COMBINATORIAL DESIGN THEORY
NORTHHOLLAND MATHEMATICS STUDIES Annals of Discrete Mathematics (34)
149
General Editor: Peter L. HAMMER Rutgers University, New Brunswick, NJ U.S.A.
Advisory Editors C. BERGE, Universitede Paris, France M. A. HARRISON, University of California, Berkeley, CA, U.S.A. V: KLEE, University of Washington, Seattle, WA, U.S.A. J. H. VAN LINT CaliforniaInstitute of Technology, Pasadena, CA, U.S.A. G. C.ROTA, Massachusetts Institute of Technology,Cambridge, MA, U.S.A.
NORTHHOLLAND AMSTERDAM
NEW YORK
OXFORD .TOKYO
COMBINATORIAL DESIGN THEORY
Edited by
Charles J. COLBOURN Department of Computer Science University of Waterloo Waterloo, Ontario Canada
Rudolf MATHON Department of Computer Science University of Toronto Toronto, Ontario Canada
1987
NORTHHOLLAND AMSTERDAM
NEW YORK
OXFORD .TOKYO
OElsevier Science Publishers B.K, 1987
All rights reserved. No part of this publication may be reproduced, stored in a retrievalsystem, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the copyright owner.
ISBN: 0444703284
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Data
C o m b i n a t o r l a l design t h e o r y / edited by C h a r l e s J. C o l b o u r n , R u d o l f Mathon. p. cm.  (NorthHolland m a t h e m a t i c s s t u d i e s : 149) ( A n n a l s o f d i s c r e r e m a t h e m a t i c s ; 34) F e s t s c h r i f t f o r A l e x Rosa. I S B N 0444703284 (U.S.) 1. C o m b i n a t o r i a l d e s i g n s and c o n f i g u r a t i o n s . 2. R o s a , Alexander. I. Colbourn. C. J. ( C h a r l e s J . ) , 1953. 11. Mathon. R. A. 111. Rosa. Alexander. IV. Series. V. S e r i e s A n n a l s of d i s c r e t e m a t h e m a t i c s ; 34. 0A166.25.C652 1987 872231 1 511'.6dc19 CIP
PRINTED IN THE NETHERLANDS
V
PREFACE Combinatorial design theory has experienced an explosive growth in recent years, both in theory and in applications. This volume is dedicated t o one individual who has played a major role in fostering and developing combinatorial design theory, Alex Rosa. Although Alex is a young man at the middle of his career, he is truly an individual worthy of such an honour. Thestrength anddiversity of his contributions t o the theory of Steiner systems over the past twenty years have directed research into new and exciting avenues, and thereby contributed dramatically t o the vibrancy of research on designs found today. His research has encompassed difference methods, colourings, decomposition and partitioning, recursive and direct construction methods, analysis of designs, and relationships with graph theory and with geometry. But Alex’s contributions extend beyond this impressive research. Virtually all design theorists have profited from Alex’s astonishing ability to organize and disseminate research information. For the past fifteen years, he has maintained a bibliography on Steiner systems, which has proved a great boon to researchers; at the same time, he has unselfishly taken much of his time t o help researchers by providing information about a vast range of queries on combinatorial designs. Alex’s willingness and ability t o help others make him a truly special individual. In 1987, Alex Rosa is celebrating his fiftieth birthday. We take this opportunity t o express our thanks to Alex for his contributions thusfar, and to honour him on the occasion of his fiftieth birthday. This volume is a collection of fortyone research papers on combinatorial design theory and related topics. They extend the current state of knowledge on Steiner systems, Latin squares, onefactorizations, block designs, graph designs, packings and coverings;they develop recursiveanddirect constructions, analysis techniques, and computational methods. Hence the collection reflects the current themes in combinatorial design theory, and captures the multifaceted nature of the field. It is not possible here to summarize all of the contributions contained in the papers of this volume, so we simply remark that collectively the papers represent advances on a number of important problems.
vi
Preface
We hope that readers of the volume find the individual papers as useful and as enjoyable as we have. Charles J. Colbourn Waterloo, Ontario July, 1987
Rudolf A. Mathon Toronto, Ontario July, 1987
vii
ACKNOWLEDGEMENTS In a project such as the publication of this volume, there are always many people without whom the project would never reach fruition. We want to thank all of the people who provided assistance. Primarily we thank the authors for such an enthusiastic response, and for helping us meet a very tight schedule. We also thank the referees; we list their names here to acknowledge their invaluable assistance: J. Abrham, B.A. Anderson, F.E. Bennett, A.E. Brouwer, J.I. Brown, C.J. Colbourn, J.H. Dinitz, R. FujiHara, M. Gionfriddo, T.S.Griggs, J.J. Harms, A. Hartman, K. Heinrich, P. Hell, M. Jimbo, D. Jungnickel, E.S. Kramer, D.L. Kreher, C.C. Lindner, S.S. Magliveras, R.A. Mathon, W. McCuaig, B.D. McKay, E. Mendelsohn, D.M. Mesner, W.H. Mills,R.C.Mullin, K.T. Phelps, V. Rodl, F.A. Sherk, D.R. Stinson, L. Teirlinck, S.A. Vanstone, and W.D. Wallis. We also thank all of those who helped with preparing the final manuscript, including many of the authors, but primarily Karen Colbourn and Zoe Kaszas. Finally, we thank Peter Hammer and the mathematics staff at NorthHolland for supporting the project and providing excellent technical assistance, as always.
This Page Intentionally Left Blank
ix
CONTENTS Preface Acknowledgements The Existence of Symmetric Latin Squares with One Prescribed Symbol in Each Row and Column L.D. ANDERSEN and A.J.W. HILTON
V
vii
1
A Fast Method for Sequencing Low Order NonAbelian Groups B.A. ANDERSON
27
Pairwise Balanced Designs with Prime Power Block Sizes Exceeding 7 F.E. BENNETT
43
Conjugate Orthogonal Latin Squares with EqualSized Holes F.E. BENNETT, L. WU, and L. ZHU
65
On Regular Packings and Coverings J.C. BERMOND, J. BOND, and D. SOTTEAU
81
An Inequality on the Parameters of Distance Regular Graphs and the Uniqueness of a Graph Related to M ~ C J A.E. BROUWER and E.W. LAMBECK
101
Partitions into Indecomposable Triple Systems C.J. COLBOURN and J.J. HARMS
107
Cubic Neighbourhoods in Triple Systems C.J. COLBOURN and B.D. McKAY
119
The Geometry of Subspaces of an S(X;2,3,v) M. DEHON and L. TEIRLINCK
137
X
Contents
On 3Blocking Sets in Projective Planes M.J. DE RESMINI
145
Star SubRamsey Numbers P. FRAISSE, G. HAHN, and D. SOTTEAU
153
Colored Packing of Sets P. FRANKL and Z. FUREDI
165
Balanced Room Squares from Finite Geometries and their Generalizations R. FUJIHARA and S.A. VANSTONE
179
On the Number of Pairwise Disjoint Blocks in a Steiner System M. GIONFRIDDO
189
On Steiner Systems S(3,5,26) M.J. GRANNELL, T.S. GRIGGS, and J.S. PHELAN
197
Halving the Complete Design A. HARTMAN
207
Outlines of Latin Squares A.J.W. HILTON
225
The Flower Intersection Problem for Steiner Triple Systems D.G. HOFFMAN and C.C. LINDNER
243
Embedding Totally Symmetric Quasigroups D.G. HOFFMAN and C.A. RODGER
249
Cyclic Perfect One Factorizations of K2n E.C. IHRIG
259
On Edge but not Vertex Transitive Regular Graphs A.V. IVANOV
27 3
A Product Theorem for Cyclic Graph Designs M. JIMBO and S. KURIKI
287
A New Class of Symmetric Divisible Designs D. JUNGNICKEL
297
2(25,10,6)Designs Invariant under the Dihedral Group of Order Ten S. KAPRALOV, I. LANDGEV, and V. TONCHEV
301
Contents
xi
On the Steiner Systems S(2,4,25)Invariant under a Group of Order 9 E.S. KRAMER, S.S. MAGLIVERAS, and V.D. TONCHEV
307
Simple 5(28,6,h)Designs from PSL 2(27) D.L. KREHER and S.P. RADZISZOWSKI
315
The Existence of Partitioned Balanced Tournament Designs of Side 4n+3 E.R. LAMKEN and S.A. VANSTONE
319
The Existence of Partitioned Balanced Tournament Designs E.R. LAMKEN and S.A. VANSTONE
339
Constructions for Cyclic Steiner 2Designs R. MATHON
353
On the Spectrum of Imbrical Designs E. MENDELSOHN and A. ASSAF
363
Some Remarks on nClusters on Cubic Curves N.S. MENDELSOHN, R. PADMANABHAN, and B. WOLK
37 1
A Few More BIBD’s with k = 6 and X = 1 R.C. MULLIN, D.G. HOFFMAN, and C.C. LINDNER
379
Isomorphism Problems for Cyclic Block Designs K.T. PHELPS
385
Multiply Perfect Systems of Difference Sets D.G. ROGERS
393
Some Remarks on Focal Graphs G. SABIDUSSI
409
Some Perfect OneFactorizations of K14 E. SEAH and D.R. STINSON
419
A Construction for Orthogonal Designs with Three Variables J. SEBERRY
437
Ismorphism Classes of Small Covering Designs with Block Size Five R.G. STANTON Graphs which are not Leaves of Maximal Partial Triple Systems D.R. STINSON and W.D. WALLIS
441 449
xii
Contents
Symmetric 2(31,10,3) Designs with Automorphisms of Order Seven V.D. TONCHEV
461
Embeddings of Steiner Systems S(2,4,v) R. WE1 and L. ZHU
465
Annals of Discrete Mathematics 34 (1987) 126 0 Elsevier Science Publishers B.V. (NorthHolland)
1
The Existence of Symmetric Latin Squares with One Prescribed Symbol in each Row and Column L.D. Andersen Department of Mathematics Institute of Electronic Syst.ems Aalborg University Strandvejen 19 DK9000 Aalborg Denmark
A.J.W. Hilt on Department of Mathematics University of Reading Whiteknights Reading RG6 2AX United Kingdom Visiting: Division of Mathematics Auburn University Auburn Alabama 36849 U.S.A. TO ALfX R O S A
O N MIS 3ITTIETU BIRTUDAY
ABSTRACT Let P be a partial symmetric n X n Latin square in which up t o n +1 entries are specified such that there is at least one specified entry in each row and column. We say exactly when P can be completed t o an n X n symmetric Latin square L . This is the first part of a proof of the symmetric Latin square analogue of the Evans Conjecture. 1. Introduction A partial n X n Latin square P is an n X n matrix in which some cells may be empty and the nonempty cells will contain exactly one symbol, such that no symbol occurs more than once in any row or in any column. P is called symmetric if whenever a cell ( i , j ) contains some symbol, then cell (j,i)also contains the same symbol.
2
L.D.Andersen and A.J. W.Hilton
Our concern in this paper is to characterize those partial symmetric nXn Latin squares with up t o n + l cells occupied, satisfying the restriction that there is at least one occupied cell in each row and in each column, which can be completed t o a symmetric nXn Latin square. Before we state our main result we recall that in a symmetric Latin square of even side, each symbol occurs an even number of times on the diagonal, and in a symmetric Latin square of odd side, each symbol occurs exactly once on the diagonal. Both these facts are easy to see. We shall call the diagonal of a partial symmetric Latin square of side n admissible if the number of symbols occurring a number of times not congruent t o n modulo 2 is less than or equal t o the number of empty cells (so that the parity can be made right for every symbol with the ‘wrong’ parity). Let t(i) denote the number of times that a symbol i occurs on the diagonal. Then, if the symbol set is {I ,...,n}, the condition for admissibility can be written.
I{i:t(i) # n (mod 2)) Is n  g
t(i).
i1
For odd n , this simply means that a diagonal is admissible if and only if no symbol occurs more than once on it. Obviously, a partial symmetric Latin square cannot be completed t o a symmetric Latin square if its diagonal is not admissible. Our main result is the following: Theorem 1. Let n 2 3 and let P be a partial symmetric Latin square of side n with admissible diagonal; suppose also that there is at least one occupied cell in each row and in each column. Let c be the number of nonempty cells of P. If c = n , then P can be completed to a symmetric Latin square of side n if and only if P is not of the form of any of the Types El, 01 or 02 of Figure 1. If c = n + l , then P can be completed t o a symmetric Latin square of side n if and only if P is neither of the form of any of the types El, 01 or 02 with a further diagonal cell filled, nor of the form of the types E 2 or L5 of Figure 1.
Remark. No doubt the meaning of the phrase ‘of the form of’ is selfevident. But, t o be formal, a partial Latin square is of the form of one of these Types if it can be transformed t o a partial Latin square of one of these Types by permuting rows, permuting columns the same way, and relabelling the symbols. It is easy to see that the partial squares in Figure 1 cannot be completed symmetrically. In types E l and 01 the symbol 1 needs an extra diagonal occurrence but cannot get it; in Types E2 and 02 it is impossible t o make the symbol 1 occur in the first row; finally we leave it to the reader t o check that Type 155 cannot be completed. The analogous problem for Latin squares with no symmetry requirement to the problem considered here was settled by Chang 161; he characterized the possible diagonals of a Latin square. A completely different proof of Chang’s result was given by Hilton and Rodger [lo]. The result in this paper is a necessary preliminary t o a more general result [5] in which we characterize the partial symmetric nXn Latin squares with up t o n + l cells occupied which can be completed. For our proof of this more general result, we need to have the result in this paper proved separately, for the general method of that paper
Symmetric Latin Squares
n
even:
Type E l
n
3
odd:
Type E 2
even
1 does not occur
Type 01
Type 02
Figure 1
Type L 5
L.D. Andersen and A.J. W . Hilton
4
fails when every row and column has an occupied cell. This general question has been known and unsolved for a number of years. The analogous more general problem for Latin squares with no symmetry requirement was settled by Smetaniuk [12]and, independently, Andersen and Hilton 131. Smetaniuk only considered the case when n  1 cells were occupied; Andersen and Hilton considered the case when n cells were occupied. Damerell 181 extended Smetaniuk's method t o cover the case when n cells were occupied. Finally, Andersen [2] dealt with the case when n +1 cells were occupied. 2. EdgeColourings of Graphs In this section we discuss briefly edgecolourings of graphs, and the relationship between symmetric Latin squares and edgecolourings of complete graphs with loops. The reason for this is that the proof of Theorem 1 is easier t o present in terms of edgecolourings. An edgecolouring of a graph G is an assignment of a colour t o each edge and loop of G in such a way that all edges and loops incident with the same vertex have distinct colours. The chromatic i n d e z CHI'(G of G is the least integer k for which G has an edgecolouring with Ic colours. Let K , denote the graph obtained from the complete graph K , by adding a loop at each vertex. It is wellknown that CHI'(K,')=n.
2
A symmetric Latin square S of side n gives rise to an edgecolouring of K," as follows. Let the vertices of Kf be vl, * * ,v,. Define the colour of the edge viwi t o be the entry of cell (i,$ of S, and the colour on the loop on vi t o be the entry of cell ( i , i )of S, i#j, lsjsn. This clearly defines an edgecolouring of K,' with n colours, and it obviously works the other way round as well; it also works for partial symmetric Latin squares and edgecoloured subgraphs of K,' . In this connection, we note that if an edgecoloured subgraph of Kf corresponds to a partial symmetric Latin square P , then the number of filled cells of P is equal t o the degree sum
lsisn,
5 d(wi)=2 IE(G)I+ IL(G)I, where E ( G ) is the edgeset of G and L ( G ) the loopset
i1
of G (a loop is not an edge; a loop on vi contributes one t o the degree d ( v i ) of (vi).We shall say that G has admissible loopcolouring if the diagonal of P is admissible. In this terminology, Theorem 1 becomes: Theorem 2. Let n z 3 and let G be an edgecoloured subgraph of K i with an admissible loopcolouring; suppose also that each vertex of G has at least one edge or loop on it, and that G spans K i . Let t =2 lE(G) IL(G) If t = n , then the edgecolouring of G can be extended t o an edgecolouring of K i with n colours if and only if G is not any of the Types e l , 0 1 or 02 of Figure 2. If t =n + I , then the edgecolouring of G can be extended t o an edgecolouring of K i with n colours if and only if G is not any of the Types e l , 0 1 or 02 with a further loop added, nor of the form of Types e2 or 9 5 of Figure 2. When G is one of the edgecoloured graphs of Figure 2, possibly with a loop added in case e l , 0 1 or 0 2 we refer t o G as being a bad case.
I+
I.
Symmetric Latin Squares
n
even.
5
8;8
Type e l
odd number o f 1's
Type e 2
n
odd.
0;7
Type 01
1 does n o t occur t=n
Type 0 2
n = 5 t=n+l
Type 95
Figure 2 In G , a loop on a vertex with no edge of G incident with it, is called a free loop. Similarly a n edge of G which is not adjacent to any other edge or loop of G is called a free edge. The letter t denotes the number of free loops of G. 3. Preliminary Result
W e make extensive use of the following result due independently to Hoffman [ll] and Andersen [I]. Theorem 3. Let G be a subgraph of K: consisting of a K: and possibly some free loops. Let G have an edgecolouring with n colours, cl, . . . ,en, and let the loopcolouring be admissible. For l < k < n let [(k) be the number of free loops of colour k. Then the edgecolouring can be completed to an edgecolouring of K ,L with el, . . . ,c, if and only if, for 1O and that the colours in question are cl, ...,c 8 . For 1 5 i < s , let Wi be the set of vertices with uncoloured loops, the vertices being incident with no edge of colour c, . We simply wish t o find a system of distinct representatives for the sets Wi. By P. Hall’s wellknown theorem [g), we must show that :E l
But if 11122 U l4$ contains all vertices with uncoloured loops, except possibly
’iEI
the vertex y in case (i) if 1112. In case (i) there are at least three uncoloured loops, so in this case, if 111=2, U W, contains at least two elements. Thus we need only coni EI
sider the case I Z I = 1 . But clearly any Wi contains an element, except if all edges have colpur c,. This is impossible in case (i), and in case (ii) it corresponds to G being of
Symmetric Latin Squares
Y
X
2 0
0
7
V
4 I I I
I I
I
Q
Q
Q
a
I
I
I
a I
I
I
I
I
I
Q
Q
43
Figure 3 Type e l or 01 with a loop added. Thus we can indeed find the required system of distinct representatives, and so we can colour a further loop for each of the colours Cl,
. . . ,cs.
We now consider various cases. Case 1. n even.
n
Case la. f? 2.
Let K:b be a subgraph of the K: containing G , and suppose that 2 all edges of G are inside K f b . Suppose that some further coloured loops have been added as described above, so that the number of loops of each colour is now even (but some loops may remain uncoloured). We now show that, with one exception, we can colour the edges of KXh so that each colour occurs at least as many times (i.e. with at least as great degreesum) in KXb as outside. We do this by colouring the uncoloured edges of K;h one by one, keeping a proper edgecolouring at each stage, and , at each L stage, using a colour which, at the stage before, occurred more times outside K,,b than inside.
L.D. Andersen and A.J.W. Hilton
8
Suppose that this process stops because there is some colour, say cl, which occurs more times outside K:fi than inside, and which cannot be placed on any uncoloured edge of K:b because at least one endvertex of each uncoloured edge is incident with an edge or loop coloured el. Let v be the number of vertices of K t f i which are not incident with edges or loops coloured e l . The number of edges between these v vertices is %v(vl), and the number of precoloured edges joining pairs of these vertices was at most % (v 1). Thus the number of edges originally available for colouring was at least %v(v1)%(v+l). Since there are v vertices in K,"p not incident with edges or loops coloured el, the're are %nv vertices in K i f i which are incident with such edges or loops. Therefore there are at least ( % n  v ) + l loops outside K e coloured e l , and thus the number of nfi. loops outside K:12 which were not coloured e l is at most v1. Hence at most v1 edges within K,'b have to be coloured with colours other than c l , and so we have
+
%v(v1)
 % ( v + l ) 5 v1,
from which it follows that v 53. If v 2 3 , then the number of loops outside K ; e not coloured c1 is at most two. But since the number of vertices at which c1 occurs is even, either v = 3 and e l occurs on % n1 loops outside K:fi, or v = 2 and c1 occurs on all loops outside K:fi. In the first case, let the colour of the last loop outside K t f ~ and cp; then the colour cp has already been ascribed to some loop of K i f i , so all nonprescribed edges coloured during the process have colour cl, contradicting that the K , has no available edges. Therefore we can assume that v = 2 and that all free loops are coloured cl. In that case, if we have given the colour c 1 t o a nonprescribed edge of K ; b , we remove it t o have 4 vertices without c 1 on them. Then 2 independent edges in this K , are not prescribed and both can be given the colour el. A similar argument holds if a loop of colour c1 was added during the first part of the proof. So we may assume that c1 is prescribed at all but 2 vertices. Then since G is not the bad case e 2 , both the loops on the two vertices which do not already have a loop or edge incident with them coloured c1 can themselves be coloured c 1 (this is the exception referred to earlier; it is exceptional as we have here t o colour loops e l rather than edges). We now colour the remaining edges of K:12 one by one, and we also colour the loops in pairs with the same colour. Since each vertex with an uncoloured edge or loop n
has at most 1 2
n
2(1)=n2
2
colours occurring on the loop or the edges incident with it, at most colours are not available for an uncoloured edge or an uncoloured pair
of loops, so this can be done greedily; i.e. one edge or looppair at a time, in any order, without ever recolouring. When this process is complete, the loopcolouring is admissible. The conditions of Theorem 3 are now satisfied and so the edgecolouring can be extended to an edgecolouring of K,.L
Symmetric Latin Squares
9
Case lb. % n is odd, .k'=Ya(n2),G is Case (ii) of Figure 3, and the loops of G ail have different colours. In this case, there are Yan loops, all of different colours, so that when the initial step of colouring uncoloured loops so that each colour appears on an even number of loops is completed, all loops will be coloured. Consider the Kk(n+2)containing all the edges of G . One colour, say cl, occurs on two loops of K:(n+2).If we show that we can colour all the uncoloured edges of Ki(n+z)in such a way that each colour apart from c, occurs on at least one edge, then, by Theorem 3, we can complete the edgecolouring to K,". The graph K s ( , + ~is) the union of the 1factor consisting of all the edges of G , n and 1 further 1factors, say F,, . . . ,F,,/,. Colour the edges of F, with some colour 2 different from c,, not used on the edges or loops of G , say c,. Let the colours different from c 1 occurring on the loops of G be c2, . . ,en/,. Suppose first that n L 1 0 . If c2 does not occur on an edge of G , colour an edge of KL(n+,)with c2. For 3 2 + ( l + %
2
We now assign the colours c3,
. ..
 to the uncoloured loops. An argument 2
similar t o the earlier one, based on Hall's theorem, can be used to show that this is possible.
For n+5 S i S n , we now attempt t o place ci on two uncoloured edges of K",3 2 2 if ci is not on any edge of G , and on one uncoloured edge of G if ci is on exactly one n5 edge of G. For 15;s, the near 1factor F2+; has n+l uncoloured edges, and 2 4 n 3 12 3 the near 1factor F,+l has uncoloured edges. For n 211 and 15;I , there4 2 2 fore, if no edge of G is coloured c , + ~ , then two edges of the near 1factor F2+i can
2+ i
be coloured c n+3 T
+
. For n 211 and l11, as e 1. The aim usually will be to modify the directed path so that one gj is lost and one gi is gained in the new Dp. The notation L1 and G l will be used to denote these elements. Most operations will also lose a second element L, and gain a second element G,. The relation between L,, G2 and D p will determine the progress of the method. For example, if DEF(P) = 2 and an operation is performed that has the property L, = G,, the process terminates with t h e sequencing.
+
I
I
Suppose now that a state P with associated (Y and Dp is given, DEF(P) 2 2 and L1 and G I have been chosen. Find i such that w = a ( i ) and w * L ,= w+. There are several important possibilities. If f' = G 1 , then apply OPERATION A. Al:
Build a new directed Hamiltonian path
w+ A2
f .*e
+a*+
+w
+*
Lefttranslate (each element of) this path by (w+)'.
This gives +
...+ (w+)l f
a new state Q with associated
(Y
+
and Dg
0 w + 1
.
+
... + ( w + )  ' w ,
Note that if
a ( j )  z = 4j+i)
30
B.A. Anderson
then [(w+)l~ = (~+)'4j+i) j)]~ so that DQ is the required modification of D p . There is no L, or G, for this operation and clearly
DEF(Q ) = D E F ( P )  2. The statement that w.G1 follows w in the directed Hamiltonian path of P means that if w = a ( i ) and w.G1 = a ( j ) ,then i < j . If f' # G, and wG1 follows to in P, then apply OPERATION B. B1: Build a new directed Hamiltonian path e
+ * a *
+w
+wG1
+  * a
f +w+
+ * * a
+(w*G1).
This gives a new state Q with associated a and D o . Note that w Gl # w+ since L , # G, and that
L, = [ ( W . G ~ )  ]  ~ ( W and G ~ )G, = f'w+. In general, this operation will be one of several attempted on a state before one is chosen. Thus, the results must be saved. There is one case, however, where the results of operation B are ignored.
B2: If L , # G2 and ( L , = L , and G, = G1) then do not save the results. This step is a precaution against the possibility of cycling occurring in the application of the process. Note that if operation B is used, then
D E F ( Q ) E {DEF(P), DEF(P)2, D E F ( P ) 4 } . The statement that w 'G, precedes w in the directed Hamiltonian path of P means that if w = a ( i ) and w  G l = a($,then j < i. If f' # G,, w *G, precedes w in P and there is a k, j Lk < i such that a(k)f'= a ( k + l ) , then apply OPERATION C. C1: Build a new directed Hamiltonian path
,
+
.+f
a ( k 1)+*+w +w *G +  .  + a ( k ) +w+
+e
+.(wGl).
There are several special cases embedded in C1. In each case, the path modification should be clear. In all cases G, = [ 4 k ) J  l W+
If w*G, # e, then L, = [(w.G1)]'. the path terminates at f
.
.
(wG,) and if w*G1= e , then L , = f'
+
and
G2: Lefttranslate (each element of) this path by [ a ( k l)]' In general, this operation will be one of several attempted on a state before one is chosen. Thus, the results must be saved. As in operation B , ignore one case. C3:
If L, #G, and (L2=L1 and G, = G,) then do not save the results.
Sequencing LowOrder NonAbelian Groups
31
Note that if operation C is used, then
DE F ( Q) E {DEF(P),DEF(P)2, D E F ( P ) 4 } . The next operation will “double” the possibilities for operations A , B and C. Suppose P , cy and D p are given. Apply OPERATION R.
R1: Consider a new directed Hamiltonian path j
+e.
+.a
If Ll and G l were chosen for P , then L,’ and G,’ can be chosen to fill corresponding roles for the new path, or, if DEF(P) > 2 other values can be chosen. Note that if
cr(i).L, = cy(i
+ 1)
then
a(i
+ l)*L,’
= a(;).
This says that when the path is reversed, even if L,’ and G;’ are chosen as the designated loss and gain, the “base point” 20 of operations A , B and C changes from cy(i) to 4 i 1) and one may expect new things t o happen.
+
R2:
Lefttranslate (each element of) this path by f  ’ .
Let rP denote the new state and note that
DEF(P) = DEF(rP). There is a small chance that the application of operations A , B , C and R t o a state will yield no choices of possible new states. In this case, apply OPERATION D.
D1:
Backtrack to the preceding state.
If Operation D is used, DEF can increase. There is a reasonable chance that the procedure t o be described will cycle if only the operations just defined are implemented. It is useful to have available a simple operation that is not easily undone by the other operations and does not drastically increase DEF. OPERATION E seems t o work quite well, although, it does not guarantee that the algorithm will always find a sequencing if one exists.
El
Modify the path e e+
+***
j*j
to e +j 4 e +
+**
4
j.
32
B.A. Anderson
Under this operation Ll = e’, L2 = (f)’f, G1 = f and G, = f’e+. DEF certainly can increase under an application of operation E.
Note that
Several comments about notation and terminology will increase the readability of the algorithm t o follow. Given a state P , the notation rA will be used t o mean “apply operation A t o r P . The term “step” will be used t o denote a counter for the number of changes in state executed by the algorithm. The phrase “first available possibility” needs t o be explained. If the algorithm arrives at a state via the backtrack operation, the algorithm is t o proceed by using the first possibility not previously used at that state, if any such possibility exists. In practice, backtracking happens rarely so the program doesn’t have t o remember very many previous states. If the algorithm arrives at a previous state by methods other than backtracking, this restriction does not apply. This is the reason for operation E. Finally, X , Y and Z are positive constants determined by the size of G. For example if 1 G I_< 20, X = 2, Y = 20 and Z = 100 work well. For IG 15 100, X = 2 , Y = 80 and Z = 500 are reasonable choices.
A Sequencing Algorithm 1.
Input an initial state.
2.
IF A or rA can be executed THEN pick the first available possibility and go t o (5) ELSE go to (3). Test all possibilities B, rB, C and rC. IF there is an available possibiliiy with L, = G, THEN pick the first available one and go to (5) ELSE IF any possibility exists THEN pick the first available one and go t o (5) ELSE go t o (4). IF D is possible THEN do it and go t o (5) ELSE failure.
3.
4.
IF new DEF = 0 THEN success ELSE IF new DEF < X and Step is a multiple of Y THEN (6) ELSE IF step >Z THEN failure ELSE go to (2). 6. Do E and go t o (2). A few observations are in order before proceeding t o a listing of results. Suppose G is a sequenceable group and the algorithm has reached a state P such that DEF(P) = 2. If G, and L , are “at random” one would expect the algorithm to find a 5.
sequencing quickly. This expectation seems to be justified by actual computation. Except for a few isolated cases, the results described below were obtained using the initial state derived from the directed Hamiltonian path 1,2,3, . . . , IG or the path 1, I G 2,3, . . . ,IG 11 (with integers 1,2, . . . ,I G I used to denote the elements of G). If troubles occurred, initial states were changed in a pseuderandom way. Of course, one could speed up the algorithm by beginning execution with an initial state having small DEF. For dihedral groups at least, this seems relatively easy t o do. Modifications of this algorithm certainly are possible. For example, if the aim is t o find 2sequencings, more operations can quickly be defined.
I,
I
Some examples may be useful. First, consider the reversing process. Suppose G = D,, the dihedral group of order 10 (see Table 1 for the multiplication table). Let P be the state with directed Hamiltonian path
Sequencing LowOrder NonAbelion Groups
33
6 7 4 2 1 0 2 3 9 8 6 + 5 + 3 + 4 .t 8 + 7 + 10 + 2 + 9. The row of numbers above P shows how P can be constructed by a collection of partial products, that is, 1  6 = 6, 6.7 = 5, 5.4 = 3, etc. From this row it is clear that DEF(P) = 2, Ll = 2 and G I = 5. Since 9' = 9 # G, operation A is not available. Since 3.5 = 2 follows 3 and 8  5 = 9 both possibilities for operation B are available. In the first case L2 = 9 and G2 = 6 while the second case gives L2 = 8 and G 2 = 3. The algorithm would therefore save both these possibilities and consider rP. 8 9 4 5 1 0 5 3 7 6 rP: 1 + 8 + 5 3 + 2 + 6 + 7 + 10 + 4 + 9.
P: 1
+

