic to the tree rooted at 0;+2 for all i, 1 5 i 5 k . Using the ( n + 1)-st coordinate to reconstruct G from H , we obtain Ti &t Ti+2 for all i, 1 5 i 5 k. If (B2) holds for G then for each odd i, all vertices in Ti have valency 1 or d. We have d c - e l ( v l ) = d - 1 and d ~ - ~ ~ =( 2. ~ k ) If (B2) holds and d > 3 then 211 is the only vertex in G - el with valency d - 1, and we can recognize e k = (vk,v l ) and hence also ( T I ,q ) , which is &t-isomorphic to the corresponding rooted tree (T3,213) in G - e3. Continuing around C , we get T; &t Ti+2 for all odd i. If (B2) holds and d = 3 then e k is the only edge in G - el with both end-vertices having valency 2 in G- e l . For k > 4,we can recognize ( T I ,211) as the subtree of (G - e l ) - e k with no vertex of valency 2, and comparing with G - e3, we get (T1,vl) &t (T3,vg). Continuing around C , we get Ti &t Z+2 for all odd i. If (B2) holds, d = 3 , and k = 4, then let 4 be a zt-isomorphism from G - el t o G - e2. In G - e l , only 211 and 214 have valency 2, and in G - e3, only 213 and 214 have valency 2; +(.I) = 212 would imply that T I ,a connected component of (G - e l ) - e4, should correspond to the connected component of (G - e3) - e2 containing T I , 212, and 214; hence, + ( q = ) 213 and TI &t T3. The theorem is proven.

By giving label 0 to all vertices of G, we obtain the unlabelled version of the theorem.

P. D. Vestergaard

430

5

The oriented version

For a connected oriented graph G we have the following theorem.

Theorem (Oriented version). Let G be a connected oriented graph. ( a ) Any two spanning trees of G are %-isomorphic

( b ) G is a tree, or ( c ) G = C(Tl,Tz,.. ,Tk) with T; 5 T;+zfor i = 1, 2, . . . , k, and the orientation of C is (i) monotone, i f k is odd, (ii) alternate, i f k is even and there exists a n i , 1 5 i 5 k, such that T; $+T;+l, (iii) either monotone or alternate, if k is even and Ti % Ti+i for i = 1, 2, ..., k.

.

Proof. That ( b ) or ( c ) implies ( c ) is easy; 3-isomorphism implies ordinary =-isomorphism, so conversely, by the first theorem, ( a ) implies ( b ) or that G is a unicyclic graph G = C(T1,Tz,.. . , T k ) with T; & T;+z. We have Ti % Ti+2, because some %-isomorphism G - el ---t G - e" will take T; t o T;+2. (i) If k is odd then C has monotone orientation, because each vertex v; will appear as central element of a tree, namely G - e i + y , and the orientations of ei-1 and e; must correspond to the orientations of e; and e;+l for all i = 1, 2, . . , k.

.

(ii) If k is even and the central element of a spanning tree of G is a vertex then C has alternate orientation, because for all i, the oriented path v. k , vl+i+p v2+;+; of G - e; is %-isomorphic to the oriented path '+ I v ~ + ~ +wl+;+4, !, vi+4 of G - e;+l by the argument of Case 2 after the proof of Proposition 2. If Ic is even and the central element of a spanning tree of G is an edge and, further, Ti $$ T;+1 then C has alternate orientation, because Ti ;% Ti+l for one i implies Ti $$ Ti+l for all i, and, as before, the oriented path vi+4, vl+;+ 4, v2+;+ k of G - ei is %-isomorphic to the oriented path v2+;+;, v ~ + ~ +vi+k : , of G - e;+l. ~

2

(iii) If k is even and T; 5 T;+1 for all i , 1 5 i 5 k, then C must have monotone or alternate orientation; this is seen by considering

Graphs whose Spanning Trees are Pairwise Isomorphic

43 1

the central edge and its two neighbouring edges on C. That both cases can occur is illustrated by taking G to be a C4 with monotone orientation or a C, with alternate orientation. The Theorem is proven.

6

0

The infinite case

The Theorem is not true for infinite graphs as demonstrated by Figure 2.

-

41

a

-

41

0

-

0

A

*I

G

Figure 2.

A. E. Brouwer has in generalization of Figure 2 produced an example of a family of graphs G, where G is a connected, infinite, locally finite graph whose spanning trees are all pairwise isomorphic, but where no reasonable characterization can be given for a graph G in the family.

P,D. Vestergaard

432

Let K be a connected, locally finite graph (it is to play the role of the triangle in Figure 2), and let v E V ( K ) be a distinguished vertex. In K , consider all spanning trees and choose as root for each of them. Partition the v-rooted trees of K: into 1 1 root-isomorphism classes: Ao, A1, . . . , A { , 1 2 1. By abuse of language, we may also denote a tree from A; by Ai. Let 5'1 = {21,52 ,...,XI} and S,' = {x;',~;', ...,21'). Then let H denote the free group generated by 5'1; i.e., H consists of 1 and words like for example x ; x ~ ~ z ~ x ~composed z ; ~ x ~by juxtaposition of symbols from Sl and 5';'; apart from xr1xi = z p f ' = 1 no simplifications are possible. Define the graph 3.1 by

+

V(3-I)= H

I

E(3-I)= { ( h , k ) h , k E H and h-lk € 5'1 U SF1} This graph is a tree, where each vertex has valency 21. For any fixed j, 1 5 j 5 I, h + zjh defines an automorphism of 3-I.

The idea The idea is to form G by disjoint union of 7-1 and K: with 1 and v identified; G will have 1 1 spanning trees, namely 7-1 enlarged with a spanning tree from A; attached to 3.1 at v , i = 0, 1, . . . , 1. The idea is to make all those spanning trees of G isomorphic by making it insignificant from which Ai a tree is attached to 7-1 a t v. This is obtained by attaching a tree of some Ai to each vertex of 7-1 making G highly symmetric. With 1 = 1, we obtain S1 = { x } , S,' = {x-'}, H = { x n I n E Z } as illustrated in Figure 3.

+

The details For brevity, we shall say that we give an h E V(31)the label j, if we to 'FI attach a copy of A j with the root v identified with h, where 3.1 and the copy of Aj are disjoint apart from v = h. Now give h E V(3.1)labels as follows. 0

If h ends in

xj

0

if h ends in

2 ; '

0

if h is the unit element 1 then h is not labelled.

then h is labelled j, then h is labelled 0, and

Graphs whose Spanning Trees are Pairwise Isomorphic

q

x3 label 1

u x2

label 1

(1

z label 1

41

1

4)

2-l

4

x - 2 label 0

label 0

The graph 7-i

The graph K: All w-rooted spanning trees of K: with distinguished vertex w

K:

w

The graph G

Figure 3.

433

P. D. Vestergaard

434

..

For j = 0, 1, . ,I, let ‘ H j denote the graph 3-1 with label j at 1, label i at words hzi ending in x i , label 0 at words hsr’ ending in a symbol from S-I. For fixed j, the map h -+ sjh is an automorphism from ‘Hj t o ‘Ho, which preserves labels. Let G be the disjoint union of 9-f and K with 1 E V(9-f) and E V ( K )identified. The graphs ‘Ho, ‘HI, . . , ‘Hi are all the possible spanning trees of G. They are all isomorphic because h + s j h takes ‘h!j to ‘Ho. The graph G may have any structure in the K-part, but all spanning trees of G are isomorphic. All vertices in the ‘H-part’of G are cut-vertices. Carsten Thomassen has the following conjecture.

.

Conjecture. A n infinite connected graph with no cut-vertex has two spanning trees which are not isomorphic to one another.

Acknowledgement I wish to thank Lars Dsvling Andersen, Andries E. Brouwer, Oliver Pretzel, and Horst Sachs for valuable suggestions.

References [l] P. Duchet, Z. Tuza and P. D. Vestergaard, “Graphs in which all spanning subgraphs with r edges less are isomorphic.” In preparation. [2] R. Fischer, “Uber Graphen mit isomorphen Geriisten,” Monatsh. f Math. 77 (1973), 24-30. [3] L. Friess, “Graphen, worin je zwei Geriiste isomorph sind,” Math. Ann. 204 (1973), 65-71.

[4]F. Harary, Graph Theory, Addison-Wesley (1969). [5] B. L. Hartnell, “On graphs with exactly two isomorphism classes of spanning trees,” Utilitas Mathematica 6 (1974), 121-137. [6] B. L. Hartnell, The Characterization of Those Graphs whose spanning Trees can be partitioned into Two Isomorphism Classes, Ph.D. Thesis, Faculty of Mathematics, University of Waterloo (1974). [7] D. Konig, “Theorie der endlichen und unendlichen Graphen,” Leipzig (1936). [8] P. D. Vestergaard, “Oriented graphs with two isomorphism classes of spanning trees.” In preparation.

Graphs whose Spanning Trees are Pairwise Isomorphic

435

[9] P. D. Vestergaard, “Graphs with one isomorphism class of spanning unicyclic graphs,’’ Discrete Math., to appear.

[lo] P. D. Vestergaard, “On graphs with two isomorphism classes of spanning unicyclic subgraphs.” In preparation.

[ll] P. D. Vestergaard, “On graphs with prescribed spanning subgraphs,” Ars Combinatorica, to appear. [12] P. D. Vestergaard, “Graphs whose one-edge deletions partition into two isomorphism classes.” In preparation.

[131 K. Wagner, Graphentheorie, Bibliographisches Institut, Mannheirn (1970). [14] B. Zelinka, “Grafu, jejichi vSechny kostry jsou spolu isomorfni,” Cas. pro PBst. Mat. 06 (1971), 33-40. Title translated: “Graphs, all of whose spanning trees are isomorphic to each other.”

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Annals of Discrete Mathematics 4 1 (1989) 437-452 0 Elsevier Science Publishers B.V. (North-Holland)

Bridges of Longest Circuits Applied to the Circumference of Regular Graphs H.-J. VOSS Section of Mat hema t ics College of Education in Dresden Dresden, GDR

Dedicated to the memory of G. A . Dirac Some results about bridges of longest circuits of 2-connected graphs are surveyed, and as an application, it is proved that a 2-connected r-regular graph, r 2 3, different from the Petersen graph, is Hamiltonian or has a circuit of length at least 2r + 4. This has been proved independently by R.-Y. Chen (for T 2 4) and also follows from work by G. Fan ( r 2 4).

1

Introduction

Undirected simple graphs with connectivity 2 2 are considered. The concept “bridge” was introduced by 0. Ore [lo] and W. T. Tutte [15]. Let H be a subgraph of a graph G. Each component K of G - H , together with all vertices of G adjacent with K and all edges of G incident with K , forms a regular bridge of H in G. An edge not in H linking two vertices of H is also called a singular bridge or a chord. The concept “bridge” was very successfully used in investigating planar graphs. It is applied in proofs of Kuratowski’s theorem (see C. Thomassen [13]) and in proofs of Tutte’s theorem: “Each 4-vertex-connected planar graph has a Hamiltonian circuit” (see W. T. Tutte [16] and C. Thomassen [14]). H. Sachs initiated the concept “bridge” to be used not only in investigating planar but also non-planar graphs. Results are obtained in the following directions (see [27]): 1. Nonseparating circuits ([12]). 437

H.-J. VOSS

438

2. The “length” of bridges of longest circuits in 2-connected graphs ([221)-

3. The circumference of bridges of longest circuits in 2-connected graphs ([221)4. The class of all 2-connected graphs with given circumference. Applications of the results to critical graphs ([28], [22]). 5. Chords of circuits in graphs with given minimum degree or with given girth (POI, ~ 5 1~, 4 1 ~, 4 1 ) . 6. Bridges of longest circuits in graphs with given minimum degree. Extensions of theorems of G. A. Dirac and 0. Ore ([17], [IS]). 7. Cubic 2-connected graphs with given circumference having maximal order ([23], [26]).

A survey of the results with respect to 2., 3., 4. and 6. was published in [18]. Here we shall deal with problems 2. and 6.! Some notation is given in $2. In $53-4 the obtained results will be stated, in $5 a theorem on long circuits in regular graphs will be proved.

2

Notation

Let B denote a bridge of a subgraph H in a graph G. The vertices of Bn H are called attaching vertices of 17, the subgraph of 0 induced by the nonattaching vertices of B (the socalled inner vertices of 0)is called the kernel off?. Obviously, the kernel of a bridge is connected. T is called an A-tree of B, if T is a tree contained in B with the property: The end vertices of T are precisely the vertices of attachment of B. We say two distinct bridges B and B‘ of a circuit L overlap if B and B‘ have the same set of precisely three attaching vertices or B and B’ have two attaching vertices P, Q E B and P’, Q’E B’, respectively, such that all four vertices are distinct and P, P‘,Q, Q‘ appear on L in this cyclic order. In the latter case B and B’ are also said to be skew. In order to describe the mutual positions of bridges of a circuit L in a graph G the overlap graph O(G : L) is introduced: The vertices of O(G : L) are the regular and singular bridges of L in G; two vertices of O(G : L) are linked by an edge iff the corresponding bridges overlap. The circumference c(G) of a graph G is the length of the longest circuits of G, where c(G) = 0 if G does not contain a circuit.

Bridges of Longest Circuits

3

439

The “length” of bridges of longest circuits in 2-connected graphs

Let B denote a bridge of a subgraph H in G with a t least 2 vertices of attachment. A path is said t o be an H,H-path iff precisely its end vertices are in H. If T is an A-tree of B with the largest number X(B) of edges among all A-trees of B then A(B) is called the length off?. A second measure x(B) of B is defined as follows: Let Al,Az,. . , A , , r 2 2, denote the vertices of attachment of B and p;j the length of the longest H , H-paths of B connecting A; and A j . Then we define

.

where the minimum is taken over all permutations il, io, . ..,i, of 1,2,...,r. Obviously, X(B) 5 x(B),and X(B) = A(B) if B is a tree. In [22] I investigated the length of bridges of longest circuits (see also [IS]). A generalized version of Theorem 1 in [18]is the following one (proved in [27]):

Theorem 3.1. Let L be Q longest circuit of Q 2-connected graph G. Let 231,. .. ,B, be s pairwise non-intersecting bridges of L in G, 3 2 s 2 1, such that the subgmph of the overlap graph O(G : L ) induced by B1, ...,Bs is Q tree. Then

i=l

Figure 1.

H.-J. VOSS

440

Figure 2. The graphs depicted in Figures 1 and 2 show that Theorem 3.1 is not true if the subgraph of O ( G : L) induced by 131,. f?, is disconnected or has circuits, respectively. There are various applications of Theorem 3.1 (see [22], [27]), one of them is the following theorem ([22]).

..

Theorem 3.2. Let B be any bridge of a longest circuit L of a 2-connected graph G with s vertices of attachment. Then

This generalizes the result c(B) < c ( G ) due to K. Hauschild, H. Herre and W. Rautenberg [7].

4

Bridges of longest circuits in graphs with given minimum degree

The order and the minimum degree of a graph G are denoted by n ( G ) and 6 ( G ) , respectively. We say that a graph G has property E T iff G is 2-connected and has 6(G) 2 r 2 3 and n ( G ) 2 2r 1. We start with the well-known theorem of G. A. Dirac [2] proved in 1952:

+

Theorem 4.1. 1’ the graph G has Er then c(G) 2 2r. In 1967 0. Ore [ll] determined all graphs with ETand c ( G ) = 2r; in 1976 I [17] found all graphs with Er and c ( G ) = 2r 1. My result was obtained by determining the structure of the bridges of longest circuits in

+

Bridges of Longest Circuits

441

+

graphs with property E, and circumference 2r 5 c(G) 5 2r 1. The same was done for graphs with E , having only odd (or even) circuits of lengths 5 2r ([17]). Also the structure of bridges of longest circuits through specified edges in 2-, 3- and 4-connected graphs with E, are determined [18]. In 1980 B. Jackson [8] proved: Theorem 4.2. If the 2-connected r-regular graph G has order n(G) 5 3r then G contains a Hamiltonian circuit.

+

+

+

+

For orders n(G) E (2r 1,2r 2) and n(G) E (2r 3,2r 4) this was proved earlier by L. Gordon (see [6]) and P. Erdos and A. M. Hobbs [3], respectively. In [4], it was proved for n(G) 5 2r c . fi,and B. Bollobds and A. M. Hobbs [l]proved it for n(G) 5 $r. Later, Y. Zhu, Z. Liu and Z. Yu [29] improved Theorem 4.2 to graphs G with n(G) 5 3r 1, the Petersen graph being the sole exception, and in [30] they proved it for n(G) 5 3r 3 if r 2 6. This was based on a recent result by G. Fan [5] saying that c(G) 2 min(n(G),3r) if either (a) G is r-regular and 3-connected, or (b) G is r-regular, 2-connected and has n(G) 5 3r 3. Condition (a) was also proved independently by H. A. Jung (see note added in proof in [5]). In [18], I proved:

+

+

+

+

(*) each 2-connected r-regular graph has a Hamiltonian circuit or contains a circuit of length 2 2r 2.

+

Erdos and Hobbs [4] gave examples of 2-connected r-regular graphs of order 3r t 5 and with c(G) = 2r 4. Answering an open question of [18, p. 2791 I shall prove here:

+

Theorem 4.3. Each 2-connected r-regular graph, r 2 3, has a Hamiltonian circuit or contains a circuit of length 2 2r 4 or is the Petersen graph.

+

In $5 this theorem will be proved. It also seems to be possible to characterize all 2-connected r-regular graphs with circumference 2r 4 having no Hamiltonian circuit. Since the writing of this paper, the paper by Fan [5] has been brought to my attention. Here it is mentioned that for r 2 4, the result of Theorem 4.3 has been proved independently by R.-Y. Chen and also follows from work in [5].

+

H . J . VOSS

442

5

Proof of Theorem 4.3

First we state a lemma which can easily be derived from Lemma 1 in [17].

