Theory and Practice of Combinatorics: A Collection of Articles Honoring Anton Kotzig on the Occasion of His Sixtieth Birthday (Annals of Discrete Mathematics)

A’(ro)=O. 0

Remarque. On peut considkrer un probleme plus general en supposant que D est forme d’entiers consecutifs distincts 2 c (c entier donne) (voir [2] pour le cas q = 1). Rogers (communication privee) a montre que le theoreme restait vrai pour de tels systemes. On peut aussi considerer des ensembles A ide tailles differentes (voir [l] et [12] pour le cas q = 1).

Un probkme combinatoire daniennes en radioastronomie I1

53

Remerciements Nous remercions particulierement D. G. Rogers pour ses remarques qui ont permis dameliorer Particle.

Bibliographie [ 13 J. Abrham, Bounds fot the sizes of components in perfect systems of difference sets, dans ce volume, pp. 1-7. [2] J. C. Bermond, A. E. Brouwer et A. Germa, Systemes de triplets et differences associees, dans: Coll. CNRS, Problemes Combinatoires et Theorie des Graphes, Orsay, 1976 (CNRS, 1978) pp. 35-38. [3] J. C. Bermond, A. Kotzig et J. Turgeon, On a combinatorial problem of antennas in radioastronomy, dans Combinatorics, Coll. Math. SOC.Janos Bolyai No. 18, Keszthely, 1976 (North-Holland Amsterdam, 1978) pp. 135-149. [4] J. C. Bermond, Problem dam combinatorics, Coll. Math. SOC.Jan& Bolyai No. 18, Keszthely, 1976 (North-Holland Amsterdam, 1978) p. 1189. [5] J. C. Bermond, Graceful graphs, radio antenna and French windmills, dans: Proc. One Day Conference on Combinatorics Open University, 1978, Research Notes in Mathematics (Pitman London, 1979) pp. 18-37. [6] C. Delorme, Two sets of graceful graphs, J. Graph Theory 4 (1980)247-250. [7] G. Farhi, These, Informatique, Universite Paris-Sud (1981). [S] P. J. Laufer, Regular perfect systems of differencesets of size five, dans ce volume, pp. 193-201. [9] K. M. Koh, D. G. Rogers, H. K.Teo et K. Y.Yap, Graceful graphs: some further results and problems. Research Report 91, College of Graduate studies, Nanyang U. Singapour (1980). [lo] K. M. Koh, D. G. Rogers et C. K. Lim, On graceful graphs: sum of graphs, Research Report 78, College of Graduate Studies, Nanyang U. Singapour (1979). [I 11 A. Kotzig et J. Turgeon, Perfect systems of difference sets and additive sequences of permutations, dans: Proc. 10th Southeastern Conference of Combinatorics, Graph Theory and Computing, Boca Raton (1979).Utilitas Math. Publ. Congressus Numerantium 24,629-636. [12] A. Kotzig et J. Turgeon, Sur I’existencede petites composantes dans tout systeme parfait densembles de differences, dans: Proc. Coll. Montreal, 1979, Annals of Discrete Math. 8 (North-Holland, Amsterdam, 1980) pp. 71-75. [13] D. G. Rogers, Addition theorems for perfect systems ofdifference sets, J. London Math. SOC.23 (1981) 385-395. [141 D. G. Rogers, A multiplication theorem for perfect systems of difference sets, Discrete Mathematics, to appear. Received 9 April 1980; revised 15 July 1980.

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Annals of Discrete Mathematics 12 (1982) 5 5 4 3

@ North-Holland Publishing Company

UN ALGORITHME DE CONSTRUCTION DES IDEMPOTENTS PRIMITIFS D’IDEAUX D’ALGEBRES SUR IF, Paul CAMION 3, rue F. Couperink 78370 Phisir, France

DCdie au Professeur A. Kotzig a l’occasion de son soixantieme anniversaire Nous donnons un algorithme de construction des idempotents primitifs de tout ideal de A = F,[XI,. . ., X,.]/(t,(X,), . . .,!,.(X,.)) dans le cas OU les lt(X,). i = 1 , . . . . r , ont des racines simples. Pour r = I , une particularisation de cet algorithme permet la factorisation de tout JCX) E IF,[X] quel que soit F, 4 impair. Toutes les operations se font dans A . Une variante de I’algorithme est alors obtenue pour q pair.

We give an algorithm for constructing the primitive idempotents of any ideal of A = F,[Xl,. . .,X,]/(t,(Xl)... ., !AXr)) in the case where all t,(X,), i = l . . . . . r , have simple roots. For r = 1, that algorithm allows the factorization of any f’(X) E F,[X] for whatever finite field IF, 4 odd. All needed operations are performed in A . A variant of the algorithm is then obtained for 4 even.

1. Introductioa 1.1. Apercu sur les algorithmes de factorisation d’un polynhme de F,[X]

Soient gl(X), . . . ,gk(X) les facteurs irreductibles def(X) E F,[X], F(X) de degrt d, et gl(al)=g2(a2)=.. . =gk(ak)=O.S’il existe i l , i2 tels que F,(ail)# Fq(ai2),on obtient un facteur de f(X) des que (f(X), h,(X))# 1, pour ht(X)=Xq’-X modf(X), t = 1,. . . , [d/2]. Sinon le plus petit t pour lequel (f(X), h,(x))=f(X)determinera le corps F,t= Fq(ai),i= 1, . . . , k. La methode de Zassenhaus [4] consiste alors a calculer pour s E IF,, pgcd(f(X), kj(X-s)), k,(X-s)z(X-s)’I modf(X), od j parcourt dans l’ordre decroissant les diviseurs de 4‘- 1. s’il existe i l , i2 tels que (ail-sy= 1 et (ai2-sy# 1, on obtiendra ainsi une factorisation de f(X). En particulier, si Ro est le sous-groupe des cards de Ft: et ai, -s E Ro, ai2-s @ Ro, une factorisation apparaitra pour j=(q‘ - 1)/2.Cette mtthode ne permet pas daboutir a coup stir. Soit, en effet, f(X)=fl(X)f2(X) E IF,(X), f l ( X ) = X 4 + X 3 + 2 et f2(X)= X4+X+2, fi(a1)=f2(a2)=0,al, a2 E FB1.On constate que a1 - s a le mCme ordre multiplicatif que a2 -s, pour s = 0, 1,2. La mtthode de Rabin [8] permet d’tviter cet inconvenient. Connaissant F,t, on calculera pour 6 E F,r le pgcd(f(X), (X - I), pour d=(qt- 1)/2 avec pour chaque choix de 6 une probabilitk proche de 1 -2-k, lorsque f(X) est separable, dobtenir une factorisation de f(X) dans F,t[X] (voir numtro suivant). Celui de plus petit degrt des deux facteurs ainsi determine sera a factoriser a son tour jusqu’ a l’obtention d u n facteur lineaire qui donne une racine de f(X) permettant de construire un de ses facteurs irreductibles sur F,. Dans la mkthode de Berlekamp, on construit une base N de

a =(g(X) I g4(X)=g(X)modf(X)). 55

56

P. Cumion

On a gq(X)-g(X)

=

n

(g(X)- s) mod f ( W

SEF4

et la factorisation complete de f ( X ) dans IF,[X] peut etre obtenue en calculant les pgcd(g(X)-s, f(X)), s E IF, g(X) E N . La mtthode que nous proposons repose sur le fait que pour g ( X ) E 28,w(X)=gd(X) pour d = (q - 1)/2, q impair, est la racine carree d u n idempotent de A = IF,[X]/(f(X)), et que l'un de (w(X)+ l,f(X)), ( w ( X ) - l , j X ) ) est un facteur non trivial de f ( X ) avec probabilite >+. Factoriser f (X), c'est construire les idtaux maximaux (ou minimaux) de l'algebre A dans le cas ou r = 1 (voir le resume). Dans le cas plus general d u n e algebre semi-simple, le probleme correspondant est celui de construire les idempotents primitifs de A , ou mieux, de tout ideal de A . L'algebre F,G ou G est un groupe abklien est le cas particulier pour A oh ti(Xi)= Xyi - 1, i = 1, . . . ,r. Dans le cas ou G est un e-groupe abelien elementaire, e premier, nous observions, [3], que les idempotents primitifs de IF,G, pour (q, e ) = 1, sont les Xg, ou s est la dimension de G sur IF, et ou E parcourt elements de la forme e-'+ l'ensemble des hyperplans de G , pourvu que q soit primitif modulo e. Les idempotents primitifs d u n ideal I = (x) de F,G sont alors ceux determines par Xg#O. les hyperplans E tels que x

xgtE

xgGE

1.2. Translations dans IF,m, carres et non-carres

Notons Ro l'ensemble Rin(Rj+I), i, j E {O, 1 ).

des carres de IF,m\IF,;

~l=IFIF,m\RouIF,.Soit A i j =

Onconsiderelespermutationsol:~-~-l,o,:5-5-' et 03:5-1 -5deIF,m\F,. Sur chaque cycle C de ol,on a Card(Aoln C ) = C a r d ( A l o n C ) .Donc Card A o l = Card A On voit alors que pour qm= 1(4), ozA1o = A 1 1 , qm-3(4), 02A01

(b- 1 = c ; h-I

qm-3(4), ~ 3 A o o = A 1 1 ;

((l-b)-l=

- I = -b-'C).

Et si

-b).

Donc si qm-3(4) les ensembles A i j sont equipotents de cardinal (qm-q)/4. Pour qmz l(4) il faut distinguer: (1) m-0(2), Card(AlouA,1)=(q"-1)/2, puisque F,*cRo, ensemble des carres de IF,*m. Card A

= Card

A l =Card A o l = (qm- 1)/4.

Ensuite Card Aoo= qm- q - 3(qm- 1)/4= (qm-4q + 3)/4. (2) rn = 1(2), alors les ensembles Aiiont donc pour cardinal (qm-q)/4. E R i , la probabilite pour que 5 -s E R j , j E {O, 1I, On voit donc que pour s E FI,: est Cgale a (a peu de chose prks pour qm= 1(4),m -0(2)). Mais ceci n'est pas verifie si I'on reste dans un cycle C de ol.Par exemple pour 5 racine de X 3 + X 2 - X + 1, 5 2 - I = ( 18, <18 - i = < 6 , p o u r q = 3 . Plus gkneralement, soit q = p , premier. A chaque cycle de o,,associons un mot

+

<

ldempotents primitit7 dideaux d'algibres sur IF,

57

circulaire dans l'alphabet {O, 1) correspondant a la succession des carres et des noncarrks dans le cycle. I1 y a (2,-22)/p+2 mots circulaires distincts dans {0, l} et le nombre de cycles de o1 agissant sur les classes de conjugues de F,m \ IF, est (p" -p)/mp, pour m premier. Cela signifie que pour m assez grand deux cycles distincts vont correspondre au m&memot circulaire et il existera deux polyn6mes irrkductibles dont le produit ne pourra pas etre factoris6 par la methode de Zassenhaus en faisant j = ( q - 1)/2 (voir 1.1) L'exemple que nous avons donne en 1.1 est plus precisement celui-ci. Soit f l ( X ) = X 4 + X3 2 et f 2 ( X ) = X 4 + X + 2 € f 3 [ X ] . Silesracinesde fl(X)sont a , , a ~ , a ~ , ~ r ~ ~ , c e f2(X)sont l l e s d e a2=a:3,a:9,a:7,a:1. al-1 =ay8;al-2=aT7. a i 3 - 1 = ~ r : ~ , a t1~ = ~ : ~ ; ; ~ , a ; eta:9 ~ , ~ sontprimitifs r : ~ eta:' et a:' sont d'ordre 20. Ici, pour f ( X ) = f l ( X ) f 2 ( X )et tout diviseur j de q" - 1= 80 on trouvera ( f ( x ) ,( X - - s ) j - 1) E {O, I}, s=O, 1,2.

+

2. L'alg4bre A = K I X 1 , . . . , X , ] / ( t l ( X l ) ., . . , t,(X,.)) L'algebre A a etC etudiee par Poll [ 6 ] . Celui-ci observe que lorsque ti(Xi) a des racines simples, i = 1, . . . ,r, alors A est semi-simple. Pour K = IF, il est immediat que t i ( X i )divise alors XYi - 1, avec (ni, q )= 1, i = 1, . . . ,r, et le treillis des ideaux de A est donc isomorphe au treillis des ideaux contenant ( t l ( X l ) ,. . . , t , ( X , ) )dans IF,[X1, . . . , X , ] / ( X l *- 1,. . . , X:.- 1) qui est elle une algebre semi-simple. K est un corps commutatif et les polynbmes ti(Xi) de degres respectifs di, i = 1 , . . . , r, sont skparables. Notons Hi= { a i l r . . ,a i , d i )l'ensemble des racines de t i ( X i )dans le corps de decomposition Ldenl.j,,.tj(Z),i=l, ..., r . L ' e n s e m b l e H l x H 2 x . * .xH,est dksignepar H e t pour h=(a, p, . . . ,y) E H , nous noterons f ( h )l'elkment f ( a , B, . . . ,y) E L, pour toutf de A. Nous notons simplement f pour f ( X l , . . . ,X , ) E A qui designera l'unique polyn6me de degre infkrieur a dien X i , i = 1, . . . ,r, de sa classe dans A .

Lemme. Soit f~ A. Alors f = O ssi Vh E H , f(h)=O. Le lemme se verifie par recurrence sur r en considerant les coefficients des termes en X , de f comme des elements de K I X 1 , .. . , X , . - l ] / ( t ~ ( X l .) .,,.t r - l ( X , - l ) ) .

Proposition 1 [ 7 ] . Pour tout h de H,l'application d@ne p a r f - f ( h ) est un morphisme la K-algi.bre A duns la K-algkbre L. Ensuite la transformation de Lagrange dkjinie par f-(f(h))hEH=f est un isomorphisme de la K-algibre A duns la K-algdbre LH. Le fait que la transformation de Lagrange soit injective decoule du lemme. Notons le prolongement de l'application f - f ( h ) a l'ensemble des parties de A . Le support de fest I'ensemble s ( f ) des h de H tels que f ( h ) # O . Le support s ( J ) d u n ideal J de A est I'ensemble des h E H pour lesquels oh(J)# (0). Desormais, K est un corps fini IF,.

(Th

Proposition 2. U n idkal J de A est minimal ssi s( J ) est minimal pour la relation &inclusion

P . Camion

58

des supports d'idtaux de A . Les supports d'idtaux minimaux de A sont deux a deux disjoints etforment une partition de H . Pour tout ideal J de A, dimKJ = C a r d s ( J ) . Soit t + 1 = q m le cardinal de L. Pour tout f de A, (f(h))'E (0, 1); f ' est donc un idempotent de A, par la Proposition 1. En particulier, f est une unite de A ssi s ( f )= H . Soit J un ideal de A tel que s ( J ) n'est pas minimal. I1 existe alors 0 i f l , fzE J et h € 4 4 tels que fl(h)#OO,fZ(h)=O, donc s ( f l ) # s ( f Z ) . Soient u et u respectivement les idempotents et fi. Alors u # v ; u-uv et u-uv sont deux idempotents orthogonaux, tout deux dans J . L'idCal J n'est donc pas minimal. Reciproquement, si s ( J ) est minimal,

f:

vf, 9 E J\10):

s(f)=s(g).

Alors f = g ( g ' - l f ) , par la Proposition 1, et J , engendre par l'un quelconque de ses elements, est minimal. Soient J1et J 2 deux idkaux minimaux distincts. Alors J I J z c J 1 nJz={O).Par la Proposition 1, s(Jl)ns(Jz)=O. Notons u l , . . . ,U k les idempotents primitifs de A et v = 1 ui. Alors s(u)=0, sans quoi l'un des ui serait tel que s(ui)cs(u),et l'on aurait uiu#O. Donc . _ < i S k ~ i =1. L'image de Lagrange (&) de (ui) a la m&me dimension sur K que (ui). Alors dimK(ui)
-zlcick

1,

Card s(ui)3 Card G'= dim, Lh = dimK(&)= dimK(ui). Mais,

Vf E A ,

Par consequent dimK(ui)=cards(ui),i = 1 , . . . , k. De meme, pour tout ideal J de A,

C uiJi(0)

uiJ=

0 ( u ~ J ) =0 u,J#{OI

(~i).

uiJf(O1

Donc dim, J = Card s( J ) .

3. Un algorithme de construction des idempotents primitifs de tout ideal J de A 3.1. L'espace de Berlekamp

Soit K = [F, un corps fini de caracteristique p > 2. L'espace 9 de Berlekamp est le Par la Proposition 1. K-espace vectoriel forme des elements g de A tels que 94'9. c'est le K-espace vectoriel des elements g de A dont les coefficients de Lagrange g(h),h E H , sont tous dans K.

Idempotents primitif:?d'ideaux d'alg8bres sur IF,

59

Proposition 3. La dimension de 9 sur K est le nombre d'idempotents primit!fs de A. Plus prbcisement, 9 est sous-tendu par les idempotents primitifs de A. Le K-espace sous-tendu par les idempotents primitifs de A est contenu dans 9,car

(1

1

liui>"=

liui, lorsque li E K , i= 1,. . . , k .

lcick

12i2k

D'autre part, V g E 9, lei,
lcick

ou les Ii sont les diffkrents coefficients de Lagrange de g qui sont dans K . On demontre de m&me:

Corollaire 1. Soit u un idempotent de A, somme de j idempotents primitifs. Alors dim, u 9 =j. En effet u9=

C

Kui

up# 0

Corollaire 2. Soit g E u 9 \ ( 0 ) .Alors w = g4- est un idempotent de A tel que wu # 0. 3.2. Calgorithme

3.2.1. Une base de 9. On construit la matrice carrCe M indexbe par les r-uples ( i l , . . . , ir)ou i j < d j , j = 1,. . . ,r dont les lignes sont les coefficients de xp . . . xgi, mod ( t l ( X l ) ,. . . , tAX,.)). L'espace orthogonal aux colonnes de M -I determine 33'. Soit N une base de 33'.

3.2.2. Une base du K-espace sous-tendu par les idempotents d'un ideal ( f ) de A. Les idempotents primitifs de ( f ) sont dans ( J ) n B .Une base de ( f ) peut &treextraite du systeme

{ x$ . . . xtrf mod (tI(X1h..

. 3

tr(Xr))lij
On formera alors une matrice M' donc les colonnes sont celles de M - I et les elements &une base de (f)i. L'espace orthogonal aux colonnes de M' determine ( f ) n 9 . Soit ( w l , . . . , w J = N ' une base de ( f ) n 9 .

3.2.3. Construction de tidempotent ginerateur u de ( f ). Nous supposons ne pas connaitre le corps de dkcomposition L. Par le Corollaire 2, ui= wf-', i= 1,. . . , I est un idempotent de ( f ) .(Pour q grand, la probabilitk que ui= u, i= 1,. . . ,I, est proche de 1.) Si u1 # u, donc si ul.f#,f, il existera une suite uil, . . . , uij, 1= il < . . . < i j , telle que la suite des ensembles U b < a S ( u i b ) , a= I , . . . ,j, est strictement croissante et U b Q js(uib)= puisque s ( u ) = U 1 < i < l s ( u i ) .

~

(

~

1

3

P. Cumion

60

Par recurrence, supposons que nous ayons construit un idempotent vf,avec s(v:Il>= alors ui,+ I = u, sera dktermine par le plus petit c> i, tel que u,u~,# u,. Lidempotent sera uf,+ u,-u~,u, dont le support est egal a U b < a + ls(uib). Notons que dans le cas ou A=lF,G, G groupe abelien, on connait le corps de t i ( Z ) .Alors u = f f , pour t=q"- 1. Dans ce cas, N' decomposition L=[F,m de est une partie maximale libre de u N . UbQ,s(Uib);

nIGi,,

3.2.4. Construction des idempotents primitifs de (u). Soient u l , . . . , ui, i idempotents u l = u. On a alors deux a deux orthogonaux avec

El

@ uja?=ua?, Isj,
Si i,2. Soit y E u j B . Alors ~ = g ( q - l ) / ~est la racine carree d u n idempotent et a donc ses coefficients de Lagrange dans {l, - 1,O). Pour chaque tirage de g dans u p , la probabilitequew= +ujest2-"+'.Si w # fuj,alorsw'=w(w+1)/2et wf'=w(w-1)/2sont deux idempotents orthogonaux de m&meque u;= w' et u; = u j - w'. Lensemble { u l , .. . , u j , u !,j , . . . , u i ) comporte alors i+ 1 idempotents deux a deux orthogonaux dont la somme est u. I

4. Factorisation d'un polyn6me f ( X ) E [F,[X] On se ramene classiquement au cas d u n polynbme a racines simples en calculant f'(X)/P~Cd(f(X), . trouvera les On se trouve dans le cas particulier ou r = 1 et f ( X ) = t l ( X l ) On idempotents primitifs de A par I'algorithme expose. uj= 1. Les facteurs irreductibles de f ( X ) sont les Soit alors pgcd(Cjii u j , f ' ( X ) ) , i = 1,. . . , k .

f'(m

zlcjck

Exemple. Soit a factoriser X 2 -2 sur [F, , [ X I . 2 etant un residu mod 17, on voit que LB=A. Ici ( q - 1)/2= 8. O n calcule dans A :

( X + I T = -1, (X+2)8=1, (X+3y=3X, (3X+1, X2-22)=X+6; X2-2=(X+6)(X-6). On a les deux idempotents primitifs 9(3X+ 1)= 1OX+9 et 9(1-3X)=9 - 1OX. Remarquons que ( X + 2)4 est racine de I'unite. ( X 2)4= - 3 X mod ( X 2- 2) donne tgalement la factorisation de X 2 -2.

+

5. Construction des idempotents primitifs d'un ideal de A ou factorisation d'un polynome (cas ou Y = 1) lorsque IF, est de caracteristique 2

S.1. Introduction Les resultats des numeros 2 et 3 restent valables lorsque q est une puissance de deux jusque et y compris le paragraphe 3.2.3.

61

ldempotents primitijs crideaux d'algPbres sur IF,

Ensuite, l'idempotent u genkrateur d u n ideal donne ( f ) de A etant obtenu par 3.2.3. la methode proposee pour obtenir les idempotents primitifs de A, u l r . . , uk tels que g I b k u I = u ne pourra pas &re appliquee. En effet, cette methode est basee sur le fait que pour q impair, pour tout c1 E IF, ~ 1 ' " ~ ) ' ' E { 1, - 1 , O ) . Nous allons exploiter ici le fait que lorsque 6, est une extension de degrt pair de IF, pour tout c1 E IF, a@EF I .,

c,

5.2. L'algorithme 5.2.1. Les donnees. Notons 0, I, y, y2 les elements de 64. Soit K = 6 , et K'= K(y). Si IF, est une estension de degre pair de F I , alors K' = K, sinon K' est l'extension de degrk deux de K. Soit u l'idempotent gknkrateur de l'ideal ( f ) donne de A et N' une base de ( f ) n a = US?.N' et u sont obtenus par les constructions donnees en 3.2.2 et 3.2.3. 5.2.2. Construction des idempotents primitifs de (u). Soient u l , . . . ,ui, i idempotents deux a deux orthogonaux dont la somme vaut u. On a alors

@

ujS?=ua.

l,<jsi

Si i < dim, US? = k, il existe alors j E [1, i ] tel que dim, u p = rang, ujN' = a 3 2. De l'ensemble ujN', on extrait une base N" de u p . Soit E le K'-espace vectoriel sous tendu par N". Si la decomposition en idempotents primitifs de u j s'tcrit U j , 1 + . . . + u j , r = u j , 1 2 2 , on a

E = @ K'uj., 1SSSI

Soit alors g E E, g = c l 6sg ct,uj,,. Notons h l et hl deux ClCments de H,tels que respectivement ~ ~ , ~ (1 het~~ ) ~=, ~ (1.hDes ~ ) lors, = g(hl)=ctl et g(hI)=aI.Puisque a, E K', a! E IF4, pour d = (Card K' - 1)/3 = (q' - 1)/3. La probabilite que ct: # a: est

qui est proche de 3. Soit w = gd. Alors w(hl )= a: et w(hr)= a:. w fluj a un coefficient de Lagrange nu1 en l'un de h l ou hl pour p E F4. Si a{ # ut, alors, pour un tel p, uJ= (w p 1 . 4 ~et ) ~ uy = uj+uJ sont deux idempotents orthogonaux. L'ensemble { u l , . . . , us, u;, . . . , ui} est formb de i'+ 1 idempotents deux a deux orthogonaux dont la somme est u.

+

+

Remarque. Lutilisation de IF4 ne s'impose pas. Ce qui importe ici est que IF; soit un sous-groupe de K'. On peut employer toutefois n'importe quel sous-groupe C de K' d'ordre (q' - l)/d de fagon que pour ct, E K', af E C , C ayant un petit ordre. Par exemple, ce qui permet de faire K'= pour q'=512, C serait le sous-groupe dordre 7 de K=IF512.

62

P. ('trrnion

6. Factorisation d'un polyn6mef(X)

6

IFAX]

Comme au numero 4, on se ramene au cas ou f ( X ) n'a que des racines simples. Notons que f ( X ) peut avoir ses coefficients dans un sous-corps de IF,. En particulier IF, peut Ctre le corps de decomposition de f(X), auquel cas l'algorithme produira une ou les racines de f(X). Si, en general, il n'est pas necessaire de connaitre le corps L de decomposition def(X), en tous cas, le corps IF,= K , corps intermediaire au corps des coefficients de j ( X ) et a L doit Ctre donne. On se trouve dans le cas ou r = 1 et f ( X ) = t l ( X l ) .Les idempotents primitifs determineront les facteurs def(X) irreductibles dans IF,[X] comme au numero 4. 6.1. Exemples

6.1.1. Soit afactoriserf(X)=X2+aX+a2dansIF,,[X],ouaest racinedeX4+X+ 1. Ici L= K =IF,, q= 16, 8 = A et K ' = K , puisque IF, est une extension de degrt pair de IF2. L'ideal dont nous voulons trouver les idempotents primitifs est A tout entier, ou l'on note la classe de X dans A , est une base N de g. donc u = 1, et { 1, Soit g = +a, tire au hasard dans 8.On calcule

xj,

x

x

( X +a)5 =a4X mod f(X). Le sous-corps IF4 de IF1, est {0, 1, a5,a"). O n calculera donc pgcd(a4X+ I , j ' ( X ) ) =

x s a ' ' ,d'ou f(X)=(X+a'')(X+a6).

Les idempotents primitifs de A sont alors (1 + 0 1 ~ x ) ~ = a ' ~ et + c1t+a~''~ ~

+ a'4X = a s + a'4X.

6.1.2. Soit a factoriser X 2 +a5'X + a s 8 dans lF64[X], ou a est racine de X6 + X + I . Nous utilisons ici la representation matricielle de A et la table des logarithmes de

Zech.

Calculons ( M x + a 1 ) 2 ' , ce qui ne donne pas de resultats. Essayons alors: ( M x ~ ' 1 =) ( M ~ x i 6 ~ r ~ ~ 1 ) ( a81)( M p M x a21)

+

'

-IM9 -

+

@3

c1 M61/-/127 @4

+

a53/-la~ a22

a2

+

M58 a48 a40/=a12

c17 a14/,

Donc ( X + M ~ ~) ~a ' " X + amod ~ ~f(X). Ici IF4={(), I , a 2 ' , a42).On calcule 1 = c ~ ' ~ X + et a ~on , constate que X + C ~ = X +nedivisepasf(X). ~ ~ ~ On calcule alors 0 1 ' ~ X + c i ~ ~=+ ~c ~l '~~' X + X a~ + a~2 .7 divise en effet f ( X ) = (X+LY27)(X+a31).

Puisque a 1 2 * a 3,a39 1 = a’, les idempotents primitifs sont alors d 2 X + a39 et a 1 z X + a 3 9 1= a I 2 x +a43.

+

6.1.3. Donnons finalement un exemple verifiable a la main de factorisation dans F,[X] oh F, est une extension de degre impair de F2.

Soit fl une racine de X 3 + X 2 + 1 et f ( X ) = X 2 + P6X + P le polynijme a factoriser dans F8[X]. Lalgorithme n’est pas signiflcatif ici, il le serait pour IF,= 6 2 oh ~ p est un grand nombre premier. ICi K ’ = K(y)= F64. Nous identifions IF8 au sous-corps de 8 ClCments de 6 6 4 par fl =a9. Comme preddemment, on calcule M x . M X 2 , MX4,MX8=M x , M X I 6 =M x 2 . O n n’obtient pas de resultat en calculant ( X + U ) ~ ’Des . lors, on calcule ( M x + a 2 1 ) 2 1= ( M x 2+a321)(MX4+ a 8 1 ) ( M x + a 2 1 ) a24 =Ia54

a14

a45

la36

a 4 ~ l1 0 .

a2

a9

a44

a16(=(19

a40 a4

1’

Donc (X+ a’)’ = a9X + 0 1 ~ ~ . On trouve alors que a 9 X + a44 + a2’ = a9X + a36,de mCme que a9X + a44+ a42= a 9 X + a s 4 divisent f ( X ) qui vaux donc ( X +a2’)(X + U ~ ~ ) = ( X + P ~+B’). )(X Puisque a9 -a4’ + a36= 1, a9X + a36= PX + P4, et PX + f14 + 1 = PX + P“ sont Ies idempotents primitifs de A.

References [ I ] E. R. Berlekamp. Factoring polynomials over finite fields. Bell System Tech. J. 46 (1967) 1853 1859. [2] E. R. Berlekamp. Factoring polynomials over large finite fields. Math. Comp. 24 (1970) 713 735. [3] P. Camion, Codes quadratiques abeliens et plans inversifs miqueliens. C.R.Acad. Sc. Paris. t. 284 (6 juin 1977). [4] D. E. Knuth. The Art of Computer Programming. Vol. 2: Seminumerical Algorithms (AddisonWesley, Reading. MA, 19711. [S] F. J. MacWilliams. The structure and properties of binary cyclic alphabets. Bell System Tech. J . 44 (1965) 303-332. [6] A. Poli. Codes dans certaines algebres modulaires. These de doctorat en Sciences. Toulouse (197x1. [7] A. Poli, Un groupe dautomorphismes dalgebre de groupe aMlien dans: Actes du Colloque ‘Permutations’ (Gauthier-Villars. Paris. 1972). [XI M. 0. Rabin, Probabilitic algorithms in finite fields. MIT/LCS/TR-213 (Jan. 1979). [9] P. Camion, Un algorithme de construction des idempotents primitifs dalgebres sur F,. C.R. Acad. Sc. Paris, t. 291, skr. A ( 1980). 479. Received 22 Augusr 1980: r c t i w i l 9 Jutluurj, 1981

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Annals of Discrete Mathematics 12 (1982) 65-78

@ North-Holland Publishing Company

KOTZIG FACTORIZATIONS: EXISTENCE AND COMPUTATIONAL RESULTS Charles J. COLBOURN Department of Computational Science, University of Saskatchewan, Saskatoon. Saskatchewan. S 7 N OWO. C a n a h

Eric MENDELSOHN* Department of Mathematics. University of Toronto. Toronto, Ontario, M S S I A l . Canadu

Dedicated to Professor A. Kotzig on the occasion of his sixtieth birthday A new combinatorial configuration called a Kotzig factorization is introduced, and its relationships to Room squares, Howell designs, perfect I-factorizations. and Kirkman triple systems are discussed. Kotzig factorizations are constructed for prime orders and for all orders up to 43. Various restrictions on the automorphism group, together with some interesting computational techniques, are employed in the construction of the small orders.

1. Definitions and related work There has been much interest in combinatorial design theory in the notions of 1-factorizations [4, 51, (double) resolvability [36, 37, 421 and Hamiltonian cycles [22, 24, 251. Perfect 1-factorizations [4, 22, 231 combine the first and third of these; Room squares [42] combine the first and second. We examine a configuration incorporating all three notions, which we call Kotzig factorizations. Recall that a l-factorization ofjhe complete graph on 2n vertices is a pair ( V F(i)), where V is a 2n-set of vertices and F(i) is a partition of the edges of the complete graph into 1-factors. A near l-factor is a set of disjoint edges spanning 2n vertices of a (2n+ 1)-vertex graph; generalizing, a near l-factorization of the complete graph on 2n + 1 vertices is a pair (V F(i)),where V is a (2n + 1)-setof vertices and F(i) is a partition of the edges of the complete graph into near 1-factors.Finally, recall that a Harniltonian decomposition of the complete (2n+ 1)-vertex graph is a pair (V H(i)), where V is a (2n + 1)-set of vertices and H ( i ) is a partition of the edges of the complete graph into Hamilton cycles. An Kotzigfactorization is a triple (V F(i), H G ) ) , where V is a (2n+ 1)-set,( V , F(i)) is a near 1-factorization, ( K H U ) ) is a Hamiltonian decomposition, and each near 1-factor F(i) intersects each Hamiltonian cycle H U ) in exactly one edge. We shall refer to the near 1-factors as colours; thus, a Kotzig factorization is a colouring of a Hamiltonian decomposition so that each Hamiltonian cycle contains exactly one edge of each colour. Since each edge has a unique colour, and belongs to a unique cycle, we may assign the edges coordinates in an n by 2n+ 1 array. More *Research supported under NRC Grant Number A7681 65

66

C . J. Colbourn. E . Mendelsohn

formally, an AK(2n+ 1)-array, or simply AK(2nf l), is an n by 2n+ 1 matrix whose entries are unordered pairs of elements chosen from a (2n+ 1)-set, for which each unordered pair appears in precisely one entry, each column contains edges forming a near 1-factor, and each row contains edges forming a Hamiltonian cycle. An AKarray leads trivially to a Kotzig factorization, and conversely. We remark that the formulation as AK-arrays brings out a similarity to Room squares and other doubly resolvable designs such as Howell designs [37], generalized Room squares [36], and Steiner tableaus [26]. Prior to pursuing AK-arrays, it proves worthwhile to phrase a more general problem. Following [28], we first define Q = { G(u)}, u = 1 (mod 2), to be a family of graphs with u vertices and u edges. We then define a Q - AK(u) array to be a (u - 1)/2 by u matrix whose entries are unordered pairs of elements chosen from a u-set, for which each unordered pair appears in exactly one entry, each column contains edges forming a near 1-factor, and each row contains edges forming a graph isomorphic to one of the { G(u)}. One can then ask the (very general) question: for which Q and u does a Q-AK(u) array exist? A sense of balance dictates that we consider the cases when Q contains just one graph for each u, and this graph is regular. Under these circumstances, Kotzig factorizations are the instance when Q(u)is a Hamiltonian cycle. At the other extreme, Q is a collection of disjoint triangles; in this case, we have a Kirkman array. Lemma 1.1. A Kirkman array exists ifand only $a Kirkman triple system does.

Proof. Let a Kirkman array be given. The edges of each row form a parallel class of triples. As every unordered pair occurs in exactly one entry of the array, the set of all triples forms a STS. Conversely, a Kirkman triple system is given on a set V= { 1, . . . ,6n + 3) with parallel classes { P(i), 1< i < 3n + 1 }. Now define an array in which the (i, j) entry is { u, u } , where { u, u, i} is the triple in parallel class PG). Each row is automatically a disjoint union of triangles, and since Kirkman triple systems are STS the columns form a Steiner near 1-factorization. 0 In this case, the existence question is completely settled: Theorem 1.2 [33]. A Kirkman triple system exists for all u = 3 (mod 6).

Returning to our original problem, we state: Conjecture 1.3. An AK(u) exists for all u = 1 (mod 2).

We note in preface that the problem has the intrinsic difficulty of dealing with Hamilton cycles, and is therefore not amenable to solution by recursive constructions involving subdesigns, pairwise balanced design constructions, or direct products. These are three of the most powerful weapons in the arsenal of graph design theory, and thus one might guess that the existence problem for Kotzig factorizations is indeed hard. This suggests one similarity with the quite different existence problem for perfect 1-factorizations, where only two infinite families (twice a prime, a prime plus one) and some sporadic results are known a t present.

Kotzig factorizations

67

The remainder of this paper deals with an infinite family of Kotzig factorizations (the primes), and some interesting computational methods with which Conjecture 1.3 has been verified for u < 43.

2. Existence and enumeration of AK(p), p prime Our primary concern is to determine for which u an AK(u) exists. In this section, we give a construction for an AK(u)when u is an odd prime, using well-known results. In order to do this, we first recall an existence proof for near 1-factorizations.

Lemma 2.1. A near l-factorization of order 2n+ 1 exists for all n. Proof. We recall the construction of the well-known near 1-factorization denoted GK(2n+ 1) (see [16] and references therein). Its near 1-factors are { ( j + i , j + 2 n + 1- i ) ) 1< i < n } ,

for O < j < 2 n

with arithmetic performed mod 2n+ 1. These near 1-factors are edge-disjoint, since in the pairs { ( i , 2n + 1 - i ) ) each difference from 1 to n occurs in precisely one pair. 0 For the second component of the Kotzig factorization, we require Hamiltonian decompositions of odd prime order.

Lemma 2.2 [I]. A Hamiltonian decomposition of prime order 2n + 1 exists. Proof. We construct a Hamiltonian decomposition of prime order 2n + 1, denoted HD(2n + 1). This decomposition has n Hamilton cycles; the i-th cycle contains all edges of the form (x, y ) for which x -y = i (or y - x = i), arithmetic mod 2n + 1. These cycles are indeed Hamiltonian, whenever 2n + 1 is prime. 0 These two lemmas provide us with Kotzig factorizations:

Theorem 2.3. For 2n+ 1 a prime, an AK(2n+ 1) exists. Proof. We claim that ({0,1,2, . . . , 2 n } , GK(2n + l), HD(2n + 1)) is a Kotzig factorization when 2n + 1 is prime. We need only show that each near 1-factor intersects each Hamilton cycle in at most one pair. To see this, note that each near 1-factor contains exactly one pair with a specified difference d, for each 1
u:=,

c‘. J . Colbourn. E . Mendelsohn

68

SP= {(a(i),b(i))),we obtain a cyclic near 1-factorization of order 2n+ 1 ( ( a ( i ) + j ,b(i)+j)1 1 < i < n } ,

for O b j < 2 n ,

arithmetic mod 2n+ 1. This near 1-factorization has an automorphism carrying i to i + 1 (mod 2n + 1 t -hence the term cyclic.

Lemma2.4. For any cyclic near l-fuctorization CF of’ prime order 2n+ 1, (i0, 1 , . . . , 2 n ) , CF, H D ( 2 n + 1)) is an A K ( 2 n + 1).

0

Proof. As Theorem 2.3.

+

Lemma 2.4 gives us a method for constructing many nonisomorphic A K ( 2 n 1). Skolem n-sequences provide near 1-factorizations of order 2n + 1 ; these sequences exist whenever n -0, 1 (mod 4 ) [30, 35, 391. Enumeration of Skolem n-sequences [2, 12, 19, 271 has suggested that the number of distinct Skolem n-sequences grows exponentially in n. Distinct sequences do not necessarily lead to nonisomorphic near 1-factorizations, but the rapid growth in the number of Skolem sequences certainly suggests a rapid growth in the number of cyclic near 1-factorizations, and hence (by Lemma 2.4), the number of Kotzig factorizations of prime order. Cyclic near 1-factorizations can also be constructed from Steiner triple systems (STS). Given an STS ( K B), its Steiner near l-fuctorization has the near 1-factors ( ( y ,4 I (x, y, 4 is in BS,

for all x in V A Steiner near 1-factorization is cyclic exactly when the STS is cyclic. Cyclic STS are discussed extensively in [lo, 111; we recall relevant results here. Cyclic STS exist for all u - 1, 3 (mod 6), 0 # 9 [31]. In fact, Bays [6] showed that when u = 6n + 1 is prime, the number of nonisomorphic cyclic STS grows exponentially in n. (Johnsen and Storer [ 2 0 ] have extended Bays’ result to non-prime orders, but for our purposes the prime case suffices.) We conclude from Bays’ result and Lemma 2.4:

Lemma 2.5. The number ofnonisomorphic AK(6n + 1 ), 6n ally in n.

+ 1 a prime, grows exponenti-

Both the existence and enumeration given here depend on the existence of HD(2n+ I), which in turn relies on the primality of 2n+ 1. To handle non-prime orders, we must employ a different Hamiltonian decomposition; this is the topic of the next section. 3. Colouring hamiltonian decompositions 3.1. Sesquirotationul Kotziy,fuc.tori=utions

In Section 2, we have shown that for a particularly nice Hamiltonian decomposition, any cyclic near 1-factorization is a colouring. Extending a cyclic near 1-factorization to a 1-factorization gives an automorphism with one fixed point and a

69

Kotzig factorizations

single cycle on the remainder. Kotzig [22] terms these 1-factorizationspyramidal; an alternate term is starter induced [3]. In this section we consider near 1-factorizations which are 1-rotational,i.e. which have an automorphism consisting of one fixed point and a single cycle; these arise from 1-factorizationswhich have two fixed points and a single cycle, i.e. which are bipyramidal[22] or even starter induced [3]. A near 1-factorization is 1-rotational if it can be given in the following form. The vertex set is (0, 1,. . .,2n-1, a), and the near 1-factors are ( ( 0 0 , i(O)+j), (i(l)+j,k(l)+j), . . . ,(i(n)+j, k(n)+j)} for O<j<2n, arithmetic mod 2n, and thefixed near 1-factor ((0,n), (1, n+ l), . . . ,(n - 1,2n- 1)). When one examines the near 1factor obtained by settingj=O, every vertex but one appears in an edge-this is aptly called the missing vertex. We will define a special class of Kotzig factorizations whose near 1-factorizations are 1-rotational; first, we specify an additional structure for the Hamiltonian decomposition. A Hamiltonian decomposition is semirotational if it can be given in the following form. The vertex set is (0, 1,. . . ,2n - 1, m}, and the Hamilton cycles are ( 0 0 , j, i(l)+j, i(2)+j,. . . ,i(2n- l)+j}, for 1 <j
+

+

(v

Lemma 3.1. If there is a sesquirotational Kotzig factorization F, A(n)), there is a sesquirotational Kotzig factorization (K F’, B(n)),and conversely.

Proof. The longest edge always belongs to the fixed near 1-factor. Hence, interchanging it with Cm -n does not change the fact that we have a sesquirotational Kotzig factorization, but does change A(n) to B(n). 0 From a given sesquirotational Kotzig factorization, we can use other transformations to obtain further Kotzig factorizations. For both 1-rotational near l-factorizations and semirotational Hamiltonian decompositions, we can map each vertex x to mx, whenever m is a multiplier, i.e. (m, 2n)= 1. This always yields an isomorphic configuration; in the event that it produces an identical one, the mapping is called a multiplier isomorphism (or automorphism) [lo].

70

C . J . Colbourn, E . Mendelsohn

Fig. 1. A(11).

Fig. 2. B( 11)

.

Kotzig factorizations

71

For Kotzig factorizations we introduce a more general concept, allowing for the independent action of multipliers on the near l-factorization and on the Hamiltonian decomposition. A Kotzig factorization (V; F , H ) is multiplier isotopic to (V; F', H ' ) if (V; F ) is multiplier isomorphic to (V; F') and ( V ; H ) is multiplier isomorphic to (V; H ' ) , where the two multiplier isomorphisms need not be identical. Examining those Kotzig factorizations which are multiplier isotopic to a given one, we typically obtain many nonisomorphic Kotzig factorizations. We remark also that A(n) and B(n) are not multiplier isomorphic for n > 5, and hence by Lemma 3.1 we can also obtain solutions which are not multiplier isotopic. Using the machinery introduced until this point, we searched for sesquirotational Kotzig factorizations by computer. For u <21, we generated all non-multiplier isomorphic solutions. For u = 23 and 25 we settled existence affirmatively. An examination of these solutions revealed an interesting pattern; given n, the missing vertex is the same for every solution. In fact, for solutions based on A(n),the missing vertex is 3k for A(4k+ 1) and k + 1 for A(4k+ 3). Solutions based on B(n) follow a similar pattern: the missing vertex is 3k for B(4k+ 1) and 3k + 1 for B(4k + 3). We have been unable to prove that every solution has this form. Nonetheless, we added a constraint in our program to reflect this fact and resolved existence for u< 33. We then constrained our program to examine only solutions containing a specified pattern of distances around the missing vertex; in this way, we ultimately found solutions for all u<41, u # 9. At this point, one might reasonably expect that sesquirotational Kotzig factorizations exist for all odd u, .u # 9, but we were unable to prove this. In summary :

Theorem 3.2. Sesquirotational Kotzig factorizations exist for all odd orders u<41 except for u =9. We here tabulate our results, giving a solution for each order. These solutions are all for A(n),and only the near l-factor containing the edge (m, 0) is given. u=5: (m, 0) (1,2), u=7: (co,0) (4, 5) (1, 3), u = 11: (a, 0) ( 6 7 ) (5,9) (2,4) (1,8), U = 13: (CO, 0) (3, 11) (1,4) (8, 10) (2,7) (5,6), U = 15: (00,0) (8, 9) (7, 12) (1, 5) (3, 6) (11, 13) (2, lo), U = 17: (m,O)(1, 2) (6, 15) (9, 13) (11, 14) (3, 8) (5, 7) (4, lo), U = 19: (00,0) (1,2) (6, 16) (9, 15) (12, 14) (3, 8) (13, 17) (4, 11) (7, lo), 0=21: (a,0) (1, 2) (10, 14) (9, 17) (11, 18) (3, 8) (5, 7) (16, 19) (4, 13) (6, 12), ~ = 2 3 :(00,0) (1,2) (12, 15) (11, 18) (10, 20) (13, 19) (4, 8) (16,21) (7,9) (5, 14) (3, 17), ~ = 2 5 :(00,o) (1,2) (13, 15) (8,23) (12,20) (14,21) (3,9) (5, 10) (19,22) (7, 11) (4, 17) (6, 161, u=27: (c0,0)(1,2)(14, 16)(8,25)(11,23)(15,21)(3, 10)(18,22)(19,24)(5,13)(9, 12) (6, 17) (4,201, u=29: (co,0)(1,2)(15,17)(14,20) (9,27) (13,25) (16,24)(4, 11)(6,lO) (7,12)(23,26) (5, 18) ( 3 , W (8, 1 9 ,

C. J . Colbourn. E . Mendelsohn

72

u =31 :

(CO, 0) (1,2) (16, 18) (15,21) (13,25) (12,29)(1 7,26) (4, 1 1 ) (7, 10)(22,27) (6, 14) ( 2 4 , W (5, 19) (3,23) (9,201, U = 33: (CO, 0) (1,2) (16,20) (10,29)(13,28)(12,30) (5,8) (21,26) (23,25)(4,14) (22,31) (6, 17) (9, 15) (7, 19) (1 1, 18) (3, 271, U = 35: (CO, 0) (1,2) (18,21) (17,23) (14,28) (16,29) (15,31) (3,12) (6, 1 I ) (8, 10)(22,32) (26, 30) (25, 33) (7, 19) (5, 24) (4, 27) (13, 20), U = 37: (CO, 0)(I , 2) (19,22)(10,33)(9,35)(12,34) (17,32)(21,30)(24,29)(26,28)(4, 16) (8, 14) (7, 18)(11, 15) (6,231 (5, 25) (13, 20) (3, 31), u=39: (CO, 0) (1, 2) (20, 23) (19, 25) (18, 28) (17, 31) (14, 37) (16, 36) (21, 33) (7, 12) (9, 11) (8, 15) (27, 35) (30, 34) (6,22) ($26) (4, 29) (3, 32) (13,24), u=41: (CO, 0) (1, 2) (5, 39) (21, 25) (12, 37) (16, 35) (15, 38) (20, 34) (24, 33) (27, 32) (29, 31) (6, 17) (11, 14) (7, 19) (10, 18) (9, 22) (4, 28) (8, 26) (13, 23) (3, 36).

The following table presents o u r enumeration results on the number of nonmultiplier isomorphic sesquirotational Kotzig factorizations based on A(n). order

3 5

7

9

II

13

15

17

19

31

number

1

I

1

0

3

1

4

8

18

120

-

3.2. Small orders

The nonexistence of a sesquirotational AK(9) based on A(9) led us to consider other methods to establish the existence of an AK(9). Hand computation quickly produced a solution. We then questioned whether the Hamiltonian decompositions of order 9 could all be coloured, if solutions other than sesquirotational AK(9) were allowed. By computational methods outlined in Section 5, we established : Theorem 3.3. There is I Hamiltonian decomposition of order 5 ; it can be coloured. There are 2 Hamiltonian decompositions of’ order 7; both can be coloured. There are 122 nonisomorphic Hamiltonian decompositions of order 9 ; I 14 can be coloured, 8 cannot be.

4. Traversing near 1-factorizations In this section we consider a dual approach: select a near 1-factorization and attempt to trauerse it, i.e. find a Hamiltonian decomposition to complete a Kotzig factorization. 4.1. Small orders

For orders 3 and 5, the near 1-factorizations are unique and can be traversed (Lemma 2.3). The near 1-factorizations of order 7 were found by examining the 1factorizations of order 8. Dickson and Safford [13], Wallis [41], and Gelling [ 17, 181 computed the 1 -factorizations of order 8; there are six of them. A near I-factorization is found by omitting one vertex and its incident edges from a 1 -factorization. In this

Korzig facforimions

73

way, we computed the seven nonisomorphic near I-factorizations of order 7. We give them here with additional information. Each column specifies a near 1 -factor; when a traversal exists, the array is an AK(7).

This is the Steiner near I-factorization of the unique STS of order 7. It is derived from Gelling’s first I-factorization of order 8, and its automorphism group has order 168. A traversal exists.

This is derived from Gelling’s second 1-factorization ; its automorphism group has order 8, and it has no traversal.

This is derived from Gelling’s third I-factorization; its automorphism group has order 12, and it has no traversal.

This is derived from Gelling’s fourth 1-factorization; its automorphism group has order 2, and it has no traversal.

This is GK(7). It is one of the derived systems of Gelling’s sixth 1-factorization. Its

C . J . Colbourn. E . Mendrlsohn

74

automorphism group has order 42. A traversal exists.

This is the second derived system of Gelling’s sixth 1-factorization. Its automorphism group has order 6. A traversal exists. Only three of the seven can appear as the near l-factorization of an AK(7). For order 9, there are in excess of three thousand nonisomorphic near 1-factorizations; this can be seen from Gelling’s computation [17] of 396 nonisomorphic l-factorizations of order 10. We expect that most of these cannot be traversed, although we have not tested them all. O u r interest is to identify those which can be traversed. Of particular interest are the Steiner near 1-factorization of order 9 and GK(9), since for order 7 the Steiner and GK factorizations can be traversed. The following is an AK(9) based on the Steiner near l-factorization of order 9: (2, 3) (1,3) (4,9) (7, 8) (2,s) (599)(196) (4, 7) (596) (4,5) (4,6) ( 5 7 ) (2,6) (3,7) (3,s) (2,9) (1,9) (1,s) ( 6 7 ) (5, 8) ( L 2 ) (3,9) ( 6 9 ) (1,7) (4,s) ( L 5 )(3,4) (8,9) (7,9) (6, 8) (1, 5 ) (1,4) (2,4) (3,5) (3,6) (2,7) We also found an AK(9)based on GK(9): (2, 3) ( L 4 ) (6, 8) (5,7) (3,7) (4,s) ( 5 9 ) ( 6 9 ) (1,2) (4,5) (3,6) (1, 5 ) (2,6) ( L 9 ) (2,9) (3,s) (4,7) (7,s) ( 6 7 ) (5, 8) (4,9) (3,9) (2, 8) ( 4 7 ) (2,4) (1,3) (5,6) (8,9) (7,9) (2,7) (1,s) (4,6) (335) (196) (235) (3,4) These two AK(9) ‘generalize’ two of the AK(7) described. The third AK(7) is based on the near l-factorization obtained by completing GK(7) to a l-factorization and then omitting one of the original vertices. Carrying this out on GK(9), we obtain a near l-factorization; unlike the case of 7, however, this near l-factorization cannot be traversed. 4.2. 2-rotational STS Our experience with small orders suggests that a promising restriction is to consider traversing Steiner near l-factorizations. Limited computational resources necessitated further constraints. A 2-rotational STS of order 2n+ 1 is an STS having an automorphism consisting of a fixed point and two cycles of length n. We represent a 2-rotational STS of order 2n + 1 as an STS with elements { 00 } u { (i, j ) I 0 < i < n, 0 < j < 1} ; we give the starter blocks which yield the entire STS under repeated application of the automorphism. j ; carrying (i, j ) to ((i + 1) mod n, j ) . Phelps and Rosa [32] have shown that a 2-rotational STS of order u exists if and

Kotzig factorizations

15

only if u = 1, 3, 7, 9, 15, or 19 (mod 24). A potential connection between 2-rotational STS and Kotzig factorizations arises in the following manner. An AK(2n+ 1) has n Hamiltonian cycles; a 2-rotational STS of order 2n + 1 has an automorphism of order n. Perhaps, given a 2-rotational Steiner near 1-factorization, we can construct a base Hamilton cycle. When acted upon repeatedly by the automorphism J this cycle would produce a Hamiltonian decomposition which traverses the near 1-factorization. Given a 2-rotational near 1-factorization of order 2n+ 1, a Hamilton cycle is a base cycle if it meets all of the following conditions: (1) n is odd; for n even, the automorphism f would repeat an edge of difference n/2 prior to repeating the other edges. (2) the cycle contains exactly one edge of the form { ( i , 0), (j, 0 ) ) with i - j = d or j-i=d, for each 1
((0, 1). (2, I), (3, I)}, ((8, O), (0, I), (5, I ) ) , { (2, O ) , (4, O), (0, I ) ) ,

A base Hamilton cycle for this system is (00, (0, 0), (1, 0), (7,0), (3,0), (5,0), (5, I), (2, (8,0), (7, I), (3. I), (2,0), (6, I), (4, 1)- (6, O), (8, I), (0, I), (4, O), (1, 1)). From this, we obtain an AK( 19). Using this approach, we are able to construct larger Kotzig factorizations than the exhaustive techniques of Section 4.1. Although the restriction to 2-rotational STS seems promising, a general pattern remains elusive.