Now L, = 5 and G l = 2. Again 9' = 9 # G1 and operation A fails. Again 3.3 = 4 follows 3 with L2 = 7 and G 2 = 8 while 6.2 = 10 follows 6 with L2 = G, = 3. This means that the last possibility yields a sequencing via operation B. 1 8 9 4 5 1 0 2 7 6 3 1 8 5 3 2 610497 The first row above is the sequencing and the second row is the collection of associated partial products. Next, consider an example of a state P o n D,. With the table for D, as in [15] or via the instructions in $3, it can be shown that the algorithm must backtrack from state P. 16 18 4 2 3 5 14 6 8 9 17 11 17 15 7 10 13 P: 1 16
[email protected] +2 +3 6 4 13 +17 10 11 4 +14 +7 +12 15 6 48. The last part of this section will be devoted t o a sample computation on D , that illustrates several features of the algorithm. 1 2 3 4 5 6 1 7 2 3 4 5 3 4 5 1 2 8 1 2 3 91 4 5 5 1 2 3 4 1 0 6 1 0 9 8 7 1 7 6 1 0 9 8 2 8 7 6 1 0 9 3 9 8 7 6 1 0 4 10 9 8 7 6 5
7 8 9 1 0. 6 5 1 2 3 4
8 9 1 0 9 1 0 6 0 6 7 6 7 8 7 8 9 4 3 2 5 4 3 1 5 4 2 1 5 3 2 1
Table 1. A Multiplication Table for D,.
Suppose that an initial state Po has been input and after a few steps, the process arrives at stage P,. 4 6 3 5 4 6 7 1 0 2 DEF(P,)=4,L,=G,Gi=g. P,: 1 + 4 + 9 + 7 + 8 4 10 + 5 + 6 + 2 + 3
34
B.A. Anderson
The implementation I have used always picks the largest available numbers for L, and G,. After testing all possibilities, the algorithm chooses the first available option, which in this case is the first possibility for operation B on P,.
4 9 5 4 6 7 1 0 2 7 P,: 1 * 4 + 7 + 8 + 10 + 5 + 6 + 2 * 3 + 9 7 rP2: 1 + 7
5
1
+8
0
+4
*
7 6 10 + 5
3 +
2
9
DEF(P2)=4,L1=7,Gl=8.
3
2 +3 +6
*
9
DEF(rP2)=4,L,=7,Gl=8.
After testing all possibilities, the algorithm chooses the first available option which is the first possibility for operation B on rP2.
8 1 0 7 6 P3: 1 + 8 + 4 + 10 * 5
2 9 3 3 DEF(P3)=4,Ll=3,Gl=5. 3 * 6 + 9 * 7 The algorithm chooses the first possibility for operation C on P3. Note that D E F falls even though L, # G , . 6
5
4
3
+2
*
2
9 3 3 7 8 DEF(P4)=2,Ll=3,G1=10. 5 + 2 + 4 + 10 + 3 The algorithm chooses the first possibility for operation C on P4 (from the second occurrence of L,). P4: 1 + 6 + 7 + 9 + 8
+
2
9 3 1 0 5 9 7 8 4 10 + 8 + 4 + 3 + 6 + 5 + 7 * 9 DEF(P5)=2,L,=Q,G,=6. The algorithm chooses the first possibility for operation B on Ps. P5:
1+2
+
2 1+2
6
4 + 9 *
5 10
3 8
1
0
5
9
+4 +3 + 6
7 5
DEF(PG)=2,L,=5,G,=8. The step that gives P6 is a multiple of Y (see the algorithm) so the algorithm chooses operation E. Note that D E F increases. P6:
+
7
+
*
5 3 6 4 5 3 1 0 5 9 DEF(P7)=6,Ll=5,Gl=8. P,: 1 t 5 + 2 + 7 * 9 + 10 * 8 + 4 * 3 + 6 The algorithm chooses an instance of operation B on P7 such that L 2  ~ , . 5 Pa: 1 + 5
4
3
9 rPa: 1 + 9
+
2
6 4
4
7 +9
*
5 3 1 0 8 10 + 8 + 4 + 6
+
9 3
DEF(Pg)=4,Ll=4,G,=7.
8 1 0 4 2 3 6 4 2 2 + 6 + 8 + 7 + 10 + 5 + 3 4 DEF(rP8)=4,L,=4,G,=7. The algorithm chooses an instance of operation C on rP8 such that L2G,. +
35
Sequencing LowOrder NonAbelion Groups
6 7 4 2 1 0 2 3 9 Pg: 1 + 6 + 5 + 3 + 4 + 8 + 7 + 10 + 2 8 9 rP,: 1 + 8 + 5
4
9
DEF(Pg)=2,Ll=2,G,=5.
0 5 3 7 6 6 + 7 + 10 + 4 + 9 DEF(rP9)=2,L,=5,G,=2. Finally, the algorithm terminates with a sequencing by choosing an instance of operation B on rPg such that L2G,. +
5
8 +
1
3 + 2
+
9 1 4 5 1 0 2 7 Plo: 1 + 8 + 5 3 2 + 6 + 10 + 4 +
+
6 +9
+
3 7
DEF(P,O)=O.
3. Results
The sequencings listed in this section can be verified by looking at the appropriate group table in [15] or, in the case of dihedral groups of order >32, constructing a table according t o the following instructions. Suppose n > 2 is a positive integer. The dihedral group D, .is the group of order 2n defined by
D, = { a i 6 j : O < i < n
l,O_<j 5 , except possibly for m €{6,10,14,18,20,22,26,28,30,34,38,42,44,52}. There are some useful PBDs derived from TDs or truncated TDs. We present some lemmas below which will be quite useful in some of our later constructions. We shall denote by B(K) the set of all integers u for which there is a PBD B ( K , l ; u ) . If ke#, then B ( K U { k * } , l ; v )denotes a PBD B ( K U { k } , l ; u )which contains a unique block of size k, and if k E K , then B ( K U { k * } , l ; u ) denotes a PBD B ( K U { k } , l ; u ) which contains at least one block of size k. For convenience, we shall define B ( k , , k 2 ,. . . ,k,.) t o be the set of all integers u such that there is a PBD B({k,,k2,...kr},l;w). Lemma 2.8. If m is a prime power and l s k < m  1 , then for O i u s m and O < u < m , we have k m + u + u € B ( k , k + l , k + 2 , u , v ) . Proof: See [ I , Lemma 3.51. It is now known [8, 201 that a T ( 5 , l ; m )exists for all positive integers m#2,3,6 with the possible exception of m=10. From this result we readily obtain our next lemma (see, for example, [ 3 ] ) . Lemma 2.9. Suppose m#2,3,6,10 and O s n s m . Then 4 m + n € B ( 4 , 5 , m , n ) . Lemma 2.10. Suppose a T ( k + 1 , 1 ; m ) exists, O4andv#6,8,9,10,12,14,15,18,19,23,26,27,30,38,39,42,50,54, 62,66,74,78,90, then v @(4,5,6,7,11).
And combining Theorem 4.4 with Lemma 4.3, we obtain the basis for most of our constructions in this section in the following theorem. T h e o r e m 4.5. If v > 4 and v # 8, 9, 10, 12, 14, 15, 18, 19, 23, then v EB(4,5,6,7,11).
We shall use PBDclosure (Lemma 3.2) and Theorem 4.5 t o obtain the main result for (3,2,1)HCOLS(2”). However, we first require some input designs in order t o apply Theorem 4.5. For the most part we give a direct method of construction of the starteradder type, where the arrays are generated cyclically in the usual manner (see, for example, (4, 131). In Figure 2 we present an example of a (3,2,1)HCOLS(Z4), and in Figure 3 we give an example of a (3,2,1)HC0LS(25), where the array is essentially of type 2421. Note that this square can be viewed as being completely cyclically generated (mod 8) from its first row, given by the vectors e = ( 4 , 2 , 1 , 6 , 4 , B , 3 , A ) and f=(7,5), and the last two elements of its first column, given by the vector g =(6,2), where the symbol 4 is used for empty cells and A, B act as “infinity” elements. In Table 1, we give additional cyclically generated (3,2,1)HCOLS(2”) for n =6,14,15, and 18, by specifying the type along with the vectors e, f and g where 4 is used for empty cells and A, B, C, etc., act as “infinity” elements. Note, for example, that type 2”8’ produces a (3,2,1)HCOLs 214 by filling in the hole of size 8 with a (3,2,1)HC0LS(24). Moreover, the type 2( 1 10’ square will produce a (3,2,1)HC0LS(216), but this is listed here only for future reference as an input design for the constructions of the next section. Before proceeding to our main result, we have the following from GDD’s:
72
F.E. Bennett, L. Wu and L. Zhu
a,.
Lemma 4.6. If n el (mod 3)and n 2 7 , then n Proof: If n 4 (mod 3) and n 2 7 , then there exists a GDD having grouptype 2" and blocks of size 4 (see [ S ] ) . We then apply Corollary 2.13 and Lemma 3.3, giving every point of the GDD weight 1, to obtain the desired result. 0 We are now in a position t o prove the main theorem of this section. Theorem 4.7. There exists a (3,2,1)HCOLS(2") for all integers n 2 4 , except possibly for n =8,9, and 12. Proof: For the proof of the theorem, we combine the examples given in this section with Lemmas 3.4 and 4.6, and Theorem 4.5, using the fact that I2 is PBDclosed. 0 5. Construction of (3,2, l)HCOLS(h")
In this section, we shall focus our attention on the (3,2,1)HCOLS(h") for h 2 3 . We commence with the following lemma.
existence
of
73
Conjugate Orthogonal Latin Squares
Table 1
Type
e
f
t.3
(173) (mod 10) (9,6,18,14,4,12,8, 2) (mod 20)
(18,20,14,16,19, 15,21,12) (mod 22)
(8,9,18,25,17,12, 14,22,16,19), (mod 26)
(2,4,12,21,14,1, 13,17,5,10) (mod 22)
Lemma 5.1. If n ~0 or 3(mod 12) and n 2 1 2 , (3,2,1)HCOLS(hnF) for h =3 and 6. Proof:
then
there
exists
a
By deleting one point from a B(4,l;n+l) we obtain a GDD with blocks of size 4 and grouptype 3°F. We then apply Lemma 3.3, giving each weight 1 for the case h =3, and weight 2 for the case h =6, t o obtain our desired result. 0 Lemma 5.2. For any integer n z 4 , n#6,12, and for h 23, there exists a (3,2,1)HCOLS(h" ). Proof: For any n>4, n#6,12, and h 2 3, h # 6, we can combine a (3,2,1)HCOLS(l"), which exists by Corollary 2.13, and a (3,2,1)COLS(h), which exists by Theorem 2.14, to obtain a (3,2,1)HCOLS(h") by Lemma 3.1. Next for a (3,2,1)HCOLS(6"),n#6,8,9,12, we combine a (3,2,1)HCOLS(2") and a (3,2,1)COLS(3) to obtain a (3,2,1)HCOLS(6") by Lemma 3.1. Finally, for n =8,9, we obtain a (3,2,1)HCOLS(6") from Lemma 5.1. This completes the proof of the lemma. 0
74
F.E. Bennett, L. Wu and L. Zhu
We are now left with the problem t o determine the existence of (3,2,1)HCOLS(h") for n 3 6 and 12, h 23. In what follows, we consider the cases (h'). We need some input separately. First of all, we investigate (3,2,1)HCOLS squares for TDs as the "base" GDDs in what follows. Lemma 5.3. There exist (3,2,l)HCOLS(3') and (3,2,1)HC0LS(2531). Proof: We give a direct construction of the startera.dder type in both cases, where the we obtain a type 3531 squares are generated cyclically. For the (3,2,1)HCOLS(3'), square with its first row given by vectors e=($,B,C,1,3,~,13,11,9,12,~,A,14,4,8) and f=(2,6,7),and the last 3 elements of its first column given by vector g =(12,8,11) and the square is generated ( m o d 15). For the type 2531 square we use e =(4,8,C19,A,$,7,B,6,2), f = (1,3,4),and g=(9,4,2)(mod 10). It is readily checked that we get the desired result. 0 Lemma 5.4. There exists a (3,2,1)HCOLS(3k46k) for k =0,1,4,5. Proof: We first consider a B(5,1;25)and, by deleting one point from a block B of this design, we obtain a GDD with blocks of size 5 and grouptype 4 ' . For k=1,4,5, we may further delete additional k points contained in another block of this GDD, which is dis'oint from B, t o obtain a GDD with blocks of size 4 and 5 and grouptype 3k4"i. We now apply Lemma 3.3, giving every point weight 1, t o obtain our desired result. 0
Lemma 5.5. If there exists a T(6,l;m) then there exist a (3,2,1)HCOLS ((2m+1)') and HCOLS((3m+1)2. Moreover, if m #5, there exists a (3,2,1)HCOLS ((3m+2) ). Proof: The proof considers each of the above three cases separately. First we consider and let B be a block. We give each point of B weight 3 and all other the T(6,l;m) points of the TD are given weight 2. We then apply Lemma 3.3, using as input designs (3,2,l)HCOLS(2'), HCOLS(3') and HC0LS(2531) t o obtain a (3,2,1)HCOLS((2m+1)6). Secondly, for a (3,2,1)HCOLS ((3m+1)'), we proceed as before except we give every point of B weight 4 and all other points of the TD weight 3. Using (3,2,l)HCOLS(3'), HCOLS(4'), and HC0LS(354')as input designs, we obtain a (3,2,1)HCOLS((3m +1)'). Finally, for a (3,2,1)HCOLS ((3m+2)'), m f 5 , we take a slightly different approach. Here we select two disjoint blocks B and B' of the TD (this is possible provided m = 5 ) and give every point of BUB'weight 4, and all other points of the TD weight 3. Then using as input designs (3,2,1)HCOLS (3'), HCOLS(4'), HC0LS(3541), and HC0LS(3442), we get a (3,2,1)HCOLS((3m +2)'). This proves the lemma. 0 Lemma 5.6. There exists a (3,2,1)HCOLS(h')for all h 2 3 , except possibly for h =5,7,13,61.
Conjugate Orthogonal Latin Squares
75
Proof: We consider the cases h even and h odd separately in our proof which follows: Case 1: h even. We write h =2m and need only consider m 23.If m #6, we combine (3,2,1)HCOLSi26) with a (3,2,1)HCOLS(m) to obtain a a (3,2,1)HCOLS((2m) ) by Lemma 3.1. If m =6 we combine a (3,2,1)HCOLS(46) with a (3,2,1)COLS(3) to get a (3,2,1)HCOLS(126). Thus we have a (3,2,1)HCOLS(h6) for all even h. Case 2: h odd. Let A={2,3,4,6,10,14,18,20,22,26,28,30,34,38,42,44,52}. We let h=2m+l and need only consider m>2. First of all, if mPA,then there exists a T(6,l;m) by Lemma 2.7, and by applying Lemma 5.5 we obtain a (3,2,1)HCOLS((2m+1)6). If m€A\{2,3,6,30}, the existence of a (3,2,1)HCOLS((2m+1)6) is established by the equations given for 2m +1 and the appropriate application of Lemmas 3.1 and 5.5 in Table 2. Consequently, we have a (3,2,1)HC0LS(h6) for all odd h 23, with the possible exception of h =5,7,13 and 61, and our lemma is proved. Table 2
Equation for 2m
9=3.3 213.7 29=3.9 +2 37=3.12 1 41=3.13+2 45=3.15 53=3.17+2 57=3.19 69=3.23 77=3.25+2 85=17.5 89=3.29 2 105=3.35
+
+
+1
Authority Lemma 3.1 Lemma 3.1 Lemma 5.5 Lemma 5.5 Lemma 5.5 Lemma 3.1 Lemma 5.5 Lemma 3.1 Lemma 3.1 Lemma 5.5 Lemma 3.1 Lemma 5.5 Lemma 3.1
We now turn our attention t o the existence of (3,2,1)HCOLS(h12),h 23. We first consider some small values of h. Lemma 5.7. There exists a (3,2,1)HCOLS(h1*)for h =3 and 6. Proof: We start with a B(4,1;37) and delete one point from a block of this design t o obtain a GDD with blocks of size 4 and grouptype 3". If we give every point weight I, we obtain a (3,2,1)HCOLS(312),and if we give every point weight 2,we obtain a (3,2,1)HCOLS(S"), by applying Lemma 3.3.0
F.E. Bennett, L. Wu and L. Zhu
76
L e m m a 5.8. There exists a (3,2,1)HCOLS(h'2) for h=4,7,8,9, and 10. Proof: For 4 < h S l O , h # 5 , 6 , we start with an R T ( h , l ; l l ) and then adjoin h new points to this TD, one point to each of h parallel classes of blocks. We then view the resulting design as a GDD with block sizes in { l l , h , h + l } and grouptype h12. Using as input designs (3,2,1)HCOLS(l"), HCOLS(lh) and HCOLS(lh+'), we apply Lemma 3.3 giving every point of this GDD weight 1 and the result follows. 0 L e m m a 5.9. There exists a (3,2,1)HCOLS(h'2) for h = 5 and 14.
Proof: To begin with, there exists a GDD with blocks of size 5 and grouptype (5)13 (see, for example, [12]). By deleting one group from this GDD, we obtain another GDD with blocks of size 4 and 5 and grouptype (5)12. We then give every point of this resulting GDD weight 1 and apply Lemma 3.3 to get a (3,2,1)HC0LS(512). Secondly, there is a GDD with blocks of size 7 and grouptype (7)13 (see [12]). From this GDD we delete one group to obtain another GDD with blocks of size 6and 7 and grouptype (7)12. We then give every point of this resulting GDD weight 2 and apply Lemma 3.3, using (3,2,1)HC0LS(26) and HCOLS(2') as input designs, to get a (3,2,1)HC0LS(1412). 0
We now need several input designs for our constructions which remain. L e m m a 5.10. There exist (3,2,1)HC0LS(1'031), HC0LS(1'041), HCOLS(1"3'), HCOLS(1"5'), HCOLS(2"8') and HCOLS(2 '10').
Proof: For a (3,2,1)HC0LS(1'031), we present a direct construction of the starteradder type. We essentially construct a square which is cyclically generated from its first row given by the vectors e=(O,C,3,9,6,2,5,A,l,B) and f=(4,7,8), and the last three elements of its first column given by the vector g =(5,4,8)(mod 10). For squares of type 1'041,1''31,1"41, and 1"5', the constructions can be found in [4,5]. For types 2118' and 211lo', the constructions are given in Table 1. 0 L e m m a 5.11. Let m > l l and suppose there exists a T(12,1;m), then there exists a (3,2, 1)HCOLS(m'2) and HCOLS((2m)'2). Proof: For m > l l , we may write m = 4 x + 5 y , where z and y are nonnegative integers satisfying O < z + y < m . Let e E{1,2} and select a group G of the T(12,l;m). For this particular group G, we shall assign a weight of 4e to x points, a weight of 5e to y points, and a weight of zero to the remaining points of G. We then assign a weight of e to all the other points of the TD. Using as input designs the squares of type 1",2" together with those given in Lemma 5.10, we obtain a (3,2,1)HCOLS((em)'2) from Lemma 3.3. This proves the lemma. 0 Lemma 5.12. There exist (3,2,1)HCOLS(1112) and HC0LS(2212). Proof:
Conjugate Orthogonal Latin Squares
77
For a (3,2,1)HC0LS(11'2) we start with a T(12,l;ll) and select a group G of this TD. On G we assign a weight of 3 to one point, a weight of 4 to two points and a weight of zero to the remaining 8 points. For all the other points of the TD, we assign a weight of 1. Using as input designs the squares of type 1",1"31 and 1114', we get a (3,2,1)HCOLS(1112), by applying Lemma 3.3. For the construction of a (3,2,1)HC0LS(2212), we take a different approach. Starting with a T(12,1;23),we first delete one block to obtain a GDD with blocks of size, 11 and 12 and grouptype (22)". We then select one group G of this GDD and on G we assign a weight of 3 to two points, a weight of 4 to four points, and give weight zero to the remaining 16 points. For all the other points of the GDD, we assign a weight of 1. We then use as input designs the squares of types 1l0,1l1,with those given in Lemma 5.10 t o obtain a (3,2,1)HC0LS(2212), which proves the lemma. 0 We are now in a position to prove
Lemma 5.13. For all h 2 3 , there exists a (3,2,1)HCOLS(h"). Proof: For our proof we consider the cases below with h 2 3 . Case 1: h odd. If (h,210)=1, then, by Lemma 2.6, there exists a T(12,l;,h)and we can apply Lemma 5.11 with h >11 and Lemma 5.12 to obtain a (3,2,1)HCOLS(h'2). If (h,210)#1, then we can express h = m  k with m E(3,5,7}and k odd. Since a (3,2,1)HCOLS(mI2) exists by Lemmas 5.75.9, and a (3,2,1)COLS(k) exists by Theorem 2.14, we obtain a 3,2,1 HCOLS(mk) by applying Lemma 3.1. ConseIfor all odd h>3. quently, a (3,2,1)HCOLS(h1 ) exists
1
Case 2: h even. If h = 2 k , k odd and (k,210)=1, we can apply Lemmas 5.11 and 5.12 to get (3,2,1)HC0LS((2/c)'2). If h = 2 k , k odd and (k,210)#1, we can express h =mk with mE{6,10,14} and k odd. Since (3,2,1)HCOLS((m)l2) exists, by Lemmas 5.75.9 and (3,2,1)COLS(k) exists, by Theorem 2.14, we may apply Lemma 3.1 to get a (3,2,1)HCOLS((mk)12). Finally, if 4 h , we can express h = 2 m k with m 2 2 and k odd. If k = l , we use the fact t h a t we have a (3,2,1)HC0LS(412)and H C 0 L S ( 8 1 2 ) and apply Lemma 3.1 as necessary. If k 1 3 , we have a (3,2,l)HCOLS(k"), from Case 1 above, which can be combined with a (3,2,1)COLS 2m to get a (3,2,1)HCOLS((2mk)'2) by Lemma 3.1. Thus a (3,2,1)HCOLS(h'2) )exists for all even h 2 3 , and the proof of the lemma is complete. 0
I
Combining the results of Lemmas 5.2, 5.6, and 5.13, we obtain the following theorem. Theorem 5.14. For any h > 3 , a (3,2,1)HCOLS(h") exists if n 2 4 , except possibly when n = 6 and h E(5,7,13,61}. In summary, we combine Theorems 4.7 and 5.14 with the necessary condition for existence of (3,2,1)HCOLS(h") to obtain our main result:
Theorem 5.15. For any h > 2 , a (3,2,1)HCOLS(h") exists if and only if n 2 4 , except possibly when nE(8,9,12} and h=2, and when n = 6 and hE(5,7,13,61}. Since the existence of (3,2,1)COLS(w) is equivalent to the existence of (1,3,2)COLS(w), we have the following corollary.
F.E. Bennett, L. Wu and L. Zhu
78
Corollary 5.16. For any h 2 2 , a (1,3,2)HCOLS(h”) exists if and only if n > 4 , ekcept possibly when nE(8,9,12} and h = 2 , and when n = 6 and h €{5,7,13,61}. 6. Concluding Remarks
It is hoped that the results of this paper will be of importance in the construction of other combinatorial designs, such as Mendelsohn designs (see, for example, [2, 141). We conjecture that n >4 is also a sufficient condition for .the existence of (3,2,1) (or (1,3,2))HCOLS(h”). At present the existence of ( 3 , 1 , 2 ) (or (2,3,1))HCOLS(h”) is under investigation by the authors, and it is hoped that results similar to those mentioned above for the other conjugates can be obtained.
Acknowledgements The first author acknowledges the financial support of the National Sciences and Engineering Research Council of Canada under Grant A5320. A portion of this work was carried out while the second author was a visiting professor at Mount Saint Vincent University and he gratefully acknowledges the kind hospitality accorded him. References F.E. Bennett, “Latin squares with pairwise orthogonal conjugates”, Discrete 111 121 131 141
I51 161 171 181
191 1101 1111
Math. 36 (1981) 117137. F.E. Bennett, “Conjugate orthogonal Latin squares and Mendelsohn designs”, Ars Combinatoria 19 (1985) 5162. F.E. Bennett and N.S. Mendelsohn, “On the spectrum of Stein quasigroups”, Bull. Austral. Math. SOC. 21 (1980) 4763. F.E. Bennett and L. Zhu, “Incomplete conjugate orthogonal idempotent Latin squares”, Discrete Math., to appear. F.E. Bennett, Lisheng Wu, and L. Zhu, “Some new conjugate orthogonal Latin squares”, J . Combinatorial Theory (A), t o appear.
Th. Beth, D. Jungnickel, and H. Lenz, Design Theory, Bibliographisches Institut, Zurich, 1985. A.E. Brouwer, “Optimal packings of K i s into a K,,,” J . Combinatorial Theory (A) 26 (1979) 278297. A.E. Brouwer, “The number of mutually orthogonal Latin squares  a table up t o order looOO”, Research Report ZW 123/7Q, Mathematisch Centrum, Amsterdam, June 1979. A.E. Brouwer, A. Schrijver, and H. Hanani, “Group divisible designs with blocksize four”, Discrete Math. 20 (1977) 110. J. De‘nes and A.D. Keedwell, Latin Squares and Their Applications, Academic Press, New York and London, 1974. J.H. Dinitz and D.R. Stinson, “MOLS with holes”, Discrete Math. 44 (1983) 145 154.
Conjugate Orthogonal Latin Squares
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H. Hanani, “Balanced incomplete block designs and related designs”, Discrete Math. 11 (1975) 255369. K. Heinrich and L. Zhu, “Incomplete selforthogonal Latin squares”, J. Austral. Math. Soc, Ser. A, t o appear. A.D. Keedwell, “Circuit designs and Latin squares”, Ars Combinatoria 17 (1984) 7990.
R.C. Mullin and D.R. Stinson, “Holey SOLSSOMs”, Utilitas Math. 25 (1984) 159 169.
K.T. Phelps, L‘Conjugateorthogonal quasigroups”, J . Combinatorial Theory (A) 25 (1978) 117127.
Robert Roth and Matthew Peters, “Four painvise orthogonal Latin squares of order 24”, J . Combinatorial Theory (A) 44 (1987) 152155. D.R. Stinson and L. Zhu, “On sets of three MOLS with holes”, Discrete Math. 54 (1985) 321328. D.R. Stinson and L. Zhu, “On the existence of MOLS with equalsized holes”, Aequationes Math., t o appear. D.T. Todorov, “Three mutually orthogonal Latin squares of order 14”, Ars Combinatoria 20 (1985) 4548. R.M. Wilson, “Construction and uses of pairwise balanced designs”, Mathematical Centre Tracts 55 (1974) 1841. L. Zhu, “Existence of holey SOLSSOMs of type 2 ” ” , Congressus Numer. 45 (1984) 295304.
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Annals of Discrete Mathematics 34 (1987)81100 0 Elsevier Science Publishers B.V. (NorthHolland)
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On regular packings and coverings JG. Bermond, J . Bond, D. Sotteau W.A. 410 CNRS, L.R.I. b i t 490, UniversitC ParisSud, 91405 Orsay Cedex, FRANCE TO AL&X ROSA O N XIS 3InIETU BIRTUDAY
ABSTRACT Let us call a regular packing of K , with Kk’s a set of edge disjoint subgraphs of K , isomorphic to K k such t h a t t h e partial graph G , called the leave of the packing, generated by the edges not covered is regular. Similarly, a regular covering of K , with Kk’s is a set of subgraphs of K , isomorphic to K k which cover all the edges of K , at least once a n d such t h a t the multigraph H , called the excess of the covering, obtained by deleting Km from the union of the Kk’s, is regular. Here we s t u d y maximum regular packings and minimum regular coverings. More exactly we determine the minimum degrees of the leaves and excesses in case of regular packings and coverings of K , with K3’s and K4’s . W e also exhibit minimum regular coverings containing maximum regular packings. 1. Introduction
A classical problem in design theory concerns the existence of ( n , k , X ) BIB designs. It is well known t h a t the existence of a ( n , k , l ) BIB design is equivalent t o the existence of a decomposition of the complete graph K , into Kk’s. It is also well known t h a t such a decomposition can exist only if n(nl)=O ( m o d k ( k  1 ) ) and nlrO ( m o d k1). These conditions have been shown to be sufficient for any k and n large enough (Wilson [ I S ] )and also for k = 3 (for a survey o n Steiner triple systems, see Doyen and Rosa [7]), k = 4 and k = 5 (see Hanani [8]). When n does not meet the necessary conditions one can either try to find the maximum cardinality of a packing of Kk’s into K , (that is the maximum number of K k ’ s included in K,), or the minimum cardinality of a covering of K , with Kk’s (that is the minimum number of edge disjoint Kk’s necessary to cover all the edges of K , at least once). Such values have been completely determined for k = 3 (see Hanani IS]) and for k = 4 (see Brouwer [5] and Mills [12,13]). Here we are interested in a slightly different problem t h a t we call regular packing or covering of K , with Kk’s. This problem was initially motivated by a problem of interconnection networks, called bus networks of diameter 1, modeled by hypergraphs. In such a network there are n processors connected to buses. Each bus contains the same number k of processors. For every pair of processors there must exist at least one
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bus containing them (diameter 1 ) . These conditions correspond t o the classical covering of K , with Kk’s.A further requirement (of regularity) is that each processor belongs to the same number of buses, which gives rise to the notion of regular covering (for details see Bermond, Bond and Sacle [3] and[2]). A regular packing of K , with Kk’S is a set of edge disjoint subgraphs of K , isomorphic to K k such that the partial graph G , generated by the edges not covered, is regular of degree dG. In the literature G is also called the leave of the packing (see 161). In other words we want t o find a regular graph G such that K ,  G can be decomposed into Kk’s. Note that a regular packing of K , with Kk’s is just a packing with Kk’s such that each vertex of K , belongs t o the same number of K k l s . We shall say that a regular packing is maximum if dG is as small as possible. Similarly, a regular covering of K , with Kk’s is a set of subgraphs of K , ismorphic to Kk which cover all the edges of K , at least once such that the multigraph H , obtained by deleting K , from the union of the Kk’S, is regular of degree d H . The multigraph H is also called the excess of the covering. In other words we want t o H can be decomposed into find a regular multigraph H of degree dH such that K , K k k . We shall say that a regular covering is minimum if dH is as small as possible. Furthermore we are interested in finding, when it is possible, a minimum regular covering of K , with Kk’s containing a maximum regular packing. That is equivalent to find whether there exist graphs G , dGregular, and H , dHregUhr, with dG and dH minimum, associated respectively t o the packing and covering such that the (dG +dH)regular graph G H can itself be decomposed into Kk’S. In other words, for that purpose, it is sufficient to find a maximum regular packing such that the dGregular graph G can be covered by Kk’S in such a way that every vertex belongs to (dG+ d H ) / ( k  l ) graphs K k ’ S (obviously a covering of a subgraph G of K , with Kk’sis a set of K k ’ S containing all the edges of G).
+
+
The following proposition is immediate
Proposition 1.1 The degree d c of the regular leave G of a regular packing of K, un’th Kk’s satisfies the folloun‘ng equalities: n 1  d c = 0 (mod k1) n ( n  l  d G ) G 0 (mod k(k1)).
The degree dH of the regular excess H of a regular covering of K , un’th K k ’s satisfies the following equalities: n  l + d H G O (mod k1) n ( n 1+dH) E 0 (mod k(k1)) From that proposition, one can easily deduce lower bounds on dG and dH
.
Here we show that these bounds are always reached in the cases k = 3 or 4 by constructing maximum regular packings and minimum regular coverings of K , with K3’s and K4’sfor any n (except for n=8 in case of a packing with K4’s). In most of the cases the covering constructed contains the packing.
a3
Regular Packings and Coverings
Let us remark that in some cases the optimum packings or coverings given in the lfterature ([5,13,14]) are necessarily regular (by counting arguments). However as we want, if possible, to give a covering containing the packing, we give the construction again.
2. Notation and basic lemmas
K , will denote the complete graph on n vertices. Usually V ( K , ) , the vertex set of K , , will be Z,, the group of integers modulo n . 0 [z1,z2, . . . , z r ] will denote the complete graph on the vertices z1,z2,. . . ,zr. In case r=2 it denotes an edge. 0 [ z 1 , z 2 ,.. . , z r ] (mod n ) will denote {[zl+i,z2+i ,...,q + i ] : i = O , l , ...,n1). 0 K ,  G denotes the partial graph obtained from K , by deleting the edges of G from K , . H denotes the multi0 If G and H are graphs on the same set of vertices V , G graph with vertex set V containing the edges of G plus the edges of H (with eventually multiple edges). 0 K,.,,,, denotes the complete bipartite graph with vertex sets of cardinality r 1 and r2. denotes the complete multipartite graph . If r l = r2 = .. = rh = r then 0 Kr,,rb,.,rI
+