Lemma 6.1. Let G be a 2-connected graph. Let C and Il be a circuit and a path of G, respectively, with C n 1T = 0. Let the end vertices of II be X and Y and let X be linked with C by an edge. Let s neighbours of Y lie on Il, Then G contains a C,C-path of length 2 s 4-2. Next we prove:

Lemma 6.2. Let r 2 3 be an integer and let G be a 2-connected r-regular graph with circumference 1 5 2r 3. Then each bridge of a longest circuit has at most one inner vertex.

+

For n(G) 5 3r-2, Lemma 5.2 is a consequence of a result by C. St. J. A. Nash-Williams [9]. Proof. The proof is by contradiction. Assume that Lemma 5.2 is not true. Then there is a graph G satisfying the hypotheses of the lemma which contains a longest circuit L and a bridge 13 of L with 2 2 inner vertices. Let K be the kernel of B. For each vertex A of K the number of A , L-edges is denoted by a ( A ) . Now let X be a vertex of I< with maximal a ( X ) . An X-path is a path of K with end vertex X. Let Y denote the second end vertex of a longest X-path II of length d (clearly, d 2 1). Let Ai (1 5 i 5 q ) , Bj (1 5 j 5 m ) , Ck (1 5 k 5 t ) denote the vertices of L adjacent to both X and Y ,only to X, and only to Y ,respectively. In [17; p. 11111 I have proved 1

If t = 0 then a ( Y ) 5 max{ -, 1). d+2 1-4

If t > 0 then 1 5 d 5 1-1 5 r - 1and 2 51 - 1 f o r d = l , q > O

{

f o r d = 1 , q = 0 and 2 5 d

5 r - 1, q 2 0.

Next I state a simple assertion: All neighbours of a second end vertex of a longest X-path p are on p U L .

(2)

Now I shall prove: If t = 0 then d 2 T If t > 0 then d = r

- 2. - 1.

(3)

443

Bridges of Longest Circuits

Proofof (3): Case 1. t = 0. Suppose 1 I d

5 T - 3. By (1) we know a ( Y ) 5

v ( K :Y ) = r - a ( Y ) 2

T -

1 >Td+2-

&.This implies 2r + 3 d+2’

where v ( K : Y ) denotes the valency of Y in K . If r 2 7 then if r 2 d t 3 we have 2rt3 d+2

dr-3 7d-3 (d+3)d-3 > max{d+2’ d t 2 d+217 d-3 = max(7 - - d + - } > d . d+2’ d+2

v ( K : Y ) 2 r---

-

1

By (2), all neighbours of Y being in K are on II. Hence II has a length 2 v ( K : Y ) > d. This contradicts the definition of d (namely, d is the length of II). Consequently, d 2 r - 2. If ( r ,d ) = (6,2) then by the same arguments we again arrive a t a contradiction. If 6 2 T 2 3, ( r , d ) # ( 6 , 2 ) then B has the structure as depicted in Figures 3 - 7 , respectively. The length of some circuit segments is indicated.

Figure 3. ( r , d ) = ( 6 , 3 ) ; c(G) = 15.

Figure 4. ( r , d ) = (6,l);c(G) = 15.

H.-J. VOSS

444

. . . . . Figure 5. (r,d) = (5,2); 12 5 c(G) 5 13.

Figure 6.

(T,

d) = (5,l); 12 5 c(G) 5 13.

Figure 7. (T,d) = ( 4 , l ) ; 9 5 c(G) 5 11. In all cases by a careful examination it can be proved: There is another bridge disjoint from B and overlapping B. Applying Theorem 3.1 we obtain a contradiction unless ( ~ , d = ) (4,1), c(G) = 11 and any such bridge b has length 1. In this case it is easy to see that there are two such chords b' and bN such that both overlap B, b' and b" are disjoint and b' does not overlap b". Applying Theorem 3.1 we again arrive at a contradiction. Case 2. t

> 0.

From (1) we know that 1 5 d then (1) implies:

5 T - 1. If 2 5 d 5 T - 2 or d = 1, q = 0

v ( K : Y ) 2 T--(Y(Y)2

T -

1 - 2d 4

-*

Bridges of Longest Circuits Since

T

445

2 d + 2 and 1 5 2r + 3 we have W ( K :Y )2

T

-

2T

+ 3 - 2d >- 2 ( d + 2 ) + 2d - 3 > d . 4

4

As in Case 1 we arrive a t a contradiction! If d = 1, q > 0 then T 2 4 because in a 3-regular graph q = 0. The assertion (1) implies

This contradicts r - 1 >

for all r 2 4. Thus (3) has been proved.

Next we state two simple assertions: Let E be the second end vertex of a longest X-path p . Let 2 be a neighbour of E on p such that (2,E ) 6 p . Let be the neighbour of 2 such that (2, E p [ Z ,El. Then is the second end vertex of the longest X-path

z z

(4)

z>

(P - (z,z))u (4El. Let p be an L , l - p a t h with ends P and Q. Then the lengths of both P,Q-arcs of L are not smaller than the one of p.

(5)

By the r-regularity of G the assertions (1) and (3) imply:

For each end vertex E of a longest X-path p~ holds : a ( E ) 5 2 and v ( K : E ) 2 T - 2.

(6)

Since p~ is a longest X-path all neighbours of E in Ii lie on p ~ By . Lemma 5.1 the graph G contains an L,L-path of length 2 v ( K : E ) 2. By (5) we conclude w ( l i : E ) 5 T - 1. Consequently,

+

v ( K : E ) 5 T - 1and a ( E ) 2 1.

(7)

Put s = w ( K : Y ) . Since II is a longest X-path all neighbours 21,. . . ,Z, of Y in K lie on ll. By (4) the vertex with (Zi, E ll[Zi, Y ]is an end has vertex of a longest X-path. (7) implies: a ( x ) 2 1, i.e., each vertex a neighbour on L.

x)

x

H.-J. VOSS

446

If each of these vertices would be adjacent only t o the same neighbour

X‘ E L of X then v(G : X’) 2 s -+ 1+ 2 2 r - 2 t 3 = r t 1. This would contradict the r-regularity of G. Therefore, TI and Y can be chosen so that: There are two different vertices X’,Y‘ E L such that (X,X’), (Y,Y’) E L.

(8)

By (8)the path (X‘,X)UIIU(Y,Y’) is an L,L-path. By ( 5 ) it has a length 5 r 1. Therefore (see also (3)),

+

T

-2 5 d 5 T -

1.

(9)

According to (9) we consider two cases: Case 1. d = r - 2. Since v ( K : Y )5 d assertion (6) implies v ( K : Y )= r - 2 and a(Y)= 2. Hence Y is adjacent to each vertex of TI and by (4) each vertex E of II - X is an end vertex of a longest X-path which has the same vertex set as II. Therefore, v ( K : E ) = r - 2 and a ( E ) = 2. Hence E and X are adjacent to all vertices of TI - E or 11 - X, respectively. This and v(G : X ) = v(K : X ) f a(X) = T and a ( X ) 2 a(Y)= 2 imply a(X)= 2 and v ( K : X ) = r - 2. Consequently, K 2 K , - 1 . If I< had more than two neighbours on L we would have x(B)1 #T contradicting Theorem 3.1. Consequently, each vertex of K is joined to both X’and Y’by an edge. This implies v(G : X‘) 2 T 2 contradicting the r-regularity of G!

+

Case 2. d = r - 1. Put p = (X‘, X)U 11 U (Y, Y’).Then p is an L , L-path of length r 1. Since L has a length 5 2r 3 by ( 5 ) there is an X‘,Y’-arc L[X’,Y’] of L having length r 1. The circuit L has only bridges with a t most one inner vertex and bridges lP with two vertices of attachment linked in B’ by a path of length r t 1. No bridge overlaps p ; this follows from Theorem 3.1 except for bridges with precisely one inner vertex intersecting p a t X‘ or Y‘.For these it can be seen by looking at the longest circuit containing p. So no inner vertex of L[X’,Y’] belongs to a regular bridge (notice that in the case r = 3 by the 3-regularity of G the vertices X’ and Y‘ are adjacent to no bridge other than B). Since G is r-regular each vertex of L[X’, Y’1-{ X I , Y’}is incident with r - 2 chords. Consequently, each vertex of L[X’, Y’]- { X I , Y’}is linked t o at least one of the vertices X’,Y‘ by an edge. we obtain the same result for p . Exchanging the roles of p and L[X’,Y’] Thus X’ and Y’are linked to at least 2r vertices of (L[X‘, Y‘]-{X‘,Y’})up.

+

+

+

447

Bridges of Longest Circuits

+

+

Consequently, v(G : X ' ) v(G : Y')2 2r 2 contradicting the r-regularity of G. 0 This completes the proof of Lemma 5.2 Now we can prove Theorem 4.3.

Proof of Theorem 4.3. Let G be a 2-connected r-regular graph with circumference 2r 3,and assume that G is not Hamiltonian. We have to show that T = 3 and G is the Petersen graph. Let L denote a longest circuit of G. By Lemma 5.2 the circuit L has only bridges with at most one inner vertex. Since G has no Hamiltonian circuit the circuit L has a bridge f3 with precisely one inner vertex. We have t o consider the following three arrangements of the vertices of attachment A l , . . . ,A, on L. They are depicted in Figures 8-10. We shall determine the sum q of the degrees of the vertices A l , ... , A , in all three cases. Since G is r-regular q = r 2 . We assume, for convenience of notation, that c(G) = 2r + 3. The proof is the same if c(G) < 2r 3.

+

+

Figure 8.

L be ( A I , B ~ , A ~.., .B, B~,,- l , A , , ~ , V 2 , V 3 , V 4 , A l )(indicated in Figure 8). By elementary methods with Theorem 3.1 it can easily be shown: Case 1. Let

(i)

V1,V4 and all Bi do not belong to a regular bridge.

(ii) Bi is not adjacent to any of (V1,Vq) and is adjacent to no Bj, for all j # i. (iii) Bi is adjacent t o at most one of the vertices V2,V3.

448

H . J . voss

(iv) Either (V1,Vq) is not an edge of G, or no Bj is a neighbour of any of

{vz, v31.

.

. .. ..

The neighbours of {Vl, V4, B1,. . ,Br-l} belong t o {Al, ,A,, VZ,V3} except that V4 may also be a neighbour of V' and vice versa. Then (i) - (iii) imply T , ( T - 1) and ( T - 3) vertices of {Al, , , A r } are neighbours of the inner vertex of B, of Bi (for all 1 5 i 5 r - 1) and of Vj ( j = 1 and j = 4), respectively. Consequently, T2

=q

2 T t ( T - 1)(' - 1)-t 2(' - 3) + 2,

where the summand 2 is caused by the condition (iv). This implies T 5 3. A special investigation of the case T = 3 leads also to a contradiction!

Figure 9. Case 2. Let L be ( A l , 4,A2, . . ., & - I , A,, %, V2,A+i,& + I , . . . ,&--I, A,, V3, V4, Vg,Al), T > s 2 1 (see Figure 9). By elementary methods with Theorem 3.1 it can easily be shown:

(i)

Vl, Vz, V3, Vs and all Bi do not belong to a regular bridge.

(ii) B; is not adjacent to V1, V,, V3, V5 and is adjacent to no B j , for all j # i, j # s. (iii) The edges joining {Vl, VZ} with {V3,Vq,V5} form a subpath of length 5 2 of the path (V3,v2,V4,Vi,V5) ((V3,V5) may be an edge of G). The neighbours of

{K, v2, V3,V5, B1,. . .,B8-1,B,+1,. . . ,Br-l} belong

to { A l , . . .,Ar,Vl,b,V3,V4,V5}. Then (i) and (ii) imply T and ( T - 1) vertices of {Al, , , . , A r } are neighbours of the inner vertex of B and of Bi

Bridges of Longest Circuits

449

(for all 1 5 i 5 r - 1, i # s), respectively. By (iii), in {V',V2,V3,V5} start 4r - 9 edges with second end in { A l , . . . , A T } .Consequently, r2

= q 1 T t ( r - l)(r - 2) $- 4r - 9.

This implies T 5 3. A special investigation of the case r = 3 leads again to a contradiction!

Figure 10.

L be ( A I , & , . . . ,Bs-1,A,,Vl,V2,As+i,Bs+i,. . . , & - I , At, V3,V4,At+l,Bt+l,...,AT,V5,V6,A1),r > t > s 2 1 (see Figure 10). By Case 3. Let

elementary methods with Theorem 3.1 it can be shown: (i)

All Vj and all Bi do not belong to a regular bridge.

(ii) Bi is not adjacent to V1,. . . ,v 6 and is adjacent to no B j , for all j i, j # s, t.

#

(iii) There is a t most one edge joining {V2k-l,Qk} with (V2h-1,fih) for all 1 5 k < h 5 3.

...

Then (i) and (ii) imply that all vertices A l , A, are neighbours of the inner vertex of L3 and of B;, for all 1 5 i 5 r - 1, i # s , t . By (iii), in {Vl, .. ,Vs} start 6r - 12 edges with second ends in { A l , . . . , A T } . Consequently,

.

r2 = Q 2 T + T ( T - 3) + 6r - 12. This implies r 5 3. A special investigation of the case T = 3 leads to the conclusion that G is the Petersen graph. This finishes the proof.

450

H e - J . VOSS

References [l]B. Bollobb and A. M. Hobbs, “Hamilton cycles in regular graphs,” Advances in Graph Theory (B. BollobAs, ed.), North-Holland, Amsterdam (1978), 138-158. [2] G. A. Dirac, “Some theorems on abstract graphs,” Proc. London Math. SOC. (3) 2 (1952), 69-81. [3] P. Erdos and A. M. Hobbs, “A class of Hamiltonian regular graphs,” J . Graph Theory 2 (1978), 129-135. [4] P. Erdos and A. M. Hobbs, “Hamiltonian cycles in regular graphs of moderate degree,” J . Combin. Theory ( B ) 23 (1977), 139-142. [5] G. Fan, “Longest cycles in regular graphs,” J . Combin. Theory 39 (1985), 325-345. [6] L. Gordon, “Hamiltonian circuits in graphs with many edges.” Unpublished report, Sydney University, Australia.

[7] K . Hauschild, H. Herre and W. Rautenberg, “Interpretierbarkeit und Entscheidbarkeit in der Graphentheorie,” 11. 2. Math. Log. u. Grungl. Math. 18 (1972), 5; 457-480. [8] B. Jackson, “Hamilton cycles in regular 2-connected graphs,” J . Combin. Theory (B) 29 (1980), 2 7 4 6 . [9] C. St. J. A. Nash-Williams, “Edge-disjoint Hamiltonian circuits in graphs with vertices of Large Valency,” Studies in Pure Mathematics (L. Mirsky, ed.), Academic Press, New York (1971), 157-183.

[lo] 0. Ore, The four-color problem, Acad. Press, New York-London (1967). [ll] 0. Ore, “On a graph theorem by Dirac,” J . Combin. Theory 2 (1967), 383-392.

[12] C. Thomassen and B. Toft, “Non-separating induced cycles in graphs,” J . Combin. Theory (B) 31 (1981), 199-224. [13] C. Thomassen, “Kuratowski’s theorem,” J. Graph Theory 5 (19Sl), 225-241.

Bridges of Longest Circuits

45 1

[14] C. Thomassen, “A theorem on paths in planar graphs,” J . Graph Theory 7 (1983), 169-176. [15] W. T. Tutte, Connectivity in graphs, Univ. Toronto Press, Toronto (1966). [16] W. T. Tutte, “Bridges and Hamiltonian circuits in planar graphs,” Aequ. Math. 15 (1971), 1-33. [17] H.-J. Voss, “Maximal circuits and paths in graphs. Extreme cases,” Combinutorics, Colloq. Math. SOC. Jcinos Bolyai, 18, Keszthely, Hungary (1976), 1099-1122. [18] H.-J. Voss, “Bridges of longest circuits and of longest paths in graphs,” Beitrige zur Graphentheorie und deren Anwendungen, Vorgetragen Internat. Koll. Graphentheorie Oberhof (DDR) (1977), 275-286. [19] H.-J. Voss, “Graphs with prescribed maximal subgraphs and critical chromatic graphs.” Comment. Math. Univ. Carolinae 18 (1977) 1; 129-142. [20] H.-J. Voss, “Graphs having odd circuits with at least k chords,” Elektron. Inforrnationsverarb. Kybernet. 16 (1980), 77-86. [21] H.-J. Voss, “Graphs having circuits with a t least two chords,” J. Combin. Theory ( B ) 32 (1982), 264-285. [22] H.-J. Voss, “Bridges of longest circuits and path coverings of labelled trees.” Periodica Math. Hungar. 13 (1982), 173-189. [23] H.-J. Voss, “Properties of 2-connected cubic graphs,” Gruphen und Netzwerke - Theorie und Anwendung, 27. Internat. Wiss. Koll. TH Ilmenau (1982), 41-44. [24] H.-J. Voss, “Hordy v ciklah grafov obhvata 2 5,” Grufy, gipergmfg i diskretnye optimizucionnye zuduc‘i, Mat. ksled. 66, $ tiinca, KiSinev (1982), 160-172, 197. [In Russian.] [25] H.-J. Voss, “Large circuits with many chords in graphs with prescribed girth and minimum degree.” Submitted to J . Combin. Theory. [26] H.-J. Voss, “On 2-connected 3-regular graphs with given circumference having maximum order.” Submitted to Mat. Issled, Stiinca, KiSinev.

452

H.-J. VOSS

[27] H.-J. Voss, Bridges in Graphs, VEB Deutscher Verlag der Wissenschaften, Berlin (in preparation). [28] H. Walther and H.-J. Voss, ober Kreise in Gruphen, VEB Deutscher Verlag der Wissenschaften, Berlin (1974). [29] Y. Zhu, Z. Liu and Z. Yu, “An improvement of Jackson’s result on Hamilton cycles in 2-connected regular graphs,” Annals of Discrete Math. 27 (1985), 237-248. [30] Y. Zhu, Z. Liu and Z. Yu, “2-connected Ic-regular graphs on a t most 3Ic-4-3 vertices to be Hamiltonian,” J . Sys. Sci. and Math. Sci. 6 (1986), 34-49.