16

C. J . Colbourn. E . Mendelsohn

5. Computational tools In the previous two sections, computational efforts were invested in small orders for two reasons. The first is to determine the spectrum for small orders, to provide a basis for conjectures regarding the existence of Kotzig factorizations. The second is to suggest patterns in solutions, which may be exploited in the resolution of further infinite families. Using computer assistance, we have determined the spectrum for ~ ~ 4 This 3 . addresses the first goal of the computational effort. In terms of the second, however, we have been unable to extract patterns for more infinite families of Kotzig factorizations. The data amassed here may yet prove useful in the discovery of such patterns. The computational tools employed were based primarily on backtracking via an orderly algorithm [9,34] using dynamic isomorph rejection. This isomorph rejection employs the subexponential time isomorphism test for 1-factorizations [7, 81 and the obvious polynomial time isomorphism test for Hamiltonian decompositions. This polynomial time isomorphism test carries over easily from Hamiltonian decompositions to Kotzig factorizations. It is inevitable that, despite additional constraints, even sophisticated backtracking will encounter a combinatorial explosion. We therefore considered a second tool for generation of combinatorial configurations: hill-climbing. This method has been successfully used by Dinitz and Stinson [14, 15, 291 for Howell designs and strong starters; however, little success has been achieved in other cases [38, 403. Unlike backtracking, hill-climbing is not normally suited to suggesting patterns for existence proofs. However, in the event that it succeeds, it can effectively be used to determine the spectrum for small orders. For this reason, we investigated the existence of Kotzig factorizations via a hill-climbing technique similar to the Dinitz-Stinson approach. Our success was minimal; we found no Kotzig factorizations which we had not already discovered by (simpler) backtracking. A modified hill-climbing approach might nevertheless prove applicable.

6. Workpoints 1. We have seen that there exist near 1-factorizations that cannot be traversed. Are

there Steiner near 1-factorizations that cannot be traversed? 2. Is it true that near 1-factorizations which can be traversed are asymptotically rare?

3. Is it true that Hamiltonian decompositions which can be coloured are asymptotically almost all of the decompositions? 4. Find another infinite class of Kotzig factorizations (possibly finding a general

pattern for sesquirotational Kotzig factorizations, or finding a Pelteschn type proof). 5., For sesquirotational Kotzig factorizations, prove that the missing vertex is given

by our formula in Section 3.1.

Korzigfactorizations

77

References W. Ahrens, Mathematische Unterhaltungen und Spiele (Teubner, Leipzig, 1901). V. E. Alekseev, Skolem method of constructing cyclic Steiner systems, Math. Notes 2 (1967)571-576. B. A. Anderson, Finite topologies and Hamiltonian paths, J. Combin. Theory 14 (1973) 87-93. B. A. Anderson, Some perfect I-factorizations, in: Prof. Seventh Southeastern Conf. Combin., Graph Theory, Computing (1976) pp. 79-91. [5] B. A. Anderson, M. M. Barge and D. Morse, A recursiveconstruction for asymmetric 1-factorizations, Aequationes Math. 15 (1977) 201-21 1. [6] S. Bays, Recherche des systemes cycliques de triples de Steiner diNerents pour N premier (oupuissance de nombre premier) de la forme 6n I , J. Math. Pures Appl. 2 (1923) 73-98. [7] C. J. Colbourn and M. J. Colbourn, Combinatorial isomorphism problems involving l-factorizations, Ars Combin. 9 (1980) 191-200. [8] C. J. Colbourn and M. J. Colbourn, The complexity of combinatorial isomorphism problems 11, in: Proc. Ninth Manitoba Conf. Num. Math. Computing (1979) pp. 159-164. [9] M. J. Colbourn, An analysis technique for Steiner triple systems, in: Proc. Tenth Southeastern Conf. Combin., Graph Theory, Computing (1979) pp. 289-303. [lo] M. J. Colbourn, Cyclic block designs: computational aspects of their construction and analysis, Ph.D. thesis, Department of Computer Science, University of Toronto (1980). [ 113 M. J. Colbourn and R. A. Mathon, On cyclic Steiner 2-designs, Ann. Discrete Math. 7 (1980)215-253. [12] R. 0. Davies, On Langford's problem, Math. Gaz. 43 (1959) 253-255. [I31 L. E. Dickson and F. H. Safford, Solution to problem 8 (group theory), Amer. Math. Monthly 13 (1906) 150-151. [ 141 J. Dinitz and D. Stinson, A note on Howell designs of odd side, Preprint (1980). [ 151 J. Dinitz and D. Stinson, A fast algorithm for finding strong starters, Preprint (1980). [16] S. Fiorini and R. J. Wilson, Edge-colourings of Graphs (Pitman, London, 1977). [17] E. N. Gelling, On I-factorizations of the complete graph and the relation to round-robin tournaments, MSc. thesis, University of Victoria (1973). [18] E. N. Gelling and R. E. Odeh, On 1-factorizations of the complete graph and the relation to roundrobin tournaments, in: Proc. Third Manitoba Conf. Num. Math. Computing (1973) pp. 213-221. [19] H.Hanani, A note on Steiner triple systems, Math. Scand. 8 (1960) 154-156. [20] E. C. Johnsen and T. Storer, Combinatorial structures in loops IV: Steiner triple systems in neofields, Math. Z. 138 (1974) 1-14. [21] D. Konig, Theorie der endlichen und unendlichen Graphen (Teubner, Leipzig, 1936). [22] A. Kotzig, Hamilton graphs and Hamilton circuits, in: Theory of Graphs and Its Applications, Proc. Conf. Smolenice (1963) pp. 63-82. [23] A. Kotzig, Groupoids and partitions of complete graphs, in: Combinatorial Structures and Their Applications, Proc. Conf. Calgary (1969) pp. 2 15-221. [24] A. Kotzig and J. Labelle, Strongly Hamiltonian graphs, Utilitas Math. 14 (1978) 99-1 16. [25] A. Kotzig and J. Labelle, Quelques problemes ouverts concernant les graphes fortement Hamiltoniens, Ann. Sc. Math. Quebec 111 (1979)95-106. [26] E. S. Kramer and D. M. Mesner, On Steiner tableaus, Discrete Math. 10 (1974) 123-131. [27] A. A. Markov, A combinatorial problem, Problemy Kibernetiki 15 (1965)263-266. [28] E. Mendelsohn and A. Rosa, On some properties of 1-factorizations of complete graphs, in: Proc. Tenth Southeastern Cod. Combin., Graph Theory, Computing (1979) pp. 739-752. [29] R. C. Mullin. Some recent developments in the computational theory of Room and Howell designs, in: Proc. Conf. Discrete Math. Anal. Combin. Comput. (1980) pp. 1-10. [3O] E. S. O'Keefe. Verification of a conjecture of Th. Skolem, Math. Scand. 9 (1961) 8&82. [31] R. Peltesohn, Eine Losung der beiden HeNterschen Differenzenprobleme, Compositio Math. 6 (1939) 251-257. [32] K. T. Phelps and A. Rosa. Steiner triple systems having rotational automorphisms, Discrete Math. 33 (1981)57-66. [33] D. K. Ray-Chaudhuri and R. M. Wilson, Solution of Kirkman's schoolgirl problem, in: Proc. AMS Symp. Pure Math. Vol. 19 (1971) pp. 187-203. [341 R. C. Read, Every one a winner, Ann. Discrete Math. 2 (1978) 107-120. [I] [2] [3] [4]

+

78

C . J . Colbourn, E. Mendelsohn

[35] A. Rosa, Poznamka o cyklickych Steinercvych systemoch trojic, Math. Fyz. Cas. 16 (1966) 285-290. [36] A. Rosa, On generalized Room squares, in: Problemes Combinatoires et Theorie des Graphes (CNRS, Paris, 1976) pp. 353-358. [37] A. Rosa. Generalized Howell designs, Ann. N.Y. Acad. Sci. 319 (1978)484-489. [38] D. P. Shaver, Construction of (0,k, I ) configurations using a non-enumerative search technique, Ph.D. thesis, Syracuse University (1973). [39] Th. Skolem, On certain distributions of integers in pairs with given differences, Math. Scand. 5 (1957) 5748. [40] M. Tompa, Hill-climbing: a feasible search technique for the construction of combinatorial configurations, M.Sc. thesis, University of Toronto (1975). [41] W. D. Wallis, On one-factorizations of complete graphs, J. Austral. Math. SOC.16 (1973) 167-171. [42] W. D. Wallis, A. P. Street and J. S. Wallis, Combinatorics: Room Squares, Sum-Free Sets, Hadamard Matrices, Lecture Notes in Math. No. 292 (Springer, New York, 1972). Received 10 June 1980; revised I August 1980

Annals of Discrete Mathematics 12 (1982) 79-86

@ North-Holland Publishing Company

CHARACTER STRINGS AS DIRECTED E W E R PATHS Karel CULIK Department o/’Computer Science, Wayne State University, Detroit. MI 48202. USA

Dedicated to Professor A. Kotzig on the occasion of his sixtieth birthday Necessary and sufficient conditions are presented for the existence of a character string for which the number of occurences of all its two-character substrings is prescribed.

1. Motivation Shevtshenko [5] generalized the LL(k) grammars introduced by Knuth [4] by characterizing character strings of length < k by certain unary predicates (for details see Mathematical Reviews 52, #4732 (1976) p. 666).The class of these predicates can be extended if the function F,(x) denoting the number of occurences of the substring y in the string x is used as the base. For example, Fub(ababc)=2,Fbu(ababc)=1, F,,(ababc)= 0, etc. The case when y is a substring of length 2 is of special importance. Assuming that A is the alphabet (of all characters) and contains m 2 1 characters, there are m2 functions &,(X) where a, b E A. If x = x 1 x 2* * * x, E A* and n 2 2 , then N string equations SE of the form FUib,(x) = ri are assumed such that ri > 0, 1
C

ri=n-l.

i= 1

The question arises as to whether or not such a system of string equations SE has a solution, i.e. if the number of occurences, ri>O, of two-character substrings is given when a string x having the required substring specification exists. At first glance, a direct answer seems difficult to obtain, but a very simple and complete answer can be obtained when a graph-theoretical point of view is accepted.

2. Graph-theoretical problem formulation Since the concept of a directed multigraph is needed as parallel edges and loops must be admitted, the concept of a directed graph as given in [11 is adopted. The term path is preferred to that of directed walk [1,3] because of its familiarity to nongraph theoreticians. Indegree and outdegree are denoted by idg and odg. Loops (slings)are admitted but not counted in degrees. There is a directed multigraph DMG(x)= (V(x), E(x), Inc(x)) associated with each 79

K . Culik

80.

character string x = x1x2 . . . x,, n 3 2, where

(i) V(x)={a; a = x i , 1 < i < n } , A is the set of all its vertices, (ii) E(x)= {x1x2,~ 2 x 3 ,... ,x,- lxn}are all its directed edges (arcs), (iii) Inc(xixi+1)=(a, b) when xi = u and xi + = b for 1 < i < n, i.e. the incidence (2) function Inc associates the initial endvertex a and the terminal endvertex h with each directed edge xixi+ (in this order). The DMG(ababc) is depicted in Fig. 1 while either the DMG(ubcbu), the DMG(babcb) or the DMG(cbabc) is depicted in Fig. 2.

Fig. I .

Fig. 2

Lemma 1. In each DMG(x) where x = x 1 x 2. . . x,, n 2 2 , there exists a directed Euler

path which is closed if x1 = x, and open if x1# x,.

Proof. If n = 2 , there exists in DMG(x) either one loop x l x l or, when x l # x 2 , one proper edge x1x2.Therefore, a directed Euler path which is closed in the first case and open in the second exists and satisfies the conditions required by Lemma 1. If n > 2 and we accept the inductive assumption for n - 1, then, when considering DMG(x), there will exist a directed Euler path in DMG(x’)where x’= x l x l . * . x,- 1 . The existence of this path indicates that DMG(x‘) is connected and that either (a) idg’(xi)=odg’(xi)for each i = 1,2,. . . , n - 1, i.e. that the existing directed Euler path is closed, or (b) the previous equalities hold except for the cases when idg’(xl)= odg’(xl)- 1 and odg’(x, - ) = idg’(x, - 1) - 1, i.e. the existing directed Euler path is open. If one considers the final vertex u, and the final edge x,- lx, in which DMG(x) may differ from DMG(x’), the following cases must be considered: (1 1 x, = x and x1 # x,- ; idg(xl ) = idg’(x,) 1 and odg(x,- 1 ) = odg’(x, - 1 ) 1 while idg(xi)=idg’(xi)and odg(xi)=odg’(xi)for i = 1, 2, 3, . . . ,n - 2. Therefore (by a wellknown theorem in graph theory) there exists a closed directed Euler path in DMG(x). ( 2 ) x,,=xl and x1 = x , - ~ .There exists a closed directed Euler path in DMG(x’) and x,- = x,, indicating that only one loop is being added. Furthermore, there is no difficulty inserting it into the existing closed Euler path, thereby, obtaining one in DMG(x). ( 3 ) x,#xl and x1 # x , - ~ ; idg’(xl)=odg’(xl)-1 and odg’(x,)- 1 and odg‘(x,-l)= idg’(xn-l)- 1 while idg’(xi)=odg’(xi)fori = 2 , 3 , . . . , n-2, and idg(xl)=idg’(xl)while 1, indicating that idg(x,- l)=odg(x,- 1), and odg(x,)= odg(x,-l)=odg‘(x,-

+

+

81

Character strings as directed Euler paths

idg(x,) - 1 in any case (either x,#xi for i = 1, 2,. . . , x,or x,= xi for some i, 1 d id n - 1). Thus, again, there exists an open directed Euler path in DMG(x). (4) x,#xl and x I = x - There exists a closed directed Euler path in DMG(x’) and either x, # xi for each i = 2, 3, . . . ,n - 1, i.e. there exists an open directed Euler path in DMG(x) having starting vertex x,- and ending vertex x, or xn=xi for some i, 2 < i < n - 1. If i = n - 1, then as in (2), only one loop is added. Otherwise, xi will be the ending vertex of the open directed Euler path in DMG(x) while the beginning vertex will remain unchanged. Possibilities (1H4) are depicted in Fig. 3 for six cases when x, is the added vertex. The full or dashed lines represent the assumed directed Euler path in DMG(x’)while the dotted line represents the added directed edge (x,- 1, x,).

Q I

Fig. 3.

3. Necessary and sufficient conditions for the existence of a solution

When considering given string equations SE: where aibi E A*, one may define other functions:

Fuibi(X)=ri>O

for 1< i < N < m 2

(3.1) From Lemma 1 and these definitions, Lemma 2 follows immediately:

82

K . culik

Lemma 2. I f x = x 1 x 2 . . . x , is a given character string, then L(a),R(a) are the outdegree, indegree of the vertex a, respectively, in DMG(x). Therefore, either (i) L(a)=R(a)for each a E V ( x ) ,or (ii) L(a)=R(a)for each a E V ( x )such that x 1 # a # x , , and L ( x l ) = R ( x l ) - 1, R(x,)=L(x,)- 1.

(3.2)

When considering given string equations SE: Fai&) = ri > 0 for 1 d i < n, one may define V= {a; a=ai or a=bi where I g i d N } independently on x (using only twocharacter substrings aibi which occur in x), and E by requesting that there be ri parallel directed edges ei,l,. . . ,ei,,i which begin in ai and terminate in bi for i= 1, 2, . . . , N , i.e. Inc(ei,j)=(ai.bi) where 1 6 id N and 1d j d r i . Thus, the directed multigraph D M G ~ = E (K E, Inc) is uniquely determined by the given string equations only.

Lemma 3. The numbers L(a),R(a)defined by (3.1) (i) and (ii) from the given string equations SE are the outdegree, indegree of the vertex a, respectively, in DMGsE. Lemma 4 follows immediately:

Lemma 4. I f the string equations SE: Faibi(x) = ri > 0 where 1 < i < N specify a character string x , then DMGsE and D M G ( x ) are isomorphic. Theorem. The string equations SE have a solution iffthe corresponding directed multigraph D M C ~ E is connected, and the functions L(a),R(b) defined by (3.1) satisfy either (3.2) (i) or (3.2) (ii).

Proof. Necessity: If the string equations SE specify a character string x , then by Lemma 1 there exists a directed Euler path in DMG(x).Furthermore, it is known that the indegrees and outdegrees, L(a) and R(a) by Lemma 2, must satisfy either (3.2)(i) or.(3.2)(ii).Obviously, D M G ( x )is connected. Sufficiency: If S E are such that D M G ~ is E connected and the functions L(a) and R(a) satisfy either (3.2)(i)or (3.2)(ii),then it is known since L(a)and R(a)are the outdegrees and indegrees, respectively, in DMGsEby Lemma 3 that there exists a directed Euler path in DMGsE.Having an Euler path ( x l ,e l ,x2, e 2 , . . . , en- x,) where xi E V and ei E E , one can suppress all edges to obtain the solution, a string x l x l . . . x,. Fig. 4 demonstrates that the problem is not trivial as there is no solution of

Fig. 4

83

Character strings as directed Euler paths

SE: Fa&)= 1 and F&(X)= 1, indicating that according to (1) x must have length 3, V= {a, b, c}. But neither condition (3.2)(i)or (3.2)(ii)is satisfied, therefore, no directed Euler path exists in this figure. Fig. 2 demonstrates that more than one directed Euler path may exist. Therefore, the string equations SE may have more than one solution. 4. Finding a solution

The following algorithm appears to be the most efficient when a computer is used for finding a directed Euler path in DMGsE, assuming that the functions L(a) and R(a)satisfy either (3.2)(i)or (3.2)(ii):

Algorithm 1. Step 1. If DMGSEsatisfies (3.2)(5),add one new directed edge (u, w) to E such that R(u)=L(u)- 1 and L ( w ) = R ( w ) - 1. If DMGsE satisfies (3.2)(i),do nothing. Let DMGl = (Vl, E l , Inc,) be the directed multigraph obtained. Eventually, w, u will be the beginning and ending vertices, respectively, of the Euler path searched. Step 2. Starting with i = 1, and, if i > 1, assuming that DMGi= (K, E i , Inti) has been obtained, one checks E i : (a) If E i # O , choose ui,o E & such that ui,o belongs to the cycle cyi- l . If i > 1, denote = u ~ ,otherwise, ~, choose ui,o arbitrarily. ui(b) Select a closed path C Y ~ = ( U ~ ,ei,l, ~ , q 1 , .. . , ei,pi,ui,pi=ui,o) in DMGi of length pi > 1 having unrepeated nonconsecutive vertices (a cycle) and containing all loops of the vertices uij. (c) Define D M G i + las follows: C + l = K - - { x i j , idg(xij)=odg(xij)=1 in DMGiand 1< j
..

9

uh+ l , q h + l = U h + 2 , 0 ,

ui-2,qi-2? e i - 2 , q i _ 2 + 1 ?

...

eh+2,13..

.)v i -

-

l , 0 7 e i - 1 , l t ui- 1 . 1 3

~ i - Z . p i - ~ - ~ i - 2 , 0 = u i - 3 . 4 i - ~ )'

. . ,u h , p h = u h - 1 , q h - , 3 e h - l , q h - , + 1 , V h - l , q h - I + 13.. . u l . . . , u h , j - 1 ) is the open Euler path compiled from c y k .

Vh,qh+l,. uh,l,

9

(b)IfUh+1,0=t)h.~~and q h < j , then ( u h , j ? e h , j + l , 01,0,.'.9 u h . 0 , . o h , j-

'

= u) is

' ' 9

Vh,qh=uh+l.O?.

'.

9

uh,j+ 1 , .

Ui-19..

' 3

..

9

..

.

3

' 1

ui- l , p i

-I-ui-

uh+l,O='h,qh?

, O ~ '~ 9. U h . 0 , .

'.

1,0=

eh,qh+l, 9

Uh.O,eh,l,

~h.ph=Uh,O=~h-i,qh_lr..

.

3

~h+l.0='h,qyeh,qh+l,~h,q~+lr...7

compiled Euler path from c y k .

If there was no edge added in Step 1, one may begin with an arbitrary vertex and compile the Euler path similarly as in (a) or (b).

K . Culik

84

Correctness proof. Step 1: It is well known that an open directed Euler path in a directed multigraph may be completed by one directed edge to a closed directed Euler path. Step 2 : The vertex U ~ , ~ = U ~ --, always exists (if i > 1) since, according to (c), only vertices of degree 2 were removed. Since, according to (b), loops are treated independently, each cycle cyk has a length 2 2 . Therefore, the selection of cycles must terminate. Step 3: All cycles obtained are mutually edge disjoint. Therefore, the paths in either (a) or (b) are Euler paths of the given multigraph. Two cases, (a) and (b) from Step 3, are depicted in Figs. 5 and 6 on simple examples where the numbering of edges expresses their positions in the Euler path. 'lrO

"1 I

"3rl

'3.2

Fig. 6 .

85

Character strings us directed Euler puths

5. Problem generalization Assuming that y is a substring of length 3, we have m 3 functions Fabc(X) where a, b, c E A and the alphabet A contains m different characters may be used for a substring specification of a character string x = x 1 x z - ” x , where na3. The N string equations SE, : Faibici(x) = ri satisfy ri > 0, 1 6 N <m3, aibiciE A 3 for i = 1, 2, . . . , N and N

2 ri=m-2.

(4)

i= 1

The question arises as to whether the system of string equations SE, has a solution, i.e. does a character string x exist which satisfies SE,. This problem may be reduced to the previous case SE = SE, as follows:

Algorithm 2. Step 1 . Two systems SE, are defined by SE,: (a) SE:: Faibi(x)=riwhere 1
M D G SE; = M D G SE:

MDGSE2

MDG

, SE2

Fig. 7.

K . Culik

86

and

MDGsE:=(J'~={~, b, c), E $ = { ( b , c), (c, a)(a, b)}). Since E l = E $ , case (b) of Step 2 must be considered. One takes (a, b) E E: and defines E z = {(a, b), (a, b), (b, c), (c, a)}, the corresponding SEz of which is Fnb(x)=2,Fbc(x)= Fca(x)= 1. Obviously SEz (and also SE; obtained by adding (c, a) instead of (a, b))has a solution as the indegrees and outdegrees satisfy (3.2)(ii). There should be no difficulty considering systems of string equations SEk where 3 < k < n with the expectation that they can be reduced to two systems S E i P l and SEiIf substrings of different lengths are to be admitted for string specification, some reasonable assumption concerning their overlap must be accepted, although it is unclear which one.

References [ 13 J. A. Bondy and V.S. R. Murty, Graph Theory with Applications (North Holland, Amsterdam, 1976). [2] K. Culik, Characterization of strings (sequencies) by directed Eulerian multigraphs, Abstracts of the International Colloquium of CNRS on Cornbinatorial Problems and Graph Theory (July 1976) Paris-Orsay. [3] F. Harary, Graph Theory (Addison Wesley, Reading, MA 1972). [4] D. E. Knuth, Top-down syntax analysis, Acta Informatica 1 (1971) 79-110. [ 5 ] V. V. Shevtshenko, On the question of deterministic methods for topdown analysis (Russian), in: Software of Electronic Digital Computers (Proc. Sem. Automation of Programming, Kiev, 1970/71), Akad. Nauk Ukrain. SSR Inst. Kibernet., pp. 86-98. Received 19 March 1981.

Annals of Discrete Mathematics 12 (1982) 87-94

@ North-Holland Publishing Company

ON MAXIMAL EQUIDISTANT PERMUTATION ARRAYS Mikhail DEZA Centre Nationul de la Recherche Scienrijique, Paris. France

Scott A. VANSTONE* S t . Jerome's College, University of Waterloo, Onrario. Canadu

Dedicated to Professor A. Kotzig on the occasion of his sixtieth birthday An equidistant permutation array A(r, I ; u ) is a t' x r array defined on a symbol set V such that each row of the array is a permutation of the symbol set V and any two distinct rows of A have precisely A common column entries. Given r and i.. one can ask for the largest u such that an A(r,2.; v) exists or find the smallest v such that an A(r,I; u ) exists which cannot be extended t o an A(r. i ;u + 1). In this paper, we consider arrays of the latter type. Such arrays we call maximal. Several classes of maximal equidistant permutation arrays are given and bounds are obtained.

1. Introduction

An equidistant permutation array (EPA) is a u x r array defined on a symbol set V such that (1) every row is a permutation of the symbol set V, (2) every pair of distinct rows has precisely A common column entries. We denote such an array by A(r, A; u). An EPA, A = A ( r , A ; u ) is called maximal if it is impossible to add 8 ( u + 1)st row to A such that the resulting array is an A(r, A; u + 1 ) . A maximal A(r, A; u) is denoted by A(r, A; u). Define R(r, A)=max{u: there exists an A(r, l ;0))

and R(r, A)= min{u : there exists an A(r, A;

0))

There are a number of results on the function R(r, A). For example, if n = r - l then R(r, A)<max{R(n+ 1, I), 2+LA/[$nlJ, n2 -5nf7, i ( r 1 + 2 ) ~ ) .

This bound is dependent on the results for R(r, 1). In this case, the following is known: 2 r - 4 6 R ( r , l ) < r 2 - 4 r + 1, r 2 6 .

+

In many cases, the lower bound has been improved. For example, R(r2 r + 1, 1)a r3 rz whenever r is a prime or prime power. These and other results on R(r, A) can be found in [l, 2, 3, 8, 10, 1 I]. In this paper, we are interested in maximal EPAs. We require several more definitions.

+

*Research supported by NSERC under grant No. A9258 87

M . Deza, S . A . Vunstone

88

A generalized Room square (GRS) is an r x r array defined on a symbol set V of cardinality u such that (1) every cell contains a (possibly empty) subset of K (2) every element of V is contained in every row and column of the array precisely once, and (3) every pair of distinct elements of V is contained in A of the cells. Such an array will be denoted by S(r, A; u). An (r, A)-design D is a collection E of subsets (blocks) taken from a finite set V of elements (varieties) such that (1') every element of Vis contained in precisely r blocks of D, (2') every pair of distinct elements of V is contained in exactly A blocks of D. If each block of D has cardinality k, D has u varieties and b blocks, then D is called a balanced incomplete block design and is denoted (u, b, r, k , 1)-BIBD. In Section 2, (r, A)-designs are used to generalize a construction for maximal EPAs which was given in [63. In Section 3, we show that this construction, applied to BIBDs, gives maximal EPAs. We conclude this section with a remark which will be useful in Section 3. Let A(r. i; u ) be an EPA with rows r , , r 2 , .. . , rv where ri is a permutation of the symbol set V= { 1,2,. . . ,r i . Form an r x r array S whose rows and columns are indexed by V For each h, 1 < h < u , in the cell (rh(j),j ) place the symbol x h , 1 < j < r . It is readily seen that S is a GRS defined on the symbol set ( X i : l < i < u ) . We call S the GRS associated with the EPA A(r, A; u). We will, at times, find it convenient to work with the GRS rather than the EPA.

2. EPAs and (r, l)-designs Let a be a permutation acting on the set N = { 1 , 2 , 3 , . . . , n ] . Define E ( a ) = { i E N :a ( i ) # i ) .

F o r a n y A ~ S , , E ( A ) = { E ( a ) :€a A J . L e tL = { l l , f2,...,l,) beasetofpositiveintegers. A(n, L ) is a set of permutations such that for all a, b E A (a # b),n - IE(a 'b)J E L. In the case where (L(= 1, A(n, L) is just an EPA. ~

Lemma 2.1. Let A = A(n, L ) such that (i) ( E ( a ) nE(b)(< 1 for all a, b E A , a # b. Then: (a) IE(A)I = 1Al; (b) E ( a - ' b ) = E ( a ) u E ( b )for all a, b E A, a # b . and hence n - [ E l uE21 E L,for all E l , E 2 E E(A), E.1f E 2 ; (c)A is a proper subset ofsome A(n, L )with property (i)iffthere exists E' c [ 1,2,. . . , n ) such that IE'I # 1, E' $ E ( A ) and n - IE'u El E L for any E E E(A).

Proof. (a)We have, ofcourse 1E(A)I < J A Jfrom definition of E ( A ) .Suppose 1E(A)I< JAl. Hence E(a)=E(b) for some a, b E A , a # b . We obtain E ( a ) n E ( b ) = E ( a ) = E ( b )con; dition (i) implies &)< 1, E(b)< 1. But neither E(a)= I nor E(b)= 1 is possible. We

M a x i m a l equidistant permutation urrays

89

have IE(a)l= IE(b)(=0, i.e. both a, b are the identity permutation. This contradicts the supposition a # b. (b) For any a, b E S , we have E(a)AE(b)~ E ( a - ' b G) E ( a ) u E ( b )where E(a)AE(b)= ( E ( a ) uE(b))- @(a)n E(b))is the symmetric difference of the sets E(a), E(b). In case (E(a)nE(b)l= O we will have E(u- ' b ) = E ( a ) uE(b) immediately. Suppose now that IE(a)nE(b)l= 1 and E ( a ) n E ( b ) =f p ) for some p E { 1,. . . ,n ) . We have a(p)#p# b(p). The case a(p)= b(p) will imply a(a(p))#a@), b(a(p))# u(p), i.e. a(p)E E(a)n E(b) and IE(a)nE(b)l>1 which contradicts condition (i). Hence, a(p)# b(p) and E(a- ' b ) = E ( a ) uE(b). (c) This follows from (a)and (b). In the next theorem we give a construction for a class of A(n, L).

Theorem 2.1. lfthere exists an (r, 1)-design D with n blocks and u varieties, then there exists an A(n, n - 2r + 1; v). Proof. Let T= {B1,B 2 , .. . ,B,) be the blocks of D. For each variety x E D, let

f x be any permutation acting on Tsuch that E ( f , ) = ( B : x E B ) . We now show that A = { f x : x E D} is an A(n, I ; u). It is clear that A satisfies property (i) of Lemma 2.1. Hence, from (b)of Lemma 2.1, E ( a - ' b ) = E ( a ) u E ( b )for all a, b ~ A ( a # b )But . E(a)=r, for all a E A and, thus, E(a-'b)=2r - 1. This completes the proof.

This result generalizes a construction given in [6]. Corollary 2.1. Let A = A(n, r) be the permutation array corresponding to a given (r, 1)design D. lf G is the GRS corresponding to A , then the blocks o f G having cardinality greater than one are precisely the blocks of the complement to D.

Proof. A block B of G of cardinality k 2 2 corresponds to a set {al,u 2 , .. . ,akj permutations of A with the property that ai(b)=a for all i, 1 < i s k and fixed elements a and b. Since k 2 2 , and t(ai,aj)=O for all i, j ( i # j ) , then a 4 E(ai),1 < i< k. Hence, the elements of D associated with a l , a2,. . . ,akdo not occur in the block B' of D associated with b. Hence, the elements of D associated with a l , a 2 , .. . ,ak form the block B . This completes the proof. We remark that the same procedure applied to an (r, 1)-design will produce an A(n, L ) where L = { n - 2 r + A + t : O
+

+

Lemma 2.2. Let A= A(n, 1) have property (i') and A u { a ' } be an A(n, 1) ,for some a' E S, \ A. Then

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(a) IE(a’)l> r - 1, (b) lE(a’)l=r-l IE(a’)l=r IE(a’)l=r 1

+

(c) IE(a’)l> r for all a E A .

.v

E(a’)nE(a)=0,foralla~A. i f f ~ E ( a ’ ) n E ( a ) ~ = l , f o r a lEl aA . iff IE(a‘)n E(a)l= 2, and t(a, a’) = 0 for all a

E A.

+ 2 implies IE(a’)n E(a)l2 2, and IE(a’)n E(a)l+ ?(a,a’)= IE(a’)- r + 1 2 3 +

Proof. Recall IE(a- a’)l= I ~ ( a ) l I ~ ( a ’ )l IE(a)n~ ( a ’ -?(a, ) l a‘). Since IE(U)I = r for all a E A and IE(a- b)l = n - 1 for all a, b E A ( a# b) then IE(a’)l=r - 1 + IE(a)n E(a’)l +?(a,a’). (a) now follows. (b) follows from the above and Lemma 1 using the fact IE(a)nE(a’)ld1 implies ?(a,a’) = 0. This completes the proof. Recall that A=A(n. 1) is maximal and denote it by A(n, I ) if A u l d ] # A ( n , I ) for any a’ E S,\A. An (r, 1)-design D with u varieties and b blocks is extendible to an (r, 1)-design D‘ with v + 1 varieties and b blocks if D is isomorphic to a restriction of D’. This is a special case of the extension given implicitly in (c) of Lemma 2.1. An A =A(n, 1) with property (i’) is strongly extendible if there exists an a E S,\A such that A u { a ) is an A(n, 1) with property (i’).

Theorem 2.2. An A(n, 1) with property (i’) is strongly extendible V t h e corresponding (r, 1)-design is extendible. Corollary 2.2. An A(n, 1; v ) with property (i’) is extendible to an A(n, 1; n) with property (i’) ifu>((n -1- 1)/2)’.

Proof. This follows from a result on extendability of (r, 1)-designs and can be found in [8]. We conclude this section with the following useful lemmas on EPAs A = A(n, 1; u ) with IAJ= u 3 2. Suppose that IE(a)l= r for any a E A . It is evident that r 2 2.

Lemma 2.3. Suppose that IE(a)l=2 for any a E A. Then either (a) I = n - 4 , E ( a ) n E ( b ) = 0 f o r a n y a , b ~ A , a # b ; o r (b) 1= n - 3, I ~ ( a= ) (1 ; or (c) l=n-3, A = ~ ( i , j ) , ( i , k ) , C i , k ) ) f o r s o ml <e i < j < k < n .

nrrEA

Proof. A consists of transpositions, i.e. cycles of length 2. The case (a) corresponds to the possibility that every pair of these distinct cycles has no element of { 1,2, . . . , n} in common. The only possibility is that every pair of these cycles has exactly one element in common. For this case Lemma (3.4) of [4] gives only two possibilities corresponding to cases (b) and (c). Remark. Using known equalities [3], max)A(n,n - 3))= n - 1, maxlA(n, n -4)) = [ i n ] we can show easily that

91

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in thecase(a)ofLemma2.1,A=A(n,n-4)iff u=L+nJ; in the case (b) of Lemma 2.1, A = A(n, n - 3) iff u = n - 1 . Also, A = A(n, n - 3) in the case (c) of Lemma 2.1. Lemma 2.4. Let A = A(n, I ;

0).

Then I # n - 1 and I 2.

Proof. It follows trivially from Lemma 2.1. In fact, suppose I= n - 2 and a a2 and a3 are three distinct elements of A . Then {a;'a2, a;'a3). is an A(n, n-2; 2) with IE(a; 'a2)[= IA(a; 'a3)1= 2, which is an impossibility. Lemma 2.5. Let A = A ( n , I ; u), u > 2 and let a' E A. Denote A'={b-'a': b E A , b#a'); then (a) A'= A(n, I; u - l), with (b) IE(a)l= n -1 for any a E A' and (c) IE(a)nE(b)l>max(+(n -I),2) for any Q, b E A', a # b.

Proof. (a) and (b) follow from the fact that Hamming distance (E(a-'b)(on S , is invariant of translation, i.e. IE(a- 'b)( = IE((ac)- '(bc))l= IE((ca)-'(cb))I. Leta, bEA',a#b. n -I= IE(a- 'b)l>, IE(a)AE(b)l = I(%)u E(b))- Eb)n E(b)l = IE(a)l+ IE(b)l- 21E(a)n E(b)l =2(n -I)-21E(a)nE(b)l.

Hence, IE(a)n E(b)l >i(n - I). Moreover, $(n - I ) > 1 (because 1 < n - 3) from Lemma 2.1 and IE(a)nE(a)l is an integer; so (c) is proved.

3. MaximalEPAs

+

In this section, we find several classes of maximal EPAs. Let LS(n I, A(n I, I; n) obtained from a latin square by adjoining I fixed points.

+

Theorem 3.1. For each positiue integer n and

I; n) be an

I, LS(n + I, I; n) is an A(n + I, I; n).

Proof. Let G be the GRS associated with LS. The nonempty cells of G contain blocks of size 1 and n. Without loss of generality, assume that the blocks of size n occur in a I x I subarray S and S occurs in the upper left corner of G and L is the latin square subarray in the lower right coi-ner. Suppose G is extendible by adjoining a new element x. If x is contained in all blocks of size n, x cannot occur in any cell of L. Hence, x cannot occur in any row or column of G which contains L. Suppose there is some block B of size n which does not contain x. x must occur once in the row of G which contains B. This row does not contain a row of L. Since x must occur with every element I times, x must occur in each row and column of L. This is impossible. Hence, G is not extendible and LS is maximal.

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This theorem implies that R(n, I ) < n -1. We conjecture that R(n, n = n -1. In the next theorem we establish a lower bound on the size of R(r, I).

Theorem 3.2. R(r, l ) > f ( r + 3 ) . Proof. Let ,4 be an A ( r , 1 : u ) with u < f ( r + 3 ) . Consider the GRS associated with A. Denote this GRS by S, its symbol set by V and let I be the cardinality (size)of a block of maximum length in S. Any element of S is contained in at most u - 1 blocks of size greater than one and, hence, must occur in at least r - u + I blocks of size one (singletons). Suppose B is a block of size 1 in SV\B= it),. . . , u , , - ~ !and B is contained in cell C o of S . Let C, be a cell of S which contains u, as a singleton and C , does not occur in a row or column containing any other C, ( j # i ) . That such a set of cells exists follows from the fact that as we select each C, we eliminate at most 2 singletons for each element of V\B not yet chosen and, also, r - u - 1 3 Z(u - 0. Permute rows and columns of S so that the cells C , ,0s i < u - I , are contained in the ( u - l + I ) x ( u - l + 1 ) subarray in the upper left corner of the array. Consider the ( r - o + l - 1 ) x ( r - u + / - 1 ) subarray E which is contained in the lower right corner I which cover the of S. We now select r - u + 1 - 1 empty cells of E , e l , e z , . . . ,e r rows and columns of E . This can, of course, be thought of as selecting a matching in the bipartite graph formed from the rows and columns of E and edges corresponding to the empty cells. Each row and column in this subarray contains at least r - t i + / - 1 - u empty cells. P. Hall’s theorem guarantees a matching (transversal) of empty cells provided for each subset X of the rows I X l < l N ( X ) l where N ( X ) is the set of columns having at least one empty cell in some row of X . We need only check sets X such that 1x12 r - 20+1- 1. Since r - 2 u + I- 1 > r! we know that if IN(X)l< 1x1 then there must be a column of E having more than u nonempty cells which is impossible. Therefore. by Hall’s theorem there is a transversal of empty cells in E . I f we now add a new symbol to each of the cells C , , O d i d v - l , and the cells e l , e z , . . . , e,,-, + I - , we get a GRS of side r and order u + 1 and, hence, the EPA corresponding to the original GRS was not maximal. This completes the proof. 1 9 ~ .

The above result leads one to ask whether every (r, 1)-design having u d r - 1 varieties is the underlying design for some GRS. This is a similar question to that posed by Erdos, Faber, and Lovasz [4] concerning the resolvability of (r, 1)-designs with v < r elements. Erdos, Faber, and Lovasz conjecture that any (r, I)-design defined on v d r varieties is resolvable. This conjecture is, to the authors’ knowledge, unsolved. Partial results exist and the interested reader is referred to [ 9 ] . Another class of maximal EPAs is given in the next theorem.

Theorem 3.3. If there exists u ( v , b, r, k , 1)-BlBD ( k 2 3 ) , then there exists un A(b, h - 2 r f l ; u). A proof of this theorem can be found in [ 9 ] . In this special case, where result stated in Theorem 3.3 was obtained in [6].

I I = b,

the

Maximal equidistant permutation arrays

4.

93

b(n,1; n )

We now consider a special class of maximal EPAs. An EPA A(n, I; u ) is called square if n = u. We are interested in 2 = A(n, I; n). For 1= 0, a latin square provides an example of an A(n, 0; n) for any n. For 1= q 2 - 4, n = 4 2 + 4 + 1 and, if there exists a finite projective plane, then there exists [6] an 2(42+ 4 + 1, 4 2 -4; 42 + 4 1). It has been shown [l] that

+

I)\ <max{l+ 2, (n-I)’+(n - I ) +

1).

If(n-1)2+(n-I)+1
and, hence, from Lemma 2.4 I A h 41 < n ~

and the square cannot exist. This will occur provided 1>n - 1 - J n + 2. Let f ( n )be the maximum value of I such that an A(n, I ; n ) exists.

Theorem 4.1. (a) f ( n )< n - 1 - J X . (b) Let n = 42 + 4 + 1 for 4 a prime power pa. Then __

n - 1- 2Jn - 1< q2 - 4 < f ( n )
Proof. Part (a) follows from the above. The lower bound in (b) follows from the results of [ 11. The upper bound follows from (a), and the fact that q2 +4 - Jq2 +4+ 3 < q 2 +4 - J42=q2.

In the case of n=7, f ( 7 ) < 3 or f ( 7 ) < 2 . From (b) of the theorem f ( 7 ) 3 2 . Hence, f(7)= 2. This completes the proof. We conjecture that the lower bound in (b) is the exact value of f ( n ) . Since there exists an 4 6 , 1; 7) [3], there exists an 4 7 , 2; 7) which is not obtained from a finite projective plane using Theorem 3.1. Thus, the extremal case for f ( 2 ) is not unique. We remark, however, that the A(7,2; 7) just given is not maximal. Another example of a square having the parameters of a square constructed in [6] but not obtainable by this construction is an 4 4 3 , 30; 43). To construct such an array by the result of [6] requires a finite projective plane of order 6. This array can be constructed as follows. By a construction given in [5], there exists an A(24 - 1 , 4 - 3; 4(4 -2)). If we take 4 = 11, we obtain an A(21, 8; 99). Deleting 56 permutations and adding 22 fixed points, produces the required array.

References [l] M. Deza, Matrices dont deux lignes quelconques coincident dans un nombre donne de positions communes, J. Combin. Theory Ser. A 20 (1976) 306-318.

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[2] M. Deza, R. C. Mullin. and S. A . Vanstone, Room squares and equidistant permutation arrays, Ars Combin. 2 (1976) 235-244. [3] M. Deza and S. A. Vanstone, Bounds for permutation arrays, J . Statist. Plann. Inference 2 (1978) 197-209. [4] P. Erdos, Problems and results in graph theory and combinatorial analysis, in: Proc. Fifth British Cornbinatorial Conference (Aberdeen, 1975), Congressus Numerantium, XV (Utilitas Math., 1976) pp. 169 192. [5] M. K . Farahat, The symmetric group as a metric space, J. London Math. SOC.35 (1960) 215 220. [6] K. Heinrich, W. D. Wallis, and G. H. J. van Rees, A general construction for equidistant permutation arrays, in: J. A. Bondy and U. S. R. Murty, Eds., Graph Theory and Related Topics (Academic Press, New York, 1979) pp. 247-252. [7] P. Lorimer, Maximal sets of permutations constructed from projective planes, Discrete Math. 25 (1979) 269-273. [8] R. Mathon and S. A. Vanstone, On the existence of equidistant permutation arrays and doubly resolvable Kirkman systems, Discrete Math. 30 (1980) 157-172. [9] J. Mitchem, O n n-coloring certain finite set systems, Ars Combin. 5 (1978) 207 212, [lo] R. C. Mullin and E. Nerneth. An improved upper bound for equidistant permutation arrays. Utilitas Math. 13 (1978) 77-85. [l I] G. H. J. van Rees and S. A. Vanstone, Equidistant permutation arrays: A bound, J. Austral. Math. Soc., to appear. [I21 S. A. Vanstone, The extendibility o f ( r . I)-designs, in: Proc. 3rd Manitoba Conf. on Numerical Math. (1973) pp. 409-41 8. [I31 S. A. Vanstone. A note on a class of maximal equidistant permutation arrays. Utilitas Math. 16 (1979) 21 7-22], Received 8 November 1980

Annals of Discrete Mathematics 12 (1982) 95- I I I

@ North-Holland Publishing Company

DOMINO SQUARES J. A. EDWARDS, G. M. HAMILTON, A . J. W. HILTON and Bill JACKSON Uniuersity of Reading, Whiteknights, Reading, U K

Dedicated to Professor A. Kotzig on the occasion of his sixtieth birthday 1. Introduction

Given an n x n chessboard (an n x n matrix of cells), can it be covered by a set of distinct dominoes (1 x 2 or 2 x 1 matrices) so that the numbers appearing in each row are all distinct and the numbers appearing in each column are all distinct? Clearly one must discard dominoes which have the same number at each end. Given a complete set of dominoes on the numbers 0, . . . , n there are (": ' ) = $ ( n ' + n ) dominoes from which one may choose. We call an n x n matrix covered in this way with dominoes based on the numbers 0,. . . ,n a domino latin square of side n (or, shortly, a domino square of side n). We give two examples in Fig. 1 to illustrate the idea. Since each domino covers two cells, and the number of cells in a domino square of side n is n2, it is clear that domino squares of side n can only exist if n is even. If n is odd we adapt the definition slightly. This time we cover all cells of an n x n matrix except for the central cell. We call such a square a domino latin square with a hole in the middle, or acronimically, a whim domino square, or a whimsy for short. We give three examples in Fig. 2.

4

0

3

Fig. I . Domino squares of sides 2 and 4.

Iu

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I' 21013

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3 1 5 1 4

I O

Fig. 2. Whim domino squares, or whimsies. of sides I . 3 and 5 95

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J . A . Edwards et al

96

Both the main and the off diagonals of the domino square of side 4 in Fig. 1 and the whim domino square of side 3 in Fig. 2 consist of distinct numbers. When this phenomenon occurs we call the squares double-diagonal. Both domino squares and whim domino squares can be treated in a somewhat more unified way as follows. Call an (n + 1) x n matrix covered by distinct dominoes based on the numbers 0,. . . , n in such a way that the numbers in each column are distinct and the numbers in each row are distinct, an (n + 1) x n domino latin rectangle, or, more shortly, an (n + 1) x n domino recfangle. An n x (n + 1) domino rectangle is defined in the obvious analogous fashion. With these each suitable domino is used, whereas with domino squares and whim domino squares there are some dominoes left over unused. Some examples are given in Fig. 3. If n is even and there is a row in an (n + 1) x n domino rectangle covered by dominoes lying ‘horizontally’, then by deleting that row we obtain a domino square of side n. The converse is obviously not true: namely, given a domino square of side n, it is not necessarily true that we can extend it to an (n I) x n domino rectangle. For example, if the domino square of side 4 in Fig. 1 were extended to a 5 x 4 domino rectangle then its last row would have to be (3, 4, 0, 2). But the dominoes 3-4 and 0-2 have already been used ! Similarly it is often possible to obtain whim domino squares of side n from (n + 1) x n domino rectangles. For example, if we delete all dominoes covering one or more cells of the bottom row of the 6 x 5 domino rectangle in Fig. 3, place the last column in the middle and the new bottom row in the middle we obtain the whim domino square of Fig. 2 of side 5. We can define some further analogous ideas. A nearly domino square of side n is a latin square in which :(n2 - n) mutually non-overlapping adjacent pairs of symbols form a set of distinct dominoes on the numbers 0, 1, . . . ,n - 1. Given a set of dominoes on the numbers 0,. . . ,n - 1, each number is paired with each other number exactly once and so, in an n x (n - 1) domino rectangle, occurs exactly n - 1 times. Thus it is easy to see that we may add a final column to any n x (n - 1) domino rectangle to form a nearly domino square of side n. We give some examples in Fig. 4 to show that nearly domino squares need not be formed this way. The completion of an n x (n - 1) domino rectangle (or of an (n - 1) x n domino rectangle) is the nearly domino square of side n formed by adding a last column (or a last row, respectively).

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Fig. 3. ( n 1) x n domino rectangles with n = 2, 3,4, 5.

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Fig. 4. Nearly domino squares of sides 4 and 5 (the non-domino entries are bracketed).

111

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Fig. 5. A cylindrical horizontal nearly domino square of side 4, a cylindrical transversal horizontal nearly domino square (i.e. a Cynthia) of size 7, and a 5 x 6 cylindrical horizontal domino rectangle (i.e. a cord).

If each domino lies within a row we call the configuration horizontal and if each lies in a column we call it oertical. If there is one which lies within a row and another which lies within a column we call it mixed. If in a configuration we allow some domino to have one cell in the first column and one in the last, then we call the configuration cylindricaf. If a nearly domino square has all its non-domino cells lying on the main diagonal we call it a transversal nearly domino square. Cylindrical transversal horizontal nearly domino squares, of which one is illustrated in Fig. 5, seem to be very pretty but hard to make. We call them Cynthias. They are very similar to Room squares and are necessarily of odd side. Cylindrical horizontal domino rectangles (or Chords), also illustrated in Fig. 5, are even harder to make.

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A pair A = ( a i j :i, j = 1,. . . ,n), B = ( b i j : i, j = 1,. . . ,n ) of latin squares is said to be orthogonal if the set { ( a i j b, i j ) :i, j = 1, . . . ,n ) contains n2 elements, or, in other words, i l = i ~andjl = J 2 . (1) We consider two natural extensions of this idea. We call a pair A = (aij),B = (bij)of the configurations under discussion weakly orthogonal (or sometimes simply orthogonal) if (1)is true. We call them strongly orthogonal if they are weakly orthogonal and also their dominoes match up, or, in other words, if when a pair of adjacent cells are covered by a domino in A then the corresponding pair of adjacent cells are covered by a domino in B. We call a pair A , B of our configurations u-orthogonal (or unequalorthogonal) if they are weakly orthogonal and aijf bij for all pairs (i, j ) . We illustrate these notions in Figs. 6 and 7. Two strongly orthogonal domino configurations are said to be t-orthogonal (or turn-orthogonal) if one may be derived from the other by turning all the dominoes round. We shall also say, more simply, that A itself is t-orthogonal. Each of the 5 x 4 domino rectangles in Fig. 7 are t-orthogonal and so are their completions. If the completion of an ( n + 1) x n domino rectangle is t-orthogonal then clearly so was the domino rectangle. One should note, however, that the converse is not true,-for it is possible for a domino rectangle to be t-orthogonal and for its completion not to be t-orthogonal. Two examples of this are given in Fig. 8. (ailjl,biljl)=(ai2j2~ bi2j2) 3

FI 3

2

Fig. 6. Weakly orthogonal pairs of domino squares of side 2, whim domino squares of side 3 , 3 x 2 domino rectangles, and 4 x 3 domino rectangles.

3

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Fig. 7. Strongly u-orthogonal pairs of whim domino squares of side 3 and 5 x 4 domino rectangles.

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5

4

3 2

3

4

Fig. 8. An 8 x 7 domino rectangle and horizontal 7 x 8 domino rectangle, both being t-orthogonal but neither having t-orthogonal completions.

2. The existence of (n + 1) x n domino rectangles, domino squares of side n, nearly domino squares of side n and whim domino squares of side n In this section we show that all these configurations exist for many appropriate values of n.

Theorem 1. For each n > 1, an (n+ 1) x n domino rectangle exists. Proof. Clearly a 2 x 1 domino rectangle exists. For each m a 2 arrange the numbers 1 , . . . ,2m or the numbers 1, 2,. . . , 2 m - 1, 2m+ 1 into m pairs (al, bl), (a2,b2), . . . , (a,,,, b,) so that bi-ai=i

( 1
This can always be done; the details of this arrangement into pairs are given in [3] for all m except for m= 1, 2, 3, 5 and 6. For these values the arrangements are: m=2; m=3; m=5; m=6;

(1,2),(3, 5 ) ; (1,2),(3, 51, (4,7); (8,9), ( 1 , 3), (4,7), (2,6),( 5 , 10); (10, 1 I ) , (2,4),(3,6),(5,9), (8, 13), (1, 7).

Fig. 9 shows an ( n+ 1 ) x n domino rectangle with n even (= 2m, say); the numbers are taken modulo n + 1 ( = 2m + 1). Fig. 10 shows an ( n+ 1) x n domino rectangle with n odd ( = 2m + 1, say); the numbers are taken modulo n + 1 ( = 2m + 2). In both cases it is easy to see that no number is repeated in any row, that no number is repeated in any column, and that no domino is used twice. This proves Theorem 1.

Theorem 2. For each n 2 1 a nearly domino square of side n exists. Proof. For n = 1 D l is a nearly domino square of side n. For n 3 2 adjoin a final column to the n x ( n- 1) domino rectangle of Theorem 1.

J . A . Edwards et ul

100

al

bl

a +I bl +I

1

a +i

a +n

1

b,,+i

a2

a

+'I

a +i

a +n bl+n 2

b2

. . . .

a

b2+l

. . . .

a +I

b +I

b2+i

. . . . .

. . . . .

. . . . .

. . . . .

a

L'

b2+n

. . . . .

. . . . .

. . . . .

. . . . .

a +n

bm

+;

+i

bm+n

Theorem 3. For each even n >,2, a domino square of side n exists. Proof. Take the (n + 1) x n domino rectangle exhibited in Fig. 9 and delete the last row. Theorem 4. For each n = 1 (mod 4), a whim domino square of side n exists. Proof. Take the (n+ 1) x n domino rectangle exhibited in Fig. 10, delete all dominoes covering one or two cells in the last row, move the last column to the middle and the last remaining row to the middle. Horizontal domino rectangles of size (n- 1) x n with n even are equivalent to balanced tournament designs whose existence was proved by Schellenburg, Van Rees and Vanstone [6]. A partial solution to the existence problem was found earlier by Hazelcroft and Leach [2] and another partial solution was found independently, later and in yet another formulation by Schonheim [7]. Thus we have the following theorem. Theorem 5. For n even, n # 4, there is a horizontal (n- 1) x n domino rectangle. The 5 x 6 horizontal domino rectangle shown in Fig. 11 has two interesting properties. One is that the last row of the completion is obtained from the first by turning the dominoes of the first row round. The second is that it is 'unique' in the sense that all the other 5 x 6 horizontal domino rectangles may be obtained from it by a combination of the following four operations: (1) Permuting the rows.

Domino squares

101 I

0 m+l

1 r + 2

Fig. 10. A (2m+ 2) x ( Z m + 1 ) domino rectangle

Fig. 1 I . The 'unique' 5 x 6 horizontal domino rectangle.

(2) Permuting the symbols. (3) Permuting the columns of dominoes. (4) Turning some of the dominoes in any given column round (so that a domino rectangle is obtained); e.g. C b 1 can be turned round without turning any other domino round; similarly the 4 dominoes &5,40,5-1, 1-4 could all be turned round without turning any other domino round. There is no 7 x 8 horizontal domino rectangle with the property that the last row of the completion may be obtained from the first by turning the dominoes of the first row around. It is however possible for 9 x 10.

Problem 1. We d o not have any general constructions for double diagonal domino squares of side n or for double diagonal whim domino squares of side n (the domino

I02

J . A . Edwards et ul.

square of side 4 in Fig. 1 and the whim domino square of side 3 in Fig. 2 are both double diagonal).