we denote Kr,,ra..,rb by K h X r 0 Let G be a graph on k vertices. A parallel class of G ’ s on n = p k vertices is the vertex disjoint union of p copies of G . It is well known that the existence of a decomposition of Khxr into Kh’S is equivalent to the existence of h2 orthogonal Latin squares of order r (or also to what is called a GD(h,l,r;hr) or T(1,r) design). Classical results are the following.
Lemma 2.1 (see Hanani [S]) (h1)r G O ( m o d 2)
Khxr can be decomposed into K3’s if and only i f
Lemma 2.2 (see Brouwer, Hanani, Schrijver [4] ) K h x rcan be decomposed into K,’s if and only i f
I
= 0 (mod 3) hr2(h1) = 0 (mod 12)
(h1)r
h 1 4 , ( h , r ) # (492) 07. (476)
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J.C. Bermond, J. Bond and D. Sotteau
Lemma 2.3 (see Bermond, Huang, Rosa, Sotteau [ l ] ) I f K r,r,r and Krjrjrjr can be decomposed i n t o graphs G , t h e n Kpr,pr,pr,qr c a n be decomposed i n t o G f o r p # 2, 6, 0 5 q 5 p (and if q = 0 for a n y p).
If K r,r,r,r and K r,r,r,r,r c a n be decomposed i n t o graphs G decomposed i n t o G f o r p # 2 , 3, 6, 10 and 0 5 q 5 p .
,then
Kpr,pr,pr,p r,qr c a n be
In [lo]Huang, Mendelsohn and Rosa introduce a problem which can be formulated as that of the existence of a decomposition of K , into graphs all isomorphic to Kk except one isomorphic t o K,. They give the necessary conditions. The following results have been proved on this problem, which will be used in the proof of the main theorem.
Lemma 2.4 (Brouwer [ 5 ] ) K , c a n be decomposed i n t o one K , and K4’s i f and only i f n n # 10, 19.
= 7 or
10 ( m o d 12),
Lemma 2.5 (Bermond, Bond [3]) K , c a n be decomposed i n t o one K,, and K4’s i f and only if n n # 7, 19, 22.
= 7 or
10 ( m o d 12),
In the section 5 we study a more particular case of this problem, since we want to have the additional property that there exists, in the decomposition, a parallel class of Kk’s on n  r vertices . 3. Main results
Theorem 3.1 If n 2 3 there exist a m a x i m u m regular packing and a minimum regular covering of K , with K3’s where t h e graph G and multigraph H are a s follows. ( m o d 6), G i s 1regular and H i s 1regular 0) if n 0 ( m o d 6), G and H are e m p t y i) if n = 1 G i s 1regular and H i s 5regular ( m o d 6), ii) if n = 2 G and H are e m p t y ( m o d 6), iii) if n = 3 ( m o d 6), G i s 3regular and H i s 3regular iv) if n = 4 ( m o d 6), G i s 4regular and H i s 2regular v) if n = 5 Moreover there exists a minimum cogering containing a m a x i m u m packing in all cases except case 0 ) Proof First, from Proposition 1.1, the given values of the degrees of G and H are lower bounds. Now we will prove that these bound are reached.
Regular Packings and Coverings Case
0)
(n=6t,
85
t 21) :
Since Kc,+, can be decomposed into K3k, we get a packing of K , by deleting a vertex and all the K3’s containing it. The leave of the packing is a perfect matching. is known (se Mills 114) for a survey on t h e subject) t h a t K , can be covered with
% I If n/3 (n1)/2 matching.
K3’s.In this case the covering is regular and the excess H is a perfect
Obviously in this case it is not possible t h a t the minimum covering contains the maximum packing since G and H are necessarily perfect matchings and the sum of two perfect matchings cannot be decomposed into K3k. Case i) ( n = 6 t + l , t 2 1 ) and case iii) (n=6t+3, ta): It is known t h a t , in these cases, t h e graph K , can be decomposed into K3’s. Case ii) (n=6t+2, t 2 l ) : The packing is obtained exactly as in o), and the leave G is a perfect matching. A regular covering of G with each vertex in three K3’s is obtained by Corollary 4.2 applied to the 3regular graph formed by the sum of G and any 2regular graph on the
same set of vertices.
ta)
Case iv) (n=6t+4, : We distinguish two subcases:
i) t = 2 r , r a If r=O then n=4 , G = K4 is itself 3regular. W e can take H = K4 since the multigraph formed by two identical K4’scan be decomposed into K3’s , [0,1,2] (mod3). If r21, K12r+4is the edge disjoint union of 3r+l vertex disjoint graphs K4’s and a K(3r+l)X4. By Lemma 2.1 K(,,+l)X, can be decomposed into K3’s. Therefore the leave G consists of the 3r+l vertex disjoint K4%.We shall take for the excess H the same graph. As the multigraph formed by two identical K4’s can be decomposed into K3k, G H can also be decomposed into K3k.
+
ii) t=2r+1, r a If r = O then n=10. A maximum regular packing consists of the following K3k: [0,3,Ql, [1,4,51, [2,0,61, [3,1,71, [4,2,811 [0,7,8], [1,8,Q], [2,Q,51, [3,5,61, [4,6,71. The leave G is t h e Petersen graph which is 3regular. From Corollary 4.2 it can be covered with each vertex in three K,’s. So we have a covering of Klo containing the maximum packing where the excess H is a 3regular multigraph. If r=l then n=22. The packing is given by t h e K3k: [0,1,3], [0,4,10] and [0,5,13] (mod 22). The leave G is %regular and consists of the edges [0,7] and [ O , l l ] (mod 22). It is the union of a hamiltonian cycle and a perfect matching so, by Corollary 4.2, it can be covered with each vertex in three K3’s. Thus we have a covering of K,, containing t h e packing with a n excess H which is 3regular. If r a , K12r+10 can be decomposed into r graphs K12’son vertex sets X i , l s i l r , one Klo,on vertex set Y and a K12, ,12,10on vertex set U X i U Y .
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From above the Klo can be decomposed into K3’s plus the 3regular Petersen graph. From case 0 ) each K,, on vertex set X i can be decomposed into K3’s plus a perfect matching Mi for 1 2, K6t+5is the edge disjoint union of t vertex disjoint K s ’ s , on vertex sets X i , one K , on vertex set Y and a multipartite graph K 6 , , , , , 6 , 5 . According t o Lemma 2.1, there exists a decomposition of K(,+,),, into K3’s.By deleting one vertex we obtain a decomposition of K6,...,6,5 into K3’s and a perfect matching M on the vertex set Xi. Each K 6 on vertex set Xi is the edge disjoint union of two vertex disjoint K3’s and a 3regular graph which consists of a perfect matching Mi and a cycle of length six Cg. K 5 is itself a 4regular graph on Y. so we have a maximum regular packing of K6,+5 with K3k, where the leave G is the 4regular graph which consists of two parts: one is Mi M C;}, on vertex set u X i , the other one is a K,, on vertex set Y. Both parts can be covered with each vertex in three K3’s , the K 5 as in case n=5 and the other one as shown in Corollary 4.3. Thus we have a covering of K6t+5 containing the packing with an excess H which is 2regular. 0
lsist,
u
{u + + u
Remark 3.2 The referee suggested to mention the following conjecture which seems very difficult. Conjecture
For every g i v e n degree d, all dregular graphs m.eeting t h e necessary conditions are leaves o f a packing of K , with K 3 k , with f i n i t e l y m a n y exceptions. Note that the case of hamiltonian cycles was solved by E. Mendelsohn [ l l ]and the general case of any %regular graph is a particular case of a result of Colbourn and Rosa (61 also obtained by Hilton and Rodgers [ S ] .
Regular Packings and Coverings
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Note that the above conjecture is in fact a special case of the following one of Nas hWilliams [ 151.
Conjecture (NashWilliams) I f G i s a graph o f order n , n
2 15, such that its number
and every vertex has a n even degree at least
3n
, 4
o f edges i s a multiple o f 3,
then G can be decomposed i n t o K,'s.
Remark 3.3 K. Heinrich drew our attention to the following strong recent result of Rees 1171 which can be restated as follows: Theorem (Rees) K,, can be decomposed i n t o a perfect matchings and p parallel classes of K3's i f and only if a+2p=6tl except for 1 and t = l or t=2 (corresponding to the nonexistence o f Nearly Kirlcman triple systems of order 6 or 12). Using this theorem we can give a shorter proof of the existence of maximum regular packings for the non immediate cases n=6t+u (u=4,5) in the following way. Let V(Kn) = X U Y with w I = 6 t , pl=u. By the theorem above the K,, on X can be parallel classes of K3's. decomposed into 2u1 perfect matchings on X and 3tu Then K , can be decomposed into the K , on Y , u1 perfect matchings on X and K,'s formed by the parallel classes of K,'s on X and the 3tu triples obtained by joining any vertex of Y t o a perfect matching of X . Thus we have a regular packing of K , with K,'s for which the leave G consists of the K , on Y plus the u1 perfect matchings on X and is clearly a regular graph of degree u1. In fact we used only a corollary of the result of Rees. We needed only t o decompose K,, into perfect matchings and triangles (Note that such a decomposition solves the conjecture mentioned in Remark 3.2, when the graph is the union of d perfect matchings).
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Theorem 3.4 I f n 2 4, there exist a m a x i m u m regular packing and a minimum regular covering of K , with K4’s where the leave G and excess H are as follows. 0) if n = 0 (mod 12), G i s 2regular and H i s 1regular (mod 12), G and H are empty i) if n = 1 ii) if n = 2 (mod 12), G i s 1regular and H i s 5regular iii) if n = 3 (mod 12), G i s 2regular and H i s 10regular iv) if n 4 (mod 12), G and H are empty v) if n = 5 (mod 12), G i s 4regular and H i s 8regular vi) if n = 6 (mod 12), G i s 5regular and H i s 1regular vii) if n = 7 (mod 12), G i s 6regular and H i s 6regular viii) if n 8 (mod 12), G i s 1regular and H i s 2regular (mod 12), G i s 8regular and H i s 4regular ix) if n 9 x) if n 10 (mod 12), G i s 3regular and H i s 3regular xi) if n 11 (mod 12), G i s 10regular and H i s 2regular except for n=8 where G i s 4regular.
Moreover there exists a minimum regular covering which contains a maximum regular packing in all cases except 0 ) and xi) and n=18 (where we don’t know whether it exists).
Proof: First, from Proposition 1.1, the given values of the degrees of G and H are lower bounds (except for n=8). Now we will prove that these bounds are reached. case 0 ) (n=12t, t>l) : We obtain a decomposition of K r z t into K4’s and a parallel class of K,’s by deleting one vertex and the edges containing it from a decomposition of K,,,+, into K4k. Thus we have a maximum regular packing with a leave G which is 2regular. A union of K3’s cannot be covered with K4’swith each vertex in only one K4. In order to be able to find a minimum covering containing the minimum packing we would need to have a decomposition of K l Z tinto Kq(s and a parallel class of cycles of length four. Such a decomposition can not exist for n=12. Maybe it does for bigger n,and we can state the following problem. Problem : Does there exists a decomposition of K I z t i n t o K,’s and a parallel class o f C4’sfor t 2 2 ?
However it is known that there exists a maximum regular packing of K,,, with K4’s where the excess H is a perfect matching. Indeed it has been proved in (12) that there exists a covering of K,,, with 12t2 graphs K4k. Since each vertex belongs t o 4t K4%, the excess H is a perfect matching and the covering is regular. case i) (n=12t+l, t 2 l ) and case iv) (n=12t+4, ta): It is known that, in these cases, the graph K , can be decomposed into K4’s. case ii) (n=12t+2, t l l ) : K 1 2 t + 2is the edge disjoint union of a perfect matching M and a multipartite graph K(6,+l)x2.From Lemma 2.2, K ( 6 t + l ) Xcan 2 be decomposed into K4’s. Let US take
Regular Packings and Coverings
89
G=M={[2i,2i+l], O si <st}. We have a covering of G with each vertex in two K,’s by taking { [2i,2i+1,2i+2,2i+3], O S i S 6 t } . So we have a covering of K12,+,containing the packing where the excess H is a 5regular multigraph. case
iii) (n=12t+3, t l l ) :
is the edge disjoint union of a parallel class of K3’s and a multipartite graph By Lemma 2.2, K ( & + l ) X 3 can be decomposed into K,’s. Let us take for G the parallel class of K3’s which is a 2regular graph. By Corollary 4.4 in section 4, we know t h a t any 6regular graph t h a t can be decomposed into K3’s can be covered by K4’s with each vertex in four K4’s . We apply this result with the 6regular graph formed by G and any two other parallel classes of K3’s. So we have a covering of K12,+3 containing the packing where t h e excess H is a 10regular multigraph. K12,+3
K(&+l)X3.
case v) (n=12t+5, ta): First we prove t h e result for n=5 and 17. If n=5 then G = K 5 and H is the ®ular multigraph formed by two copies of K,. Indeed, the graph G + H can be decomposed into K,’s by taking [0,1,2,4] (mod 5). If n=17, the packing is given by the following K4k: [0,1,4,6] (mod 17). The leave G is formed by the edges [0,7] and [0,8] (mod 17). It is the union of two hamiltonian cycles and can be covered with each vertex in four K,’s by [0,2,5,9] (mod 17). Now let n=12t+5, with t>2. By Lemma 2.4, K12t+7can be decomposed into one K , and K4’s, if t f l . We take such a decomposition and delete two vertices of the K7. We obtain a decomposition of K I 2 $ +into , a K,, two parallel classes of K3’s (on the 12t other vertices) and K4’s. So the leave G we obtain consists of two parts, one is a K , and the other one is a 4regular graph on 12t vertices, which is the union of two parallel classes of K3’s. The K , can be covered as in t h e case n=5. By Corollary 4.4 in section 4, any 6regular graph t h a t can be decomposed into K3’s can be covered by K4k, with each vertex in four K,’s . W e apply this result to the ®ular graph formed by the part of G on 12t vertices and any other parallel class of K3’s on the same set of vertices. So we have a covering of K12t+scontaining the packing where the excess H is a 8regular multigraph. case vi) (n=12t+6, ta): First we prove t h e result for n = 6 and 18. If n=6 then G = K 6 is itself 5regular and H is a perfect matching. The covering is given by t h e K4’s:[0,1,2,3], [0,1,4,5], [2,3,4,5]. If n=18, then a regular packing can be obtained by taking [0,1,3,8] (mod 18). An optimal covering (which is regular in t h a t case) is given in [12] by Mills. W e note t h a t this covering does not contain any maximum regular packing and we don’t know if such a covering does exist. Now let n=12t+6, with t a . The graph K12t+6 is the edge disjoint union of a parallel class of K 6 and a multipartite graph K(2t+l)
[email protected] Lemma 2.2, K(2t+l)&5 can be decomposed into K,’s. Therefore we obtain a leave G which is a parallel class of K6’s. We cover each K 6 of G as in the case n=6 and the excess H is a perfect matching.
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case vii) (n=12t+7, t > O ) : First we prove the result for the cases n=7, 19. If n=7 then G = K 7 is itself a ®ular graph. We can take H = K 7 since the multigraph formed by two copies of the same K , can be decomposed into K4’s as follows: [0,1,2,4] (mod 7 ) . If n=19, a maximum packing is given by [0,1,3,8] (mod 19) and the leave is a 6regular graph G which is an edge disjoint union of K3’s : [0,4,10] (mod 19). So by Corollary 4.4, G can be covered by K4’swith each vertex in four K4k. Now let n=12t+7, with t > 2 . By Lemma 2.5, K12t+10 can be decomposed into one Klo and K4’s, if t#1. We take such a decomposition and delete three vertices of the Klo. We obtain a decomposition of K12t+7 into a K,, three parallel classes of K3’s (on the other 12t vertices) and K4k. So the leave G we obtain consists of the vertex disjoint union of a K , and a 6regular graph on 12t vertices, which can be decomposed into
K3’s. The K7 can be covered as in the case n = 7 . By Corollary 4.4 in section 4, the part of G which is ®ular on 12t vertices can be covered by K4’s,with each vertex in four K4’s. So we have a covering of K 1 2 t + 7 containing the packing where the excess H is a 6regular multigraph.
viii) (n=12t+8, ta): First we prove the result for the case n=8. If n=8 then there exists no regular packing of K8 with K4’swith a leave G 1regular. Indeed, if such a packing exists K8  G should be decomposed into four K4’s with each vertex in two K4’s which is easily seen t o be impossible. A maximum regular packing of K 8 is obtained with a 4regular leave G by taking two disjoint K4’s. A minimum regular covering of K8 with a 2regular excess H is given by the following K4’s(found in [13]): [l,2,3,419[192,5,6]t[1,2,7,8],[3,4,5,6],[3,4,7,81,[5,6,7,8] This covering contains the maximum regular packing since it contains two vertex disjoint K4k. Now let n=12t+8, with t y l . The graph K 1 2 t + 8 is the disjoint union of a perfect matching and a multipartite graph
[email protected]+4)~2. From Lemma 2.2, if t 2 1, K @ t + 4 ) X , can be decomposed into K4’s.Let G be the matching {[2i,2i+l] : i=O,1, ...,6t+3}. It can be covered with each vertex in one K4 with the following K4’s : {[4i, 4i+l, 4i+2, 4ii31 : i = O , l , ...,3t+l}. Note that the excess H is the vertex disjoint union of cycles of length 4. case
case ix) (n=12t+9, t > O ) : It is well known that K12t+13 can be decomposed into K4k. If we choose a K4 and delete its four vertices, we get a decomposition of K,2t+ginto four parallel classes of K3’s and K4’s. So the leave G is the 8regular graph formed by four parallel classes of
K~’s. From Coroilarv 4.5, we have a covering of G with each vertex in four K4’s. So we have a covering of K,,,,, containing the packing where the excess H is a 4regular multigraph. 
I
Regular Packings and Coverings case x) (n=12t+10,
91
t>O) :
If t=O then n=10. The packing consists of the following K4k: [072,8,Ql,I1,3,9,51, [2,4,5,6], [3,0,6,7), [4,1,7,8] The leave G is the Petersen graph which is a %regular graph. In order to cover G with each vertex in two K4k, we add the following K4k: [0,1,5,71, [1,2,6,8], [2,3,7,9], [3,4,8,5], [4,0,9,6] If t = l then n=22 . Let V(K,,) = 2, X Zl1. A maximum regular packing is given by the following K4k: {[~~,~~,~~,~+~~,(~,~+~),(~,~)],[(0,~+9),(1,~),(1,~+~),(~, [(O,i),(O,i +5),(l,i+6),( l,i+Q)] : O3. Therefore K12,,,, ,12,11 can be decomposed into one parallel class of K3’s on 12t vertices and K4’s. Each K,, is the union of three vertex disjoint K4’s and an 8regular graph. Let us take for leave G the 10regular graph which consists of two parts, one is formed on 12t vertices by these 8regular graphs plus the parallel class of K3’s, the other one is the Kll. Thus we have a maximum regular packing. The existence of a minimum regular covering has been proved in [13] where it is proved t h a t there exists a covering with (t+1)(12t+11) K4’s with each vertex in 4t+4
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graphs K4’s. Therefore the excess H is 2regular and the covering is regular minimum. Notice t h a t in this last case we don’t know whether there exists a minimum regular covering which contains a minimum regular packing. 0 Remark 3.5 The proof of the theorem suggests a lot of decomposition problems. The most general one would be a problem analogous to the conjecture of remark 3.2, i.e. t h a t all dregular graphs meeting the necessary conditions are leaves (respectively excesses) of a regular packing (respectively covering) of K , with K4k. First interesting cases are when the leaves or excesses are hamiltonian cycles or more generally regular graphs of degree two, which generalize Mendelsohn’s [11] or Colbourn a n d Rosa’s [6] result. For example we state t h e following conjectures. Conjectures
K,,, minus a h a m i t t o n i a n cycle K,,,,, m i n u s a h a m i l t o n i a n cycle K,,,,, p l u s a h a m i l t o n i a n cycle K,,,,,, p l u s a h a m i l t o n i a n cycle
c a n be decomposed i n t o K,’s
One can also ask for a generalization of Rees’ theorem. Conjecture
K I z t c a n be decomposed in.to CY parallel classes of K,’s and p parallel classes of K4’s i f a n d only if 2~+3/?=12tl (with a f i n i t e n u m b e r of exceptions). If we don’t want parallel classes of K4’s we obtain the problem introduced by Huang, Mendelsohn and Rosa [lo]. A related problem is considered in section 5. Note t h a t the problem of decomposing K , into CY perfect matchings and p parallel classes of K,’s for n=O ( m o d 4) is easy to solve (essentially because a parallel class of K4’sis the union of 3 perfect matchings). 4. Some lemmas
We state here some lemmas used in the proofs of t h e main theorem. The results are not necessarily best possible, and they are given in the form used in section 3.
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L e m m a 4.1 : i) Let G be an r(r1)regular graph on n vertices that can be decomposed into K,’s. Then G can be covered with K,+,’s, in such a way that each vertex belongs t o r+l graphs K,+, ’s.
ii) Let G be an r(rl)+lregular graph on n vertices that can be decomposed into K,’s and a perfect matching. Then G can be covered with K,+,’s, in such a way that each vertex belongs to r+l graphs K,+, ’s. iii) Let G be a n r(rl)+2regular graph on n vertices that can be decomposed into K,’s and a 2regular graph. Then G can be covered with K,+,’s, in such a way that each vertex belongs t o r+l graphs K,+, ’s. Proof First we remark that in case iii) the 2regular graph is the vertex disjoint union of cycles, and one can give an orientation to the cycles so t h a t every vertex is the origin of only one arc. Any vertex of G is contained in r graphs K,’s of the decomposition , so the total number of K,’s is n . Let us add one vertex of G to each of the Kr’s in order to transform i t into a Kr+l,with each vertex of G added exactly once. Obviously after t h a t transformation we have n graphs K,+,’s, with each vertex of G belonging t o r + l of them. In order to do t h a t and to have all the edges of G covered we need a bijection f between V ( G )and the set of the n Kr’s, such t h a t for any vertex x of G:  in case i): x does not belong to f ( x ) .  in case ii): if [ x , y ]is an edge of the matching, then y € f ( x ) .  in case iii): if [ x , y ]is the arc having x as origin in the 2regular oriented partial graph of G , then y E f ( x ) . In order to show that such a bijection exists we define a bipartite graph, having as stable sets of vertices the K,’s and the vertices of G. We put a n edge between a vertex of G and a K , if the vertex can be added to the K,. The result is a regular bipartite graph (of degree nr, r and r respectively in cases i), ii) and iii)). A wellknown corollary of the KiinigHall theorem states that a regular bipartite graph admits a perfect matching. The matching defines the bijection we need. 0 We state now some corollaries of this lemma, which are used in section 3, in these terms. Corollary 4.2 : Let G be a 3regular graph containing a perfect matching. Then G can be covered by K,’s, in such a way that each vertex belongs to three K3’s. Corollary 4.3 :
Let G be a 4regular graph containing two disjoint perfect matchings. Then G can be covered by K3’s,in such a way that each vertex belongs to three K,’s.
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Corollary 4.4 : Let G be a ®ular graph that can be decomposed into K3’s. Then G can be covered by K4k, in such a way that each vertex belongs to four K4k. Corollary 4.5 : Let G be a n 8regular graph that can be decomposed i n t o K,’s and a 2regular graph. Then G can be covered b y K4k, in such a way that each vertex belongs to four K4k. 5. A m u l t i p a r t i t e g r a p h decomposition problem
We propose t o study the following problem: Problem For what values of h and r can the multipartite graph K4,4,..,4,r on 4h+r vertices be decomposed into K 4 k ? If the multipartite graph K4,4,,,,,4,r on 4h+r vertices can be decompmed into K4’sthen necessarily
I
h = O (mod 3) r E 1 (mod 3) h2r/L2+1
We obtain the first two necessary conditions by divisibility arguments and the third one by counting the edges covered by the K4’scontaining an element of the rset. We can remark that Lemma 2.2 proves that the above necessary conditions are sufficient for the case r = 4 (this case can also be seen as a corollary of the well known theorem of Hanani [8], which says that K , can be decomposed into parallel classes of K4’sif and only if n = 4 (mod 12)). Here we will prove that the above necessary conditions are sufficient for the case r = 10. We conjecture that the necessary conditions are sufficient in general, except eventually cannot be decomposed into for a few values (for example it can be shown that K4,4,4,1 K4’s).
Regular Packings and Coverings
95
Theorem 5.1 K424,.,.,4,10 c a n be decomposed i n t o K,’s i f and only i f h , t h e n u m b e r o f 4vertex parts, satisfies h
G
0 ( m o d 3)
{ h a 3 or equivalently
K , c a n be decomposed i n t o one Klo, a parallel class o f K,’s on t h e other n10 vertices and K,’s i f and o n l y i f n E 10 ( m o d 12), n # 22. or also, K , c a n be decomposed i n t o a parallel class of K,’s, 10 parallel classes o f K3’s and K,’s i f and only i f n = 0 ( m o d 12), n # 12. In what follows we will use the three alternative forms. One can easily see, on the third formulation of the theorem, that it is a refinement of Lemma 2.5.
To prove the theorem we will need some composition lemmas, for which we will use the following remark:
Remark 5.2 K,,,,,,, c a n be decomposed i n t o parallel classes o f K,’s, for t#2,3,6,10. Proof It is well known that Kt,,,,,t,tcan be decomposed into K5’s if and only if t#2,3,6,10 (see [ S ] ) . Given such a decomposition we can delete the five vertices of a stable set of K t , , , f , t , tEach . deleted vertex gives rise to a parallel class of K4’sin K,,,,,,,. 0 We have the following composition lemmas. Lemma 5.3 Let t z u , i f K , , , and K12uc a n be decomposed into one parallel class o f K,’s, 10 parallel classes of K3k a n d K4’st h e n so do K48, and K48t+12u. Proof The graph K48t+12u is the edge disjoint union of four K l z t ’ s ,one KlZuand the mulThe K,,,’s and the K12ucan be decomposed into 10 tipartite graph K12t,12t,12t,12t,12u. parallel classes of K 3 , one parallel class of K 4 and K4’s by hypothesis. By Lemma 2.2, K,,,, and K,,,, can be decomposed into K,’s. Therefore, according to Lemma 2.3, K12t,12t,12t,12t,,2u can be decomposed into K4’s for u 5 t. Therefore I(,,,+,, can be decomposed into 10 parallel classes of K3’s , one parallel class of K,’s , and K4k. 0
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96
Lemma 5.4 If K12t+9 can be decomposed i n t o 10 parallel classes of K3’s and K4’s t h e n K48t+36 can be decomposed into a parallel class of K , ‘s, 10 parallel classes of K3’s and K4’s. Proof K,,,,,, is the disjoint union of four K,2,+g and the multipartite graph K12t+9,12t+9,12t+9,12t+9. By hypothesis the Ki,t+g can be decomposed into 10 parallel classes of K3’s and K4k. As 12t 9 # 2,3,6,10, by applying Remark 5.2, there exists a decomposition of K12t+9,12t+9,12t+9,12t+9 into parallel classes of K,’s. 0
+
Lemma 5.5
I f K,,,,,
can be decomposed i n t o 7 parallel classes of K,’s and K,’s t h e n K4,,+1, can be decomposed i n t o a parallel class of K,’s, 10 parallel classes of K,’s and K4’s. Proof
I
Let X = X l u X 2 u X 3 u X 4 with B, = 12t+4. By Remark 5.2 the multipartite graph K12t+4,12~+4,12t+4,12t+4on vertex set X can be decomposed into a parallel class of K4’s and K4’s as 12t+4 # 2,3,6,10. Choose a particular K,, {x1,22,%3,24}with xi=;. Then the edges of the K48t+12 constructed on X  {x,, z2,x3, z,} can be partitioned into the four K12t+3 on vertex set Xi  z,,each decomposable into 7 parallel classes of K,’s and K,’s by hypothesis, the K4’s of K12t+4,12t+4,12t+4,12t+4 not containing xi for any i and the K3’s obtained from the K4’s containing one of the vertices xi when deleting it. The K3’s of this decomposition can be partitioned into 10 parallel classes of K3’s on 48t +12 vertices in the following way: 6 parallel classes, each of them being the union of parallel classes on each X ,  xi, 4 parallel classes obtained for i = 1, 2 , 3, 4 by taking, for each i, the parallel class (Xi xi) still unused on Xi  xi and by adding the parallel class of K,’s on
u
obtained from the K4’s containing xi after deletion of xi. 0
i#i
Proof of Theorem 5.1. First the given conditions are necessary as we saw before. To prove that they are sufficient, by using Lemma 5.3, we only have to prove t h a t K , can be decomposed into a parallel class of K 4 k , 10 parallel classes of K3’s and K 4 k , for n=12t with t = 2,3,4,5,6,7,9,11,13,17,21in order t o finish the proof. From Lemma 5.4 a n d Lemma 2.5 we get the result with t = 7 , l l . From Lemma 5.5 and Lemma 2.4 we get the result with t = 5,9,13,17,21. In what follows we will give ( t = 2,3,4,6). 0 The case
t
the direct constructions for the remaining cases
= 2:
For n = 24, the decomposition of .K,4 into a parallel class of K4’s and 10 parallel classes of K3’s is as follows. Let V(K,,) = Z6XZ,; the vertices are labeled (i,j)with i = 0, 1,..., 5 and j = 0, 1, 2 , 3. The decomposition is given below:
97
Regular Packings and Coverings
for j for j for j for j for j for j for j for j for j for j
= 0,1,2,3 = 0,1,2,3 = 0,1,2,3 = 0,1,2,3 = 0,1,2,3 = 0,1,2,3 = 0,1,2,3 = 0,1,2,3 = 0,1,2,3 = 0,1,2,3
for i = 0,...,5 The case t = 3:
For n = 36 the decomposition of K36 into 10 parallel classes of K3’s and 5 parallel classes of K,’s is as follows. Let V(K3,)= Z4XZg;the vertices are labeled ( i , j ) with i = 0, 1, 2, 3 and j = 0, 1,..., 8. The decomposition uses t h e known decomposition of K , into 4 parallel classes of K3’s ([16]). We form 8 parallel classes in the following way: for each t = 0, 1, 2, 3 we take 2 parallel classes on t h e 27 vertices ( i , j ) , i # t with two parallel classes of the K , constructed on the vertices ( t , j ) . The last two parallel classes are formed with the two unused classes on each K,. for for for for for for for for for for
j = 0,...,8 j = 0,...,8 j = 0,...,8 j = 0,...,8 j = 0,...,8 j = 0,...,8 j = 0,...,8 j = 0,...,8
i i
= 0,1,2,3 = 0,1,2,3
for for for for for
j j j j j
= 0, 8 = 0, 8 = 0, 8 = 0,...,8 = 0, 8
and and and and and and and and
a class from ( 3 , j )
a a a a a a a
class class class class class class class
from from from from from from from
(3,j) (24 (2,j) (1,j) (1,j) (0,j) (0,j)
..., ..., ...,
...,
The case t = 4: For n = 48 the decomposition of K,, into 10 parallel classes of K3’s,a parallel class of K4’s and K,’S is as follows. Let V(K4,) = Z4,.
JC.Bermond, J. Bond and D. Sotteau
9s Classes of K 3 k [i,i+16,i+32] [i,i+l,i+5] [i,i+l,i+5] [i,i+l,i+5] [i,i+2,i+13] [i,i+2,i+13] [i,i+2,i+13] [i,i+7,i+26] [i,i+7,i+26] [i,i+7,i+26]
for i = 0,...,15 for i = 0 (mod 3) for i = 1 (mod 3) for i = 2 ( m o d 3) for i = 0 (mod 3) for i 1 (mod 3) for i = 2 (mod 3) for i = 0 (mod 3) for i = 1 (mod 3) for i = 2 (mod 3)
Class of K4k: [i,i+l2,i+24,i+36]
K4’s: [i,i+3,i+2O,i+30] [i,i+6$+14,i+39]
for i = 0,...,11
for i = 0,...,47 for i = 0,...,47
The case t = 6:
For ‘u = 72 the decomposition of K72 into 10 parallel classes of K3’s,a parallel class of K4’s and K,’s is as follows. Let V(K72)= Z72. Classes of K3’s: [i,i+24,i +48] [i,i+l,i+5] [i,i+l,i+5] [i,i+l,i+5] [i,i+2,i+10] [i,i+2,i+10] [i,i +2,i +lo] [i,i+7,i+20] [i,i +7,i +2o] [i,i+7,i +20]
for i = 0,...,23 for i = 0 (mod for i = 1 ( m o d for i = 2 (mod for i = 0 (mod for i = 1 (mod for i = 2 (mod for i = O (mod for i = 1 (mod for i = 2 (mod
Class of K4k: [i,i+18,i+36,i+54] 3) 3) 3) 3) 3) 3) 3) 3) 3)
for i = 0,...,17
K4k: [i,i+3,i+12,i+41] [i,i+ll,i+26,i+51] [i,i+14,i+3OIi+49] [i,i+6,i+33,i+50]
for for for for
i = 0,...,71 i = 0,...,71 i = 0,..,71 i = 0,...,71
Acknowledgement This research was partially supported by P.R.C. Math. Info. References 1. JC. Bermond, C. Huang, A. Rosa, and D. Sotteau, Decomposition of complete graphs into isomorphic subgraphs with five vertices, Ars Combinat. 10 pp. 2112. 3.
4. 5.
254 (1980). JC. Bermond, J. Bond, and JF. Sacle, Large hypergraphs of diameter one, i n Graph Theory and Combinatom’cs,Proc. Coll. Cambridge, 1083, pp. 1928 (1984). JC. Bermond and J. Bond, Combinatorial designs and hypergraphs of diameter one, in Proc. First ChinaUSA Conf. on Graph Theory, Jinan,June 1086, (1987). A.E. Brouwer, H. Hanani, and A. Schrijver, Group divisible designs with block size four, Discrete Math. 20 pp. 110 (1977).
A.E. Brouwer, Optimal packings of K,’s into a K,, Journal of Comb. Th., ser. A 26 pp. 278297 (1979).
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6.
C.J. Colbourn and A. Rosa, Quadratic Leaves of Maximal Partial Triple Systems, Graphs and Combinatom'cs 2, pp. 317337 (1986).
7.
J. Doyen and A. Rosa, An updated bibliography and survey of Steiner systems, pp. 317349 in Topics on Steiner systr;ms, C.C. Lindner and A. Rosa ed., Annals of Discrete Math., 7, (1980).
8.
H. Hanani, Balanced incomplete block designs and related designs, Discrete Math. 11 pp. 255369 (1975).
9.
A.J.W. Hilton and C.A. Rodger, Triangulating nearly complete graphs of odd order, i n preparation, (1986).
10. C. Huang, E. Mendelsohn, and A. Rosa, On partially resolvable tpartitions, pp. 169183 in Theory and Practice of Combinatorics, A. Rosa, G. Sabidussi and J. Turgeon ed., Annals of Discrete Math., 22, (1982). 11. E. Mendelsohn, On (Near)Hamiltonian Triple Systems and related factorizations of complete graphs, Technion Report MT655,(1985).
one
12. W.H. Mills, On the covering of pairs by quadruples. I, Journal of Comb. Th., ser. A 13 pp. 5578 (1972). 13. W.H. Mills, On the covering of pairs by quadruples. 11, Journal of Comb. Th., ser. A 15 pp. 138166 (1973). 14. W.H. Mills, Covering designs I: Coverings by a small number of subsets, Ars Combinat. 8 pp. 199315 (1979). 15. C.St.J.A. NashWilliams, An unsolved problem concerning decomposition of graphs into triangles, Technical report, University of Waterloo, (1969). 16. D. K. RayChaudhuri and R. M. Wilson, Solution of Kirkman's schoolgirl problem, pp. 187204 in Proc. of Symp. in Pure Math., vol I S Combinatom'cs,Amer. Math. SOC. Providence (1971). 17. R. Rees, Uniformly resolvable pairwise balanced designs with blocksizes two and three, Preprint, Queen's University, Kingston, (1986). 18. R.M. Wilson, An existence theory for pairwise balanced designs, 111: Proof of the existence conjectures, Journal of Comb. Th., ser. A 18 pp. 7179 (1975).
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Annals of Discrete Mathematics 34 (1987) 101106
101
0 Elsevier Science Publishers B.V. (NorthHolland)
An inequality on the parameters of distance regular graphs and the
uniqueness of a graph related to M23. A.E. Brouwer & E. W. Lambeck
Tech. Univ. Eindhoven, The Netherlands
To Alex Rosa on his jiftieth birthday
We give an inequality on the parameters of distance regular graphs, and show the geometric consequences of equality. As an application we show that there is a unique distance regular graph with intersection array { 15,14,12;1,1,9}. 0. Introduction.
If r is a graph and y is a vertex of r, then let us write r;(y) for the set of all vertices of r at dis6 tance i from y, and T(y) = TI(y) for the set of all neighbours of y in r. We shall also write y to denote that y and 6 are adjacent. The graph r is called distance regular with diameter d and intersection array (bo, ' ' . ,bdI;CI, ' ' . . C d ) if for any tW0 VWtiCeS 7, 6 at distance i we have I r , + I ( y ) n r(6)I = bi and I r,,(y) n r(6)1, = ci (0 Q i Q 4. Clearly bd = c0 = o (and c 1 = 1). Also, a distance regular graph r is regular of degree k = bo, and if we put a . = k  b , ci then I r,(y)n r(6)I = a, whenever d(y,6) = i. We shall also use the notation k; = 1 T,(y) I (this is independent of the vertex y). For basic properties of distance regular graphs, see BIGGS[I]. Distance regular graphs of diameter 2 (ix., strongly regular graphs) are very common, but not so many of larger diameter are known. Often, but not always, such graphs turn out to be uniquely determined (is., determined up to isomorphism) by their intersection array. Our aim in this note is to present an inequality for the parameters of distance regular graphs, and use the geometric information obtained from the proof of this inequality in the case of equality to show that there is a unique distance regular graph with intersection array {15,14,12;1,1,9). Let us first show that there is at least one such graph. Indeed, let r be the graph with as vertices the 506 blocks of the Steiner system S(5,8,24) not containing one fixed symbol, where two blocks are adjacent whenever they are disjoint. From the parameters of S(5,8,24) one sees that two blocks have distance 0,1,2,3 in r when they meet in 8,0,4,2 points, respectively, and it is easy to check that r is distance regular with intersection array (15,14,12;1,1,9) and distance distribution diagram