Annals of Discrete Mathematics 41 (1989) 453-472 0 Elsevier Science Publishers B.V. (North-Holland)

On a Standard Method Concerning Embeddings of Graphs K. Wagner Wodanstrafie 57 5000 Cologne 91

R. Bodendiek Institute of Mathematics and its Didactics PH Kiel Kiel, FRG

FRG

Dedicated to the memory of G. A . Dirac We present theoretical evidence for a conjecture about the class of graphs which are not embeddable into the torus or the so-called spindle-surface GI. The settling of the conjecture has been reduced t o a finite search, which has been carried out by the authors since the writing of the paper, and the conjecture has been proved.

1 Let

Critical graphs F be a closed orientable or non orientable surface or the spindle-surface

61, that means the sphere (or the plane), the torus or the pretzel in the first case mainly, the projective or Klein's bottle in the second case. The spindlesurface 61 arises from the torus F1 by contracting one parallel of latitude to exactly one point, called the singular point of 91. As is well-known the theorem of Kuratowski can be thought of as the starting-point of the socalled embedding-problem of graph theory. It asks the question whether a graph can be embedded into a given surface or not. The theorem of Kuratowski answers this question in case of the plane Fo. It says: A graph' can be embedded into Fo if, and only if, it contains no subdivision of 1-5 or 1-3,s. The corresponding problem in case of the projective plane has been solved in [2]. The number of the so-called Kuratowski graphs of the torus or of Klein's bottle is very large. Therefore it is advantageous t o introduce new partial orderings > i (i = 2,3,4)to simplify the investigations. > I is the well-known subdivision-relation.

'In the remainder of this paper, except where otherwise noted, all graphs mentioned are assumed to be finite and undirected.

453

454

K . Wagner and R. Bodendiek

This introduction depends on the elementary relations R; (i = 0,. . . ,4) being defined in [3]. Ro and R1 can be thought of as removing an edge or an isolated vertex of a graph or as contracting an edge e = ( a , b ) of a graph where at least one of the two vertices a or b has degree 2 a t most. Rz means contracting an edge e‘ = ( c , d ) where the endpoints c and d of the edge e’ are incident with at least three edges of the graph. R3(G) = H means that H arises from G by replacing a trihedral {v} * { a , b , c } of G by triangle (a,b,c,a). Finally, &(G) = H means that H arises from G by replacing a double trihedral by a double triangle’. Corresponding t o > 1 one can define the partial orderings > 2 , > 3 , > 4 . G > i H (i = 2,3,4) if, and only if, G = H or there exists a sequence of graphs GI, Gz, . . . ,G, ( n 2 2) with G1 = G and G, = H in such a way that Gj+l = Ri,(Gj) for each j = 1 , 2 , . . . ,n - 1 and i j E {0,1,. . . ,i}. The set of all graphs being >i-minimal and being not embeddable into 3 is called the minimal basis of 3 relative to > i . This minimal basis is denoted by M ( T , > j ) or briefly Mi(T).The following inclusion chain holds for every surface T :

As it has been shown in [3] that all graphs of M l ( 3 ) can be explicitly constructed with the help of the graphs of M4(T) for each surface T it is sufficient to determine all graphs of M4.F). Subsets of M4(61) and M4(31) are given in [l].Since every graph G being non-embeddable into 3 is not embeddable into TO,G contains a subdivision of K 5 or K3,3 on account of the theorem of Kuratowski. These facts can be used for the solution of the embedding-problem of 61 and 3 1 . So it has been shown in [l]that the sets of all graphs being elements of M4(&) or of M4(.?”) and containing a subdivision of K 5 , consists exactly of three or five graphs each. These eight minimal graphs are depicted in [l].So M4(&) and M4(&) are determined explicitly if, and only if, the sets of all graphs being elements of M4(61) or M4(31) and containing a subdivision of K 3 , 3 are explicitly known. The same holds for MI(&) and Ml(F1). 10 graphs of M4(91) and 23 graphs of M4(31) have been found so far [l].Meanwhile the authors have succeeded in determining two other >r-minimal graphs, so we know 12 graphs of M4(61).

As M4(projective plane) has exactly 12 graphs the interesting question arises whether the minimal basis M4(&) contains more elements than the minimal basis of the projective plane. This paper deals with this question. 2See Figure 10 in the paper “The Fascination of Minimal Graphs” by R. Bodendiek and K . Wagner in this volume.

On a Standard Method Concerning Embeddings of Graphs

455

We want to give a contribution t o the answer. As mentioned above we would like to determine all graphs G being elements of M4(G1) or M 4 ( F l ) and containing a subdivision U ( K 3 , 3 ) of K 3 , 3 . Therefore we at first define the concept of a minimal subdivision c ( K 3 , 3 ) as a subdivision Of K 3 , 3 whose number of vertices is minimal. Then we consider the two sets M " ( F ) and M^"(.F)of graphs ( n 6 No) with the help of the minimal c ( K 3 , 3 ) . M"(F) can be thought of as the set of all those graphs G of M 4 ( F ) containing an U ( K 3 , 3 ) , and so a minimal @ ( K 3 , 3 ) too, and having the property that all relative components3 of G in relation to every 6 ( 1 1 3 , 3 ) of G have order 5 n. On the other hand @(F) can be thought of as the set of all those graphs G of M 4 ( F ) containing a minimal c ( . K 3 , 3 ) and having the property that all relative components of G in relation to at least one 6 ( K 3 , 3 ) of G have order 5 n E No. As the investigations of these two sets are very extensive we have to 4 confine ourselves to the investigations of M (F)and z o ( F ) which are obviously identical. That is to say we shall investigate the set of all those graphs G of Md(.F) which arise from a minimal U ( K 3 , 3 ) by only adding edges (= relative components of G in relation to 6(IL13,3) of order 0 ) . In order to simplify the discussion we consider a partition of the set of all edges which can be added to 6 ( K 3 , 3 ) into four subsets called A-, B - , Cor D-subsets. The elements of these four subsets are called A - , B-, C- or D-edges. So we can reduce the description of all graphs of @(GI) and of 4 M (F') to a finite number of cases. We conjecture that @ ( G I ) = { K c } and @ ( T I ) = 0. This new method is interesting for the following reasons: 1. We confine ourselves to investigations of those imal numbers of vertices.

U ( K 3 , 3 ) having

a min-

2. We only use such properties being fulfilled by all minimal graphs G

independent of the choice of the surface F. Especially we want to mention the so-called triangle-property (or briefly A-property). It says: If A is a triangle of G (consisting of three edges of G), then all vertices of the triangle have valency 2 4.

So the investigations will be carried out in such a way that we shall look for properties of the minimal 6 ( K 3 , 3 ) of minimal graphs G and of the relative components of G in relation to f i ( K 3 , 3 ) without specializing the 3The concept of relative component of a graph G in relation to a subgraph of G is defined exactly in [4].

K . Wagner and R. Bodendiek

456

surface T . Confining ourselves to T = 91 or T = F1 we obtain very nice results. Together with those results a way how to study the sets r ’ ( F )and z ( F ) for each i E {1,2} will be indicated. The next chapters show the surprising result that only the minimality of 6 ( K 3 , 3 ) leads to the minimal graphs of @(F) in the case of T = GI or F = F1.

2

Definitions and notations

Let G be a minimal graph of M 4 ( J ) with G 1 U ( K 3 , 3 ) . Then G contains a U ( K 3 , 3 ) with a minimal number of vertices. A minimal subdivision U ( K 3 , 3 ) of G will always be denoted by ? ( K 3 , 3 ) = 6. The two triples of the main vertices of 6 will be denoted by { a l , a2, a3) and { b l , ba, b 3 ) . Now let k = ( a , b ) be an edge of G which does not belong to 6 and whose end vertices a and b belong to 6. Since 6 is minimal a and b do not belong to the same main path a; ...bj of 6 ( i , j E {1,2,3}). Considering k we consequently have to discuss the following four possibilities:

I

I

Ui

I

Figure 1. The vertex a is a main vertex of and the vertex b is inserted into a main path of 6,We say briefly that b is an inserted vertex in opposite to a main vertex of 6. If a = a1 and if b belongs to the main path a ; . . .b j with i # 1 (Figure 1) we call the main path a1 . . . b j ( j E { 1,2,3}) the projection of k. This projection of k has necessarily got only one edge (i.e., the main path a1 .bj contains no inserted vertex) so that a1 and b j are adjacent, for otherwise ? would not be minimal. In this case 1, k is shortly called an A-edge of G in relation t o 6.

..

On a Standard Method Concerning Embeddings of Graphs

457

Figure 2. 2. The two end vertices of k = ( a , b ) are inserted vertices of two adjacent main paths of (Figure 2). If these main paths are a1 . . .bl and a1 . . . b2 for example the two end vertices a and b of k are adjacent So there exists a triangle in G t o a1 because of the minimality of spanned by the three vertices a, a l , b. In this case 2, k is briefly called a B-edge of G in relation to I?.

6

6.

Figure 3. = ( a ,b ) are .aserted vertices c two non3. The two end vertices o adjacent main paths of U. In this case 3, k is called a C-edge of G in relation to 0.Figure 3 shows three C-edges being possible.

4. At last let k = ( a ,b ) be an edge of G whose end vertices a, b are equal to two vertices of the same triple of main vertices of 6.In this case 4,

K . Wagner and R. Bodendielc

458

Ic is shortly called a D-edge of G in relation to D-edges of G in relation to 0 being possible.

@. Figure 4 shows all

Figure 4.

Now let kl = ( a , b ) and k2 = (a’,b’) be two A-edges of G in relation to the same Then we define: k1 and k2 are crossing if, and only if, there are two vertices of { a l , a2, a 3 ) and two vertices of the set { b l , b2, b 3 } , for example u1, a2 and b l , b2, in such a way that the end vertices of Icl and Ic2 are four different vertices of the (subdivided) quadrilateral4 V = (a1 . bl . . . a 2 . b 2 . . .u1) of and that the two pairs of end vertices of kl and k2 are separated on V. Figure 5 shows three examples of crossing A-edges. It follows immediately from Figure 5 that two A-edges Icl = ( a , b ) and k2 = (a’,b’) cross if, and only if, there exists an embedding of @ into the projective plane SO that kl and k2 are crossing in one of the three subdivided quadrilaterals of the 0being embedded into the projective plane. In order to approach the determination of the minimal graphs G of M*(.T) it is useful to introduce some subsets of M 4 . T ) with the help of the concept of the relative component of G in relation to @.Let r ( F ) be the set of all those graphs G of M4(F) containing an U ( K 3 , 3 ) and thus a minimal 0 ( 1 { 3 , 3 ) , too, and having the property that all relative components of G in relation to every 5 ( 1 { 3 , 3 ) of G have got order 5 n E No. Let zn(F) analogically be the set of all those graphs G of Md(.T) containing an U ( K 3 , 3 ) and thus a minimal 0 too, and having the property that all

c.

..

..

‘V is not only a quadrilateral of

c

6,but even a 2-cell of 6.

On a Standard Method Concerning Embeddings of Graphs

a’ = b l

459

b3

b

a1 = a

a2

b’ b3

b2

Figure 5 . relative components of G in relation to at least one order 5 n.

of G have got the

Figure 6. At first it is obvious that the inclusion M”(F) Z n ( F )holds for every 4 n E No and that especially in case of n = 0 the equation A4 (F)= Z0(.F) holds. The minimal graph G‘ = 1 * 3 * 1 of M4(&) (& = projective plane) being given in [l]shows that the equation M””(.F) = M’n(F) is not true for every n E No. From Figure 6 we see that G‘ contains two ~ the two relative components of order 1 in relation to U ( K 3 , 3 ) = 1 1 3 with triples of vertices {ul,u2, u 3 ) and { b l , b ~b 3, } , and that therefore G belongs h

K. Wagner and R. Bodendiek

460

$(F;).

Furthermore, Figure 7 shows that G can also be thought of as to consisting of the minimal fi(K3,3)= K3,3 with the two triples of vertices {u1,u2,u3) and { b l , b ~ , b 3 } and a relative component of order 2. Therefore G' does not belong to @($I). As a consequence is a proper subset of @(TI).

xl(&)

Figure 7. With the help of the subsets r ( T )and M?I(T)the determination of the graphs G of M*(T)with G 2 can be decomposed into the following four parts: 1. Determining all graphs of p(.F).

2. Determining all graphs of

@(F) or sl(T).

z2(T),

3. Determining a11 graphs of v2(.F) or

4. Determining all the other graphs of M4(T).

3

On the minimal basis of the spindle-surface and the torus Fl

This chapter deals with the determination of all graphs of @ ( T ) in case of T = G1 or T = T I . At first we shall prove the following proposition.

On a Standard Method Concerning Embeddings of Graphs

46 1

(1.1). Let G be any graph with the following properties: 1. G contains a n U(K3,3), and thus a minimal 8(K3,3).

6 with 6 G by adding A-, B - or D-edges. G does not contain any 5 with two crossing A-edges.

2. G arises porn every

3.

4. There is no vertex of valency 2 in G. 5. If G contains a vertex y of valency 3 with the neighbours then {y1,y2,y3} is a n independent set of vertices in G.5 Then follows for every minimal a) Every main path of

y1, y 2 , y3,

5 of G:

6 contains at most one inserted vertex.

b) If a main path, for example we can consider a1 . . . b l , contains an inserted vertex x , then a1,x, bl are vertices of valency 4 in G.

Figure 8.

Proof. The proof will be divided in three: I: G contains an A-edge k = ( a l , x ) , and x has been inserted into the main path a2 ... b2 (Figure 8). Then we shall show that b2 and cc are adjacent; i.e., that the main path b2.. , a2 of $ contains no further inserted vertex between b2 and x. Now we assume that there is a vertex x' being 'Notice that a graph G E @ ( T ) always fulfills the conditions of (1.1) if G contains neither C-edges nor two crossing A-edges in relation to any U of G. The last condition of (1.1) says that every vertex of a triangle of G has valency 2 4. h

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inserted into the main path b2 .,. a2 between b2 and x (Figure 8). Then there must necessarily exist an A-edge or a B-edge of G, denoted by k‘, being incident to 5’. At first we shall prove that k’ cannot be a B-edge. Now let k’ be a B-edge. Because of the minimality of the end vertex of Ic’, being different from x’, belongs either to the main path a 2 . . . bl or to the main path u 2 . . . b3 and is an inserted vertex in each case. In both cases the vertices 2’ and a2 must be adjacent. This is a contradiction. Hence k’ is an A-edge. Then the second end vertex of k’, being different from x‘, must be equal to bl or b3 or a3 or a l . The first two possibilities cannot be true, because G would contain two crossing A-edges. Because of the minimality of the third possibility can be excluded. So we obtain k‘ = ( a l ,5’).This holds for every inserted vertex x‘ of the path b 2 . . .z. Therefore we can assume that x’ is adjacent to b2 and belongs to b 2 . . . x and that the three edges ( b z , z’), ( u I , ~ ’ )(a1,bz) , are spanning a triangle of G. Then it follows from condition 5 and from the footnote 5 that another A- or B-edge of besides k‘ has t o be incident to 2’. Thus we obtain a contradiction since b‘ is only edge of this kind. Hence b2 and x are adjacent. This proves I.

6

Figure 9. Now the assertion a) of (1.1) follows easily; i.e. 11: Every main path of U contains one inserted vertex at most. For assume that the main path b 2 . . a2 contains two inserted vertices x and 5’ (Figure 9). According to I z has via A-edges at most the neighbours bl or b3 (Figure 9). Likewise x‘ has via A-edges at most the neighbours a1 or a3. Similar facts can be concluded in case of B-edges being incident to z and 2’. This gives a contradiction to the minimality of Finally the conclusion b) of (1.1) will be proved. It says: 111: If a A

.

c.

On a Standard Method Concerning Ernbeddings of Graphs

463

main path a ; . .. b j of U contains one, and only one, inserted vertex x, then x,ai,bj are vertices of valency 4. Because of the conditions x is incident to the least two edges beside the edges (x,a;) and (z,bj). By minimality there cannot be three edges being incident to x a t the same time. From this follows immediately that x is a vertex of valency 5 4. According to condition 5 x is a vertex of valency 2 4. So we obtain that the valency of x is 4. Then there are two A-edges or B-edges being incident to x (Figure 10). h

Figure 10. Then G always contains two new minimal l7(1<3,3) of G where b j or ai are inserted vertices. Then it easily follows from the first part of the proof 0 that a; and b j are vertices of valency 4. I1 and 111 prove (1.1). Now we shall introduce the following notation because of (1.1): 1. Let

c" ( n E No) be the set of all graphs G containing a a

c(

1<3,3)

U ( K 3 , 3 ) and

and fulfilling the following properties:

All vertices of G have valency 2 3. For each vertex a of G of valency 3 the following holds: The set of the three neighbours of a are independent; i.e., each two of the three neighbours of a are not adjacent in G. All relative components of G in relation to every 5 have order 5 n. ( n E No) be the set of all graphs G containing a 6 ( 1 < 3 , 3 ) 2. Let and fulfilling the two properties a) and b) of 1 and the further property c'): All relative components of G in relation to at least one 6 ( K 3 , 3 ) C G have got order 5 n.

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Figure 11. Then the following facts are true: If a graph G E Cocontains a C-edge k l = ( a l ,b l ) in relation to at least one i? C G, the graph x8 = 6 U kl C G can be embedded into the projective plane, as shown in Figure 11. The two sets {a2,a3,a4} and {bZ,b3,b4} are the two triples of main vertices of 6. The end vertex a1 of kl belongs to the main path b 2 . . a4 of the other Furthermore we end vertex bl of kl belongs to the main path a 2 . . . b4 of can conclude from the minimality of that G does not contain a B-edge in relation to this i? and that every other A-, C- or D-edge of G always connects two vertices of two disjoint paths belonging to one of the four quadrilaterals of x8 being %cells. The following proposition can also be concluded from the minimality of

.

6

c.

6,

i?;

(1.2). If there are two crossing edges k' and k" of G E Co i n one of the four quadrilaterals of x8 being a-cells, k' and k" are necessarily the two diagonals of this quadrilateral, and it does not contain any other edges of

G. This statement leads us to the following definition: If a quadrilateral V being a 2-cell of x8 contains only two crossing edges of G,we call V of diagonal type. The following theorem can be added:

On a Standard Method Concerning Embeddings of Graphs

465

(1.3).