Problem 2. We do not have a general construction for whimsies valid for n = 3 (mod 4). 3. The construction of t-orthogonal domino configurations In this section we give some methods of constructing t-orthogonal domino rectangles, domino squares and whim domino squares.

Theorem 6. I The completion of any (n+ 1) x n horizontal domino rectangle is t orthogonal (necessarily n must he even). Proof. I Let A be a horizontal ( n+ 1) x n domino rectangle and let B be obtained from A by turning all the dominoes round. Let A* and B* denote the completions of A and B respectively. Then A* and B* will have the same final column. Also A and B will be u-orthogonal since in A each i occurs exactly once in a domino with j f i . It then follows that A* and B* are t-orthogonal to each other. Theorem 7. For each even n 2 2, there is an even ( n + 1 ) x n domino rectangle whose completion is t-orthogonal, and there is also a t-orthogonal domino square qf’side n. Proof. Let A be the ( n + 1) x n domino rectangle exhibited in Fig. 9. Then, by Theorem 6, the completion of A is t-orthogonal, To obtain a t-orthogonal domino square of side n, simply delete the last row from A . This proves Theorem 7. Theorem 8. I f n

-

+

1 or 7 (mod 8) there is an (n 1 ) x n t-orthogonal domino rectangle.

Remark. The completion of the ( n+ 1 ) x n domino rectangle we construct below is not t-orthogonal. Proof. The theorem is true if n = l . Let n 2 7 and let m = i ( n + l ) . Then m - 0 or I (mod 4). Then the numbers 0,. . . , 2 m - 1 may be arranged into m pairs (u,, b l ) , ( a z ,bz), . . . ,(a,, b,) so that bi-ai=i

(ldidm).

For m = 5 this is shown in the proof of Theorem 1, and for all other values of m it is shown in [3] (or [8]). The ( n + 1) x n domino square with m even is shown in Fig. 12 (the numbers being taken modulo 2m). Notice that b, + i is missing from row 2i + 1 and a, + i is missing from row 2i + 2. Therefore the rectangle obtained by turning the dominoes round is a domino rectangle. The other details of the proof are as in the proof of Theorem 1.

Problem 3. For n 3 3 or 5 (mod 8) we do not know how to construct an ( n+ 1) x n t-orthogonal domino rectangle.

103

Domino squares

a b, +m

bm

a +m+l 1

d

1

ti

a +m+i 1

a +m+l 2

bl+m+l

bl +i bl+m+i

7

. . . .

bl +I b2+m+l

a +i b 2 + i

2

a +m+i 2

b2+m+i

. . . .

. . . .

. . . .

. . . .

. . . . . . . . . . . .

a

m-I

+Ibm-l+l

am-l +m+l

bm-l

+m+l

am-,+i

bm-l+i

a +I

bm+l

a +i

a m _ , + m + i bm_l+m+i

b,+i

a

a +m-

. . . . al + m - I

bl+m-I

a2 + m - I

b2+m-l

a +Zm-I

bl+2m-I

a2+2m-1

b2+2m-1

1

I

. . . . .

. . . . . . . . .

. . . . . .

m-I

+ m - I bm_l+m-l

a +2m-1 b m _ , + 2 m - l m-I

b +m-

I

Fig. 13. A t-orthogonal whim domino square of side 9.

Problem 4. For n odd we do not know how to construct a pair of (n+ 1) x n domino rectangles whose completions are t-orthogonal to each other. Probably for n small they do not exist. We think for larger n they very likely do exist. It is easy to see that if they exist they have to be mixed. Theorem 9. If n = 1 (mod 8) there is a ?-orthogonal whim domino square of side n. Proof. Take the t-orthogonal (n+ 1) x n domino rectangle constructed in Theorem 8, discard all dominoes overlapping the last row, move the last column to the middle and the last row to the middle (see Fig. 13).

J . A . Edwards et al.

104

Problem 5. For n = 3,5 or 7 (mod 8) we do not know how to construct a t-orthogonal whim domino square of side n. 4. The construction of complete sets of mutually orthogonal n x (n - 1) domino rectangles

It is well known (see, for example, [ 5 ] ) that one cannot have more than n - 1 mutually orthogonal latin squares of side n. The argument that proves this fact may also be used to prove that one cannot have more than n - 1 mutually orthogonal n x (n - 1) domino rectangles. A set of n - 1 mutually orthogonal n x (n - 1) domino rectangles is called complete. We show now how to construct such a complete set when n = p", r 2 1 and p is an odd prime.

Theorem 10. If q = p", where p is an odd prime, there is a set of q - 1 t-orthogonal domino rectangles whose completions form a complete set of mutually orthogonal nearly domino squares of side q. Remark. None of the set of q - 1 domino rectangles we construct may be obtained by turning the dominoes of another around. The example in Fig. 14 illustrates the construction when q = 5. Before proving Theorem 10 we state and prove the following lemma, which was kindly pointed out to us by D. G. Rogers.

Lemma 1. Let q = pr where p is an odd prime and let s = i ( q - 1). The non-zero elements . . . ,cq- ofGF(q) can be arranged into s pairs ( a l , bl), ( a z ,bz),. . . ,(a,, b,) such that the q - 1 numbers f ( b l -al), f(bz -az), . . . , -t(b,-a,) are all distinct.

cl,

1 0

' 1 2

4 1 3 1

3

2

3

4

2

3

4

0

4

3

2

2

3

0

4

4

0

2

1

4

2

0

Fig. 14. A set )f 4 mutually orthogonal completions of 5 x 4 t-orthogonal domino rectangles.

Domino squares

105

Proof. The elements of GF(q) may be represented as vectors ( f l , . . . ,fi) with each component being the integers modulo p and addition being vector addition. Let (ai, b i ) run over all distinct sets {(fi,. . . , J., ( p - f l , . . . , p - f i ) } , where not all of f l , . . . ,fi are zero. Then {ai- bi, bi -ai} runs over all sets {(p- 2f1, . . . ,p - 2J), (2S,, . . . ,2f,)};these are easily seen to be distinct, and the lemma follows. Proof of Theorem 10. It is well known [ 5 ] that if the elements of GF(q) are denoted by co=O, clr c 2 , . . . ,c,- 1, then the set of matrices ID"), D"), . . . ,D t q - ' ) } , where LYe)=(d$)) and d!;)=cic,+c,

(i, j = O ,..., q - l ; e = l , . . . , 4 - 1 )

form a complete set of q - 1 mutually orthogonal latin squares. We show that the same permutation may be applied to the columns of each of these latin squares so that the first q - 1 columns of each form a q x (q - 1 ) domino rectangle whose completion is t-orthogonal. Applying the same permutation to the columns of each of a set of mutually orthogonal latin squares does not destroy the orthogonality (see [S]), so the set of completions of domino rectangles that we finally produce remains a complete set of mutually orthogonal latin squares. Let s = i ( q - 1) and let (al, bl), . . . ,(a,, b,) be the arrangement of the non-zero elements of GF(q) into disjoint pairs of Lemma 1 . Then let the permutation of the columns of each of D"), . . . ,D(4- ') be such that, in the first row of each, ai and bi are placed in columns 2i - 1 and 2i respectively (1
Problem 6. It seems quite probable that there is some way of proving the existence of more than two mutually orthogonal q x (q - 1) domino rectangles when q is a power of two, but we have not been able to see how. 5. A direct product construction

It is well known [ 5 ] that, given sets { A l , . . .,A,} and { B l , .. . , B , } of mutually orthogonal latin squares of side n and m respectively, a set { C1, . . . ,C,} of mutually orthogonal latin squares of sides nm can be constructed. We show in this section that this construction can in certain cases be adapted to domino rectangles. Let n and m be odd positive integers and let A and B be horizontal n x (n - 1) and m x (m - 1) domino rectangles respectively. We show how to construct from A and B a horizontal nm x (nm- 1) domino rectangle A @B, the direct product of A with B. Let A* = ( a i j )and B* = (b,,) be the completions of A and B respectively. We first form an nm x nm latin square X of the form of Fig. 15. In this each element aijof A* is replaced by a latin square Aij of side m, where each A i j is a copy of B* formed by replacing each element b,, of B* by the ordered pair (aij, bJ. It is easy to see that this matrix X is a latin square.

J . A . Edwards et al.

106

Fig. 15.

Foreach iand kwith l < i < n a n d l
..

9

CG,D!,k, . . .

9

DTk)

where each C and D is a column of length m. We then permute the columns of each such submatrix so as to produce the submatrix Now, for each such i and k and for each I with 1 < I


and of submatrices may be paired off into the dominoes

and

respectively, where 1

[

(a, b)

(a', b')

I

with a fa' and b # b' is formed exactly once. For each i, 1
[

(ai.n, b p . 2 2 - 1 )

(ai.n,b p . 2 2 )

1

where 1< p < m and 1< I < f ( m - 1). Let the latin square obtained after all these permutations have been carried out be described by a modification of Fig. 15 in which the new submatrices are A: instead of Ai,. Then all elements are paired into dominoes except those of the last column of each Aij. Now permute the columns of the whole matrix so as to move all n2 of these last columns of the submatrices A: to the last n columns of the whole matrix, keeping

107

Domino squares

. .

Laln,

blm)

(aln,

bZm)

.

(aln.

bmm)

.

(ajn,

blm)

.

(ajn,

bZm1

.

Lajn, bmml

. . .

Lann,

blm)

( a n n . bZrn1

Lann,

bmm)

their relative order the same and the relative order of all the other columns the same. Call the latin square obtained Y. Then all pairings into dominoes described above remain extant in Y. The final n columns of Y are indicated in Fig. 16. We now permute the rows of Y so that the final n columns of Y are rearranged into the following form :

Here E v (1 s v
J . A. Edwards er al.

108

Thus E , is a copy of A* with each element ai,jreplaced by (ai3j,bv,m).Therefore each element of each E v , except for those elements in the last columns, may be paired off into dominoes of the type

1

~ . _ _ _ _ _ ~ _ _ _ _ _ ~

(a;.26 - 1,

bv.m)

(ai,263 b v , m )

1

where 1 sh
A =

0

1 1 2

4

0

EI= 6 c r

A @B =

Fig. 17. A @ E .

4

109

Domino squares

Now we are able to state and prove the analogue of the result on mutually orthogonal latin squares mentioned above.

Theorem 11. Let n and m be odd positive integers and let [ A , , . . . , A,) and { B l ,. . . , B , f be two sets of s mutually orthogonal horizontal domino rectangles of sizes n x ( n - 1 ) and m x ( m - 1) respectively. Then ( A , @Bl, A 2 @ B 2 , .. . , A , @B,f is a set of s mutually orthogonal horizontal domino rectangles of size nm x (nm - 1). Remark. Since each A i@Bi is horizontal, it necessarily has a t-orthogonal completion.

Proof of Theorem 11. Let the set of latin squares of the form of X in Fig. 16 which may be obtained from A l , B , ; A 2 , B 2 ; .. . ; A,, B, respectively be denoted by { X , , . . . , X , ; . It is easy to see (and well known) that X I , . . . , X , are mutually orthogonal. The subsequent manipulations are the same for each X iand consist entirely of permutations of rows or columns. It is easy to see (and again well known) that this does not affect orthogonality, so Theorem 11 now follows. Using Theorem 11 it follows that if I = p : l p ; * . . . p:l, where p , , . . . ,p , are distinct odd primes and a , , . . . ,a, are all positive, then there exist at least minl ,
6. Concluding remarks-Cynthias and Chords As we have seen, quite a number of interesting problems arise in trying to construct approximations to latin squares out of a set of dominoes. One problem which is almost certainly amenable, but which is probably fairly hard, is to show the existence of Cynthias of side 2n + 1 for all n 20,n # 2. One approach to this problem is to adapt the starter-adder method used so successfully for Room squares [4]. Let G be the additive cyclic group of integers modulo 2n 1. By a Cynthesizer in G we mean a set X = [(x,, y , ) , (x2, y 2 ) ,. . . ,(x,, y,,)} of ordered pairs of elements of G such that (i) the elements x,, y , , x 2 , y,, . . . ,x,, y , comprise all the non-zero elements of G, (ii) the differences f ( x i - y i ) (1 < i < n ) comprise all the non-zero elements of G, generating each exactly once, and (iii) the elements

+

x l + l , y 1 + 2 , x2+3, y 2 + 4 , . . . , x , + 2 n - 1 , y , + 2 n

comprise all the non-zero elements of G. Let C = ( c i j )be the following (2n + 1 ) x (2n + 1) matrix. The first row of C is (0,y n . x,, y,-

1, Xn- 1 , .

. . y,, 9

XI

1

with y i and xi (1 < i < n ) 'paired off as a domino. Row i + 1 of C is obtained from row i

J . A . Edwards et al.

110

by the rule ci+1,

+

= ci,

+1

(1 < i < 2n, 1< j < 2n

+ 1)

and if ci,j is paired with c i , j + lin row i then ci+I , j + is paired with ci+ 1 , j + in row i + l ( l < i < 2 n , l < j < 2 n + l ) ; thesubscriptsherearemeant tobereadmodulo2n+l.

Theorem 12.tThe matrix C ( X ) constructed as above from the Cynthesizer X is a Cynthia of side 2n 1.

+

Proof. It follows from properties (i) and (ii) together with (2) that each domino is used exactly once. Similarly it follows from (i) and (2) that no symbol is repeated in any row. To see that no symbol is repeated in any column we first observe that (2) provides a rule for obtaining column j + 1 from column j (1 <j<2n + 1) and it is sufficient to show that no element occurs twice in column 1. But column 1 is the transpose of ( 0 , x l + 1 , y l + 2 , x 2 + 3 , y z + 4,..., xn+2n-l,y,+2n) in which, by property (iii), no element is repeated. This proves Theorem 12. It is fairly easy to see that there is no Cynthia of side 5. Cynthias of sides 1 and 3 are trivial. A Cynthia of side 7 is displayed in Fig. 5. This corresponds to the Cynthesizer (XI,yl, x2, yz, x3, y3)=(2,6,3, 1,4,5). Another Cynthesizer for 2n + 1 = 7 is (3,4,6, 1, 5, 2). Cynthesizers for 2n+ 1 =9, 11 and 13 are, respectively, (2, 3,7,4, 1, 5,6, 8), ( L 6 , 2, 3, 5 9 , 7-4, 8, lo), (4, 1,9, 5, 2, 8, 3, 1L6, 7, 10, 1% and so, using these, we can construct Cynthias of sides 9, 11 and 13

Problem 7. We do not have any general methods for constructing Cynthesizers and we do not have any other method for constructing Cynthias.

Fig. 18. A 7 x 8 Chord.

Domino squares

111

Problem 8. It would be pleasing to find a Cynthia with the property that the two terms of each of the dominoes can be gathered into one or other of the two cells to produce a Room square. The corresponding problem of constructing Chords seems to be rather more difficult. All we have at the moment are some examples. Some are given in Fig. 5 and a further one, size 7 x 8, is given in Fig. 18.

References H. Hanani, A note on Steiner triple systems, Math. Scand. 8 (1960) 154156. J. Hazelcroft and J. Leech, A tournament design problem, Amer. Math. Monthly 8 (1977) 198-201. A. J. W. Hilton, On Steiner and similar triple systems, Math. Scand. 24 (1969) 208-216. R. C. Mullin and E. Nemeth, On furnishing Room squares, J. Combin. Theory 7 (1969) 266272. H. J. Ryser, Combinatorial Mathematics, Carus Mathematical Monographs (Wiley, New York, 1963). P. J. Schellenberg, G. H. J. van Rees and S. A. Vanstone, The existence of balanced tournament designs, Ars Combin. 3 (1977) 303-318. [7] J. Schonheim, Unpublished manuscript. [8] Th. Skolem, On certain distributions of integers in pairs with given differences, Math. Scand. 5 (1957) 5768. [9] Th. Skolem, Some remarks on the triple systems of Steiner, Math. Scand. 6 (1958) 273-280. [l] [2] [3] [4] [5] [6]

Received 8 May 1980.

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Annals of Discrete Mathematics 12 (1982) 113-116

@ North-Holland Publishing Company

SOME PROBLEMS ON ADDITIVE NUMBER THEORY P. E R D ~ S Hungarian Academy ofsciences, Budapest, Hungary

Dedicated to Professor A. Kotzig on the occasion of his sixtieth birthday Denote by f ( n )the largest integer k for which there is a sequence 1 < a l <** .
f(n) = n ‘I2

+ O(I).

(1)

The conjecture seems to be very deep and I offered long ago a prize of 500 dollars for a proof or disproof of (1).The sharpest known results in the direction of(1) state [ 5 ] n1/2-n1/2-c

(2) In several papers Abrham, Bermond, Brouwer. Farhi, Germa, Kotzig, Laufer, Rogers and Turgeon considered the following somewhat related problem. Let m, n l , . . . , n,, c be positive integers. Let A = { A 1 , . . .,A,,,] be a system of sequences of integers < f ( r ~ ) < n ” ~ + n ” ~1.+

A 4 i = ( a i , l < . . . < u i , n i ! , i=l,. . . , m

(3)

and let Di={aj,j-aj,k

1 1< k < j < n i , \

(4)

be the difference set of Ai.The system S = ( D i , . . . , D,,1 is called perfect for c if the set D = i= 1

uy= Di consists of the integers

();

Clearly, for a perfect system, the representation o f t in the form (4) must be unique. The authors proved several interesting results on these sequences [l, 2 , 3 , 4 , 7 , 8 , 9 , 10, 123, but many interesting unsolved problems remain. Put

N=

f (;).

i= 1

J. Abrham proved in [l] that, for every perfect system, m > a N , where a>O is an absolute constant. The best value of a is not yet known, though Kotzig has some plausible conjectures.

I noticed some time ago that the method that Turan and myself used to get an 113

P . Erdos

I14

upper bound for f ( n ) can make some useful contribution to the study of perfect systems. In particular I prove the following:

Theorem. Assume that the integers (4) are all distinct and are all in [ l , N ] , and that Di, nDi, = 8 for all 1 < il < i2 <m. Then, to every E >0, there is an q >0 so that, for N > NO(&,q), if ID1 >( 1 &)N/2then m > qN.

+

Let me first explain the relation of this result to our old result with Turan. Our result with Turan states that if m= 1, then (DI <(1+ o( 1))N/2,and our theorem states that if ID1 > ( 1 +&)N/2then m>qN, i.e. the number of sequences must be large. Perhaps the following problem is of some interest. Let l < a l < ai-a,,

* * *



l < j < i < k l , l
are all distinct. Determine or estimate

as accurately as possible. Our theorem gives g ( N )< ( 1 + o( 1))N/2and trivially

Is it true that

+

Perhaps ( 5 ) is too optimistic. It might be of some interest to investigate max(kl k2). Clearly max(kl + k 2 ) > f ( N ) but it is not clear if max(kl +k2-f(N))+oo. Now we prove our theorem. The proof will be very similar to our old proof with Turhn. Let t = to(&,m) be large but fixed, i.e. t is independent of N . To prove our theorem it clearly suffices to consider the sequences Ai satisfying IAi(>t (i.e. ni>t). To see this, observe that the contribution of a sequence ( A i l s t to D is at most (;); thus one needs qlN of them to significantly change the size of D. We will only consider the integers 1 < m < N/t1’2. Clearly every such integer m has at most one representation of the form ai,j-ai,k, 1
I,= [x, x + N/t’”]. Consider the set of all the differences

Some problems on additioe number theory

115

Since, by our assumption, the ai,j-ai,k (1 GiGrn, 1 < k < n i ) are all different, an integer m=ai,j-ai,k can occur in (6) for at most N t - 1 / 2 - r n intervals I,, i.e. both ai,jand ai,k must be in I,. Thus rn either does not occur in (6), or it occurs N t - ' l 2 -rn times. Observe that N

ni(x)=niNt-112

(7)

x=l

since each aiVjE Ai occurs in N t form U i ,j - U i , k ,

X
1/2

<( l i , j <

intervals I,. Now the number of differences of the X

+N t -

1/2

clearly equals .

Now

$

i= 1

.

ti:?

is minimal if the ni(x)are as nearly equal as possible. Thus from (7) we obtain by a simple computation for every fixed 6, if t > to(@,

$

( [ w ; " ~> ]( 1)- 6 ) -Nn! 2t

x = 1 ti:))..

Thus from (8)

On the other hand, since rn can occur at most N t - 1 / 2-rn times in (6) we obtain

Thus from (9) and (10) m

n:

N>(l-6) i=l

Now by our assumption

PI=

f (;)>(l+e)?,

N

i= 1

or

... n!>(l+&)N i=l

which contradicts (11) for sufficiently small 6, and hence our theorem is proved. It is not difficult to deduce the result of Abrham from our theorem. If c = 1 this is

P . Erdos

116

obvious. If c = o(N)then since

c

Ni- 1/2 ... I=1

c+

I=(l+o(l))

NI-

c

I

I=c

nothing has to be changed in the proof. If c > q N then (13) of course no longer holds. But if c > qN everything is trivial. To see this observe that if al a2< . . . qN, we immediately obtain k < 1 + l/q, thus m >q N immediately follows.

-=

+

-=

References [l] J. Abrham, Bounds for the sizes of components in perfect systems of difference sets, this volume, pp. 1-7. [2] J.-C. Bermond, A. E.Brouwer and A. Germa, Systkmes de triplets et difftrences ass&&, Probltmes Combinatoires et Thtorie des Graphes, Colloq. Internat. Orsay 1976, Colloq. C.N.R.S. No. 260 (Paris, 1978)pp. 35-38. [3] J.-C. Bermond and G. Farhi, Sur un probltme d’antennes en radioastronomie 11, this volume. [4] J.-C. Bermond, A. Kotzig and J. Turgeon, On a combinatorial problem of antennas in radioastronomy, in: Combinatorics, Proc. 5th Hungarian Combinat. Colloq. 1976 (North-Holland, Amsterdam, 1978)pp. 139-149. [5] P. ErdBs and P. Turan, On a problem of Sidon in additive number theory and on some related problems, J. London Math. Soc. 16 (1941) 212-216. [6] H. Halberstam and K. F. Roth, Sequences, Vol. 1 (Oxford Univ. Press, Oxford, 1966). [7] A. Kotzig and J. Turgeon, Perfect systems of difference sets and additive sequences of permutations, Proc. 10th S.-E.Conf. Combinatorics, Graph Theory and Computing, Boca Raton 1979, Vol. 2, Congr. Numer. 24, pp. 629-636. [8] A. Kotzig and J. M.Turgeon, Sur I’existence de petites composantes dans tout systeme parfait d’ensembles des diffbrences, Ann. Discrete Math. 8 (1980)71-75. [9] P. J. Laufer, Regular perfect systems of difference sets of size 4, this volume, pp. 193-201. [lo] P. J. Laufer and J. M. Turgeon, On a conjecture of Paul Erdos for perfect systems of difference sets, Discrete Math., to appear. [ll] B. Lindstrom, An inequality for &sequences, J. Combin. Theory 6 (1969)21 1-212. [I23 D. G . Rogers, Addition theorems for perfect systems of difference sets, J. London Math. SOC.23 (1982)385-395. Received 15 September 1980; revised 15 May 1981.

Annals of Discrete Mathematics 12 (1982) 117-123

@ North-Holland Publishing Company

ON ALMOST BIPARTITE LARGE CHROMATIC GRAPHS

P. ERDOS, A. HAJNAL and E. SZEMEREDI Hungarian Academy olSciences. Budapest. Hungary

Dedicated to Professor A. Kotzig on the occasion of his sixtieth birthday

0. Introduction

In the past we have published quite a few papers on chromatic numbers of graphs (finite or infinite), we give a list of those which are relevant to our present subject in the references. In this paper we will mainly deal with problems of the following type: Assuming that the chromatic number ~ ( 3of)a graph $ is greater than K , a finite or infinite cardinal, what can be said about the behaviour of the set of all finite subgraphs of $. We will investigate this problem in case some other restrictions are imposed on 3 as well. Most of the problems seem difficult and our results will give ) be arbitrarily large while the just some orientation. The results show that ~ ( 4 8 can finite subgraphs are very close to bipartite graphs. It is clear from what was said above that this topic is a strange mixture of finite combinatorics and set theory and we recommend it only for those who are interested in both subjects. Finally we want to remind the reader the most striking difference between large chromatic finite and infinite graphs which was discovered by the first two authors about fifteen years ago [4]. While for any k < o there are finite graphs with z(Y)>k without any short circuits, [ l ] , a graph with x ( ~ ) > K ~ ohas to contain a complete bipartite graph [k, K ' ] for all k < w. Hence such a graph contains all finite bipartite graphs, though it may avoid short odd circuits. Our set-theoretical notation will be standard as for graph theory we will use the notation of our joint paper with F. Galvin [3] with some self-explanatory changes.

1. The standard examples and the ordered edge graph

Definition 1.1. For each ordinal a, for 2 < k < o,and for 1< i
is the graph with V#(a,k, i)= Calk, where E(a, k, i ) is defined by the following stipulations. For X E Calk we write X = { x o , . . . , xk- 1 ) with xo< . . . < x k - and for X , Y€[aIk {X, Y } EE(a,k,i) iff y j = x i + j f o r j < k - i - l . We call Y0(a,k, 1)= %(,a,

k) and we call this graph the k-edge graph on a. 117

P . Erdos et al.

I18

Definition 1.2. For each ordinal a, for 3 < k < o and for 1 < i < k - 1, Y1(a, k, i)= (Vl(a, k, i), &(a, k, i ) )

is the graph with Vl(a,k, i)= Calk where for X, Y E [.Ik

x, YEEl(a, k , i )

iff

xi
<Xk-2
We call Yl(a, k , i ) the k, i-Specker graph on a, and just as in the case above we omit i for i= 1. The above graphs are standard examples of large chromatic graphs, we list some properties of them.

Lemma 1.1. (For the proofs see [3] and [4]). (a) x(Yo(a, k , i ) ) > K prooided a-(2k)f: and, as a corollary of this, for all x(Y0(@,k, i ) ) > K prooided

K),o,

a2(expk-,(ic))+ f o r a l l 2 < k < o , l
+ 00

X(Yl(n,k, i))-

if n+

+ 00.

(c) 4B0(a,k ) does not contain odd circuits of length 2j+ 1 for 1< j < k - 1. (d) Y1(a,k ) does not contain a complete rc-graphfor 3
Definition 1.3. For a graph Y = ( K E) and for an ordering < of V we define the ordered edge graph OE(48, <)=(V(Y, <),E(Y, <))of Y for the ordering <, as follows V(48,<)=E. For X E [ V ] write ~ X = { x o , x l } where xo<xl. For X , YEE put { X , Y} E E(Y, <)iff either x l = y o or

yl=xo.

It is clear from the definition that the 2-edge graph Yo(a,2) is the ordered edge graph of the complete graph with vertex set a for the natural ordering of a. More generally the following is true:

Lemma 1.2. For each a, and for 1 < k < o there is a well-ordering

can be any ordering of

min X <min Y

[.Ik

satisfying

3 x
Lemma 1.3. Let Y be a graph with x(Y)<2" for the vertices. Then x(OE(Y, < ) ) < 2 ~ .

Proof. Let Y be Y =

( x E). Then

K),

1 and let

< be any ordering of

V is the union of 2" independent sets. Since the

Almost bipartite large chromatic graphs

119

complete graph of 2" vertices is the union of K bipartite graphs, there are disjoint partitions V = A i u B i of V for i < K such that E c U i < , [ A i , Bi]. We now partition each [Ai, Bi] into the union of two sets independent in OE(Y). Let Ci= { {xo,x , ~ E} [Ai, Bi]: x j E Ai} for j < 2 . Clearly Ci contains no edge of OE(Y) and CbuCi=[Ai, Bi] for i < K . In what follows log denotes the 2 basis logarithm and log'k) is the k-times iterated logarithm function.

Corollary 1.4. If Y is a subgraph of n uertices of some Yo(a,k ) for k >,2 then x(Y)
log"-"(n)

for some ck>O.

To close this section we mention the first problem we cannot solve.

Definition 1.4. For each (infinite) graph Y= ( V , E) let f&)=max{X(Y(A)): A c V ~ l A l = n } .

Problem 1. For what functions f : w + o is it true that for all cardinals is a graph 48 with x(Y)> K and f$(n)
K>O

there

Corollary 1.4 shows that 10g'~)(n)is such a function for all k w there is a graph with x(48)> K all whose finite subgraphs are in $ then 9 must contain all finite subgraphs of Yo(o, k) for some k
2. Omitting vertices of subgraphs DefinitiOn 2.1. Let Y = ( V , E) be a graph; fsl'(n)=min{max{lZI: Z c A ~ Z i s i n d e p e n d e n t ) .A: c l / ~ l A ( = n ) , f&~)=min{max{lZI: Z c A r \ Y ( Z ) i s bipartite): A c V ~ ' l A l = n ) . Clearly fi(n)>if&n) holds. We are interested in the problem if these functions can be large for a graph of large chromatic number. The following theorem contains our main information about this.

120

P. Erdos et al.

Theorem 1. For all E > 0 and for all K there is a graph Y with ~ ( 4 6>) K such that f A n ) >,( 1 - E ) n

holds for all n < w .

Proof. By Lemma 1.1(a)we only have to show that for Yk = Yo(@,k ) f:,(n) >,( 1 -2/k)n.

Let A c [0Ik,( A (= n, for some 2 < k < w and for some n. We prove, by induction on n, thqt there is a set Z c A, [Zl a(1 - 1/2k)n such that Z spans a bipartite graph of Y(o, k ) . The statement is trivial for n = 1. Assume n > 1 and that the statement is true for all n k n . ForxEWlet S ( X ) = { X E A x : ~ X } , a n dP ( x ) = { X ~ A : x = x ~ f ol O and IP(x)l>,(1-2/k)lS(x)J. Now the graph induced by P(x) can be shown to be bipartite bythefollowingpartition:Ao={X ~ P ( x ) : x = x ~ ~ i i s e v e n ) , (AX, ~= P ( x ) : x = x ~ r \ i is odd}. A o u A l =P(x) and A j contains no edge of Y(w, k) for j < 2 . By induction, there is a subset Q c A\S(x), IQI >,( 1 - 2/k)(n - (S(x)J),which induces a bipartite graph of Yo(o, k). Since no edge of Yo(w, k) joins a point of P(x) to a point of A W x ) , Z = P(x)uQ satisfies the requirements of the theorem.

,

The theorem yields that there is a sequence E,+O so that for some Y with x(Y)=w, n( 1 - E J . We do not know how fast E, can tend to 0. A result of Folkman gives information for the case $ n - f d ( n ) is bounded. This says that if i n - f J ( n ) < k then x(Y) < 2k 2. Here we at least know that the situation is different for graphs with chomatic number >o. &I)>

+

Lemma 2.1. !fx(Y)>w then there is an E > O such that f$(n)<($-E)n.

hf. Clearly Y contains HIvertex disjoint copies of some C Z i + for , some i < w . The union of m copies has cardinality rn(2i+ I ) and contains no free set larger than mi. Problem 2. Does there exist a graph Y and c > O such that x(Y)=w,, lYl=ol and f A n ) >cn. Here we only have the very little information that the Specker graph Y l ( o l , 3) does not have this property. To see this it is sufficient to prove:

Tbeorem 2. Let Y = Y,(w, 3). Then

Proof. We define an B c . [ n l 3 with lqa n log n/8 log log n such that for all M c 9, M independent in 3 ' , IM(<4n holds for all sufficiently large n. This will prove the result. Let ai = [log nilfor i < io = [log n/log log n ] . Then ai E n for i < i o . Let X ( t , i ) = ( t , ? + a z i ,t + a 2 i + l ) and B={X(t, i ) : X(t, i)cn}. Then ( B ( > , nlog n/8 log log n provided n is large enough. Let M c $ JMI34n. We prove that M contains an edge of

Almost bipartite large chromatic graphs

Y.Let Mi= { t : X(t, i ) E M ) . Then

121

JMil= IMI 24n. Let

N i = { t € M i : 3 t ' ~ M r'+azi
We claim that [Nil2 [Mil- 2n/log n. We can choose elements to, . . . ,tj- of Mi so that M i c U { [ t l , t r + a 2 i + l ) :l < j ) where the above intervals are pairwise disjoint. Now M i \ N i c IJ{ [trt tr+azi+ I]\(tl+a2i, tl+a2i+ 1): l<j>. Since a2i 2 -<-, a ~ i + l logn

/Mi- Nil <2n/log n. It follows now, that

c INiI 2 C IMiI

i < io

-

i < io

2n log log 3 4 n -

2n >n. log log n

-~

Then there are i l < i2 with NilnNi,#O. Let t E NilnN i , . Then for some t' E Mi,, t ' + ~ ~ ~ , < t < t ' + a ~X(t', ~ , +il) ~ ,E M. On the other hand X(t, i 2 ) E M i , c M . Now t + a ~ i 2 > t ' + a ~ i 2 > t ' + ~ 2since i l + i [log n2i1+1]<[logn 2 ' 2 ] if i l < i 2 and n is large enough. Then X(t', i l ) and X(t, i2) are adjacent in Y.

3. Omitting edges of a subgraph

Delinition 3.1. Let ,f&n)=max{min{lE'1: ( A , [A]'nY\E')

is bipartite): ACT/ ~ l A ( = n } ;

more generally for 2
k') for k' < k .

It is also clear from the definition that n -f;(n)< 2f&1). This in view of Lemma 2.1 implies that for any given graph Y with x(Y) > w there is an E > 0 such thatfi(n) 3 En. This contrasts again with the situation for finite graphs. L. Lovasz recently informed us that, as a generalization of a theorem of T. Gallai, he can prove the following theorem: For 2 < r < o there is a j n i t e graph Y with x(Y)>r+2 andf&)=O(n'-'"). We describe his example: Let rn be even. Let V be the set of r-dimensional lattice points mod(rn),(xo, . . . ,x, - (yo, . . . ,y, - ) are connected if for some i < r [ x i- yil = 1, and x j = y j for j # i , j < r . This graph satisfies the requirements for large enough rn. In case of infinite chromatic graphs we are again left with the examples Yo(cc, k).

P . Erdos e l al.

I22

Theorem 3. (a) For all K > W there exists a graph with

~ ( Y ) > K and f & 1 ) < 2 n ~ ’ ~ (b)V E > Othere i s an r < O such that for all K > O there exists a graph with x ( S ) >K undf;(n, r)
By Lemma 1.1 it is sufficient to prove

Theorem 3.A. (a) If Y= Yo(w, 2), t h e n f & 1 ) < 2 n ~ ’ ~ . (b) I f Q= Y(w k ) for some 3 < k 0 there is an r < O such [hat f:(O,

r)
Proof. (a) Let V c [ o l 2 , IVI=n. Put V ( x ) = { { x y, ) : {x, y ) E V ) , v ( x ) = ( V ( x ) I forx E O . Fore E Klet Q(e, V‘)= {e’E V’:{e,e’}E 48). Let A = { x E O : u ( x ) 2 n ” 2 } .Then (Al
Iu

Theorem 4. Let 2 < k < o . Y is said to have property P ( k ) iffor all q > O there is an r < w w i t h j i ( n , r ) < n l + k - l + oIf. Y=(K E ) has property P ( k ) and < is any ordering of V, then OE(Y, <)has the property P(k + 1).

Proof. Let r(q)=min(r: f&,ri
f&&,

s) < n

l + ( k + l -1+q.

Given q>O, first choose q k q , q ’ > O . Then choose q”>O so that (1 - ( k + l)-’-q’). ( 1 + k - ’ + q “ ) = 1-6 for some 6>0. Let 1 be an integer with 1
E

V let E’(x)=

Y(e, E”)= {e‘ E E”: {e, e’) E OE(3, <)}. Let A = { x E V : e ’ ( x ) a n ( k + l ) - l f ” Then ‘ ) . l A l < n l - ( k + l ) - l - q ’ . We partition the set E‘ into three pieces. Let E j = { e E E‘: Ie-AI=j} for j < 3 . For each e E E‘, I3(e, E2)l< 2n(k+1)-1+q’, hence omitting 2n’+(k+1)-1+q’ edges. E 2 is independent and no edge

Almost bipartite large chromatic graphs

I23

joins any point of E 2 to any other point of E'. The subgraph spanned by El in the ordered edge graph is clearly 2-chromatic, Eo G[A]'. Since (A1
Problem 3. Assume Y has chromatic number w . Can f:(n) slowly? Can it be at most log n or log(k)(n)for k < w ?

tend t o infinity very

Referem [l] P. Erdos, Graph theory and probability, Canad. J. Math. I I (1959) 34-38. [2] P. Erdos, O n circuits and subgraphs of chromatic graphs, Mathematica 9 (1962) 17C175, [3] P. Erdos, F. Galvin and A. Hajnal, On set-systems having large chromatic number and not containing

prescribed subsystems, in: Colloquia Math. SOC. Janos Bolyai l c l n f i n i t e and finite sets (Keszthely, Hungary, 1973) pp. 425-513. [4] P. Erdos and A. Hajnal, On chromatic number of graphs and set-systems, Acta Math. Acad. Sci. Hung. 17 (1966)61-69. [5] P. Erdos, A. Hajnal and S. Shelah, On some general properties of chromatic numbers, in: Colloquia Math. Soc.Janos Bolyai 8-Topics in Topology (Keszthely, Hungary, 1972) pp. 243-255. Received 19 December 1979.

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Annals of Discrete Mathematics 12 (1982) 125-127

@ North-Holland Publishing Company

COVERING GRAPHS BY EQUIVALENCE RELATIONS Peter FRANKL Centre National de la Recherche Scientflque, Paris, France

Dedicated to Professor A. Kotzig on the occasion of his sixtieth birthday Let K.-C. denote the graph we obtain from the complete graph on n vertices by omitting the edges of a hamiltonian cycle. Following a conjecture of Duchet we prove that the minimum number of equivalence relations needed to cover this graph tends to infinity with n.

1. Introductioa

Let Y= ( K E) be a finite simple graph with vertex set V and edge set E . For an equivalence relation R on I/ we denote its graph by Y(R), i.e. the edges of Y(R) are the unordered pairs (x, y) for which x is in relation with y. Obviously Y is the graph of an equivalence relation if and only if it is the vertex-disjoint union of complete graphs.

De5nition. Denote by eq(Y)the minimum number t such that there exist t equivalence relations R 1 , R z , . . . , R, such that E(Y)=E(Y(Rl))u . .. uE(Y(R,)) (in this case we say that the equivalence relations R1,. . . , R, cover 9). Of course we have eq(K,) = 1, eq(Y)
Theorem. Suppose that for a positive integer r we have n 2 3t !. Then eq(K, - C.) 2 t .

2. The proof of the theorem

For a positive integer r let K z r- F z , denote a graph on 2r vertices which we obtain by deleting the edges of a 1-factor from K z r . Proposition 1. For 3r
P . Frank1

126

Proof. Let x l , . . . ,x ~. . ~. ,x,, in this order, be the vertices of the missing hamiltonian cycle in K,-C,. Then xl, x2, x4, x5,. . . ,x ~ x ~~ span -~ a~- K 2~r~- F 2 r . 0 If .@ is an induced subgraph of Y, then trivially eq( &)' tion 1 yields:

<eq(Y)holds. Thus Proposi-

Proposition 2. Let r = [n/3]. Then eq(K,-C,)2eq(K2, - F 2 r ) . Now we will prove that eq(K2,- F Z r )is not too small.

Proposition 3. Let X = {xl,. . . ,xr}, Y= {yl,. . . ,y r } be two disjoint r-element sets. Suppose thatfor some positiue integer s and equivalence reiations R1,.. . , R, the union ofE(Y(Rl)), . . . , E(Y(R,))contains the pair ( x i , y j ) ifand only ifi#j, i.e. it intersects the complete bipartite graph K(X, Y ) in K(X, Y) - F z r . Then r 2 t ! implies s 2 t . Proof. The statement is trivial for t = 1. We apply induction on t . For an equivalence relation R on X u Y let us choose a set Z=Z(R) such that (i) Z c X or Z c T: (ii) the restriction of Y(R) to Z is the complete graph on 2, (iii) ( Z (is maximal subject to these constraints. Let us suppose now that the equivalence relations R1,. . . ,R, satisfy the assumptions of Proposition 3 and IZ(Rl)l21Z(R2)2 * . . 2lZ(Rs)l. We distinguish two cases. (a) \ Z ( R l ) I < ( t - 1 ) ! - - l . Let us look at the edges (xl, y j ) , j = 2 , . . . , t ! . If (x, y i , ) ,. . . ,(xl, yi,) are covered by N R i ) , then yi,, . . . ,yiq are all in the same equivalence class, thus q< (Z(Ri)(< lZ(Rl)l. As each pair (xl, yj), j = 2, . . . , t !, is contained in at least one equivalence relation we deduce s((t- I)! - 1 ) 2 t ! - 1 , yielding s 2 t , as desired. (b) IZ(R1)>(t - 1 ) ! By symmetry reasons we may assume Z ( R l ) c X , { x l , .. . , x ( , - ~ , ! )sZ(R1).Let us set X 1= {xl,. . . ,x ( ~l-p ) , Yl = { y l , . . . , y ( * -l,!). As the classes of an equivalence relation are pairwise disjoint we deduce that 3 ( R 1 )contains none of the edges of the complete bipartite graph K ( X l , Yl). Thus we may apply the induction hypothesis to X1, Yl and R 2 , . . . , R,, and we infer s - 1 2 t - 1, i.e. s 2 t . 0 As Proposition 3 implies eq(K2,!- F 2 * ! ) 2 t ,using Proposition 2, for n 2 3t! we obtain eq(K, - C,)> t , which completes the proof of the Theorem.

3. Concluding remarks Remark 1. One might think that eq(K,-C,)is monotonic in n. However, this is not the case as eq(K5- C 5 ) =3, eq(K6- C,)= 2.

Covering graphs by equivalence relations

127

Remark 2. The theorem gives a lower bound on eq(K, - C,)of the order of magnitude log n/log log n. The best upper bound we can prove is 3 log,n. It would be interesting to know which one is closer to reality.

Reference [l] P. Duchet, Representations, noyaux en thkorie des graphes et hypergraphes, These de doctorat dEtat, Universite Paris VI (1979). Received 22 August 1980; revised 12 November 1980.

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Annals of Discrete Mathematics 12 ( I 982) 129- 140

@ North-Holland Publishing Company

ANALOGUES FOR TILINGS OF KOTZIG’S THEOREM ON MINIMAL WEIGHTS OF EDGES* Branko GRUNBAUM Unioersity oj’Washingron. Seattle, WA 98195, U S A

G. C. SHEPHARD Unioersity of East Anglia, Norwich NR4 7TJ. England

Dedicated to Professor A. Kotzig on the occasion of his sixtieth birthday

1. Iatroductioa

For any edge E in a graph G, the weigh w(E)is defined to be the sum of the valences of the two vertices at the ends of E . The weight of’G is defined to be the least value of w(E) for all edges E in G. The remarkable theorem of Kotzig [ 141 referred to in the title of this paper states that the graph of vertices and edges of every convex polyhedron (3-polytope) has weight 13 or less. Since, by Steinitz’ theorem (see Griinbaum [7, Section 13.11, or Barnette-Grunbaum [ 11) graphs of convex polyhedra coincide with finite, 3-connected, planar graphs, Kotzig’s theorem can be regarded as a theorem about such graphs. It can be shown that there exist infinitely many graphs of this type for which the weight is precisely 13. One such graph is shown in Fig. 1; it is constructed from the graph of the regular icosahedron by dividing each triangular cell into three triangles. This process, or its geometric equivalent of erecting a tetrahedral ‘tent’ on each of the triangular faces of a polyhedron, will be used several times in the constructions described later in this paper. Here we investigate analogous results for plane tilings, but before stating these it is necessary to introduce some terminology. By a (plane) tiling Y = { T 1 i E 1 ) we mean a family of plane sets T , called the tiles of Y, which covers the plane without gaps or overlaps. More precisely, (i) the union of the sets 7;. is the whole plane, (ii) the interiors of the tiles are pairwise disjoint, and (iii) each tile is a closed topological disk. With each tiling f we associate an infinite graph, the graph of T,which has as vertices the one-point intersections of two or more tiles, and as edges the arcs common to pairs of tiles; tiles which have a common edge are said to be adjacent. We shall restrict attention to tilings Y in which: (iv) each tile is adjacent to at least three tiles, (v) the graph of f is 3-connected, and *This material is based upon research supported by the National Science Foundation under grant No. MCS77-01629 A01. 129

I30

B. Grunbaum, G.C . Shephard

Fig. 1. A 3-connected planar graph of weight 13. The heavily drawn edges indicate a Schlegel diagram of a regular icosahedron; each of its twenty triangular faces is subdivided into three triangles by the thin edges.

(vi) 9 is locally finite, that is, each bounded set in the plane meets only a finite number of tiles of F. These properties imply that the intersection of any two tiles of F is either empty, or a vertex of 9, or a single edge of F. It is our intention to search for analogues of Kotzig’s Theorem for graphs of such tilings. However, it is easy to see that no assertion about minimum weights of edges can be made which is valid for all tilings. In Fig. 2 we show a tiling in which each vertex has valence 7 and hence each edge has weight 14. An obvious modification of the construction shown in this diagram enables one to construct tilings with every vertex of valence k, and hence every edge of weight 2k, for arbitrarily large values of k.

Fig. 2. An example (with k = 7) of the construction of tilings in which each vertex has valence k and therefore each edge has weight 2k. The vertices are situated on concentric circles.

Analogues for filings o i Kotzig’s theorem

131

It follows that any assertion about weights of graphs can only be true for tilings of suitably restricted kinds. In Theorem 1, stated below, we shall consider periodic tilings. A proof of this theorem will appear in Section 2, and the final section of the paper will be devoted to discussing variations, extensions and open problems related to Kotzig’s theorem. A tiling 9is called periodic if there exist non-parallel translations t l and t 2 which are symmetries of 9, that is, each of these translations maps 9onto itself. For each periodic tiling 9 there exists a finite region n (a period parallelogram) such that f

Fig. 3. Two examples of periodic tilings, with one period parallelogram indicated in each; the first tiling (from Burrows [3]) is triangulated. In the notation explained in Section 2, the first tiling has p=p3=68, e = 102, v=34, v4=5, v 5 = 8 , v6=12, v7=4, v e = 1 , v9=4, and the second has p=p4=u=u4=2. e = 4 .

132

B. Griiiihaum. G. C . Shepharti

Fig. 4. A periodic tiling in which each edge has weight 15 or 24. This tiling is often designated [3.122], to indicate that each tile has one vertex of valence 3 and two vertices of valence 12. I t can be considered as dual to the Archimedean tiling (3.12*),in which each vertex is incident with one triangle and two 12-gons.

consists of repetitions (through translations pt + q t z , where p and q are integers) of the part of 9which lies in I7. Many familiar tilings (such as the three regular tilings, by equilateral triangles, squares, and regular hexagons) are periodic, and more complicated examples appear in Fig. 3. In each diagram a period parallelogram is indicated, and it is clear that for any periodic tiling it is possible to choose a period parallelogram I7 so that the boundary of I7 does not contain any vertex of y. A rigorous proof of this fact is easily constructed; if the required condition is not satisfied by a given period parallelogram, we need only apply to it a suitable small translation. Our first result can now be stated. Theorem 1 (Kotzig’s theorem for periodic tilings). The graph ofeach periodic tiling y has weight ut most 15. Moreover, ij‘the weight is exactly 15 then y is isomorphic (combinatoridly equivalent) to the tiling [3.12’] shown in Fig. 4. It will be observed that [3.12’] is obtained by dividing each triangular tile of the regular tiling by equilateral triangles into three smaller triangles.

2. Proof of the theorem We begin by recalling some concepts from the theory of tilings. For further information on this and related topics see Griinbaum-Shephard [l l , 121. The boundary of a tile T in a tiling 9 consists-for some integer k with k 2 3ofexactly k edges and k vertices where k is the number of adjacents of 7:For simplicity, when k = 3 we shall refer to T as a triangle, without implying that it is a polygon-its edges may be arbitrary curves. If all the tiles in a tiling y are triangles, then we say that 9 is triangulated.

Analogues for tilings of Kotzig’s theorem

133

Let I7 be a period parallelogram of a periodic tiling Y,such that I7 contains no vertex of y on its boundary. Denote by u the number of vertices of in fl, and by u j the number of j-valent vertices in (j= 3, 4, 5 , . . .). Let e, p and p k denote the numbers of edges, tiles, and tiles with k adjacents in fl respectively. In determining the numbers p and P k fractions of a tile that add up to a whole tile must be counted as such, and a similar remark applies to edges in determining the value of e. (In Fig. 3 we indicate, for each tiling, all the values of u, u j , e, p , and P k which are non-zero.) It is well known that the following relations hold (the first four are trivial consequences of the definitions):

u - e + p =O. The last of these is the Euler relation for periodic tilings (see [l 1, Sections 3.3 and 3.71 and [12]). By simple manipulations of these equations it is easy to deduce that

and so, in particular,

+

( k - 6)Uk< 303 2114+ U s , k31

(1)

with equality if and only if the tiling is triangulated. After these preliminaries we can start with the proof of Theorem 1. Let be a periodic tiling such that every edge of 9 has weight at least w. The first stage of the proof is to show that there exists a periodic triangulated tiling f* in which every edge has weight at least w. If f is already triangulated we put Y*= Y;otherwise let T be a tile of f with four or more adjacents. Denote by V , , V,, V3, V4 four consecutive vertices of T labelled in order, and let the valences of these vertices be w , , w,, w3 and w4 respectively. Because edges V,V, and V3V4 each have weights at least w we have w 1 w2 2 w, w3 w 4 2 w, and so w w , + w3 + w4 2 2w. We deduce that either

+

wl+w32w

+

+

or w z + w 4 2 w .

(2)

Hence we can ‘split’ the tile T by introducing an extra edge (either V,V3 or V2V4 according to which of the inequalities ( 2 )holds) and this new edge will have weight at least w + 2. Since the weight of every other edge is either increased or left unchanged we deduce that the new periodic tiling (obtained by this tile-splitting process applied to T and all the corresponding tiles in the translates of fl that make up Y)has every edge of weight at least w. The procedure of tile-splitting can be applied repeatedly so long as any tiles with four or more adjacents remain; eventually, after a finite number of steps, we arrive at the required triangulated tiling Y*. This argument shows that in order to establish the theorem it will suffice to consider only triangulated tilings, and from now on we shall restrict attention to this case. Let 5 be such a tiling and suppose that the weight of y is at least 15. Since no edge can join two vertices of degree 3, it follows that the number of edges in Y incident with

134

B. Griinbaum, G. C. Shephard

a 3-valent vertex is exactly 3u3. Now let V be a vertex of valence k 2 12 and consider the k edges of f that meet at I/: It is impossible for two consecutive edges (by which we mean two edges belonging to the same tile T ) to lead from I/ to vertices of valence 3, for if this were the case, one edge of the triangular tile T would have weight 6. Hence at most [:k] of the edges meeting at V can lead to 3-valent vertices, and we deduce that

(Here, as usual, [x] denotes the greatest integer not exceeding x.) Similarly, the number of edges incident with either a 3-valent or a 4-valent vertex is exactly 3u3+4u4 and so

By considering edges incident with vertices of valence 3,4 or 5, we obtain analogously

Multiplying these inequalities by 5, 3 and 2 respectively and adding yields

where the final equality is a consequence of (1). Rearranging we obtain

from which we deduce that under the stated conditions (every edge of the triangulated tiling has weight at least 15)each term on the right must be zero. Hence u k = o unless k=3,4, $ 6 or 12. Moreover, since equality must hold in (6), it must hold also in (3), (4) and (5) as well; hence u4=u5=0. As not all vertices can be 6-valent, there are some 3-valent and some 12-valent vertices. By (3), of the 12 edges leading from a 12valent vertex, precisely six must lead to 3-valent vertices, and thus the other six must lead to vertices of valence 12. Hence u6=0 and the tiling must be combinatorially equivalent to the tiling [3.122]. Deletion of any edge from this tiling would leave some other edges of weight less than 15; it follows that all periodic tilings with weight 15 must be combinatorially equivalent to that of Fig. 4. We have shown that every periodic tiling with weight at least 15 contains edges of weight 15. Hence no periodic tiling can have weight greater than 15, and the proof of Theorem 1 is complete.

3. Remarks and open problems Kotzig's theorem was first published in 1955 (in [14]), but remained relatively inaccessible and not widely known. Expositions of the theorem and of variants and generalizations can be found in Griinbaum [8,9, 101 and JucoviC [13]. It is not hard to see that many of these have analogues valid for tilings. For example, trivial modifications

Analogues f o r rrlrtiys of Kotxy’s theorem

135

of the reasoning in Section 2 (which is modelled on Kotzig’s original proof) yield:

Theorem 2. If f is a periodic tiling with no 3-valent vertices then the weight of f is at most 12. Moreover, ifthe weight of f is exactly 12 then f is combinatorially equivalent either to the tiling [4.8’] or to the tiling [63] shown in Fig. 5. Theorem 3. If f is a periodic tiling with no vertices of valence 3 or 4,then either f is combinatorially equivalent to the tiling [63] or else the weight of f is at most 11. Other known generalizations of Kotzig’s result for polyhedra concern the numbers ejk,where ejk is defined as the number of edges in a graph, having one endpoint of valencej and the other of valence k . It is not hard to show that if a finite 3-connected planar graph G is a triangulation, and if no vertex of G has valence less than 5, then

Fig. 5. Two periodic tilings with no 3-valent vertices, in which each edge has weight at least 12. The first tiling is usually denoted by [4.8’], the second by [63].

136

B. Grunbaum, G . C . Shephard

Fig. 6 . A finite graph in which 2 e S S+ eS6= 56t60.