2
6
In fact r is distance transitive under the action of M23. (For undefined terminology, see, e.g., BIGCS[ 11 and CMRON & VAN LINT[4].) Any two blocks at distance two determine a tetrad (4
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set), and hence a sextet (partition of the set of 24 symbols into six tetrads such that the union of any two of these is a block of S(5,8,24)). One of the tetrads of the sextet contains the forbidden symbol, and the pairwise unions of the remaining five tetrads give us ten blocks, forming a Petersen subgraph of r. Thus: any two vertices of r at distance two are contained in a unique (geodetically closed) subgraph of r isomorphic to the Petersen graph. In this note we show that r is the unique distance regular graph with intersection array {15,14,12;1,1,9). Our strategy will be, given some graph r with these parameters, to first construct Petersen subgraphs, and next embed the graph as a subgraph in the (unique) near hexagon on 759 vertices, the graph on all the blocks of S(5,8,24) (where disjoint blocks are adjacent). [Added in proof A.A. Ivanov infoms us that he and S.V. Shpectorov had done some work on the characterization of this graph, and in particular also had proved the existence of Petersen subgraph.] 1. Aninequality.
'Ibeorem. Suppose Then
is a distance regular graph and a,
# 0 for some
i, 1 < i
< d Put ad+ I
= 0.
(1)
ai+lbi a,Ic, bi+ci 1, then ci G ai
ailci +a, '
an4 for i < 4 equality implies ai+l = ai.
 
Proof. Count Ctuples (a,B,y,6) with a B, y 6, d(y,a) = d(y& = i, d(6,a) # i. Clearly, there are vkiai(bi ci) such Ctuples. We 6nd the inequality
+
vkiai(bi
+ ci) G vkiaf + vki+lq+lc,+l + vkilailbil
by covering this collection of Ctuples with the three sets
I I Ia
((a,B,yJ) a
{(a.B,yJ) a ((a,B,yJ)
  
0, + I,d(y,a) = i ) ,
B,y
a,d(y,a) = d(y,B) = i 4 6 , B ) =
B,y
6,d(6,4 = d(&B) = i
B,y
6 , d ( b ) = 4 , 8 ) = i  l,d(y,a) = i } .
+
This proves the inequality in (i); those in (u), (iii) follow similarly by restricting d(a,6) to be i 1 and i  1, respectively. Equality in (i) is equivalent with: if a 8, y 6, d(a.y) = i, then the three distances d(a,S), d(B,y), d @ J ) are not all equal. With d(y,B) = d(S,B) = i f 1 this implies ai,l < a, (varying 6). Similarly, we find a,l < ai in case of equality in (i), but comparing with (i) yields the opposite inequality, so ai  I = ai. Likewise a, + I = ai in case of equality in (iii). 0
 
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Parameters of Distance Regular Graphs
Remarks. Clearly, some more geometric information is available in case of equality in (ii) or (ii). A special case of this inequality has already been given by IVANOV,IVANOV& FARADJEV [6];we have been told that also N o m [7] contains a weak form of this result. 2. Construction of Petersen subgraphs. Lemma. Let r be a graph with h = 0, p = 1, a 2 = 2, and suppose that each induced hexagon with at least one pair of vertices at distance 3 contains three such pairs. Then for each vertex y of the subgraph r2(y) is the disjoint union o f i k 2 hexagons, and hence any two vertices of at distance 2 are contained in a (unique) Petersen subgraph.
r
r,
 
Proof. Fix 00 E I‘ and let for a E r2(w), Z be the unique common neighbour of 00 and a. From = 2 it follows that r2(w) is a+on of polygons; ... q be such a polygon; clearly g 2 5. Since d ( a j , a j + l )= 2, the vertex a i f l has twoeighbours in r2(a,),say w and &. Let y, be the common neighbour2f a, and pi. Since d(n,,ai+l) = 2, it follows from the a; a, yi pi ai+l 01 that also d ( w , y ; ) = 2. hypothesis applied to the hexagon 00 z }= . aj+2 &n w a a v e the triangle But then y, E ( a i  l , a j + l }and pi E ( ~ ~ ~  ~ , a If, + pi a,+la;+la;+Z, contradicting X = 0. Thus 8, = so that ai+l = ajz (i EZ,). Now q a3, a3,+l al for all j E Z,, so that there are only two points a3j,and g = 6. 0

a2
     
 

Now assume that r has intersection array (15,14,12;1,1,9}. Then c 3 = 9, a2 = 2, a 3 = 6 and we have equality in (iii) (and hence in (i)) of the theorem in the previous section. The information about the w e of equality stated there immediately implies that the hypothesis of the Lemma about hexagons is satisfied, and hence the conclusion of the Lemma holds: any two vertices of r at distance two are contained in a unique Petersen subgraph of r. 3. 7he equivalence relation on the edges.
Let r be a distance regular graph with intersection array (15,14,12;1,1,9}, and let P be a Petersen subgraph of r. Write r j ( P ) for the set of vertices of r at distance j from P, and r ( P ) = r l ( P ) . Let us study the structure of r around P. 1. For evev y E r ( P ) we have I r(y) fl P I = 1 and I T(y) i l r ( P ) I = 0. (For: there are no triangles or 4gons in r; every 5gon is in a unique Petersen subgraph and P is geodetically closed.) 2. Ify E r2(P), then r2(y) n P does not contain 2claws. (For suppose yI, y2, y3 E r2(y) n P such that y1 y2 y3. Then y and y2 determine a unique Petersen subgraph P ’ , which contains r(yz) n F2(y). But now yl, y2, y3 E P n P ’ , contradiction.)
 