1. X, does not contain two adjacent quadrilaterals of diagonal type. 2. If x8 contains two non-adjacent quadrilaterals of diagonal type, the other two quadrilaterals of x8 are empty, and the graph G 2 x8 is necessarily equal to the graph G12 E M4(projective plane) as shown in [l]. Summarizing we can state the following:

c"

Theorem 1. If the graph G E contains a C-edge in relation to at least one 5 C G, either G = G12 E M4(projective plane) or x8 = 6U kl G G contains one quadrilateral of diagonal type at most. It follows immediately from Theorem 1 that:

Theorem 2. If G E @(GI) or G E @(F1) then G does not contain a n y C-edge in relation to any minimal 5 G G. Proof. Since the graph GI2 E Md(projective plane) is embeddable into 61 and into 3 1 it follows that G # Gl2. If G contains a C-edge kl it follows from Theorem 1 that x 8 U bl G contains at most one quadrilateral of diagonal type. Therefore G is embeddable into G1 and into F1. This is a 0 contradiction to the choice of G. Hence G contains no C-edge. We shall now prove another interesting result:

-

G E Cocontains two crossing A-edges k l , k2 in relation to at least one 61 2 G and if G does not contain any C-edge in relation to any @ of G, G is embeddable into G1 and into 3 1 and G does not contain any B-edges in relation to 61 (that is to say it contains A-edges and D edges only). (1.4). If a graph

Proof. Because of the definition the A-edges k l , Ic2 must cross in a quadrilateral of 61. There are only two possibilities depicted in Figure 12. But in the first case we obtain a C-edge because the graphs being shown in Figure 13 are isomorphic. Therefore we can assume that 61 is as being depicted in Figure 14. At first we shall prove step by step that there are not any edges in the quadrilateral of 51 being hatched. We can conclude from Figure 15 (thick lines represents edges in Figure 15 and in the following figures) and from the minimality of U1 that as well the vertices a2,bs as the vertices bz,a3 are adjacent. h

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K. Wagner and R. Bodendiek

Figure 12.

bl

a2

b3

Figure 13.

bl

a2

b3

Ul

Figure 14.

On a Standard Method Concerning Embeddings of Graphs

46 7

Figure 15. Figure 16 similarly shows that the vertex a1 is not incident to any A- or B- or D-edges in the hatched quadrilateral. The vertex bl fulfills the same property. Furthermore it follows form Figure 16 that there is no A-edge (us, y) if y belongs t o the path a1 . . .z'.

Figure 16. Finally Figure 17 shows that there are not any B-edges being incident to a3 and b3. Consequently there are at most A-edges as shown in Figure 18. Furthermore it follows from Figure 19 that there is not any A-edges being incident to ul,for the two graphs shown in Figure 19 are isomorphic. It can be similarly proved that there are no A-edges being incident to b1. Figure 20

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K. Wagner and R. Bodendiek

Figure 17. shows that there are no A-edges being incident to minimality of 61,

a3

or

83

because of the

Figure 18.

So we can sum up: G contains at most A-edges being incident to a2 or b2 and belonging to the two quadrilaterals being adjacent to the main path a2 . b2 (including the four diagonals of these two quadrilaterals). There are no other edges of G. Figure 21 once more shows all edges belonging to G at most. In this case we must notice that the A-edges can cross each other or can cross the

..

On a Standard Method Concerning Embeddings of Graphs

469

Figure 19. D-edges. Hence a part of the proposition has been proved.

Figure 20. Obviously it remains t o show that G is embeddable into 61. In order U (k1,kz) into to prove this we shall a t first embed the graph GI = 61 and notice that the graph GI - a3 is planar (Figure 22). The graph GI - a3 is shown by thick lines in Figure 22. The neighbours b l , ba, b3 of a3 are represented by little circles. According to the embedding criterion of 61 given in [l]we only have to notice that b l , bz, b3 are lying on the boundaries of two countries. If we add two copies of a3 and three suitable edges ( u s , b l ) , (u3, b z ) , (u3, b3) and furthermore all edges of G (see the Figures 21 and 22),

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we obtain a planar graph whose two points a3 must be identified. This 0 shows the embedding of G into 61 and proves finally (1.4).

A x

5'

a2

IU

Y

Figure 21. According to the proposition (1.4) and Theorem 2 every graph G of M (GI)or @(.Fl) fulfills all conditions of the proposition (1.1). Therefore we can give the following summary:

4

Theorem 3. For every graph G E

?(GI)

U

g ( &the)following holds:

1. G does not contain a C-edge in relation to any

8 C G. h

2. G does not contain crossing A-edges in relation t o any U & G. h

3. Every main path of every U contains at most one inserted vertex in relation to every i 7 5 G. 4. If a main path of @ contains an inserted vertex, this vertex and the two end vertices of this main path (that is to say all the three vertices of the main path) are vertices of valency 4.

On a Standard Method Concerning Embeddings of Graphs

471

Theorem 3 provides evidence for the conjecture that @(GI) = { K s } and @ ( T i ) = 0. The conjecture has been reduced to a finite search and has been proved by the authors (not using a computer) since this paper was written.

Figure 22.

References [l] R. Bodendiek, H. Schumacher and K. Wagner, “Uber Graphen auf Flachen und Spindelflachen,” Festschrift K. Wagner: Graphen in Forschung und Unterricht (R. Bodendiek, H. Schumacher and G. Walther, eds.), Franzbecker-Verlag (1985), 18-47. [2] R. Bodendiek, H. Schumacher and K. Wagner, “Die Minimalbasis der

Menge aller nicht in die projektive Ebene einbettbaren Graphen,” J . reine angewandte Mathematik 327 (1981), 119-142.

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K . Wagner and R. Bodendiek

[3] R. Bodendiek, H. Schumacher and K. Wagner, “Uber Relationen auf Graphenmengen,” Abh. Math. Sem. Univ. Hamb. 51 (1981), 232-243. [4] R. Bodendiek and K. Wagner, “Zum Basisproblem der nicht in die projektive Ebene einbettbaren Graphen 11,” J. Combin. Theory 17 (1974), 249-265. [5] A. T. White, Graphs, Groups and Surfaces, North-Holland, New York (1973); revised edition 1984,

Annals of Discrete Mathematics 4 1 (1989) 473-486 0 Elsevier Science Publishers B.V. (North-Holland)

Construction of Critical Graphs by Replacing Edges W. Wessel Karl- Weiers traa-Institute of Mat hematics Academy of Sciences of the GDR Berlin, GDR

Dedicated to the memory of G. A . Dirac We discuss criticality with respect to properties and property types in general, and we construct new critical graphs from old by replacing edges by other graphs. In particular, the property type thickness is investigated.

1

Critical graphs

It is one of the outstanding achievements of G. A. Dirac to have introduced the concept of criticality in graph theory. In order t o get an idea of some class of graphs characterized by a certain property it will often be more useful to know its critical members than only to know the extremal ones with respect to some invariants. The critical graphs constitute the “boundaries” of the classes, exceeded by performing some corresponding operation. The concept was introduced by G. A. Dirac [7], [8]in 1952 in connection with investigations of vertex colouring. A critical graph in the classical sense is a graph in which each proper subgraph has a smaller chromatic number than the graph itself. G. A. Dirac and many other authors have found a lot of properties of these graphs since then. But the concept was also transferred to many other areas of graph theory and investigated there, sometimes retaining its name, but often given other designations (maximal or minimal, forbidden, saturated, essential, irreducible, prefix: hypo, and others). In [22]and [24]the unifying term “critical with respect to the property p and the operation op” and the three others obtained by replacing “property” by “property type” and/or “operation” by “operation type” are 473

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474

proposed. In this paper we restrict our interest to the operation type “turning to some proper subgraph”, but we shall consider different properties and property types.

2

Properties, criticality

Generally we use the terminology of F. Harary [15]. The union G U H and intersection G n H of two graphs G and H are defined by their corresponding vertex and edge sets. By a property of a graph we denote (cf. [24]) any quantified distinctive mark of it like “chromatic number 4”, “planar”, “degree 1 of its vertex s” or an expression obtained by putting together some marks with connectives “or” and “and”, like “chromatic number 5 a”, “girth 5 and planar”, whereas the distinctive marks themselves like “chromatic number”, “planar or nonplanar”, “degree of its vertex s”, and sets of them are called property types. So property types are equivalence relations on the set of graphs, and the corresponding properties are determined by the corresponding equivalence classes and unions of them. A property determined by exactly one equivalence class is called an elementary property (of the type in question). A property type is called additive if its elementary properties have been weighted by integers in sucli a way that the weight of the property of the disjoint union of two graphs equals the sum of the weights of the properties of these graphs. E.g., the crossing number v and the vertex covering number a are additive, whereas the thickness 8 and the chromatic number x are not. If a graph has a property p of type t , but none of its proper subgraphs has p, then it is called p-critical and t-critical in this paper. E.g., the complete 5-graph is (v = 1)- and ( 6 = 2)-critical and so v and 8-critical, too, and the 5-cycle is (a = 3)- and (x = 3)-critical and so Q- and 2-critical, too. If a graph has a property p of type t, but loses p by the addition of the edge 2/21, then it is called uvtcritical with respect to p and t. If a graph loses its property p of type t by removal of the edge st, then s t is said to be p-critical and t-critical. So a graph with property p of type t is p-critical (t-critical) only if each of its edges and each of its vertices (whose criticalities are defined analogously) is p-critical (t-critical). The converse is true if any subgraph of a subgraph without p itself is without p. Splitting a vertex of a graph G into (at least two) new vertices means removing it (and the edges incident with it) from G and joining each of its

Construction of Critical Graphs by Replacing Edges

475

former neighbours with exactly one of the new vertices by an edge, each new vertex being joined at least once. Let G and H be two disjoint graphs and uv E E( G) , u’,v’ E V ( H ) . If the graph GH is obtained from G - uv and H by identifying u and u’ and identifying v and v‘, then it is said to be obtaind from G by replacing its edge uv by the graph H(u’,v’). Generally H(u’,v’) will denote the graph H with distinguished vertices u’ and v‘.

3

Replacing an edge

In 1953 G. A. Dirac [9]has given a method for deriving two ( x = Xo)-critical graphs from any such graph having connectivity 2. In 1963 and 1964 T. Gallai [13] and G. A. Dirac [12] showed that also the inversion of this method can be used for constructing new such graphs from given ones. Their result generalized earlier ones due to G. A. Dirac [lo], [ll]and G. Haj6s [14] (see also G. Ringel [19]). In our terminology the first mentioned results of Dirac and Gallai can be stated in the following way:

Theorem 1 (Dirac, Gallai). Let G and H be two disjoint ( x = ~ 0 ) critical graphs ( X O 2 3), u v E E ( G ) , w E V ( H ) , and let H‘ be a graph with chromatic number < xo obtained from H by splitting w into the vertices u’ and v’ ( $ V ( G ) ) .Then the graph GHI obtained from G by replacing u v by H’(u’, d)is ( x = Xo)-criticaE, too. Conversely, every (x = Xo)-critical graph with connectivity 2 can be obtained in this way.

-

This characterization theorem has found its counterparts in the areas of vertex-covering (T. Gallai, see B. Andrdsfai [l]; W. Wessel [21]) and edge-colouring (I. T. Jakobsen [17]). We state the two results as Theorems 2 and 3 (x’ denotes the edgechromatic number, also called the chromatic index):

Theorem 2 (Gallai, Wessel). Let G and H be two disjoint a-critical graphs, uv E E ( G ) , w E V ( H ) , and let H’ be a graph obtained from H by splitting w into the vertices u’ and v’ ($ V ( G ) ) . Then the graph G H ~ obtained from G by replacing uv by H’(u’, v’) is a-critical, too. Conversely, every a-critical graph with connectivity 2 can be obtained in this way.

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476

Theorem 3 (Jakobsen). Let G and H be two disjoint (x' = xb)-critical graphs, which are not stars, uv E E(G),w E V ( H ) ,d e g u t d e g w 5 xbi-1, and let H' be a graph obtained from H by splitting w into the vertices u' and v' ( 4 V ( G ) )with deg v' = 1. Then the graph GHI obtained from G by replacing u v by H'(u', v') is (x' = xb)-critical, too. Conversely, every (x' = xb)-critical graph which is separable by two independent edges can be obtained in this way (then deg u = 2 or deg w = 2). Notice that Theorems 1and 3 are concerned with criticality with respect t o certain properties, but Theorem 2 is concerned with criticality with respect t o a property type. Indeed it seems to be impossible to find a general interesting assertion of the considered type if these criticality types are exchanged in one of the theorems. This fact seems to be caused by the additivity and nonadditivity of Q and of x and x', respectively. This seeming connection will be respected in our further investigations. The attempt to find results analogous to those of the theorems above for other property types and the knowledge of the nonplanar-critical graphs by the famous theorem of K. Kuratowski [18] lead us to reformulate the construction rule to replacing an edge of the given graph G by the other given graph H itself. The cited theorem can be stated as follows:

Theorem 4 (Kuratowski). Let G be a ( 6 = 2)-critical graph, uv E E ( G ) , and let H be a path disjoint from G, the endvertices of H being u' and v'. Then the graph GH obtained from G by replacing uv by H(u', v') is ( 6 = 2)critical, too. Conversely, every ( 6 = 2)-critical graph with connectivity 2 can be obtained in this way. More specifically, it can be obtained from G = K 5 or G = K 3 , 3 by one or more applications of the construction. Notice that the former construction rule is not applicable here since the cycles obtained from H by joining or by identifying u' and v' are not (0 = 2)-critical.

4

Adjoint properties

In order to construct p or t-critical graphs from given such graphs by replacing an edge by some graph H , we must know suitable graphs H .

Construction of Critical Graphs by Replacing Edges

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Every such graph H must have a property p‘ of some type t’ which ensures that the compound graph has a fixed property of type t ; so t’ will have a special relation to t. H must be p’-critical and t‘aitical, and one can expect that the exceptional importance of the vertices u‘ and v’ of H identified with the endvertices of the replaced edge in the construction will be reflected in p‘ and t’. We call a property p‘ adjoint t o a property p, if p’-criticality of any graph H(u‘, w’) not containing u’v’ implies p-criticality of the edges of H in every graph GH obtained from a p-critical graph G by replacing some edge uw of it by H ( d , v’). Analogously an adjoint property type is defined. Notice that if a property p’ is adjoint to a property p , then their types are not necessarily in the same relation (cf. the following examples). Define the ( s ,t)-chromatic number Xs,t(G)of a graph G with s, t E V ( G ) as the minimum n for which G has a colouring by n colours assigning the same colour t o s and t if s t $ E(G). Then we can deduce from Theorem 1 that the property X s , t = xo (2 3) is adjoint to the property x = 20. There {s, t } = {u’, w’}. Notice that the graph H‘ used there is u‘w‘+critical with respect to = xo. Conversely, we can prove the first part of Theorem 1 by independently showing that X s , t = xo is adjoint to x = X O , that sttcriticality of a ( X s , t = Xo)-critical graph H with respect to Xs,t = xo implies (x = Xo)-criticality of the edges of G in G H ~and , that H’ is ( ~ ~ t= , Xo)-critical ~ , and u’v’tcritical with respect to xzlt,vt= XO. Notice that X s , t is not adjoint to x. Define the (s,t)-covering number G~,,~(G) of a graph G with s, t E V ( G ) as the minimum n for which G has a vertex cover, which in case s t $ E(G) either consists of n - 1 vertices # s, t or of n vertices including s,t. Then we can deduce from Theorem 2 that the property type a,,t is adjoint to the w’}. Notice that again the graph H‘ property type a. There {s, t } = {u’, used there is u’w’tcritical with respect to a , ~ , and ~ t we can prove the first part of Theorem 2 by the method indicated above. Define the (s,t)-chromatic indez &(G) of a graph G with s, t E V ( G ) as the minimum n for which G has an edge-colouring by n colours assigning mutually different colours to the edges incident with s or t if s t 4 E(G). Then we can deduce from Theorem 3 that the property “X:,t = XO, deg s = deg t = 1” is adjoint to the property x’ = xb. There { s , t } = { u ’ , ~ ’ } .The graph H‘ used there is v’v’+critical with respect to the given property, and we can prove the first part of Theorem 3 in case deg w = 2 by the method indicated above. {&, deg s, deg t } is not adjoint to x’. In the last section we shall follow the path, gone in the indicated proofs of the first parts of Theorems 1 - 3, for finding a corresponding result for ~

~

1

,

~

’

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478

another property type. Our decision for constructing either new p-critical or new t-critical graphs from given ones is guided by the experience gained in Theorems 1 - 3 (cf. the remarks there): Given ap-critical (in the nonadditive case) or t-critical (in the additive case) graph G , search for a graph H(u’,v’) such that the graph GH obtained from G by replacing its edge uv by H(u’, v’) is also p-critical or t-critical, respectively.

For that in the nonadditive case

-

give an adjoint property p‘, - decide if sttcriticality of a p’-critical graph H with respect to (the same) p‘ implies p-criticality of the edges of G in GH (if no, start again; if yes, continue), - give a p’-critical graph H and decide if it is stfcritical for some s t 6 E ( H ) with respect to p‘, too (if no, repeat or start again; if yes, end). The additive case is handled in exactly the same way, only replacing “property” by “property type” and p’ by t’. There are many properties and property types not allowing our method for different reasons, and others only allowing trivial results. But there are interesting cases, too, suitable for investigation, e.g. thickness and outerthickness which are nonadditive, and crossing number and genus which are additive ( J . Battle, F. Harary, Y . Kodama and J. W. T. Youngs [3]).

5

Thickness

The thickness 0 is a nonadditive property type. Define the (s,t)-thickness 0,,,(G) of a graph G with s , t E V ( G )as the minimum number of mutually edge-disjoint planar subgraphs whose union is G and which all have s and t in different components except for the subgraph containing s t (if existing). In the proofs of the following two lemmas let G, H , and GH be given as in the definition of adjointness; u’, v‘ are identified with u, v , respectively. Lemma 1. Ou,v = d is an adjoint property to 8 = d .