4ess+eS6260.(In [8, p. 4041 it was stated that 2 e 5 s + e S6 2 6 0We . are indebted to Dr. Steve Fisk for pointing out that this is erroneous-a counterexample appears in Fig. 6.) Other relations between the numbers ejkappear in [13] and [lo]. At present no analogous results for tilings are known, though there seems no doubt that such relations exist. A maximal matching A in a graph G is a family of disjoint edges such that each edge not in A has at least one endpoint in common with an edge of A. Kotzig’s theorem for finite graphs clearly implies that no 3-connected planar graph has more than 12 pairwise edge-disjoint maximal matchings. Similarly, Theorem 1 implies that no periodic tiling can admit more than 14 pairwise edge-disjoint maximal matchings. However, while the bound 12 for finite graphs is attained (an example is given in [9] and reproduced in [4]) it is not hard to verify that the bound 14 cannot be attained for any periodic tiling. The least upper bound is not known; it is either 12 or 13. There exist generalizations of Theorem 1 which do not require periodicity. By a limiting argument we can show that strongly balanced tilings have weight at most 15, and it seems likely that all normal tilings have this property as well. (For the definitions of these concepts see [11, Chapter 31 or [121. The proof of Theorem 1 we have given cannot be modified to deal with the latter case since, in general, Euler’s theorem for tilings does not hold for normal tilings (see [11, Section 3.31). Let M be any 2-dimensional manifold in the Euclidean space E 3 . By a tiling I = { &’ I i E I > of M we mean any family of sets in M which satisfy conditions analogous to those labelled (i) to (vi) in Section 1 of this paper. (More precisely, in (i) the union of the tiles must be M, and in (ii) the ‘interiors of the tiles’ must be taken relative to M.) These conditions no longer imply that the intersection of any set of tiles is connected; such an intersection, if non-empty, will consist of a finite number of points (uertices) and arcs (edges). In case of a compact manifold M condition (vi) implies that the numbers of tiles, edges and vertices are finite. The graph of the tiling 5 is 2-cell embedded in M, and it may contain multiple edges, but no edge or vertex is incident more than once with the same tile. As before, a tiling is called a triangulation if every tile has three adjacents. It is immediate that Kotzig’s original theorem also applies

Anaioyuesjor tilings oj Koiziy’s theorem

I37

to tilings of the 2-sphere. Similarly, by taking a period parallelogram of a periodic tiling 5 and identifying the opposite edges in the usual manner, we can deduce that Kotzig’s theorem for periodic tilings applies also to tilings of the torus. (This result was mentioned already in [8].) For a compact oriented manifold M , of genus g (sphere with g handles) we suspect that the following holds:

Conjecture. Each tiling of M , has weight at most 13 + 29. However, it seems that the known proofs of Kotzig’s theorem cannot be modified to establish this. Evidence supporting this conjecture is provided by the fact that for each g there exist triangulations of the sphere with g handles with weight 13+2g. These can be constructed in the following manner. We start with a regular icosahedron with 12 vertices, each of valence 5, and note that four triangular tiles (faces) contain, between them, all 12 vertices. (These four faces are shaded in Fig. 7(a).)We remove these triangles and then complete the manifold by adjoining two handles (or tunnels), each triangulated as in Fig. 7(b). The lettering indicates how one of the handles is to be fitted. This yields a sphere with two handles (of genus 2) and all vertices of valence 7. There are still twelve vertices and we can find four triangles which contain them all, namely pairs of ‘opposite’ triangles (such as ACD and BFE) on each of the two handles. Deleting these four triangles and adjoining two handles as before yields a manifold of genus 4 with every vertex of valence 9. Proceeding in this way we can obtain tilings of manifolds of genus g with each of the 12 vertices having valence 5 +g, for each even value of g. For odd values of g the analogous result can be obtained by adding handles to the toroidal polyhedron shown in Fig. 8. Four triangles which contain all 12 vertices, are indicated in the diagram. Now we divide each triangular tile into three triangles as described above. The

Fig. 7. In the construction described in the text two pairs of shaded triangles of the regular icosahedron shown in (a) are deleted: each pair of triangles is replaced by a ‘tube’ consisting of six unshaded triangles as shown in (b). This construction is topological. that is. the triangles can be deformed and bent as convenient.

138

B. Grunbaum. G . C . Shrphuril

result is a triangulated manifold of genus g with every edge of weight either 2(5 + g) + 3 or 4(5 +g). This establishes our assertion. If we restrict attention to those tilings of manifolds which are normal (in the sense that the intersection of any family of tiles is a connected set) or t o tilings which can be realized polyhedrally (that is, so that each tile is a planar polygon) then the problem

Fig. 9. The Csaszar polyhedron, which is of genus 1 and has 7 vertices of valence 6 .

Analogues for tilings of Kotzig’s theorem

Fig. 10. The Schulz polyhedron, which is of genus 2 and has 12 vertices of valence 7

Fig. 1 1 . The Brehm polyhedron, which is of genus 3 and has 10 vertices.

139

140

B. Griinbaum. G . C . Shrphurd

becomes much more difficult; it seems likely that in general the above conjecture does not hold in these cases, although it does for small g. This is easily verified for g= 1 by subdividing the triangles of a Csaszar polyhedron (see Fig. 9; compare Csaszar [ S ] , Gardner [ 6 ] ) into three triangles each. (The Csaszar polyhedron is a triangulated polyhedral torus, with 7 vertices of valence 6 each.) But there also exist polyhedral triangulations of manifolds with g = 2 and g = 3 of weights 17 and 19. For g = 2 Schulz [151 has established the existence of a triangulated polyhedron P with 12 vertices and with every vertex of valence 7; it is constructed by joining together two copies of the Csaszar polyhedron (see Fig. 10). If each triangular face of P is replaced by three triangles, the resulting polyhedron has genus 2 and weight 17. Brehm [2] has described the construction of a triangulated polyhedral tiling of a manifold of genus 3, with 10 vertices, each of valence 8 or 9 (see Fig. 11). Splitting each of its triangular faces into three triangles leads to a polyhedral tiling of weight 19, as required by the conjecture. No results for g 2 4 seem to be known.

Added in Proof (February 1982) In a forthcoming paper “Extending Kotzig’s Theorem” Joseph Zaks has disproved the conjecture on page 137 and established interesting generalizations of Kotzig’s theorem to various kinds of tilings of 2-manifolds of genus 2 2.

References [ I ] D. W. Barnette and B. Grunbaum. On Steinitz’s theorem concerning convex 3-polytopes and on some properties of planar graphs, in: G. Chartrand and S. F. Kapoor, Eds.. The Many Faces of Graph Theory, Lecture Notes in Mathematics No. I10 (Springer. Berlin-Heidelberg-New York, 1969) pp. 2740. [2] U.Brehm, Polyeder mit zehn Ecken vom Geschlecht drei, Geom. Dedicata 11 (1981) 119-124. [3] R. Burrows, GRIDesigns to Color, No. 8 (Price-Stern-Sloan, Los Angeles, 1980). [4] M. Capobianco and J. C. Molluzzo, Examples and Conterexamples in Graph Theory (North-Holland, New York, 1978). [5] A. Csaszir, A polyhedron without diagonals, Acta Sci. Math. (Szeged) 13 (1949) 140-142. [6] M. Gardner, On the remarkable Csaszar polyhedron and its applications in problem solving, Scientific American 232 (5)(1975) 102-107. [7] B. Griinbaum, Convex Polytopes (Interscience, London, 1967). [8] B. Grunbaum, Acyclic colorings of planar graphs, Israel J. Math. 14 (1973) 3 9 W 8 . [9] B. Grunbaum. Polytopal graphs, in: D. R. Fulkerson, Ed., Studies in Graph Theory, Part 11, Math. Assoc. of America Studies in Math., Vol. 12 (1975)pp. 201-224. [lo] B. Grunbaum, New views on some old questions of combinatorial geometry, in: Colloq. Internaz. Teorie Combinatorie (Rome 1973) Vol. 1 (1976)pp. 451-468. [I I] B. Griinbaum and G. C. Shephard. Tilings and Patterns (Freeman, San Francisco, 1982). [I21 B. Grunbaum and G. C. Shephard, The theorems of Euler and Eberhard for tilings in the plane, Resultate Math. (1982) to appear. [13] E. Jucovit, Strengthening of a theorem about 3-polytopes. Geom. Dedicata 3 (1974) 233-237. [14] A. Kotzig, Prispevok k teorii Eulerovskych polyedrov (in Slovak; summary in Russian), Mat.-Fyz. casopis Slovensk. Akad. Vied. 5 (1955) 101-113. [ 151 C. Schulz, Geometrische Realisierungen zellzerlegter 2-Mannigfaltigkeiten in euklidischen Riumen. in: 2. Kolloquium uber diskrete Geometrie, Salzburg (May 1980) pp. 207-215.

Received 8 August 1980.

Annals of Discrete Mathematics 12 (1982) 141-154

@ North-Holland Publishing Company

SETS OF INTEGERS WHOSE SUBSETS HAVE DISTINCT SUMS

Richard K . GUY Unicer,\ity o j C u l g u r y , Alhertu, Cunuda

For Anton Kotzig, founder of the Bratislava school of graph theory

1. Introduction

The subsets of the set of integers {2': O < i < k ) all have distinct sums. P. Erdos [6-121 has asked for the maximum number, rn, of positive integers a l < a 2 < . . .
O<j
are different. In particular, is it possible to have rn = k + 2, when x = 2k? We answer this last question affirmatively. Erdos notes that all the sums are less than mx, so that 2"-1
(1)

and it follows that rn
(2)

where the logarithms, here and in Theorem 3, are to base 2. Erdos and Leo Moser [6] give a result in which the second term in (2) is reduced by half. A proof of this is incorporated as Theorems 2 and 3. We are unable to decide whether the set of numbers defined by (7) and (10)is best possible in any sense, but we conjecture that the answer is affirmative. Erdos and Turan [13] considered the corresponding problem in which sums of pairs of the ai, with repetitions allowed, are required to be distinct, i.e. a i + a j ( i < j ) all distinct. If a h< n 6 ah+ where h = &n) and Wn)is the maximum &n) for a given n, they show that (2- 112 --E)n< o

(~ cn ) + 0(~1/4).

(3)

In an addendum [14], Erdos notes that the Singer [17] difference sets show that the upper bound in (3) is the right result, except perhaps for the error term. Lindstrom [15, 161 has improved this so that the right member of (2) may be replaced by n'/Z+n'/4+ 1. 141

R . K . Guy

142

2. An upper bound for m

Theorem 1. I f a l < a2 < . . . < a m are positive integers whose subsets have distinct sums, then m

1 ai>2"-1 i=l

Proof. The 2"' - 1 non-empty subsets of the ai have distinct positive integer sums, so the largest total is at least 2" - 1. Equality obtains just if ai= 2'- '. Theorem 2 (Leo Moser). Under the hypothesis of Theorem 1, m

i=l

with equality again holding just

if

a i= 2'-

Proof. Consider the sum of the squares of the 2" quantities +a1+az+

. . . +am.

(4)

The product terms cancel in pairs, so that m

On the other hand, the quantities (4) are distinct, different from zero, and of the same parity, so that the sum is at least 12+(- 1)'+ 3' + ( -3)'+

. . . +(2" - 1)2+(1-2"')2 =$2m-1(22"'- l),

and the result follows. Comparison of Theorems 1 and 2 tempts one to conjecture m

but its falsity is shown by the set of six numbers, 11, 17, 20, 22, 23, 24, whose subsets have distinct sums. The sum of their fourth powers is 1104035, but the right member of(5) with m = 6 is 1118481.

Theorem 3. With the notation of the Introduction, m < l o g x + i l o g l o g x+1.3 where the logarithms are to base 2.

Proof. Since the ai are distinct

(~22)

143

Sets of integers whose subsets have distinct sums

so Theorem 2 gives

34m- 1)<mx2 and 4" < 3mx2.

Take logs to base 2, and note that (1) implies m < log m have m <2 log x and

+ log x, and, since m <x, we

2 m < 2 log x+log 3 m < 2 log x+log log x+log 6 and the result follows since log 6 < 2.6.

3. 'Lhe Conway4uy sequence [4,5] To find a set of k + 2 numbers not exceeding 2k whose subsets have distinct sums, define an auxiliary sequence by uo = 0, u1 = 1 and u.+ = 2u, - u . - ~ , n >, 1,

(7)

Table 1. The auxiliary sequence u.

1

1

0

0

1

2 3

2 4

0 1

0 1

2 2

4 5 6

7 13 24

1 2 4

I 2 3

3 3 3

7 8 9 10

44 84 161 309

4 7 13 24

3 4 5 6

4 4 4 4

11 12 13 14 15

594 1164 2284 4484 8807

24 44 84 161 309

6 7 8 9 10

5 5 5 5 5

16 17 18 19 20 21

17305 34301 68008 134852 267420 530356

309 594 1164 2284 4484 8807

10 11 12 13 14 15

6 6 6 6 6 6

22 23 24 25 26 27 28

1 051905 2 095003 4 172701 8311101 16 554194 32 973536 65 679652

8807 17305 34301 68008 134852 267420 530356

15 16 17 18 19 20 21

7 7 7 7 7 7 7

29 30 31 32 33 34 35 36

130 828948 261 127540 521 203175 1040 31 1347 2076 449993 4144 588885 8272 623576 16512273616

530356 1051905 2 095003 4 172701 8311101 16 554194 32 973536 65 679652

21 22 23 24 25 26 27 28

8 8 8 8 8 8 8 8

37 38 39 40 41 42 43 44 45

32958 867580 65852 055508 131573 282068 262885436596 525249670017 1049459028687 2096841 607381 4 189538625877 8370804628178

65 679652 130 828948 261 127540 521 203175 1040 311347 2076449993 4144 588885 8272 623576 16512 273616

28 29 30 31 32 33 34 35 36

9 9 9 9 9 9 9 9 9

R . K . Guy

144

where r = (JZ),the nearest integer to 4%.Table 1 shows the values of this auxiliary sequence for 1 < n ~ 4 5To . obtain successive values of u, (column 2), double the preceding value and subtract the corresponding entry, u , , - ~(column 3). Note that these entries are repeated whenever n - r (column 4) is a triangular number. This occurs when n is also triangular. We prove a series of lemmas which lead to the existence and value of the limit of uJ2".

Lemma 1. u,, is strictly increasing with n.

Proof. u l > u o a n d ( 7 ) m a y be w r i t t e n u , + l - u , = u . - u , - r ( n ~ l ) a n d s i n c e u , > u . - , for n = 1, the result follows by induction. Lemma 2. 0
(na0).

Proof. uo=O, so u,aO by Lemma 1. From (7) u . + ~<2u. ( n 2 1) and since u,<2"-' for n = 0 and 1, the result follows by induction. Lemma 3. un/2" is a decreasing function of n for n 2 1; strictly decreasing for n 2 4 .

Proof. uJ2" = i for n = 0, 1, 2. From (7), un+1 -5 -2n+1 2n

un-r

2n+1'

and since u,-,>O for n a 3 , it follows that ~ , , + 1 / 2 " ~ ~ < u J 2 " .

Theorem 4. As n+ co,u,/2" tends to u limit, u, where O S u <

i.

Proof. From Lemma 3. In order to calculate 01, write un/2"=un.In the range p1 ( m + l ) + l
%, - m

gives

+

- 1.

Sum this over the range (8) and we have %n+ l ) ( m + 2112 + 1 =urn(,+

1 )/2

m(m

+ 1 -2-(m+2J

+ 1)/2

1

ui.

i = m ( m - 1)/2

write rn + j - I for rn and sum this from j

= 1 to j =

p,

~ ( r +np ) ( m + p + 1 )/2 + 1 =am(,

+ 1 )/2 + 1 -

1 2j= 1

c

(m+ j ) ( m + j - 1 )/Z

P

(m+j+l)

ui.

(9)

i = ( m + j- I ) ( m + j- 2 ) / 2

Since u2,=20950O3 x 2 - 2 3 < $ Lemma 3 implies that u < u , < i for n 3 2 3 . Hence, for m > 8 , the inner sum in (9) lies strictly between u ( m + j ) and 2rn+j),and the last

145

Sets of integers whose subsets have distinct sums

term itself lies between 2-m- 1a(m+ 2 -(m + p + 2)2-p) and 2-"+(m

Keepmfixedandletp-,co,anda=a,,,+l,,2+1 and (m+2)2-"-'. Put m = 26 and we have a352 -28 x 2 - 2 9 < a <

a352

1 +28 x 2 - 2 7

+2 - ( m +p + 2 ) 2 - 9

-flwhereflliesbetweena(m+2)2-"-'


x2-29,

so that x2-29 with an error of less than 2-29. For n2352, a,, does not differ from a in the first seven decimal places. Hence the upper estimate for a, based on a,>a, is much closer than the lower, based on an<:. So a=ajS2/(l+28 ~ 2 - ~ ' ) , with an error ofless than 2-48. A computer calculation gave

so that

..., ~1~~~=0.235125333862141 a=0.23512524581118.. . .


enables us to find k + 2 numbers less than 2k whose subsets have The fact that a distinct sums. Since a > it is not possible to find k + 3 such numbers by this method. We conjecture that it is not possible by any method. 4. Sets of numbers with distinct sums of subsets

Define ai=uk+ 2 -uk

(10)

t 2 -i

We conjecture that distinct subsets of the set { ai I 1 < i
+

u, + = u,,+ (u, -u, - I ) which is greater than u, (u, - u. - 2) provided r > 2 , i.e. provided n 2 4 . So, for n 2 4 , u.>un-1+u.-2 implies that u n + l > u . + u n - l r so the result follows by induction, since Table 1 shows that it is true for 1 < n<4.

Proof. By (7),

R . K. Guy

146

Proof. It is true for n = 4, 5 and 6. Suppose it is true for n = k 2 6. Then, on the one hand k+ 1

c

k

ui=uk+l+

c

ui>Uk+l+uk+l

i=O

i=O

by the induction hypothesis, so the sum is greater than 2#k + -uq = # k + 2, where q = k + 1 - (J(2k + 2)) is positive for k 2 6 . On the other hand, the inductive hypothesis implies that the sum is less than Uk+l+Uk+l+Uk-z, which is equal to U k + z + U q + U k - 2 G U k + z + U k - 1 by Lemma 4, provided q < k - 2, which is true for k B 6.

Theorem 5. If two subsets of {ai}have equal sums,then there are two subsets of {ui} with equal sums and equal cardinalities. Conversely, i f there are two subsets of { ui>with equal sums and cardinalities, then there are two subsets of {ai>with equal sums.

Proof. Suppose (Uk + Z -uil) + (uk + 2 - U i 2 ) + ' ' ' + (uk + 2 -ui,)=(uk+ Z - U j , ) + ( u k + Z - # h ) + * * * + ( u ~ -+ uj,). ~ We may assume (i) that the two sets are disjoint, else we couId cancel common terms; (ii) that il < i2 < . . . < is, j , < j 2 < * . <j,; and (iii) that s 2 t. Then (s - t ) # k +

2 =Mi,

+ + #i2

* ' '

+ui,-(#j,

+ + + uj,). Uj,

(11)

* ' *

+

If:,'

The right member of (11) is less than ui, which is less than # k + #k - by Lemma 5, and so s - t < 2. I f s= t , then (11) gives subsets of {ui>with equal sums and equal cardinality s=t, while if s=t+ 1, (1 1) again gives such subsets, each of cardinality s = t + 1. Conversely, if ui, uiz * . uis= u j , uj, . . . ujs,then, for any n > max(i,, j s ) , (u, - u J + (u, -ui2)+ . * * (2.4, -Mi,) =(u, - u j , ) + (u, - u j * ) + . . . (u, -ujJ Since we appealed to Lemma 5, our proof is only valid for k + 1 2 4. However, the theorem is vacuously true for small cardinalities. By Lemmas 1 and 4, there are no equal singletons and no pairs with equal sums. Lemmas 6 and 7 concern triples and quadruples.

+ + + +

+ +

+

+

Lemma 6. There are no distinct triples of the ui with equal sums.

Proof. The result will follow if we show that u,+12u,+un-1 +u,-z

(n~2).

(12)

We can verify this from Table 1 for 2GnG7 (with equality for 3GnG6). For n>7, r > 3, n - r < n - 3 and (7)with Lemma 1 gives u,+ > 224, - u, - 3. The inequality (12) follows inductively on adding this to u, 2 u, - + u, - z + u, - 3.

Lemma 7. There are no distinct quadruples of the ui with equal sums.

147

Sets of integers whose subsets have distinct sums

Proof. This can be checked from Table 1 if no value of i is greater than 11, while for n 2 11, the result follows from u , + ~ > u , + u , - ~ + u , - ~ + u , - (~n 2 l l ) . This is proved as in Lemma 6, since nail implies r > 4 , n - r < n - 4 u,+ 1 > 2u, - u,- 4.

(13) and

These results can clearly be extended, but at the cost of increasing amounts of computation. Some further results help us to reduce this.

If T,=is(s+ l), s>O and O
Tbeorem6.

(If s= 1 or 0, interpret the empty or ‘less than empty’ sum on the right as 0 or

- 1 respec-

t iuely.)

Proof.,The theorem may be verified from Table 1 for s = 0,1,2 and 0 < t <s + 2. Note that the result for t = s + 2 is the same as that for t = 0 and s + 1 in place of s, if we add uT,+,to each side, since T , + 1 + s + 2 = T , + 2and T , + ~ f 2 = T , + ~ +So l . we assume inductively that (14) is true for some s > 2 and some t, O
+,

+f

(O t <s

+2

+ l)

+ t, may be written

by (7), which, for n = T,+ UT,+

=uT,

+ f + 1 = 2uT,

+

+f - U T , + f - 1 (

< t <s + 2).

This completes the proof.

Lemma 8. If s>O, and with the convention of Theorem 6, S

1

uTi < $ u T , +

1

+ U T , -,+ 2 ) .

i=2

+

+ 2).s = 1 :0 <$(2 2).s = 2: 4 <$7 Assume that the theorem is true for s = u >,3. Then

Proof. s= 0: - 1 <$1 u+ 1

+ 4).s = 3 :4 + 24<$44 + 13).

V

C uTi=uT,+,+, i 1 U T ~ < ~ T , + ~ + $ U ~ , + ~ + U ~ , - ~ + ~ ) i=O =O =$2uT,+~+uT,+1

+UT,-,+2)

=$urY+,+ 1 + u T , + u T , + <$~T,+I+ 1

1

+U T , - , + 2 )

by (7) with n = T,+

+uT,+ 2 )

by (12)Withn=T,+l,provided T,-l+2
u a 3 , s o theresult

R . K . Guy

I48

Lemma9. If O > T , + ~t h, e n ~ ~ = , , - , u i < u , + 1 . Proof. ,Thecasess=O, 1,2,3 are Lemmas 1,4andinequalities(l2),(13).Ifu= T,+I Theorem 6 with t = 1 and Lemma 8 give

+ I,

< U T , I + 2 + ~1 ( U T +, 1 + U T , + 2 ) - uTs + 1 +

UT, + I

+2

provided T , - 1 + 2 < T , + 1, which is true for s 2 l . Now suppose the lemma holds for v = w > T,+ Then

c

c

w+ 1

W

uj=Uw+l

-u,-,+

i=w-s+ 1

Ui<2UW+1 -Uw-s
i=w-s

by (7), since w > T,+1 , and the result follows by induction.

'Iborem 7 . If s>O and 1 < t T,+t- 1

uT,+f+l>

1

i=O

<s+

c

2, then, with the same convention as in Theorem 6,

S

ui+

uTi.

i=2

Proof. The theorem may be verified from Table 1 for 0 <s < 2 and 1 < t <s + 2. If the theorem is true for given values of s and t, then it is true for the same s and for t 1 in place oft, since, to increase t by one we add u ~ , + -~ u+ ~~ * + ~to + the left member of (16), and we add u ~ , to + ~the right member, and the latter is less than the former by Lemma 4. If the theorem is true for some s 2 2 and for t = 1, then it is true for s + 1 and t = 1, since, to increase s by one when t = 1 we add uT, + - uT,+ t o the left member of (16), while we increase the right member by an amount

+

by Theorem 6, i.e. by an amount U T , + ~ + ~ - ( ~ T , + -~ u+~I , + ~ ) + c : = 2UTi'which Lemma 8, less than

by

by (121, s 2 1; this in turn is less than uT,+ I + - uT,+ provided T,+ - 12 T,+ 2, i.e. provided s 2 2, so the theorem is proved.

149

Sets of integers whose subsets have distinct sums

Theorem 8. If there are two sets of the ui with equal sums, and the largest member of either set is uTS+ I + t + where 1d t d s 2, then the other set contains at least s 2 members, including the s 1 members ui, T, t 1 d i < T, + t .

+ ++ + Proof. If the other set did not contain these s + 1 members, then its sum would be at +

+

most T A - 1

S

T.+t- 1

by Theorem 6, and this is less than u ~ ,+ ,t +~ by Theorem 7. Also by Theorem 6, the sum of the s+ 1 members

+ ~must , contain at least one by Lemma 8, and since this is less than u ~ , + ~ +the~ set more member if there is to be equality of sums. This completes the proof of the theorem.

Note that Theorem 8 is not vacuous, since there are sets of the ui with equal sums, e.g. 7 = 4 + 2 + 1, 1 3 = 7 + 4 + 2 , 2 4 = 13+7+4 and 44=24+13+7. These sets do not have equal cardinality, though after adding uo=O to the first member, the cardinalities differ only by one. More generally, Theorem 6 exhibits sets of the ui with equal sums, whose cardinalities are s + 2 and s or s + 1. For example 1164+ 594 + 309 + 161+ 84 = 2284 + 24 + 4 ( + 0). In Theorems 9 to 13, on the other hand, we will assume the following conditions, which include equal cardinality, so we conjecture that these theorems are only vacuously true. Conditions C and D are of 'minimal criminal' type.

Condition A. There are two sets of the ui with equal sums. Condition B. These two sets have the same cardinality, c. ConditionC. Of such pairs of sets we choose one with the least possible greatest member, u,,+ 1 , and write n in the form T,+ + t , where 1< t d s + 2. Note that if we stay in this interval, formula (15 ) holds good, but if we stray outside it, some care is needed. For example, if t = 1.or 2, uT,+ I = ~ U T , +, 1~- 2 - u T , + ~ - z . Condition D. Among pairs of sets satisfying A to C, choose one with least value of c. This condition implies that the two sets are disjoint. Lemmas 1, 4, 6 and 7 imply that c 2 5 .

R . K . Guy

150

We call the set containing u.+ the major set and the other the minor set. If we use subscriptsjand i respectively for the two sets, then Theorem 8 implies that ( i ) contains (i: T,+t+l
Theorem 9. Under conditions A to D, u T , + [ - belongs to the minor set.

Proof. By Theorem 8 we have U i , +ui,+

' . . + u ~ , + t +1 + u ~ , + t + ~ '+.

'

+UT,+,

+t=uj,

+ uj2+

' ' '

+ u T , + ,+ t + 1 .

(17)

Substitute for by formula (15), cancel u ~ , + , from + ~ each side, and add u ~ , + ~to-each side and we will have produced two sets of ui with equal sums, but with a smaller largest member, uT,+, than before, contrary to condition C , unless u ~ , + ~was - already present on the left of (U), i.e. in the minor set.

Theorem 10. Under conditions A to D, the minor set does nor contain all the s + 4 members ui, T , + t - 2 < i < T , + , + t . Proof. If the minor set contained these s + 4 members, its sum, by Theorem 6, would be at least S

uT, +

,+ + i + 1 u T i + u T , + - 1 + u T , + t

t

t - 2,

i=2

while the sum of the major set would be at most T.+t-3

which, by Lemma 5, is at most u ~ , + , ++~u+~ ~, + ~ - ~ + u ~Thiscontradictscondi,+~-~. tion A.

Theorem 11. Under conditions A to D, uT,+[belongs to neither set. does not belong to the major set. If it did, then the sum Proof. We first show that of the major set would exceed uT,+, + + uT,+ while the sum of the minor set is less than

1

i=T,+t+l

Ts+I+~

T.+t- 1

Ts+~+f

ui+

1

ui=

i=O

1

C uTi-uT,+t+

i=2

1

ui

i=O

i=T,+t S

=uT,+,+t+1+

Ts+t- 1

ui-uT,+r+ T,+t- 1

C

i=O

ui

by Theorem 6,

151

Sets of integers whose subsets have distinct sums

by Lemmas 8 and 5, and

U~,+~>~(U~,+~+~U~,+,-~ since +U~,_~+~),

UT,+ I -UT, +t - 3 3 U T , + t - I

+U T , +

by ( 12)

I -2

(f31)

>,uT,+uT,-L 1

by (7).

>~(uT,+l+uT,_,+2)

This contradicts condition A. + ~ to the minor set, its sum, by Theorems 8 and 9 On the other hand, if u ~ , belongs is at least Ts+I+~

1

S

ui=uT,+l + I

i=T,+t- 1

+1 +

1

uTj+ uT,+f - I

i=2

by (14), while the sum of the major set is at most

c

i=O

and

Ts - 1 + f - 3

s- 1

T,ff-2

uT,+l+f+l+

1

~i=~T,+,+f+l+~T,+f-I+ uT,+ i=2

1

ui?

i=O

T, - 1 + I - 3

c

uT,>

ui

i=O

by Lemma 5, unless t = s + 2. So for the remainder of the proof we assume that t = s + 2 and we next show that the major set contains the s + 1 members ui where T,< i < T, s. For suppose one of these ui is not in the major set. Then even if this were the least, uT,, the major set would have sum at most

+

Ts+s

1

~ T , + ~ + s + 3 + i=T,+1

T,- 1

ui+

1 ui

i=O

s- 1


1

uTj - uT,

i=2

+uT, +uT,

~

3

by Lemma 5, while the minor set has sum at least +s+

TS +

2

S

1

Ui=U~,+s+i+~T,+~+s+3+ uTi. i=T,+s+ 1 i=2

contradicting condition A. Now from (14) with t = s + 2 we have T,

+1 +s+

1

2

c Y

ui=uT,+,+s+3+

i=T,+s+ 1

uTi+uT,+s+I

i=2

+

and, with s reduced by 1 (since T,= 'I- s) T,+s

1

i=TS

s-1

ui=uT,+s+l+

1

uTi

i=2

The major set contains ui for i = T,+ + s + 3 and those ui on the left of (19), while the minor set contains those ui on the left of ( 18). From (1 8) and (1 9) the sums of these

R . K . Guy

152

Table 2 (s. t)

T,+,+r+l UT,+,+i+I

(*) (**)

(2, 3)

(294)

(3, 1 )

(3,2)

(53)

(3,4)

(3,5)

10 309 302 7

11 594 578 10

I2 1164 1172 -1

13 2284 2272 19

14 4484 4435 56

15 8807 8687 127

16 17305 17048 264

+ +

terms differ by uT,.So we may replace these 1 (s 1) terms in the major set and these s 3 terms in the minor set by a single term uT,in the minor set and produce two sets still satisfying condition A, and condition B with cardinality reduced by s + 2, contradicting condition C . This completes the proof of Theorem 11.

+

We are assuming that 1 d t ds + 2. Cases at either end of this interval are sometimes critical or exceptional so in order that statements should be substantial it is necessary that s be not too small. Lemmas 1,4,6 and 7 imply that the cardinality c of each of the two sets is at least 5, so that 10or more distinct ui occur: T,+ t 1 2 9 and (s,t ) 2 (2,3). In fact we can see that T,+ t + 1 2 17, i.e. that s 2 4 , since if (s,t ) takes the values of Table 2 then T , + l + t + 1 and u T , , l + I + lare as shown, and the sum of the 1 +(s+ 1 ) members,u,i=T,+t-1, T , + t + l < i < T , + l + t , that weknowareintheminorset, is given in row (*), so that if we make up the cardinalities of the two sets to 5 by adjoining the smallest possible members uo=0, u1 = 1, u2 = 2, and u3= 4 to the major set (and the largest possible member u4 or u5 to the minor set in the first two cases) the sum of the major set will be in excess of that of the minor by the values of row (**), while the unused ui with which we can attempt t o balance the sums and cardinalities are those for 4 < i < T , + 1 + t - 6 (e.g. the empty set when (s, t ) = ( 2 , 3 ) ) and it can be verified that these d o not suffice. So henceforth we can assume that s a 4 .

+

++

Theorem 12. If s 2 4 and 1 ~ t d sthe , minor set contains ui for T,+ 1 d i d T,+t I f t = s + l ors+2,itcontainsuifor T , + 2 < i d T , + t - l .

- 1.

Proof. The theorem is vacuously true for t = 1, and it is true for t = 2 by Theorem 9. Next assume 3 G t < s. If any of the ui of the assertion were missing, then the sum of the known members of the minor set would be at most Ts+I+~

1

i=T,+f+l

T,+t-1

ui+

1

i=T,+2

ui=

Sets of integers whose subsets have distinct sums

I53

so that the sum of the known members of the minor set would be deficient by i=T,-l+f-I

i=2

i=2

and, since t < s and T,- +s- 1 = T,- 1, by Lemma 8 this is at least UT, + I

+ U T , + UT,

-1

- $uT, + 1 +u T , 1

-I

+ 2 + u T , - I + 1 + uTs

-2

+ 2)

(21)

and by (7) with n = T,, ( 1 3) with n = T, - 2 and Lemma 5, this is greater than lTzo ui, the sum of all the remaining ui which could be adjoined to the minor set to make up for its deficiency. Similarly, if t = s + 1 or s + 2, and ui for i = T,+ 2 were missing from the minor set, the sum of its known members would be deficient by (20)with the upper terminal of the first sum replaced by T, 2. Much as before, this is greater than ui, which is all that is available to make up for the deficiency.

+

2:;

Theorem 13. The minor set contains u,. where x = 'l-if t = 1 (or 2 ) and x = T,- 1

if 2 < t < s + 2 .

+t -2

Proof. If u, is not in the minor set, transfer u ~ , -+ from ~ the minor to the major set and balance by adjoining u ~ , and + ~ u, to the minor set. (If u, was in the major set, - are in the major it can now be cancelled from both sets.) Now uT,+ I + and set and may be replaced by 2 ~ ~ , +one ~ +copy ~ , of which can be cancelled from both sets, yielding a pair of sets with smaller largest member, contrary to condition C. + f

5. Inconclusion

The results of this paper have been known for twenty years, at least to John Conway and the author, and perhaps to others. We are unable to conclude that definition (10) yields a set with distinct sums of subsets for arbitrarily large k , although this writer conjectures that this is so. In spite of occasional claims to the contrary, the more general problem of showing that m = k + 0 ( 1 ) remains open; Erdos offers $500 for settling this.

Acknowledgements A Killam Resident Fellowship at the University of Calgary allowed time for this and other research. I also gratefully acknowledge the support of a grant from the National Science and Engineering Research Council of Canada.

References [I] G . S. Alojan, in: Proc. Sem. Specialized Electronic Simulating Machines and Simulators, Inst. Kibernet. Akad. Nauk Ukrain. SSR, Kiev (1969)pp. 59-63 [ R Z Mat 1970 +6V366].

I54

R . K . Guy

[2] V. S. Bludov and V. I. Uberman, A certain sequence of additively distinct numbers (Russian), Kibernetika(Kiev) 10(5)(1974) 111-115 [MR 53 ~ 3 3 2 ; Z b l291. . 100431. [3] V. S. Bludov and V. I. Uberman, On a sequence of additively differing numbers, Dopovidi Akad. Nauk Ukrain. SSR Ser A (1974) 483-486.572 [MR 50 #7007; Zbl. 281.10030]. [4] J. H. Conway and R. K. Guy, Sets of natural numbers with distinct sums, Notices Amer. Math. SOC. 15 (1968) 345. [5] J. H. Conway and R. K. Guy, Solution of a problem of P. Erdos, Colloq. Math. 20 (1969) 307. [6] P. Erdos, Problems and results in additive number theory, in: Colloque sur la Theorie des Nombres, Bruxelles, 1955 (Liege and Paris, 1956) pp. 127-137 (especially p. 137, with L. Moser). [7] P. Erdos, Problem 319, 27:9:56, The New Scottish Book, H. Fast and S. Swierczkowski, Wroclaw, 1946-58. [8] P. Erdos, Problem 209, Colloq. Math. 5 (1957-8) 119. [9] P. Erdos, Some unsolved problems, Michigan Math. J. 4 (1957) 291-300; problem 1 1 on p. 294. [lo] P. Erdos, Some unsolved problems, Magyar Tud. Akad. Mat. Kutato. Int. Kozl. 6 (1-2) (1961) 221-254 (especially p. 226). [ I 11 P. Erdos, Problem 30 in: Quelques problemes de la theorie des nombres. Monographies de I'Enseignment Math. No. 6 (Geneva, 1963) 81-135 (101-102). [ 121 P. Erdos, Some problems in additive number theory, Amer. Math. Monthly 77 (1970) 619-621. [I31 P. Erdos and P. Turan, On a problem of Sidon in additive number theory and on some related problems, J. London Math. SOC.16 (1941) 212-215. [14] P. Erdos, Addendum to [13], J. London Math. SOC.19 (1944) 208. [15] B. Lindstrom, On a combinatorial problem in number theory, Canad. Bull. Math. 8 (1965) 477-490. [I61 B. Lindstrom, Om ett problem av Erdos for talfoljder, Nordisk Mat. Tidskrift 16 (1-2) (1968) 29-30, 80. [I71 J . Singer, A theorem in finite projective geometry and some applications to number theory, Trans. Amer. Math. SOC.43 (1938) 377-385. [I81 P. Smith, Problem E 2526*, Amer. Math. Monthly 82 (1975) 300. Solutions and comments: 83 (1976) 484. [191 V . I. Uberman, The Conway-Guy conjecture and the density of almost geometric sequences (Russian). [20] V. I. Uberman, On the theory of a method of determining numbers whose sums do not coincide (Russian), in: Proc. Sem. Methods of Math. Simulation and Theory of Elec. Circuits, Izdat. Naukova Dumka, Kiev (1973) pp. 76-78,203 [MR 51 ~ 7 3 6 3 1 . [21] V. I. Uberman, Approximation of additively differing numbers, in: Proc. Seminar Methods Math. Simulation & Theory Elec. Circuits, Naukova Dumka, Kiev, 1 1 (1973) 221-229 [MR 51 #7363; Zbl. 309.680481. [22] V. 1. Uberman and V. 1. Sleinikov, A computer-aided investigation of the density of additive detecting number systems, Akad. Nauk Ukrain. SSR, Fiz.-Tehn. Inst. Nizkih Temperatur, Kharkov (1978) 60 PP. Received 28 September 1980.

Annals of Discrete Mathematics 12 (1982) 155-168

@ North-Holland Publishing Company

ANALOGUES OF THE SHANNON CAPACITY OF A GRAPH Pavol HELL* and Fred S. ROBERTS** Rutgers Unioersity. New Brunswick. N J 08903, U S A

Dedicated to Professor A. Kotzig on the occasion of his sixtieth birthday

1. Introduction

This paper was stimulated by analogies between two graph-theoretic notions which arise from apparently unrelated applied problems in information theory and operations research. The graph theoretic concepts are the capacity of a graph and the ultimate chromatic number. The notion of capacity arises in a natural way from the problem of transmitting information through a noisy channel (Shannon [42], Berge [4], Roberts [33]; cf. also Rosenfeld [35,36], Hales [19] and McEliece and Posner [28]). There has been a recent surge of interest in the problem of computing the capacity. This has been highlighted by the development of significant new techniques by Lovasz [27] that can be applied to obtain general upper bounds on the capacity, and yield the exact value of the capacity for many graphs, including the long-standing open case of the pentagon. Further improvements and related work can be found in the papers by Haemers [18], Li and Li [22], and Schrijver [37]. The notion of ultimate chromatic number is related to the problem of assigning mobile radio frequencies to vehicles operating in zones (Gilbert [ 161, Roberts [34], Opsut and Roberts [30, 3 11). In fact, the problem of assigning frequencies motivates the study of multicolorings, which have also engendered rather widespread recent interest-see for example the papers by Barany [ 11, Bollobas and Thomason [S], Chvatal, Garey, and Johnson [7], Clarke and Jamison [8], Garey and Johnson [13], Geller [14], Geller and Stahl [15], Lovasz [26], Scott [39], and Stahl [43]. Multicolorings have since been applied to a variety of practical problems, dealing with fleet maintenance, task assignment, traffic phasing, etc. (See [30,3 11 for a summary.) Multicolorings lead naturally to the concept of ultimate chromatic number, as introduced by Hilton, Rado, and Scott [20]. There are similarities between the concepts of capacity and ultimate chromatic number. We shall relate them explicitly, by making them part of a general framework which will allow us to bring together, summarize, and extend several scattered results and methods. We shall prove no new results about capacity or ultimate chromatic number, except to interrelate them within our general structure. We do point out *Dr. Hell acknowledges the support of NSF Grant No. MCS80-0301 I to Rutgers University. Current address: Simon Fraser University, Burnaby, B.C., V5A IS6 Canada. **Dr. Roberts acknowledges the support of the Air Force Office of Scientific Research under contract 80-0196A to Rutgers University and of Rutgers University under its sabbatical leave program.

I55

P. Hell, F . S . Roberts

156

other parameters arising from our framework, which may turn out to be equally interesting. We also introduce and briefly investigate a notion of degree of perfectness of a graph. 2. Product numbers

The vertex-set of a graph G is denoted by V ( G ) ,the edge-set by E(G).The independence number a(G) is the largest number of vertices in an independent set of G ; the clique number o ( C ) is the largest number of vertices in a clique (maximal complete subgraph) of G ; the chromatic number x(G) is the smallest number of independent sets of G whose union is V ( G ) ;the clique covering number O(G)is the smallest number of cliques of G whose union is V(G). Let G and H be graphs. The strong product G - H (called the normal product by Berge [4]), the lexicographic product G H , and the disjunction G v H are defined on the vertex-set V ( G )x V ( H )as follows: 0

E ( G . H ) = { ( g , h ) ( g 'h'): , ( g g ' E E ( G )& h h ' ~ E ( H ) ) o r ( g = g ' & hh' E E ( H ) )or (gg' E E(G)& h = h ' ) ) , E(G0 H ) = ( ( 9 ,h)(g',h'): (gg'

E(G v H ) = ( ( 9 ,h)(g',h'): gg'

E E(G))or E

( g = g ' & hh' E E ( H ) ) } ,

E(G) or hh' E E ( H ) ) .

It is easy to see that if @ represents any of the three products -,0 , v , then a(G @ H ) > a(G)a(H), w(G @ H ) 2 w(G)(H), x(G @HI GX(GIX(H),and O(G @HI G O(G)O(H). Let G"= G - G . . . . .G, G(")= G o G o . . . 0 G, and G["]= G v G v . . . v G, n times. The above inequalities suggest the following definitions: as(G)=sup "J,

aL(G)=sup Va(G(")),

UD(G)=SUP Va(G["]),

ws(G)=sup"J,

WL(G)=sup "J(G'"'),

(%(G)=sup nJw(G["l),

Xs(G)=inf "J(G"),

XL(G)=inf "J(G'")),

XD(G)=inf "J(@"),

&(G)= inf nJO(Gn),

O,(G)= inf nJO(C(")),

O,(G)= inf .Je(crnl).

-_ __

The parameter as is the Shannon capacity, and the parameter Os was also studied by Hales [19] and (inter alia) McEliece and Posner [28]. We observe that a(G)=w(G), X(G)=O(c), -= G vH, and m = c o H . Therefore as(G)=%(G), os(G)=~x,(G), xs(G)= OD(@, M G ) = X D ( ~ ) ~ , L ( G ) = ~ L (and G ) ,X L ( G ) = ~ L ( G . There are, of course, many products to which the parameters can be applied (and there are many other parameters); we have tried to select ones for which the resulting problems appear to be the most interesting. The strong product leads to the notion of capacity, the lexicographic product is related to multicolorings, and the disjunction was chosen because of its complementary relation to the strong product. Some other standard products, such as the Cartesian product and the conjunction, appear to yield only trivial concepts with the given parameters.

Analogues of the Shannon capacity ofa graph

157

3. Ultimate and fractional numbers Let N = { 1, 2 , . . . } and n E N. Let a,,(G) denote the largest number of vertices of any induced subgraph H of G K, with clique number w(H)a,(G)+cr,(G), wm+n(G)>, WAG)+ W A G ) , X m + n(G)G X m ( G ) + X,,(G), and Om + .(G) ,na(G), 2 nw(G),X,,(G)< nX(G), and BAG)
0

K).

The ultimate chromatic number xu was introduced by Hilton, Rado and Scott [20], and also studied by Chvatal, Garey and Johnson [7], Bollobas and Thomason [5], and others. Because of the superadditivity of a,,, W , and the subadditivity of x,,,B,,, the suprema and infima in the above four definitions may be replaced by limits (cf. Fekete [lo] or Polya and Szego [32], problem 98). We shall point out below that in fact each ultimate number is attained for some n, and can be calculated by a linear program. (For xu this was first observed by Lovasz [23], Scott [39], and Clarke and Jamison [8]. Let 2' = {0, 1,2, . . . }. Let G be a graph and let V= V(G).Let 9 denote the set of all maximal independent subsets of G, and V the set of all cliques of G. Associate with each induced subgraph H of G OK, an integer vector X with coordinates xu, u E r/: in which xuis the number of vertices i of K, such that (u, i ) E V ( H ) .It then follows from the definitions that a,(G)=max

{ Ex,,: uov

and similarly, w,,(G)=max

{

I I

E x , < n foraIlCE(e,x,EZ+ foralluEV , uec

y,,: ~ y , < n f o r a l l I ~ 9 , y u ~ Z + f o r a l l.u ~ V UEV

uel

It is easy to see [43] that an n-tuple coloring of G corresponds to a family of independent sets of G, with repetition allowed, such that each vertex of G is covered precisely n times. (Simply let the i-th set consist of all vertices using the color i.) Associate with

P . Hell. F . S . Roberts

158

each family of independent sets of G an integer vector Z with coordinates z I , I E 9, in which zI is the number of times I appears in the family. It again follows from the definitions that

C IEJ

and

zI:

1 zl>n

I

for all u E V, z I E Z + for all I € 9 ,

I=

-

-

--

.

is seen analogously. There are obvious linear programs associated with the last four integer linear programming formulations of the numbers a,, on,x,, and 8,. We shall call the numbers they define thefractional independence, clique, chromatic, and clique-covering numbers. They are given by: aF(G)=max

i

r

(oF(G)=max XF(G)=min

C xu
1 y,: C y u < 1

iC i1

for all

c E V, x,aO

for all u E

UEC

usv

IEJ

BF(G)=min

xu:

ueC

for all I €9, y,ao for all u E

uel

zI:

1 zI>l

I I I I v

v

,

,

foraIIuEV,zI>O f o r a l l I E S ,

I=

wc: C wc> 1 for all u E V, w c a O for all

c Ev

.

cn, The number aF is also known as the Rosenfeld number, and has served as a useful upper bound on the capacity as before Lovasz [27]. (Rosenfeld [35] studied aF and proved that as(G)
4. Degreesofperfectness

We define a graph G to be n-perfect if x , ( H ) = ~ , ( H for ) each induced subgraph H of G. Clearly, 1-perfect graphs are precisely the perfect graphs in the usual sense (cf. Berge [3], Lovasz [23,24,25], Fulkerson [ l l , 121, Golumbic [17]).

,

~

159

Analoyues of rhe Shannon capacity of a graph

Proposition 1. Every graph is n-perfect for some n E N .

.

Proof. According to the last two paragraphs of Section 3, xV(H)= XF(H) = WF(H) = w d H ) for every H. Since every ultimate number is attained, for each induced sub, d H ) = o,(H)/rn. Note graph H of G there are integers k, rn such that xU(H)=z k ( H ) / k w and ) Xsn(H)/sn<Xn(H)/nfor any that the subadditivity of x,,implies that X ~ . ( H ) S S X , , ( H s E N . Hence if X V ( H ) = X , ( H ) / then ~ , XV(H)=X~,(H)/~~ for any s E N ; a similar argument applies to wu. Let n be a common multiple of all the k and rn arising from the induced subgraphs H of G. Then X,(H)/n = w,(H)/n for all induced subgraphs H of G. 0 It follows from part of the above proof that Xsn/sn<x,/n. Similarly, osn/sn>o,,/n. Thus, if a graph is n-perfect, it is also sn-perfect, because for every induced subgraph,

However, a graph which is n-perfect need not be ( n + 1)-perfect, cf. Section 6. We define the degree ofperfectness of G to be the least n such that G is n-perfect. Evidently, perfect graphs have degree of perfectness 1. Lovasz [24, 251 proved that the complement of a perfect graph is also perfect (as conjectured by Berge [3]; cf. also Fulkerson [12]). This is no longer true for ti-perfect graphs with n > 1, as we shall show in Section is 2. The former 6 that the degree of perfectness of C 2 k + is k, while that of example also shows that there are graphs of arbitrarily high degree of perfectness.

c,,,

5. Relations amoag the numbers

The main conclusion of this section will be the perhaps unexpected fact that there are, up to complementation, only five distinct numbers among those we have defined, namely: c(, tlS, &, aF, and 8. In particular, we shall show that c(

=

-

= c 1 ~= WL

_

_

=0 s = W


= M u = 8u=

OF = 8s = 8~= 1~= & = XF = x u = a u = aj~

<e=x, where x ( G ) = ~ ( c ) and , so on. To prove all of the above equalities and inequalities (each of which can be strict, as we shall remark below), it is enough to prove the relations among the unbarred numbers, as the effects of complementation were observed in Sections 2 and 3. Furthermore, we can ignore the ultimate numbers, as the equality of ultimate and fractional numbers was also established in Section 3. That the duality theorem of linear programming implies U F = & was noted in Section 3 as well. The remaining relations will be verified by the following three propositions.

P. Hell. F . S. Roberts

160

Proposition4. For every graph G, as( G)
Lemma A. For every graph G, &(G)<&(G).

Proof. This result was first proved, in a more general context, by McEliece and Posner [28], using techniques of information theory, game theory and probability. We present a self-contained and elementary equivalent of their proof. Let

i1

T=max

tc-t>,O

t:

for all v E r!

c=

and

LflC

wc: C wc-1 forall

W=min

1t c = l ,

t,>,~

U E

r! wC>O for all

cw

I

for all C E % , t20 .

C&

I

CE%.

Note that W= OF; cf. Section 3. We may assume that G # K N (otherwise the lemma holds), and hence T c 1. We begin by observing that W T = 1. Indeed, if wc(CE %) is an optimum solution of the second program, then t = 1/W, tc= (l/W). wc (C E @ is a feasible solution of the first program, so T 2 1/W ( >O). Similarly, if t , tc (C E %) is an optimum solution of the first program, then wc=(l/T).tc (C E %)' is a feasible solution of the second program, and W < l/T, proving WT= 1. For the remainder of this proof, let t , tc (C E %) be an optimum solution of the first program; thus CcBVtC2t for all v E r! CCEYPtC= 1, T= t . and 8 F = l/t. Let v be the number of vertices of G, and ( = l/v. Next we show that, for every positive n. B(G")s n

- log(log(1 - t") + l .

By the definition of the strong product, the set X of cliques of G" is precisely the set of all Cartesian products K = C1 x . . . x C , with each Ci E % (the set of cliques of G); we define t K= t c , . . . . -tc,. Note that

-

and, by a similar calculation, cannot be covered by n-

log 4 +l=r log(1 - t")

cKav t K = t " for each vertex

u E V". Assume that G"

Analogues of the Shannon capacity of a graph

161

cliques, i.e. that O(G")> r. Then

= 1,

where the hypothesis O(G")> r is used to obtain the inequality. Consequently, since CKov

tK2tn,

1<

1(1

UEV"

K $ u t ! K > . = e UEV"

(1-c

Keu tKy<(1-ffl).'"".

Taking logarithms, we find r.log(1 -t")+n-log v20.Since t < 1, we find r$n.

log 5 log(1 - t") '

which is a contradiction. We complete the proof by showing that inf

log(1 - t")

t

(Since log( 1 - t") and &(G) = l / t , the lemma follows.) Since, for 0 < x < 1, x + log(1 - x) <0, we have - log(1 - t") 2 t", and

P . Hell. F . S. Roberts

162

(The limit may be evaluated by applying 1'Hopital's rule to the logarithm of the argument and using the fact that t < 1.) 0

Lemma P For every graph G, &(G)
Proof. Let a,=B(G'")). It is easy to see that B(G H)=B,(G) for n=B(H). (This also follows by taking complements from the similar observation x(G H)= zn(G) for n = X ( H ) made by Stahl [43] and based on a result of Geller and Stahl [lS].) Hence a, + 1 = ean(G)y and 0

0

for all n EZ'. Since Bu(G),(B,(G))" by induction on n. There0 fore, BF(G)=BU(G)
It is easy to see that every graph is isomorphic to some orthogonality graph, and Lovasz defines 9(G) as the smallest value of an orthogonality graph isomorphic to G. I f x = ( x l , . . . ,xk)andy=(y1,. , . ,yl),letxo y = ( x l y l , . . . , x l y l r . . , x k y l r . . ,xkyI).Itis easy to check that (x y).(x'o y')=(x-x')(y.y'); it then follows that if G is isomorphic to an orthogonality graph on vertices ui(i= 1 , . . . p) and H is isomorphic to an orthogonality graph on vertices w j (j=1,. . . ,q), then G - H is isomorphic to an orthogonality graph on vertices ui w j(i = 1, . . . ,p , j = 1, . . . ,q). With these concepts, simple computations show that 9(G.H)<9(G)9(H)and a(G)<9(G). Therefore as(G)=sup ;/a(G")< sup m ) < 9 ( G ) . It is left to show that 9<&. The coualue of an orthogonality graph with vertices ul, u 2 , . . . , u p E R" is the maximum over all unit vectors u E R"' of the sum ( U - U ~+) ~. . . + ( U . U , ) ~ . It is a surprising fact proved by Lovasz [27] that 9(G) can be alternately defined as the largest covalue of an orthogonality graph isomorphic to G. We are now able to deduce the following two lemmas, inherent in the work of Lovasz. 0

0

163

Analogues of the Shannon capacity of u ymph

Lemma C. F o r any graphs G and H , 9(G v H ) > 9(G)9(H). Proof. Let ~ ( G ) = ( u - u , ).~. .++ ( U . U , ) ~ and ? ( H ) = ( t ~ w , ) ~. .+. + ( t - w J 2 , where is isomorphic to an orthogonality graph on ui ( i-= 1,. . . p) and H is isomorphic to an orthogonality graph on w j (i=1 , . . . 4 ) . Then G - H = G v H is isomorphic to an orthogonality graph on ui w j ( i = 1, . . . p , j = 1, . . . q), and

c

~

0

= 9(G)9(H ) .

0

Lemma D. F o r any graph G , 9(G)
c.

k

(u*u1)2+(u'Vz)2+. . . +(U.U,)'=

1 C (u.U)'
0

j = l vslj

Now we are in a position to see that 9<eD.For any graph G, Lemmas C and D imply that ~

&(G)= inf"JB(G'"')>,inf"J9(Gtn1)~inf;/S(G)" = 9(G).