3. If y E T 2 ( P )then r2(y) n P is either a 4coclique or it induces 3K2 as a subgraph of P. Proof. By 2. there are only 10 possibilities left for the isomorphism type of the subgraph of P induced by rz(y) fl P , namely (i) 3K2, (ii) 2Kz K1, (iii) 2K2, (iv) K2 2K,, (v) K2 + K1, (vi) K2, (vii) 4K1, (viii) 3K1, (ix) 2KI, (x) K 1 .(Note that in case (viii) there are two essentially distinct embeddings in P.) Write Q = P \ (P n T2(y)) = P n r3(y). For each vertex S of Q we find c3 = 9 points in r(y) n T2(S). If we put ea = I r(6) n (P \ Q ) I , then ea of these points lie in r ( P ) , and the remaining 9  es are in r 2 ( P ) . Put A ( 6 ) = T(y) n r2(6)n r 2 ( P ) . Now suppose Q contains a path S1 S2 S3; then by 2. the set A ( S I ) n A ( & ) n A ( S 3 ) must be r 2 ( P ) I = 15  a, then the ineempty. In particular, if we put a = I P \ Q I so that I r(y) fl 2a holds. This quality 27  ea,  ea,  ea, < 30  2a, or, equivalently, 3 ea, e$ e& observation will kill most of our ten possibilities. Indeed, suppose we have situation (i). Then we can find a path 61 82 8, in Q such that e = (es,,e$,es,) = (2,1,2), but a = 5, contradicting the above inequality.
+
+
 
+ + +
 
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Similarly, in cases (i),(iv),(v) we find a = 4, e = (2,1,0) and a = 4, e = ( l , l , l ) and a = 3, e = (1,0,0), respectively, a contradiction in eacli case. In case (vi) thisargument is not strong enough, but we can find a K 1 , 3in Q,say with vertices 80, S , , 82, 83, where 8, is the vertex of degree 3, and there must be a pair i, j in 1 , 2 , 3 such that A (8,) n A (8,) n A (8,) # 0 , again a contradiction. The cases (viii)(x) can be ruled out by similar arguments, but since we shall not need the absence of these, we do not give the details. 0 We define a relation / on the edges of r by: el X e2 if and only if there is a Petersen subgraph P containing them, and where e l U e2 induces a subgraph isomorphic to 2K2 (it., where each vertex of e l has distance 2 to each vertex of e2). Observe that X restricted to a Petersen graph is an equivalence relation with equivalence classes of size 3. 4. The relation X is on equivalence relation. Prcklf. Suppose e l l e i and el / e 3 . If e l , e2 m d e3 are in one Petersen subgraph, then by the above observaticn el / e l . So assume el,e2 C P I and e2, e3 C P 2 , and put el = (yl,yz}. Now it fcl!cws from 1. that d(yi,P2) = 2 (i = 1,2) and e2 C r 2 ( y i ) n P 2 (i = 1,2). By 3. we have e 3 C rl(y,) r; P 2 (z = 1,2) acd therefore e l and e3 determine a unique Petersen subgraph P 3 containing them, and e l C, e 3 induces 2K2 as subgraph of P J . This means that e , X e 3 . 0
No!e that ihere are precisely 253 equivalence classes in r, with 15 edges in each equivalence class. (Indeed, each d g e is m 7 Petersen graphs, and hence equivalent to 14 other edges.) These equivalence classes will be the missing points of our near hexagon. 4. Embedding in a near hexagon. A near pobgon is a partial linear space such that given a point and a line, there is a unique point on the given line closest to the given point. Here ‘closest’ refers to distance in the collinearity graph. A nenr hexagon is a near polygon of diameter 3. For the theory of near polygons, see BROUWER & WILBRINK[3].
Continuing with the assumptions of the previous section, we construct a partial linear space ( X , 9 as follows. The point set X is the union of the set X , of vertices of r and the set XIof equivalence classes of edges. The set of lines e consists of the 3sets ( y , S , E } for each edge e = (yJ} of r, where E is the equivalence class of e. Let us check that this linear space is a regular near polygon of diameter 3, with parameters (s + 1,t2 1,t 1) = (3,3,15).
+ +
First of all,we really have a partial linear space: an equivalence class E of edges and a collinear vertex y determine uniquely an edge e = (7,s)E E. Next, the partial linear space is connected, since r is connected. Note that r is an induced subgraph of the collinearity graph of (X, e), and that XIis a coclique. Let us look at the parameters. We have I X O I = 506, I X I I = 253 and hence 1 X I = 759. Each point is on t 1 = 15 lines, and each line contains s 1 = 3 points. In order to show that r 2 + 1 = 3, we must show that any two points 5, q at distance 2 have 3 common neighbours. If 5, q E X,, then they determine a Petersen subgraph P, and the three common neighbours are their common neighbour in r and the two equivalence classes containing edges on ihem. If 5 E X,, q E XI,then q contains an edge { a $ } , where 5 a. Now a, p and 5 are contained in a Petersen subgraph P, and any common neighbour of 5 and q must be in X, (since XI is a coclique), and therefore must be one of the three neighbours of on the three edges that 7 has in P. Finally, if [ , q E X I , then 5 contains an edge ( a , p } and q contains an edge { a , y } ; again a, p, y are contained in a Petersen subgraph P, and any common neighbour of 5 and q must be a vertex 8 at distance two from each of a, /3 and y, and therefore (by 2.) in P; but in P we see three common neighbours of 5 and q. This proves that any two points at distance 2 have precisely three common neighbours.
+
+

105
Parameters of Distance Regular Graphs
Now, let us check the near polygon axiom. Let L be a line, say L = { y , 6 , E } with E E XI, and let 5 be a point with d ( [ , L ) = 1. If ( E, then 5, y and 6 are contained in a Petersen subgraph of r and ( has distance two to the edge (y,6}. Otherwise, if ( E X,,,then ( is adjacent to either y or 6, but not both, since r does not contain triangles, while if 5 E X I and ( y. 6 then 5 = E. Thus, the near polygon axiom is satisfied in this case.


Next suppose that d ( ( , L ) = 2. If ( E XO and d((,y) = d(.&S) = 2, then [,y and 6 are contained in a Petersen subgraph, and ( is on an edge parallel to {y,S), so that 5 is collinear with E, contrary to our assumption. If ( E X o and d((,y) = d([,,E) = 2, then E contains an edge (a$}, where [ a, but now a,b, y, 6 and 5 are all contained in one Petersen subgraph of r, and we find ( 6, contradiction. If ( € X I and d([,y) = d((,6) = 2, then ( contains two edges (.$} and {s.~) where y a and 6 5. But these two edges are contained in a Petersen subgraph P , and this situation contradicts 1. Finally, if 5 E X I and d([,y) = d(5,E) = 2, then 5 contains two edges (aJ) and ( 5 , ~ where ) y a and, for some 0 E Xo, { q , O , E ) is a line. But then the edge {a,y)is contained in rz(V), so there is a Petersen subgraph P of r containing 7 , a and y, and therefore also p, I, 6,O. It follows that d((,6) < 2, contradiction. Thus, the near polygon axiom is dso satisfied in case d(5.L) = 2.



Given a line L, there are 3 points on L, 3.2.(151) = 84 points at distance one to L, and 84.2.( I5  3) / 3 = 656 points at distance two from L. But 3 84 + 672 = 759, so (X,f) has diameter 3 and each point is at distance at most two from any l i e . This concludes our verification: we do have a regular near polygon of diameter 3 and with the parameters stated. But BROWEE i2] shows that there is a unique such near polygon, namely that with as points the blocks of the Steiner system S(5,8,24), and as lines the triples of pairwise disjoint blocks. Our last task is to identify r as a subgraph of the cohearity graph of this near polygon.
+
Lemma. The near polygon constructed above contains precisely 24 cocliques of size 253, numek the 24 sets of blocks of S(5,8,24) containing any given symbol.
Proof. Let C be such a coclique. Sir.= there are 3795 lines, and each point is on 15 lines, it follows that C meets each line in precisely one point. In particular, if Q is any quad, then C f! Q is an ovoid in Q (for terminology, cf. BROUWER [2]), that is, the set of five blocks containing a fixed tetrad T. Each block in C meets T, for blocks disjoint from T are disjoint from a block containing T. If Q ’ is another quad, we find another tetrad T’. If T f’ T’ = 0 , then there is a block on T 23 disjoint from T’, contradiction. Thus, the set of 1771 = ( 3 ) tetrads found by varying Q is a linked system, and by a theorem of ERDOS,KO & RADO [5] this collection of tetrads consists of all tetrads on a fixed symbol. It follows that all blocks in C contain this symbol. 0 Thus, starting from an arbitrary graph r with intersection array (15,14,12;l,l,9} we find r as the complement of a 253coclique in the collinearity graph of the near polygon on 759 points, and we have shown that r is uniquely determined. References 1.
2.
3. 4.
Bigs, N.L., Algebraic Graph Theory, Cambridge Tracts in Math. 67, Cambridge University Press, Cambridge (1974). Brouwer, A.E., The uniqueness of the near hexagon on 759 points, pp. 4760 in: Finite Geometries, Lecture Notes in Pure and Applied Math. 82 (ed. N.L. Johnson, M.J. Kdaher & C.T. Long), Marcel Dekker, New York, 1982. MR 84d51021 (MR 82j05038) Brouwer, A.E. and H.A. Wilbrink, The structure of nearpolygons with qua&, Geometriae Dedicata 14 (1983) 145176. Cameron, P.J. and J.H. van Lint, Graphs, Codes and D e s i p , London Math. SOC. Lecture Notes 43, Cambridge Univ. Press, Cambridge (1980).
106 5.
6.
7.
A.E. Brouwer and E.W.Lambeck Erdk, P., Chao KO, and R. Rado, Intersection theorem for systems of finite sets, Quart. J. Math. Oxford (2) 12 (1961) 313320. T ~ ~m eEn e~m E T E I I ~ ~ Ivanov, A.A., A.V. Ivanov, and LA. Faradjev, ~ E C ~ ~ O ~ ~ O  rpa+bl 5, 6 E 7 @reprint) = Distancetransitive graphs of valency 5, 6 and 7, Eur. J. Combinatorics (to appear). USSR Comput. Maths. Math. Phys. 24 (1984) pp. 6776, &umd ViEisl. Mat. i Mat. Fiz. 24 (1984) pp. 17041718. Nomura, K., An inequali!y between intersection numbers of a distance regular graph, J. Combinatorid Th.(B) (submitted).
Annals of Discrete Mathematics 34 (1987) 107118 0 Elsevier Science Publishers B.V. (NorthHolland)
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Partitions into indecomposable triple systems Charles J. Colbourn and Janelle J . Harms Department of Computer Science University of Waterloo Waterloo, Ontario, N2L 3G1 CANADA TO A L E X ROSA O N UIS
3I3TIETU BIRTUDAY
ABSTRACT The indecomposable partition problem is to partition the set of all triples on v elements into s indecomposable triple systems, where the i t h triple system has index Xi and XI+ * . * +A, = v2. A complete solution for v S l 0 is given here. Extending a construction of Rosa for large sets, we then give a v + 2v+l construction for indecomposable partitions. This recursive construction employs solutions to a related partition problem, called indecomposable nearpartition. A partial solution to t h e indecomposable nearpartition problem for v=10 then establishes t h a t for every order v = 52' 1, all indecomposable partitions having X i = 1,2 for each i can be realized.
1. The problem A triple system of order v and index 1, denoted TS(v,X) is a pair (V,B). V is a set of elements, and B is a collection of 3subsets of V called blocks or triples; every 2subset of elements appears in precisely X of these triples. W e assume throughout this paper t h a t there are no repeated blocks; t h a t is, blocks are distinct 3subsets. A triple system is indecomposable if there is no proper nonempty subset B'CB for which (V,B') is a triple system. The full triple system of order 'u consists of all h u b s e t s of a vset, and hence is the unique TS(v,v2) without repeated blocks. A partial triple system PTS(v,X) relaxes the requirement t h a t each pair appear in exactly blocks to a requirement t h a t each pair appear in at most X blocks. Mesner [11] asked the following question. Let A = {Xl,,..,X,) be a partition of the . . * X, = v2. Can one partition the blocks of the full integer v2, i.e. X, design TS(v,v2) into s classes BI,...,B, so t h a t for each 14;u{cqv}. Adding B , to this covering produces a B[3,3;n+l] with N(oq)=FIUQ’. As in theorem 5.1, we now simply exchange edges in GN(+ and N(oq)G. 0 Since in this section we deal only with simple graphs, theorem 2.10 ensures t h a t every cubic graph with a component of at least eight vertices has a 2PFZ; hence, we need only handle simple ;ubic graphs with each component of size four or six. Every such cubic graph has a 1 factor (a K1,3and n4/2 disjoint edges); we prove a somewhat stronger statement t h a n t h a t needed here: Lemma 6.2: Let G be a cubic multigraph on n& (mod 6) vertices having a l*factor. Then G is a neighbourhood of a B[3,3;n+l]. Proof: Let F be a I*factor of G and Q=GF. By theorem 4.2, there is a PB[3,l;n] with leave F , and by theorem 4.5 there is a PB[3,2;n] with leave Q ; their union is a PB[3,3;n] with leave G . By adding a new element oq we get a B[3,3;n+l] with N(oq)=G, as required.
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7 . Cubic Multigraphs, n s 4 (mod 6)
We do not have a characterization of cubic multigraphs having 2PFZ’s, and hence are unable to use theorem 6.1 directly. Nevertheless, we are able to circumvent the need for such a characterization. Lemma 7.1: Suppose t h a t G is a cubic multigraph on n 3 (mod 6) vertices, and t h a t G is constructed by replacing an edge { a $ } not in a 3bond of G I by adding a bowtie. Then G is a neighbourhood in a B[3,3;n+l]. Proof: Form G‘ by removing all 3bonds from GI, and let (H,H‘) be a P F Z for G’ (by theorem 3.2); suppose without loss of generality t h a t a , b m . Now the bowtie added and {z,y}. Of replaces t h e edge {.,b} by edges {.,u},{b,u},{u,v}, v,z},+,y},{z,!/}, these edges, place {u,v} and both copies of {z,y} in H a n d the remainder in H‘ (removing { a $ } in the process). Now we proceed exactly as in theorem 6.1, with u,u,z,y serving t h e same roles. 0
I
Next we handle cases resulting from addition of a 2bond: Lemma 7.2: Suppose t h a t G is a cubic multigraph on n=l (mod 6) vertices, constructed from a cubic multigraph G’ by adding a 2bond, a n d adding any number of 3bonds. Further suppose t h a t G’ has a 1PFZ. Then G is the neighbourhood of a B[3,3;n+l]. Proof: Let (H,H‘) be a 1PFZ for G’, and suppose t h a t a 2bond is added by replacing t h e edge { a , d } by { a , b } , { b , c } , { b , c } , { c , d } . If { a , d } is a path of length one in H , we replace it by { a , b } , { c , d } and add { b , c } twice to H’; now we proceed as in theorem 6.1 with u , b , c , d serving the roles of u,u,z,y. If, on the other hand, {u,d} appears in the middle of a path of length three of more in H , we replace it by { b , c } twice, and add { a , b } , { c , d } to HI. Interchanging the roles of H and H‘ enables us to proceed as before. Finally, suppose t h a t { u , d } appears at the end of a path of length two or more in
H , and t h a t d is an endvertex in this path. Replace { u , d } by { a , b } , { c , d } in H , and add { b , c } t o H’ twice. Now since (H,H’) is a IPFZ, there is a path of length one { r , s } either in H or in H’. If i t is in H , apply the method in theorem 6.1 with r , s , c , d playing t h e roles of u,v,z,y. If it is in H’, interchange H a n d H’ a n d apply theorem 6.1 with r , s , b , c playing the roles of u,w,z,y. In all cases, this application is straightforward; one need only ensure t h a t when H‘ is completed to a quadratic graph, the edge {z,y} isn’t added, and also t h a t when H is completed to a quadratic graph, the only repeated edges are {u,v}and {z,y}. These conditions are easy to ensure. 0 In addition, theorem 6.1 is easily adapted to handle any cubic multigraph having a simple component on eight or more vertices. This, together with lemmas 7.1 and 7.2, leaves only the following as possible components:
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C.J. Colbourn and B.D. McKay
All of these graphs have a 1factor, and all but the three in the first row have a l*factor. If a cubic multigraph not handled by previous methods has any component other than those in the first row, it is handled by lemma 6.2. What remains is only those cubic multigraphs having components depicted in the first row. All such cubic multigraphs are 3edgecolourable, and so we establish one more lemma: Lemma 7.3: Let G be a 3edgecolourable cubic multigraph with n=l (mod 6) vertices having some component other than a 3bond. Then G is a neighbourhood in a B[3,3;n+l] unless G=B1. Proof: Let Fl,F2,F3 be a 3edgecolouring of G , with some edge {z,y} appearing in F,, but in neither F2 nor F3; this is always possible since G is not entirely %bonds. Now let Q=F2UF3U{{z,y},{z,y}}, and form a B[3,2;n] with leave Q using theorem 4.7. This yields a PB[3,2;n+l] with N(oo)=F+JF3 which leaves {3c,y} uncovered. Next form a covering with X=1 having N(oo)=F, and covering {z,y} three times. Combining these two yields a B[3,3;n+l] with N(+=G. 0 One final case remains: G contains only 3bonds. In this case, we use the following lemma of Hanani IS]: Lemma 7.4: A group divisible design on w elements with blocks of size 3, groups of size 2, and X=3, exists whenever v d (mod 2), u#4. 0 The triples from such a GDD form the blocks of a PB[3,3;v] whose leave is w / 2 3bonds, which gives a B[3,3;u+l] with N(oo) being all 3bonds, so long as u is not 4. In the sequence of lemmas developed in sections 57, we have dealt with all cubic multigraphs except three: B1, B3, and two 3bonds. In each of these three cases, the graphs are not neighbourhoods, and we prove a simple lemma to show this. Lemma 7.5: Let G be an nvertex eedge graph having a minimal edge cutset of size c separating G into components of size s and ns. If G is a leave of a PB[3,3;n],
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Proof: The left hand side counts edges which remain “inside” the set of s elements or the set of ns elements. Now triples use either three of these inside edges, or they use one inside edge and two edges crossing from one set t o the other. Hence the number of such edges must exceed twice the number of “cross” edges which remain. 0
For B1, we have n=4,e=6,s=2,c=2; if B1 is a leave (or a neighbourhood), we require by the lemma 6 6  4 2 2[12  21, which is not satisfied. For B3, we have n=6,e=g,c=l,s=3 and for two 3bonds we have n=4,e=6,c=O,s=2; in both cases, the requirements of the lemma fail.
+
In summary, we can state the main characterization theorem. Theorem 7.6: A cubic multigraph G on n vertices is an element neighbourhood of a B[3,3;n+l] unless G is B1, B3, or two 3bonds. 0 8. Conciuding Remarks
Theorem 7.6 lends strong evidence in support of a conjecture that every Aregular multigraph meeting numerical conditions and the conditions of lemma 7.5 are realized as neighbourhoods in triple systems; in fact, theorem 7.6 provides the major necessary tool to prove this conjecture for all simple graphs [I]. Perhaps the major reason for interest in the results here is that they provide graphtheoretical tools for addressing problems involving leaves and neighbourhoods.
Acknowledgements Thanks t o Alex Rosa, and to the referee, for helpful comments. The first author also acknowledges the financial support of NSERC Canada.
References C.J. Colbourn, “Simple neighbourhoods in triple systems”, submitted for publication, 1986. C.J. Colbourn, M.J. Colbourn, J.J. Harms, and A. Rosa, “A complete census of (10,3,2) block designs and of Mendelsohn triple systems of order ten. 111. (10,3,2) block designs without repeated blocks”, Proc. Thirteenth Manitoba Con f. Num. Math. Comput. (1982) 211234. C.J. Colbourn, M.J. Colbourn, and A. Rosa, “Completing small partial triple systems”, Discrete Math. 45 (1983) 165179. C.J. Colbourn and A. Rosa, “Element neighbourhoods in twofold triple systems”, J. Geometry, to appear C.J. Colbourn and A. Rosa, “Quadratic leaves of maximal partial triple systems”, Graphs and Combinatorics 2 (1986) 317337. H. Hanani, “Balanced incomplete block designs and related designs”, Discrete Math. 11 (1975) 255369. A.J.W. Hilton and C.A. Rodger, “Triangulating nearly complete graphs of odd order”, in preparation.
C.J.Colboum and B.D. MCKGY
136
[8] R.A. Mathon and A. Rosa, “A census of Mendelsohn triple systems of order nine”, Ars Combinatoria 4 (1977) 309315.
[9] C.E. Shannon, “A theorem on coloring the lines of a network”, J. Math. Phys. 28 (1949) 148151.
[lo] J.E.
Simpson, “Langford sequences: perfect and hooked”, Discrete Math. (1983) 97104.
44
[ll] J. Spencer, “Maximal consistent families of triples”, J . Comb. Theory 5 (1968) 18.
Annals of Discrete Mathematics 34 (1987)137144 Elsevier Science Publishers B.V. (NorthHolland)
0
137
The Geometry of Subspaces of an S(X;2,3,v) Michet Dehon De'partement de Mathe'matique Universite' Libre de Bruxelles 1050 Brussels BELGIUM Luc Teirlinck Division of Mathematics (Algebra, Combinatorics, and Analysis) Auburn University Auburn, Alabama 36849 U.S.A. TO A L E X R O S A O N MIS 313TIETU BIRTUDAY
ABSTRACT W e define some linear spaces on the set of all proper subspaces of a triple system S(x;2,3,v). The connected components of these linear spaces are projective spaces of order 2 and punctured projective spaces of order 3, i.e. projective spaces of order 3 from which a point has been deleted. W e show how these connected components can be used t o find affine and projective factors in S(x;2,3,v). 1. Introduction
A %covering is a pair (S,p) where S is a set whose elements are called points, and is a family of subsets of S called blocks, such t h a t any two distinct points of S are contained in at least one block and every block contains at least two points. We allow repeated blocks. Where no confusion is possible, we denote (S,p) simply by S. A su6space of a 2covering S is a subset S, of S such t h a t any block having at least two points in S, is contained in S,. Clearly, any intersection of a family of subspaces is a subspace. If (S,,B)is a 2covering and if S, is a subspace of S , we denote by p IS,the family of all blocks of S contained in S,. A connected component of a 2covering S is a connected component of the graph obtained on S by joining x and y if and only if x and y are contained in a block B with 123. Every connected component is a subspace. A linear space is a %covering in which any two distinct points x and y are joined by exactly one block, denoted by xy. The blocks of a linear space are usually called lines.
p
M.Dehon and L. Teirlinck
138
Let X,k and v be integers, X>o, k>3, v a . A &design S(X;2,k,v) is a 2covering blocks and every block contains exactly k points. Clearly, if S , is a subspace of (S,p), then (S,,p IS,) is an S(X;2,k, IS, If X=1, we write S(2,k,v) instead of S(l;P,k,v). (S,P), IS I=v, such that every pair of distinct points is contained in exactly
I).
Let S be an S(X;2,3,v) and let X ( S ) denote the set of all proper subspaces of S. Let L3(S) denote the set of all 3subsets {Sl,S2,S3}of X ( S ) such that s,ns,=s,ns,=s,ns, and S , U S , u S , = S . Let L 2 ( S )denote the set of all 2subsets of X ( S ) that are not contained in an element of L3(S). P u t L ( S ) = L 2 ( S ) U L 3 ( S ) . Clearly, ( X ( S ) , L ( S ) is ) a linear space. In section 2, we prove that the connected components of ( X ( S ) , L ( S )are ) projective spaces of order 2 (possibly of dimension 0 or 1). If A and B are sets, AAB = ( A  B ) U ( B  A ) denotes the symmetric difference. If A , B C S for some set S , put A + s B = S  ( A A B ) = ( A n B ) U ( S  ( A U B ) ) . 2 . The structure of L ( S )
Let S be an S(X;2,3,v). Lemma 1: If S , is a subspace of S and if pESS,, then the number of blocks containing p and disjoint from s, equals X(v2 IS, 11)/2. Proof: The point p is contained in X(vl)/2 blocks of S . For every point x of S,, there are X blocks containing { p , x } and every block through p has at most one point in S,. Thus there are XlS,I blocks through p intersecting S , and X(v1)/2X(S11 = X(v2 IS, 11)/2 blocks through p disjoint from S,. 0 Lemma 2: If S , € X ( S ) , we have IS, lv2 with equality holding if and only if every block joining a point of SiS, and a point of S,(S,nS,) has its third point in S,S,.
(2)
IS,nS,
I> (v1+2vsv)/2
with equality holding if and only if
vq=v3.
(3) S, is a subspace if and only if v1=w2=v3 if and only if v1=v2 and (3v 12' )/2Proof:
IS,nS,I
=
Let p ES,S,. Counting, in two different ways, the number of blocks containing p , one point of S3(Sl~S,)and one point of S,S,, yields (1). . with (l),this yields (2). Clearly ISlnS2I = ( ~ 1 + ~ 2 + ~ 3  ~ ) / 2 Together
If S, is a subspace, then (1) yields
~ 1 = ~ 2 = ~ 3 .
this implies S,ftS,). We will prove that S3 is a Assume ~ 1 = ~ 2 = ~ 3(Obviously, . subspace. As S,nS2 is a subspace, any block containing two points of S,nS, is contained in (S,nS2)CS3. Any block joining a point in S,nS, and a point of
139
GeometSy of Subspaces of an S(x;2,3,u)
s,(s,nS,)
must have its third point in S3. Indeed, if this third point would lie in S, (S2), the block would have exactly two points in S1 (S,), contradicting our assumption that Sl (S,) is a subspace. Finally, any block joining two points of s3(s,ns,) must have its third point in S,. Indeed, if this third point would lie in S,S, (S2S1), then by ( 1 ) we would have v 3 h 2 (",>I). Thus S3 is a subspace if and only if ~ 1 = ~ 2 = ~ 3Clearly . q = v 2 and ISlnS, = (3v,v)/?. 0
I
W~=TJ~=V if~ and
only if
Theorem 1: Every connected component of ( X ( S ) , L ( S )is ) a projective space of order 2.
Proof': We first prove that the relation on X ( S ) defined by SlS, if and only if Sl=S2 or the line joining Sl and S, has three points, is an equivalence relation. Obviously, is reflexive and symmetric. Assume that SlS2S3. By lemma 3, we have IS, I= IS, I= IS, I. If S,=S,, then S,S3. Otherwise, let {S2,S3,S4}EL(S).If S, E{S2,S3,S4},then S,S3. Otherwise, {s,ns,, s,ns3, s l n s 4 } ~ ( s l )Thus . Is,ns31 = Is,ns,I = and