Proof. Assume d(GH) 5 d - 1 and partition GH into d - 1 plane subgraphs. This induces corresponding partitions of G - uv and H . Because

Construction of Critical Graphs by Replacing Edges

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of O,,,(A) = d, u and v must belong to the same component in at least one of these plane subgraphs of H . Therefore in at least one of the considered plane subgraphs of G - uv u and v must lie on the boundary of some face. So we can add uv there without producing a crossing. This yields a partition of G into d - 1 plane subgraphs in contradiction to B(G) = d (recall that G is assumed to be ( 0 = d)-critical). Thus O ( G H ) 2 d. Because of the (6' = d)-criticality of G there is a partition of G - uv into d - 1 plane subgraphs. The (O,,, = d)-criticality of H implies, that for every zy E E ( H ) the graph H - zy can be partitioned into d - 1 plane subgraphs each having u and v in different components. Combining both these partitions shows 8(GH - zy) 5 d - 1 for every s y E E ( H ) , so completing the proof. 0

Lemma 2. uv+criticality of a (O,,, = d)-critical graph H with respect to O,,, = d implies ( 0 = d)-criticality of the edges of G in G H . Proof. O(GH)2 d can be taken from the previous proof. Let zy E E(G - uv).Because of the (0 = d)-criticality of G there is a partition of G - s y into d - 1 plane subgraphs one of which contains uv. The uv+criticality of H implies, that H + uv can be partitioned into d - 1 plane subgraphs one containing uv and each of the others having u and v in different components. Combining both these partitions (and cancelling U V ) shows e(GH - s y ) 5 d - 1 for every s y E E(G - UV), so completing the proof. The proofs of these lemmas can be completed in a straightforward manner to a proof of the following theorem implying them:

Theorem 5. A graph G and a graph G H , obtained from it by replacing its edge uv by a graph H(u', v') which is (O,,, = d)-critical and uv+critical with respect to Ou,v = d, either both are ( 0 = d)-critical or both are not ( 0 = d)-critical. The third step on our path of constructing &critical graphs is executed by the following theorems.

Theorem 6. A (e,,, = d)-critical graph G is also st+critical with respect to O,,t = d, if the distance between s and t in G is 2. Proof. Let T be a common neighbour of s and t in G. Partition G - rs into d - 1 planar subgraphs with s and t in different components and embed the graph containing r t into the plane in such a way that r t and s lie on

480

W. Wessel

the same face. Then rs and st can be added there without producing a 0 crossing, hence proving 0,,t(G st) 5 d - 1.

+

Theorem 7. The graph 1<2d-1,(2d-3)2, d > 2, is (0,,t = d)-critical and st+critical with respect to 0,,, = d, if s and t belong to its first vertex class. Corollary (cf [22], [23]). Let G be a (0 = d)-critical graph, uv E E ( G ) , and H = 1<2&1,(2&3)2) d > 2, u',v' elements of the first vertex class of H . Then the graph GH obtained from G by replacing uv by H ( d , d ) is (0 = d)-critical, too.

Notice that every path with endvertices s and t is (O,,, = 2)-critical and sttcritical with respect to 0,,t = 2, and so can be used for constructing (0 = 2)-critical graphs from given ones (cf. Theorem 4). Proof of Theorem 7, Let the vertex classes of H = 1<2&1,(2&3)2 be A = {S,t,al,a2,...,a2d-3} and = {b,bl,b2,...,b2(d-l)(2d-4)}. A planar bipartite graph with n vertices consisting of a t least two components has a t most 2(n-3) edges. Thus, ifagraph with (2d-1)+(2d-3)2 vertices is the union of d - 1 such graphs, then it must have less than (2d - 1)(2d - 3)2 edges and so is different from H . Because of s t $! E ( H ) this implies O,,t(H) 2 d. For proving the (0+ = d)-criticality of H it is sufficient to show that 0,,t(H - zy) 5 d - 1 for every x y E E ( H ) . For that first partition B - b into the subsets Bi = {b2(d-1)(i-1)+19 b2(d-l)(i-1)+2, , h ( d - ~ ) i } , i = 1 , 2 , . ,2d - 4.

.

B2

\

..

/ Figure 1.

B1

Construction of Critical Graphs by Replacing Edges

48 1

Then compose the graphs GI, G2,. . . ,Gd-l as shown in Figure 1; there - B; (and B; - a j ) means that aj is joined with every vertex of B; by an edge. These graphs are planar and together contain each edge a j b k , 1 5 j 5 2d - 3, 1 5 k 5 2(d- 1)(2d - 4). Now represent each GI, 1 5 I 5 d - 2, in the plane as sketched in Figure 2 and arrange that the sets of the vertices bk lying on the exterior faces of the d - 2 graphs are mutually disjoint. This is possible since every set Bi has 2(d - 1) > 2(d - 2) vertices and the Bi's are mutually disjoint. aj

Figure 2. Then for every Bi all vertices but two lie on an exterior face of these graphs. Let BI be the set of the two exceptional vertices of B;. Next replace the graphs GI and Gd-1 of Figure 1 by the graphs Gi and G&-, analogously presented in Figure 3; there B;\B: is the set of vertices of B; not belonging to Bi.

Figure 3. These graphs are planar, too, and together with Gz, G3,. . . ,Gd-2 still contain each edge a j b k . But this new graph family has the additional

W. Wessel

482

property that you can represent its members in the plane in such a way that every vertex bk lies on the exterior face of one of Gi, G 2 , G3, .. ,G d - 2 and, moreover, on the exterior face of Gh-l. Given these plane representations, insert the vertex s into each of the exterior faces of G i , G a , G 3 , . . . , G d - z and the vertex t into that of G&-l and join them with the vertices bk on the respective face by edges. By construction the enlarged graphs are planar, too, and additionally contain all edges bks and b k t . Because of d > 2 there is a vertex b' in B I \ B ; . In all but one graph of the last family b' has degree 2; the exceptional graph is that one which contains the edge b's. So, if we insert the vertex b into some face containing b' in the plane representation of each of the considered graphs, then it can be joined by edges there with all neighbours of b' (without producing a crossing) except for one of s and aj (in the exceptional graph) which can be prescribed. The d - 1 graphs constructed now are planar, they together contain all edges # bs or # baj (for some j ) , and each of them contains only one of s and t. After adding the other one of s and t as a second component each time, we have shown by construction that B,,t(H - zy) 5 d - 1 for every zy E E ( H ) , since bs and baj cover all nonsymmetric cases of an edge x y of H . So H is (Bs,t = d)-critical. 0 Theorem 6 completes the proof of Theorem 7.

.

The following graphs are basic ( 0 = d)-critical graphs for producing infinite families of such graphs by means of the corollary of Theorem 7 : K2d-1,2(d-1)(2d-3)+1,

2

(L. W. Beineke [4]); K4d-5,4d-5,

d 12

(L. W. Beineke [4],I. Z. Bouwer and I. Broere [ 5 ] , A. M. Hobbs and J. W. Grossman [16]);

+

I<2 ( d - 1)+ r ,2( d- 1) (2d- 3)

/T9

T

a factor of 2d - 3

(I. Broere [S]); IC3(d-1),3(2d-3) IL',d,2(d-1)2+1

- Kd-2,1, d 2 2; - KlJ, d L 3

(W. Wessel [22], [23]).

Construction of Critical Graphs by Replacing Edges

483

In the case d = 3 we also know the graphs KS (J. Battle, F. Harary and Y. Kodama [2], W. T. Tutte [20]) and Ii10- K 3 (W. Wessel [25]). In this case we can even show directly: Theorem 8. The graph Kg - P3 obtained from K g by the removal of the edges of a path of length 3 whose inner vertices are s and t , is (Os,t = 3)critical and st-critical with respect t o BSlt = 3.

Corollary. The graph GH obtained from a (0 = 3)-critical graph G by replacing its edge sit’ by the graph H = (Iig - P3)(s,t) of Theorem 8 is (0 = 3)-criticaZ, too,

Acknowledgement I would like to thank the referee and B. Toft for their critical and helpful remarks (again showing “boundaries”). Added in proof. J. SirSfi and P. HorSk (“A construction of thicknessminimal graphs,” Discrete Mathematics 64 (1987), 263-268) have given a further method for constructing infinite families of (0 = d)-critical graphs with connectivity 2 from given critical graphs for every d > 2. The method could be reformulated as a construction by replacing edges, then yielding a strong means for finding suitable graphs H .

References [l] B. AndrSsfai, “On critical graphs,’’ Theory of Graphs, International Symposium Rome 1966, Paris, New York (1967), 9-19.

[2] J. Battle, F. Harary and Y. Kodama, “Every planar graph with nine points has a nonplanar complement,” Bull. Amer. Math. SOC. 70 (1964), 618-620. [3] J. Battle, F. Harary, Y. Kodama and J. W. T. Youngs, “Additivity of the genus of a graph,” Bull. Am. Math. SOC.68 (1962), 565-568. [4] L. W. Beineke, “Complete bipartite graphs: Decomposition into planar subgraphs,” Seminar of Graph Theory (F. Harary, ed.), Holt, Rinehart and Winston, New York (1967), 42-53. [5] I. Z. Bouwer and I. Broere, “Note on t-minimal complete bipartite graphs,” Canad. Math. Bull. 11 (1968), 729-732.

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[6] I. Broere, “Families of t-minimal complete bipartite graphs,” Tydskrif Natuurwet. 9 (1969), 177-180. [7] G. A. Dirac, “Some theorems on abstract graphs,” Proc. London Math. SOC.(3) 2 (1952), 69-81. [8] G. A. Dirac, “A property of 4-chromatic graphs and some remarks on critical graphs,” J. London Math. SOC.27 (1952), 85-92. [9] G. A. Dirac, “The structure of Ic-chromatic graphs,” Fund. Math. 40 (1953), 42-55. [lo] G. A. Dirac, “Circuits in critical graphs,” Monatsh. f, Math. 5 9 (1955), 178-187.

[ll] G. A. Dirac, “A theorem of R. L. Brooks and a conjecture of H. Hadwiger,” Proc. London Math. SOC. (3) 7 (1957), 161-195. [12] G. A. Dirac, “On the structure of 5- and 6-chromatic abstract graphs,” J . reine angewandte Math. 214-215 (1964), 43-52. [13] T. Gallai, “Kritische Graphen. I,” Publ. Math. Inst. Hungar. Acad. Sci. ( A ) 8 (1963), 165-192.

[ 141 G. Hajbs, “Uber eine Konstruktion nicht n-farbbarer Graphen,” Wiss. 2. Martin-Luther-Univ. Halle- Wittenberg, Math.-Naturw. R. 1 0 (1961), 116-117. [15] F. Harary, Graph Theory, Addison-Wesley, Reading, Mass. (1969). [16] A. M. Hobbs and J. W. Grossman, “A class of thickness-minimal graphs,” J. Res. Nat. B. Standards ( B ) 72 (1968), 145-153. [17] I. T. Jakobsen, “Some remarks on the chromatic index of a graph,” Arch. der Math. 2 4 (1973), 440-448. [18] K. Kuratowski, “Sur le problbme des courbes gauches en topologie,” Fund. Math. 15 (1930), 271-283. [19] G. Ringel, Furbungsprobleme auf Fliichen und Graphen, Berlin (1959). [20] W. T. Tutte, “The non-biplanar character of the complete 9-graph,” Canad. Math. Bull. 6 (1963), 319-330. [21] W. Wessel, “Kanten-kritische Graphen mit der Zusammenhangszahl 2,” Manuscripta Math. 2 (1970), 309-334.

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[22] W. Wessel, Eigenschaften, Operationen, Kritizitat von Graphen, Diss. B, Techn. Hochschule Ilmenau (1982). [23] W. Wessel, “Thickness-critical graphs - a generalization of Kuratowski’s topic,” Graph Theory, Proc. Conf. Lagbw, Poland 1981, Lecture Notes in Math. 1018,Berlin (1983), 266-277. [24] W. Wessel, “Criticity with respect to properties and operations in graph theory,” Finite and Infinite Sets, 6th Hungar. Combin. Colloq., Eger, 1981, Colloq. Math. SOC.JAnos Bolyai 37, Amsterdam (1984), 829-837. [25] W. Wessel, “The non-biplanar character of the graph 1<10-1<3,” Akad. Wiss. DDR, Karl-Weierstrafl-Inst. Math., Preprint (to appear).

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Annals of Discrete Mathematics 4 1 (1989) 487-496 0 Elsevier Science Publishers B.V. (North-Holland)

A Brief History of Hamiltonian Graphs R. J. Wilson Faculty of Mathematics The Open University Milton Keynes, England and Department of Mathematics The Colorado College Colorado Springs, Colorado, USA Dedicated to the memory of Gabriel Dirac A graph is hamiltonian if it contains a closed cycle passing through every vertex. In this paper we outline the history of hamiltonian graphs from the early studies on the knight’s tour problem to Gabriel Dirac’s important paper of 1952.

1

The knight’s tour problem

The earliest known example of a hamiltonian-type problem is that of the knight’s tour problem on a chessboard: Can a knight visit all sizty-four squares of a chessboard just once and return to its starting position? Chessboard problems of this kind have a long history (see [17]), and solutions are known to exist as far back as the 14th century. A solution to the knight’s tour problem is given in Figure 1. Methods of solution for the knight’s tour problem were given at the end of the 17th century by De Montmort and De Moivre. The first systematic approach to the problem, however, seems to have been due to Euler, in a paper [7] written for the Academy of the Royal Society of Berlin in 1759. Euler had been interested in chess problems for some time and corresponded with both Goldbach and Bertrand on the subject. In this paper Euler described various methods for solving the problem and discussed the analogous problem for chessboards of different shapes and sizes. In particular, he showed that no solution is possible for chessboards with an odd number of squares. 487

R . J . Wilson

488

I 28 39 16 -41 64 29 --32 17 42 -19 30 55 -66 L 18

110149 I 6 4 121 140 I13 136 127 61 48 60

6 3

Figure 1. The next significant advance came in 1771, when Vandermonde wrote a paper [20] on problhes des situation for the Acaddmie Royale des Sciences in Paris. Among the subjects discussed in this paper was a systematic general method for the knight's tour problem, in which the square in row a , column b was represented by the symbol :, and a knight's move corresponded to a pair of symbols (: to oraf2). Vandermonde referred to Euler's 1759 paper, remarking that where2'that great geometer presupposes that one has a chessboard to hand, I have reduced the problem to simple arithmetic.'' An extract from Vandermonde's paper appears in [3]. In the 19th century the problem continued to attract much attention, and was discussed at length in Rouse Ball's book on recreational mathematics [17] and in a three-volume work by Jaenisch [9]. Particular interest centred around finding solutions with some extra property, such as symmetry. An interesting example of a special solution (due to Jaenisch) is that given in Figure 1, where the successive knight's-moves give rise to a magic square whose rows and columns all sum to 260.

iii

2 The work of Kirkman The Reverend Thomas Penyngton Kirkman (1806-95) was rector of the parish of Croft-with-Southworth in Lancashire for about fifty years, His parish duties were not onerous, and he spent much of his spare time working on mathematics. He was a fine mathematician whose significant contributions to both group theory and the theory of block designs have been much underrated. A detailed appraisal of his work is given in [2].

A Brief History of Hamiltonian Graphs

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Among his mathematical interests was the study of polyhedra (or polyedm, as he preferred to call them). In particular, he was fascinated by the following question: For which polyhedra does there exist a closed cycle passing through every vertex? In [lo] he gave a sufficient condition for the existence of such a cycle, but this proved to be incorrect. He also showed that any polyhedron with even-sided faces but an odd number of vertices has no such cycle. We can obtain such a polyhedron if we “cut in two the cell of a bee,” giving the diagram in Figure 2. Nowadays we observe that since such a diagram is a bipartite graph of odd order, and since any cycle must visit the two partite sets of vertices alternately, no cycle through every vertex can exist. Kirkman’s argument was not as simple as this.

Figure 2. In 1880, Tait [18] conjectured that every trivalent polyhedron has such a cycle. This conjecture intrigued Kirkman who remarked [ll]that it “mocks alike a t doubt and proof.” If it were true one could deduce that every such polyhedron is 3-edge-colourable, and hence obtain a simple proof of the four colour theorem. (At the time Tait made his conjecture, Kempe’s ‘proof’ of the four colour theorem was still regarded as correct.) However, in 1946 Tutte [19] produced an example of a trivalent polyhedron which contains no hamiltonian cycle (see Figure 3).

3

Hamilton’s icosian game

Kirkman was not the only mathematician interested in cycles on polyhedra. William Rowan Hamilton’s involvement arose from a suggestion from his friend and biographer John T. Graves that the subject might be relevant to his work on quaternions and non-commutative algebra.

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Figure 3. Hamilton’s interest soon led to the invention of the icosian calculus [8], a system consisting of the symbols L , K and X and the relations L~

= K’ = X5 = 1

It follows from these relations that if p = L p5 = 1,

Xp2X = pXp,

= LK.

and K = ~

X K , then

and (X3p3XpXp)2= 1.

By interpreting X as ‘turn left’ and p as ‘turn right’, Hamilton used this last relation to obtain a closed circuit of faces on an icosahedron. This leads, by duality, to a cycle passing through all the vertices of a dodecahedron (see Figure 4). K

W

Figure 4.

S

491

A Brief History of Hamiltonian Graphs

Hamilton became increasingly intrigued by these ideas, and invented the icosian game, a solid or flat dodecahedron with holes at the vertices and pegs to be inserted into the holes according to certain instructions (see [3]). The object was to find, by experimentation, the number of hamiltonian cycles starting with a given set of vertices. The game was marketed under the title A voyage round the world, with the vertices B , C, D, ..., Z standing for Brussels, Cunton, Delhi, . . . , Zanzibar. Further studies into Hamilton’s icosian game were made by Hermary (see [15]) and A. Kowalewski ([13], [14]). Hermary showed that any hamiltonian cycle can be obtained as the boundary of the flat figure obtained by unfolding the dodecahedron (see Figure 5). Using this representation he showed that the number of possible hamiltonian cycles with a given number k of starting vertices is given by the following table:

k

I

cycles

I

1

2

3

4

5

...