This completes the proof of Proposition 4.

Remark. Each of the inequalities a
5

and so the last two inequalities can also be strict. Before leaving this section we note that when a(G)=B(G)(in which case G is sometimes called weakly a-perfect), all five numbers discussed in this section coincide, i.e.

P. Hell, F . S . Roberts

164

x = as = OD = C ~= F 8. Similarly, when w(G )= x ( G ) (G weakly y-perfect), all the complementary numbers coincide, i.e. w =cq,= us= O F = x. The observation that under weak x-perfectness as = a was originally made by Shannon [42] and was important

both historically and practically. The observation that under weak y-perfectness to suggest that there could be similarities of the kind studied in the present paper.

xu=x was made by Roberts [33] 6. Oddcycles

In this section we discuss the various numbers in the case of odd cycles C2k+l, k >,2. (The results are summarized in Table 1 .) Odd cycles have been of central interest in our subject (Baumert et al. [2], Stein [44], Shannon [42], Stahl [43], Berge [4], Hales [19], Lovasz [27], Rosenfeld [35, 361, Roberts [34] and KO [21]). A primary reason for this is that every graph G without odd cycles (i.e. a bipartite graph) is perfect, hence cc-perfect and y-perfect, and hence has a(G)= ccs(G)= OD(G) = c(F(G)= O(G) and w ( G )= cq,(G) = x s ( G) = OF( G) = x( G). Proposition 5. For every n E N and k >,2,

Proof. The last two formulas are proved in [43] (also in [21]) and [34], respectively. xu subject to xu E Z + (for each u E v(c2k+ and Recall that C(n(CZk+1) is max Cue" x , + x , < n (for each uw E E(C2k+l)); cf. Section 3. Let I be an independent set of k vertices of c 2 k + 1, and let x, = rn/21 for u E I and x, = Ln/2] for v 6 I . Then x, is a feasible solution of the above program, and 4,(C2k+ krn/21 +(k + 1)Ln/2] = nk + Ln/2]. On the other hand, summing up all the constraints xu+ x,
Value

Value

k

3

k

k

1 1 2 + - 2+k k

?

1

k+i k+i

2

1 1 2+- k + l k k

2+-

?

2

2

'?

k+i

k+$

1

2+k

1

2+-

k

k+i k+i

Analogues of the Shannon capacity of a graph

165

integer. The formula for mn(C2k+1) is proved analogously. Let n = q k + r , q E Z + , 0
+

cvEi

+

Corollary 1. The degree of perfectness of CZk+ is k, and the degree of perfectness of c , k + is 2 . Proof. Clearly Xn(C2k+1)=W,(CZk+I ) if and only if Ln/kJ= 1 + L(n - l ) / k J , i.e. when n is a multiple of k . Since each proper subgraph H of CZk+ is bipartite, x.(H)= m,(H) for all n E N. Therefore C2k+ has degree of perfectness k . (Also note that CZk+ is k-perfect but not k + 1-perfect.)Similarly, e , ( c 2 k + 1)=M,(C2k+ if and only if n is even and B,(H)=a,(H) for every proper subgraph H of c 2 k + Thus the degree of perfectness of c 2 k + is 2.

o

Of course, the numbers in Corollary 2 are easily calculated directly in the fractional versions MF. OF, WF, XF. From Corollary 2 and the trivial numbers t l ( c z k + l ) , W(Czk+l),X(CZk+I ) , e ( c 2 k + I), we can use the main result of Section 5 to fill in all of Table 1 except those entries labelled ?. These remaining values, O I S ( C ~(the ~ + capacity of CZk+ % ( c Z k + (the capacity of C2k+l), XS(CZk+l),and e D ( c 2 k + l ) = X S ( C 2 k + l ) , appear to be difficult to evaluate. Some partial information is summarized below. (Additional information about a(C$k+l)may be found in Stein [44], McEliece et al. [29] and especially Baumert et al. [2].)

+

Proposition 6. (a) a(c:k+ = k2 Lk/2J ; (b) o(C\;]+ ,)=4 for all k 2 3 (and w(CkZ1)=5); (c)X(c:k+l)'5 for all k 3 2 ; (d)B(C\?+,)=kZ+k+l for all k 3 3 (and 0(C521)=5);

166

P . Hell, F . S. Roberts

Proof. Part (a)is the result of Hales [19]. Part (b) is proved as follows: let the vertices of C2k+ be 0, 1 , . . . ,2k and assume K is a complete subgraph of C\$l+ with 5 vertices including (0,O).If two vertices of K agree in one coordinate, we may assume they are (0,O) and (0, 1); then all other vertices of K have first coordinate 1 or 2k, and at least two agree in the first coordinate. Without loss of generality K contains four vertices (0,0), (0, l), (1, x), and (I, x + 1); no other vertex of C\?+ is adjacent to all four of these vertices. Therefore no two vertices of K agree in a coordinate. Without loss of generality K contains four vertices (0, 0), (1, a), (2k, a - l), and (b, I). Since (1, a) is adjacent to (b, 1 ) and afO, a#2, b#O, we must have b = 2 ; then the fifth vertex of K must be (b+ 1, 2k)=(3, 2k). Similarly, ( 1 , a) is adjacent to (3,2k) and a#O, thus a=2k- 1. On the other hand (b, 1)=(2, 1) is adjacent to (2k, a - 1) and a - 1 fo,hence a - 1 = 2 and a=3. We conclude that k = 2 and 2k+ 1 = 5 . Since 0(C5’~)=~1(C[52~)=tl(C:)=~1(C:)=5 by (a), and since ~ ( C \ 2 k ] + ~ ) 2 ~ ( C ~ k +(b) ~ ) is ’ =proved. 4, To prove (c),note that since for any graph xcr 2 1 VI,

Thus it only remains to show that each c$k+can be 5-colored. The function c(i,j ) = 2i+j (mod 5) is easily seen to define a 5-coloring of C : . Let h : {0, 1 , . . . ,2k)+ {0, 1,2,3,4) be a function defined by h(O)=O, h(l)= 1 , . . . ,h(4)=4, h(5)=h(7)= . * * = h(2k- l ) = 3 , and h(6)=h(8)= . . . =h(2k)=4. A 5-coloring of C$k+lis obtained by letting c(i,j)=2h(i)+h(j) (mod 5). Indeed, if i, i‘ are adjacent in c 2 k + then h(i), h(i’) are adjacent in C , ; thus if ( i , j ) ,(i’, j ’ ) are adjacent in C:k+ then (h(i),hv)), (h(i’),h(j‘)) are adjacent in C: and c gives them different colors. (We have “extended the definition of c via the homomorphism h”.) Since for any graph, 00 2 I Vl,

for k >,3 (by (b)).Thus we have O(C\’J+ 1 ) 2 k2 + k + 1 . In order to prove (d), it remains to find k2 + k + 1 cliques covering C\’J+ 1 . Such a covering was constructed by Farber ~91. Lovasz [27] proved that c(s(C5)=J 5 ; since c5= C5, we also have %(C,)= 43. We already have xs( C , ) < JX(C:)= J5 by (c). Moreover,

and so xs(C,)= JS. Hence also e,(C5)=xs(c,)=xs(C5)= Lovasz [27] calculated

43,proving (e).

~

Part (f)followsfromtheinequalities ccs(G)2Ja(G2)(togetherwith(a)), andcrs(G)<9(G).

Analogues of the Shannon capacity of a graph

167

Part (g) follows from the relations % ( G ) > o ( G ) , %(G)=as(G), as(G)<9(G), and 8(c2k+ I)=(% 1)/9(c2k+I ) = 1 l/cos(X/(2k+ I)), [27]. G ) XS(G)<XF(G). Part (h) follows from the inequalities x ~ ( G ) > ~ (and Finally, part (i) follows from the inequalities $(G)>9(G) and &(G) <&(G). 0

+

+

We close by observing that the work of Stein [44]on sum-free sets in abelian groups can be used for calculating X(C;k+l). In particular, it suggests the following 49coloring of C: : let c(x, y, z, u, u) = (y + z + 2u + 3u, x + y + 32 + u + 6u) (mod 7). It is easy to verify directly that c is a coloring. Hence xS(C7)<

v@.

Acknowledgements The authors wish to thank M. Farber, R. Giles, L. Lovasz, R. Opsut, and M. Rosenfeld for helpful comments and suggestions.

References [l] 1. Barany, A short proof of Kneser’s conjecture, J. Combin. Theory Ser. A 25 (1978) 325-326. [2] L. D. Baumert et al., A combinatorial packing problem, in: Computers in Algebra and Number Theory, SIAM/AMS Proc. No. 4 (1971) pp. 97-108. [3] C. Berge, Farbung von Graphen, deren samtliche bzw. deren ungerade Kreise starr sind, in: Wiss. Z. Martin-Luther-Univ. Halle-Wittenb-rg, Math-Natur. Reihe (1961) pp. 114-1 15. [4] C. Berge, Graphs and Hypergraphs (American Elsevier, New York, 1973). [5] B. Bollobas and A. Thomason, Set colorings of graphs, Discrete Math. 25 (1979)21-26. [6] V. Chvatal, P. Erdos and Z. Hedrlin, Ramsey’s theorem and self-complementary graphs, Discrete Math. 3 (1972) 301-304. [7] V. Chvatal, M. R. Carey and D. S. Johnson, Two results concerning multicoloring, Annals of Discrete Math. 2 (1978) 151-154. [S] F. H. Clarke and R. E. Jamison, Multicolorings, measures and games on graphs, Discrete Math. 4 (1976)241-246. [9] M. Farber, Personal communication (1980). [lo] M. Fekete, Uber die Verteilung der Wurzeln bei gewissen algebraischen Gleichungen mit ganzzahligen Koeffizienten, Math. Z. 17 (1923)228-249. [ 111 D. R. Fulkerson, The perfect graph conjecture and the pluperfect graph theorem, in: Second Chapel Hill Conf. on Combinatorial Math. and its Appl., Chapel Hill (1969) pp. 171-175. [ 121 D. R. Fulkerson, Blocking and antiblocking pairs of polyhedra, Math. Programming l(1971) 168-194. [13] M. R. Carey and D. S . Johnson, The complexity of near-optimal graph coloring, J. Assoc. Comput. Mach. 23 (1976) 43-49. [ 141 D. Geller, r-tuple colorings of uniquely colorable graphs, Discrete Math. 16 (1976)9- 12. [15] D. Geller and S. Stahl, The chromatic number and other functions of the lexicographic product, J. Combin. Theory Ser. B 19 (1975) 87-95. [161 E. N. Gilbert, Unpublished technical memorandum, Bell Telephone Laboratories. Murray Hill, New Jersey (1972). [17] M. C. Golumbic, Algorithmic Graph Theory and Perfect Graphs (Academic Press. NY, 1980). [ 181 W. Haemers, On some problems of Lovasz concerning the Shannon capacity of a graph, IEEE Trans. Inform. Theory 25 (1979) 231-232. [19] R. S. Hales, Numerical invariants and the strong product of graphs, J. Combin. Theory Ser. B 15 (1973) 146-155.

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[20] A. J. W. Hilton, R. Rado and S. H. Scott, A ( < 5)-colour theorem for planar graphs, Bull. London Math. SOC.5 (1973) 302-306. [21] C. S. KO, Personal communication (1978). [22] R. Li and W. Li. Independence numbers of graphs and generators of ideals, Combinatorica 1 (1980) 55-61. [23] L. Lovasz Minimax theorems for hypergraphs, in: Hypergraph Seminar, Lecture Notes in Mathematics No. 41 1 (Springer, Berlin-New York, 1974) pp. I 1 1-126. [24] L. Lovasi Normal hypergraphs and the perfect graph conjecture, Discrete Math. 2 (1972) 253-267. [25] L. Lovasz, A characterization of perfect graphs, J. Combin. Theory Ser. B 13 (1972)95-98. [26] L. Lovasz, Kneser’s conjecture, chromatic number. and homotopy, J. Combin. Theory Ser. A 25 (1978) 319-324. [27] L. Lovasz, On the Shannon capacity of a graph, IEEE Trans. Inform. Theory 25 (1979) 1-7. [28] R. J. McEliece and E. C. Posner, Hide and seek, data storage, and entropy, Ann. Math. Stat. 42 (1971) 1706- I7 16. [29] R. J. McEliece, E. R. Rodemich and H. C. Rumsey, Jr., The Lovasz bound and some generalizations, J. Combin. Inform. Systems Sci. 3 (1978) 134-152. [30] R. J. Opsut and F. S. Roberts. I-colorings, I-phasings, and I-intersection assignments for graphs, and their applications, Mimeographed Notes, Dept. of Math., Rutgers University (1980). [31] R. J. Opsut and F. S. Roberts, On the fleet maintenance, mobile radio frequency, task assignment, and traffic phasing problems, in: G . Chartrand et al., Eds., The Theory and Applications of Graphs (Wiley, New York, 1981) pp. 479-492. [32] G. Polya and G . Szego, Problems and Theorems in Analysis (Springer, New York, 1972). [33] F. S. Roberts. Graph Theory and its Applications to Problems of Society, CBMS-NSF Monograph No. 29 (Society for Industrial and Applied Mathematics, Philadelphia, PA, 1978). [34] F. S . Roberts, On the mobile radio frequency assignment problem and the traffic light phasing problem, Annals New York Acad. Sci. 319 (1979)466-483. [35] M. Rosenfeld, On a problem of C. E. Shannon in graph theory, Proc. Amer. Math. SOC.18 (1967) 315-319. [36] M. Rosenfeld, Graphs with a Large Capacity, Proc. Amer. Math. SOC.26 (1970) 57 - 59. [37] A. Schrijver, A comparison of the Delsarte and Lovasz bounds, IEEE Trans. Inform. Theory 25 (1979) 425-429. [38] A. Schrijver, Association schemes and the Shannon capacity: Eberlein polynomials and the ErdosKO-Rado theorem, to appear. [39] S. H. Scott, Multiple Node Colorings of Finite Graphs, Ph.D. Dissertation, Department of Mathematics, University of Reading, England (March 1975). [40] J. J. Seidel, Strongly regular graphs with ( - I, I, 0) - adjacency matrix having eigenvalue 3, Linear Algebra Appl. 1 (1968) 281-298. [41] P. Seymour, On multicolorings of cubic graphs, and conjectures of Fulkerson and Tutte. Proc. London Math. SOC.38 (3) (1979) 423-460. [42] C. E. Shannon, The zero-error capacity of a noisy channel, IRE Trans. Inform. Theory 2 (1956)8- 19. [43] S . Stahl, n-Tuple colorings and associated graphs, J. Combin. Theory 20 (1976) 185-203. [44] S. K. Stein, Free sets in abelian groups, Linear Alg. Appl. 17 (1977) 191- 195. 1

Received 30 October 1980; revised 5 February 1981

Annals of Discrete Mathematics 12 (1982) 169-183

@ North-Holland Publishing Company

ON PARTIALLY RESOLVABLE ?-PARTITIONS Charlotte HUANG Department of Mathematics. Carleron University, Orrawa, Ontario. Canada

Eric MENDELSOHN* Department of Marhemarics, Universily of Toronto. Toronto, Ontario. Canada, M S S I A l

Alexander ROSA** Department ofMathematica1 Sciences, M c M a s t e r University, Hamilton, Ontario, Canada. L8S

4K I

Dedicated to Professor A. Kotzig on the occasion of his sixtieth birthday

A t-(u, K, I)-partition (briefly, a t-partition) is a pair (r!B) where V is a u-set and B is a collection of ki-subsets of V (called blocks) with ki E K such that every t-subset of V is contained in exactly A blocks of B. (A t-partition is sometimes called a t-wise balanced design). The elements of K are referred to as the block sizes. Given a t-(u, K, &partition (r! B), a parallel class is a subset of B which partitions K We introduce the following definition: A partially resoloable t-partition PRP t-(P,S , u ; M ) is a t-(u, K, 1)-partition such that (i) K = P u S , PnS=8, (ii) IP1= IM(, (iii) If P = { p l , . . . ,p q ) , M = {ml,. . . ,mq], the blocks of size p i can be partitioned into mi parallel classes. Here t 2 2 , and p , sat for every p E P,s E S. In this paper we will be concerned primarily with the case IP(= 1, IS1= 1. We will denote the corresponding partially resolvable t-partitions by PRP t-(p, s, u ; m ) provided P = { p ) , S = {s). Although these designs are interesting on their own they also arise naturally in many instances, for example, when considering embeddings of Steiner triple systems, Doyen and Wilson [7], recursive constructions for Steiner quadruple systems, Lindner and Rosa [14], and other occasions. In Section 2 we discuss necessary conditions for the existence of PRP t - ( p , s, u ; m ) and present some general results and constructions. Section 3 deals with the existence of PRP t-(p, s, u ; m)in the two ‘smallest’ non-trivial cases, i.e. when t = 2, p = 2, s = 3, *Research supported by NSERC Grant No. 7681. **Research supported by NSERC Grant No. 7268. 169

C . Huang, E . Mendelsohn, A . Rosa

I70

and t = 2, p = 3, s = 2, respectively. Section 4 discusses the case of PRP 2-(2,4, u ; m). The paper concludes with two examples of partially resolvable 3-partitions. Some of the results in this paper have been presented at the 6th British Combinatorial Conference at Egham in July, 1977.

2. Necessary conditions and general results Thus a PRP t-(p, s, u ; m ) is a t-(u, { p , s), 1)-partition containing exactly mu/p blocks of size p which can be partitioned into m parallel classes of blocks. Observe that our definition allows to occur blocks of size s only or blocks of size p only. In the former case m=O and the PRP is a Steiner system S(t, s, u) while in the latter case, PRP is a resolvable Steiner system S(t, p , u ) provided p > t , and a 1-factorization of the complete t-uniform u-hypergraph (that is, a resolvable ‘complete’ Steiner system S ( t , t, u)) provided p = t. In general, we have

if p = t , these conditions reduce to

0< m

<

( ).

We have the following necessary conditions for the existence of a PRP t-(p, s, u ; m ) :

Condition (2) is obvious: if there is at least one parallel class of blocks of size p then p must divide u. Conditions (3) and (4) are obtained by counting the total number of blocks of size s, and the total number of blocks of size s containing a given element, respectively. Next we present some general constructions for PRPs with t = 2. A parallel class of blocks of size 2 will be called for brevity a 1-factor. For undefined terms we refer the reader to Hanani [lo]. In what follows we will assume that the trivial necessary conditions (1)-(4) are satisfied.

Lemma 2.1. (a) Ifthere exists a group divisible design GD(s, 1, p ; u ) then there exists a

PRP 2-(p, S, U ; 1). (b) Ifthere exists a GD(s, 1, p ; u) and a PRP 242, s, p ; m) then there exists a PRP 242, s, u ; m).

Partially resolvable t-partitions

171

Proof. Obvious: (a) take the groups to form a parallel class of blocks of size p ; (b) put a copy of a PRP 242, s, p ; m) on each group, and take the union of the m one-factors from the various groups to form the m one-factors of the PRP 2-(2, s, u ; m). Lemma 2.2. Let k = O (mod 2) and let m = O (mod k - 1). If there exists a resolvable BIBD(u, k, 1) then there exists a PRP 2-(2, k, u ; m). Proof. Obvious: take m/(k - 1) parallel classes of the resolvable BIBD and ‘dismantle’ them; each of them yields k - 1 one-factors. Lemma 2.3. If there exisis BIBD(u+m, k, 1) containing a sub-BIBD(m, k, 1) then there exists a PRP 2-(k- 1, k, u ; m). Proof. Remove the m elements of the sub-BIBD(m, k , 1) from the blocks of the BIBD(u+m, k, 1); each of these m elements ‘induces’ a parallel class of blocks of size k - 1. Lemma 2.4. Let tu f 1 (mod 2). If there exists a PRP 2-(2, s, u ; m ) then there exists a PRP 2-(2, s, t u ; m + (t - 1 ) ~ ) .

Proof. It is well known that if t o is even there exists a resolvable group divisible designGD(2,l,u; tu)(withgroupsofsizeu). PuttingoneachgroupoftheGD(2,1,~;to) a copy of a PRP 242, s, u ; m)yields a PRP 2-(2, s, tu; m + ( t - 1 ) ~ ) . Corollary. Ifthere exists a PRP 2-(2, k, u ; m ) then there exists a PRP 242, k, 2u; u+m). Lemma 2.5. If there exist a PRP 242, k, u ; m),a BIBD(t, k, 1) and a transversal design T [ k , 1; u] (cf: [lo]) then there exists a PRP 242, k, t u ; m). Moreover, if k =O (mod 2), the BIBD(t, k, 1)contains q parallel classes and the T [ k , 1; u] contains p parallel classes (i.e. is a T,[k, 1 ; u], cf: [9]), then there exists a PRP 242, k, t u ; m + v(k - 1)) for v =0, 1,. . ., pq. Proof. Consider a set V= {uijli= 1, 2,. . . ,t ;j = 1, 2, . . . ,u } of t u elements, and let the elements of the BIBD(t, k, 1) be &, i= 1,2,. . . , t , with &= { u i j ( j = 1,2,. . . ,u } . For any block of this BIBD, say, { V,, V,, . . . , v k i , put a copy of the transversal design T [ k , 1;u] on the set k I V,, with the r s as groups. For each i= 1,2,. . . , t , put a copy of the PRP 242, k, u ; m) on the set b, and take a union of the m one-factors of the sets b to form the m one-factors of the set V. Clearly, the~. result is a PRP 2-(2, k, tu; m) on the set K If { V,, . . . , Vk) is a block in a parallel class of the BIBD(t, k, l), then independently V, can be replaced by k - 1 each of the p parallel classes of T,[k, 1 ;u] on the set one-factors (as k is even). Combining these 1-factors for each block of the parallel class into 1-factors of r! we obtain up to p ( k - 1) one-factors for each parallel class of the BIBD(t, k, 1).

u,=

ut=

C . Huang, E . Mendelsohn, A. Rosa

172

Lemma 2.6. l j there exist a PRP 242, k , u ; m) and PRP 242, k, t ; tm+ t - 1).

a

T [ k , 1; t ] then there exists a

Proof. Let {ul, u 2 , . . . , u v ) be the set of elements of a PRP 242, k , u ; m), and let U i ={ui}x { I , 2 , . . . , t } for i= 1,2,. . . ,u. For each block of size k, say, { u l , u2, . . . ,u k } of the PRP 2-(2, k, u ; m), put a copy of T [ k , 1 ; t ] on the set U,, with the sets U , as groups. Let F be any of the m one-factors of the PRP 2-(2, k, u ; m). If, say, {up,u,} is an edge of F then take the complete graph K Z fon vertices U p u U,, and decompose it into 2t - 1 one-factors. Take a union of these 1-factors obtained for distinct edges of F to form 2t - 1 one-factors of U = Ui. There are still m - 1 one-factors of the PRP 242, k, u ; m) remaining. If {u,, u,} is an edge in such a 1-factor G, decompose the complete bipartite graph on U,uU, (where the subgraphs induced by U , and U , are null) into t one-factors; combine these 1-factors obtained for distinct edges of G into t one-factors of U . Since there are m - 1 such 1-factors G, this yields t(m - 1) one-factors of U . Thus altogether we get 2t - 1 + t ( m - 1)= tm + t - 1 onefactors, and consequently a PRP 242, k, t u ; tm+ t - 1) on the set U .

u:=

uy=

Lemma 2.7. Let k a 3. A PRP 242, k , 2 k ; m) exists if and only if either (i) k = 3, m = 1, or (ii) m = k, or (iii) m = 2k - 1. Proof. Clearly, we must have m > 0. In a PRP 242, k , 2 k ; m) an element belongs to exactly r = (u - 1 - m)/(k - 1)= (2k - 1 - m)/(k- 1) blocks of size k. On the other hand, the total number of blocks of size k is b = ur/k = 2r= 2(2k - 1 -m)/(k - 1). If m < k then b= 2(k - 1 k -m)/(k - l), i.e. 2 < bG4. Assuming b= 3, we get r= 3/2, an impossibility. Assuming b = 4, we get r = 2, m = 1. For k = 3, a PRP 2-(2, 3, 6; 1) does indeed exist (cf. Theorem 3.3 below). For k > 3, assume w.1.o.g. that one of the blocks is {ul, u 2 , . . . , u k } and the unique 1-factor is F = { {ui,wi} i= 1,2,. . . ,k ) . There must be another block B containing u1 and another block B’ containing u2, and B and B’ must be distinct. However, (BnB’I2 2 for k > 3, a contradiction. Let now m a k, i.e. k G m G 2 k - 1. This implies 0 s b G 2 , i.e. either (i) b=O and m = 2k - 1, or (ii) b = 1 and m = (3k - 1)/2, or (iii)b = 2 and m = k. The case (ii) is clearly impossible while in cases (i) and (iii), the corresponding PRP obviously exists, as in these cases to find a PRP amounts to finding a 1-factorization of K 2 k , and of K k , k , respectively. In a PRP 242, k , u ; m) with OGmGu-1 and k a 3 , each element belongs to r =(u - 1 -m)/(k - 1) blocks of size k, the total number of blocks of size k is b = ur/k = u(u - 1 -m)/(k(k - 1)). The set of these blocks will be denoted by B = {Ill,. . . ,Bb). The set of rn one-factors will be denoted by F = { F 1 , . . ,F,,,}. In the following lemmas we investigate the existence of PRPs for some specific values of these parameters.

+

I

Lemma 2.8. Let r s 1. T h e n the corresponding PRP 242, k , u ; m)always exists. Proof. If r = 0 then m = u - 1 ;the PRP 242, k , u ; u - 1) is then the same as a l-factorization of K , . I f r = 1 then m = u - k , b=v/k. Put u = kt, t 2 1 hence b= t and m = u - k = k(t - 1). Since a PRP 2-(2, k , k ; 0)( = BIBD(k, k , 1)) always exists, we get from Lemma 2.4 that there exists a PRP 2-(2, k, t k ; k(t - 1)).

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Lemma 2.9. Let r = 2 and b < 8. I f a PRP 242, k, u ; m) exists then it is one of thefollowing :

(i) (ii) (iii) liv) (v)

(vi) (vii) (viii)

PRP 2-(2, 3,6; l), PRP 242, 4, 12; 5), PRP 242, 4, 14; 7), PR P 2-12, 3, 12; 7), PRP 2-12, 4, 16; 9), PRP 242, 5, 20; 1 l), PRP 2-(2,6, 24; 13), PRP 2-(2, 7,28; 15).

Proof. First of all, when r = 2 and a PRP 242, k, u ; m ) exists then b> k + 1 ;indeed, if {al,a2, . . . ,a t ) is one of the blocks, there must be another block of size k containing ai,i = 1, 2, . . . ,k, and for i # j, the block containing aiis distinct from that containing aj. Further, when k is odd, k 2 3; then since k I u, say u = 2kt, we get b = 2u/k = 4t. Ifb=4then k=3(case(i)).Ifb=5 then k = 4 b u t aPRP2-(2,4, 10;3)doesnotexist (cf. Lemma 4.8 below). If b = 6 then k = 4 (case (ii)). If b = 7 then either k = 4 (case (iii))or k=6; in the latter case we get u=7.6/2=21. a contradiction as u must be even. If b = 8 then either k = 3 (case (iv)) or k = 4 (case (v)), or k = 5 (case (vi)) or k = 6 (case (vii)),or k = 7 (case (viii)). Let us remark that all the corresponding P R P s except possibly the last one d o actually exist. (Note that a P R P 2-(2,7,28; 15)exists if and only if T8, the complement of the triangular graph T8 has a 1-factorization. For the cases (i)-(v), see Sections 3,4 below. The authors also succeeded in constructing a PRP 2-(2, 5, 20; 11) and a PRP 2-(2,6,24; 13). The following lemma is stated without proof (the proof is similar to the proof of Lemma 9 but tedious).

Lemma 2.10. If there exists a PRP 242, k, u ; m ) then either k < r + m - 3 or m = 1, b=u=k(k- l)+2. 3. The case t =2, [ p , s) = (2, 3) In this section, we consider the smallest two non-trivial cases of PRPs, namely, when t = 2 and (I) p=2, s = 3 , and (11) p=3, s=2. I. In the former case, a P R P 2-(2,3, u ; m)is a PBD(u, { 2, 3), 1) having exactly mu12 blocks of size 2 which can be partitioned into m parallel classes (1-factors). Here O < m < u - 1, and the conditions (2)-(4) become (5) Ifm=Othenu=l or3(mod6). (6) If rn > 0 then u = O (mod 2) (which in turn implies m = 1 (mod 2)),and, moreover, (6.1) if rn 2 1 (mod 6) then u = O or 2 (mod 6), (6.2) if m = 3 (mod 6) then u = O or 4 (mod 6), (6.3) if m = 5 (mod 6) then v = O (mod 6). LetZ,=(O, 1, . . . , u - l ) ; l e t , f o r e a c h a ~ Z , , a # O ,

C . Huang, E . Mendelsohn, A . Rosa

174

~a={x,y)(x,y~Z,,x-y=+a(modu)J,

and let, for a, b EZ,, a # b , a, b#O, r o * , , b = { ( x , y ) I x , y E Z , , x - y = + a o r +b(modu))

It is a simple observation that if a is odd and u is even then racan be partitioned into is a parallel class. We will also need the two parallel classes ( = l-factors); also, rOi2 following lemma:

Lemma 3.1 (Kotzig [12]). If u is even, and (a, b ) = 1 (i.e. a and b are relatiuely prime) then r;,b can be partitioned into four one-factors. With this lemma, we can prove the following:

Lemma 3.2. I f m - 5 (mod 6 ) and u = O (mod 12) then there exists a PRP 2-(2,3, u ; m). Proof. Let u=12t. Determine v = ( u - l -m)/6. If v = O (i.e. m = u - I ) there can be no blocks of size three, and a PRP 2-(2,3, u ; u - 1) is the same as a l-factorization of the complete graph K , ; this certainly exists. Thus assume v>O; then v E { 1, 2,. . .2t - 11. Let { { p i , q i ) ( i =I, 2 , . . . , v} be an (A, v)-system if v ~ 01 (mod , 4),and a (B, v)-system if v = 2 or 3 (mod 4) (cf. [18]).Form the set of triples B defined as follows

{ (0, i, i + p i + v) mod u, i = 1 , 2, . . . , v)

B={uo. i , i + 1 + p i + v ) m o d u , i = 1 , 2 , . . . , v)

if v EO, 1 or 3 (mod 4), ifv=2(mod4).

Further, let G be the following set :

I

r3v+l, r l ~ ~rfv+4,3v+s,. + ~ . ~ . . ,~ r*6t-2,6t-1t + ~ L ~ if v -0 (mod 4),

r:v+1,3v+2,

r:v+3.3v+4,

'

..

7

r61

r%t-2,61-1?

if v = 1 (mod 4),

G =< rv+lr

r 3 v + l , r 3 v + 3 , r*3v+4,3v+S,

r*3v+6,3v+7r..

.

9

r*6t-2,6t-l,

r6t

if v = 2 (mod 4), r3v,

r3v+23

r3v+3.3v+4,

r;v+S,3v+6,.

..

7

r*6t-2,6t-17

r 6 t

if v ~3 (mod 4). Clearly, for each r. in G, either a is odd or a equals 4 2 , and for each l-& in G, a and b are relatively prime. Thus, by Lemma 3.1, each raand each r&yield two and four parallel classes respectively; let F be the set of parallel classes obtained from G . A simple counting argument shows that there are m parallel classes in F . We claim that (Zb,B u F ) is a PRP 2-(2,3, u ; m). To see this, one has to show only that each 2-subset of Z , belongs to precisely one block of B u F . This verification proceeds by showing that for each a+O (mod u), the pairs { x , y } with x - y = + a (mod u ) belong to exactly one block of B u F ; this is tedious but straightforward. Now we are in position to state the following theorem.

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175

Theorem3.3. A PRP 242, 3, u ; m ) exists if and only if conditions ( 5 ) and (6) hold. Proof. It has been stated already that (5) and (6) are necessary. A PRP 242, 3, u ; 0) (i.e. when (5) holds) is nothing else than a Steiner triple system S(2, 3, u) which is well known to exist if and only if u 3 1 or 3 (mod 6). Let now one of (6.1),(6.2) be satisfied. Then u + m r 1 or 3 (mod 6) and u m 2 2m 1 . By [7], there exists an S(2, 3, u + m ) with a subsystem S(2, 3, m ) ; applying Lemma 2.3 yields a PRP 2-(2, 3, u ; m). Thus it remains to consider only the case when (6.3)holds. If m = 5 (mod 6) and u = 6 (mod 12) then a PRP 242, 3, u ; m ) exists by Lemma 2.10 of [7], and if m r5 (mod 6) and u-0 (mod 12) then a PRP 2-(2,3, u ; m ) exists by our Lemma 3.2, and the proof is complete.

+

+

Remark. Although the preceding proof relies heavily on the results of Doyen and Wilson [7] on embeddings of Steiner triple systems, an alternative direct proof along the lines of Lemma 3.2 not using those results can be presented but is longer. 11. I f t = 2 , p = 3 , s = 2 , we haveO<m<(u- 1)/2,and thenecessaryconditions(2)-(4) become (7) If m=O, u is arbitrary; if m>O then u-0 (mod 3). If a PRP (2-(3, 2, u ; ( u - 1)/2) exists we must have u = 3 (mod 6) and there are no blocks of size 2. Such a PRP is the same as a Kirkman triple system of order u ( = a resolvable S(2, 3, u ) ) ; Kirkman triple systems of order u have been shown to exist for all u = 3 (mod 6), [ 17). Applying now Lemma 2.2 (i.e. ‘dismantling’the parallel classes of a Kirkman triple system one by one, and replacing each triple {x, y , z } of a dismentled parallel class by three blocks {x, y } , {x, z } , { y , z } of size 2) we obtain that thereexistsaPRP2-(3,2,u;m)forallu~3(mod6),andallmsuch thatO
The 1-factor is

1

{cot, co,~u{{i, i+41) i = O , 1 , . . . ,401

mod 82.

Half of a base parallel class is as follows: 751, 19, 26, 481, (14, 37, 81), {8, 32, 77}, { 16, 27, 52). { 19, 28, 54}, (5, 12, 39), { 17, 22, 711, (0, 38, 70}, (20, 21, 72). ( 1 , 3, 64,), (2, 6, 181, (4, 10, 241, (15, 25, 33).

{ ~ O I 35, ,

The other half if obtained by replacing each i above by i + 41 modulo 82, and replacing co by co2. The 41 parallel classes are then obtained by developing the base parallel class mod 82.

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C . Huuny, E . Mendelsolin. A . Rosu

For the sake of completeness, we can state now: Theorem 3.4. .4 nearly Kirkman triple system of order u exists (mod 6) and u >, 18.

if and only if u ~0

Returning to the case of PRPs, we have: Theorem 3.5. A PRP 2-(3, 2, u ; m) exists (12, 5).

if and only i f ( 7 ) holds, and (u, m)#(6, 2),

Proof. It only remains to be shown, first, that a P R P 2-(3,2,6; 1) exists-a trivial fact, and second, that a P R P 2-(3, 2, 12; m) exists for m= 1, 2, 3,4. It was shown, in effect, in [13] that a PRP 2-(3,2, 12; 4) exists, and applying Lemma 2.2 yields the remaining designs. 4. Thecaset=2,p=2,s=4 The necessary conditions (2)-(4) in this case become (8) I f m = O t h e n u = l o r 4 ( m o d 12), (9) If m > 0 then (i) u -m = 1 (mod 3), and (ii) v -0 (mod 2), or, more precisely, either (ii.a) u rO (mod 4), or (ii.b) u = 2 (mod 4) and m = 1 (mod 2). In other words, (9.1) if m =O (mod 6) then u=4 (mod 12) (9.2) if m E 1 (mod 6) then v -2 (mod 6) (9.3) if m = 2 (mod 6) then v = O (mod 12) (9.4) if m = 3 (mod 6) then v = 4 (mod 6) (9.5) if m -4 (mod 6) then u = 8 (mod 12) (9.6) if m = 5 (mod 6) then u =O (mod 6). It is well known that condition (8) is also sufficient (see Hanani [lo]). Although we are unable to obtain necessary and sufficient conditions for the existence of PRP 2-(2, 4, u ; m)’s, we gather in this section various results concerning the existence of these P R P s and formulate the existence conjecture. First, we note that Lemmas 2.4,2.5,2.8all apply to this case. Further, we have a few direct constructions. Lemma 4.1. Let u = 4 (mod 12). A PRP 2-(2,4,u ; m) exists if and only if m = O (mod 3). msu-1.

Proof. Necessity is obvious. Sufficiency follows from Lemma 2.2 since a resolvable BIBD(v, 4, 1) is known to exist for v = 4 (mod 12) [ll]. Lemma 4.2. A PRP 242, 4, u ; 1 ) exists if and only if u = 2 (mod 6), u f 8 .

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Proof. See Brouwer and Schrijver [4]. Lemma 4.3. If u z 0 or 6 (mod 24) then there exists a P R P 2-(2,4, u ; u/2 + 2).

Proof. Let V=Zu12x { 1, 2). Since u/2+ 1 = 1 or 4 (mod 12), we may put on V;.= Zu12x { i } a P R P 2-(3, 4, 4 2 ; 1) obtained from an S(2,4, 4 2 ) by omitting an element from all blocks containing it (cf. Lemma 2.3). Let the single parallel class of triples be, w.l.o.g., { { ( 3 q ) i , ( 3 q + l ) i , ( 3 q + 2 ) i ) ( q = O ,I , . . . , ( ~ - 4 ) / 6 ) , i = l , 2 .

Define now a set of I-factors as follows:

The set 9={ F, F‘, F”, F1,F 2 , . . . ,Fui2-,) together with the blocks remaining from the two S(2,4, u/2)’s obviously forma a P R P 2-(2, 4, u ; u/2+2).

Lemma 4.4. (a) If u = 2 (mod 24) there exists a P R P 2-(2, 4, u ; 42). (b) ff u = 8 (mod 24) there exists a PRP 2-(2,4, u ; u/2 + 3t) for t = 0, 1, . . . ,(u -2)/6.

Proof. Since u/2= 1 or 4 (mod 12) there exists an S(2, 4, u/2). Let V=Z,,, x { 1, 2). Put a copy of S(2, 4, u/2) on each I(= Zu12x ( i ) , and decompose Ku12,u12 (with bipartition V= Vl u V2)into u/2 one-factors. When u = 8 (mod 24), there exists a resolvable S(2, 4, u/2), and applying Lemma 2.2 if necessary yields a P R P 242, 4, u ; u/2+3t) for t=O, 1 , . . . ,(u-2)/6. Lemma4.5. If u - p = O (mod 3), u ( u - p ) = O (mod 12), p f 6 and there exists a P R P 2-(2,4, p ; m) then there exists a P R P 2-(2,4, u ; m). Proof. It was shown in [ 5 ] that a GD(4, 1, p ; u ) exists whenever the conditions on u, p given in the statement of the lemma are satisfied. The rest follows from Lemma

2.l(b).

Lemma4.6. If there exists a P R P 2-(2, 4, u ; m) with m > O then there exists a P R P 2-(2,4, 2u;u+m-3).

proof. Let V = Z , x { 1,2). Put a copy of P R P 2-(2,4, u ; rn) on each V;.= Z , x { i } . and

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C. Huang, E . Mendelsohn, A . Rosa

let, without loss of generality, Fi={{O, li), ( 2 , 3i), . . . , { ( ~ - 2 ) i , ( ~ - l ) i } } ,i = l , 2 , be one of its m one-factors. We use the 1-factors F', F 2 to obtain the following set of blocks of size four (which will henceforth be called quadruples): 11, 0 2 , 123, {21, 31, 22, 32), . . . , {(U-2)1, (V-l)i,

(0-2129

(u-l)2;';'.

What is now left of the complete bipartite graph K U +on V= V, u V2 can clearly be decomposed into u - 2 one-factors; combining the remaining m - 1 one-factors of the PRP 2-(2,4, u; rnys on the Vs into 1-factors of V yields a PRP 2-(2,4, 2u; u + m - 3).

Corollary. There exists a PRP 2-(2, 4, u; u-4) for u = O (mod 4).

Lemma 4.7. There exists a PRP 2-(2,4, u; u -7) for u -0 (mod 4), u > 12.

Proof. Let V = Z u i 2x { 1,2}. Let B beaset ofquadruplesdefined byB= { {01,11,02,j2) mod 4 2 ) where j > 1 is the smallest integer such that (j, u/2) = 1. The remaining edges of V,= Zu12x { i ) can be decomposed into - 3 one-factors (for example, the edges of Kui2on &, i = 1,2, used in B have been chosen so that the two 1-factors into which they can be decomposed can be embedded in a 1-factorization isomorphic to GK,,,, cf. [16]). Taking the union of these - 3 one-factors of &, we get - 3 one-factors of K Further, define

iu

iu

iu

F , = { { i l , ( i + r ) 2 ) I i = 0 ,1, . . . , u/2-1}

mod42

iu

for r E Zu,2\{0, - 1,j, j - 1) ; there are -4 such 1-factors.Thus we have altogether u -7 one-factors of V; together with the blocks of B they yield a PRP 2-(2,4, u ; u - 7).

Lemma 4.8. There exists no PRP 2-(2,4, 10; 3).

Proof: If a PRP 2-(2, 4, 10; 3) exists it must contain 5 quadruples. Assume first that there are two disjoint quadruples, say { 1, 2, 3, 4) and {a, b, c, d ) . Let A, B be the remaining two elements. Then one can form one more quadruple, say {A, B, 1, a } but no further quadruple. Thus the existence of two disjoint quadruples is impossible, and so any two quadruples intersect (in exactly one element, of course). W.1.o.g.: let 11, 2, 3, 4}, (4, 5, 6, 7}, (7, 8, 9, l} be three of the quadruples, and let x be the remaining element. Then x must occur in the remaining two blocks, and so, again w.l.o.g., the remaining two blocks are { x , 2,5,8} and { x , 3,6,9>,and the whole configuration is essentially unique. Its complement, i.e. the set of pairs that do not occur in any of the five blocks, is, however, the Petersen graph which cannot be decomposed into three 1-factors, and so a PRP 2-(2,4, 10; 3) does not exist.

Lemma 4.9. There exists no PRP 2-(2,4, 12; 2).

Proof. If a PRP 2-(2,4, 12; 2) exists it must contain 9 quadruples and each element occurs in three quadruples. Assume that there are two disjoint quadruples, say, { 1, 2, 3, 4) and {a, b, c, d}. Any of the further quadruples can contain at most one

Partially resolvable t-partitions

k.79

number, and at most one small letter, and therefore at least two of the remaining four elements A, B, C , D.Thus we can form at most 6 further quadruples, falling one short of the required 9. Consequently, there can be no two disjoint quadruples, and any two quadruples intersect. Assume that the blocks containing the element 1 are { 1, 2, 3,4), { 1, 5, 6, 7) and { 1, 8,9, 10). There can be no quadruple containing both remaining elements 11 and 12 since if such block were {a, b, 11, 12) then there could be no third quadruple containing a. It follows that w.1.o.g. there are three further quadruples (2, 5, 8, ll}, (3, 6,9, 111, (4, 7, 10, 11). This forces either (4, 6, 8, 12) or {4,5,9, 12) to be a further quadruple; let this quadruple be, w.l.o.g., {4,6,8, 12). Then there must be two more quadruples containing 12, say C, D.If 2 E C then 3,5 E D,and 7 E C , 10 E D,and 9 E C so that the two quadruples are C = { 2,7,9, 12), D = { 3,5, 10, 12). Thus the configuration of 9 quadruples is unique (up to an isomorphism). However, its complement is 4K3 which cannot be decomposed into two 1-factors, and thus, a PRP 2-(2,4, 12; 2) does not exist.

Lemma 4.10.There exists a PRP 2-(2,4, 14; 7).

Lemma 4.11. There exists a PRP 2-(2,4, 18; 5). Proof. V=Z18; B = { { O , 4, 10, 15) mod 18). Form the graph rt,2(cf. Section 3); l-t,2is decomposable into four 1-factors. A fifth 1-factor is { {0,9>mod 18).

Lemma 4.12. There exists a PRP 2-(2, 4, 18; 11).

Proof. V=Z9 x { l , 2); B = { { O , , 41, 12, 32) mod 9); the base 1-factor {{01,31), {I1, {41,61),{OZ, 621, {L 721, {32, 42), {51, 12), { 7 1 , 5 ~ ){81,82) , mod 9) yields nine 1-factors. The remaining two 1-factors are { { O , , 22) mod 91, and { {01,42) mod 9).

L e m m a 4.13. There exists a PRP 2-(2,4, 20; 4). Proof. V = Z z o ;B = { {0,5,10,15),(0, 1,3,9>mod 20). The complement ofthe blocks in B (= the set of ‘unused pairs) is the graph rx,,(cf. Section 3) that is decomposable into four 1-factors. Lemma 4.14. There exists a PRP 2-(2,4, 20; 7). Proof. Take an affine plane AG(2,5)of order 5, and remove the elements of a block B from all the remaining blocks. Consider the graph induced by all the edges ( = pairs) of blocks of the parallel class containing B (excluding B itself) and of the blocks of another parallel class. This graph can be decomposed into seven 1-factors.

L e m m a 4.15. There exists a PRP 2-(2,4,20; 10).

180

C . Huang. E . Mendelsohn, A. Rosa

Proof. V=Zlo x { 1,2); B = { { O , , 51,02,52,){01,61, 22,42)mod 10).In Vl = Z l o x { I), the pairs with differences 1,2 and 3 have not been used up in B; these yield six 1-factors of Vl. In V,, the pairs with differences 1, 3 and 4 have not been used in B, and they likewise yield six 1-factors of V2.Taking the union of these six 1-factors of we get six 1-factorsof I/: Additional four 1-factors of V are yielded by the unused ‘mixed differences 1, 3, 7 and 9.

vs,

Lemma 4.16. (a) There exists a PRP 2-(2,4, 22; 3). (b) There exists a PRP 2-(2,4,22; 9).

Proof. v = z l l X { l , 2 ) . (a) B = { {01,41, 32,821, {01,31, 51, 12), {81, 22, 32, 10,) mod 11). The differences not occuring in B are the pure difference 1 in Z l l x { 11, the pure difference 2 in Z1 x {2), and the mixed difference 0. Thus the complement of the above blocks is the generalized Petersen graph GPG(11,2) which is decomposable into three 1-factors (see [ 6 ] ) . (b) Take the design given in (a), and ‘dismantle’ the blocks of the orbit determined by the base quadruple {01,41, 3,, 82); the resulting graph is decomposable into six I-factors (this can be seen, e.g. as follows: the pairs with difference 4 in Z1 x { 11, the pairs with difference 5 in Z l x { 21, and the pairs with mixed difference 1 form a GPG(11,4) which is decomposable into three 1-factors; the pairs with mixed difference 3, -3, -4 yield further three 1-factors). Lemma 4.17. There exists a PRP 242, 4, 22; 15).

Proof. V = Z l l x { l , 2 ) ; B = { { 3 1 , 5 1 , 0 21,) mod 11).The 1-factorsareobtainedas follows: form GPG(11, 2) by taking the pairs with pure difference 1 in Z l l x { l ) , with the pure difference 2 in Z1 x { 21, and with the mixed difference 0; this yields three 1-factors. Further, use the pairs with pure difference 3 (4 and 5, respectively) in both Z1 x { 1) and Z1 x 12) and with mixed difference 1 (8 and 9, respectively) to form GPG(11, 1). Each of those three graphs yields three further I-factors. The remaining three 1-factors are obtained by taking all pairs with mixed differences 6, 7 and 10, respectively. Lemma 4.18. There exists a PRP 242, 4, 24; rn) for rn = 5 , 8, 1 1.

Proof. The following nearly Kirkman (2, 4)-system appears in [19]: V=(Z,u{ 0 0 ) ) x Z 3 and the parallel classes are {{(a 0), ,(4, O), (2, l), (1,2)) mod (-, 3), {(3,0), (5,0), (6, O), (0, 1) mod (-, 3)) mod (7, -)}. The complement of this design is 8K3. However, if one adjoins to it the pairs of one parallel class, the resulting graph is decomposable into five 1-factors. Thus a PRP 2-(2,4, 24; 5 ) exists (in fact, the quadruples are also partitioned into parallel classes although this is not required). The P R P s for rn = 8 and rn = 1 1 are then obtained by applying Lemma 2.2 to one, and two parallel classes of quadruples, respectively, each parallel class yielding three 1-factors. Lemma 4.19. There exists a PRP 242, 4, 24; 2).

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181

Proof. V=Z12 x { 1, 2); B consists of {01,41, 22, 32), {01,&, 51, 92), {OI, 12, 52, 82), {O,, 61, 02,62) mod 12. The only pairs that do not appear in blocks of B are those with pure difference 1 in Z1 x { 1) and with pure difference 2 in Z1 x { 2) ;the corresponding graph is clearly decomposable into two 1-factors. Lemma 4.20. There exists a PRP 242, 4, 26; 7).

Proof. V=Z13 x {1,2);B=({O1, 61, 12,52), {OI, 21,31,02), {01,32,42, 62) mod 13). The pairs not occuring in B are those with pure difference 4 or 5 in 2 1 3 x { l), with pure difference 5 and 6 in ZI3x {2},and with mixed difference 4 , 6 or 11. The graph induced by these pairs is easily decomposable into seven 1-factors. Lemma 4.21. There exists a PRP 2-(2,4, 26; 19).

Proof. V=Z13x{l,2);B={{01,61,12,52)mod 13).Thecomplement ofBcontains pairs with pure differences 1, 2, 3 , 4 , 5 in Z I 3x { 1); with pure differences 1,2, 3, 5, 6 in 2 1 3 x 121, and with mixed differences 0,2,3,4,6,7,9,10,11. The graph induced by these pairs can be decomposed into 19 one-factors (for instance, by forming arbitrarily five generalized Petersen graphs GPG(13, t ) that would yield altogether 15 one-factors, and getting the remaining four 1-factors by using the pairs with the remaining four mixed differences). Theorem 4.22. Ler 0 ~ 2 8A. PRP 2-(2, 4, u ; m ) exists (9) are satisfied, and (u, m)#(8, I), (10, 3), (12, 2).

if and only if conditions (8),

Proof. The nonexistence of PRP 2-(2,4, u ; m)for (u, m)=(8, l), (10, 3), (12,2) follows from Lemmas 4.2, 4.8 and 4.9, respectively. Lemmas 2.1, 2.2, 2.4-2.6 of Section 2 and Lemmas 4.1-4.7 imply that a PRP 2-(2,4, o; m) with u, m satisfying the conditions of the theorem exists always except possibly when (u, m)=(14, 7), (18,5),(18, 1 l), (20,4), (20, 71, (20, 101, (22, 3), (22,9), (22, 1 9 , (24, 2), (24, 9, (24, 8), (24, 1 I), (26, 7), (26, 19). These cases are covered by Lemmas 4.10-4.2 1. Conjecture. A PRP 2-(2, 4, u ; m ) exists if and only if conditions (8), (9) are satisfied and (v, m)#(8, I), (10, 31, (12, 2).

5. Concluding remarks The case of PRP 2-(3,4, u ; m)’s appears to be more difficult. It follows from [5, 101 that a PRP 2-(3,4, u ; 1) exists if and only if u rO or 3 (mod 12) and a PRP 2-(3,4, u ; 4) exists if and only if u rO or 9 (mod 12). Several further isolated designs of this type appear in the literature, e.g. in a recent paper, Brouwer [3] constructs a PRP 2-(3,4, 24; 7) (as well as several examples of what amounts to PRP 2-((2, 3), 4, u ; { 1, m))’s, cf. Section 1).To construct PRPs with t >, 3 is certainly even more difficult. We present

C . Huang, E . Mendelsohn, A . Rosa

182

here only two examples: Example 1. PRP 3-(3,4,9; 16). V= Z9; the quadruples are ({0, 1, 3 , 4 ) mod 9}. Parallel classes of triples are: (1 parallel class), (3 parallel classes), mod 9 (3 parallel classes), (9 parallel classes).

Example 2 (R. Mathon [15]). PRP 3-(3, 5, 15; 7).

v=z,,;

8=(1{0,1, 2, 3, 12), (0, 1, 5, 6, 141, (0, 1, 7, 8, 131, {0, 1, 9, 10, 111, (0, 2,4,6,9), {O, 2, 5, 11, 131, (0, 2, 7, 10, 14), (0, 3,4, 10, 131, (0, 3, 5, 8,9), (0, 3, 6, 7, 111, {0,4, 5, 7, 121, (0,4, 8, 11, 141, {0, 6, 8, 10, 12), (0, 9, 12, 13, 141, { I , 2, 4, 7, l l ) , (1, 2, 6, 10, 131, (1, 2, 8,9, 141, (1, 3,4,6, 8 ) , { 1, 3, 5, 7, lo), ( I , 3, 1 1, 13, 141, { 1,4, 5, 9, 13}, {1,4,10,12,14], {1,5,8,11,12), {1,6,7,9,12), {2, 3, 4, 5, 141, (2, 3, 7, 9, 131, (2, 3, 8, 10, 11), ( 2 , 4, 8, 12, 13), (2, 5, 6, 7, 81, (2, 5, 9, 10, 121, (2, 11, 12, 13, 141, (3,4, 9, 11, 121, (3, 5, 6, 12, 131, (3, 6, 9, 10, 14), (3, 7, 8, 12, 141, (4, 5,6, 10, l l ) , {4,6, 7, 13, 141, (4, 7,8,9, 101, ( 5 , 7,9, 11, 14), ( 5 , 8, 10, 13, 141, (6, 8,9, 11, 131, {7, 10, 11, 12, 13)).

For the 7 parallel classes of triples, take any of the 240 resolutions of the following S(2, 3, 1 5 ) ( = PG(3,2)): ((0, 1,4), (0,2, 8), (0, 5, 10) mod 15).

Acknowledgement The authors thank the referee for valuable comments and suggestions.

Note added A recent paper by G. Stern and H. Lenz, “Steiner triple systems with given subspaces; another proof ofthe Doyen-Wilson theorem”, Boll. Un. Mat. Ital. A 17 (1980) 109-1 14 came to our attention recently. Their proof is similar in spirit to the proof of our Lemma 3.2.