(~IS,~W)/?
S,S3.
Thus if P is a connected component of ( X ( S ) , L ( S ) then ) , all lines of P have three points. This means that P is a subspace of the projective space of order 2 defined on all proper subsets of S by +s. 0 3. Generalized Mooreproducts with projective multiplier
A transversal design T(X;k,l) is a triple (S,P,G)where S is a klset, p is a collection of Icsubsets of S called blocks, and G={A1, ...,Ak} is a partition of S into 6subsets called groups, such that
(i) DnA, I = 1 for all
[email protected] and i=l,...,k, and (ii) if z w ,y w j , l (p1)/2 if and only i f KM c a n be decomposed i n t o graphs o n p+l vertices with k edges in s u c h a w a y t h a t every vertex belongs t o 51 p(p1) 2r graphs o f t h e decomposition ( w h e r e M = (51) p+l + l J
+
Proof: Let K , be (Kl,8, k)coloured, let C be the set of colours used and let c E C With the convention that the sums and the maxima are over C unless other
.
wise indicated we have: (1)
C e c = m ( m  1 ) / 2 since each edge appears in exactly one monochromatic subgraph and all edges are coloured.
(2)
m(s1) graphs.
(3)
m1
2 Em,
5 2(s1)
since no vertex lies in more than 51
c eC CmC
from ( 1 ) and (2).
monochromatic sub
P. Fraisse, G. Hahn and D. Sotteau
156 (4)
Cec 5 Em,
max
e, m,
with equality if and only if
(5)
e, e, = maxm, m,
for all c E C
.
since for each c , e , can be written as if t,=O
+ t,, 0 I t , < s,, and since m, 2 each c , 9,1 m = { 7
}5
+
sc(scl) 2t, 2(9,1)
But this is easy: recall that e,
5k
it suffices t o show that for
=
ax{
p 1 2
+ r and s,
’
p (p 1)+2r 2(p+1)
1.
5 p . Then consider two cases;
for each c: (a) s, = p or max{
s,i 2
s,(sc1)+2t,
’
2(sc+1)
s,1 (b) s,
(pl)/2
we have equality in (4) if and only if
+
with
r edges. 2 In order t o have equality in (6) we must also have equality in (3) and thus in (2) which says that every vertex lies in exactly s1 graphs of the decomposition. 0
Corollary 1 For any s 2 2 and k
22,
+ m) +2
sr(K1,, , k ) 5 ;(sl)(l
Proof:
In the proof of the Theorem we have from (1)(4) m1
Now e,
5 m,(m,
This implies that m,
5 2 (s1)
1) / 2
for every c
2 ( 1+
vm)
max
e,
m,
, and, therefore, and
/2
ec
5 m,
rn; 1
 m  2 e, 2 0
+dTz as 4
e,
is
always less than or equal to k . 0
Corollary 2 Let
s 2 2 ,
enough,
P 25k=(2)+r,
or (ii) p
5 5 , then
sr(K1,, , k ) = (sl)(pl) (and only if for
r #
Osrs'' 2 +2
if
.
If (i) p > 5
(sl)(s2)
G
and s is big
0 (mod p )
P 1 ). 2
Proof: The conditions necessary for the existence of a decomposition of K , K,'s are well known :
into
m(m1) = 0 (mod p(p1)) m1 0 (mod p1) Indeed such a decomposition yields a Steiner psystem.
=
Now Wilson's theorem [11] says that the necessary conditions are also sufficient if " for P 1 5 . 1 , the necessary conditions reduce t o Letting m = (sl)(pl) (sl)(s2) = 0 (mod p ) and " m big enough " becomes " s big enough " if k is fixed. 0
m is "big enough". In particular it is known that "big enough" is "at least p
+
P. Frakse, G. Hahn and D. Sotteau
158
Corollary 3 Let s
>_ 2 ,
sr(K13
' 3,
S'(K1,s
9
4)
{
if and only if s otherwise.
= 25
< 251
52 s
mathequality i f sr(Klj, , 5)
+2
5 5 (s1) 2
E
1 or 2
( m o d 3)
s = 1 or 2 ( m o d 3). with equality i f a n d o n l y
i f K5[,1),+j+1 c a n be decomposed i n t o graphs o n 4 vertices and 5 edges wi h each vertex in 51 graphs. = 351 if and only if s = 1 or 2 ( m o d 4) Sr(K1,s 3 6, = Sr(K1,6 9 7, < 3s2 otherwise. 16 sr(K1,,, 8) 5 (s1) 2 un'th equality i f and only 5 i f Klq8l b + c a n be decomposed i n t o graphs o n 5 vertices with 8 e ges with each vertex in 81 graphs. 18 9) 5 (s1) 2 with equality i f a n d o n l y sr(K1,, 5 i f K1qsl)b+ c a n be decomposed i n t o K 5 minus one edge with every vertex in 51 graphs. = 482 if and o n l y if s = 1 or 2 ( m o d 5 ) Sr(K1,6 7 lo) = sr(K1,8 11) 5 4 5  3 otherun'se. sr(K1,, , 12) 5 452 with equality i f s = 1 or 2 ( m o d 5 ) .
+
d
i_
+
t
f
This is an immediate consequence of the Theorem and Corollary 2 for p
55 .
0
Proposition (1) (2) (3)
sr(Kl,, 3) = 291 sr(K1,, , 4) = 2s 5 sr(K1,, 5) = (s1) 2 5s   2 5 w(K1,, , 5) 2
f o r all
sGO s .
+2
i f and o n l y i f
for
5s
51 2
s
i s odd.
i f s i s even.
(4)
sr(K1,, 6) = sr(Kl,, , 7) = 352
(5)
16 sr(K1,, 8) = (s1) 5
+2
( m o d 3).
for
i f and only i f s
s = O or 3 ( m o d 4)
=1
(mod 5)
.
.
Proof (1) is proven in (61
(2) For k = 4 , if s = O ( m o d 3) then 251 E 5 ( m o d 6). It is known (see [lo]) that K6t+5 can be decomposed into K 5 ' s and K 3 's. Therefore K2s1 can be decomposed into C , 's (cycle of length 4 ) and K 3 's (since each K 5 can be
Star SubRamsey Numbers
159
decomposed into two K 3 ’s and one G4 ). If we colour each graph K 3 or C4 of this decomposition of K2s1 with a distinct colour we get a (K1,+, 4)colouring. This, together with Corollary 2, proves the claim.
+
5 (mod 4) then (sl) 1 = 1 (mod 5) In [I] 2 it is shown that K5t+l can be decomposed into graphs isomorphic to Q (where Q is a K4 minus one edge) in such a way that each vertex has degree 2 (resp. 3 ) in exactly (sl)/%! graphs of the decomposition. This and the Theorem give the result. If s is even, then from Corollary 3 we get the upper bound on s r ( K 1 , ,, 5 ) . To get the lower bound it is sufficient t o notice that we always have ~ K I ,, k) , > S ~ ( ~ K I,, k) ,~ Here 81 is odd so we have 5s S ~ ( K ~ ,, 5) , =~ (s2) 2 and therefore sr(K1,, , 5 ) 2   2 2 2 (3) For k = 5 , if s
E
+
(4) For k = 6 , if ~ r ( K 1 ,,, 6) 5 382 .
s
1 or 3
.
.
= 0 or 3
.
(mod 4) we know already from Corollary 2 that
+
If s = 4t+3 , t 2 0 , we have 3s  3 = 12t 6 and it is known that Klzt+6 plus a perfect matching on the same set of vertices can be decomposed into K4’s(see [9]) in such a way that every vertex belongs t o 4t+2 = s1 graphs K4 of the decomposition. Colour all these K4’swith different colours and then delete the additional perfect matching. Now no colour has been used more than six times in this colouring of K3s3 and only 81 colours appear at each vertex, so we have a (K1,, , 6)colouring. Thus sr(K1,,, 6) 2 352 , and we have equality. If s = 4t+4 , t 20 , consider the complete graph G’ of order 382 = 12t+10 . Brouwer showed [4] that if t 2 1 then G’ can be decomposed into graphs all isomorphic t o K4 except one isomorphic to K7 . In this decomposition every vertex belongs to 81 graphs K4 except seven vertices which belong each to 93 graphs K4 and one K , . Colour each K4 with a different colour and consider the graph G obtained by deleting one of the seven special vertices together with the inherited colouring. Since K 6 plus a perfect matching can be decomposed into K4’s (see [9]) the colouring of G can be completed by adding a perfect matching A4 t o the complete graph induced by the six special vertices, colouring each K4 of such a decomposition by a distinct colour and deleting the perfect matching M . It is easy t o see that the result is a (K1,, , 6)colouring of K 3 8  3 , which implies that sr(K1,,, 6) 2 382 if s = 0 (mod 4), 8 # 4 . In the case s = 4 , a (K1,4 , 6)colouring of K , can be obtained as follows. Let the vertices of K 9 be the elements of 2, and colour each of the following K4 with a different colour : (1,3,6,8) , (2,4,6,0), (3,5,7,0) . Consider , for i = 1, 4, 7, the three graphs with vertices i , i+l, i+2, i+3, i+7 and 6 edges ( i , i + l ) ,( i + l , i + 2 ) , (i+2,i+3) , (i+3,i) , (i+3,i+7), (i+7, i+l) and colour each of them with a different new colour. It is easy t o check that only three colours appear at each vertex and that the whole graph is coloured. So sr(K1,4 , 6) 2 10. Since any (G , k)colouring is also a (G , k‘)colouring for k’ > k , the above shows that s r ( K l , , , 7) 2 392 when s E 0 or 3 (mod 4) and we have equality in that case as well.
P.Fraisse, G. Hahn and D. Sotteau
160
( 5 ) For k = 8 , let s
16
=1
+
( m o d 5) (necessary condition for (sl) 2 to be 5 16 an integer); then (sl) 1 = 1 ( m o d 16) . It is known that for any t , K,,,+, 5 can be decomposed into K , minus two non adjacent edges in such a way that each vertcx belongs to 5t graphs of the decomposition (see [2]). This with Corollary 3 proves the claim. 0
+
Remark 1: Using the Theorem, Corollary 2 and ‘Lhe fact that we have srWl,8+1
9
k) 2 sr(K1,, k) + 1 9
we can get the following bounds for sr(K1,,, k )
Let s 1 2 , T h e n , if 6) p (sQ+l)(p1)
P 2 I / ~ = ( ~ ) + O r ,I r L 
.
’’2
s =q
(mod p ) , q 2 2 . > 5 and s is big enough, or (ii) p 5 5 , w e have
+4 I
4K1,8
7
k)
5
and
+
(sI)(pI) I if r < (p1)/2 (sl)(pl) + 2 if = (p1)/2
+
 + + .
Indeed, sr(K1,,, k) 2 sr(K1 + 2 , k) 9  2 , and, as s q 2= 2 ( m o d p ) , we have, from Corollary 2, S ~ ( K ~ , , ,k) ~+ =~ (sq+I)(pl) 2 This gives the lower bound. The upper bound is already in the Theorem and Corollary 2.
k
For clarity and completeness we collect the values or bounds of sr(K1,, , k) for in Figure 1.
5 12
Star SubRamsey Numbers
161
Values of s
Upper bound
2s
1
s
8
qs1) 5
I
+2
=1
(mod 2)
s = 1 (mod 5)
3s  2 9
3s  2
4s  2 10, :
4s
I
s
4s  5 4s  8
4s  3 4s  3
 11
4s  3 4s  2
4s
l2
5
4s
4s  8 4s  11
4s
2 2
= 1 or
(mod 5) (mod 5) (mod 5 ) 2 (mod 5 ) (mod 5 ) s = 4 (mod 5) s = O (mod 5 )
s =4 s r O s = 1 or s r3
4s  2
Figure 1 :Values or bounds f o r s r ( K 1 , , , k ) for k
5 12
Remark 2 :We take this opportunity to correct a minor omission in [7] 1 of that paper proves that
, , ~= 7 except that .sr(K,,, , 2 ) = 4 and s ~ ( K ,7) The proof provides colourings of K,,, for
2 (mod 5 )
s ~3
. Proposition
.
m = [ (3+?)]
1
in all
other cases except the cases k = 8 and Ic = 9 . These two cases should fall in case b), where m = 3n+l ; the construction, however, needs n 2 3 . The missing colouring is given in Figure 2. Clearly if K7 can be coloured with k = 8 then it is coloured for k = 9 as well.
P.Fraisse, G. Hahn m d D. Sotteau
162
Figure 2
Acknowledgements Research of the first and third authors is supported by P.R.C. Math. Info.; research of the second author is supported by NSERC Canada under grant number AO199.
References 1. J.C. BERMOND and J. SCHONHEIM, Gdecomposition of K , , where G has four vertices or less, Discrete Mathematics 19 pp. 113120 (1977). 2. J.C. BERMOND, C. HUANG, A. ROSA, and D. SOTTEAU, Decomposition of complete graphs into isomorphic subgraphs with five vertices, Ars Combinatoria 10 pp. 211254 (1980). 3. J.A. BONDY and U.S.R. MURTY, Graph Theory with Applications. 1976. 4. A.E. BROUWER, Optimal Packings of K , ' s into a K,,, Journal of Combinatorial Theory A 26 pp. 278297 (1979). 5. P. ERDOS, Some remarks on the theory of graphs, Bull. Amer. Math. Soc. 53 pp292294 (1947). 6.
7.
G. HAHN, Some Star AntiRamsey Numbers, Proc. 8th SE Conference on Combinatorics, Graph Theory and Computing, Congressus Numerantium X I X , Utilitas Mathematica, pp. 303310 (1977). G. HAHN, More Star SubRamsey Numbers, Discrete Mathematics 34 pp. 131139 (1981).
8.
G. HAHN and C. THOMASSEN, On path and cycle subRamsey numbers, Discrete Mathematics 62 pp. 2933 (1986).
Star SubRamsey Numbers
163
W.H. MILLS, On the Covering of Pairs by Quadruples I., Journal of Combinatorial Theory A 13 pp. 5578 (1972). 10. R.M. WILSON, Construction and Uses of Pairwise Balanced Designs, Combinatorics Part I, Math. Centre Tracts 55, Math Centrum Amsterdam, pp. 1841 (1974). 11. R.M. WILSON, Decompositions of Complete Graphs into Subgraphs Isomorphic t o a Given Graph, Proc. 5th Brit. Comb. Conference 2975, Congressus Numerantium XV, Utilitas Mathernatica, pp. 647659 (1976). 9.
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Annals of Discrete Mathematics 34 (1987) 165178 0 Elsevier Science Publishers B.V. (NorthHolland)
165
Colored packing of sets P. Frank1
cms Paris, FRANCE 2. Fzjredi
AT&T Bell Laboratories Murray Hill, NJ, 07974 U.S.A. TO A L E X R O S A O N X I S 3I37IETU BIRTHDAY
ABSTRACT Let M be a family of tsets on {1,2, . . . ,k}. A family 3 of ksets on v elements is called a (v,k,M)packing if for all F E 3 there is a copy of M , M F such t h a t the tsets of F corresponding to M F are covered only by F. Clearly, 171 5 b ] / I M
1,
and if M is the complete thypergraph
then we obtain the usual definition of the (partial) Steinersystem. The main result of this for every fixed k and M the size of the
/lM
largest Mpacking is
1, whenever v +m
1. Preliminaries. Packings and near perfect matchings
Let X be a velement set, X = {1,2,
...
denote the collection of all ksubsets of X by
b].
X . A family of subsets of X is just a subset subset of
TE
Pc
k] k]
A Steinersystem S
there is exactly one B
=
For an integer k , 0
PnP’l < t
5k 5v
we
while 2x denotes t h e power set of I t is called kuniform if it is a
S ( v , k , t ) is an S C
E S with T C B.
is called a (v,k,t)packing if
.
such t h a t for every
Obviously, IS I =
I]/lj
holds. A
holds for every pair P,P‘ E P. R6dl
P. Frankl and Z. Furedi
166 [16] proved that
holds for every fixed k,t whenever u
+ OQ
This theorem was generalized by Frankl and R6dl (131. To state it, we recall some definitions. For a family of finite sets 3 and an arbitrary set A the degree of A is defined by d e g 3 ( A ) =: I{F € 5A C F } I . For A = { a } we set d e g r ( a ) =: degF({a}), the usual definition of the degree of an element. A matching M of 3 is a subfamily of E M . The largest carpairwise disjoint members, M C 3,M f l Td = 0for all dinality of a matching is denoted by
v(3). Clearly, for 3
4 7 ) i ./k.
[:I,
(Frankl and R6dl 131) For every such that if 3 C
E
> 0 and
(2)
k there exists a 6
and every degree of 3 is almost d (i.e., ( d e g 3 ( x )  d
for every x E X ) and for every x , y E X we have d e g T ( { x , y } )
v(3) > holds for u
> 0 and a vo = u0(k,e)
I 5 ~d holds
< d / ( l o g ~ )then ~
;
(16)
> uo.
For a family of sets
5
C 9 the subset A C X is an own
and d e g $ ( A ) = 1, i.e., A is contained only in G. Hence if and only if every G
€5
has
5
C
of G E 5 if A C G is a (u,k,t)packing
own tsubsets. The aim of this paper is to con
struct such families 3 in which for every F E 3 the family of own tsubsets of F is is* morphic to a given tuniform hypergraph U . Such a family is called an U  p a c k i n g . If U is the complete thypergraph on k elements, an Upacking is just the usual ( u , k , t ) packing. The existence of large Upackings is proved in Chapter 2 and 5. The proof is probabilistic, the main tool for the construction is (2). In Chapter 3 we give an application solvin (at least asymptotically) the question: what is the maximum size of a family 3 C
I]
such that none of the members is contained in the union of r others.
Colored Packing of Sets
167
2. Upackings and colored Upackings Let M be a family of tsets over k elements.
Suppose t h a t 3 C
I = 21
and for every F € 7 there exists a copy of M on F (i.e., MF C
k]
KJ, ( 1
MF
where
x M).
If every tset T E M F is covered only by F (i.e., degF(T) = 1) t h e n we call 3 a (w,k,M)packing (or, briefly, Mpacking). Clearly,
E.g.,
I
the
following
u {2jl,2j}:
7 = {2il,2i}
Definition
k/71 =
x:
2.1.
kl
exists
ily (v2/8)
I
/K = k, c =
1
 lM
I
fix
and
a
partition
is
a
coloring
x ( T ) = 0 for T EM. The family 3 C
k1
is
In
other
words,
this
k,Mtpacking if coloring
a
C U . . . U C,
is isomorphic to
K
k],
1
+ O(v).
, Bl= v
holds for every two F,# € 7 ,and
= Co U
TF: F
< j 5 "/2
15 i
. U{T,}.
* *
(21,
pn#l 5 t
(i) (ii) there
C
. . . ,c } with
{OJ,
called a colored
M
Let
M U {T,}U +
3 is a (v,4,C4)packing of size (w2/8)
family
of
such t h a t for T E a
+ O(w): 3 = lS
t
tsets
of
X
with
c+l
colors
such t h a t for every F E 3 the induced coloring of
x, especially
3
the
(2i 1,2i}
Co n
M
1
we have T
colored
u {2j,2j+l}:
M
M . (I.e., there exists a n injection
E C x ( T F ( ~ )E.g., . ) the following fam(v,4, C,)packin 1
1
5 i < j < w/2
of
size
W e prove t h a t
the upper bound in (3) for the size of 3 is essentially the best possible.
Theorem 2.2. For every given k and M the size of the largest colored (u,k,M)packing is (10(1)) k \ / \ N
Iwhen Y  0 0
In the proof we will use the following
P.F’rankl and 2.Furedi
168 Suppose n
Lemma 2.3.
ksets (i) and (ii)
> n , ( k ) , k > t 2 1, D 5 n 2 b .
Then there exists a family of
S on the nelement set N such that
ISflS‘ 1 5 t ldegS(T)D
holds for every two S ,S‘ E S
I 0 ( u f  ’ ) . In general,
In
the
cases
(4)
and
(5)
the gap may be much larger: Let K be the
graph with vertexset {u,b,c,d} and edges { u , d } , {b,c>, {b,d), {c,d}. Proposition 2.4. Suppose that morphic to K . Then
U
is a graph, and that it has an induced subgraph isc
169
Colored Packing of Sets 3. rcoverfree families
We call the family of sets 7 rcoverfree if Fo (Z F,U . . . UF, holds for all Fo,F1, . . . ,F, E 7 (Fi # F j for i # j ) . Let us denote by f, ( n , k )the maximum cardinality of an rcoverfree family 7 c
k1
, INI
= n.
Let us set t =
integer part). An (n,k,t)packing is rcoverfree, hence by (1)
On the other hand every F € 7 has an own tsubset. Indeed, F can be covered by r tsets, F = T,U *
UT,, Ti
* *
E
M
.
If for every i degF(Ti;.)> 1 then F is covered by r
others, which is a contradiction. This yields
Proposition 3.1. ([7]) For fixed k and r
exists and is positive. In the next theorem we determine the value of c , ( k ) , or at least we show that the calculation of it is a finite problem depending only on k. Definition
Let
3.2.
m ( k , t , l )= max{IN
1
N C
members}.
For k
(2kDlogn)'F) < . nk
Hence
Proposition 4.1. Prob(3U
E
ltY1]
1 with degF(U) 1 3 k ) < . t +1
Proof: We can choose a set U in [ t ; l ] distinct ways. Then we can choose 3k ksets n t 1 through U in [' vo(k). Let p = vh and let 2, be a random variable for every T E Prob(ZT = i) = p for i = 1 , 2 ,
Prob(2T = 0) = 1
1)
kl
with distribution
. . ., c
and
 cp .
In other words we color randomly and independently the tsets of X . Recall t h a t
c
=

lU
I. Let C i
=: {T: 2, = i}, then for i
21
E(~ci~)=pk],
and by (9) B o b ( IICi
Call a set S
E S wellcolored
I p
1
I > d') < 2e" .
(13) C
(with respect to the coloration {C
restriction of the coloration to
k1
is isomorphic to
x.
Define
an
lU
wellcolored) = g p'(1p)
tuniform
11c
A
x into
. with
klcl the choice'of p and D we have
if the
Denote the set of wellcolored
ksets by W . If g is the number of nonisomorphic embeddings of
P~O~(S is
. . . ,C c})
vertexset
lU
D/h].
By
174
P. Frank1 and Z. Furedi
d >vo4
(14)
Proposition 5.1. Let T E Co,then
I> . . . ,S,
< e” .
Prob( ldegA ( T )  d Proof:
Consider
the
Prob(Si E W )= g p c (1  c )
sets
S,, S,,
ES
with
T CSi.
ks
T ECo,
I U . v k ] . Moreover these events are independent,
hence by (9)
VT)< 2eu* < e’* . IdegA (T)  1 I > vh VT. 0
Prob( IdegA ( T I 1 I > vh However Idegg ( T )  d
I > v0.3 implies
This proposition and (13)yield that there exists a choice of {Co, which
holds for all i
2 1, and for every T E C ldegA ( T )  d
But
by
C1, . . . ,C,} for
d
(14)
and
for
.
I5
every
T , # T2, T I ,T2 E
k]
we
have
degI({Tl,T2})_< 1 < d / ( l ~ g v ) ~ .Thus we can apply (2) to A . This gives that whenever v coo v ( A ) 2 (10(1))
Finally, a matching
M
l C o I / l ~ I = (14))
in A gives a colored Ucover,
1 t
/IN
I.
7 = {S E s:
6. Proof of 2.4 Let 3 be a (v,/c,U)packing and let g denote the set of “crowded” edges, i.e., {i,j}is uncovered or degF({i,j}) 2 2). For a vertex z E X denote by f, the number of F E 3 for which the vertex corresponding t o a in F is x. Then
9 = {{i,j) either
The main point is that
Colored Packing of Sets
Indeed, if 2 = aF and {aF,bF} Moreover,
U F ,{uF,cF}
f, 5n
175
U p then, clearly { b F , c F }
1
# {bp,cp}. (17)
holds for all z EX because {aF, dF} is an own edge. 14I 2 ."/2/2 fi I.
Finally (15)(17) give
7. Proof of Theorem 3.4 Construction. Let does not contain 1+1
N be a family of tsets over the kelement set K such that N members and suppose that IN I is maximal, i.e.,
IN I = m ( k , t , l ) . Let
let 3 be a colored (n,k,U)packing with size
We claim that 3 is an rcoverfree family. Suppose on the contrary, that FO C F l U * UFr. A s ponFi 15 t and pol= ~ ( t  1 ) 1+1, we have at least 1+1 4 such that l(4\(Flu * * n Fo( 2 t . Then ponFiI = t must hold and thus we have at least 1+1 disjoint sets FonFi such that FonFi E N , a contradiction.