30

20

10

6 or 4

4 or 2

...

Kowalewski related the icosian game to the so-called Buntordnungspmblem. In these papers he was the first to present the Petersen graph as the graph whose vertices are pairs of numbers from the set {1,2,3,4,5} with vertices joined by an edge whenever the corresponding pairs are disjoint (see Figure 6 ) . T

S

12

m

45

34

G

H

15

Figure 5.

4

23

Figure 6.

Lucas’ R6cr6ations Mathgmatiques

In Volume 2 of his four-volume work on recreational mathematics [15], Lucas described a number of problems which are related to hamiltonian

492

R . J . Wilson

cycles in graphs. The most well-known of these (attributed by Berge [l]to Kirkman) is as follows: Les Rondes Enfantines: A number of children dance around in a circle. How can we arrange the dances so that each child dances next to every other child just once? To solve this problem he noted first that the total number of children must be odd (2k 1, say), and that the arrangement of children in each dance corresponds to a hamiltonian cycle in K Z k + l . A solution to the problem is obtained by finding k edge-disjoint hamiltonian cycles in K 2 k + l . Lucas found these cycles by placing 2k of the vertices symmetrically in a ring, and the remaining vertex on a diameter (see Figure 7). The first hamiltonian cycle is the zig-zag pattern shown. The remaining cycles are then obtained by fixing the ring of vertices and rotating the zig-zag pattern. For k = 5, this yields the hamiltonian cycles

+

a a a a a

f

b cde c e bg e g c i g i e k i k gj

fg h d i f b k d c j b e h c

i j ka k hj a j f ha hd fa f b da

(shown) (one step) (two steps) (three steps) (four steps)

h

Figure 7.

Figure 8.

Another problem of a similar type is the following: Les Rondes Alterne'es: A number of girls and boys dance around in a circle with boys and girls alternating. How can we arrange the dances so that each girl dances next to each boy just once? We illustrate the solution when there are six boys ( a ,b, c , d , e , f) and six girls (a', b', c', d', e', f'). Lucas first drew the star-shaped diagram shown in

A Brief History of Hamiltonian Graphs

493

Figure 8. To obtain the required cycles he rotated the star inside the circle, giving the solution

b‘ c c’ d d’ e e’ f f’ a f’ c a’ d b’ e c’ f d’ a a c’ b d’ c e’ d f’ e a’ f b’ a

a a’ b a e‘ b

(shown) (two steps) (four steps)

The solution for a general even number of girls and boys is similar. The analogous question for an odd number of girls and boys was answered by Dirac in 1972 [6].

5

From 1890 t o 1940

In the following fifty years there were a number of results of greater or less significance. We mention four of them briefly: (i) In 1891, Brunel wrote a short note [4]in which he considered the graph with 64 vertices, corresponding to the squares of a chessboard, and 168 edges, corresponding to the possible knight’s-moves. He then related a solution of the knight’s tour problem to the determination of the 168 edge-currents in a related electrical network. (ii) In 1934, RQdei [16] proved the important result that every tournament has a directed path passing through every vertex. Surprisingly, the natural extension of this result, that every strongly-connected tournament has a hamiltonian cycle, was not proved until 1957. (iii) The whole subject of traversability was given a new lease of life in 1936 with the appearance of DQnes Konig’s classic text Theorie der endkichen und unendkichen Graphen [12]. Eulerian and hamiltonian graphs were discussed in detail, and the term “hamiltonian” was established for all time. Gabriel Dirac was a great admirer of Konig’s book. (iv) In 1939, Vdszonyi [21] showed the existence of two-way infinite hamiltonian paths in the n-dimensional square lattice L,, for each n 2 1; the case n = 2 is shown in Figure 9. He also found a sequence of knight’s moves covering each vertex of L , just once.

R. J. Wilson

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4-

Figure 9.

6

Dirac’s 1952 paper

We conclude this brief survey by outlining Dirac’s fundamental paper of 1952 [ 5 ] . He had just completed his doctoral thesis for King’s College, London, and the paper contains many of the important results in this thesis. It is divided into two parts, with part A devoted to properties of cycles of graphs, and part B applying these results to the colouring of graphs. Many results which are now in the ‘folk-lore’ of the subject made their first appearance in this paper. We now state the main results, with brief comments on each: Theorem 1, If a graph G has no cut-vertex, and has a path of length 1, then G contains a cycle of length greater than

a.

Dirac’s proof of this theorem used Menger’s theorem. He remarked that the bound can be improved to 2 8 , but that this is much harder to prove. Theorem 2. If G is a finite graph with deg(v) 2 d for each vertex v , then G contains a cycle of length at least d f 1.

A similar result holds for infinite graphs with no cut-vertex. Dirac used Theorem 2, and the fact that every vertex-critical graph has minimum degree at least k - 1, t o deduce that every k-chromatic graph contains a cycle of length at least Ic, for k 2 3. The most well-known result in the paper is the following, now called Dirac’s theorem:

A Brief History of Hamiltonian Graphs

495

Theorem 3. If G is connected, with deg(v) 2 d for each vertex v, and i f G has at most 2d vertices, then G is hamiltonian. Using Theorem 3, Dirac proved that, for k 2 3, a b-chromatic graph with at most 2k - 2 vertices has a k-chromatic hamiltonian subgraph, and that a critical k-chromatic graph with at most 2k - 1 vertices is hamiltonian. Another consequence of Theorem 3 is that, if x ( G ) = k, then G contains I i k or a cycle of length at least k 2. Moreover, if 0 5 r 5 k - 1 and x(G) = k, then G contains or a cycle of length at least b r 2.

+

+ +

Theorem 4. If G is 2-connected with deg(v) 2 d for each vertex v, and i f G has at least 2d vertices, then G contains a cycle of length at least 2d. The proof of this also used Menger’s theorem. Finally, using Theorem 1 and a counting argument, Dirac proved the following result: Theorem 5. Let G be a 2-connected graph with n vertices. Then the length of a longest cycle tends to 00 as n + 00, provided that every vertex-degree remains bounded. Since 1952 much progress has been made in the study of hamiltonian graphs. Details of some of this work can be found elsewhere in this volume.

References [ l ]C. Berge, Graphs and Hypergraphs, North-Holland, Amsterdam, 1973.

[2] N. L. Biggs, “T. P. Kirkman, mathematician,” Bull. London Math. SOC.13 (1981), 97-120. [3] N. L. Biggs, E. K. Lloyd and R. J. Wilson, Graph Theory 1736-1936, Clarendon Press, Oxford (1976) (paperback edition 1986). [4] G . Brunel, “Sur la relation qui existe entre le problkme de la determination de la distribution electrique dans un systhme de conducteurs lindaires et la question du saut du cavalier sur l’dchiquier,” M4m. SOC. Sci. Phys. Nat. Bordeaux (4) 2 (1891), Extr. Proc.-Verb. xxxiv-xxxvi. [5] G. A. Dirac, “Some theorems on abstract graphs,” Proc. London Math. SOC.(3) 2 (1952), 69-81. [6] G. A. Dirac, “On Hamilton circuits and Hamilton paths,” Math. A n n . 197 (1972), 57-70.

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R. J. Wilson

[7] L. Euler, “Solution d’une question curieuse qui ne paroit soumise B aucune analyse,” Mem. Acad. Sci. Berlin 15 (1759)) 310-337.

[8] W. R. Hamilton, “Memorandum respecting a new system of roots of unity,” Phil. Mag. (4) 12 (1856), 446. [9] Major Jaenisch, Applications de l’Analyse Mathdmatique au Jeu des Echecs, 3 vols., St. Petersburg (1862-63).

[lo] T. P. Kirkman, “On the representation of polyedra,” Phil. Trans. Roy. SOC.London 146 (1856)) 413-418. [ll] T. P. Kirkman, “Question 6610, solution by the proposer,” Math. Quest. Solut. Educ. Times 35 (1881), 112-116. [12] D. Konig, Theorie der endlichen und unendlichen Graphen, Akademische Verlagsgesellschaft, Leipzig (1936); reprinted Chelsea, New York (1950). [13] A. Kowalewski, “W. R. Hamiltons Dodekaederaufgabe als Buntordnungsproblem,” Sitzungsberichte Akad, Wiss. Wien 126 (1917), 67-90. [14] A. Kowalewski, “Topologische Deutung von Buntordnungsproblemen,” Sitzungsberichte Akad. Wiss. Wien 126 (1917), 963-1007. [15] E. Lucas, Re‘crdations Mathimata’ques, 4 vols., Gauthier-Villars, Paris (1882-94). [16] L. RQdei,“Ein kombinatorischer Satz,” Acta Litt. Sci. Szeged 7 (1934), 39-43. [17] W. W. Rouse Ball, Mathematical Recreations and Problems of Past and Present Times (later entitled Mathematical Recreations and Essays), Macmillan, London (1892). [18] P. G. Tait, “Note on a theorem in geometry of position,’’ Trans. Roy. SOC.Edinburgh 29 (1880)) 657-660. [19] W. T. Tutte, “On hamiltonian circuits,” J. London Math. SOC.21 (1946), 98-101. [20] A.-T. Vandermonde, “Remarques sur les problkmes de situation,” Mdm. Acad. Sci. (Paris) (1771), 556-574. [21] E. Viszonyi, “Uber Gitterpunkte des mehrdimensionalen Raumes,” Acta Litt. Sci. Szeged 9 (1939), 163-173.

Annals of Discrete Mathematics 41 (1989) 497-498 0 Elsevier Science Publishers B.V. (North-Holland)

Erinnerungen an Gabriel Dirac in Hamburg E. Witt Rothenbaumchaussee 67 Zi 4 2000 Hamburg 13

FRG Dedicated to the memory of G. A . Dirac Personal recollections about Gabriel Dirac in Hamburg 1959-1962.

Gabriel Dirac war 1959-1962 wissenschaftlicher Assistent bei Professor Schmetterer am neugegriindeten Institut fur Stochastik, damals ein Teil des Mathematischen Seminars. Ich besorgte Dirac ein schones Arbeitszimmer in der obersten Etage, mit einer groBen Plattform vor dem Fenster. Das Seminar war kurz vorher umgezogen, und ich war gerade dabei, meine Unterlagen zu ordnen und alte Briefe zu sortieren, da besuchte mich Herr Dirac in meinem Arbeitszimmer. Ich zeigte ihm einen Brief vom Februar 1945,eine Denunziation einer Frau Eiermann uber Alexander Aigner (jetzt Professor in Graz) und mich iiber regimefeindliche AuDerungen, die wir “in einer Zeit, da Kinder zu Helden werden” gemacht hatten, ein Schreiben, das Professor Franz zum Gluck abgefangen hatte. Als Dirac dies gelesen hatte, taute er auf und wir waren von d a an gute Freunde. Als er spater einma1 horte, daD ich in Princeton wahrend des Eichmann-Prozesses schlecht behandelt wurde, entschuldigte er sich dafiir! In den nachsten Sommerferien iiberraschte mich Gabor mit einer Post aus Hammerfest in Norwegen. Er hatte sich vorgenommen, Europa genau kennenzulernen, und er habe jetzt mit dem nordlichsten Punkt angefangen. Beigefugt war ein Briefoffner aus Rentierknochen. Als mein Assistent Leptin, unser “Prbident” der Arbeitsgemeinschaft (Kolloquium) nach Heidelberg berufen wurde, schlug ich Gabriel fur diesen Posten vor. Er kam jeden Dienstag nachmittags zwei Stunden vor der Sitzung an, um personlich mit 6 eigenen Tauchsiedern und verschiedenen Topfen heif3es Teewasser zu machen, er wollte es allen Teilnehmern gemutlich machen. Unsere Freundschaft war mehr personlicher Art, wir trafen uns haufig zum Mittagessen oder zum Kinobesuch (er schltzte z. B. sehr “Der General” mit Buster Keaton).

497

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E. Witt

Im Herbst 1959 iiberkam Gabor eine plotzliche Begeisterung fur Modellfliegerei. Er nahm mich mit zu einer Werkstatt der Zigarettenfabrik Reemtsma, in der Jugendliche in ihrer Freizeit Miniflugmotore bastelten. Fast alle kannte er beim Vornamen. Etwas spater nahm er mich rnit zu einem Modellflugtag. Es war sehr spannend, besonders als ein groi3es schones Flugexemplar sich immer hoher schraubte und dann unkontrolliert iiber einem Wald fur immer verschwand. Ich traf Gabor haufig mit Alexander Gazdagh, seinem ungarischen Freund. Dieser war (fur meine Begriffe) ein Finanzgenie, wufite iiber alle Aktienkurse Bescheid, und ich stellte spater fest, dafi er Gabor damit angesteckt hatte. Spater kam bei Gabor noch das Kunstinteresse hinzu, bzw. eine erstaunliche Mischung von Kunst und Finanzen. Um ein geerbtes Fernrohr auszuprobieren, fur das sich Artin’ sehr interessierte, bat ich Gabor, uns durch sein Zimmer auf das flache Dach des Mathematischen Seminars zu lassen. Und so standen wir eines Nachts zu dritt auf dem Dach und Artin zeigte und erklarte uns den Andromedanebel. 1962 fuhren wir beide zur Tagung nach Halle und dort sahen wir uns gemeinsam den Film “Soldat Schweik” an - Gabor lachte so gerne. Bald darauf mufiten wir in Hamburg Abschied nehmen, er hatte eine Stelle an der Universitat Ilmenau gefunden. Ich brachte ihn mit seinem Gepack zum Flughafen, wo schon seine Frau auf ihn wartete. 1963 waren wir beide in Canada, er in St. John’s und ich in Hamilton, und ich konnte erreichen, dafi er zu einem Vortrag eingeladen wurde. Von da aus fuhr ich ihn mit dem Auto nach Buffalo, er sagte mir, wie ich mich an der Grenze verhalten sollte, denn er hatte keinen Pa& Aber alles ging gut. Spater lud er mich wiederholt ein nach Aarhus. In einem Brief von Gabor vom 10. Dezember 1980 schreibt er u. a.: Wir horten im Radio, dafl Hamburg -16% hatte, und da dachte ich an Dich und an Alexander, das war sicherlich unangenehm, und ich hoffe, dai3 solches nicht wieder geschieht! - Solche Fiirsorge eines fernen Freundes ist wohltuend!

‘Emil Artin (1898-1962) war 1925-1937 und ab 1958 Professor in Hamburg, 19381946 in Bloomington und 1946-1957 in Princeton. Er reformierte 1926 die Klassenk6rpertheorie mit seinem Reziprozitatsgesetz.

Annals of Discrete Mathematics 4 1 (1989) 499-514 0 Elsevier Science Publishers B.V. (North-Holland)

Hamilton Paths in Multipartite Oriented Graphs C.-Q. Zhang Department of Mathematics a n d Statistics Simon Fraser University Burnaby, B.C., Canada

Dedicated to the memory of G. A . Dirac In this paper, the following results will be shown: 1. There is a Hamilton path and a cycle of length at least p - 1 in any regular multipartite tournament of order p ; 2. There is a longest path uo,. . . ,ut in any oriented graph such that d-(uo) + d + ( u t ) 5 t .

1

Introduction

An n-partite tournament is an oriented graph in which the vertex set V can be separated into n parts Vl, . . . ,V,, such that there is no arc between any pair of vertices in the same part and there is exactly one arc between any pair of vertices in different parts. In this paper, we will show the following results:

Theorem 1. There is a cycle of length at least p - 1 in any regular n-partite tournament of order p . Corollary 1. There is a Hamilton path in any regular n-partite tournament. When n = p , the graph is a regular tournament. A direct corollary of [l] shows that it has a Hamilton cycle. When n = 2 the graph is bipartite, [3]and [5]have shown that it has a Hamilton cycle. So we may assume that n 2 3. In [4],Thomassen conjectured that there is a Hamilton cycle in any oriented graph of order at most 3k in which the indegree and the outdegree of each vertex is at least k. Although a counterexample has been found

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C.- Q. Zhang

500

to this conjecture (see [2], p. 9), the author still thinks the conjecture is true when this graph is 3-partite. Note, this graph will be k-regular, of order 3k and complete in the sense of 3-partite, hence, it is a regular 3partite tournament. Moreover, the author wishes to give a more general conjecture. Conjecture 1. There is a Hamilton cycle in any regular n-partite tournament. Before the proof of Theorem 1, we will give a result which concerns arbitrary oriented graphs.

Theorem 2 . Let G be an oriented graph. Let t be the length of the longest path in G. Then there must be a longest path u1,. , ,ut+l in G such that

.

This theorem is a generalization of some theorems in the papers [3] and [6] about longest paths in oriented graphs.

2

Terminology

Let G = (V,A) be an oriented graph, where V is the vertex set of G and A is the arc set of G. Denote an arc from vertex z to vertex y by (2,y). Let ID(.)

= {Y E V ( D ): ( Y , 4 E A(G)),

O D ( 4 = {Y E

v / ( q: ( Z , Y ) E A(G)}

and

= ID(.> u O D ( Z : > , where D is a subgraph of G. When V(D)= V(G), we omit the subscript D and write I ( s ) ,O(z) and N ( z ) . IfV(D) = {Yl,...,Yr}, let M

Z

)

JJ’(4 = {Yitl

E V(D): Y i E ID(.)}

and

O,’(S)

= (yi-1 E V(D): yi E o D ( s ) }

where subscripts are to be read modulo r . Let

[D’,D’‘] = ((2,y ) E A(G) : 2 E V ( D ’ ) y, E V(D”)},

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501

where D’and D” are subgraphs of G. Let G be an n-partite graph, and TI,. .. ,T, be the n parts of the vertex set. If 2 E T;, we denote Ti by T ( z ) . If P = 211.. . u p is a path, denote the subpath u j . . .uj by uiPuj for l
3

The proof of Theorem 2

Proof of Theorem 2. Assume that G is an oriented graph which does not satisfy the theorem. Let t be the length of a longest path in G. Let

P = {P : P = v1. . .vt+l is a longest path of G}. Let

T ’ ( P )= max{i : vi E I(vl)}, r”(P) = max{t - i 2 : D; E O(vt+l)}, r ( P ) = max{r’(P),r”(P)}, r = max{r(P) : P E P}.