Parrially resolvable t-partitions

183

References R. D. Baker and R. M. Wilson, Nearly Kirkman triple systems, Utilitas Math. 1 I (1977)289-296. A. E. Brouwer, Two new nearly Kirkman triple systems, Utilitas Math 13 (1978)31 1-314. A. E. Brouwer. Optimal packings of K4)s into a K., J. Combin. Theory ( A ) 26 (1979) 278-297. A. E. Brouwer and A. Schrijver. A group-divisible design GD(4, I. 2: n ) exists if and only n = 2 (mod 6).n # 8 (or: The packing of cocktail party graphs with K,'s), Math. Centrum, Amsterdam, ZW 64/76. A. E. Brouwer. A. Schrijver and H. Hanani. Group divisible designs with block-size four. Discrete Math. 20(1977) 1-10, F. Castagna and G. Prins. Every generalized Petersen graph has a Tait coloring, Pacific J. Math. 40 (1972)53-58. J. Doyen and R. M. Wilson, Embeddings of Steiner triple systems, Discrete Math. 5 (1973) 229-239. R. Guy, Private communication (1977). M. Hall, Jr., Combinatorial Theory (Ginn-Blaisdell, Waltham, MA, 1967). H.Hanani. Balanced incomplete block designs and related designs, Discrete Math. 1 I (1975)255-369. H. Hanani. D. K. Chaudhuri and R. M. Wilson, On resolvable designs, Discrete Math. 3 (1972) 343-357. A. Kotzig, Unpublished manuscript (1978). A. Kotzig and A. Rosa, Nearly Kirkman systems, in: Proc. Fifth S.E. Confer. Combinatorics, Graph Theory and Computing, Boca Raton 1974 (Utilitas Math., Winnipeg, 1974) pp. 607-614. C. C. Lindner and A. Rosa, Steiner quadruple systems a survey, Discrete Math. 22 (1978) 147-181. R. Mathon, Private communication (1979). E. Mendelsohn and A. Rosa, On some properties of I-factorizations of complete graphs, in: Proc. Tenth S.E. Confer. Combinatorics, Graph Theory and Computing, Boca Raton, 1979 (Utilitas Math., Winnipeg, 1979)pp. 739-752. D. K. Ray-Chaudhuri and R. M. Wilson, Solution of Kirkman's school-girl problem, in: Proc. Sympos. Pure Math. 19 (AMS, Providence, RI. 1971) pp. 187-203. A. Rosa, Algebraic properties of designs and recursive constructions, in: Proc. Confer. Algebraic Aspects of Combinatorics, Toronto (1975) pp. 183-200. P. Smith, PBIB designs and the Kirkman problem, Unpublished manuscript (1974). Received 16 February 1981

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Annals of Discrete Mathematics I2 (1982) 185- I92

@ North-Holland Publishing Company

DECOMPOSITIONS OF COMPLETE GRAPHS DEFINED BY QUASIGROWS Donald KEEDWELL U n i t w s i f j qf Surr6.j. G u i k / / i r t / .S u r r c , ~G U 2 S S H . U . K

Dedicated to Professor A. Kotzig on the occasion of his sixtieth birthday 1. Iaboduction

In 1970, A. Kotzig [6] introduced the very fruitful concepts of partition groupoid and partition quasigroup (usually abbreviated to P-groupoid and P-quasigroup respectively). Every partition of the complete undirected graph K, on n vertices into edge-disjoint circuits defines and is defined by a P-groupoid of n elements. (n is necessarily odd, since the degree of each vertex of K, has to be even if K, is to permit a circuit decomposition.) This groupoid is a quasigroup if and only if the graph partition has the additional property that, if there is a circuit which has ui, uj, u k as consecutive vertices, there is not a circuit which has ui, uh, u k as consecutive vertices for any O h # uj. Ofmost interest for many purposes are those decompositions of K , for which all the circuits have the same fixed length h. For example, those in which the circuits are edge-disjoint triangles correspond to Bruck quasigroups and are co-extensive with the Steiner triple systems. In [ 5 ] , the latter concept has been generalized to that of uniform P-circuit designs. These, like the Steiner triple systems, are associated with Room squares. Decompositions for which all the circuits are Hamiltonian are associated with latin squares which have cyclic transversals. In the present paper, we prove (1) if n = p";py . . . p?, where p l , p 2 , . . . ,p r are primes such that each of pol' - 1, py - 1, . . . , p? - 1 is divisible by 2h and h is odd, then there exists a P-quasigroup of order n which defines a decomposition of K, into a set of edge-disjoint circuits of length h, each having h distinct vertices; (2) the analogue for a directed graph of a P-quasigroup is an idempotent quasigroup; (3) if n = p : i p y . . . p:r, where p l , p 2 , . . . ,pr are primes such that each of p:l - 1, py - 1, . . . ,p y - 1 isdivisible by h and h > 2, then there exists an idempotent quasigroup of order n which defines a decomposition of K,*into a set of edge-disjoint circuits of length h, each having h distinct vertices; (4) for many (and probably for all) odd integers n (> 3), there exists an idempotent quasigroup of order n which defines a decomposition of K,* into a set of n edge-disjoint directed circuits of length n - 1 each having n - 1 distinct vertices. Finally, we investigate the particular case of decompositions of K,* which are defined by symmetric idempotent quasigroups. 2. Decomposition of K ,

Definition. A groupoid (G,

a )

which has the properties (i) a . a = a for all a E G ; (ii) I85

186

A.

D.KeedweII

a # b implies a # a - b and bf a - b for all a, b E G ; and (iii)a . b = c implies and is implied by c - b = a for all a, b, c E G is called a P-groupoid (partition groupoid).

In any P-groupoid ( G , .) we have (i) the number of elements is necessarily odd, and (ii) the equation x. b = c is uniquely soluble for x.

DeMtion. A P-groupoid which is also a quasigroup is called a P-quasigroup. Every P-groupoid or P-quasigroup of order n defines a decomposition of the complete undirected graph K , into a set of edge-disjoint circuits. Conversely, any such decomposition defines a P-groupoid (or P-quasigroup). These definitions and observations are due to Kotzig [ 6 ] . Detailed explanations may also be found in Denes and Keedwell [l]. An edge-disjoint circuit decomposition of K , corresponds to a P-quasigroup rather than a P-groupoid if and only if it has the following property P (see [4]):

Definition. If AB and BC are a pair of adjacent edges of a circuit in a decomposition of K , into edge-disjoint circuits and there does not exist any vertex D (Ddistinct from B ) such that A D and DC are adjacent edges of a circuit then the decomposition is said to have property P. Theorem 1. I f h > 2 is odd and 2hl+ 1 is a power of an odd prime, then there exists a P-quasigroup of order 2hl+ 1 which defines a decomposition of the complete undirected graph K 2 h 1 + into a set of edge-disjoint circuits of length h, each having h distinct vertices.

Proof. Let 2hl+ 1 = pr, p an odd prime, and let w be a generating element of the finite field GF[p'] so that w2h'=I , ws# 1 for 1 <s < 2hl. We label the vertices of K Z h 1 +by the elements of the field. Then, as vertices of a circuit we take the h vertices: u, u+w'(l

-w2'),

u+o'(l

-w4'),

. . . , u+w'(l -w2(h-1)1),

where t = O , 1, 2 , . . . ,I- I and u is an arbitrary element of the field G F [ p r ] . These vertices are all distinct. The differences between the labels of adjacent pairs of vertices of the above circuit are fw'(I -w2'), ~ w ' + -w2'), ~'(I -w2'), . . . , f ~ ' + ~ ( ~ - l ) ' (-w2'). f Write x = 1 -w2'.Then, since wh'= - 1 and since h - 1 and h 1 are even integers, we may + ( h + 1 ) I x, . . . ) re-write these differences as w'x, o ' + ~ ' x ,o ' + ~ ' x ,. . . , + ( h - l)lx, w r + 2 ( h - 1 ) 1 x, 0' + hlX( + ( h - 2)1X. + ( h + 2)' x,, . . , w f + ( 2 h - l ) l x,w'+'x,w'+3'x,. . . , - ofx), As t ranges through its permitted set of values, all non-zero elements of G F [ p r ]occur exactly once among these differences. Consequently, as u ranges through GF[p'], each edge of K2h1+1 occurs exactly once among the 1(2hl+ 1) circuits so generated. It remains to show that the edge-disjoint circuit decomposition of + thus defined has property P. Consider the differences between the labels of the pairs of nearly adjacent-vertices of the circuit. (Two vertices which occur in the same circuit are called nearly-adjacent if just one other vertex of the circuit lies between them.) These differences are

+

Decompositions of complete graphs

187

fw'(l-w4'), f 0 ' + 2 * ( 1 -w4'), . . . , +wr+2(h-1)1 (1 -a4').If we put y = 1 -w4', we may express these differences in a similar way to those between adjacent vertices. (Note that h must be odd.) As before, when t ranges through its permitted set of values, all non-zero elements of GF[p'] occur exactly once among the differences. Hence, when u ranges through GF[p'] any assigned unordered pair of nearly adjacent vertices , just once among the circuits. This is a sufficient condition for the of K Z h l + occurs decomposition to correspond to a P-quasigroup.

Since, as may easily be verified, the direct product of two P-quasigroups of orders rn and n which decompose K, and K, respectively into sets of edge-disjoint circuits of length h, each having h distinct vertices, is a P-quasigroup of order rnn which decomposes K,, into a set of edge-disjoint circuits of length h, each having h distinct vertices, the result (1) of the introduction is an immediate corollary of Theorem 1.

Remark 1. If n is an odd integer, n = 1 or 3 mod 6, then there exists a P-quasigroup of order n which defines a decomposition of K, into a set of edge-disjoint circuits of length 3, each having three distinct vertices. The P-quasigroup is the Bruck quasigroup associated with the Steiner triple system which exists for each such value of n. This is defined by (i) a i . a j = a k if (ai, aj, ak) is a triple of the system and (ii( ai.ai=ui for all symbols ai. Remark 2. Each circuit decomposition of K,, n odd, which has property P is equivalent to a one-factorization of K,+l.-In the case that such a decomposition of K , has the property that no vertex occurs more than once in any circuit of the decomposition and that the circuits are all of equal length h, we shall call the decomposition a uniform P-circuit design. In particular, Steiner triple systems are uniform P-circuit designs of block size h = 3. Perpendicular uniform P-circuit designs are ones which correspond to pairs of perpendicular one-factorizations. Such a pair of perpendicular uniform P-circuit designs is equivalent to an embedded Room square of side n (Keedwell [ 5 ] ) . The particular case of perpendicular Steiner triple systems is well known (Mullin and Nemeth [8], also Rosa [9]).

3. Decomposition of K:

Let us consider now the analogous decomposition problem for the complete directed graph K,*. The analogue for directed graphs of property P for undirected graphs, we shall call property P*. Thus,

Defhition. If Axand B d a r e adjacent directed edges of a circuit in a decomposition of K,* into edZ-disjo2 circuits and there does not exist any vertex D (Ddistinct from B) such that AD and DC are adjacent directed edges ofa circuit then the decomposition is said to have property P*.

188

A . D. Keedwell

The analogue of a P-quasigroup turns out to be any idempotent quasigroup. We have the following theorem :

Theorem2. Every idempotent quasigroup of order n defines a decomposition of the complete directed graph K,* into a set of edge-disjoint directed circuits with property P*. Conversely, any decomposition of Pnintpa set of*edge-disjoint directed circuits with property P* and with the property that A B and B A are not consecutive edges for uny choice of vertices A and B defines a corresponding idempotent quasigroup.

Proof. Let labels {a, b, c, . . . of a set Q of cardinal n be attached to the vertices A, B, C , . . . of K,*, where (Q,.) isagivensempotent quasigroup. Define directed circuits as follows: if a . b = c then AB and BC are adjacent edges of the same circuit. Since the equations x - b = c and b - c = z are uniquely soluble for x and z in (Q, ), the directed edge BC can belong to only one circuit. Since the equation a -y = c is uniquely soluble for y in (Q,.), property P* holds. Moreover, since the elements a, b, c in the equation are all distinct if a # b, the sequence of equations a . b = c, b - c = d , c-d = e, . . . , must eventually reach an equation of the from s1 * t =a. If t l .a = u1# b, then we shall get a further sequence of equations t l - a = u l , a.ul =vl, u l - v l = w l , . . . , until we reach an equation s2-t2=a, where s2#sl in virtue of property P* and so t 2 # t l . I f t 2 . a = u 2 # b then weshallget furtherequationst2-a=u2,a.u2=v2, u2-v2=w2,.. . , until, if t , is the unique solution of the equation y.a= b and s, is the unique solution of the eauatkn szr. = a, we finally + reach - -the equation sr-t , = a. Then the dir%d ed es AB, BC,:CDg . . . , S X , T I A , AU1, U,Vl, ..., A 3 2 , U2V2, V2 2, . . . , SJ,, T,A form a circuit. For the converse, define a. b = c if A X and B C are adjacent directed e d g e s 2 the same circuit. This gives a well-defined groupoid (Q, .). Since the directed edge BC can belong to only one circuit, the equations b - c= z and x - b = c are uniquely soluble for z and x respectively when b and c are distinct. SinceAhe circsts have property P* the equation a - y = c is uniquely soluble for y. Since A B and BA are not consecutive edges of a circuit for any choice of vertices .4 and B, a and c are distinct and we also have z # b, x # c. Since K,* has no loops, z # c, x # b and y # a. So, if we define a - a= a for all a E Q, (Q,.) becomes an idempotent quasigroup.

-

+-

ml,

5,2

Corollary. If the idempotent quasigroup is a P-quasigroup then the decomposition consists of pairs of oppositely directed circuits having the same edges and vertices. Theorem 3. If hl+ 1 is a power of a prime and h > 2, there exists an idempotent quasigroup of order hl+ 1 which defines a decomposition of the complete directed graph Kf,+l into u set of edge-disjoint directed circuits of length h, each having h distinct vertices. Proof. Let hl+ 1 = pr, p prime, and let w be a generating element of the finite field G F [ p ' ] so that mh'= 1, as#1 for 1 <s
Decompositions of complete graphs

189

where t = O , 1,2, . . . , I - 1 and u is an arbitrary element of the field GF[p']. The differences between the labels of adjacent vertices of this ordered circuit are d ( 1 -a+), 0'+1(1 -d), 0 ' + 2 1 ( 1 -mi), . . . , m'+(h-l)l( 1 -d). As t ranges through its permitted set of values, all non-zero elements of G F [ p r ] occur exactly once among these differences. Consequently, as u ranges through GF[p'], each directed edge of K & + occurs exactly once among the l(hl+ 1) circuits so generated. Consider now the differences between the labels of nearly adjacent vertices of the ordered cycles. These are Since h > 2, w21# 1 and so, as t and u range through their permitted sets of values, all non-zero elements of GF[p'] occur exactly once among these differences. Consequently, any assigned ordered pair A, C of nearly adjacent vertices of K f i + l occur just once among the cycles. This is equivalent to stating that property P* holds and so, from Theorem 2, it follows that the statement of the present theorem is true. A direct product construction similar to that used to obtain our result (1) from Theorem 1 gives the result (2) of the introduction as a corollary of Theorem 3.

Remark 3. If n is an integer, n =O or 1 mod 3, n # 6 , then there exists an idempotent quasigroup of order n which defines a decomposition of the complete directed graph K,* into a set of edge-disjoint directed circuits of length 3, each having three distinct vertices. Mendelsohn [7] has proved that, for all such values of n, a decomposition of K,* into a set of edge-disjoint directed circuits of length 3, each having three distinct vertices, exists. Let a, b, c be the labels of the vertices of one of these directed circuits. We define a - b = c , b . c = a, c - a= b. Since the directed edge 2 occurs in only one circuit, there does not exist any directed circuit a, d, c with d # b. It follows easily that, if Q denotes the set { a , b, c, d, . . . of vertex labels and we define a . u = a for all a in Q, then (Q, ) is the required idempotent quasigroup. For a detailed investigation of certain particular types of quasigroup which define decompositions of K,* of this special type, see also [ 2 ] .

-

Theorem 4. I f n is an integerfor which a set of three mutually orthogonal latin squares can be constructed by the column method, there exists an idempotent quasigroup of order n which defines a decomposition of the complete directed graph K,* into a set of n edge-disjoint directed circuits of length n - 1, each having n - 1 distinct vertices. proof. It is shown in [3] that a sufficient condition for the existence of a set of three mutually orthogonal latin squares of order n is that there exist a group G of order n whose non-identity elements go, g l , . . . ,g n T 2can be ordered in such a way that each non-identity element occur exactly once among the differences g ; lgi+1, i = O , 1, . . . , n -2, and that each non-identity element occur exactly once among the differences g ; ' g i +2, i = 0,1, . . . ,n - 2, where all suffix addition is taken modulo n - 1. The method of construction for orthogonal latin squares of which this is a special case was called

I90

A. D. Keedwell

'the column method in [3] and was originally used to prove the existence of triads of orthogonal latin squares of order 15. If we use the elements of G to label the vertices of K,* and define the circuits as the following n ordered sequences of vertices: gg0,g j g l , . . . , gd,,- ( j = O , 1, . . . , n - l), where go, g 1, . . . , g n - is the (primary) ordering that has the abovementioned properties and g n - l is the identity element of G, then it is easy to see that the circuits so defined will be edge-disjoint (because the products g ; ' g i + are all different in the primary ordering) and will satisfy property P* (because the products g ; 1 g i + 2 are all different in the primary ordering). The result of the theorem follows.

Corollary. If n is an integer which satisfies the condition of Theorem 4 (in particular, if n is a prime power, n > 2) then there exists an idempotent latin square L of order n whose off-diagonal elements can be decomposed entirely into disjoint cyclic partial transversals of length n - 1. I f each of these n cyclic partial transversals is indexed by the symbol which does not occur in it, then the indexing system defines an idempotent latin square which is orthogonal to L. Remark 4. For n = 15 and for n = 21 there exist a large number of orderings of the cyclic groups C15=gp{x: x15=e) and CZ1=gp(x:x Z 1= e ) which satisfy the conditions of Theorem 4. It is extremely probable that such orderings exist for the cyclic groups of every odd order except 9 though the author has been unable to prove this. For n = 15, there exist 32 such orderings, excluding mirror images, and for n = 2 1 a much larger number. For n= 15, the lexicographically first and last are 1,2,6,3, 14, 9,12,11,4,10,8,13,5,7;and1,14,10,12,9,2,5,6,11,3,13,4,8,7respectively,where numbers represent powers of x . For n = 2 1, the lexicographically first is 1,2,4,3,9, 17, 1 1,6, 19, 10,7,5,14, 18, 8, 15, 20, 13, 16,12. We consider finally the particular case of decompositions of K,* which are defined by symmetric idem potent quasigroups.

Theorem 5. Every symmetric idempotent quasigroup defines a decomposition of the completedJrecteddraph into a set of edge-disjoint directed circuits with t h L p r o p e 2 that, if AB and BC are consecutive directed edges of a circuit, then so are BA and A C .

Proof. The occurrence of A X and B C as consecutive directed edges implies that ab= c, where a, b, c are the labels attached to the vertices A , B, C of the graph respectively. Then ba=c and the stated result follows.

Theorem 6. For any odd integer n, the groupoid (Q, * ) deJned on the set of residue classes modulo n by the relation r . t - i ( r + t ) mod n is a symmetric idempotent quasigroup (associated with the P-quasigroup (Q,o ), where r 0 s = 2s -r mod n). When n = uh(h> 2), where uh is the odd integer defined by the recurrence relation ui= 2'- - u i - for i= 3, 4,. . . , and u 2 = 1, then the quasigroup (Q,.) defines a decomposition of the complete directed graph K,* into a set of edge-disjoint circuits of length h having the property of the preceding theorem and each having h distinct vertices.

Decompositions of complete graphs

191

Proof. The equation 2x= r + t mod n has a unique solution for x when and only when n is an odd integer, so (Q,-)is a groupoid. Moreover, the relations r l .t = r2-t *rl = r2 andr.tl = r . t 2 * t l = t 2 both hold whennisodd,so(Q, .)isaquasigroup. Itisobvious that it is both symmetric and idempotent. Now let the vertices of K,* be labelled by the elements of Q and suppose that f i isanarbitraryedgeofK:. Wehaver-t=i(r+t), t . ~ r + t ) = b ( r + 3 r ) , i ( r + t ) . b ( r + 3 t ) = 33r + 5 t ) =3u3r + u4t), where u3 = 2 + 1 , u4 = 2 + 3: and, generally, 1 1 1 2'-3 (ui- 3r ~ ~ - ~ (uit ) 2r- +vu i - l t ) = ji-l (ui- l r + uit),

+

-

where ui=2ui-2 + u i - for i = 3,4,. . . , and u2= u1 = 1 . Thus, u i + u i - =2(ui- + u ~ - ~ ) for i = 3 , 4 , . . ., and U ~ + U ~ =It ~follows ~ . that U ~ + U ~ - ~ = ~ 'for - ' i = 3 , 4, . . . . So, modulo uh, we have 0 + uh - = 2h- I and U h + + 0 = zh. Hence, modulo uh,

and

It follows that the h vertices with labels 1 r , t , i ( r + t ), ~ r + 3 t ) , . . . F, ( U i - i r + u i t ) , . . . (

1 ~ b h - z r + ~ h - l t )

form a directed circuit of length h and having h distinct vertices. We note that u3 = 3, u4= 5, u5 = 1 1 ,

u6

= 21, u, = 43, us= 85, and so on.

Example. Since u4=5, the complete directed graph K 3 has a decomposition into quadrangles with property P* and also with the property stated in Theorem 5. This decomposition is defined by the following symmetric idempotent quasigroup: 1 0 1 0 3 3 1 1 4 4 2 2 0

2 1 4 2 0 3

3 4 2 0 3 1

4 2 0 3 1 4

T 0 1 2 3 4

References [I] J. Denes and A. D. Keedwell, Latin Squares and their Applications (Akademiai Kiado, Budapest; English Universities Press, London; Academic Press, New York; 1974).

192

A. D. Keedwell

[2] J. W. Di Paola and E. Nemeth, Generalized triple systems and medial quasigroups, in: Proc. Seventh South Eastern Conf. Combinatorics. Graph Theory and Computing. Congressus Numerantium XVII, (Utilitas Math., 1976) pp. 289-306. [3] A. D. Keedwell, On orthogonal latin squares and a class of neofields, Rend. Mat. e Appl. 25 (5) (1966) 519-561. [4] A. D. Keedwell, Some connections between latin squares and graphs, in: Atti dei Convegni Lincei 17, Colloquio Internazionale sulle Teorie Combinatorie, Roma, 3-1 5 settembre 1973 (Academia Nazionale dei Lincei, Roma, 1976) pp. 321-329. [5] A. D. Keedwell, Uniform P-circuit designs, quasigroups and Room squares, Utilitas Math. 14 (1978) 141- 159. [6] A. Kotzig, Groupoids and partitions of complete graphs, in: Combinatorial Structures and their Applications, Proc. Calgary Internat. Conf. Calgary, Alta., 1969 (Gordon and Breach, New York, 1970) pp. 215-221. [7] N. S. Mendelsohn, A natural generalization of Steiner triple systems, in: Computers in Number Theory, Proc. Sci. Res. Council Atlas Sympos. No. 2, Oxford, 1969 (Academic Press, London, 1971) pp. 323-338. [8] R. C. Mullin and E. Nemeth, On furnishing Room squares, J. Combinatorial Theory 7 (1969)266-272. [9] A. Rosa, On the falsity of a conjecture on orthogonal Steiner triple systems, J. Combinatorial Theory Ser. A. 16(1974) 126-128. Received 3 June 1980

Annals of Discrete Mathematics 12 (1982) 193-201

@ North-Holland Publishing Company

REGULAR PERFECT SYSTEMS OF DIFFERENCE SETS OF SIZE 4 AND EXTREMAL SYSTEMS OF SIZE 3 Philip J. LAUFER College militaire royal de Saint-Jean, St. Jean. Quebec. Canada, JOJ 1RO and C.M.R.A.. UniversitP de Montreal, Montreal, Quebec, Canada, H3C 357

Dedicated to Professor A. Kotzig on the occasion of his sixtieth birthday Using the technique of 'central diamonds' suggested by A. Kotzig, we resolve the problem of determining all perfect (6, 5, 1)-systems and a special class of perfect systems for the case (8, 5, 1).

A combinatorial problem in radioastronomy describing the distribution of antennae [3,4], gave rise to the notion of perfect systems of difference sets [3]. Because of their importance both from a theoretical and an applied point of view, perfect systems of difference sets have attracted the attention of many mathematicians (see references).

2. Perfect systems of dmerence sets In this paper we use the definitions and notation of [3]. Let m, nl, n2,. . . ,n, and c be positive integers. Let A = A of sequences of integers

A

3,.

. . , A, be a system

A i = ( a i l < a i , < . . .
and let Di={aij-uikI l < k < j < n i )

be their difference sets. We say that the system S={D1,

D2,.

. . , D,)

is perfectfor c if

D = D ~ v D ~ * v* * vD,

{

= c, c

+ 1, . . . ,c - 1 + c i= I

Each Di is called a component of the system. The size of Di is ni - 1. A perfect system of difference sets for c is called regular if n l = n2 = . . . =n,=n, i.e. if its components are all of the same size. We then speak of a perfect (m, n, c)-system. It is convenient to display the differences in components of perfect systems as difference triangles (see [S]). If A = { a i l< a i 2< . . * < uini;,the difference triangle of *Research supported by Grant FCAC-EQ-539 and CRAD Grant 3610-592 193

Ph. .I. Laufer

194

A i is

... ... ..

where d f j ) = a i , j + k - u i j The size of a component is the number of rows in its difference triangle, or equivalently, the number of elements in the first row of its difference triangle.

3. Techniquesemployed Given the positive integers nl, nz,. . . ,n, and c, the corresponding problem of perfect systems involves both the existence and the determining of their explicit form. Determining the form of each of the systems clearly resolves the problem of existence. Both of the above problems are extremely complex. Most attempts by use of a computer were unsuccessful. The difficulty of finding a perfect system lies in the enormous number of possible ways of selecting rn sets of n elements chosen from among rn(Z) numbers. In [3], the number of possible cases is reduced by using the equations k

2

j= 1

n-k df;-k’=

2 dff’;

i = l , 2 , . . . , rn.

j= 1

Even with this simplification the number of possible solutions still remains large, rendering the problem of finding the corresponding perfect systems intractable. Results concerning the existence of regular perfect systems of size 2, 3 and 4 can be found% [2, 3,4, 5, 7, 1 1 , 12, 131. In [3] it is shown that the size must be less than 5 and for size 4 the number of components must be even. Moreover, in a system of size 4 we also have rn>,4c-2 [3, Prop. 2.11. Proposition 5.1 of [3] states there is no perfect (2, 5, 1)-system. A. Kotzig and J. Turgeon used an exhaustive elimination process to show that no perfect (4, 5, 1)system exists. For size 4, the difference triangle of a component is

Perfect systems of diflerence sets

dji)

dj:)

d!:)

195

d!:)

One of the first approaches considered in trying to determine regular perfect systems of size 4 was to examine the lower triangles

d!:)

dl:’ and

d!:’

dp

dp

dj:’

But the number of cases to be considered was too large even for the fastest available computers. Following a suggestion of A. Kotzig we instead considered the differences dj?), df:), d!:) and d!:’ which we call the ‘central diamond of the corresponding difference triangle. In every central diamond we have

+

d::’ d!:’

= d!:’

+ d!:).

If we require that the elements of the six central diamonds form an integer interval then we have 152 possibilities for the (6,5,1) case. Once a six-tuple of central diamonds has been obtained, we can calculate d(’)-d(4)-d!:) 14 - 1 1

and d$:)=d$) -&);

the remaining differences are then found by trial and error. The above procedure yields two perfect systems, namely numbers 1 and 2 of Table 1. It has later been modified to work without the requirement that the elements of the central diamonds form an integer interval; this has led to 73 more perfect systems.

4. Perfect (6,5,l)-systems Using the technique of central diamonds we determined all (6, 5, 1)-systems. In order to simplify their listing, we give only the first row of the difference triangles representing the six components. This suffices because each difference triangle is completely determined by its first row.

5. Perfect (8,5,1)-systems Table 2 contains all (8, 5, 1)-systems that satisfy the following condition:

{d!;),d!?, d!:’ I i = 1,2, . . . , 8 ; j = I, 2, 3,4} = { 1, 2, . . . ,48).

Ph. .ILauler .

I96

1

-7 3 4 5 6 7 8 9 10

II 12 13 14 15 16

17 18

19 20 21 77 --

23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51

Triangle I

Triangle 2

Triangle 3

Triangle 4

141225 9 16 I I 31 2 1021 20 9 18 1623 3 91431 6 1611 26 7 1 1 I430 5 18 14 17 I I 18 16 17 9 15 I3 22 10 17 1028 5 6 2 5 24 5 8 2030 2 102622 2 I5 I330 2 93013 8 10 1731 2 19 731 3 10 I732 I 1721 16 6 15 1922 4 8 I234 6 1430 12 4 14 1032 4 19 1321 7 132421 2 10 9 33 x 20 I326 I 1031 14 5 18 1522 5 17 9 2 7 7 19 8 2 4 7 82026 6 18 I I 26 5 8 2 8 I9 5 1227 17 4 13 7 2 8 12 I4 I324 9 12 740 I 13 1927 I 14 I I 23 I 2 10 1229 9 172411 8 I5 20 I 2 13 71336 4 I8 I722 3 14 5 2 9 I2 10 9 3 4 7 16 12 19 13 102324 3 19 I720 4

15 1624 4 5 20 30 4 16 8 3 2 3 9 1333 4 131626 4 14 1031 4 72031 I 1620 19 4 102023 6 17 1620 6 13 7 3 7 2 19 17 15 8 19 17 18 5 16 9 2 8 6 I2 6 3 4 7 18 1723 1 152021 3 1032 16 I 1321 19 5 9 3 1 15 4 I I 21 18 9 21 7 2 9 2 21 7 2 9 2 20 I I 26 2 14 9 35 I 19 7 2 9 4 131530 I I 2 1824 4 15 1721 6 14 6 3 2 7 I I lo35 3 13 1231 3 1724 15 3 13 I9 20 17 18 14 16 I 1 19 731 2 8 3 3 I7 1 1032 16 I 8 I728 5 15 6 2 9 8 19 I 3 1 8 71730 5 141527 3 I9 I I 26 3 14 17 16 I 2 I I 833 7 I6 1522 6 1423 17 5 18 2 3 5 4 20 131 7 16 5 2 6 12

10 1329 6 8 2 8 15 6 5 23 26 4 14 I721 6

33320 1 10 1423 9 1325 17 2 51930 2 81133 5 15 1722 3 15 1821 2 1025 13 9 I I 2421 I 5 24 25 2 14 16 18 8 1321 18 4 14 1524 3 51829 4 I4 I720 5 14 531 6 12 7 3 0 6 82718 2 14 1 6 2 2 4 8 24 20 5 17 16 14 10 1326 16 I 5 27 20 3 62721 I 5 22 25 4 10 I528 3 17 1224 3 9 2 2 I9 6 8 2026 3 17 19 10 I I 15 5 2 4 1 3 15 6 2 9 7 12 7 3 6 2 9 25 22 1 10 I231 3 1324 8 10 I I I626 3 I2 635 3 18 4 3 2 2 71133 5 5 22 28 2 I I 21 23 2 18 1320 6 6 I828 5 I I 1922 5 13 16 15 12 92718 2 12 6 30 8 8 22 23 3 14 1522 4 10 I329 2

17 732 2 8 I236 2 12 17 19 9 8 2 9 I5 6 8 I928 3 I I 838 I I I 1229 6 16 1427 1 10 I625 6 13 17 15 12 4 2 9 21 3 10 1529 3 18 I I 23 5 12 I I 28 5 I I 935 2 10 I823 7 I2 835 3 1411 24 9 10 I5 24 9 1330 8 7 I I 2617 3 8 2 7 I7 5 20 1421 2 81529 5 16 2 3 3 1 1925 16 8 12 1822 6 I I I7 20 10 I I I4 23 10 14 16 24 4 I3 I737 I 1520 14 9 19 4 3 2 2 21 8 2 6 2 I4 1621 6 9 25 20 3 1324 17 4 19 8 18 13 9 2 8 16 5 1021 23 4 21 2 3 2 3 20 10 24 4 8 2425 I 11 21 25 1 7 I733 I 1228 16 2 82221 I

Triangle 5 71127 I3 22 I7 7 2 1 I2 10 15 20 10 I 8 2 2 5 I828 10 3 34 3 27 24 1425 13 I2 9 3 1 3 1931 91126 1221 13 1421 19 I I 827 7 2026 83213 14 1521 71823 I4 I326 2 23 28 51827 1322 18 92516 16 15 18 16 18 14 63211 10 736 13 9 3 0 3 2 3 28 41931 2 I438 91327 1721 I2 15 629 3 22 28 10 I425

I I 2019 I 1 2015 14 1722 92915 16 4 3 3 7 25 22 1722 14 1029 9 62621 13 4 3 5 13 1524 I I 29 9 61330 1425 9

5 1 6 8 3 1 6 2 2 3 1

7 9 I 9 2 1 4 6 3 I 3 1

3 6 6 5 2 4 2 I 1

4 3 4 I 5 4 9 2 3 3 1

2 6 2 3 3 5 5 6

Triangle 6 82217 7 1226 11 22 I4 7 2 9 11 I2 1520 9 2 1 13 8 I622 5 733 7 532 7 23 14 9 1521 3 35 10 4 737 731 8 10 1622 I I 22 I 2 9 1622 91324 I2 3 2 8 I I I33 7 631 1022 15 82611 12 17 18 83010 93011 71822

2 3 1 1 1

6 4 1

4 4 4 2 I 3 1

4 4 6 8 2 5 4 6 5 2 1 4 162111 3 Ill224 I 4 3 1 12 I 82514 2 5 I822 4 16 5 2 9 I 635 8 2 91723 2 63011 5 922J5 6 7 I525 5 10 1326 3 I2 4 2 6 10 7 2 6 10 6 6 2 8 I4 I 10 2 3 4 4 734 8 I 8 1824 1 5 936 1 I I 1023 7 16 4 2 9 2 10 I521 6 I I 25 9 8 71133 2

197

Perfect systems of difference sets Table I. First row (d$). d!;), d$), 4;)) of (continued)

52 53 54 55 56 57 58 59

60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75

Triangle 1

Triangle 2

Triangle 3

16 1032 2 I7 1526 2 15 1721 7 I223 16 9 112325 1 12 1624 8 91727 7 Ill530 4 13 1231 4 17 7 3 1 5 19 6 2 3 12 12 19 24 5 12 I7 20 1 1 16 19 22 3 91826 7 15 8 2 8 9 9 2024 7 1422 19 5 1228 15 5 I520 I9 6 101725 8 I 1 7 3 2 10 15 I034 I 21 5 3 3 1

24 12 17 6 I1 933 6 92722 1 21 6 3 1 1 12 I921 6 18 7 3 0 4 21 1023 5 19 8 20 I2 91430 6 1334 3 9 101827 4 141128 6 21 2 3 2 4 11 I726 5 16 5 3 6 2 2027 7 5 1 1 3 3 5 10 101532 2 22 1 3 4 2 17 9 32 1 18 4 3 6 1 23 5 3 0 1 18 2 2 7 1 1 18 17 11 12

1425 15 3 19 3 2 7 8 I4 I030 3 7 1036 4 14 15 18 10 22 17 14 5 13 6 3 6 3 1421 18 5 20 1 2 7 10 I4 1626 2 16 8 32 2 22 4 2 3 9 8 1035 5 20 1 2 9 8 12 22 13 11 I6 I031 1 21 4 3 2 1 20 129 8 9 1032 7 I 1 236 8 1231 3 I 1 1624 14 3 531 17 4 14 I620 7

Triangle 4

Triangle 5

I I 830 7 12 1321 10 12 19 20 5 Ill924 2 17 7 3 0 2 I3 29 9 6 14 8 2 4 1 1 71731 2 15 24 16 2 82718 4 13 7 2 6 1 1 1030 16 1 I427 I5 1 14 10 23 9 19 23 8 6 1322 18 3 1326 15 2 I327 12 4 I I 1427 4 5 2423 4 521 28 2 82915 4 73013 6 8 2 9 I5 4

9 I328 5 41436 1 13 16 18 8 14 8 2 0 1 3 1322 16 4 11 1033 2 1525 12 4 13 1622 3 5 I729 3 15 1023 6 33614 1 13 20 18 3 7 19 25 3 43613 2 15 1029 1 11 I424 6 8 19 23 5 61728 3 16 17 13 8 18 16 14 7 16 19 13 7 1221 13 9 14 8 2 4 9 6 25 22 2

Triangle 6

4 27 20 71624 1 1 35 2 5 2 9 15 936 5 3 20 26 21829 9 136 11 8 2 6 11 21 19 9 2 1 17 8 735 13 9 2 4 72712 4 2 8 17 4 2 9 17 I2 18 16 I 1 726 61826 10 1228 91523 62025 12 1623 9 32 10

I 5 4 3 3 1 1 6

7 1 5 2 6 6 3 2 6 9 3 3 6 2 3 3

Remarks. Our computer program applied to the (4,5, 1) case confirmed the results of Kotzig and Turgeon: there are no perfect (4,5, 1)-systems.We also applied it to the case (6, 5, 2), and found that there is no such system. Combining our results with those of Kotzig and Turgeon [8], we obtain the following: perfect systems (6.5", 5, $"+ 1)) and (8.5",5, $5"+ 1)) exist for every positive integer n. This means that regular perfect systems of size 4 with 2r components exist for any infinite number of positive integers r . The notion of a 'split' perfect system of difference sets is introduced in [131 where it has an important role in addition theorems which are useful in the construction of large perfect systems from smaller ones. It was noted there that nothing was known about (m,5, c)-systems with proper splits. Rogers (private communication) has now checked the computer print outs for (6, 5, 1)- and (8, 5, 1)-systems and finds that none have proper splits, although this leaves open whether there are (m,5, c)-systems with proper splits. (It is now known [6] that, while there are (6, 4, 3)-systems, none have proper splits, thus giving a counterexample to Conjecture 3 in [131.) 6. Perfect (2c - 1,4,c)-systems It is shown in [3] that a necessary condition for the existence of (m,4, c)-systems is that m >,2c - 1. This inequality is best possible since (2c - 1, 4, c)-systems have been

Ph. J. Laufer

198

~~

1

-7 3 4 5 6 7 8 9 10 II 12 13 14 15 16 17 18 19 20 21 71 --

23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46

Triangle 1

Triangle 2

Triangle 3

Triangle 4

Triangle 5

Triangle 6

191539 7 161841 5 20163311 16233011 122141 6 92640 5 171938 6 201935 6 142238 6 83829 5 221935 4 212332 4 19242611 20272211 161743 4 1425 30 I I 241833 5 16 25 28 1 1 152630 9 72644 3 16282313 191835 8 17193311 72644 3 122144 3 19282211 13332212 142537 4 73238 3 21 2230 7 16282610 172140 2 132144 2 112240 7 172041 2 42945 2 82544 3 33045 2 202230 8 82645 1 72844 1 53242 I 44431 I 44134 1 33937 1 33343 I

162827 8 132337 6 182134 6 132438 4 152238 4 14282710 202823 8 I22538 4 17292310 202231 6 16302112 15243010 92240 8 14282413 9263717 19 18 38 4 14253010 2023 32 4 181740 4 181343 5 64229 2 93038 2 132138 7 18222910 16223110 53139 4 232128 7 152036 8 14233111 33638 2 202132 6 201635 8 92240 8 211641 I 13223410 15222814 152241 1 172136 5 22947 I 133031 4 182333 4 211931 7 222129 6 231931 5 181536 9 152623 14

122141 4 152531 7 123127 8 181440 6 173123 7 151643 4 133224 9 13342llO 192427 8 121844 4 13272810 13183512 181542 3 191640 3 102045 3 62743 2 152041 2 13 21 42 2 14332011 162528 9 181537 8 14282313 42647 I 201735 6 72343 5 I22045 I 34529 I 102245 I 201343 2 I I 3424 9 24530 I 15302210 16143611 54327 2 44230 I 113523 8 102631 9 16183210 142136 4 182327 9 143225 6 15262511 15252611 132037 7 161240 7 171242 4

131742 5 I2203510 153023 9 12223310 202524 8 171241 7 72741 2 15292211 121845 2 14272313 83725 7 92041 7 122935 1 54130 1 222425 6 103622 9 113722 7 7 2245 3 34528 I 11233210 102039 7 102241 3 182327 8 123627 1 171939 I 162527 8 14262511 162631 2 91844 4 1432 I7 12 I22233 7 44425 3 122337 4 13193410 212623 5 I23620 7 202823 4 141340 6 151832 9 141540 7 121738 9 121644 4 161738 3 163022 6 133522 4 9 1839 8

112531 9 143321 8 42546 1 192821 8 11292610 192524 8 16212910 183023 5 92539 3 113226 7 15292011 171634 8 161438 6 73823 6 152328 8 202823 3 62146 1 17 18 33 5 122232 6 122435 4 192134 1 161147 1 143224 5 151634 9 202527 2 182623 6 161538 4 133418 9 15193010 182627 1 111839 4 Ill837 9 182427 6 152030 9 151439 6 182524 6 182129 6 93520 8 101739 7 172033 5 162034 5 172924 3 93623 5 12173211 113224 6 133520 5

181437 6 17 9 4 4 4 103819 7 34525 1 141343 3 112338 1 111546 1 34326 1 161540 I 192821 3 22447 1 22645 I 101739 7 101539 9 I2292111 123221 8 121636 8 12 1440 6 192425 2 153019 8 Ill443 4 153121 4 161237 6 Ill442 4 131534 8 103815 9 83919 5 192921 3 172921 5 8 I542 5 133515 8 141341 6 192628 1 172823 3 112724 9 191339 1 13173211 192922 I 162823 3 163219 6 111040 8 82233 6 101439 7 91539 8 142723 8 192130 2

-

Triangle 7 Triangle 8 1038202 1119393 1328222 720315 919395 1333183 I230223 832177 1137137 1035169 934185 I127223 1334205 1232188 1314402 7 17 35 5 1332179 1037 159 1036167 1729211 I226243 1233177 1015393 1332212 1124266 1321307 1017376 1228237 I235166 1335 1 6 4 1725243 I219345 I015385 1224256 I236163 1034179 1234195 Ill5394 1134195 1012422 1527242 1035149 I230198 1038183 1729205 1028247

326231 222271 537143 935152 216341 630202 535144 936142 521284 215391 636143 637145 428212 417342 534181 13 34 15 I 419313 8 19 30 1 813375 614382 932175 620295 931202 830195 937144 314402 932182 611385 828241 1031 1 9 6 937145 726231 732173 838144 832187 516383 738142 1225247 1312376 1128213 1330193 1334182 1334182 1426252 1034192 1134166

Table 3. Extremal Systems which cannot be. generated by additive sequences of permutations

1. (9, 4, 5):

57 58 47 42 49 36 16 31 1 1 21 28 8

56 55 54 43 37 40 39 45 33 19 24 13 22 23 10 15 25 14

53 52 46 32 48 35 18 30 5 20 26 6

51 44 34 17 21 7

50 41 38 12 29 9

68 67 71 69 66 70 60 49 58 45 53 42 56 43 59 39 51 50 22 38 11 20 31 19 26 30 13 23 35 10 25 28 14 27 32 7 62 61 65 63 64 47 44 55 40 57 41 54 46 52 48 24 33 8 16 36 12 17 37 9 18 29 15 21 34 6 78 84 79 82 83 81 80 3. (13. 4. 7): 71 58 70 55 63 53 60 56 66 48 69 50 62 61 26 45 13 30 33 20 27 43 12 25 35 21 32 34 14 18 44 17 28 41 9 l(ll.4.6):

73 72 77 74 75 76 57 54 61 46 59 52 64 51 68 41 65 49 31 36 10 29 39 8 24 40 11 22 37 15 19 38 16 23 42 7 4. (15.4, 8):

97 96 92 93 94 95 91 90 82 67 80 63 75 58 71 65 81 64 79 56 72 57 78 53 30 52 15 33 47 16 31 50 14 29 42 23 35 40 18 36 43 13 34 38 19 37 41 12 88 85 87 89 84 86 70 62 69 68 66 61 77 59 73 60 76 54 21 48 20 27 39 22 25 45 17 32 44 10 26 51 8 24 49 I 1

83 74 55 28 46 9

Table 4. Extremal Systems which can be generated by additive sequences of permutations

1. (13.4, 7):

84 83 82 81 80 79 78 71 58 68 54 70 55 62 53 69 48 61 56 64 47 26 45 13 29 39 15 27 43 12 28 34 19 32 37 11 23 38 18 31 33 14 77 76 75 74 60 57 66 46 59 51 67 29 20 40 17 30 36 10 24 35 16 25 42 7

2. (15, 4, 8):

73 65 52 21 44 8

72 63 50 22 41 9

97 96 95 94 93 92 91 82 67 80 63 81 64 77 58 71 61 79 55 70 65 30 52 15 33 47 16 31 50 14 36 41 17 32 39 22 37 42 13 26 44 21 89 88 86 85 84 87 78 53 68 59 75 62 69 66 73 60 76 57 23 46 20 35 43 10 28 40 19 24 51 11 25 48 12 27 49 8

83 74 54 29 45 9

90 12 56 34 38 18

Perfect systems of difference sets

201

found for c= 1, 3, 4, 7, 10 and some other values derived from these by means of certain special sequences of permutations (see [8] and [12]). It is conjectured that these extremal systems exist for all c, c # 2 (compare Conjecture 2 in [133). The method of central diamonds is especially suitable for the study of this extremal case since the argument in [3] may be developed to show that the central diamonds contain a consecutive run of large differences (that is, in the terminology of [13], a (2c - 1,4, c)-system has a split at 5c - 2. In this way all (2c - 1,4, c)-systems have been foundforc=5 and6(thereare96and 2144ofthem respectively)andmany(2c- 1,4,c)systems have also been found for c = 7 and 8. Some examples are given in Tables 3 and 4 above. Note that this work also shows that the special sequences of permutations discussed in this context in [8] and [ 121 do not exist for c = 5 and 6 but do exist for c = 7 and 8, although not all (2c - 1,4, c)-systems arise in this way (as is brought out in the examples given in Table 3).

[ I ] J. Abrham, Bounds for the sizes of components in perfect systems of difference sets, Chapter 1 of present volume, pp. 1-7. [2] J.-C. Bermond. A. E. Brouwer and A. Germa, Systemes de triplets et diNerences associees, in: Proc. Colloque C.N.R.S. Problemes combinatoires et theorie des graphes, Orsay (1976) pp. 35-38. [3] J.C. Bermond, A. Kotzigand J. Turgeon, On a combinatorial problem of antennas in radioastronomy, in: Proc. 18-th Hungarian Combinatorial Colloquium (North-Holland, Amsterdam, 1976) pp. 135-149. [4] F. Biraud, E. J. Blum and J. C. Ribes, On optimum synthetic linear arrays, IEEE Trans. Antennas Propagation 22 (1974) 108- 109. [5] F. Biraud, E. J. Blum and J. C. Ribes, Some new possibilities of optimum synthetic linear arrays for radioastronomy and astrophysics. Astronomy and Astrophysics 41 (1975) 409-41 3. [6] F. Biraud and D. G . Rogers, Some more perfect systems of diNerence sets, to appear. [7] A. Kotzig and P. J. Laufer, When are permutations additive? Amer. Math. Monthly 85 (1978) 364-365. [8] A. Kotzig and J. Turgeon, Perfect systems of difference sets and additive sequences of permutations, in: Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory, and Computing, Boca Raton 1979, Vol. 11, pp. 629-636. [9] A. Kotzig and J. Turgeon, Sur I'existence de petites composantes dans tout systeme parfait densembles de diNerences,in: Combinatorics 79, Proceedings of the Joint Canada-France Combinatorial Colloquium held in Montreal in June 1979, Annals of Discrete Mathematics No 8 (North-Holland, Amsterdam. 1980) pp. 71-75. [lo] P. J. Laufer and J. Turgeon, On a conjecture of Paul Erdos for perfect systems of difference sets, to appear in Discrete Mathematics. [ l l ] P. J. Laufer, Some new results on regular perfect systems. A.M.S. Abstracts (February 1980) p. 214. [I21 D. G . Rogers, A multiplication theorem for regular perfect systems of diNerence sets, to appear in Discrete Mathematics. [I31 D. G. Rogers, Addition theorems for perfect systems ofdiNerence sets, J. London Math. Soc. 23 (1981) 385-395. [I41 J. Turgeon, An upper bound for the length of additive sequences of permutations, Utilitas Mathematica 17 (1980) 189-196. Received I5 October 1980

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Annals of Discrete Mathematics 12 (1982) 203-209

@ North-Holland Publishing Company

A NOTE ON ONEFACTORIZATIONS HAVING A PRESCRIBED NUMBER OF EDGES IN COMMON C. C. LINDNER* Auburn University, Auburn, AL36849, U S A

W. D. WALLIS** University of Newcastle, Newcastle, NSW 2308, Australia

Dedicated to Professor A. Kotzig on the occasion of his sixtieth birthday 1. Introduction

A 1-factor of the finite set V is just a set of 2-element subsets of V which partition K (In what follows the 2-element subsets of V will be called edges.)A lfactorization is a pair (K F) where I/ is a finite set and F is a collection of 1-factorsof V which partition the set of all edges of K We will also say "F is a 1-factorizationof V".The number (VI is called the order of the 1-factorization (K F) and, of course, the spectrum for 1-factorizations is precisely the set of all even positive integers. In fact, in [ 3 ] it is shown that limu+mN(u)= co,where N(u) is the number of nonisomorphic l-factorizations of order u. Now let (K F) and (K G) be 1-factorizations where IV(=u and F={F1,F 2 , . . . , F u - l }and G = { G I , G 2 , .. . ,G u - l } . We will say that F and G have k edges in common if and only if k = CYr: IFinGi(. It is not difficult to see that a necessary condition for two 1-factorizations(K F) and (K G) of order u to have k edges in common is for

(3,

k E { O , 1,2,. . . , u = u ( u - ~ ) / ~ } \ { u - ~ , u - ~ , u - ~ , u - ~ } .

In what follows we will denote by J [ u ] the set of all k such that a pair of l-factorizations of order u having k edges in common exist and by Z [ U ] = { O , 1,2,. . . . , u = u ( u - ~ ) / ~ } \ { u - ~ , u - ~ , u - ~ , u - ~ } .

The object of this note is to give a complete solution to the intersection problem for 1-factorizations by showing that 4 2 1 = {I}, 4 4 1 = {O, 2,6}, J [ 6 ]= (0, 1, 2, 3, 5, 6, 7,9, 15}, J [ u ] = I [ u ] for all even u 2 8. *Research supported by NSF Grant MCS 80-03053and by a grant from the Internal Research Assessment Committee, University of Newcastle. **Research supported by an ARGC Grant. 203

C . C . Lindner, W. D. Wallis

204

As we shall see, the following result due to Webb [6] substantially reduces the amount of labor involved.

Tbeorem 1.1 (Webb[6]).If ~ - 1 > 7 a n d i s o d d a n d k ~ I [ u ] \ ( O , 1,2, . . . , u-2},then there exists a pair of idempotent commutatiue latin squares of order u - 1 agreeing in exactly k cells on or above the main diagonal. 0 The following connection between 1-factorizations and idempotent commutative quasigroups is well known. Let ( V * , be an idempotent commutative quasigroup of order u - 1 (u - 1 is necessarily odd of course) based on V* = { 1, 2, . . . , u - 1) and set V= (0) u V*. If we define 0 )

F i = {[0,i ] } u {[x, y ] 1 x y = y o x=i, x # y } 0

for all i E V*,

then F = {Fl, F 2 , . . . ,F u - l ) is a 1-factorization of I/: In view of Webb's Theorem, in order to show that J[u] = I[u] for all u > 8 we need show only that {0,1,2, . . . ,v - 2) c J [ u ] . The technique of proof is recursive and so we must begin by determining J[u] for some small values of u. In our case 'small' means u=2, 4, 6, 8 , 10, 12 or 14.

2. The sets J[u] for small u The proof of the first lemma is trivial and so it is stated without proof.

0

Lemma2.1. 5[2]={1} andJ[4]={0,2,6).

In what follows if F and G are any 1-factorizations of V= {O, 1, 2,. . . , u } we will denote by I(F, G) the number of edges they have in common, and if a is any permutation on { 1, 2,. . . ,u> by Fa the 1-factorization F (or G as the case may be) with the subscripts on the 1-factors renamed by a.

Lemma 2.2. 5[6]

= (0,

1, 2, 3, 5, 6, 7,9, 15).

Proof. Let F and G be the 1-factorizationsof {0, 1,2,3,4,5} given in Table 1. Then I ( F , H)= k where k and H are as shown in Table 2. Since any pair of 1-factorizationsof (0, 1,2,3,4,5}must be isomorphic to (F, G) or 0 (F, F), see [ 5 ] for example, it is easy to verify that no other case occurs.

Lemma 2.3. J [ 8 ] = I [ 8 ] .

F=

01 23 45

02 14 35

03 15 24

04 13 25

05 12 34

C=

01 23 45

02 15 34

03 14 25

04 12 35

05 13 24

A note on one-factorizations

205

Proof. By Webb's Theorem we need only show that {O, 1, 2, 3, 4, 5, 6) s J [ 8 ] . So, let F and G be the 1-factorizationsof {0, 1, 2, 3,4, 5, 6, 7) given in Table 3. Then I ( F , H)= k where k and H are as shown in Table 4. 0

Lemma 2.4. 5[10]=1[10]. Proof. Tables 5 and 6 are self-explanatory ( I ( F , H) = k).