+
 Ucl))
Upper bound. Let F0 be an rcoverfree family. Let 30be the sets with small own U I5 t1 such that degF(U)= 1). Clearly parts, i e . , 30= {F E 7: 3U C F , I
,
17~1s ltZlJ.
Consider an F E7F0 and let
NF be
the nonown parts of F with t
( \
elements, i.e., N F = {T E
Propoclition 7.1. members.
KJ
:3$
E 3 such that F n F / 3 T}.
If F E7F0 then
NF
does not contain 1+1 pairwise disjoint
.. .
Prooj: Suppose for contradiction that T,, ,TI+, E NF with IUII: I = (I+l)t. Let P = {T,, . . ,T,+,, S,,. . ,Sr1,} be a partition of F such that ISi I = t1. Then for each P E P there exists an Fp €3, Fp # F with P C Fp. Hence F C u { F p : P E P}, a contradiction. 0
.
I]
.
Now Proposition 7.1 implies that least
lUFl5 m ( k , t , l ) , i.e., every F €77, contains at
 m ( k , t , l )own tsubsets. Hence
176
P. Frank1 and Z. Furedi
A slightly more complicated argument gives t h a t for n
> no(k)
8. Final Remarks
Actually, using the argument in Chapter 5 we can prove t h e following stronger statement.
k]
Theorem 8.1. Let M be a family of tsets on {1,2,
3c
,
(whenever w (i)
I= w
+ 00)
PFnF’I5 t
. . . ,k}.
There exists a family
of size
with the following properties:
for all distinct F,F’
E3
(ii) For every F E 3 there is a permutation of its elements F = (zl, . . . , z k )such t h a t whenever PFnF’I= t and F’ = (yl, . . . , y k ) and FnF’ = { x i l , . . . ,zit} then xi, = yi, for 1 5 a 5 t , and {il, . . . ,it} 6 U. I t remains open whether we can suppose in this theorem t h a t the orderings o n each F E 3 can be obtained as a restriction of an ordering of X . Another open problem arises from t h e fact t h a t our proof is probabilistic. It would be interesting to give other (“real”) constructions. It is not necessarily hopeless, e.g., N. Alon [I] pointed out t h a t an exponentially large rcoverfree family can be obtained using a recent explicit construction of J. Friedman [14] of certain generalized Justensen codes.
Colored Packing of Sets
177
References (11 N. Alon, Explicit construction of exponential sized families of Lindependent sets, Discrete Math. 58 (1986), 191193.
[2] B. Bollobb, On generalized graphs, Acta Math. Acad. Sci. Hungar. 16 (1965), 447452. 13) B. Bollobb, D. E. Daykin and P. ErdGs, On the number of independent edges in a hypergraph, Quart. J. Math. Oxford (2) 27 (1976), 2532. [4] H. Chernoff, A measure of asymptotic efficiency for tests of an hypothesis based on the sum of observations, Ann. Math. Statist. 23 (1952), 493507. 151 P. Erdzs, A problem of independent rtuples, Ann. Univ. Budapest 8 (1965), 9395. [6] P. ErdGs, P. Frankl and Z. Fiiredi, Families of finite sets in which no set is covered by the union of two others, J. Combin. Th. A 33 (1982), 158166. [7] P. ErdGs, P. Frankl and Z. Fiiredi, Families of finite sets in which no set is covered by the union of r others, Israel J. Math. 51 (1985), 7989. [8] P. ErdGs and T. Gallai, On maximal paths and circuits of graphs, Acta Math. Acad. Sci. Hungar 10 (1959), 337356.
[g] P. Erd&, C. KO and R. Rado, An intersection theorem for finite sets, Quart. J. Math. Oxford (2) 12 (1961), 313320.
ilO] P. Erd6s and E. Szemere'di, Combinatorial properties of a system of sets, J. Combin. Th. A 24 (1978), 308311. [ll]P. Frankl, A general intersection theorem for finite sets, Ann. Discrete Math. 8 (1980), 4349.
(121 P. Frankl and Z. Fiiredi, Extremal problems concerning Knesergraphs, J. Combin. Th. B 40 (1986), 27G284. [13] P. Frankl and V, RGdl, Near perfect coverings in graphs and hypergraphs, Europ. J. Combin. 6 (1985), 317326. [14] J. Friedman, Constructing O ( n logn) size monotone formulae for the Lth elementary symmetric polynomial of n Boolean variables, Proc. 25th Symp. on Foundation of Comp. Sci., Florida 1984. [15] A. Re'nyi, Probability theory, NorthHolland, Amsterdam 1970. [16] V. RGdl, On a packing and covering problem, Europ. J. Combin. 6 (1985), 6978.
This Page Intentionally Left Blank
Annals of Discrete Mathematics 34 (1987) 179188
0 Elsevier Science Publishers B.V. (NorthHolland)
179
Balanced R o o m Squares from Finite Geometries and their Generalizations R. FujiHara Institute of Socioeconomic Planning University of Tsukuba Tsukub a, JAPAN
S.A. Vanstone Department of Combinatorics and Optimization University of Waterloo Waterloo, Ontario, N2L 3G1 CANADA TO AL&X R O S A O N MIS 3I3TIETU BIRTXDAY
ABSTRACT Room squares have been extensively studied and their existence has been completely settled. The balanced Room square problem appears to be much more difficult. The existence problem for these designs is far from complete. In this paper we show how to construct a certain class of balanced Room squares from finite geometries and use recursive constructions for the geometries to produce infinitely many new balanced Room squares. The paper generalizes the concept of a balanced Room square to Kirkman squares with larger block size. Using finite geometries infinitely many balanced Kirkman squares are constructed. 1. Introduction.
A Kirkman square KSk(w)is a.n r
xr
u1
array ( r = ) k1
defined on a rset V such
that (1) each cell of the array is either empty or contains a ksubset of V. (2) each element of V is contained in precisely one cell of each row a n d column of the array. (3) the subsets in t h e nonempty cells form the blocks of a
(w,k,l)BIBD.
The spectrum for KSk(w)has not yet been completely settled. In fact, we are at present far from the complete solution. The following results are known.
180
R. FujiHara and S.A. Vanstone
Theorem 1.1. (Mullin and Wallis 171). There e z i s t s a K S 2 ( v )if a n d only if v is a positive integer, v 0 (mod 2) and v # 4 or 6. Theorem 1.2. (Rosa and Vanstone [a]). There e x i s t s a n integer wo such t h a t for all v > v,, and = 3 (mod 6) there is a KS3(v). It is conjectured that for all v 2 27, v = 3 (mod 6) there exists a KS,(v). In this paper we are interested in KSk(v)’s with additional properties. These require several more definitions. A K S k ( v )is said t o be ordered if each cell containing a ksubset K of the point set V is replaced by an ordered ktuple on the elements of K . We denote such an array by OKSk(v). Let D be an OKSk(v). For each row (column) of D we form k blocks B,,B2, . . . ,Bkwhere Bi consists of all elements in the ith coordinate of each ktuple in this row (column). The set of all blocks taken over all rows (columns) of D is called the row (column) design of D. Let D * be a K k ( w ) . If there exists an OKSk(v)D formed from D*, as the underlying array, such that the row (column) design of D is a BIBD then D * is called a row (column) balanced Kirkman square and is denoted RBKSk(v) (CBKSk(v)).If D * is both row and column balanced then D* is denoted RCBKSk(v). If D* is a RBKSk(v)and a CBKSk(v)and both have the same underlying 0KSk.v) then D* is called a strongly balanced Kirkman square and is denoted SBKSk (v). A RBKSk(v) or CBKSk(v)in the case k = 2 is commonly referred t o as a balanced Room square and denoted BRS(vl). A number of articles (see for example [S], [lo])have been concerned with the existence of BRS(v1) but the spectrum is far from determined. The concepts of RCBKS,(v) and SBKS2(v) have not been considered before. In this paper we approach the problem through the use of finite geometries. The next section introduces the required geometrical ideas. We conclude this section with an example of a SBKS2(8).
The row and column designs are (8,4,3)BIBDs.
Balanced Room Squares
181
2. Skew and Hyperplane Skew Resolutions.
A skew class in AG(n,q), the affine geometry of dimension n and order q , is a set of lines, no two distinct lines are parallel, which partitions the point set of the geometry. A skew resolution of AG(n,q) is a set of skew classes which partitions the line set of AG(n,q). Skew resolutions were introduced in [5] for an approach to constructing packings in PG(n,q). In order to see how skew resolutions can be used to construct Kirkman squares we require another definition. Let D be a (w,k,l)BIBD and let R and R' be two resolutions of t h e block set. R and R' are said t o be orthogonal if n R'I 5 1 for every resolution class R ER, R' ER'. It is not difficult to see t h a t a (w,k,l)BIBD with two orthogonal resolutions implies the existence of a KSk(w)and conversely. The points and lines of AG(n,q) constitute a (w,k,l)BIBD where u = q", k = q. If we denote t h e parallel resolution by P and if AG(n,q) has a skew resolution S then P and S are orthogonal resolutions of the BIBD. Section 3 will also require the concept of a hyperplane skew resolution. Let H , be the hyperplane at infinity for AG(n,q) so t h a t AG(n,q) plus H,gives PG(n,q). If S is any set of lines in AG(n,q) we let be t h e set of points in H,which are not incident with any line of S . A skew class T of AG(n,q) is said to be a hyperplane A skew resolution skew class if r ( T ) is the point set of a hyperplane of H , S = {S1,S2, . . . ,S,} is said to be a hyperplane skew resolution (HSR) of AG(n,q)if
r(S)
(i)
Si is a hyperplane skew class for 1 5 i
5 r.
(ii) r ( S i ) # r ( S j ) ,for all i # j. Hyperplane skew resolutions exist [6]. It is known ([4]) t h a t any skew resolution in AG(3,q) is an HSR. 3. Balanced Kirkman Squares from Finite Geometries.
In this section we apply t h e concepts of t h e previous section in order to construct various types of balanced Kirkman squares. Theorem 3.1. If there exists a skew resolution in AG(n,q) then there exists RBKS,(q").
r!
Proof. Consider t h e (q",q,l)BIBD associated with the points a n d lines of AG(n,q). Let P = {P,,P2 ,...,P,} and S = {S,,S2, . . . ,Sr},r = (q"l)/(ql), be t h e parallel and skew resolutions respectively. Form an r X r array A with the rows indexed by the parallel classes Piand the columns by the skew classes Si. In cell (Pi,Sj) place the subset Pi n Sjwhich is either the empty set or a line of t h e geometry. Clearly, A is a KS,(q"). Associated with the parallel class of lines Pi is a parallel class of q hyperplanes Hi = {H/,H?, . . . ,Ha} such t h a t H? contains precisely one point from each line in Piand the Hi's are distinct. There are many ways to associate Hi with Pi.The easiest way is the following: Let be the line of Piwhich is incident with the origin, in the sense of linear algebra. Then there is a unique linear subspace F which is We can take the parallel class of hyperplanes the orthogonal complement of containing F as Hi. Order a ksubset in row Piof A so t h a t the j  t h coordinate point
e
e.
R. FujiHara and S.A. Vanstone
182
is on the hyperplane H,3. The resulting ordered KS,(q”) is a row balanced Kirkman square whose row design is the design formed from the points and hyperplanes of AG (n1 ,q ). Theorem 3.2. If there exists a hyperplane skew resolution in AG(n,q) then there exists a RCBKS,(q”).
Proof. Since a HSR is also a skew resolution we can apply the construction given in the proof of Theorem 3.1 t o get a RBKS,(q”). We now prove that the array A is also column balanced but perhaps with a different ordering placed on the ksubsets. To do this we need only prove that for each skew class Si there exist q parallel hyperplanes H/,H,?,...,HBsuch that each H i contains precisely one point from each line of Si. Let e,,e,, . . . ,e, ( t = qnl) be the lines of a hyperplane skew class Si and let H , be the hyperplane at infinity for AG(n,q). Let
Q = {x EH,:
e n H,=
z for some
e
ESi}.
The points in H Q \, are the points of a PG(n24) which we denote by T . There are precisely q 1 hyperplanes of PG(n,q) which contain T; of course, H , is one of these hyperplanes. Let H’ be any one of the remaining q hyperplanes. Each t i , 1 5 i 5 t , meets H‘ in a point which is not in T. This completes the proof.
+
In order to apply the above results we need to know about the existence of skew resolutions and HSRs in AG(n,q). Theorem 3.3. (FujiHara and Vanstone 14)). There exists a H S R in AG(3,q)for all q a prime or prime power. Corollary 3.3. There exists a RCBKS,($) for all q a prime or prime power. Theorem 3.4. (FujiHara and Vanstone IS]). There exists a H S R in AG(2’1,q) for all q a prime or prime power and for all integers i 2 2. Corollary 3.4. There exists a RCBKS,(qt) t = 2’ power and for all integers i 2 2.
 1 f o r all
q a prime or p r i m e
More existence results for skew resolutions will be given in the following section. 4. Balanced R o o m Squares.
The existence of B K S k ( v ) for k = 2 is of particular interest since it is the first case considered by other researchers and its spectrum is not yet settled. As indicated in the introduction, these arrays are commonly referred t o as balanced Room squares. We again make use of the concepts from section 2. The following result on skew resolutions in AG(n,2) is originally due t o R.D. Baker [2] and independently proved by B.A. Andersen (11 although neither phrased the result in terms of skew resolutions.
Balanced Room Squares
183
Theorem 4.1. There exists a skew resolution in AG(2m+1,2) for all integers m 21.
Corollary 4.1. There exists a RBKS2(22m+1)for all integers m
2 1.
It was previously unknown whether or not a skew resolution exists in an affine geometry of even dimension. We answer this question in the affirmative.
Theorem 4.2. There exists a skew resolution in AG(4,2). Proof. It is enough to display one skew class in AG(4,2). Instead we list all skew resolutions in AG(4,2) which admit an automorphism of order 15.
We associate the points of AG(4,2) with the points of GF(24) generated by the primitive irreducible polynomial f(z)= x4 z3 1. Let CY be a root of f ( x ) . If {d,$} is a line of AG(4,2) we abbreviate the notation to {i,j}. Also, we define cu" = 0. A parallel class of lines in AG(4,2) is
+ +
Po = {{d,l+d}: i = mor 0 5 i
5 14).
The other 14 parallel classes are
Pi= {{d+J,$(l+d)}: i
= mor 0
5 i 5 14}, 1 5 j 5 14.
Explicitly,
Po = {{0,~,{1,12},{2,9},{3,4},{~,10},{6,~},{7,~3},{11,14}}. A skew class is represented by a list of 7 distinct integers (alra2,...,a7) such that 1 5 ai 5 14 and
+ a1 = {l+a,,12+a1} {2,9} + a2 = {2+U,,g+a2} {3,4} + a3 = {3+a3,4+a3} {5,10} + a4 = @+a4,10+a4} ( 6 4 ) + a5 = {6+a5,8+a5} (7713) + a6 = {7+a6,13+a6}
{1,12}
{11,14}
+ a7 = {ll+a7,14+a7}
is a partition of the nonzero residues modulo 15. Such a class will generate a skew resolution under the action of the cyclic automorphism of order 15.
R.FujiHara and S.A. Vanstone
184
al
a2
a3
1 9 5 1 1 1 4 2 2 3 5 2 3 6 1 6 2 11 4 1 3 4 5 14 4 1 0 6 6 3 7 4 7 1 10 8 1 10 Q 7 14
a5
a4
1 9
1 1 7 9
1 1 1 11 7 3
a6
4
3 1 1 5 13 5 2 2 3 14 13 6
a7
7 5
8 11 9 14 1 0 8 9 12 1 0 1 3 6 13 5 1 7 1 0 2 4 13 9 12 13 8
It is a simple matter t o check that there is no HSR in AG(4,2) which admits an automorphism of order 15. Recently, the following results in AG(5,2) were obtained.
Theorem 4.3. (Stinson and Vanstone 1111). I n AG(5,2) there exist 179 inequivalent skew resolutions which admit an automorphism of order 31 and 26 HSRs admitting an automorphism of order 31. Corollary 4.3. There exists a RCBRS(31). Another recent result is stated in the next theorem. Theorem 4.4. (Stinson and Vanstone [12]). There exists a skew resolution in A G ( n , 2 ) for n = 6,8,10 and 14. These give rise to new balanced Room squares.
Corollary 4.4. There ezist BRS(2"1) for n = 6,8,10 and 14. The following recursive constructions give rise to infinitely many new balanced Room squares. Theorem 4.5. (FujiHara and Vanstone [5]). If there e h s t s a skew resolution in A G ( m , q n ) and a skew resolution in A G ( n , q ) then there exists a skew resolution i n A G (mn,q). Theorem 4.6. (FujiHara and Vanstone [5]).If there exists a skew resolution in AG(m+l,q") and i f there ezists a skew resolution in A G ( n , q ) then there exists a skew resolution in AG(rnn+l,q). For hyperplane skew resolutions we have the following result. Theorem 4.7. (FujiHara and Vanstone [S]). If there exists a HSR in A G ( m , q " ) and a H S R in A G ( n , q ) then there exists a H S R in A G ( m n , q ) .
Balanced Room Squares
185
5. Strongly Balanced Room Squares.
Strongly balanced Kirkman squares were defined in section 1 and an example of a strongly balanced Room square was displayed in the same section. There has so far been very little work done on SBRS(w1)'s and few results on their existence are known.
Theorem 5.1. I f q i s a prime power and q
= 3 (mod 4) then there exists a SBRS(q).
Proof. Index the rows and columns of a q X q array A with the elements of GF(q). Let (Y be a generator for the nonzero elements of GF(q). Consider the set of ordered
pairs
o 5i 5 e l .
{(di,di+l):
In row g, column g
+ c~ +
o 5 i 5 ,Q 32
i+l
place the ordered pair ( d i + g , d +g) for all g EGF(q). In row g, column g place the ordered pair ( o q g ) where 00 GF(q) for all g EGF(q). A is an OBRS(q). It is easily checked that A is in fact a SBRS(q). Theorem 5.1 shows the existence of a SBRS(31). An alternate way t o construct such a design is through finite geometries. We illustrate the technique. For notation and a detailed explanation the reader is referred t o [ l l ] . A base class for the parallel resolution in AG(5,2) is
Po = {2i{{5,2),{3,29),{26,28}}: 0 5 i
k4+3k3+k2+1. Then uk+l has a partial parallel class containing at least blocks. 0 k+2 For the systems STS(w) and SQS(v) this result implies that: Theorem 1.1: In every S(2,3,v) with ~ 2 4 5 there , are at least (v1)/4 pairwise parallel triples.
M. Gionfriddo
190
Theorem 1.2: In every S(3,4,v) with ~ 3 1 7 2there are at least ( v  2 ) b pairwise parallel quadruples. In another theorem of [3], C.C. Lindner and K.T. Phelps have proved that in the statement of Theorem 1.1 (for STS(v)) we can replace the inequality v>45 by v>g, except possibly the cases w=19 and v=27. Theorem 1.3: A Steiner triple system of order v>s has a partial parallel class containing at least (v1)/4 blocks, except possibly for v19 and w=27. In [5] and [6], G. Lo Faro has proved that the statement of Theorem 1.1 is true also for v=l9 and v=27. Theorem 2 ( G . Lo Faro): In every STS(19) there are at least five painvise parallel blocks, and there exist Steiner triple systems of order 19 with a partial parallel class containing exactly five blocks. Regarding the Steiner systems STS(v), after C.C. Lindner and K.T. Phelps, D.E. Woolbright [7] and A.E. Brouwer [l]have obtained the following results. 3v Theorem 3 (D.E. Woolbright): In every S T S ( w ) there are at least 7 parallel 10 blocks. 0 This improves the result of LindnerPhelps for ~ 2 1 3 9 . Theorem 4 (A.E. Brouwer): In every STS(v) there are at least
fu_sv2/5) parallel 3
blocks. 0 This result applies for every admissible ~ 2 1 2 7 .It improves the result of Woolbright for large values of w (from 950010000). In what follows, for any number a , la] denotes the maximum integer l a and 1. denotes the least integer >a. 3. New results
In this section we prove that in every Steiner system S(lc,k+l,v) with Ic23 there exist at least [(w+2)/2Ic] pairwise disjoint blocks. This result is general, and for v3 there exist at least l(w+2)/2k] parallel blocks. Proof: Let (S,/3) be a Steiner system S(Ic,Ic+l,v) with k>3. Let be a family of parallel blocks of this system, such that if P = 6 then ISPI >(kl)InI+2(Icl). This
u
b En
n
implies that w L2Ic III 1+2(kl). In what follows, let I=t. We prove that in the system there exists a family II' of parallel blocks, such that lII'l> 1. The result is immediate if there exists a block contained in SP. Therefore we suppose that for each
[email protected], 6gSP. We observe that, for a fixed set of Ic1 distinct points A1,...,AklESP, if R = (SP){Al,...,Akl}, then there exists an
In
Pauwise Disjoint Blocks in a Steiner System
p;(kli)+h(kl)+k1
191
1
.
Let XiiER such t h a t lC(Xij)=aij and let Y,<ER such t h a t 1C(Yf;)=b/,. Case 1: At first, suppose t h a t ai,
[email protected]$R) for each i=l,...,r . It follows t h a t
IsP l(kl)
k1
=
C p;*i+kr.
11 k1
Since ISP IL(kl)t +2(kl) and t = h + r
+ C p i , in this case i t follows: 11
k1
Cpii+kr
2 (kl)t+k1
k1
=(kl)fi+(kl)r+(kl)~pi+kl, 11
i1
and hence r
k2
2 C pi(kil)+h(kl)+(kl). i1
Consider the injection 4:R'+P, where R' = {Xi,ER:i#l}, such t h a t for all element of q R ) satisfying t h e condition
XijER', d(Xij) is t h e { X I1 rXl,k1 ,xij ddxij )}Ep. 7.
192
M. Gionfriddo
If I? is the family of the blocks {Xll,...,X1jkl,Xij,&f,j)}and L = {ci.: i = 1 , ...,h , i=l,...,k + l } U {b;l :i=l,...,pl} U {alk}, it follows that Irl = k(r1) and I = ( k + l ) h + p l + l , with
Irl= k(r1)
c
k2
2
= krk
p i (kci1)
+hk(k1) +k22k >
i1
( k + l ) h + p , + l =)L 1, where we have considered that for each k>3 i1
> h(k+l)
hk(k1)
> 1.
k2k
Then it is possible to find an element zWL such that {Xll,...,Xl,kl,41(z),z}@.Further, there exists at least an element y E$(R), y#z, with z and y belonging to the same block b,, of II. If
n' = n{bzy}U then
{{X1l ,...,Xl , k  l t ~  l ( z ),z},{A1,...,Akl,~l(Y)rY}}
n' is a family of parallel blocks of p with ln'I> III I.
Case 2: Now suppose that there is at least one element
ai,k+l
such that
{'il,...,ui,k+l}~lc(R).
We can suppose that {'il,*.*tui,k+l}~lc(R)
for each i = l ,...,r', and {ail,*,'ik)CIC(R))
and ai,k+lE q R )
for each i=r'+l, ...,r . If r & ! ,consider the injection p:R"+P, where k1 R I I=
such that for all k1
(xll,,x 1I
If
and
I?'
9
I T 1
{xijm:(i,i)+(i,i) ,...,( 1 , ~ 2  1 ) , ( 2 , 1 ),...,(
is the element of ?I(R)satisfying the condition xii,fi(xii)}@
X i j w " , p(Xij)
x21,*vx
1 , 1%
k1
is the family of blocks
>
2 ~ 9 1 ) )
Pairwise Disjoint Blocks in a Steiner System
193
if follows t h a t
i1
i1
=
(kl)r+(k1)h
+c (kil)Pi k2
i1
2 k+l 2
h(k21) ; (k1)Z 2 2
Epi(kli)+ i1
> (k+l)h+p,+l
=
L’I+l
where we have considered t h a t k1
t =r+h+CPi i1
k 23. So it is possible to find at least two distinct elements x‘,zl’ belonging to two distinct blocks of I?’
such t h a t x~,x‘~€PL’. Since zl#z”, we can suppose t h a t x’#a find an element y €$(I?),
l2,, I
k1
.
In this situation, it is possible to
y#x’, with z1 and y belonging to the same block bzy of
It follows that there exists a family TI’ of parallel blocks with If k2
r=1,
then
r’=r=l.
c pi(kli)+h(kl)+kl III I.
then there exists a family of parallel blocks It follows that, if t = 11, 2k v +2 having size t ‘ = t + l = 2k From Theorem 5 it follows that in every Steiner quadruple system there are at least 11 v+2 pairwise disjoint blocks. Therefore we have the following inequalities: 6 SQS(.) V . number of parallel blocks (small values) 4 1 8 2 10 2 14 2 2 16 2 3 20 2 3 2 4 22 26 2 4 28 1 5 32 >5 34 2 6 38 2 6 40 2 7
I].
In particular, for the systems SQS(l6) we have the following interesting open problem: is it true that in every Steiner quadruple system of order 16 there are four painvise disjoint blocks, or that there exists an SQS(l6) with exactly three pairwise disjoint blocks?
Acknowledgements Lavoro eseguito nell’ambito del GNSAGA e con contributo del M.P.I. (40%, 1985)
Pawwise Disjoint Blocks in a Steiner System
195
References
A.E. Brouwer, “On the size of a maximum transversal in a Steiner triple system”, Canadian J . Math. 33 (1981) 12021204. I21 H. Hanani, “On quadruple systems”, Canadian J . Math. 12 (1960) 145157. 131 C.C. Lindner and K.T. Phelps, “A note on the partial parallel classes in Steiner systems”, Discrete Math. 24 (1978) 109112. 141 C.C. Lindner and A. Rosa, “Steiner quadruple systems: a survey”, Discrete Math.
111
22 (1978) 147181.
G. Lo Faro, “On the size of partial parallel classes in Steiner systems STS(l9) and STS(27)”, Discrete Math. 45 (1983). (61 G. Lo Faro, “Partial parallel classes in Steiner systems S(2,3,19)”, J . Infor. Optimization Sci. 6 (1985) 133136. 171 D.E. Woolbright, “On the size of partial parallel classes in Steiner systems”, Ann. Disc. Math. 7 (1980) 203211. [51
This Page Intentionally Left Blank
Annals of Discrete Mathematics 34 (1987)197206 0 Elsevier Science Publishers B.V. (NorthHolland)
197
On Steiner Systems S(3, 5, 26) M.J. G r a n n e l l , T.S. G r i g g s a n d J.S. P h e l a n School of Mathematics and Statistics Lancashire Polytechnic Preston, PR1 2TQ UNITED KINGDOM TO JIL EX ROSA ON HIS 3I37IETII BIRTUVAY
1. Introduction Denote a Steiner system by S(t,lc,w)where the parameters have their usual meanings. In contrast to the situations where t = 2 and k = 3 , 4 , 5 , 6 or t = 3 and lc=4 where the spectrum of w for which a Steiner system exists is either completely or substantially determined, very little seems to be known concerning systems S(3,5,w). In this case the necessary admissibility conditions give w =2,5,17,26,41,50 ( m o d 60). The system S(3,5,5) is of course trivial and it has long been known [7] t h a t S(3,5,17) is unique. For v = 2 6 there are two realisations in the literature. One was constructed by Hanani [5], and the other occurs as a derived subsystem of the S(5,7,28) given by Denniston 121, all such derived subsystems being isomorphic due to t h e double transitivity of the automorphism group. There appears to have been no published investigation as to whether the Hanani and Denniston systems are isomorphic. The only other known result is a construction also due to Hanani [4] which produces an S(3,5,4v3) from an S(3,5,w). Hence there exists an S(3,5,65) but the existence of an S ( 3 , 5 , v ) for the intermediate cases w =41,50,62 as well as for an infinite number of other values of is still undetermined. In this paper we present a method by which with the aid of a computer we hope t h a t further Steiner systems S ( 3 , 5 , v ) may be constructed. W e illustrate the method by applying it to produce a third realisation of an S(3,5,26). Finally we determine t h a t the three S(3,5,26)’8 are isomorphic thus raising the question of whether this is the unique Steiner system with these parameters. 2. Theory
A Steiner system S ( t , k , w ) which has an automorphism of order Y is said to be cyclic. I n this case the base set V is usually represented by the set of residue classes modulo v and the automorphism by t h e mapping i + i + l ( m o d w). W e will foIlow this throughout the remainder of this paper. The Steiner system may be constructed from orbits under the cyclic group C, generated by the above mapping and each orbit may be characterised by a cyclically ordered difference ktuple < d,, d,, . . . ,dk> where d,+d,+ dk=w. In the same realisation any orbit which is stabilized by the
....+
198
M.J. Grannell, T.S. Griggs and J.S. Phelan
mapping i +i (mod v ) is said to be symmetric and a cyclic Steiner system, all of whose orbits are symmetric is said t o be Scyclic. In (31, two of the present authors proved that if S ( t , k , v ) is Scyclic then t  3 and if in addition k is odd then v=k22k+2. Hence when k = 5 then v17 and indeed the unique S(3,5,17) is Scyclic [3]. For v=26, the system S(3,5,v) can not be Scyclic but the theory of such systems may still be utilized t o advantage. All orbits of 5element sets under C2, are of length 26 and precisely 10 of these are required to construct an S(3,5,26). Each cyclic &element set orbit contains 10 cyclic %element set orbits. If a cyclic &element set orbit is symmetric then its characteristic difference quintuple may be written in the form where 2a+2b+c=26. Two of the cyclic 3element set orbits contained in this have difference triples and . Since there are precisely 12 difference triples of the form G , z , 2 6  2 z > i.e. for z = l , 2 , ....12 there are at most 6 symmetric cyclic orbits contained in S(3,5,26). We call a cyclic Steiner system in which the maximum number of cyclic orbits are also symmetric, a maximally Scyclic system or MScyclic. The computational advantage of constructing an MScyclic Steiner system over an ordinary cyclic system is that the total number of symmetric cyclic orbits is far less than the number of all cyclic orbits. This facilitates the partial construction of the system using symmetric cyclic orbits only which may then be able t o be completed with nonsymmetric cyclic orbits. We now construct an MScyclic S(3,5,26) and prove that this is the unique MScyclic S(3,5,26). 3. Construction of partial systems
The Appendix lists all difference quintuples of symmetric cyclic 5element set orbits under C26. For use as an orbit in the construction of a Steiner system the 10 difference triples of the cyclic 3element set orbits contained in each of the 5element set orbits must be distinct. In the case of the system S(3,5,17) if the difference quintuple was of the form it was shown in (31 that this implies that all of the following conditions must hold.
a#b b #c 2a+b#c a #e a#b+c 2a#b+c b #2a In the case of S(3,5,26) the same conditions apply as well as one other namely (viii) c#2a which is automatically satisfied when v=17 because then c must be odd. Those orbits listed in the first column of the Appendix which are unsuitable because they do not satisfy all of the above have the reason given alongside and the others, 36 in all, are numbered. For the latter orbits the second column gives the 2 values x, namely a and a + b , of the difference triples O , x , 2 6  2 x > of the cyclic 3element set orbits of this form which they contain. The third column gives the difference triples of 4 of the other cyclic %element set orbits contained, the other 4 being obtained as the reverse triples.
199
Steiner Systems S(3,5,26)
The next stage is t o find collections of 6 of these orbits which form a partial cyclic S(3,5,26). Although this is feasible by hand the tedious nature of the task makes it more suitable for computer. Programs were written by two of the present authors independently and the results obtained showed that there are only 2 such collections. These are given below in the same format as in the Appendix. Also given are the difference triples of the cyclic %element set orbits which are not contained in any of the 6 symmetric cyclic 5element set orbits and remain to be contained in nonsymmetric cyclic 5element set orbits if either of the partial systems are to be completed.
Partial System 1
Difference
Difference
Other difference triples
triples quintuple
(together with their reverse triples)
<x,r,2%2x>
1
4
2
7
3
12
5
6
8
11
9
10
Difference triples remaining
together with reverse triples of all the above.
M.J.Grannell, T.S. Griggs and J.S. Phelan
200
Partial System 2
Difference
Difference
Other difference triples
triples quintuple
1
(together with their reverse triples)
6
2
9
3
8
4
5
>
7
10
11
12
< 1,2,23>
.
together with the reverse triples of all the above. 4. Completion of partial systems
It remains t o ascertain whether either of the partial systems can be completed to form a cyclic S(3,5,26). We deal with each in turn.
Partial System 1 Since difference triple remains, one of the 4 nonsymmetric cyclic 5element set orbits must have a difference quintuple of the form . By considering other remaining difference triples of the form < l , z , y > it then follows that c=6,10,11,15 or 21. But the latter two can be rejected since the difference triples and are already contained in symmetric orbits. Similarly e = l l is impossible because the difference triple is contained in the orbit with difference quintuple . Hence c = 6 or 10. Similarly e =2,8,12,13,17 or 23. But by considering the remaining difference triples of the form < 3 , x , y > the possibilities reduce to e = l 3 or 17. Since c + d + e = 2 3 and d > l , this leaves the orbit having difference quintuple as the only possibility. By a similar argument the second of the 4 nonsymmetric cyclic 5element set orbits has the reverse difference quintuple . The remaining 2 orbits can now be deduced in a similar manner using the difference triples still remaining and it is
Steiner Systems S(3,5,26)
201
easy t o show that they must have difference quintuples and .
Partial System 2 The argument follows the same reasoning as for the first partial system. By considering difference triple which remains there must be a nonsymmetric Further cyclic 5element set orbit with difference quintuple . consideration of the remaining difference triples of the form < l , z , y > gives c = 3 , 5 , 6 , 8 or 11 but the remaining difference triples of the form eliminate all of these possibilities except c =6. Hence d e = 12 and again by considering the remaining difference triples of the form < I , z , y > there are just two possibilities namely e = 7 and d = 5 or e =10 and d =2. However the difference quintuples and can be eliminated as in both cases the orbits contain the cyclic 3element set orbit with difference triple which is already contained in the orbit with difference quintuple . Hence this partial system can not be completed t o a cyclic S(3,5,26).
+
Summarising we have proved that there exists a unique MScyclic S(3,5,26). It may be constructed from the set of residue classes modulo 26 under the action of the cyclic group C,, generated by the mapping i 4 1 (mod 26) and consists of 10 orbits, 6 of which are symmetric and 4 nonsymmetric. The orbits are characterized by the following difference quintuples.
+
A :
B :
C :
D :
E :
F :
G :
H :
I :
J :
Finally we note that a primitive of 26 is 7 and that the action of the mapping i 7i (mod 26) on the above orbits is as follows.
Thus the system is stabilized by the group of mappings i   t a i + b (mod 26) where 6 = 0 , 1 , 2 , ....25 and (a,26)=1, being formed by two orbits of this group, one consisting of the 6 symmetric cyclic orbits and the other of the 4 nonsymmetric cyclic orbits.
202
M.J. Grannell, T.S. Griggs and J.S. Phelan
5. Isomorphism testing
In [I], Cameron states the following theorem.
Theorem: Suppose a Steiner system S(3,k,w) has the property that given any block
B of the system, the blocks which intersect B in precisely two elements form an S(2,k2,vk) on the complement of B . k=105,v =557026.
Then k=5,w=26 or k=23,w=5084
or
He continues by saying that no systems satisfying the above property are known. For the MScyclic S(3,5,26) constructed above a program was written to determine whether it had the property. For each block of the system there are 70 blocks which intersect it in two elements. Removing these two elements leaves 70 triples which give rise to 210 pairs of elements. The program counted, for each block of the system, how many pairs occurred once and how many occurred twice among the total 210 pairs. It is easy t o see that no pair can occur three times and to satisfy Cameron’s property, there must be 210 pairs occurring precisely once for each block considered. The results obtained are that for each of the 156 blocks in a symmetric cyclic orbit, 126 pairs occur once and 42 occur twice whilst for the remaining 104 blocks, 132 pairs occur once and 39 occur twice. Hence the MScyclic S(3,5,26) does not have the required property nor indeed seems to have any similar structure. However, the figures given above are an invariant of the system which can be used in helping t o determine whether it is isomorphic t o the S(3,5,26)’s constructed by Denniston and Hanani. In [2], Denniston constructs a Steiner system S(5,7,28) as the union of 2 orbits under the group of 7element subsets of GF(33) U {m} with x3 = x 2 as the irreducible polynomial. He gives starter blocks for the two orbits as
+
(04 0 , 1 , 2 , x , x
{q0 , 1 , x2
+ 1 , x + 2)
and
+ 2 x , x 2 + 22 + 2 , 2 2 2 , 2 2 2 + 22)
By determining all blocks of this system which contain the pair {q0}, another task for the computer, a derived S(3,5,26) was obtained and found t o have the same invariant numbers with respect to Cameron’s property as the MScyclic S(3,5,26). The element x is a primitive in GF(33) and further investigation leads to establishing that the mapping i +XI from the set of residue classes modulo 26 t o GF(33) \{O) with the above irreducible polynomial is an isomorphism from the system constructed in this paper to the derived system of Denniston’s S(5,7,28) through {q0). Hanani 151 constructs an S(3,5,26) on the Cartesian product 2 ( 2 ) X 2 ( 1 3 ) where Z ( n ) denotes the set of residue classes modulo n.
He gives the following starter blocks {(O,O), (1, 01, (1,22), (1,22+4),( 1 , 2”+8)}, x = 0,17 2 , 3 ,
and
((0,2y), (o,2y+6), (I,()), (1, 2y+2), (17 2y+8)), Y = 0,1 , 2, 3 , 4 , 5 which under the mapping ( i , j ) + ( i + l , j + l ) (mod (2,13)) form an S(3,5,26). It is easy t o establish that the mapping ( i , j ) + 13i + 2 j (mod26) is an isomorphism from the Hanani system t o the one constructed in this paper.
Steiner Systems S(3,5,26)
203
6. Concluding Remarks
The obvious question t o ask is whether the S(3,5,26) dealt with in this paper is the unique Steiner system with these parameters. It would be particularly interesting if this were the case as all derived S(2,4,25) subsystems of it are isomorphic. Since it is known that the number of nonisomorphic S(2,4,25)’s 2 6 [6] we would have an example of nonisomorphic Steiner systems on the same parameters which are not either all derived or all not derived. To the present authors’ knowledge no occurrences of this are known. It would also be of interest to construct MScyclic S(3,5,u)‘s for other values of o, in particular u=41,50,62.
References
121
P.J. Cameron, “Extremal results and configuration theorems for Steiner systems”, in: Topics on Steiner systems (ed. C.C. Lindner and A. Rosa), Ann. Discrete Math. 7 (1980) 4363. 8 (1976) R.H.F. Denniston, “Some new 5designs”, Bull. London Math. SOC.
131
M.J. Grannell and T.S. Griggs, “Scyclic Steiner systems”, Ars Combinatoria
141
H. Hanani, “A class of 3designs”, Combinatorial Mathematics (Proc. Conf. Canberra 1977), Lecture Notes Math. 686,Springer, Berlin (1978) 3446. H. Hanani, Technion preprint (earlier version of 141). A. Ja. Petrenjuk, “0 postroenii neizomorfnykh blokskehm s porno&& izografi6eskikh naborov blokov”, Kirougrad Inst. S.Kh. Masinostra.,
I11
[51
PI 171
263267. 16A (1983) 173188.
Kirovgrad (1983) 16 pp. E. Witt, “Uber Steinersche systeme”, Abh. Math. Sem. Uniu. Hamburg 12 (1938) 265275.
M.J. Grannell, T.S. Griggs and J.S. Phelan
204
Appendix Symmetric cyclic 5element set orbits under
Difference quintuple
a=b
b =2a
c26.
Difference triples
Other difference triples
<x,x,262x>
(together with their reverse triples)
1
1
4
2
1
5
3
1
6
4
1
7
5
1
8
1
6=c
6
7
c =2a
8
a=b
9
b =2a
10
1
11
2
3
2
5
2
7
10
2a+b=c
11
2
9
12
2
10
c =2a
a=c
13
3
4
14
3
5
a=b
15
3
7
16
3
8
Steiner Systems S(3,5,26)
b =2a
c =2a
>
k & a where
and X > l ,
BsI] and every
The members of B are blocks. The k(u,k,l) design
we
a t(u,k,X) design t o be a pair
kl
.
(X$)
contained in precisely X members of B. our definition excludes repeated t o as the complete design with block
A. Hartman
208 size k
into two parts of equal
In this paper we consider the problem of partitioning size, ea h o which is the block set of a 2(v,k,X)
X=
2 k2.
design. Clearly we must have
This problem is a special case of the problem of the existence of large
sets of disjoint designs. A large set of disjoint t(v,k,X)
designs is a partition of
into n parts, each of which is the block set of a t(v,k,X) design where
=