+

for any P E P for any P E P for any P E P

+

Obviously, 3 5 T 5 t 1. Let P = u1 . . . ut+l such that u,. E I(u1).

I. To show that r 5 t , and hence there is no cycle of length t + 1 in G . If T = t + 1, then P can be extended to a cycle of length t + 1 . Hence, ui+l . . . u t + l u i . . . ui is a longest path in G, for each i = 1,.. . , t + 1, this implies that II(ui+l)I

+ lO(u;>l 2 t + 1,

i = 1 , . . . ,t

+ 1 (mod t + 1)

C.-Q. Zhang

502

by assumption. Then

i=l

So there must exist u; such that IN(ui)l 1 t+ 1. Since ui+1 . . .ut+lu1. . . ui and u; . . ut.+lul.. .u;-1 are both longest paths in G, N ( v ; ) 2 V(P) \ { u ; } , which implies that IN(.i)l L t ,

.

a contradiction.

11. Obviously, O(ut+l) U I(u1) 2 V ( P ) . Now let C = Q = ur+l...ut+l = 21...za. Here, T a = t 1.

+

+

111. TO show that OC(Z,) # 8. If not, lO(za>l IIV(Q)\{za)l= a - 1 , and T

- 2. Thus,

10(za)l+ II(u1)l 5 a - 1

+

T

11(.1)1

u1

.. . u T q ,

i IV(C)\{ul, w}I =

- 2 = T + a - 3 = t - 2,

which contradicts the assumption.

IV. TO show that Q'(z~) n OC(Z,) = 0. Otherwise, let u; be a vertex in the intersection; now Uj-1z1QzatiiCUi-1 is a cycle of length t + 1, which contradicts I. V. Choose distinct vertices ui E I(z1) and u j E O ( z a )such that

Relabelling V ( C ) ,if necessary, such that

say, we will have b 2 a. Otherwise, on the path P = u1.. . u,z1.. . x,, we have that

T"(P) 1 t - ( b + 1 ) + 2 = t - b +

1> t - a + l

=T,

503

Hamilton Paths in Multipartite Oriented Graphs which contradicts the definition of T .

VI. Let Q’ = UlCUb and D = U b + l c U , . Here, V ( P )= V(D)UV(Q‘)UV(Q). VIL Consider the cycle C‘ = ub+lDu,xlQxa. By a similar proof to IV,

VIII. Considering the two longest paths P = ulCuTxlQxa,and P’ = xl&xaub+lcub,we have that

Since u, E I ( u l ) , ~ ‘ ( p ’ ’2) definition of T .

T

t 1 { x i , . . . ,xa}l >

T,

which contradicts the

XII. Since $ 1 6 I Q ( z ~ I)l Q, ( X 1 ) 1 5 a - 1. Similarly, I O Q ( X ~5) ~a - 1, IIQl(u1)15 b - 1, 1oQl(ub)l5 b - 1. XIIL Now, the inequality (1) becomes 2(t

+ 1) 5 2 ( T - b + 1) + 2(“ - 1)+ 2 ( b - 1) = 2 ( T + a - 1) = 2t,

a contradiction.

4

0

The proof of Theorem 1

Proof. Assume that G = (V,A) is a k-regular n-partite tournament of order p in which there is no cycle of length p - 1 or p . Hence, we let n 2 3.

C.-Q. Zhang

504

Let the parts of V(G)be TI,. ., ,T,. By the definition of n-partite tournament, ~ ( v =) ~i for v E Tj, j = I,. . ,n,

U

.

i#j

and

I

uTi1

. . ,n.

= 2k for j = 1,.

i#j

Hence

UT;l = p - 2k for j = 1,.. .,n,

lTjl = IV(G)l - I

i#j

so

p=

2nk < 3k (n - 1) -

2 3. Part one. To show that the subgraph outside a longest cycle is an indebecause n

pendent set.

.

Proof: Let C = u1. . u,u1 be a longest cycle in G. Let W = G \ V(C). Let P = v1 . vt be a longest path in W satisfying Theorem 2. Obviously, r = IV(C)l 2 k t 2, (by Lemma 2 in [3]),and IWI 2 2.

..

I. O c ( v t ) # 0 and Ic(v1)# 0.

Otherwise, let Oc(vt) = 0 and vt E TI. i). Since P is the longest path in W , O(vt) E V ( P ) . Let {vjl,. . , ,v j k } = O(vt). Then { D j ~ , . . , , v j k , v t - l } n T l =0. ii). If V ( C )n TI $ 0, let u1 E TI.Then u, $ T I ,and (u,, vt) E A(G), and {Vjl,. . ,v j k ,~ t - 1 ) C O ( U ~ otherwise, ), the cycle u,vtvjlPvsulCuT would be longer than C, where v, E I(u1) n {vjl,. . . , v j k , wt-l}. But

.

k=

IO(u1)l

L I{vjl,*+*,vjk,vt-l)l = k t 1,

a contradiction. iii). If V(C)n TI= 0. Since Oc(vt) = 0, V ( C )g I(wt) \

r = IV(C)l 5 lI(wt)l- 1 = k - 1

+

which contradicts r 2 k 2. iv). Similarly, Ic(v1)# 0.

{vt-1).

So

Hamilton Paths in Multipartite Oriented Graphs

505

II. Since c is a longest cycle, ~ ; + ' ( q )n O c ( v t ) = 0. III. W e claim r 2 2k. By I, O c ( v t ) # 8 and Ic(v1) # 0. If IOc(vt) u Ic(vl)( 2 2, then choose two distinct vertices u; and u j of C such that ui E Ic(vl) and u j E Oc(vt) with j - i as small as possible (mod r). Without loss of generality, let i = r , i 1 = 1 and j = s 1, that is u, E I(wl), u,+1 E O ( v t ) and {ul,. .. ,u,} i l [I(v1) U O(vt)]= 8. If I0c(vt)u Ic(vl)l = 1, without loss of generality, let (u,.} = Oc(vt) = Ic(v1). Then

+

+

(211,.

.

*

,us}

n [I(%) u O(vt)l = 0

where s = r - 1. Here, s 2 t, otherwise, the cycle uTvlPwtuS+lCu, would be longer than C. By 11 and I;'(vl) U Oc(vt) C_ (u,+1,. . ,u,., ul}, we have

.

by Theorem 2 and s 2 t ,

r 2 2k - (t - 1)i(t - 1) = 2k. IV. W e claim I(v1) C V ( C )and O(vt) E V ( C ) . Assume that Ip(v1) # 0, and vh E I p ( q ) . IV(P)I 5 k because IV(C)l 2 2k and p 5 3k. Hence, all of I c ( q ) , O c ( q ) , Ic(vt)and O c ( v t ) are not empty. Choose u; E Ic(v1) and uj E Oc(v1) such that j - i is as small as possible. By avoiding a cycle longer than C, j - i > l(mod r ) . And { U j + 1 , . . . ,ui-1) C T(v1). Relabelling, we can assume 01 E Tl and assume (uT,~1),(v1,u2)E A(G), 211 E Ti. Since u1 E TI and v2 6 TI, 0 2 E N(u1). If ( 2 1 2 , ~ E) A ( G ) , the cycle uTv1v2u1Cu,. would be longer than C. If (u1, w2) E A ( G ) , the cycle ulv2Pvhvlu2Cul would be longer than C. Similarly, Op(wt) = 0. V. W e claim t 5 1, and hence W is an independent set. Assume that t 2 2 and v1 E T,, vt E Tb. i) I;"(v1) \ I ( q ) # 8. Otherwise, Iz'(v1) 2 Ic(vt). By IV, IIc(v1)l = Ic. But k = IJ(vt)l 2 IIc(vt)l t I{vt-1}1 2 I g ( v 1 ) l t 1 = k 1, a contradiction. ii) Let ui+l E I$'(q) \ I(wt). ui+1 E Tb because, otherwise, ui+l E O(vt), and the cycle u;vlPvtu;+lCu; would be longer than C.

+

506

C.-Q. Zhang

iii) Choose { u j , , . . ,uj} as a maximum interval with initial vertex such that

ui

on C

{ui,.. uj } c I ( w ~u>I;'(w~). a

(Note, vj must exist because Oc(v1) # 0). Here, U j + l $? I(v1) U I&'(wI) and U j E I&"(v1) because of maximality. Since uj @ N ( v l ) , uj E T, and uj+1 E O(v1). Obviously, uj E I(v2) because uj-1 E I(v1). And {ui+l,.. . ,uj} n O(vt) = 0 because C is longest and t 2 2. iv) Let C' = uivluj+lCui = y1 . . .yT!yl and P' = ui+lCUjv2PVt. Here, O(vt) n V ( P ' ) = 0 because O(vt) G V ( C )and {ui+l,.. . ,uj} n O(vt) = 0. Let = \ Tb = ( 2 1 , . . . , z a } . Here, IOx(vt)I = k. v) We have that IX(ui+l) n Ozl(vt) = 0. Otherwise, let yc = zb E Ix(Ui+l)n O,'(vt) and let zb+l = yc+l or yc+2 E O x ( v t ) . Then the cycle C" = Ycui+lP'vt(y,+l)yc+zC'yc would be longer than C. vi) SO IX(ui+l) E X \ O i l ( v t ) . Since X = Nc~(ui+l), and O ~ l ( v t )G

x v(c')

OX(Ui+l),

k = IO(ui+l)l 2 IOx(ui+l)l+ lOP!(ui+1)12 IOx(vt)l+ 1= k

+ 1,

a contradiction.

Part two. In this part, we will find a cycle C' = u1.. . uT-lul of length r - 1, ( r is the length of the longest cycle in G) such that there is a vertex w' outside C' belonging to Ti, and all other vertices outside C' belong to Ti, and il # i ~ .

Proof: Let C be a longest cycle, W = G \ V(C). By part one, W is an independent set, by T ( W )we denote the part which all w E W belong to. Obviously, T ( W )n V(C) # 0 because N ( W ) = G \ T ( W )G V(C).

I. To show that, for any longest cycle C = u1 . . . uTul, there must be three successive vertices u;, ui+1,ui+2 on C such that they are not in T ( W ) . If not, for any pair of nearest vertices u i , uj E T ( W )n V(C), we would have Ij - il 5 3(mod r ) . Then

T

2 1.5(V(G)\ T(W)I = 3k 2 p ,

so C is a Hamilton cycle.

Hamilton Paths in Multipartite Oriented Graphs

507

II. Let C = u 1 . . . u,u1 be a longest cycle, w $! V ( C ) .For ui E V ( C ) n I ( w ) , let uj

= REI(w, u;,C)

if {ui, .. . , u j } n O(W)= 0, uj-1 E I ( w ) , uj E T ( w ) , uj+i E O(w). ( u j is the right end point of a maximal interval on C which contains u; but is disjoint from O(w)). Similarly, for ui E V ( C )n O(w),let ~h

if

= LEO(w, ui,C)

..., u i } n I ( w ) = 0, uh+1 E O(w), uh E T ( w ) ,uh-1 E I ( w ) . There must be a longest cycle C = u1. . . u r u l and a vertez w

{uh,

III. outside C and a vertez u j o n C such that T(w)= T ( u l ) and either (a). { u j , . . . ,uT} n [0(w) u O(u1)] = 0 and u j - i , ~ j E I(u1) n I ( w ) , uj+l E T(w);or (b). {u2,. . . , u j } n [ I @ ) u I ( u l ) ] = 0 and u j , ~ j +E ~O(W)n O ( u l ) , uj-1 E T ( w ) . We assume that the assertion is not true. By I, for any longest cycle C , we can label it so that C = vTwl . . . v,. and or, v1, v2 $! T ( w ) . Let T(w) = TI. Obviously, for q , w 2 $! V ( C ) ,either v,,v1 E O(w1) or v1,v2 E I(w1), and either or, v1 E O(w2) or v1,v2 E I(w2). i). If v1,v2 E I ( w l ) n I ( w 2 ) . Either 03 E I(wl)nl'(w2) or v3 E T I . So we can relabel C such that v1, v2 E I(w1) n I(w2), 03 E TI. Let v p = REI( w1, v1, C) and vq = REI(wz,v1, C), let p 5 q. The cycle wlvp+l . . . v,v1 . . . vp-lwl = u1 . .u,u1 with w = w2, uj = 212 satisfies the assertion (a). A similar proof holds for v,.,v1 E O(wl) n 0 ( w 2 ) . ii). We can assume that ~ 1 , 2 1 2E I(w1) and v,,v1 E O(w2). Among all longest cycles labelled as above, we choose C such that the subscript p of v p = REI(w1, v1, C) is as small as possible. Here, v1 and 212 E I ( w l ) , v3 E TI. On the new cycle C' = vp-lwlvp+l . . . vp-l, by considering the pair of vertices 202 and ' u p , and avoiding the solved case in (i), w l and v2 E I('up). Let 'up' = REI(vp, v1, C'). If p' > p or = w1, C' = 'up-lwlvp+l . . . vp-l = u,u1 . ..u, and w = ' u p , u j = v2 satisfies the assertion (a). So let p' < p . But this contradicts the choice of p and C.

.

IV. Now, we can prove the assertion required in this part. Let C be the longest cycle and w E W = V ( G )\ V ( C )such that (a) of IIIis satisfied. When j = r , C*= u,.-lul.. . uT-1 and w' = u,. satisfy the assertion. Since uj+1 E T(u1) and u, f T ( u l ) , j # T - 1. So we will only deal with the case of j 5 r - 2. Let C' = u j - 1 ~ 1 . ..uj-1 and P' = u j + l . . .uTw and Q' = V ( G )\ V(C') and W' = Q' \ V ( P ' ) = W U { u j } \ {w}. Here, Oq(w) = 0, by choice of

C.-Q. Zhang

508

C, hence, IOct(w)l = k. Let X = V ( C ' )\Ti = {xi,. . . , z a } . If I x ( U j + l ) n OX1(w) = 0, then I x ( U j + l ) E X \ OX1(w). Hence O i l ( w ) C O x ( U j + l ) But since

uj+2

EO~t(~j+l),

k = Io(Uj+l)l=

IOx(uj+l)l

t IO)dt(Uj+l)l 2 I O X ( W )t~ 1 = k t 1,

a contradiction. If zc E I x ( U j + l )

n oil(w), let U d = ZC E I x ( U j + l ) and U d + i or = zc+l E Ox(w). When u d + l E Ox(w), the vertices of W' outside the longest cycle u d u j + l . .. U r w u d + l . . .uj-1211 . . u d would belong t o two different parts T(w) and T ( u j ) , which contradicts Part one. When Ud+2 E o x ( W ) and '1Ld+l E TI, the Cycle ' t b d u j + l . . .U r w u d + 2 . . . U j - 1 U 1 . . . ud and the vertices of W'U { ~ d + ~outside } this cycle satisfy the assertion, here u j = w'. Ud+2

.

Part three. We willfind two disjoint cycles D' and D" such that ID' U D"I = r t 1, and there ezists x1 E D' such that either V ( D " ) n I ( q ) = 0 or V(D'') n O ( z 1 ) = 0, and V ( G )\ [ V ( D ' )U V ( D " ) ]E T(z1).

.

Proof: Let C* = 81.. V , . - ~ V ~ be the cycle we found in Part two, let W* = G \ V ( C * )and wo E W' such that wo E T2,W* \ {wg} T I . I. Since W' \ {WO} E N(wo), assume that (w0,w) E A(G) for some w E W' \ {wo}. We will show that W' \ {wo} E O(w0). (The proof holds symmetricly if (w,wo) E A(G) for some w E W* \ {WO}). Otherwise, let w' E I(wo),P* = w'wow. Since V(G)\[V(C*)UV(P*)]= W' \ {w',wo,w} E TI, O(w) U I(w') C V ( C * ) . Let X = V ( C * )\ TI = ( ~ 1 , .. . , 5 2 1 ~ - 1 ) . Obviously, X = V ( G )\ [TI U {wo}].Hence 1x1 = 2k - 1. IF(w')n Ox(w) # 0, because, otherwise,

2k - 1 = 1x1 2 I$(w')[

t IOx(w)l 2 2k.

Let zc E Iil(w')n O x ( w ) and let Vd = zc E ~ x ( w and ) Vd-2 or vd-1 = zc-l E Ix(w'). Then the length of the cycle V d - 2 ( D d - l ) W ' W O W V d . . .vd-2 would be r t 1 or r -/- 2, a contradiction.

II. We claim ~c+.'(zo) n ~ p ( w =) 0,for any w E W* \ {wo}.

If not, let Vi be a vertex in this intersection, then the vertices W*\ {w} outside the longest cycle V i - l w V i . . . Vi-1 would belong to two different parts TIand T(wo), which contradicts Part one.

III. We claim ~cf.'(wo) n o ~ ( w = ) 0,for any w E W*\ {wg}. If not, let vi be a vertex in this intersection then ~ ~ - 1 w O w 2. .1.~v ; - ~ would be a cycle of length r t 1.

Hamilton Paths in Multipartite Oriented Graphs

509

IV. We will show that O;[t(w) n T2 # 0. Icl(wo)nO;[t(w) = 0 because of 111.Hence, Ic+(wo)E V(C*)\OC:(w). If o,,'(w) n TZ = 0,~ ; [ t ( w ) g 0 p ( w 0 ) . Since ~ ( w5 )v(c*), l~;.'(w)l = k. But

k = IO(wo)l 1 IOc*(wo)l+ l{w}l L IO;!(w)l+

1 =k

+ 1,

a contradiction.

V. Let w; E O$(w) r l T2. w; E N ( w ) because wi E T2. Also, w; $ I ( w ) because of II. Hence w; E O(w).

VI. By 11,let V j = LEO(w, wi, C*), (see Part two, II.), that is, wj E TI and W j - 1 E I ( w ) and {wj,. . . ,wi} n I ( w ) = 0. Hence, { w j , . . . ,wi} \ 21 ' g O(w). Let Q = {wj, . . ,q},Here, w j + l E O(w).