0

Table 2

k

H=

0 1

2 3 5 6 7 9 15

F(12345) G(143X25) G(15X24) F(1234) G(24X35) F(123) G F(12) F

Table 3

F=

01 23 45 67

02 13

46 57

03 12 47 56

04 15 26 37

06 17 24 35

05 14 27 36

07 16 25 34

'=

01 24 36 57

02 14 37 56

03 16 27 45

04 12 35 61

05 17 26 34

06 13 25 47

Table 4

0 1 2 3 4 5 6

F(1234567) G(264753) G(1765432) G(276543) G(4756) G(1234567) G(67)

Table 5

Fi

FZ

F3

F4 F5

01 23 F = 48 57 69

02 17 34 59 68

03 04 05 06 16 18 14 13 28 27 29 25 45 39 38 49 79 56 67 78

F6

F7

07 15 24 36 89

Fs Fs 08 19 26 35 47

09 12 37 46 58

G=

GI

G2

G3

G4

G5

Gs

G7

Gs G9

01 26 34 58 79

02 18 37 45 69

03 17 29 48 56

04 13 28 59 67

05 16 24 39 78

06 14 27 35 89

07 19 25 38 46

08 I2 36 49 57

09 15

23 47 68

07 15 23 46

C . C . Lindner. W. D.Wallis

206

Table 6 k 0 1 2 3 4 5 6 7 8

H= G(162738495) G(17395X2846) G(17362845) G(1735)(2846) 6(1735)(29846) G(19)(28)(37)(46) 6(1829)(37)(46) G(69)(78) G(6789)

Table 7

F=

01 25 3T 49 58 67

02 13 45 5T 69 78

03 15 24 65 7T 89

05 19 28 31 46 TJ

04 17 26 35 85 9T

06 1J 2T 39 48 57

07 12 35 4T 59 68

08 14 23 5J 6T 79

09 16 25 34 75 8T

OT 27 36 45 9J

OJ 1T 29 38 47 56

OT 12 39 46 58 75

OJ 18 23 4T 57 69

18

Table 8

G=

01 29 34 5J 68 7T

02 16 3T 45 79 85

03 19 27 45 56 8T

04 15 2T 38 67 9J

06 14 25 37 5T 89

05 IT 26 3J 49 78

07 13 25 48 65 9T

Table 9

k 0 1 2 3 4 5 6 7 8 9 10

H=

F(123456789TJ) G(lJT98765432) G(3456789TJ) G(456789TJ) G(586TJ97) G(67J978) G(7T8J9) G(789TJ) G(9TJ) G(789J) G(TJ)

08 17 24 36 59 TJ

09 1J 28 35 47 6T

A note on one-$actorizations

207

Lemma 2.5.J[12]=1[12].

0

Proof. Tables 7 , 8 and 9 are self-explanatory (T= 10 and J = 1 1;I(F, H)=k). Lemma 2.6.J[14]=1[14].

Proof. Tables 10, 1 I and 12 are self-explanatory (T=10, J = I 1, Q = 12, and K = 13 ; I(F, H)=k). Table 10

02 15 36 4T 79 8K JQ

03 19 26 47 5J 8T QK

04 1K 2T 37 58 6Q 9J

05 12 35 48 69 7K TQ

06 18 23 4Q 59 7T JK

07 lQ 29 34 5K 6T 85

GI

G2

G3

G4

GS

G6

G7

01 25 38 45 6Q 79 TK

02 1J 3Q 49 56 7K 8T

03 18 2Q 4K 5T 67 9J

04 15 29 3K 63 78 TQ

05 14 26 3T 7Q 89 JK

06 IQ 25 37 45

07 19 2K 36 48 5Q TJ

01 25 39 F = 4K 68 7Q TJ

08 16 2K 3T 45 75 9Q

09 13 27 45 56 8Q T K

OT 1J 24 38 SQ 67 9K

0J IT 2Q 35 49 6K 78

OQ 17 2J 3K 46 5T 89

OK

GQ

GK

14 28 3Q 57 65 9T

Table 11 ~

H=

8K

9T

Gr

08 13 2T 47 59 6K

JQ

Table 12 k

H=

0 I 2 3 4 5 6 7 8 9 10 11 12

F( 123456789TJQK) G(1KT74X2Q963J85)

G( lQTJ98765432) G(lQKJT98765432) G( 1T98765432) C( 198765432) C( 18765432) G(lQT859765432) G(9KQJT) WQJK) G(JKQ) WQWJK) G(QK)

09

17 24 35 58 6T QK

OT 1K 28 35 4Q 69 75

OJ 12 39 46 5K 7T 8Q

OQ 16 23 4T 57 85 9K

-

OK 1T 27 34 5J 68 9Q

C. C. Lindner, W.D.Waiiis

208

3. ”be main theorem Now that we know J[u] for u=2,4,6,8, 10, 12 and 14 the proof of the following theorem goes quickly.

Theorem3.1. J[2]={1),3[4]={0, I[u] for u >, 8.

2,6),J[6]={0, 1,2, 3, 5,6,7, 8, 15},andJ[u]=

Proof. In view of the results in Section 2 we need only show J[u] = l[u] for u >, 16. Now we can always write’u = 2u or 2u + 2 where u is even. In [13 Allan Cruse has shown that if n and m are odd positive integers and n>,2m+ 1 then there exists an idempotent commutative quasigroup of order n containing a subquasigroup of order m. This translates into 1-factorizations as: if x and y are even positive integers and x 3 2y, then there exists a 1-factorization of order x containing a sub-1-factorization of order y. Now let F) be a 1-factorization of order u containing the sub-l-factorization (V*, F*) of order u, where u=2u or 2u+ 2. We can assume F = I F , , F z , . . . , F,-,) and F * = { F : , F: ,..., F : - l } where F t s F i f o r i = l , 2 . . . , u-1. By Webb’s Theorem weneedonlyshowthat {0,1,2,.. . , u-2) t J C u ] . Nowsinceu>,8, {0,1,2,..., u-2) GJ[u]and so let k E (0, 1, 2,. . . ,u -2) and ( V * , A*) and (V*, B*) be a pair of 1-factorizations intersecting in exactly k edges. Further, let (V, A) and (K B) be the 1-factorizations obtained from (K F) by unplugging (V*, F*) and successively replacing it with (V*, A*) and ( V * , B*). Finally, let C = {Cl, Cz,. . . ,C , - 1 } be the 1-factorization of V defined by: (i) Ci=(Fi,\FZ)uBi, i = 1, 2,. . . , u - 1, where a is any derangement on {1,2, . . . , u-1}andB*={B1,B2 , . . . , B,-,);and (ii) C j = F j p , j = u, u+ 1, . . . ,u - 1, where is any derangement on {u, u 1,. . . ,u - 1). It is immediate that (r!A) and ( K C) have exactly k edges in common, completing the proof.

(v

+

4. Remarks Apart from being of interest in itself,Theorem 3.1 has some quite useful applications one of which is the construction of pairs Steiner triple systems having a prescribed number of triples in common. In [4] C. C. Lindner and A. Rosa gave a complete solution to this problem. Theorem 3.1 can be used to give a very simple direct solution to this problem for half of the spectrum for Steiner triple systems; namely, all Steiner triple systems of order u= 2u+ 1 3 15 where u is the order of a Steiner triple system. This is achieved via the use of the standard ‘u to 20 + 1’ construction for Steiner triple systems (see [4] for example) and the well-known fact that a pair of disjoint Steiner triple systems of order u exist for every 0 2 7 [2] as follows: If k E (0,1, 2,. . . ,u(u - 1)/6} use the u to 2u + 1 construction with a pair of disjoint Steiner triple systems of order u and a pair of 1-factorizationsof order u 1 intersecting in k edges. If

+

A note on one-factorizations

209

use the u to 2u+ 1 construction with the same Steiner triple system and a pair of 1factorizations of order u 1 intersecting in k - u(u - 1)/6 edges.

+

References [ 11 A. Cruse, On embedding incomplete symmetric latin squares, J. Combin. Theory Ser. A (1974) 18-22. [2] J. Doyen, Constructions of disjoint Steiner triple systems, Proc. Amer. Math. SOC.32 (1972) 409-416. [3] C. C. Lindner, E. Mendelsohn and A. Rosa, On the number of I-factorizations of the complete graph, J. Comb. Theory Ser. B20 (1976)265-282. [4] C. C. Lindner and A. Rosa, Steiner triple systems having a prescribed number of triples in common, Canad. J. Math. 27 (1975) 1166-1175. [5] W. D. Wallis, Room squares of side five, Delta 4 (1973) 32-36. [6] T. M. Webb, On idempotent commutative latin squares with a prescribed number of common entries, Ph.D. Thesis, Auburn University (1980). Received 3 September 1980.

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Annals of Discrete Mathematics 12 (1982) 21 1-221

@ North-Holland Publishing Company

SYSTEMES DE TRANSITION ET COLORIAGES EULERIENS Tewfik LOBBOS Universitd de MontrPal, Canada

En hommage au Professeur A. Kotzig a l’occasion de son soixantieme anniversaire Dans cet article, nous caracteriserons les decompositions rotatives sur un groupe aMlien d’ordre impair et nous en deduirons, entre autres choses, des formules permettant de calculer le nombre et la longueur des promenades determinees par ces decompositions. Les rtsultats preckdents seront utilids pour une nouvelle contribution a la solution du probleme des coloriages eulbriens (ou, de faqon bquivalente, des P-quasigroups eulbriens).

1. Introductioo

Tous les graphes que nous considererons sont finis, non-orientbs, simples et sans boucles. Un facteur quasi-liniaire d’un graphe est un facteur dans lequel un des sommets est isole et tous les autres sont de degrC un. Le graphe complet K , admet une dtcomposition 9 en facteurs quasi-lineaires si, et seulement si, m est impair. Soit m = 2 n + 1. Alors 9est constitub de 2n+ 1 facteurs et les sommets isolbs de facteurs distincts sont diffkrents. Si x est le sommet isole d u n facteur de 9, on denote ce facteur par 4x. Ainsi 9= {4x1 x est un sommet de K,} (reprtsentation canonique de 9).Rappelons que dans unfacteur lintaire tous les sommets sont de degre un; K , admet une decomposition en facteurs lineaires(1-factorisation)si, et seulementsi, m est pair. Une decomposition de K , en facteurs soit linkaires, soit quasi-linbaires, est un coloriage de K , (lesfacteurs sont appelks couleurs).Deux coloriages 9et 9‘ de K , sont isomorphes s’il existe une bijection de l’ensemble des sommets de K , transformant les argtes de mcme couleur dans 9 en des argtes de m6me couleur dans 3”.Pour memoire, mentionnons qu’a partir d’un coloriage de K z n +1, on peut dtfinir un coloriage de K 2 n + 2en adjoignant un nouveau sommet a ceux de K2,,+et en le liant a x dans 4x Une promenade P (closed line, closed walk) de longueur k de K , est une suite alternbe de sommets et d’ar6tes XO, A09 XI, Al,

. . . xk-1, 9

Ak-1,

xk

telle que A iest une ar6te de K , d’extrtmitts xiet xi+ (OG i < k), A i # A j (pour i # j ) et xo = xk. On ecrit P = (xo,xl,. . . ,xk). Les promenades obtenues de P par une translation de ses sommets ou par une inversion de leur ordre sont considtrees comme egales a P . Une partition de K , en promenades [6] est un ensemble 9 de promenades de K , tel que chaque argte de K , appartient a exactement une promenade de 9; *Le contenu de cet article est extrait de la these de doctorat [8] redigee par I’auteur sous la direction du Professeur A. Kotzig. 21 1

212

7: Lobbos

nous imposerons aussi a 9 la condition suivante [4]: quels que soient les sommets x, y (xfy), il existe un unique sommet z tel que x, z, y apparaissent successivement dans une promenade de 9 (nous dirons que [x, z, y ] est une transition de 9). K, admet une partition en promenades si, et seulement si, m est impair. La classe des coloriages de IS2,,+ et la classe des partitions de K 2 n + len promenades sont en correspondance biunivoque [l, 61: {x, y } ~ & e [ x , z, y ] est une transition de 8.On voit aistment que deux coloriages isomorphes engendrent respectivement le mtme nombre de promenades de mgme longueur. Les ensembles de sommets que nous considtrerons sont munis dune structure de groupe. Soit donc (G, + ) un groupe. Posons G* = G\ (0).Soient X une partie de G, Z un ensemble de parties de G et y un eltment de G. Les translations de X et de Z par y sontdtfiniesparX+y={x+y(x ~ X ) , E + y = { z + y l ZEE}. Legraphecomplet sur G est Klcl,les sommets de ce graphe ttant les elements de G. Soit G un groupe abklien dordre impair. Soit L un facteur quasi-lintaire du graphe complet sur G et soit D I'ensemble des diffkrences entre les deux extrtmitts de chaque artte de L. Si D = G*, L est unfacteur parfait (sthrter) de G. Les translations d u n facteur parfait constituent un coloriage du graphe complet sur G appelt dkcomposition rotative sur G. On voit aistment que fx, Y } € 42

* { x f u , Y + u> E

#Z+"

(pour x, Y , z, u E GI.

On vtrifie immtdiatement que { { -x, x) I x E G * ) est un facteur parfait de G. C'est lefacteur trivial (patterned starter). La dtcomposition rotative correspondante est aussi appelte trioiale.

2. Systdmes de transition La construction suivante gtntralise un critkre dtmontre par Kotzig et Turgeon [7]. En voici dabord un exemple. Sur la Fig. 1 les lignes continues constituent un facteur parfait I. de ZI3, dans lequel0 est isolt. La figure comporte deux cycles. Suivons l'un dentre eux en partant d u n sommet x quelconque dans la direction de - x, et notons un sommet sur deux. Recommenqons avec I'autre cycle. Nous obtenons (par exemple): (1 4 2) (3 6 - 5). On constate que les valeurs absolues des nombres obtenus cornprennent chacun des 0

-4

5

-----I :: Fig. 1.

Systtmes de transition et coloriages eulkriens

213

nombres 1,. . . , 6 exactement une fois. Les sommes (modulo 13) de deux termes successifs (cycliquement)ont la mime proprikte: (1 4 2 3 , (3 - 4 6 - 5 '-* ). Nous allons dabord montrer que ces suites (ou cycles) caractkrisent le facteur L. Soit G un groupe abClien dordre 2n+ 1. Soit I-= {rl,. . . , r,}un ensemble de suites d'tltments de G* :

rv=(&)u(;I)-.u&)), pour v = l , .

. . , t.

Posons I:v= (&) + up

... u g - 1 +u q

+

+

ug) u p )

et I:={ X I , . . . ,I:r).On appellera les rvdes cycles de transition de G ([7]: transition sequence) et r un systime (de cycles) de transition de G si, pour tout x E G*, I'un de x ou - x apparait exactement une fois dans et l'autre pas du tout, et I'un de x ou -x apparait exactement une fois dans I: et l'autre pas du tout. En d'autres termes,

l{x, -x) n

rv= {x, - x > n L) I:, = 1.

Dans rv,les operations sur les indices se font modulo k v + 1. Si l'on remplace I'un des r vpar un cycle obtenu de r, par une translation des indices, le nouvel ensemble de cycles est encore un systeme de transition. I1 en est de mgme si on remplace r vpar son opposk: (

-uc) -&-

. . . - u p - u(0y)).

Par abus de langage, nous considererons qu'un cycle de transition r,,est Cgal aux cycles obtenus soit de r vsoit de son oppose par translation des indices. Donc (a b . . . c d ) = ( -d

-C

*

*

- b -a)=(d u b . . . c).

(Nous definissons ainsi une relation &equivalencesur I'ensemble des cycles de transition de G). Par contre, les cycles (a b . . . c) et (c . . . b a ) seront considtres comme inkgaux. Comme n elkments de G apparaissent dans r, on a C (kv+l)= kv+t=n. isv
1QVQt

Thhr&ne 1. Les dkcompositions rotatives sur un groupe abklien G dordre impair et ses systimes de transition sont en correspondance biunivoque. Dhwstratioa. Etant donne un systeme de transition

={

rl,. . . , r,}de G , on pose

L = { { - - u l " ) , ~ l IO l
I: Lobbos

214

Rtciproquement, soit 9 une decomposition rotative sur G. Pour x E G*, soit a ( x ) le sommet adjacent a x dans c$o (alors a ( x ) E G* et a a ( x ) = x ) . Pour chaque x E G*, dkfinissons une suite d'klbments de G* de la faqon suivante: xo = x, xifl=a(-xi)

pour i>O.

La suite est cyclique: soit K le plus petit entier i pour lequel il existe j=j(i) telquejO), donc k,= k, et Ryest obtenu de R, par translation des indices modulo k, 1. De plus, si I'un des elkments de R, est I'inverse (pour I'opkration de groupe) de Pun des elements de 0,. alors R, R R,. En effet,

+

xi= - y j * x i + , = a ( - X i ) = a ( y j ) = a a ( - y j - l ) =

-yj-1,

donc xI= - y i + j - l

(1)

(pour 1 2 0 )

et R, est obtenu de I'oppose de 0, par translation des indices. I1 decode de ce qui predde que si un element x de G* apparait dans un cycle R, de A, alors ni x , ni - x n'apparaissent dans un cycle de A different de R,. Montrons maintenant que x et - x n'apparaissent pas tous deux dans R,. Pour cela, supposons que x = y i = - y j . Si i + j est pair, alors y(i+j)/2= - y i + j - ( i + j ) / ~ = - y ( i + j ) / ~par (I), done y ( i + j ) / ~ = O , ce qui est impossible.Si i + j est impair, alors y(i+j- 1112 = - yi+ j - ( i + j - 1)/2 = -Y(i+ j + 1 ) / 2 par (1). Posons I = ( i + j - 1)/2. Alors yl yl + = 0 et yI a( -y I )= 0, ce qui est impossible puisque yl+ a( - y l ) est la difference entre les deux extremitks de I'arete { - y l , a( - y J ] de c$o. Donc, pour tout x E G*, l'un de x ou --x apparaft exactement une fois dans A et l'autre pas du tout. Montrons que A satisfait a la seconde condition de la dtfinition d u n systeme de transition. Soient x , y E G* tels que R,, 0, E A, et supposons que X ~ + X ~ + ~f = ( ~ ~ + y ~ +Alors ~ ) . x i + a ( - x i ) = k ( y j + a ( - y j ) ) . Or, { - x i , a ( - x i ) > , { - y j , a( - y j ) } E c$o et les differences entre les extremitks de chacune de ces aretes sont respectivement tgales a x i +a( - x i ) et y j a( - y,). Donc { - x i , a( - x i ) ) = { - y j , a( - y j ) } . Si - x i = - y j , alors x = y et i=j, puisque R,, R, €12. Si - x i = a ( - y , ) , alors - X ~ = Y ~ +ce~ qui , est impossible. Donc A est un systeme de transition de G. Par abus de langage, nous identifierons les cycles equivalents modulo R. I1 reste a montrer que la correspondance est biunivoque. Soit un systeme de transition de G. Si x=uIy) et y=uIy' 1, alors { - x , y } EL, donc y =a( - x ) et Tv= (uIy)uIy!l...)=(xy...)=R,EA . Par consequent, r r A et donc T = A . Soit Y une decomposition rotative sur G et soit { x , y } E c$o. Alors y = a ( x ) et on obtient successivement ( - xy . . .) E A, { x , y } E L, c$o C L et c$o = L, ce qui termine la demonstration.

+

+

+

Systimes de transition et coloriages euliriens

215

'lWorhe2. Soient 9 une decomposition rotative sur un groupe abelien G d'ordre 2n+1, r son systime de transition et 9 la partition en promenades induite par 9; pour v = 1,. . . , t , soient A,= MI") et m, Pordre de Av duns G. Le cycle r, entoutes de longueur m,(k,+ l), et chaque promenade gendre (2n+ l)/m, promenades de 8, de 9 est engendrte par Pun des cycles de r.

xO,
Demonstrath. Si x [x, x

+

E G,

MIV), x

1G v G t et 0 G i G k,, alors

+ + u!? UIV)

11

est une transition de 9; en effet, { - U!"),u!") j E 40,donc { - u!'), u!")1} + ( x + u!')) E 4x+uy~, c'est-a-dire { x , X + U ! ' ) + U ~ " ) ~ } E 4x+u!~, . Reciproquement, toute transition [a b cl de 9 est de la forme (2), car {a, c ) E donc {a- b, c - b} E 40,d'oG ( - a b c -b . . .) E r; la conclusion decoule du fait que [a, b, c ] = [ a , a + ( - a + b ) , a + ( - a + b ) + ( c - b ) ] . T vdefinit (2n l)(kv 1) transitions de la forme (2) (choix de x E G et de O G i< k,), donc r definit ainsi

+

+

C

+

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

(k,+ 1)=(2n+ l)n 1<

l$VhI

V d I

transitions, qui est prkcisement le nombre des transitions de 9. Ces dernikres titant toutes de la forme (2), les transitions (2) sont toutes differentes. On vtrifie aisement que les promenades de 9 sont de la forme P = ( X , X " ~ ~ ' , X + U ~ ' ' + U, '. ~. .. ,) x + A , , x + A , + u b y ) , . . . , ' X+

Ah,+ uby)+ . . . +MI'")

ou x E G, 1 < v < t , O < f < k v et ou ilest un entier; la longueur de Pest kgale a A(k,+ 1) I 1. Montrons que I = k,. Notons d'abord que x AAv uby)+ . . . uj")= x , donc x LA, uby) * * . ul? = x - uIv) et [ x - uj'), x , x uby'] est une transition (car ce sont trois sommets successifs de P);on en deduit: { x - uj'), x uby)} E &, { - uj'), u$"} E 40, u$'= a( -ul")= ul!) 1, I 1 z0 (mod k v + 1) et I= k,. La longueur de P est donc 6gale a l ( k , + l)+(k,+ l)=(A+ lXk,+ l), c'est-a-dire il'(k,+ l), ou ,?'=A+ 1. De plus, 1'est le plus petit entier positif K tel que K A ,=0, autrement dit, 1'=m,. Donc r, engendre des promenades de longueur mv(k,+ 1). Puisque chacune de ces promenades utilise rn,(kv+ 1) transitions, le nombre des promenades engendrtes par r, est 6gal a (2n lXk, + l)/mv(kv I)= (2n + l)/mv.Cela termine la demonstration.

++ + + +

+ +

+

+

+

+ +

+

+

+

En particulier, si G est un groupe cyclique, alors r, engendre pgcd(2n 1, A,) promenades de longueur (2n+ l)(k,+ l)/pgcd(2n+ 1, A,).

Exemple. Le facteur parfait de h9 illustrk sur la Fig. 2 determine les deux cycles de

transition rl =(3) et Tz= (1 4 - 2). Alors A l = 3 = A2 et ml = 3 = mz. Le cycle rl engendre trois promenades de longueur 3 : (0 3 6 0), (1 4 7 l), (2 5 8 2).

Le cycle Tzengendre trois promenades de longueur 9 : (0153486720), (1264507831), (2375618042).

7: Lobbos

216

Proposition. Soit Y la dtcomposition triviale de G et soit A un sous-ensemble maximal de G* ne contenant pas deux tltments de la forme a et -a resp. Alors l- = { ( x )1 x E A ) . Rtciproquement, si r est un systime de transition dont tous les cycles sont de longueur un, alors 9 est la dtcomposition triviale. (Immediat). Les promenades engendrkes par la decomposition triviale sont des cycles. Si G =, ,Z ou p est premier et c( est un entier positif, alors I’ordre des 616ments non-nuls de G est p , et r engendre pa- ‘ ( p a- 1)/2 cycles de longueur p .

TMoreme 3. Deux groupes abtliens G, G’ d’ordre impair sont isomorphes si, et seulement si, leurs decompositions triviales sont isomorphes. D6monstration. La condition est tvidemment necessaire. Pour montrer qu’elle est suffisante notons d’abord que deux groupes abkliens finis sont isomorphes si, et seulement si, ils contiennent le mime nombre d’ekments de mkme ordre [2, p. 821. Notons aussi que I’ordre de x est egal a l’ordre de - x . Supposons que G G‘. Soient

+

r = { ( x )I x E A }

I

et r‘= {(x’) X‘ E A’)

les systbmes de transition triviaux de G et G‘ resp. I1 existe un entier positif 8 tel que A (resp. A’) contient r (resp. r’) elements d’ordre 8, et r f r ’ . Chacun de ces elements engendre (2n+ l)/8 promenades de longueur 8, et les autres elements de A (resp. A’) engendrent des promenades de longueurs diffkrentes de 8. Donc r (resp. r‘)determine r(2n+ l)/8 (resp. r’(2n+ l)/8) promenades de longueur 8, et r(2n + l)/8 # r’(2n+ 1)/8.

3. Coloriageseuleriens Un coloriage de Kz,+ est eultrien s’il induit une partition de K 2 ” + se reduisant a une seule promenade (promenade eultrienne). Kotzig [ 6 ] a propose le probleme suivant: trouver les valeurs de k pour lesquelles il existe un coloriage eulerien de Kk.I1 a note qu’un tel coloriage existe lorsque k = 3 ou 7 (et qu’il est alors unique, a isomorphisme pres), mais qu’il n’en existe pas pour k = 5. Par la suite, Keedwell [5] a demontrk l’existence d’un coloriage eulerien d’ordre k pour une infinite de valeurs de k de la forme 4r + 3. Hilton et Keedwell[3] ont ensuite ttendu ce resultat a d’autres valeurs de k de la forme 4r + 3 et aussi de la forme 4r 1. Kotzig et Turgeon [7] ont etabli un critere permettant de verifier si une decompo-

+

SystPmes de transition er coloriages eulkriens

217

sition rotative sur Ez,+ est eulerienne (il en a deja ete question au debut de la Section 2). (Kotzig-Turgeon). Soit 8 une dkcomposition rotative sur H2"+1. Alors Y est eulkrienne si, et seulemerrt si, le systime de transition correspondant se rkduit a un seul cycle rl et ml=2n+ 1. -me

Ce resultat dtcoule immtdiatement du Thtoreme 2. Ajoutons qu'une dtcomposition rotative sur un groupe non-cyclique n'est pas eulerienne. Dans le meme article, Kotzig et Turgeon dtfinissent une serie de decompositions rotatives sur Z2, + ( n> 3) et dkmontrent que ces decompositions sont eultriennes si, et seulement si, 2n + 1 $0 (mod 7). Compte tenu des rksultats obtenus par Hilton, Keedwell, Kotzig et Turgeon, les valeurs de k=2n+ 1 pour lesquelles l'existence d u n coloriage eulerien reste a demontrer sont: k ~ 2 (mod 1 28) et k-511 (mod 2380). Kotzig a conjecture que la serie illustre sur la Fig. 3 reglerait les cas k-35 et 49 (mod 84). O n peut dkmontrer que les decompositions de cette strie sont eultriennes dam les quatre cas suivants: k-l(mod12) et kfl3(mod156), k-7 (mod 12) et k f 7 (mod 841, k E 9 (mod 12), k $21 (mod 36) et k f 2 1 (mod 84), k = 11 (mod 12).

Nous nous contenterons dktudier ici les deux seuls cas qui contribuent a la solution du probleme &existence des coloriages eultriens, a savoir k-1 ou 11 (mod 12). Dans ces deux cas, r est constitue d'un seul cycle.

n=3

54 3 Ti n=5

3 n.4

n=6

n=7

Fig. 3.

218

7: Lobbos

Cas I . Pour 2n+ 1 = 1 (mod 12) (n=O (mod 6)), posons:

1

si 1
Alors r comprend les elements suivants: 1, 2, . . . , n/2, (n+6)/2, (n+ 12)/2, . . . , n, -(n-2), -(n-5), . . . , -(n+2)/2, (n + 4)/2, (n+10)/2, (n+16)/2, . . . , n-1. Sommes successives ui + ui + 1 : 3,5, . . . , n - l -(n-2) -(n-8), -(n-14), 2

. . . , -4

8, 14, . . . , n-4 1

-(n-6) -(n-12), n On peut voir que

1
+

+

-(n-18),

. . . , -6

i=n.

r est un systeme de transition de Z2"+

Cas I I . Pour 2n+ 1 = 11 (mod 12) (n = 5 (mod 6)), posons:

ui=

I

i

3i-n+l 3i-3n (n + 3)/2 3i-2n-1

si l
+

Alors r comprend les tlements suivants: 1, 2, . . . , (n-1)/2, (n+5), (n+11)/2, . . . , n, -(n-2, -(n-5), . . . , -(n+ 1)/2, (n + 3)/2, (n+9)/2, (n+15)/2, . . . , n-1.

219

SystPmes de transition et coloriages euliriens

Sommes successives ui + ui+ : 3, 5, . . . n-2 -(n - 1) -(n-7), -(n-13), 2 8, 14, . . . , n-3 )

. . . , -4

1 -(n - 5) -(n- ll), -(n-l7), n

1 < i < ( n - 1)/2, i = (n - 1)/2, (n+1)/2
+

. . . , -6

r est un systeme de transition de ZZn+1. TMoreme4. Tous les systimes de transition dbjlnis dam le cas I1 engendrent des promenades eulkriennes.Les systimes du cas I en engendrent si, et seulement si, 2n 1 $0 (mod 13).

+

Dt5monstration. Dans le cas I, on a 2n + 1 = 1 (mod 12) et

1

A=

i+

15 iS n / 2

1

( n + 2)/2 < i < 2 4 3

+(n+4)/2+ -[lnn+2] - 2 2 2 +

+

1

( 5 n + 1 2 ) / 6 S i Sn

[:(:

2n:3

(3i-3n-1)

(Zn+3)/3
(3i-2n-1)

-(:

- --__

-

n +2 2

[:(5,._ - _ _ - )( -

1

(3i-a)+

+I

2-

2n+3, 5n 3 6

2n+3)-(5+1 3

- _2n+3+ __

n+2 2

~

+ 1) n]

l)n]

3

=in(, + 2)+&n(7n + 6) - in2 +$n(3n+ 2) -in' +$n+4)+&(n-6)(11n+12)

+

=&(3n2 + 6n 7n2+ 6n -4n2 +9n2 +6n -4n2

+ 12n +48 + 1InZ- 54n -72)

= 3 2 2 n 2 -24n -24)=&(l InZ- 12n - 12).

Donc 12A = 1InZ- 12n - 12 = 1l n 2 + 12n = n( 1In + 12)G n(n + 7) zz - n(n -6). nombres 2, 3 et n Ctant relativement premiers avec 2n + 1, on a

Les

pgcd(A, 2n+ l)=pgcd(12A, 2n+ l) = p g c d (~ -6 , 2 n + 1). Posons d =pgcd(n-6, 2n+ 1). Alors il existe des entiers a et b tels que n -6=ad et 2n+l=bd. Donc n=ad+6, bd=2n+1=2(ad+6)+1=2ad+13, d ( b - 2 ~ ) = 1 3 et 2n+ l $ O (mod 13).

220

7: Lobbos

Dans le cas 11, on a 2n + 1 = 11 (mod 12) et

+

c

(3i-3n)+(n+3)/2+

( 2 n + 2 ) / 3 < i < ( 5 n - 1)/6

1

( 5 n + 1 1)/6< i b n

(3i-2n-1)

- ( T2n-1 - T + l ) ( n -nl +) ]l

-(T

5n-1

-

2n+2 -+ 3

5n+11 6 =3n-l)(n+l)+&(n+1)(7n+l)-3n+l)(n-l)

+3n +

+ -3n +

+

+

U(3n 1) l)n+$n 3 ) + 3 n - 5)(n 1) - 2 ,1 [(n+~)(3n-3+7n+1-4n+4+9n+3-12n+11n-55)+12n+36] =&[(n

+ 1X14n -50)+

12n + 361 = 3 7 n 2 - 12n -7).

Donc 12A= 7n2 - 12n -7 -7n2 +2n =n(7n +2) -n(n - 1). Les nombres 2, 3 et n etant relativement premiers avec 2n + 1, on a

+

pgcd(A, 2n + 1)= pgcd( 12A, 2n 1)= pgcd(n - 1,2n + 1). Posons d = pgcd(n - 1, 2n + 1). Alors il existe des entiers a et b tels que n - 1 =ad et 2n+ 1 = bd. Donc n=ad+ 1, bd=2(ad+ 1)+ 1=2ad+3, d(b -2a)=3. Donc d = 1 ou 3. Mais pgcd(3,2n+ 1)= 1, doh d = 1 et r engendre une promenade eultrienne. Les valeurs de k pour lesquelles on ne connaissait pas de coloriage eulerien sont k = 2 1 (mod 28) et k=511 (mod 2380). Autrement dit, k = 2 1 , 4 9 , 7 7 (mod 84) et k=511,2891,5271 (mod 7140). Le thkoreme prCcCdent regle le cas ou

k =49 (mod 84) et k $0 (mod 13) et la cas oh kG2891 (mod 7140). La question est donc ouverte dans les cinq cas suivants:

Systemes de transition et coloriages eulkriens

22 1

k =21,77 (mod 84), k -637 (mod 10921, k=511,5271 (mod 7140). Le cycle de transition suivant engendre une promenade eultrienne de K 2 : - 3 5 -9). La plus petite valeur de k pour laquelle la question est encore ouverte est donc 77. ( - 1 2 - 8 -4 7 6 10

BiMiographie [l] J. Ddnes et A. D. Keedwell, Latin Squares and their Applications (Academic Press, New York and London, 1974). [2] L. Fuchs, Infinite Abelian Groups, Vol. 1 (Academic Press, New York, 1970). [3] A. J. W. Hilton et A. D. Keedwell, Further results concerning P-quasigroups and complete graph decompositions, Discrete Math. 14 (1976) 311-318. [4] A. D. Keedwell, Some connections between latin squares and graphs, in: Atti del Colloquio Internazionale sulle Teorie Combinatorie, Roma, 1973 (Academia Nazionale dei Lincei, Roma, 1976) pp. 321-329. [5] A. D. Keedwell, Row-complete squares and a problem of A. Kotzig concerning P-quasigroups and Eulerian circuits, J. Combinatorial Theory Ser. A 18 (1975) 291-304. [6] A. Kotzig, Groupoids and partitions of complete graphs, in: Combinatorial Structures and their Applications, Proc. Calgary Conf. 1969 (Gordon and Breach, New York, 1970)pp. 215-221. [7] A. Kotzig et J. Turgeon, Quasigroups defining Eulerian paths in complete graphs, J. Combinatorial Theory Ser. B, a paraitre. [8] T. Lobbos, Sur les coloriages des ardtes des graphes complets, These de doctorat, Universite de Montreal (1978). Received 25 April 1980; revised 12 February 1981

This Page Intentionally Left Blank

Annals of Discrete Mathematics 12 (1982) 223-238

@ North-Holland Publishing Company

REGULAR AND STRONGLY REGULAR SELFCOMPLEMENTARY GRAPHS

I. G. ROSENBERG C.R.M.A., Universite de Montreal, Canudu

Dedicated to Professor A. Kotzig on the occasion of his sixtieth birthday The graphs given in the title are studied using boolean techniques. First we investigate the change of certain simple graphical parameters (like the number of triangles through a vertex) of a regular self-complementary graphs under a complementing permutation. Next we describe the selfcomplementary graphs on 4k + 1 vertices belonging to a given permutation in terms of certain zeroone parameters. Such regular graphs are determined by natural integer parameters satisfying certain inequalities. Finally strongly regular graphs are considered. For a permutation with a single cycle of length 4k the corresponding strongly regular graphs are determined by a system of 2k2 quadratic equations in 2k2 + k zero-one unknowns.

0

The selfcomplementary graphs (sc-graphs) present an interesting but still quite large and varied family of graphs. The regular sc-graphs (rsc-graphs) possess more specific and elegant properties than sc-graphs and the strongly regular sc-graphs (srsc-graphs) form an intriguing and somehow elusive class of very special graphs. In [ll] A. Kotzig proposed to study rsc-graphs and srsc-graphs and formulated several problems. The purpose of this paper is the description and enumeration of such graphs of small size with a given complementary permutation. This is achieved mainly by boolean techniques (for references see [8]) which seem to be well adapted for this task and whose application in graph theory is of some interest on its own. Thus, for example, the tedious case by case search for srsc-graphs is replaced by routine handling of zero-one equations. We start from the basic and natural ideas that appear throughout the literaturesome of then even in Sachs [18]-but often for our purposes we put them in more definite and explicit form. First, following an idea of A. Kotzig, we investigate the change of certain simple parameters (like the number of joint neighbours of two vertices, the number of triangles through a vertex, etc.) of an rsc-graph under complementation. Results of this type could simplify the search for a complementing permutation of a regular or strongly regular graph. Here we apply simple boolean techniques to the zero-one entries of the adjacencency matrix. From this point on all results refer to sc-graphs on 4k + 1 vertices with a fixed complementing permutation. Without loss of generality we assume that besides a single fixed point the cycles C iof the permutation have length 4di ( i = 1, . . . ,n) where hidivides di + for i = 1, . . . , n - 1. For later use we reduce the highly redundant aggregate of zero-one entries of the adjacency matrix to an independent set of zero-one parameters describing uniquely 223

224

I . G . Rosenberg

every sc-graph. These binary parameters lead naturally to ‘even’ and ‘odd integer parameters eijand oij (1 < i < j < n). The degrees of vertices are then expressed in terms of these parameters and 6 . . . ,6, yielding thus a description of degree sequences of sc-graphs with a given complementing permutation. In particular, we obtain that an sc-graph is regular if and only if the parameters are integer solutions of a simple linear inequality system. For such parameters the number of the corresponding (labelled) graphs is given. In particular, it is shown that every permutation of our kind is a complementing permutation of quite a few rsc-graphs. For permutations with at most two cycles we can list all parameters but otherwise the situation seems to be quite complex, even in the extreme case of 6 = . . . = 6, = 1 (all cycles of length 4) and we present only an inequality for the parameters. These results bring forth an interesting connection between the enumeration problem and the problem of describing the integer solution set of the corresponding inequality system. Strongly regular graphs arise naturally in finite geometries, block designs, group theory, etc. (see e.g. [lo, 141).The graph determined by quadratic residues (called also Paley graph) is not only one of the best examples of strongly regular graphs but also an sc-graph. The strongly regular sc-graphs, srsc-graphs for short, are rather elusive for it is known that on less than 37 vertices all srsc-graphs are isomorphic to the Paley graph and examples of other graphs are known only for 37, 41 and 49 vertices [l]. In [ l l ] Kotzig suggests to study the srsc-graphs for their own sake. Applying the above boolean techniques we express strong regularity in terms of a system of 0-1 quadratic equations whose solutions determine the labelled srsc-graphs. The main drawback of this approach is the fact that we must analyze separately the cases corresponding to different cycle structures of c. For simplicity we discuss only the two extreme cases. For permutations with a single cycle half of the conditions holds in every rscgraph and the graph is determined by a system of k + 1 quadratic equations in 2k + 1 zero-one unknowns. As an application without much trouble the complete solution is given (in a parametric form) for srsc-graphs on 9, 13 and 17 vertices. The other extreme case is a permutation with k four-cycles. Here the strong regularity is described by 2k2 quadratic equations in 2k2 + k zero-one unknowns. We express the restriction on v,, in terms of the parameters eij and v i j for 3k special vertex pairs. In general the corresponding system of quadratic equations is quite clumsy but, since the above examples could be solved so easily, there is hope that in combination with other methods, with more effort and computer help we could get solutions for bigger k. Even partial solutions could be used to reduce the number of cases to be searched. The strong regularity of Paley graph rests on Jacobsthaler’s formula for the quadratic sum

of Legendre symbols. In a certain sense higher sums of this type reduce to the cubic ones [16]. It would be interesting to see whether something similar holds for the srsc-graphs and to what extent graph properties leading to cubic formulas in xu,,

Regular and strongly regular selfcomplementary graphs

225

determine the srsc-graphs. Another problem is a nice description of the parameters e i jand oij for rsc-graph for permutation with k four-cycles. The author is much indebted to A. Kotzig for suggesting the topic and for stimulating conversations and to the referee for valuable comments. The support provided by the Natural Sciences and Engineering Research Council Grant A9 128 and the Ministere de 1'Education du Quebec FCAC grant EQ-539 is gratefully acknowledged. 1

A graph means a finite simple graph (i.e. loopless non-oriented and without multiple edges). A graph G = (r!E ) is self-complementary (sc-graph) if it is isomorphic to its complement G=(T/: E), i.e. if there is a permutation of the vertex set V carrying each edge (non-edge) on a non-edge (edge). The permutation 4 is said to be complementary and G to belong to 4. Let CGdenote the set of complementary permutations of an sc-graph G. It is easy to see and well-known that CGuAut G is a subgroup of the symmetric group on V in which CGhas index 2, i.e. C, = CG,CG0 CGSAut G, and Aut G CG0 Aut G r C G .In particular the odd and even powers of 4 E C G belong to CGand Aut G, respectively. The sc-graphs form a large family of quite diverse graphs. More special and interesting is the family of regular sc-graph (rsc-graphs).It is well known that every rscgraph G has 4 k + 1 vertices, each vertex of G has degree 2k [18], G has diameter 2 [ll] and possesses a Hamiltonian path [2]. In this section, following an idea of A. Kotzig (Lemma 3 below) we derive some simple properties of rsc-graphs. In Lemmas 1-6 below, G is an rsc-graph on 4k + 1 vertices belonging to a permutation a. For simplicity a(v) is denoted by 0'. Let [x,,] be the zero-one adjacency matrix of G (i.e. xu,= 1 if [u, u] E E and xu,= 0 otherwise) and for x real set X = 1 - x.To start out, let p,, denote the number of vertices adjacent to the vertex u but not adjacent to the vertex u. We have: 0

Lemma 1. The number puuis invariant under a.

Proof. Clearly p u u = ~ y x u y Xwhere , y y runs through V\{u, u ) . By the same token x,,,X,,, = X,.~.X,,,~, where z runs through V \ { u', v'}. Using = Xuy and Xu,y,= x U ywe obtain pufu,=

cy

c,

Pdd

=

c

~uyxu= y

P",.

Y

Finally p,, = p,, in a regular graph.

0

Note that we have, in fact, proved p,,.,. = p,, in every sc-graph. In concrete instances Lemma 1 could be used to prove that a given permutation is not complementing. Next let v,, denote the number of joint neighbours of distinct vertices u and v.

Lemma 2. For distinct vertices u and v V,,",

- v,,

= 2xuu- 1.

I. G . Rosenbery

226

0

Proof. Use v,, = 2k -p,, -xu,, and Lemma 1.

Let t, denote the number of triangles passing through v . The following lemma, proved by Kotzig [121 by a direct argument, can be also proved by the method used in Lemma 1. The last statement can be obtained also from [13,10.32(a)].

Lemma 3. For every vertex v t,+ T,,,= 2k(k - 1).

In particular, there are exactly k(k - 1) triangles through the fixed point of o and G contains exactly 3 4 k + l ) k ( k - 1) triangles. Let I , denote the number of vertex pairs ( y , z ) such that [ v , y ] E E, [ y , z ] E E and [v, z ] $ E .

Lemma 4. For every vertex v

A, + A,, = 4k2. Proof. There are 2k(2k - 1 ) pairs ( y , z ) such that [v, y ] E E, [ y , z ] E E and z # u. We deduce Iu=2k(2k - 1 ) - 2 ~ ,and the statement follows from Lemma 3. 0 For distinct vertices u and v let p,,, denote the number of length 3 paths between u ahd v .

Lemma 5. For distinct vertices u and v p,,,

+ p,,,,, = 2k(2k - 1) + 2 - 3 ( ~ , ,+, v,,,,).

Proof. By definition puu= ~ v z x , y x y z x zwhere , we sum over two-element subsets { y , z } of V\{u, v } . As in the proof of Lemma 1 we have p,,.,,. X,yXyzX,,. Using X = 1 - x we obtain

=cy,z

p,,

+ p.,,, = c 1 - ex,),- cxyz

-

+

e x z " + ~ x , y x y z cx,yx,,

+c x y z x z ,

where the summation range is the one above. Now Z l = ( 4 k - 1)(4k-2),

C~,,=C~,,=(4k-2)(2k-~,,).

Next xyzis twice the number ofedges incident to neither u nor v, i.e. Cxy,=2((EI -4k+x,,)=2(4k2 -3k+x,,). Finally

~ X , ~ X ~ ~ = X X ~-~,,)(2k ~ X ~ , -= 1)( -~ v,,,, ~ ~ x u y x z=u(2k - Vuur and the result follows by direct computation.

0

Regular and strongly regular selfcomplementary graphs

227

Finally, let E.,, denote the number of joint neighbours of three distinct vertices u, u, w. Applying the above proof technique one can obtain:

Lemma 6. For three distinct vertices u, v and w Euuw

+

EU,U,W,

= vu,

+ + v,, + 2(x,u + xu, + xu,) vuw

-X , , X ~ ,

- x,,x,, - x,,x,,

- 2k - 2.

2 The zero-one parameters xu, used in Section 1 are not mutually independent; in fact, as we shall see, they are highly redundant. For this we select 6 E CGin an appropriate way. Since we are mostly interested in regular sc-graphs we restrict our attention to the sc-graphs with 4k+ 1 vertices. Camion [ 2 ] has shown the existence of 0 E CG such that 02' is the identity for some integer t > 1 . It is not difficult to see that such 6 has a unique fixed point f and its cycles C1,.. . , C , may be arranged so that ICi\= 2"' where the integers ai satisfy 1 < a , < * . .
I = {(O, i, 0): 1 < i < n>u {(i,j , I ) : 1 < i < j < n, O
For (i,j, I) E I we abbreviate xciocjlto xijl.Thus xoio=1 signals the fact that there is an edge between the fixed point of 0 and the chosen starting point cio. Similarly, for i < j the zercl-one parameters xijldescribe the presence of edges between the starting point cio of Ci and the first 46, vertices of C j while xiilcharacterize the edges from cio to the subsequent 2di vertices of Ci. We show now that for a given 0 E Pk the binary parameters (xu: (i, j, I) E I) determine uniquely the sc-graphs belonging to 0.

hopition 1. For 1 V1= 4k + 1 and 0 E Pk there is a bijection between the family of sc-graphs belonging to 0 and the 2"' binary vectors x = (xijl:(i, j , I ) E I ) . In particular, for such a vector x the corresponding sc-graph G = ( K E ) is determined as follows: For l < i < j < n set (i) x~~~~~ P + xoio (mod 2), (ii) xcipcjq - p + x i j l (mod 2), where r r q - p (mod 4di), O
0

such that [cia, c j l ] E E iff xijl= 1.

I . G.Rosenberg

228

We show that G satisfies (i) and (ii). The condition (i) follows from the definition and coo fixed point of 6.For (ii) we consider first the case i=j, r -4 - p (mod 2) where 0 < r < 2di. Now [cia, cir]E E iff xiir= 1. Applying 0'' we get [cip ciq] E E iff xiir+ p is odd. The case i <j is quite similar. Sufficiency: Note that (i) and (ii) completely determine a graph G inasmuch as they tell us whether (u, v) E E or [u, v ] 4 E for every pair of distinct vertices u and v. The proof that G is an sc-graph belonging to 0 is also almost immediate from (i) and (ii).

0 It is worth noting that the 2"' sc-graphs are only distinct as labelled graphs (i.e. isomorphisms are possible). To illustrate: to x = (xijI:(i,j , I ) E I ) assign x' defined by (i) ~ i , i , 2 1 + 1 = X i , i , 2 1 + 1 for 1
Corollary 1. The map 4 is an isomorphism of G onto G'

Proof. We get &G) by drawing G on a transparent paper and turning it over. Using (i) and (ii) it can be verified that &G)= G'. 0 Now we express the degrees of vertices in terms of x. By definition it is enough to know the degrees of cio for i=O, . . . ,n. For this set 2di- 1

Zdi- 1

1=0

1=0

( 1
e l l , . . .,en,,, e12, 0 2 1 , . . . , G-i,,,,

On-1.n.

Proposition 2.Let G be an sc-graph belonging to o E Pk.Then the j x e d point of degree 2k and for 1 < i
0

has

Regular and strongly regular selfcomplementary graphs

and 4k -di,respectively. In particular, G is regular ifand only eii=ai+2

C h j - C (ehi-ohi)-6;'

j>i

h
229

if

hj(eij+oij) j>i

(i= 1,. . . , n).

Proof. The degree of coo is known [181and follows from Proposition l(i). Let 1
Similarly for i<j the number o f edges between cio and C j is 6;'6, (eij+oij)proving (3i). Finally, (4,)is obtained from (3,)and di= 2k = 2(d1 . . . 6,). 0

+

+

For example, if n has a unique cycle (of length 4k) the fixed point has degree 2k and the even and odd vertices have degree k +el and 3k -el 1. By varying eij and oj within (0, 1,. . . ,26,} (1
3 From now on we discuss exclusively regular sc-graphs belonging to 0 E Pk. We transform the equations (4)into an inequality system by eliminating eii. The integer vector 5=(e12,012,.. .9en-1,n90n-1,n) with coordinates O<eij, oij<26, is called a paramater vector of G. We have

Propositioo 3.T h e parameter vector of a regular sc-graph belonging to 0 E Pk satisfies

( i = l , . . . , n)

O<eij<26,

0<0~~<26~

(1
I. G . Rosenberg

230

exactly (7)

rsc-graphs belonging to (T with the parameter vector 5.

Proof. Inserting (4,) into O<eii<2di we obtain (5,). For the converse let 5 satisfy (5)-(6). In ( lij) we have

independent binary choices for xijp(i<j, p<4di). From (2,) we get e..=x - oio + Xi.i.26, (mod 2).

(8i)

it

Let eii be odd, eii=2e'+ 1. The equation ( S i ) has then two solutions (0, 1) and (1, 0) and for both of them (2,) reduces to

(tf)

possessing binary solutions. After these values have been chosen we must select the di zero-one values x ~ , ~ , ~ ,. . . ,xi,i,2ai1. Since t i = 1 we have altogether the factor

Next let eiibe even, eii=2e'. From (8,) we have xoio= xi,i,2ai.Now (2,) has (ely;iiJ independent binary solutions, hence altogether (aiil) solutions each of which can be complemented by 26i binary choices of X i i l , ~ i i 3 , .. . ,X i , i , z a i - 1. 0 To illustrate consider the simplest case of (T with one cycle. We have n = 1, d1 = k, ( 5 , ) is vacuous and (4,) reduces to el = k. Setting k' =ik we get a1= k -2k'. It follows

,

that there are exactly

rsc-graphs belonging to (T with one cycle. We list the two simplest cases k = 1, 2. Example 1. There are 4 rsc-graphs on 5 vertices belonging to CT E P1 (all 4 isomorphic to the 5-cycle C5).

Example 2. There are eight 4-regular sc-graphs on 9 vertices belonging to o E P 2 withasinglecycle. Thegraphcorresponding toxolo=x114=0,x111=xl12=x113= 1

231

Regular and strongly regular selfcomplementary graphs

is the srsc-graph formed by 9 triangles, such that every pair of triangles meets exactly in one vertex.

To settle the existence problem of rsc-graphs belonging to a given (T E Pk we choose the ‘average’ values eij=oi,=di (1
Corollary 2. Let ~ ~ = $ l --( l)di)(i= 1, . . . , n) and let least

+ ...

+E,.

There are at

rsc-graphs belonging to (T E Pk, Proposition 3 reduces the enumeration problem of rsc-graphs belonging to a given o E Pk to the integer solvability of (5)-(6). For n = 2 we have the following description.

Corollary 3. Let e 1 2 ,0 1 2 be the parameter vector of an rsc-graph belonging to (T E Pk with two cycles. Then OSel2, o12<2d1and the distance of both e12and o 1 2from d1 does not exceed $6:+6;)&’. In particular, for 6, = d 2 ( = + k ) the vector 5=e12,o12 is characterized by

5 ~ ( 0 ,. . , k)’, Proof. Set y=6:6;’.

(e12-0121<$k,

el2+ol22+k.

(9)

The conditions (5,) and 52) are:

2h1- y ~ e ~ ~ + o ~ 2 < 2 6 ~ + y -62<e12-012< ,

-62.

Adding and subtracting we obtain le12-611<6’ and 1 0 ~ ~ - 6 ~ 1 < 6Finally ’. (9) is a restatement of (5)-(6). 0

Example 3. Let k = 2 and a1 = d 2 = 1. The solution set of (1)-(6) is (0, 1, 2}2\{0, 2}2. Choose 5 = (1, 0). Then (4)yields el = 2 and ez2=0, whence by (7) there are 9 rscgraphs belonging to (T and 5. The corresponding vector x has x 0 1 0 = x 1 1 2 =I , x020=x 2 2 2= x l Z l=0, x120 x 1 2 2= 1. Choosing x 1 = x 1 2 2= x 2 2 1= 1, x l z O= 0 we obtain a graph isomorphic to the graph of Example 2.

+

For n>3 we do not know how to describe the parameters even in the simple case n=3 or h1 = . . . =6,= 1. The full description of the integer solution set of a linear inequality system is, in general, a difficult task (although it is not difficult to verify whether a given vector solves (5)-(6)). In this direction we combine the inequalities ( 5 ) to produce a simple bound. Write (5i) as

+(n - 1)6,,.Adding all ( loi)we obtain 21 - k < C ((6; ‘ 6 + l)eij+ (6; ‘6 - l)oij)< 22 + k .

and set 1 = h2 + 2d3+ . . . i<j

I. G.Rosenberg

232

In the uniform case d1 = . .. =6, using h i = kn-', 21= k(n- 1 ) and integrality we get jk(n-2)6

eij<$kn. i<j

Moreover, subtracting (10,) from (11 ) we obtain '2k ( n - 2 ) - k n P ' <

c

l=si<j
eij+

oij
Example 4. Let 6' = .. . =6,= 1. We determine the solutions of ( 5 ) having e i j = o i j for all 1 < i < j < n. The inequality (5)becomes

leading to Ceij=n-i

(l
eijE{O,

1,2} (1
j>i

Thus for n = 3 we have 3 solutions e12= o 1 2E (0, 1,2},el3 = o 1 3= 2 -e12, eZ3= oZ3= 1.

4

As we have seen even the family of rsc-graphs belonging to a given d E Pk is still large and varied. Kotzig [113 suggested to study in detail the strongly regular scgraphs (srsc-graphs). Recall that a regular graph is strongly regular if there are 1 and u , such that each edge of the graph is on exactly A triangles and nonadjacent vertices have exactly /A common neighbours. The srsc-graphs are much more special than the rsc-graphs. For example they can exist only for k such that in the canonical decomposition of 4k + 1 no prime congruent to 3 (mod 4 ) appears with an odd exponent and their 0-1 adjacency matrices have 3 eigenvalues 2k, - 1 & J 4 k + 1) with multiplicities 1 , 2k, 2k [9]. The condition for strong regularity maybe slightly weakened. The equivalence of (i)-(iii) seems to be known and was communicated to us by Kotzig [12].

3

Lemma7. T h e following conditions are equivalent for an rsc-graph belonging to d E

Pk

(i) G is strongly regular, (ii) euch edge is on exactly k - 1 triangles, (iii) every pair of non-adjacent vertices has exactly k joint neighbours, (iv) vCiocjr=k-xijrf o r a l f ( i , j , l ) E I .