.

VII. Since 200 E T2 and O(W)and 111.

wj

E 2'1, wo E N(vj).

(w0,wj) E

A(G) because of

wj+1 E

VIIL Igf'(w0) C TI. Let w, E Q n I(w0). Since (w0,wj) E A(G) and wi E T2, j 1 5 s 5 i - 1. And, w8+1 6 O(w)because of III and VI. Hence, %+l E Tl. IX. W e will show that 16'(w~) \ I(.;) # 8. If not, let I$'(wo) E I(wi). Let W" = W* \ { w o , ~ } . (Note that

+

wll

# 0.)

i). Either 21;-2 $ I(w0) or I(.;) n W" # 0. Otherwise, let w ; - ~ E I(w0) and I ( q ) n W" = 0. (i.e., W" O(w;)). Let C" = w;-2woww~+1.. .wi-2 = 21.. . zr-1tl and P" = wi-lwiw', where w' E W" O(wi). Here wi-1 E 21' because of VIII. Using the same argument as I a contradiction will follow. We have assumed in IX that ii). We claim IIQuw"(u)O)I< IIQ~WO(W;)~. Izl(wo) C IQ(w;). From I, Iw~(w0)= 0. From i), either 34-2 !$ I(w0) or Iw#t(wi)# 8. If wi-2 6 I(wo), but w;-l E I(wi), we get that IQ(w;)2 I,$'(wo) U {w;-1}. Hence, lIQ(ui)l

If Iw+%)

2

IIQ(w0)l

+ 1= IIQuW"(wO)(-k 1.

# 0, IIQUWtl(vi)l

> IIQ(Vi)l 2

lIQ(wO)l = IIQUW'j(wO)I*

+

iii). Let C" = wj-lww;+l, . , wj-1 = u1. ., u8ul (where s = r - i j - 1) and Po = wowj.. w;. Let X = V ( C o )\ T2 = (21,.. . ,xa}. iv). w e claim Ii'(w0) n Ox(w;) = 8. Otherwise, let Zb be a vertex in this intersection and let uc = 2 b E O(w;), u,-1 or u,-2 = zb-1 E I(w0). If

.

C.-Q. Zhang

5 10

.

uC-l E I( wo),uc-l, Wovj ., . v;uc . . uc-l would be a cycle of length T i1. If 2 b - 1 = uc-2 E I ( w o )and uc-l E Tz,the vertices of W”U{uc-l} outside the longest cycle uc-2wovj . . . v;uc . . uc-l would belong to two different parts, which contradicts Part one. v). Since Ii’(w0) n Ox(v;) = 0, Ox(v;) 2 X \ I$’(wo) and Ix(vi) 2 IJ’(w0).That is (Ic~(vi)l2 IIcO(w0)l. By (ii), k = II(v;)l = IIc0(v;)l \IQuw”(~);)~ > (Ico(wo)It IIQ”Wff(Wo)l = II(w0)l = k, a contradiction.

.

+

X . Let v,+1 E I;’(wo) \ I(v;). By VIII, vs+l E 2’1. Thus, v,+1 E N ( v i ) because v; € T2. Hence vd+l € O ( v i ) . Let D’= wvi+l.. v,wow = 21 . . . z , q (where w = 21) and D” = v;v,+1.. .vi = y1 . . . YbYl be two disjoint cy1. And cles and W” = V ( G ) \ [V(D’) U V(D“)]. Here, a i- b = T V(D/’>\ TI c O(z1).

.

+

Part four. Final step. Let D’and D“ be the cycles found in Part three, and let W” denote G \ [V(D’)u V ( D ” ) ] .Let ( D ’ n T l ) + l ={z;+llz; ~ V ( D ’ ) n T ~ , m oa}, d and

( D ” n Tl)-l = {yi-lIyi E V ( D ” ) n TI, mod b } . I. We will show that [V(Dlj), V ( D ’ )\ ( U n T1)+’] = 0. By induction. Since V ( D ” )\TI O(z1), we assume V ( D ” )n I ( z h )= 0 and prove that V(D”)nI(zh+l) = 0 when X h $! 2’1, or V ( D ” ) n I ( z h + 2 ) = 0 when zh E T I ,for h = 1,...a - 1. When zh $ TI, to show that (y;,zh+l) @ A ( G ) , for any yi E V ( D ” ) . If Yi+l $ T ( z h ) , by the assumption, (%Yi+l) E A ( G ) . so (Yi,%+l) 4 A ( G ) , because, otherwise, y;zh+l. shyi+l yi would be a cycle of length T t 1. If yi+1 E T ( z h ) # TI, then y;+2 4 T(zh). By the assumption, (sh,yi+2) E A ( G ) . So (yi,zh+l) $! A ( G ) , because, otherwise, the vertices W”U {yi+l} outside the longest cycles y;zh+l. .zhy;+2 . . y; would belong t o two different parts 2’1 and T ( z h ) ,which contradicts Part one. When zh E TI, to show that (y;,zh+2) @ A ( G ) , for any y; E V(D’l). Here, z h + l $ TI, because zh E TI. If y;+l E TI, then y;+2 4 TI. By the assumption, (zh,y;+2) E A(G). So (yi+l,zh+l) @ A ( G ) ,because, otherwise, yi+lzh+l . . .zhyi+2 . .y;+1 would be a cycle of length r + 1. Since yi+l E TI and zh+i C Ti, (zh+i,yi+l) E A ( G ) . Thus, (yi,zh+2) @ A ( G ) , because, otherwise, y;zh+2.. . zh+ly;+l . . .y; would be a cycle of length T 1. If yi+i $! TI, then Y;+I E N ( z h ) . By the assumption, ( s h , y ; + ~ ) E A ( G ) . So (y;,zh+2) @ A ( G ) , because, otherwise, the vertices W”u { ~ h + outside ~ }

..

. ..

.

.

.

+-

Hamilton Paths in Multipartite Oriented Graphs the longest cycle y;xh+2 . ..xhyi+l which contradicts Part one.

511

. .. yi would belong to two different parts,

11. W e will show that [V(D”)\ (D”n T1)-l, V(D’)] = 0. Since V(D”)\T~ G O ( x 1 ) ,we assume [(V(D“)\(D”~T~)-~]~I(Z~) = 0, and prove that [V(Dt’)\(D’W’l)-l]nI(~h+l)= 0, for h = 1,. . . , u - l . For suppose that (yj,Xh+l) E A(G). Here, yi+l 6 any yi E V(D”)\(D%T1)-’, TI. Then (xh,yi+l) $ A(G), because, otherwise, xhyi+l ., .y i ~ h +. .~. xh would be a cycle of length r 1. Hence, either T ( x ~ =) T(y;+l) or (yi+l, xh) E A(G). And (xh,yi+2) 4 A(G),because, otherwise, the vertices W”U {yj+l} outside the longest cycle 2hyi+2 . . . yish+l . .. X h would belong to two different parts TI and T(yi+l), which contradicts Part one. Thus, either T ( z h ) = T(yj+2) or (yi+2,xh) E A(G). Since T ( z h ) = T(y;+l) and T ( z h ) = T(yi+2) could not happen at the same time, either (yi+l,sh) E A(G) or ( ~ i + 2 , x h )E A(G). BY I, xh-1 E TI,and ( x h - i , ~ i + i ) E A(G), because y;+l $! 2’1. And, zh $ 2‘1 because xh-1 E TI. Then the vertices W“ U {xh} outside the longest cycle zh-lyi+l . . . y i ~ h +. .~.zh-lwould belong to two different parts, which contradicts Part one. Now, we have seen that if (yj,zj) E A(G), then both xj-l,yi+l E TI.

+

111. Let W1 = {w E W”lO(w)n V ( D ’ ) # 0}, and W2 = {w E W”lI(w)n V(D”) # 0). W e will show that Wl n W2 = 0. If not, let w’ E W1 n W2 and (yj,w‘), (w’yxj) E A(G). (x;-l,yj+l) 4 A(G), because, otherwise, ~ j - l y j +,~. yjw‘xi . . . xi-1 would not be a cycle of length r 2. Hence, either (yj+l,xi-l) E A(G) or T(yj+l) = T(x;-l). Also, ( x i - 2 , yj+i) 4 A(G), because, otherwise, xi-2yj+1. . .yjw‘xi.. . xi-2 would be a cycle of length r 1. Hence, either (yj+l,xj-2) E A(G) or T(~j+i= ) T(5i-2). If ( ~ j + i , x ; - ~E) A(G) , by I and 11,yj+2, xi-2 E Ti. However, (xi-2, y j + l ) E A(G) because of I and 2;-3 4 TI and yj+l $ 2-1, a contradiction. If T(yj+l) = T(xj-l), then T(yj+l) # T(xi-2). And, TI because hence (yj+1,5i-2) E A(G). By I a n d 11,yj+2,5;-3 E TI. xi-1 ~ j + 2E TIand yj+l $ TI and T(xi-I) = T(yj+l). By 11,(zi-1, ~ j + 2 E) A(G) because yj+2 E 2’ and 6 TI. Then z;-lyj+2.. .yjw’z,. . .x;-1 would be a cycle of length r 1.

.

+

+

+

IV. Finally, we obtain a contradiction by counting the arcs from D“ to D‘. (i). By the symmetry of DUW1 and D”UW2, we can assume I v ( B l ) u W ~ l5 IV(D”) U W2l. Let E’ = V ( U )U W1 and E” = V ( D ” )U (W” \ W1). And, let e = IE’I, and f = IV(D’) n T1I. Here, 2f 5 e 5

P- < 1.5k, 2 -

C.-Q. Zhang

5 12 (because 2 f 5 1D’I 5 IE‘I = e ) and

f5

2k IT11 = (n - I)’

(iii). By counting, I[E’,E’]I I k - regulari ty ). So

and I[V(G),E’]I= ke (because of the

I[E“,E’]l = I[V(G),E’]I- I[E’,E’]I 2 ke -

e ( e - 1) 2 .

(iv). By (ii) and (iii), we have that

We now consider two cases. (v). Case one. e I k . Let + ( e ) = ke - c(e-1) . # ( e ) = k - e + !j > 0, because e 5 k . So the minimum of this increasing function is at e = 2f, by (i). That is 21cf 5 4 ( e ) . NOW,

2k

Hamilton Paths in Multipartite Oriented Graphs (because f 2 0)

2k(because f 5

A)

513

2k 2k t l < f <(n - 1 ) '

(n- 1)

2k+1<-

4k (n - 1)

'2k,

(because n 2 3), a contradiction. (vi). Case two. e 2 k t 1. 2 e e 1 Let $(e) = ke - 9 (which is less than ke - -k$ ). #(e) = k - e 5 0 which implies that 6 is a decreasing function, and the minimum value of 6 is at e = 1.5k, by (i). Let +(f) = -f' t f. +'(f) = -2f t

[A] which implies that II,has maximum value at f = A.Thus

5 - -[ ( n k l ) ] ' +

1.5 - 1.125 <

[A]

(;)'

(because n 2 3), giving 0.375 < 0.25, a contradiction.

The proof of the theorem has been finished.

Acknowledgement Many thanks are due to Professor Brian Alspach for his guidance.

&

514

C.-Q. Zhang

References [l] B. Alspach, “Cycles of each length in regular tournaments,” Canad. Math. Bull. 10 (1967),283-286. [2]J. C. Bermond and C. Thomassen, “Cycles in digraphs - a survey,” J. Graph Theory 6 (1981), 1-43. [3]B. Jackson, “Long paths and cycles in oriented graphs,” J. Graph Theory 6 (1981), 145-157. [4]C. Thomassen, “Long cycles in digraphs with constraints on degrees,” Survey in Combinatorics, Proc. 7th British Combinatorial Conf., London Math. SOC.Lecture Notes 38,Cambridge University Press (1979), 211-228. [5] C.-Q. Zhang, “The longest paths and cycles in bipartite oriented graphs,” Journal of Mathematical Research and Exposition 1 (198l),

35-38. [6] C.-Q. Zhang, “Paths and cycles in oriented graphs,” Kexue Tongbao 26

(198l), 865-868.

Problems

A problem session was held during the conference a t Sandbjerg, Denmark, 2-7 June 1985, in memory of G. A. Dirac. The following is a list of the problems presented by the participants and later submitted in writing. Edited by W. T. Trotter, Department of Mathematics, Arizona State University, Tempe, Arizona, USA.

1. Let G have 2n vertices and minimum degree 6. If n necessarily have a bipartite [$]-factor? (Haggkvist)

5 6, does G

2. Let G be r-regular and have 2n vertices with n 5 T . Does G necessarily have a bipartite l r / 2 - 2 f i l o g rJ-factor? (Haggkvist) 3. Let G be an eulerian graph on n vertices and minimum degree 6. If G has lOOOk edges and 6 2 n/10, can G necessarily be decomposed into k cycles each of length lOOO? (Haggkvist)

4. Let G be a plane 4-regular multigraph. Suppose that G has a cycle decomposition S so that for every pair e , f of adjacent edges in a face of the drawing, we always have e and f belonging to distinct cycles in S. Call a subset S' C S a parallel class when any two cycles in S' have no vertices in common. Jaeger proved that if S can be partitioned into three parallel classes, then x(G) 5 3. Koester gave an example where x ( G ) = 4 and S consists of five mutually intersecting cycles. Question a): If S can be partitioned into 4 parallel classes, what can be said about x(G)? Is there a connection with the 4CT? Question b): If G is of the type described above, is G necessarily 1factorizable (conjectured yes by Jaeger)? (Problems by Sachs and Fleischner) 5 . Is it true that if x ( G ) 2 4, then G contains a subdivision of I i , in which each of the six paths has an odd number of edges? (Toft) 5 15

W. T. Trotter, ed.

516

6. Let G be an edge-critical graph with respect to chromatic number. Is it true that every pair of vertices is joined by an induced path on an odd number of edges? This is not true for vertex critical graphs in general, but every critical imperfect graph has this property. (Duchet and Meyniel) 7. For k 2 4, does there exist a graph G so that x ( G ) = k and x(G - z) = k - 1 for every x E V ( G ) ,but x(G - e ) = k for every edge e E E(G)? (Dirac, Toft)

8. Let H be a k-uniform hypergraph with chromatic number 3 and edges A l , A2, . . , A,. Let f ( H ) be the largest t so that it is possible to choose a set ( e l , e2, , , . ,e,} of two-element sets (edges) which form an ordinary graph of chromatic number t , and for every i, there exists a unique j so that e; c Aj. Then let f ( k ) = min{ f ( H ) 1 x ( H ) = 3 ) . It is easy to show f(k) 5 2k - 1. We cannot even prove f(k) 2 4. (Erdos and Nesetfil)

.

9. Suppose x and y are adjacent vertices in G and both G - x and G - y have A-free 2-factors. Does G necessarily have a A-free 2-factor (or a 2-factor with a t most one A and if this A exists, it contains the edge xy)? There is an obvious analogue for cycles of 4 edges rather than triangles. However, there are counterexamples for cycles of k edges when k > 4,e.g., the Petersen graph when k = 5 . (Hell and Kirkpatrick).

10. Let f l ( n )be the largest t so that there exists a graph G on n vertices so that the triangle degrees (number of triangles incident with a vertex) are distinct and all at least t . Estimate f l ( n ) . Let f2(n) be the largest t so that there exists a graph G on vertices v1, v2, ..., vn with respective triangle degrees A,, A2, . .., An so that I A; - Aj I 2 t whenever i # j . Estimate f2(n). Does there exist a regular graph in which all triangle degrees are distinct? Note: All known constructions for graphs with distinct triangle degrees include vertices with triangle degree 5 10 and pairs with I Ai - Aj I 5 10. (Problems by Erdos and Trotter) 11. The mean distance in a graph on n vertices is the sum of the distances between pairs of vertices divided by n(n - 1)/2. Let c be the least number so that for every graph G, there exists a vertex x so that: mean distance in G - x 5 c. mean distance in G Prove that c = 4/3 and that the extremal case is a cycle. The best result t o date is due to E. Gyori who showed that c < 3 - fi.(Winkler and Hell)

Problems

517

12. Let K be the least number so that for every graph G , there exists a partition of the vertex set V = V1 U V2 so that Vl and V2 induce subgraphs G1 and Gz of G so that the mean distance in both G1 and GI is at most Ir‘ times the mean distance in G. Note that I< does not exceed the constant c in Problem 11. The complete graph shows ‘ h 2 1. The Robertson-Wegner graph, a (5,5) cage, shows that K 2 404/399. This is the best known lower bound on K. (Winkler) 13. Let Q be a class of graphs with x(G) even for every G E 6. How difficult is it t o decide whether x ( G ) 5 4 for graphs in Q? Comment: The question x ( G ) 5 2 is in P, and the question x ( G ) 5 6 is NP-complete. (Kratochvil and NeSetfil)

14. Let G = ( V , E ) be a graph embedded on a rectangular grid and let a, b E V . How difficult is it to decide whether there exists a path from a to b so that the arcs corresponding to edges of G are noncrossing? (NeSetfil) 15. Let G = (V,E ) be planar, directed, and di-eulerian (indegree = outdegree for every z E V ) . Let F C E. Say that a circuit C is F-good if IC n FI = 1. Conjecture: There are IF1 pairwise edge-disjoint good circuits (j there is no set of IF1 - 1 edges covering all good circuits. Remark: True if each face is a directed circuit. (Frank)

16. Suppose G is connected and contains 2p vertices v1, v2, . . . , v2p and p(2p - 1) paths P;j so that P;j is a path from v; to vj whenever 1 5 i < j 5 2p. Suppose further that any two of these paths are disjoint whenever they do not share an endpoint. Conjecture: G is contractible into K p . (Duchet)

Added in proof. The decision problem of Problem 14 has recently been shown to be NP-complete by Anna Lubiw (Waterloo); see also I. Kratochvil and J. NeSetEl, “Planar subgraphs of topological layouts,” to appear. (Personal communication by J. NeSetEil.)

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