Proof. The equivalence of (i)-(iii) can be easily proved using Lemma 3 and Lemma 2. Also (i)* (iv) follows from the matrix equations X J = 2kJ, X 2 + X = k(J + I ) where X is the 0-1 adjacency matrix and J is the matrix with all entries 1. For (iv)*(i) apply Proposition l(i).

Regular and strongly regular selfcomplementary graphs

233

The condition vcoocio= k - xoiomay be expressed in terms of the basic parameters.

Lemma 8. The condition vco,cio= k - xoio is equivalent to

1 (XOhOehii-zOhO(2ah

-0hi))

h
+ Xoioeii + ioiodi + 6;

j>i

6,{Xojoeij

+

%ojoOij)=

k

(i= 1 , . . . , n).

Proof. Let h
1. By Proposition l(ii) the joint neighbours of coo and cio on Ch are ch,21 such that x ~ ~ , ~ ~ ~ ~1. Thus ~ = there x ~ are , ~exactly , ~ ~ehi~joint - ~ ~ = neighbours o f coo and cio on Ch and we obtain the term XOhOehi in (12,). The other terms are obtained in a similar fashion. (For example, coo and cio have exactly eii- 1 0 and hi joint neighbours on C i depending on xoio= 1 and xoio= 0.) XOhO=

The checking of Lemma 7(iv) seems to be fairly complicated in general and therefore we consider only the two extreme cases n = 1 and 6 = . . . = 6 , = 1. Let cr have a unique cycle (of length 4k). We can simplify the notation by setting f=coo and ci=cli (1
x2i + 1 = z - z i - 1

x2i = x - 2 i,

(13) for all integer i. With this notation we have xcoci=x i for all i. The regularity condition el = k (derived at the end of Section 2) is now

The strong regularity conditions are automatically satisfied for the pairs { f, c i } and {ci,c j } with i # j (mod 2):

Lemma 9. Let G be an rsc-graph belonging to t s with a single cycle. Then for every i and every odd 1, the vertices ci and c i + l have exactly k -xci,ci+ljoint neighbours.

Proof. According to Lemma 7(iv) it suffices to consider i=O. For O
+

,,= 1. Now the fixed point is never

4k -

VC0Cl

=L 291 + P p=l

Separate the summation range into 7 sets: [l, 2k -q and the even and odd numbers from [2k-1+1, 2k-11, [2k, 4k-I] and [4k-1+1, 4k-1). Let l'=$l-l) and m = k - l'. Applying first (13) and then X = 1 - x obtain 2k-1 'COCl=

1

1'

zpxp+ I

p= 1

+ 1 ( Z Z k -qXZkq= 1

2q

+ 1 + 22k - 2q + l X 2 k - Z q + 1 + I )

m ... - 1 .

+ r1 (X4k-l-ZrX4k=O

2r

+ i 4 k - I - 2 r - 1 X 4 k - 2 r - 1)

I. G.Rosenberg

234 I'

+

c

( 2 4 k - 2sX4k- 2 s + l + % 4 k - 2 s + l X 4 k - 2s-t 1 + I )

s= 1

2k-I

=

c

1'

1

x&p+l+

(-f2k-2q~2k+2q-1+~2k-2q+lX2k+2q-l-I)

p= 1

+

q=l

21'

2k-1 (-1)IXZk-tX2k+t+l+m-

c

1

(-X2~+l-~Zu+l+l)

u=o

21'

I'

+w = l

m- 1

xuxu+I+

u=o

1

t=

c

x1-2w+

(-1)zxzxI-z.

z=1

Noting that 21'

1

21' (-l)tX2k-tX2k+t-I=

1

(-1)zxzxI-p=0~

z=1

1=1

direct simplification yields the required result v,,,~ = k - xl.

0

Now we can formulate a strong regularity criterion.

Proposition 4. Let G be an sc-graph belonging to a permutation with one cycle. Then G is strongly regular if and only if the corresponding binary parameters y, x l r . . . , x 2 k satisfy (14)and for I = 1, . . . , k Zk-2

2

1-1 xixi+21+2

i= 1

c

(-l)i(XiXZ1-i+X2k-21+iX2k-i)

i= 1

where

Proof. It is obtained by a computation similar to the preceding one.

0

Regular and strongly regular selfcomplementary graphs k= 1

For k = 1 obviously every rsc-graph is strongly regular. We consider the cases k=2 and k=3.

Example 6. Let k = 2. The condition (14) implies x4= y and xz = j.Next (&) is 4x1x3=2 -2xz -y -x4 -2x1 -2x3

which simplifies to 2 ~ ~ =xx1 3 + x3 and x1= x3. The condition (16) is now 2(x1x3+xzx4)=1+x1+x3-y-x2

holding trivially for xz=y,

x4=y,

x3=x1.

(20)

Thus (20) is a necessary and sufficient condition for an srsc-graph on 9 vertices belonging to one cycle 0. These 4 labelled graphs (out of 8) are pairwise isomorphic. Indeed the selection of the initial vertex co allows us to have y= 1 and the two graphs (corresponding to x1= O and x1= 1) are isomorphic by Corollary 1 from Section 2.

Example7. Letk=3. From(14)wegetx6=yand ~ 4 = X 2 . Next (&)is (17): 4( -X1X5 +xzxq)= 3 +2x3 -y -xg -2(x1 +x3 +x5) which reduces to x1+x5 -2x1x5= 1 and

~5=X1.

Next (15,) is (19):

2(xzj-2x1x3 -x3jz1)= 2 -xz - 2Xz - y -2x3

which gives xz=y. Finally (151) is (18):

+

2(X1x3 yy+

+ x i + 31 -y -y + 2x3

X3X1 +jJ)= 1

which is identically true. Thus xz=y,

x4=y,

x5=%19

x6=y

describe the 8 srsc-graphs on 13 vertices belonging to 0 with one cycle. To detect the nonisomorphic ones fix again y= 1 and x1= O (choice of initial point and Corollary 1). The two graphs left (corresponding to x3= 0 and x3= 1) are both isomorphic to the quadratic residue graph (identify f with 0, ci with ( 6 + ~ (i= ~ ) 1,~ . . . , 12) and [u, 01 E E e u - u is a square mod 13). Directly an isomorphism is provided by the

236

I . G.Rosenherg

map leaving even vertices fixed and sending odd vertices ci to the opposite vertices CI

+6.

Example 8.. Let k = 4. For simplicity set right away y = 1 . Parity in (14) yields then x g= 1 and we have x2 x 4 x6 = 1. Note that this implies ~ 2 x = 4 x2x6 = x4x6 = 0.

+ +

Now (17) is 4( -x1x7 - x,x,)=4

-2x4

-

2 -2(x1+ x3

+ xs +x7)

which can be written as (xS

-X~)~+(X~-XI)’=X~.

(21)

Next ( I 8) gives 2(xlX~+X3X7+X4-XIX3-X~X7)=1-X~-X4-X~=0

or, equivalently, (xs -x3)(x, -x1)=x4.

(22)

For x4= I clearly (22) contradicts (21). Thus x 4 = 0 and Xg=x2. Now conditions (15,) and (16)simplify to 2(x1x7-X1X5 -x3x7)= 1 -x2 -x3 -x5,

(23)

2 ( x ~ X ~ + X ~ X ~ + X ~-2+x1 X 7 ) = +X2+2X3+2XS+X7.

(24)

Here parity requires x5=X2 -x3 +2x2x3,

x7 =xI

+ x2 -2x1x2.

(25)

By direct check one can verify that for y = x g= 1, x4= 0, x6 = X 2 and xs, x7 given by (25) the equations (21)-(24) hold (e.g. by using x5 - x 3 = X z ( l -2x3), x 7 -xl = x2(1- 2x3), (1 - 2x3)’ = 1 and the zero-one nature of the x;s). Thus we have obtained 8 srsc-graphs. There are 8 more corresponding to y = O and consequently altogether 16 rscs-graphs on 17 vertices belonging to IJ with one cycle.

Finally we turn to the other extreme case of x with k four-cycles. The following necessary condition captures the strong regularity property for certain pairs of vertices expressible in terms of the parameters e i j and o I j . For 1
Proposition 5. Let G be an rsc-graph belonging to IJ with k four-cycles and such that cia] E E for all i = 1 , . . . , k . Then the pairs of vertices {coo, c i O ) (cia, , c i l ) and {cia, c i z )( i = 1,. . . , k ) satisfy the strong regularity condition ifand only if

[coo,

C ehi+ h
j>i

eij=k-$(3+(-l)k),

(26,)

Reyulur und stronyly rryulur selJi~omplementuryyruphs

237

for i = 1 , . . . , k .

Proof. Necessity: For xoio=Si= 1 (i= 1 , .

. . ,n) the conditions ( IZi) and (4,)become

Eliminating eii we obtain (27,). Now set Pij=xijoxij2 + x i j l x i j 3and by a direct computation obtain

Noting that xij0xij2 # O iff eij=2 and similarly xi,lxij3#0 iff oij=2 and using eii= xoio+ xiiz= 1 + xii2transform vcioci2 = k - xii2into 2ui=2

C ohi-2i+k-eii+2. h
Thus k =eii (mod 2), or, equivalently, e . . = q 3 + ( - 1 ) k) . 11

2

(32)

Now (26,) and (29,) follow from (30)-(32). To prove (28;)compute

Set

3

sij=

C xijpxi.j,p+l p=o

Direct verification shows S i j = eijoijand (32)can be easily turned into (28,).This whole derivation being reversible the converse is also true. 0

238

I . G . Rosenberg

References I. C. Bussemaker, R. A. Mathon and J. J. Seidel, Tables of two-graphs, Report Techn. Univ. Eindhoven, 79-Wsk-05 (1979). P. Camion, Hamiltonian chains in self-complementing graphs, Cahiers Centre Etudes Rech. O@r. I7 (1975) 173-184. C. R. J. Clapham, Potentially self-complementary degree sequences, J. Combin. Theory Ser. B (1976) 75-83. C. R. J. Clapham and D. J. Kleitman, The degree of sequences of self-complementary graphs, J. Combin. Theory Ser. B (1976) 7-74. D. J. Curran, On self-complementary strongly regular graphs, Res. rep. CORR 78/23, Univ. of Waterloo (July 1978). R. A. Gibbs, Self-complementary graphs, J. Combin. Theory 16 (1974) 106-123. J. M. Goethals and J. J. Seidel, Orthogonal matrices with zero-diagonal, Canad. J. Math. 19 (1967) 1001- 1010. P. L. Hammer and S. Rudeanau, Boolean Methods in Operations Research and Related Areas (Springer, Berlin, 1968; French translation: Dunod, Paris, 1970). [9] D. G. Higman, Characterization of families of rank 3 permutation groups by the subdegrees I, Arch. Math. 21 (1970) 151- 156. [lo] X. L. Hubaut, Strongly regular graphs, Discrete Math. 13 (1975) 357-381. [ I 13 A. Kotzig, Selected open problems in graph theory, in J. A. Bondy, U. S. R. Murty, Eds., Graph Theory and Related Topics (Academic Press, New York, 1979) pp. 258-267. [I21 A. Kotzig, Personal communication. [ 131 L. Lovhsz, Combinatorial Problems and Exercises (North-Holland, Amsterdam, 1979). [I41 R. Mathon, Symmetric conference matrices of order p q 2 + 1, Canad. J. Math. 30 (1978) 321-331. [I51 G . Ringel, Selbstkomplementare Graphen, Arch. Math. 14 (1963) 354-358. [I61 I. G. Rosenberg, Sums of Legendre symbols I, I I (Czech) Sb. Vysoke UEeni Tech. Brno (1962) pp. 183-190,331-334; MR 33 #4026. [I71 S. Ruiz, Note: On a problem of A. Kotzig, Preprint, Universidad Catolica de Valparaiso (1979). [181 H. Sachs, Uber-selbstkomplementareGraphen, Publ. Math. Debrecen 9 (3--4)(1962) 270-288. Received 28 October1980; revised 19 March 1981.

Annals of Discrete Mathematics 12 (1982)239-242

@ North-Holland Publishing Company

CONSTRUCTION OF ADDITIVE SEQUENCES OF PERMUTATIONS OF ARBITRARY LENGTHS Jean M. TURGEON* DCparternent de Mathkmatiques et de Statistique, UniuersitC de Montrial, Canada

Dedicated to Professor A. Kotzig on the occasion of his sixtieth birthday Let r and n be positive integers. Let X = [x,, . . . , x,] be an ordered set of distinct integers. For . . ,n, let X‘”= [xy), . . . ,x!”] be a permutation of X. An ordered set XI’), . . . , XI“) of such permutations is called an additiue sequence ofpermutations of degree r if, for every subsequence of consecutive permutations of the set, their vector-sum is again a permutation of X. The number n is called the length of the sequence. Until now, no additive sequence of length greater than 4 was known. We present here a simple construction for additive sequences of length p - 1, where p is prime. If another length, say n, is desired, one can take the first n permutations of an additive sequence of length p - 1, where p is some prime greater than n + 1.

j = 1,.

Let r and n be positive integers. Let X = [xl, . . . ,x,] be an ordered set of distinct integers. For j = 1, . . . ,n, let X‘j’= [xi’), . . . ,x y ] be a permutation of X. An ordered set X“’,

x‘2’,. . . , X ( ” )

of such permutations is called an additive sequence of permutations of degree r if, for every subsequence of consecutive permutations of the set, their vector-sum is again a permutation of X . The number n is then called the length of the sequence and the set X is the basis of the sequence. The cardinality of X is of course the degree of the permutations. Additive sequences of permutations were first introduced in relation with perfect systems of difference sets (cf. [9, lo]), but were also considered for their own intrinsic interest, either with integer intervals as bases (cf. [3, 8, 111) or with arbitrary bases (cf. [l, 2, 4, 5, 71). But until now no additive sequence of length greater than 4 was known. In the present paper, we present a simple construction for additive sequence of length p - 1, where p is a prime. If another length, say n, is desired, one can take the first n permutations of an additive sequence of length p - 1, where p is some prime number greater than n + 1. In the sequel, we shall call ‘bijection of X , rather than ‘permutation of X , a 1-1 mapping of X onto itself, reserving ‘permutation’ to denote an ordered set or vector. A bijection can be decomposed into cycles or cyclic factors (usually called ‘cyclic permutations’).

1x1

*Research supported by FCAC (EQ-539) and CNRC (A-4089). 239

240

J. M . Turgeon

Theorem 1. Let p be a prime number, p > 2. Let A = [ a l ,. . . ,a p - be afinite sequence of positive integers such that the partial sums of consecutive elements of A are all different:

Let D- 1

For j = 1, . . . ,p and k = 1, . . . ,p - 1, let j+k-1

cJ@)=

1

i=j

ai,

where the sums of indices are taken modulo p. Let X = { c j ( k )I j = 1,. . . ,p ; k = 1,. . . , p - 1 ) with its elements taken, say, in their natural order. Let f : X+X be defined by

f (CXk))=cj + k(k).

(4)

Then the sequence X“), . . . , X ( p - ’ ) defined by X(l)=x;

X(”)=[f(h-l)(X(ll)),. , f ( h - l ) ( X (p(p-i))], l) , ,

h=2,. . . , p - 1 ,

(5)

is an additive sequence of permutations of degree p ( p - 1 ) and length p - 1.

Proof. The element cJ{k)of X is the sum of k cyclically consecutive elements of A, starting with aj. So cXk) is characterized by its start a j and its length k. For k= 1,. . . ,p - 1, let Dk be the set of all elements of length k of X :

&= {cj(k)I j = 1, . . . , p } The sets Dk from a partition of X and every aj is the start of exactly one element of each Dk; k= I , . . . ,p - 1. By (2), cj(k)= - C j + k ( p - k). So, by ( l ) ,X contains exactly p ( p - 1) elements: those of the set described in ( 1 ) and their opposites. To prove that the function f : X - r X defined in (4) is a bijection, it suffices to note that, since p is prime, the restriction off to Dk is a cycle of length p : (cl(k), Ck+l(k),C Z k + l ( k ) , . . c(p-l)k+l(k)). ’

9

(6)

The mapping f is the product of p - 1 such cycles and (5) is thus a sequence of permutations. Let X‘”’, . . . , X @ + [ ’)- be any subsequence of t consecutive permutations of (5): 1d s d p - 1 ; 1
Additive sequences of permututions

241

defined by 1-1

g(cAk))=cAk)+

1 f‘(cj(k)).

i= 1

By (3) and (4), we have (7)

g(cj(k))=cj(tk),

so that g maps the element with start a j of Dk into the element with start a j of Dtk. Since p is prime, the subsets D,,form a partition of X . So g is a bijection of X and the sum of any t consecutive permutations of (5) is again a permutation of X “ ’ : the sequence (5) is additive. Thanks are due to A. Kotzig, who suggested the basic idea of the construction. Would another application off make (5) a longer additive sequence? No, since the cycles ( 6 )are of length p , in view of the following.

Theorem 2. If an additive sequence of permutations of length n i s defined by repeated applications of a unique bijection, then every cyclic factor of this bijection contains more than n elements or exactly one [ 12, Theorem 2.41. For every prime number p the sequence [l, 2, 22,. . . , 2p-2, 1 - 2 p - 1 -

1

satisfies the conditions (1) and (2). So do also the lines 2, 5 and 7 of the ‘Colombs triangle’ together with the opposite of their sums (cf. [6, p. 343). Other examples are provided by certain components of perfect systems of difference sets.

Acknowledgements The author has benefited from discussions with Jaromir Abrham and Douglas Rogers.

References [I] J. Abrham and A. Kotzig, Bases of additive permutations with a given number of odd elements, Utilitas Mathematica 18 (1980) 283-288. [2] J. Abrham and A. Kotzig, Generalized additive permutations of cardinality six, in: Proc. Eleventh Southeastern Conference on Combinatorics, Graph Theory and Computing, Boca Raton, March 1980, Congresus Numerantium 28 (1980) pp. 175-185. [3] J. Abrham and A. Kotzig, Skolem sequences and additive permutations, Discrete Mathematics 37 (1981) 143-146. [4] J. Abrham and A. Kotzig, Cubic graphs and certain pairs of additive permutations, to appear id:

Proc. Twelfth S.E. Conf. on Combinatorics, Graph Theory and Computing. [ 5 ] H. Desaulniers, Sur un probleme de Kotzig sur les permutations additives, Rapports de recherche de l’U.Q.T.R. No. 11 (1980) 6 p.

242

J . M . Turgeon

161 S. W Golomb, How to number a graph, in: R. C. Read, Ed., Graph Theory and Computing(Academic Press, New York, 1972) pp. 23-37. [7] A. Kotzig, Existence theorems for bases of additive permutations, in: Proc. Eleventh Southeastern Conference on Combinatorics. Graph Theory and Computing, Boca Raton, March 1980,Congressus Numerantiurn 29 (1980) pp. 573-577. [8] A. Kotzig and P. J. Laufer, When are permutations additive? Amer. Math. Monthly 85 (5)(1978) 364-365. [9] A. Kotzig and J. M. Turgeon, Perfect systems ofdifference sets and additive sequences of permutations in: Proc. Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing, Boca Raton. April 1979. Congressus Numerantium 24 Vol. I1 (1979) pp. 629-636. [101 D. G. Rogers, A multiplication theorem for regular perfect systems of difference sets, to appear in Discrete Mathematics. [I 11 J M. Turgeon, An upper bound for the length of additive sequences of permutations, Utilitas Mathematica 17 (1980) 189-196. [ 121 J. M. Turgeon. Additive sequences of permutations generated by perfect systems of difference sets, to appear. Receioed 22 April 1981

Annals of Discrete Mathematics 12 (1982) 243-253

@ North-Holland Publishing Company

COUNTING ROOTED TRIANGULATIONS W. T. TUTTE Uniuersiry o] Waterloo. Canada

Dedicated to Professor A. Kotzig on the occasion of his sixtieth birthday This paper is about planar enumeration, I would like to contribute to this collection of essays not some minor improvement in some special corner of the theory, but a fairly general exposition of what the theory is and what it tries to do. Not too general either; a single strand of argument should suffice to make clear the essentials, without obscuring them behind a clutter of varied examples. So I propose to write about rooted planar triangulations and how to enumerate them, ignoring, as far as the argument permits, all other kinds of rooted or unrooted planar maps.

1. Introduction

A planar triangulation is a planar map in which each face is triangular, that is bounded by a simple closed curve made up of exactly three edges of the map. The outer face, like each of the inner ones, has to be triangular in this sense. In the most general kind of planar triangulation it is possible for two distinct edges to join the same two vertices. But the digon they enclose must be dissected into triangular faces. It is not possible however for an edge to be a loop, that is to have two coincident ends. By specialization we arrive at the strict triangulations, in which no two edges have the same pair of ends. Then we can define a simple triangulation as a strict one in which there is no separating triangle. A separating triangle is defined by three edges of the triangulation making up a circuit, there being at least one vertex of the triangulation inside the circuit and at least one outside. Analogously we could define a strict triangulation as one with no separating digon. Fig. 1 exhibits three planar triangulations: (a) is not strict, (b) is strict but not simple and (c) is simple. To root a triangulation is to choose one outer edge as the ‘root-edge’, and one end of that edge as the ‘root-vertex’.In diagrams we usually represent the root-edge as an edge with an arrow, the arrow being drawn from the root-vertex to the other end. Whether we are considering general, strict or simple rooted planar triangulations we may want to know how many combinatorially inequivalent ones there are with 2n faces, counting the outer one. It is easy to see that the number of faces must be even. We say that two rooted planar triangulations are combinatorially equivalent if they are related by a 1-1 correspondence of their elements, which maps vertices onto vertices, edges onto edges and faces onto faces, which preserves incidence relations and the outer face, and which preserves the root-edge and root-vertex. A planar triangulation has three outer edges, and so allows six choices of root-edge and root-vertex. But because of symmetries of the figure the six choices may not yield six combinatorially distinct rooted triangulations. For example the first triangulation 243

244

W. 7: Turte

Fig. 1

of Fig. 1 can be rooted in only three essentially distinct ways, and the third triangulation in only one. But the second does give rise to six distinct rooted maps. In the next Section we discuss the enumeration of general rooted triangulations. In order to get the necessary equations it is found necessary to extend the theory to a wider class of map, the class of ‘near-triangulations’. A ‘non-degenerate’ near-triangulation is a planar map in which each inner face is triangular. The outer face is not necessarily triangular, but we do require it to be bounded by a simple closed curve. It is permissible for the outer face of a nondegenerate near-triangulation to be a digon. It is convenient to define also a single ‘degenerate’ near-triangulation. This has just one edge and just two vertices, the latter being the ends of the edge. The remainder of the plane constitutes the only face, naturally counted as the outer one. We root near-triangulations by the same rule as for triangulations. The two ways of rooting the degenerate near-triangulation are equivalent by the symmetry of the figure. So there is essentially only one degenerate rooted near-triangulation. Given a rooted near-triangulation N we write m ( N )for the number of edges incident with its outer face, and t ( N ) for the number of inner faces. In the degenerate case we write accordingly t ( N )= 0. But because the single edge is now doubly incident with the outer face we modify the definition of m ( N ) and write m ( N ) = 2 . Figs. 2 and 3 give examples of near-triangulations.

Counting rooted triangulations

245

2. Afunctimal equation Our enumerative theory for general rooted planar triangulations makes use of two generating series q = q ( x , z ) and h(z), the first in two independent variables x and z and the second in z alone. We define q by the equation

1Xm(N)Zf(N).

q(x, z)=

(1)

N

Here the summation is over all (combinatorially inequivalent) rooted near-triangulations N . It is clear that there can be only a finite number of such figures with a given value of t(N). Hence the expression on the right of (1) is a well-defined power series. Indeed the coefficient of zk in it is merely a polynomial in x . The coefficient of xazbin q is clearly the number of rooted near-triangulations N such that m ( N ) = a and t ( N ) = b. We define h(z) as the coefficient of x 2 in q. We can therefore write h(z)=

1

Z'(N),

N

where the summation is now restricted to those rooted near-triangulations N for which m ( N )= 2. Among these we note the degenerate near-triangulation, which provides h with a unit constant term. There is a 1- 1 correspondence between the rooted non-degenerate near-triangulations N satisfying m ( N ) = 2 and the general rooted triangulations T (Fig. 2). From such an N we get the corresponding T by deleting from its graph the non-root outer edge. We then have t ( N ) = 2 n ( T ) , where 2n(T) is the total number of faces of the triangulation T. We can now rewrite (2) as h(z)= 1

+I T

(3)

z2"(T).

Fig. 2.

W . T Tutte

246

Accordingly the number of rooted general triangulations with 2n faces is the coefficient of z2"in h (n>0). We now try to obtain a functional equation for 4 and h. First we separate the rooted near-triangulations N into three classes, numbered I, I1 and 111. We then attempt to evaluate the contribution to 4 of each class separately, in terms of the functions q and h. We denote these contributions by C(I), C(I1)and C(II1) respectively. Finally we add the three contributions and equate their sum to 4. Class I consists solely of the degenerate rooted near-triangulation. Evidently (4)

C(1)=x2.

Let N be any non-degenerate rooted near-triangulation. Let V be its root-vertex and let W be the other end of its root-edge. Then the root-edge is incident with a single inner face V W X . We assign N to Class I1 if X is incident with the outer face, and to Class I11 otherwise. Fig. 3 shows a typical member N of Class 11. Let us now rub out the root-edge in this diagram. We then see N fall apart into two near-triangulations N 1 and N 2 . We can describe these as rooted, taking their rootedges to be the edges V X and X W respectively of the old inner face V W X . V can be the root-vertex of N1,and X that of N , . We may find that N1 or N , (or both) is degenerate. It is clear that we can choose N so as to get any desired rooted near-triangulations as N1 and N 2 . Moreover if we know N1 and N 2 we can reconstruct N uniquely. We can therefore evaluate C(I1) as a sum, over all ordered pairs (Nl, N , ) of rooted neartriangulations N1 and N 2 , in the following way.

C(II)= x - 1zq2. Fig. 4 shows a typical member N of Class 111. If we now delete the root-edge, merging the old inner face VW X with the old outer face to make a new outer face, we obtain a new near-triangulation N'. We can call this rooted, taking the edge V X as its root-edge, and V as its root-vertex.

v

w

Fig. 3.

Counting rooted triangulations

241

Fig. 4.

It is clear that when N' is given the structure of N is uniquely determined. Moreover any rooted near-triangulation M can appear as N', provided only that m ( M ) is at least 3. Accordingly we can write

=I

~(111)

X m ( M ) - 1 Zr(M)+ 1

M

where M is restricted as stated. Thus C(III)= x - ' z ( q - x2h). We now get our functional equation by writing

+ + = x 2 + x - 'zq2 + x -'z(q - x2h).

4 = C(1) C(I1) C(II1)

It is convenient to write the equation as follows. zq2

+(z - x)q + x 3 -x2zh =0.

(7)

Equations of this kind are of common occurrence in the theory of planar enumeration. Basically they are quadratics in an unknown function q of two variables x and z, but they are complicated by the presence of an unknown function h of z, bearing some simple relation to q. In this case h is the coefficient of x 2 in q.

3. Solution of the equation We can begin our study of Equation (7) by noting the possibility of a term-by-term solution for q and h in powers of z. First we write q and h as power series in z as follows. q = 40 +q1z q2z2 * . . , (8)

+

+

h = ho+ h l z + h2Z2 + * . . .

(9)

By the definitions of q and h each coefficient qj must be a polynomial in x with integral

248

W. 7: Tutte

coefficients,and each coefficient h j must be an integer. Moreover hj must be zero when j is odd.

Let us substitute from (8) and (9) in (7) and then pick out the coefficients of zo. We find that

This of course represents the contribution to q of the degenerate near-triangulation. Taking the coefficient of x 2 in qo we find that ho = 1. We can now proceed to the coefficients of z1 in (7). From them we obtain q;

+ qo - x q ,

- X 2 h o = 0.

Since qo = x 2 and ho = 1 it follows that q 1 = x3, and therefore hl = 0. We expected hl to be zero. q1 is the contribution to q of the simplest triangulation of all, the one defined by a single circuit of 3 edges. This triangulation has just two faces, the inside and outside of the circuit. We can carry on in this way, getting q2 and h2 from the coefficients of z2 in (7), q3 and h3 from the coefficients of z3, and so on as far as time and patience permit. A solution of this kind is not usually considered satisfactory; we want information about hln for all values of n, both large and small. But the process is of theoretical importance; it provides us with an existence theorem asserting that Equation (7) has a unique solution for the pair of power series q and h. In some analogous cases researchers have made more practical use of the term-byterm solution. In the present example they might determine h2, as far as, say, h16. With luck they might guess the general formula for h2, from these numerical results. Assuming this formula they could solve (7) as an ordinary quadratic for q. If one solution for q came out as a genuine power series, with no negative indices, they need go no further. Because (7) has only one solution in power series they would know their result to be correct. Many of the earlier papers on planar enumeration solve their equations in this way. We can however solve (7) by a method involving no guesswork. We first write the equation as (2zq(x, Z )

+ z -x

) =~D,

(11)

where D=(z-x)’

-4zx3+4z2x2h.

If 5 is a power series in z such that 2zq(5, z )

+ 2 - 5 =0,

(13)

then it follows from (1 1) that D and dD/dx both vanish when 5 is substituted for x . We thus have two equations from which we can determine the two unknowns 5 and h. From h we can get q by solving (7) as an ordinary quadratic, taking that root which is a genuine power series. In practice we solve our two equations not for 5 and h in terms of z, but for z and h in terms of 5. The results can be expressed as follows.

Counring rooted triangulations

249

z = t(1 - 257, z2h= t2(1 - 35'). We can eliminate t2 from these two equations by using Lagrange's Theorem. We then find that

So the number hln of rooted general triangulations with 2n faces is as follows (see (3)): 2"(3n)! h 2 n - ( n + 1)!(2n+ l)!' The method described above raises a few questions of rigour. In particular we ought to insert an existence theorem stating that there is a power series 5 in z satisfying (13). It is not difficult to do this by exhibiting a procedure for the term-by-term solution of (13), whereby the coefficients in t are obtained in terms of the coefficients in q. Next one can imagine queries as to whether we have unjustifiably assumed the convergence of t, or even of q and h. The answer is that we can avoid such questions by working only in the algebra of formal power series. I maintain that Lagrange's Theorem is really a proposition of this algebra, and should be proved as such.

4. Stricttriangulations

It is possible to give a theory of strict triangulations analogous to the above theory for general ones. This is done for example in [4]. But it is now simpler to deduce a formula for strict triangulations from the results already obtained for general ones. Let us write

where S denotes any strict triangulation and 2 4 s ) is the number of its faces. Given a strict triangulation S we can imagine a cut being made along one of its edges, the two sides of the cut being separated slightly so as to enclose a digon, and the digon being filled with some near-triangulation N satisfying m ( N )= 2. The resulting figure is a triangulation T of the general kind. For example the first diagram in Fig. 1 can be obtained in this way from the strict triangulation of two faces only. We can regard T as rooted, having the same root-vertex and root-edge as S. If it is the root-edge of S that is cut, then the new root-edge is taken to be the side of the cut incident with the outer face. Let us suppose that the above operation is applied to all the k = 3 4 s ) edges of S. We enumerate the edges as A A2, . . . ,Ak.For each suffix j we distinguish one end of A j as 5,and one side of the resulting digon as Ei We denote by N j the near-triangulation used to fill this digon. We can describe N j as rooted, arranging that the rootvertex is vj and the root-edge Ei. We may want to leave some edge A j uncut. We can do so, explaining this in terms

W. 7: Tutte

250

of the above theory by saying that A j has indeed been cut and that the degenerate near-triangulation has been inserted into the resulting digon. The net result of the operation is a general triangulation I; uniquely determined by the sequence ( N l , N 2 , .. . , N k ) .We say that T is derived from S . We can check that any general triangulation T can be derived from some strict S in this way, and that this S is uniquely determined. To find S we make a list of those separating digons of T that are separated from the outer face by no other digons. For each such digon we delete the structure in its interior and identify its two sides so as to form a single new edge. The result is S. We now study the contribution C(S)to the sum in (3) of the general triangulations T derived from a given strict triangulation S. Initially we can write C(S)= Z 2 n ( S ) + r ( N ~ ) + r ( N z ) + . .+, t ( N k )

1

where the slim is over all sequences ( N l ,N 2 , . . . , Nk)ofk = 3 4 s ) rooted near-triangulations N j . This formula is conveniently rewritten as C(S)= Z2n(S)h3n(S) (18) by (2). Hence

S

h = ~ ( z h ~ ’ ~ by ) , (17).

(19)

The problem of enumerating rooted strict triangulations is now reduced to that of solving the functional equation (19) for the power series g , h being known. To solve (19) we first adopt Z = ~ has our ~ new / ~basic variable. Using (14) and (15) we express Z 2 and g ( Z ) , that is h(z), in terms of the parameter 8 = t2.The results are as follows.

We now have to eliminate 8 from these two equations. The problem can be simplified by a change of parameter. We put

and have 2 2

=~

-A),

3 ( 1

g(z) = 4 3 -2 4 .

An application of Lagrange’s Theorem now determines the coefficient g z n of Z 2 ” in g ( Z ) ,for n > 0, as follows. g2n =

2(4n - 3) ! n!(3n- l ) !

This is the number of combinatorially distinct rooted strict triangulations with 2n faces.

Counting rooted triangulations

251

5. sipletrirrngulrrtioas We can go from the strict to the simple triangulations in much the same way as we went from the general triangulations to the strict ones. Consider a rooted simple triangulation U , with n( U ) > 1. We can imagine one of its inner faces being filled with some strict triangulation X . For example the second diagram of Fig. l can be obtained from the tetrahedral triangulation by such an operation. Let us suppose that the operation is applied to all the k = 24s)- 1 inner faces of U . We enumerate these inner faces as F1,FZ,. . . , Fk.For each suffix j we distinguish one vertex of F j as 5 and one edge of F j , incident with 5,as E,. We denote by X j the strict triangulation used to fill F j . We can treat X j as rooted, arranging that the root-vertex is 5 and the root-edge E,. We may want to leave some inner face F j of U unsubdivided. We can do this, saying that we have inscribed in it the strict triangulation of 2 faces. As with the other inner faces of U the outer face of the inscribed triangulation is ignored. The net result is the conversion of U into a rooted strict triangulation 7;determined . . . , X k ) . We say that T is derived from U. uniquely by the sequence (Xl, Xz, We can check that any strict triangulation 7; with n(T)>1, can be derived from some simple U , with n( V)> 1, in this way. To find U we make a list of those separating triangles of T that are not separated from the outer face by any other triangles. For each such separating triangle we delete the structure in its interior and treat the triangle as a new inner face. The result is U,and we note that U is uniquely determined by T Necessarily we have n(U)> 1 since the boundary of the outer face of T cannot be a separating triangle. We take note of the exceptional character of the simple triangulation J of two faces only. We can indeed subdivide its inner face so as to get a strict triangulation T satisfying n(T)> 1. But if we now submit T to the procedure for recovering a simple triangulation U , from which T is derived, we obtain a U for which n( U )> 1 and which is therefore different from J. We write our enumerating function for rooted simple triangulations as F(z)=

z~"'~), U

where U can be any rooted simple triangulation satisfying n(U)> 1. We now study the contribution C ( U )to the sum in (17) of those rooted strict triangulations that are derived from a given rooted simple triangulation U satisfying n(U)> 1. Initially we have C(u)=czl- k + Zn(X,)+ Zn(X2)+." + Zn(Xk) where k = 2n(U)- 1 and the sum is over all sequences (Xl, Xz, . . . ,x k ) of rooted strict triangulations X j . By (17) we can write the formula as follows: C(U )= z{z - 1(g - 1)}2"'U' - 1.

Allowing for the rooted triangulation of two faces we can now write (17) as z2F{z- '(g(z) - l)} g(z)-l=zZ+ g(d- 1

(23)

252

W. 7: T i m

The problem of enumerating rooted simple triangulations is now reduced to that of solving the functional equation (24) for F, the function g - 1 of z being known. By analogy with our procedure for (19) we take t=Z-l(g-l) as our new basic variable. The functional equation can now be written as t 2 = (9 - 1)

+ F( t ) .

(25)

We can get parametric equations for F by using (21), with Z replaced by z, to express t and F ( t ) in terms of A. We find that f ( t ) = ( l -A)(l-21)

(26)

and t 2 = r 3 ( 1 -n)(i -2112,

where f(t)=F(t)-t2.

By analogy our next step should be to look for a simplifying change of parameter. I have used the transformation q=-

-A 1-21’

finding that

and t2=

-q-3(1

-rl).

It is possible to solve these equations, for f as a power series in t 2 , by using Lagrange’s Theorem. This is done in [4], and the result is somewhat simplified in [ 5 ] . But the formula is more complicated than (16)and (22),and not helpful when we need numerical results. However it is transformed in [5] into a simple recursion formula. Letf, denote the coefficient of t2” inf, so that when n is 2 or moref, is the number of rooted simple triangulations of 2n faces. Let us also write S, = 8n - 5. It is found that (30) for any non-negative integer n. Starting with the observation that fo = 0 we can use (30)to calculate the coefficientfj up to as large a value o f j as we please. There is a more direct way of getting to Equation (30).It uses (28) and (29) to verify thatf satisfies the differential equation (8t2+32)-+(5-44-’)f=5+7q. df do2) It then evaluates q by applying Lagrange’s Theorem to (29):

Counting rooted triangulations

253

Equation (30)can now be deduced from (31)and (32)by equating coefficients of t2"' in (31). The purpose of the present paper, as stated in the summary, is now achieved. More information about the enumeration of triangulations and near-triangulations can be found in papers of W. G. Brown and R. C. Mullin [ l , 2,3].

Refereaces [I] [2] [3] [4] [5]

W. G . Brown, Enumeration of triangulations of the disk, Proc. London Math. SOC. 14 (1964) 746-768. R . C. Mullin, Enumeration of rooted triangular maps, Amer. Math. Monthly 71 (1964) 746-768. R . C. Mullin, On counting rooted triangular maps, Canad. J. Math. 17 (1965) 373-382. W. T. Tutte. A census of planar triangulations, Canad. J. Math. 14 (1962) 21-38. W. T. Tutte, A census of planar maps, Canad. J. Math. 15 (1963) 249-271.

Received 30 January 1980.

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Annals of Discrete Mathematics 12 (1982) 255-263

@ North-Holland Publishing Company

NON-HAMILTONIAN SIMPLE PLANAR GRAPHS Joseph ZAKS University of Haija, Israel

Dedicated to A. Kotzig on his sixtieth birthday The purpose of this paper is to present various results on the said topic, some of which were announced in [24], while others extend earlier results [22, 231 of the author. In Section 1 we present a non-Hamiltonian simple 3-connected planar graph in which the largest face is a 'I-gon; Section 2 extends earlier results on non-Hamiltonian simple 3-connected planar graphs in which there are only two types of faces; Section 3 deals with a curious property of a family of so&e simple 3-connected planar graphs having only quadrangles and hexagons; we finish by two short remarks, one on a theorem of A. Kotzig and another on covering graphs by their Hamiltonian circuits.

1

Let G(m)denote the family of all simple ( = 3-valent) 3-connected planar graphs in which the face of largest number of vertices is an m-gon. Various non-Hamiltonian members of G(8) are known, due to Grinberg [9], Tutte [20], Grunbaum [ll] and Zaks [22]. It had been conjectured by Griinbaum-Walther [14] (see also Ewald [3], Goodey-Rosenfeld [8] and Goodey [7]) that all the members of G(7) and all the members of G(6) are Hamiltonian. While the case of G(6) is still open, we settle the case of G(7) in the following theorem :

Theorem 1. There exists a non-Hamiltonian member of G(7) having 54 vertices. Strengthening this, we have:

Here p(G) is the shortness coefficient of a family G of graphs, defined (see [14]) as lim inf h(G,)/o(G,),taken over all sequences (G,),"= of members of G, with u(G,)+ co as n + m , and where h(G,) is the size of a longest circuit in G ; p* is defined in a similar way for the size of longest paths. Following Bosak [13 (see also [17,9,6,2,20,213) let an edge E of a graph G be called an a-edge if E belongs to every Hamiltonian circuit of G. We need the following lemma:

Lemma 1. If a graph G consists of two subgraphs H I and H1. which are connected as shown in Fig. 1, and if (7; 8) is an a-edge of G, then so is the edge (11; 12). 255

256

J . Zaks

Fig. I .

Proof of Lemma 1. Suppose that ( 1 1; 12) is not an a-edge of G ; then there exists a Hamiltonian circuit C of G which contains the edges (5; 6) and (17; 18)and which does not contain the edge (11; 12); this implies (4; l l ) u ( l l ; lO)cC, since 11 EC.( 5 ; 17) 4 C , since otherwise C will be disconnected; therefore (4; 5)u(16; 1 7 ) c C ; likewise, (10; 16) @ C, because otherwise C will be disconnected, hence (9; 10)u(15; 16)cC. (3; 4) e C , because (4; 5)u(4; l l ) c C , hence (2; 3)u(3; 15)cC, implying that (14; 15) $ C , hence (8; 14)u(13; 14)cC. (2; 9) 4 C , because otherwise C will be disconnected, hence (1 ; 2)u(8;9 ) c C , implying that (7; 8) 4 C ; this is a contradiction to the assumption that (7; 8) is an a-edge of G ; hence ( 1 1; 12) is an a-edge of G and the proof is complete. Proof of Tbeorem 1. The non-Hamiltonian member Go of G(7) is shown in Fig. 2; it can be considered a modification -of the smallest-known non-Hamiltonian simple polytopal graph on 38 vertices, due independently to Lederberg [171, Basak [ 11 and Barnette (see [l l]), which modified Tutte's [18] graph on 46 vertices. Clearly Go E G(7). To show that Go is non-Hamiltonian, it follows by Tutte's [ 181 argument (see also [lo]) that the edges El and E 2 are a-edges of G o ; by Lemma 1 it follows that E 3 and E4 are also a-edges of G o ; it then follows, as in the case of the Bosak-Lederberg-Barnette's graph, that Go is non-Hamiltonian. This completes the proof of Theorem 1. Proof of Tbeorem 2. Define a sequence (H,),"= as follows; H = Go (of Fig. 2); H,+ is obtained from H , in the following way: pick three vertices of H, at which a quadrangle meets two pentagons, so that all the nine faces are different (as are the vertices

Fig. 2.

257

Non-Hamiltonian simple planar graphs

Fig. 3

6

Fig. 4.

V,, V2 and V3 in Fig. 2 ) ; replace each one of them as shown in Fig. 3 and replace each one of the new inner vertices w by a copy of Go - V,, as shown in Fig. 4. Clearly v(H,)=62 198(n- 1); A simple path in H, cannot contain all the vertices of the three new copies of Go - V , , for otherwise it will have no end points in at least one copy of Go - V,, having just two end points; therefore it misses at least one vertex in that copy; it follows that h*(H,)<62+( 198 - 1Xn - l), and since H,E G(7)for all n, it follows that p*(G(7))< 1 ; here h*(G)denotes the size of a longest path in G. This completes the proof of Theorem 2.

+

2

Following [23], let G(n, k) denote the family of all the simple 3-connected planar graphs having just n-gonal and k-gonal faces. Extending an earlier result [22] in which we showed that G(5, 8) contains non-Hamiltonian members, we [23] proved that p(G(5, k ) ) < 1 holds for all k 2 I I , and p*(G(5, A))< I holds for all k > 1 I except possibly for the values k = 14, 17,22 and k = 5n 5 for all n 2 2. All members of G(3,6) and of G(4,6) are Hamiltonian by [6,7]. We show here the following:

+

Tbeorem 3. p*(G(5, 8))< 1 .

"heorem 4. p*(G(5,9))< I . We will present here Ewalds (in private communication) examples of members of G(5,5n 4) and of G(5,5n 5) which admit no Hamiltonian paths, for all n 2 2.

+

+

Proof of Theorem 3. The proof is similar to the previous one, using the non-Hamiltonian member of G(5,8) of [22] and the fact that it has many vertices at which three pentagons meet; attaching two such copies as shown in Fig. 5 yields another member of G(5, S), so it follows that p*(G(5, 8))< 1. We omit the details.

J . Zaks

258

Fig. 5 .

P

v Fig. 6

For the proof of Theorem 4,that p*(G(5,9))<1, we need the following:

Lemma 2. The graph P of Fig. 6 is non-Hamiltonian.

First proof of Lemma 2.' Assume that P has a Hamiltonian circuit C ;from the possible behaviour of C in each one of the three blocks of pentagons, touching the outer face, as proved in [S], one gets easily a contradiction. We omit the details, and present a

Seed proof of Lemma 2. Assume P has a Hamiltonian circuit C ; P has seven 9-gons besides pentagons, hence by Grinberg's ([9], see also [111) criterion it follows that there exist nonnegative integers p$ and p $ such that pb + p $ = 7 and p $ = p: (mod 3 ) . Therefore { pb, p $ ) = { 2, 51, and the circuit C has two 9-gons on one of its sides and five of them on the other side; however, every five 9-gons of P separate one of the three outer blocks of pentagons from the other two, hence C is not Hamiltonian. This completes the proof of Lemma 2.

Proof of Theorem 4. Let (P,),"_ be defined as follows: PI = P of Fig. 6; P , , is obtained from P, by replacing the ten marked vertices of the graph Q of Fig. 7 with copies of P,, in which a vertex where three pentagons meet is removed, as shown in Fig. 8. P , E G(5,9)for all n,because the ten vertices of Q were so chosen as to meet each one of the hexagons of Q exactly once. Every simple path in P , misses at least one vertex of at least eight copies of P , - which do not contain the end points of the path; it follows that p*(G(5,9))< 1, and the proof of Theorem 4 is complete. Ewalds example of members of G(5,5n+ 4)and G(5,5n+ 5 ) which admit no Hamiltonian paths are reproduced here in Figs. 9 and 10, respectively; for the meaning of the

Non-Hamiltonian simple planar graphs

259

Q

@+a-(Fig. 7

Fig. 8

Fig. 9

Fig. 10.

260

J . Zaks

notation of the blocks 3 , 5 n - I, 3 , 5 n - 1, and the reasons which imply that these graphs admit no Hamiltonian paths, see [23]. In Fig. 10, one starts with the graph of the dodecahedron.

3 Let Q denote the family of all simple 3-connected planar graphs having just quadrangles and hexagons and so that the six hexagons appear in three pairs having each an edge in common. All the members of Q are well known, since they are obtainable from the collection of all simple planar multigraphs having just digons and hexagons, fully described in [lS], by removing one edge of each one of the digons. We have the following curious property, which might be used in settling Barnette’s conjecture:

Theorem 5. I f a graph G i s in Q and it has an odd number of hexagons, and if C is a Hamiltonian circuit in G , then card(E(C)n{El, E2, E3))=0 (mod 2), where El, E2 and E3 are the edges of G in common to two quadrangles.

Proof of Theorem 5. Suppose card(E(C)n(E,, E l , E3))=3; the graph G’, obtained from G by adding two 2-valent vertices on eachof El, E 2 and EJ,is clearly Hamiltonian and the number p6(G’)=p,(G)+6 of hexagons of G’ is clearly odd, while G’ has no other types of faces. It follows from Grinberg’s criterion that there exist nonnegative integers pk and p z such that pk + p‘; is odd and 4pk = 4&, which is impossible. If the Hamiltonian circuit C contains just one of these edges, then it follows from the

Fig. 11.

Non-Hamiltonian simple planar graphs

26 1

proof of Grinberg’s theorem that both p i and p i are odd, and p i + pi =6 and 2pi + 4p; = p i 4pz. If, say, p i = 1 and pE = 5, then it follows that p; = p ; + 2, which is impossible since p; p%is odd; if p i = p i = 3, then it follows that p b = pz which is again impossible. This completes the proof of Theorem 5.

+

+

An example of a graph to which Theorem 5 is meaningful is given in Fig. 11; we had an interesting experience asking colleagues and students to prove the assertion of Theorem 5 concerning this example, without mentioning p6. 4

We turn next to the problem of covering graphs by their Hamiltonian circuits, which is possible if and only if the graph has no b-edges, edges belonging to no Hamiltonian circuits. The prism over a (2k + 1)-gonis the union of at least k + 1 of its Hamiltonian circuits, but not fewer (prove it); in fact, it is not hard to prove the following:

Theorem 6 . Let G denote the family ofall the n-valent 3-connected planar graphs which are the union of their Hamiltonian circuits,for n = 3,4 or 5 ; there exists no k for which every member of G is the union ofalready k of its Hamiltonian circuits. Every 4-connected planar graph is Hamiltonian, due to Tutte [191. We wish to raise, jointly with P. Erdos, the following problem:

Problem. Is there a k for which every 4-valent 4-connected planar graph is the union of k of its Hamiltonian circuits? Is there a k(n) for which every 4-connected planar graph, having all of its valences to be at most n, is the union of already k(n) of its Hamiltonian circuits? 5

We close with a remark on a beautiful theorem of A. Kotzig ([ 161, see also [121 and [131)which states that every planar 3-connected graph has an edge in which the sum of the degrees of its end points is at most 13. Denote by ej,kthe number of edges of a graph which have a j-valent and a k-valent end point. Grunbaum [121 proved that

“If G is a triangulation of the sphere with at least four vertices and ifj + k < 12 implies ej,k= 0, then e3,10>,60’. Actually, Griinbaum’s [121 proof can be simplified, avoiding the many iterations, as follows: where v l 0 is the number of 10-valent vertices, is mentioned by Kotzig [ 161 (see also [I21 and [13]); another inequality, follows from Kotzig’s general observation that a k-valent vertex can have at most

J. Zaks

262

[$I 3-valent neighbors (if the graph is not K4).It follows that 5e3,lo>, 120+ 15U,o>, 120+3e3,10,hence 2e3,,0>,120 or e3,10>,60.This simplification is implicitly included in Jucovii.‘s [4] and was remarked by one of my students. Added io proof 1. Theorems 1 and 2 were improved and extended, as follows: (a) p(G(5,7))< 1 (P.J. Owens, Discrete Math. 36 (1981)227-230). (b) G(5,lO)contains non-Hamiltonian members (P.J. Owens, Shortness parameters of families of regular planar graphs with three types of faces, Discrete Math., to appear). (c) G(4, k) contains non-Hamiltonian members, for all odd k, k > 11 (H. Walter, Discrete Math. 33 (1981) 107-109); it is also mentioned there that the shortness exponent of G(4, k) is less than one for all odd k, k 2 17 (to appear). (d)p*(G(3,k)) 1 for all k, 8
-=

References [ I ] J. Bosak. Hamiltonian lines in cubic graphs, in: Proc. Intern. Seminar on Graph Theory and its Applications. Rome (1966) pp. 35-46. [2] J . W. Buttler. Hamiltonian circuits on simple 3-polytopes, J. Combinatorial Theory B 15 (1973) 69- 73. 131 G . Ewald, Some results on polytopal graphs, Conference on Geometry, Haifa, 1975. J. of Geometry 7 (1976)6- 8. [4] E. Jucovit, Strengthening of a theorem about 3-polytopes, Geometriae Dedicata 3 (1974) 233-237 [S] P. R. Faulkner and D. H. Younger. Non-Hamiltonian cubic planar maps, Discrete Math. 7 (1974) 67- 74. [6] P. R. Goodey. A class of Hamiltonian polytopes, J. Graph Theory 1 (1977) 181-185. 171 P. R. Goodey. Hamiltonian circuits in polytopes with even sided faces, Israel J. Math. 22 (1975) 52-56. [8] P. R. Goodey and M. Rosenfeld, Hamiltonian circuits in prisms over certain simple 3-polytopes, Discrete Math. 5 (1973) 389-394. 191 E. J. Grinberg, Plane homogeneous graphs of degree three without Hamiltonian circuits, in: Latvian Math. Yearbook 4 (Izdat. “Zinatne” Riga, 1968)pp. 51-58 (in Russian). [ 101 B. Grunbaum, Convex Polytopes(Interscience, New York, 1967). [ I l l B. Griinbaum, Polytopes, graphs and complexes, Bull. Amer. Math. SOC.76 (1970) 1131-1201. 1121 B. Griinbaum, New views on some old questions of Combinatorial Geometry, in: Colloq. Intern. Teorie Combinatorie, Rome (1976) pp. 451 -468. [I31 B. Griinbaum, Polytopal graphs, in: R. D. Fulkerson, Ed., Studies in Graph Theory and its Applications, pp. 201 -224.

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[I41 B. Griinbaum and H. Walther, Shortness exponents of families of graphs, J. Combinatorial Theory A 14 (1973)364-385. [ 151 B. Griinbaum and J. Zaks, The existence of certain planar maps, Discrete Math. 10 (1974)93-115. [I61 A. Kotzig. Prispevok k teorii Eulerovskych polyedrov, Mat.-Fyz. Casopis Slovensk. Akad. Vied. 5 (1955)101-I13 (in Slovak, Russian summary). [ 171 J. Lederberg, Systems of organic molecules, graph theory and Hamiltonian circuits, Instrumentation Res. Lab. Report No. 1040, Stanford Univ., CA (1966). [I81 W. T. Tutte. On Hamiltonian circuits, J. London Math. SOC.21 (1946)99- 101. [I91 W. T. Tutte, A theorem on planar graphs, Trans. Amer. Math. SOC.82 (1956)99-1 16. [ZO] W. T. Tutte. Non-Hamiltonian planar maps, in: R. C. Read, Ed., Graph Theory and Computing (Academic Press, New York, 1972)pp. 295-301. [21] J. Zaks, Pairs of Hamiltonian circuits in 5-connected planar graphs, J. Combinatorial Theory 21 (1976)116- 131. [22] J. Zaks, Non-Hamiltonian non-Grinbergian graphs, Discrete Math. 17 (1977)317-321. [23] J. Zaks, Non-Hamiltonian simple 3-polytopes having just two types of faces, Discrete Math. 29 (1980)87-101. [24] J. Zaks, Few Hamiltonian results, Notices Amer. Math. SOC.25 (1978)p. A-270; abstract 753-A39. Received 15 September 1979; revised 16 January 1980.

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Annals of Discrete Mathematics Previous Volumes in this Series Vol. 1:

Studies in Integer Programming edited by P.L. HAMMER, E.L. JOHNSON, B.H. KORTE andG.L. NEMHAUSER 1977 viii + 562 pages

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Advances in Graph Theory edited by B. BoLLoBAS 1978 viii + 296 pages

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Discrete Optimization, Part I1 edited by P.L. HAMMER, E.L. JOHNSON and B. KORTE 1979 vi + 454 pages

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