When n = 2 we refer to the problem as the halving of The usual approach t o the problem of large sets of disjoint designs is to fix t , k, and X and determine necessary and sufficient conditions on v for the existence of a large set. Note t h a t if n=ab then the existence of a large set of disjoint t(v,k,X) designs implies the existence of a large set of disjoint t(v,k,aX) designs. The existence of large sets of 1designs with X = l has been solved by Baranyai [2] for all values of k. Baranyai's theorem also implies the solution for all X, and we shall prove this corollary later in the paper. The existence of large sets of 2designs with k=3 and all 1 has been effectively decided in the papers of Teirlinck [10,11] and Lu [8,9]. In ill] Teirlinck also gives large sets of disjoint 3(v,4,X) designs for all v a ( m o d 3 ) and the smallest possible value for X. In another remarkable paper, Teirlinck [12] constructs large sets of t(v,t+l,X) designs for infinitely many values of v and all t when X=(t+l)!'"+'. The halving problem has been investigated by Alltop [l]where he shows that the complete design with block size k on 2k points can be partitioned into two 2designs if and only if k is not a power of 2. This result follows from his more general theorem t h a t if t is odd and the complete design with block size k on 2 k + l points can be partitioned into two tdesigns, then the complete design with block size k + l on 2k+2 points can be partitioned into two t+ldesigns. A beautiful solution
halving problem is given by Denniston [6] where he
constructs a partition of
two 4designs with k1=12. In this paper we gen
eralize Denniston's technique and apply it t o t h e halving problem for 2designs. Another solution to the halvin roblem has recently been found by Kreher a n d Radziszowski [7]. They partition
into two 6designs with
1=14. A solution t o the
halving problem for 2designs with k=3 can be found in Dehon (51, where he constructs 2(v,3,X) designs without repeated blocks. The only other example of a large set of disjoint designs of which we are aware is the partition into 2(13,4,1) designs given in 141 by Chouinard.
209
Halving the Complete Design
In this paper we sol e the halving problem for 2designs when k=3 or 4. We also show how to partition
kl
into two 2designs for higher values of k and infinitely
many values of WI. In Section 2 we define the concept of elegant matrices, deduce some of their properties, and show how they can be applied t o the halving problem. In Section 3 we give constructions for families of elegant matrices, and in Section 4 we apply the constructions to the halving problem. Section 3 also contains the proof of the corollary to Baranyai’s Theorem. 2. Halving With Elegant Matrices
Let I , denote the set {O,l12,...,nl}, and let M=(vzI,J)be an whose rows are indexed by the members of
.
I:].b]
01 matrix
and whose columns are indexed by
A matrix M with the above dimensions and row and column labels
will be referred t o as an (n,i,j)matrix. For subsets X , Y of I , we define a(M,X,Y) by
that is, a ( M , X , Y )is the number of 1’s in the submatrix whose row indices contain X and whose column indices contain Y. ‘By convention we take the empty sum to be zero. The matrix M will be called elegant if it has the following properties:
~(M,{Z},b))= 4 M , b ) , { z } ) for all z,Y%* a((M,X,@)= a(M,Y,@) for all X , Y E
a(M,@,X) = a(M,@,Y) for all X , Y E
k1 b1
(El)
.
.
(E3)
The constant a(M,X,@) when Rl12will be denoted by r 2 ( M ) ,and we denote the constant a(M,@,X) by c,(M). These constants are not independent, since if we count the total number of 1’s in two ways we obtain the following identity:
and hence
A. Hartman
210
Note that elegant matrices are generalizations of designs since an elegant matrix M with j=O is just the block incidence vector of a 2(n,i,r2(M))design. Two operations which preserve elegance are the transpose and the complement. If
A4 is an elegant ( n , i , j matrix then MT is an elegant (n,j,i)matrix with r2(MT)= c 2 ( M )and c2(MJi=
is the all 1's (n,i,j)ma r'x the
also elegant with r2(JM) =
JM
is
17) b 1 i ]  c 2 ( M ) .
and c2(JM) =
Hence (JM)T is elegant with
We now ,show how t o use elegant matrices to solve the halving problem. Let M' = ( V X ~ , ~ )i=O,1,2, , ...,[(k1)bJ be a set of elegant (n,ki,i)matrices. (The notation x] enotes the greatest integer less than or equal t o x). In the case when both k and
tk>\
are even, we define MkB by
,
where J and 0 are the all 1's and all 0's square matrices of side
 [h]respectively.
It is a simple matter to check that M k B is an elegant (n,k/2,k/2)matrix. Now let 1)/2] and we have a set of k + l elegant matrices.
Mk'= ( J  M f T ,for i=0,1,2,
...,I(&
Theorem 2.1: If the matrices M' defined above satisfy the equation
then there exists a partition of P;l2] into two 2(2n,k,$
Proof: We construct the following set of blocks.
p = {(AX{O})U(BX{1}):m~,p= 1, It is sufficient to show that (InX12,p)is a 2(2n,k,y that
I1
p
I2tG21
) designs.
i=0,1,2 ,...,k}
) design, since this implies
is also the block set of a 2design with the same parameters. We first
211
Halving the Complete Design
[;IF]
count the blocks, by counting the 1’s in M’. Since M k  i = ( J  M ’ ) T , t 1’s in
M
plus
the
number
of
Mk’
1’s in
is
precisely
u ber of
for
i=0,1,2 ,..., [(kl)D]. The number of 1’s in MkF is
The number of blocks containing the pair of points (z,O),(y,l) is given by
when k is odd. When k is even we also have t o add in
[
acMk/”,{.),{Y)) = i  1 1 1
n 1
k
I
2
In either case, using (El), we see that
k
The number of blocks containing the pair of points (z,O),(y,O) is
C r2(M’)and
the
i0
number of blocks containing the pair of points (z,l),(y,l) is
C c2(M.).Hence to comi0
k
plete the proof, it is sufficient t o prove that c ~ , ( M ‘ ) + c , ( M= .) equations (2.2) and (2.3) we have for k odd:
i0
When k is even, we must add
on the right hand side of both equations. Hence in either case
A. Hartman
212
Theorem 2.1 asserts that if we can construct elegant matrices for i=0,1,2, ...,\(k1)/2] which satisfy equation (2.4) then we can halve the complete design on 2n points withblock size k. Equation (2.4) contains only [(k+1)/2] independent variables since, Mk' is constructed from M' and equation (2.1) gives the relationship between r 2 ( M ' )and c2(M). The most convenienb form of equation (2.4) is the following
The proof that equations (2.4) and (2.5) are equivalent is tedious and is omitted. The interested reader can verify the truth of the assertion by using equations (2.1), (2.2), and (2.3) and standard combinatorial identities. Before proceeding with general constructions for elegant matrices, we first give some examples of the construction described in Theorem 2.1. Example 2.1: Let n be an odd integer, and let k=3. Let Lz,y.be a symmetric idempe tent Latin square, for example Lz,y = (x+y)/2(mod n). (Division by 2 is welldefined, since n is odd.) Now consider the matrices @ = J , the all 1's (n,3,0)vector, and define M' by m{,,y),{z) 1 = 1 if and only if Lz,y= z.
It is easily verified that
r2(Mo)= n2 r2(M') = 1
c2(@) = o c2(M') = 0
r2(M2)= 0 r 2 ( ~ 3=)o
c ~ ( M=~o)
[email protected])
= n1
and to verify (El) we note that
so in either case ~(M1,{~},{~})=~(M1,{y},{x}). The block set of the design is then
P
=
u ;z=L,,,} u
{(",o)(Y,~)(z,o):x#Y#z#x}
{(.
,O)(Y,O)(Z,l) :Z#Y
{(Z,O)(Y
,l)(z,l):Y#Z;x#Ly,z}.
Example 2.2: Here we give another construction for k=3. Let n r 1 , 3 (mod 6 and let (
[email protected]) be a Steiner triple system of order n (i.e. a 2(n,3,1) design). Define d a n d M' by
213
Halving the Complete Design m i , @ = 1 if and only if
[email protected] m{z,yl,(z} 1 = 1 if a n d only if ze{z,y}
In this example we have
r2(MO)= 1 r 2 ( ~ '=) n2 r2(M2)= 0
T~(M =~ o)
CZ(MO) =0
4~') =o c2(M2) = 2
c ~ ( M=~n3 )
and to verify (El) we note t h a t
so in either case a(M',{z},{y}) = a(M',{y},{z}). 3. Constructing Elegant Matrices
Ideally one would like to determine necessary and sufficient conditions on the parameters n , i , j and r 2 ( M )for the existence of an elegant (n,i,j)matrix M. As noted in the previous section this is equivalent to the existence problem for 2  ( n , i , r p ( M ) ) designs without repeated blocks when j=O. This problem is very difficult and has only been solved in the case i = 3 IS]. In this section we construct infinite families of elegant matrices and leave the general existence problem for further research. Because the operations of complementation and transposition preserve elegance it is sufficient t o consider ; n l , he existence of elegant matrices with n>&2a and
O ’ X be integers and let O g l n be an integer. We define the symmetric set of d integers Sym(d) by if d iseven f(dl)/2} if d is odd.
**.&/2}
Sym(d)= {O,fl,B,f3,

* *
Note that ISym(d) I=d and that s ESym(d) if an only if s ESym(d). If gcd(n,i)=l we define the average u(I), of an isubset, solution to the equation
i a(Z)=Cz
Z, of 1, t o be the unique
(mod n).
z€I
Now we define the average segment A(I,d) of size d about an isubset to be the set A(I,d)={u(I)+s
(mod n ) : s ESym(d)}.
Finally we define the (n,i,l)matrix Avg(n,i,d) by Awg(n,i,d)I,(,)=l
if and only if
z€A(l,d).
Since j=1 we have c2(Avg(n,i,d))=0. Since the row sum of Awg(n,i,d) is d for all rows of the matrix we have
Halving the Complete Design
215
To verify that condition (El) holds for A v g ( n , i , d )we note that it is sufficient t o verify the following proposition. Let f(x,y,s)denote the number of isubsets, I , such that zEI and u(I)=ys. Proposition f(x,y,s)=f(y,x,s) If the claim is true then we have
c
c
f(z,y,s)=
s ESym(d)
f(y,r,S)=
s ESym(d)
c
r(ll,Z,S)
s ESym(d)
since s ESym(d) if an only if s ESym(d). Hence a(Avg(n,i,d),{~},{y})=a(Avg(n ,i,d),{v},{z}). To prove the Proposition we note that I’=(z+y)I is a bijection between the then yE(z+y)I, and if ys=u(I) then objects being counted. That is, if z+s=a((x+y)1). This completes the proof that A v g ( n , i , d ) is an elegant matrix. The matrix M’ of Example 2.1 is an example of an elegant matrix from this family. We would also like t o have a construction for elegant (n,i,l)matrices when g c d ( n , i ) # l , however a general method has thus far eluded us. Nevertheless we have the following constructions for 2 5 i C 5 . We say that an isubset is averagable if its sum is congruent t o zero modulo g = g c d ( n , i ) . If I is averagable then the equation defining a(1) has g solutions of the form a+kn/g, with OO and k,,k,, * . . ,k, are integers satisfying O g i < n for all i, together with an r x s matrix A = ( u ; , ~ of ) nonnegative integers satisfying
Halving the Complete Design
217
r
C ki
ai,j=n, for all j .
i1
(Since C r C a
[ki n1 I]')
ilj1
Baranyai's Theorem: Given a datum ((ki),(ai>j))on n there exist sets X i , j (I l contains each member of I , precisely p  1 times. Hence each Y j contains each member of I, precisely p(pX)=X times. We now construct the Baranyai family of elegant matrices. Let n>i>k>2 integers and let X,m be positive integers satisfying And
(mod k ) ,
be
I::]& (mod A),
for some integer c . Let Y j be the partition of the complete design with block size k constructed in the proof of Corollary 3.1, and let e an arbitrary bijection from the set of isubsets of I, to the integers from 0 to
Construct Yr t o be the set of
ksubsets of I , defined by a1
reducing the subscripts modulo
x
k1
. We
now define Bar(n,i,k,X,m) to be the
Halving the Complete Design
219
(n,i,k>matrix given by
The basic idea behind the construction is t o make each row of the matrix the incidence vector of m disjoint Afactors of the complete design with block size k and ensure that the column sum is constant. The row sum of Bar(n,i,k,A,rn) is mXn/k for all rows, and the column sum is c for all columns, and hence
To verify that Bar(n,i,k,X,rn) satisfies (El) we note that ~(Ba~(n,i,k,~,m),I,{~})=rn~ for all I and for all x. Hence
An example of an elegant Baranyai matrix with parameters Bar(6,3,2,1,4) can be constructed as follows. Let Yl={el,e2,e3}, Y2={e4,e5,e6}. * . Y5={el3,eI4,el5} be a lfactorization of the complete graph on 16. Label the columns of Bar(6,3,2,1,4) by the edges el,e2, . . . eI5 in order. Then the construction defined above gives
4. Halving the Complete Design
In this section we apply the methods of the preceeding sections t o construct partitions of the complete design with block size k into two 2designs. We first derive necessary conditions for the existence of a partition into two tdesigns. Lemma 4.1: If there exists a partition of the co plete desi w’th block size k into two tdesigns i=0,1,2,
with parameters
. .  ,t.
b1)
t  ( v , k , z kt
then
12)
is even for all
Proof: The number

counts the number of blocks in each part which contain a
fixed isubset of I,, and hence must be an integer. In section 2 we gave a method for constructing such a partition when t2 even. The following lemma gives a method for the case when v is odd.
and v is
220
A. Hartman
It])
Lemma 4.2: If there exists a partition of t e complete design with block size k into two designs with parameters t  ( w , k , T kt plete
)t:!. desF1
block
size
k1
and there exists a partition of the com
into
two
designs
with
parameters
) then there exists a partition of the complete design with block
t (v,k1, 2 k1t
size k into two designs with parameters
Proof: We construct the new block sets from the old ones, adding the new element (w) to f the blo ks f size k1. The tsubsets of I, occur precisely eac 2 kt
LI9 ts[bsets
[
+y k ''  1 4 )=$lzATt] of,
1
v(t1) 2 k1(t1)
lI,,rl
1hi;h
_ _ '+'' 2
kt
times in each part of the new partition, and the contain
the
point
w
occur
precisely
times.
We now focus our attention on the cases where t=2 and k is small. Theorem 4.3: (Dehon [5]) A partition of the complete design with block size 3 into 1 two designs with parameters 2(w,3,(~2)) exists if and only if w a (mod 4). 2
Proof: Necessity follows from Lemma 4.1, and sufficiency is proved in Example 2.1. Note that in Example 2.1 we constructed a 2(2n,3,nl) design containing a 2(n,3,n2) design for all odd values of n>l. Using the more general constructions of the previous section we obtain the following interesting embedding theorem.
Theorem 4.4: The block set of a 2(n,3,h) design can be embedded in the block set of a 2(2n,3,nl)
design for all odd n > l .
Proof: Using the construction given in Theorem 2.1, take @=Des(2,n,3,X)
and
M'=Awg(n,Z,nlh).
We turn now t o the cme where k=4.
Theorem 4.5: A partition of the complete design with block size 4 into two designs 1 with parameters 2  ( ~ , 4 ,  ( ~  2 ) ( ~   3 ) )exists if and only if v S 2 or 3 4 Proof:
(mod 8).
Necessity follows from Lemma 4.1. By Lemma 4.2 and Theorem 4.3, it is sufficient to prove the existence of a partition when v S 2 ( m o d 8). Let w=2n and use the con3 struction of Theorem 2.1 with i"=O and M1=Awg(n,3,(n1)). Since n = l (mod 4) 4
Halving the Complete Design
221
the existence of these matrices is given in the previous section, both when gcd(n,3)=1 and when gcd(n,3)=3. When k X we have only partial results, despite the fact t h a t the number of integer solutions to Equation 2.5 increases rapidly. This is because the constructions for elegant matrices given in t h e previous section do not have a sufficiently wide range of values for the constant rP(M).When k=5 the necessary conditions of Lemma 4.1 are t h a t w d , 3 or 4 (mod 8), but we are only able to find constructions for the following result. (Note t h a t the case w=10 is covered by Alltop's Theorem D [I])
b?]
Theorem 4.6: There exists a partition of he complete design with block size 5 into
) if w €(l0,11,12,18,19,20,34,35,36,58,59,60} two designs with parameters 2(w 5 "2 or w=2,3 or 4 (mod 24), or w=12,42,43,44,60,66,67 or 68 ( m o d 72). Proof: As before we rely on Lemma 4.2 and the previous theorem to reduce the problem ( m o d 24), and w=12,42,60 or 66 (mod 72). Let w=2n and to w=10,18,34,58 w& except when w=34,58. In these cases we let @=Des(2,17,5,5) and we take @=O, @=Des(2,29,5,45) respectively. The 2(17,5,5) design is constructed by taking the orbit under the action of the affine group of the 5set comprising 0 and the 4 t h roots of unity in GF(17). The 2(29,5,45) design is constructed similarly, adding two full affine orbits of 5subsets to the orbit of 0 and the 4 t h roots of unity in GF(29). The constructions for M' a n d & are given in Figure 1.
Figure 1. Constructions for k=5 When k=5 and w=12 Denniston [6] has given the following construction. Let . . ,Y5 be a onefactorization of K6,i.e. a partition of t h e 2subsets of 1 6 into five 1(6,2,1) designs. For each 2subset, E , of I , define f ( E ) to be the unique index i such t h a t Emi.Now define the (6,3,2)matrix Den by lTnE I=o or DenT,E=l if and Only if ITnE I=1 and f(E)=f(TE).
YI,Yz, *
Denniston showed t h a t with @=O,
M1=In1(6,4,1,{1})
and M2=Den the design
222
A. Hartman
constructed by the technique of Theorem 2.1 is in fact a 4(12,5,4) design. Generalizing this construction t o give other partitions into 4designs appears t o be difficult. Another interesting problem is t o generalize the construction of Denniston's matrix to give another family of elegant matrices. Theorem 4.6 implies the existence of a 2(11,5,42) design without repeated blocks. Applying Alltop's Theorem B [l]to this design yields a 3(12,6,42) design. Thus we have also succeded in halving the complete design with block size 6 on 12 points into 3designs. This result is not new since Kreher and Radziszowski's construction [7] implies a solution t o the halving problem for 4designs with v=12 and k=6. Returning now t o the halving problem for 2designs we note that when k=6 the necessary conditions of Lemma 4.1 are that v=2,3,4 or 5 (mod 8), and we have the following constructions. Theorem 4.7: There exists a partition of the omplete design with block size 6 into 2n2 ) if nE{5,10), or n = l (mod 8), or 2 [ 4 1 n a 9 (mod 40), or n ~ 1 4or 74 (mod 80), or n d or 6 (mod 16), but nf150 or 210 (mod 240).
two designs with parameters 2(2n,6,
Proof: When n=5 the result follows from Theorem 4.5, by taking the complement of each block in a partition of the complete design with block size 4 on 10 points. In the The constructions for M' and M2 are given in Figure 2. other cases we take @=O. Note that a construction for for the matrices Avg(5m,5,5d) with m a mod 6 would also give solutions for the cases where n ~ 1 5 0or 210 (mod 240). n 10 8m+l 40m +29 80m+14 80m 6 16m +2 16m +6
M' Int (10,5,1,{0}) Avg(8m+1,5,5rn) Avg(40m+29,5,5m +3) Avg(80m+14,5,15m+2) Avg(80m 6,5,5m 1) Avg(l6m +2,5,5m) Avg(16m+6,5,15m +5)
Id2
Int (10,4,2,{1}) Bar (8m +1,4,2,2,2m) Bar (40m +29,4,2,2,18m +12) Bar(80m+14,4,2,1,68m+lO) Bar (80m6,4,2,1,76m 8) Bar (16m +2,4,2,1,3m) Bar (16m +6,4,2,1,4m +2)
Figure 2. Constructions for k=6 Other partitions of the complete design with block size 6 are constructible using the results of Theorem 4.6 and Lemma 4.2. For k=7 the necessary conditions are vk2,3,4,5 or 6 (mod 8), and we have the constructions for all admissible v=2n with 5 5 n 5 4 7 , gcd(n,b)=l, and n#11,25. For k8 the necessary conditions are vS!,3,4,5,6 or 7 (mod 16), and we have the constructions for vE{38,50,66}. All of these constructions have @=O, M' an Average matrix, and M,M3 are Baranyai matrices.
Halving the Complete Design
223
5. Conclusions and Open Problems
We have given necessary and sufficient conditions for the halving of the complete design with block sizes 3 and 4 into 2designs and infinitely many constructions for block sizes 5 and 6. We conjecture that the necessary conditions of Lemma 4.1 are sufficient for the existence of a halving of the complete design into tdesigns. This appears t o be a difficult conjecture, but when t=2 the methods of this paper appear to make the problem tractable. An important step in the proof of this conjecture is the construction of new families of elegant matrices. Another important problem raised in this paper is the determination of necessary and sufficient conditions for the existence of elegant matrices. All the elegant matrices constructed here (with the exception of the Design family) have the property that they have constant row and column sums. This property is important when aiming for higher values of t , but may be unnecessarily restrictive when t =2. Acknowledgements
I would like t o express my gratitude to Alex Rosa and Earl Kramer for drawing my attention to the work of Kreher, Radziszowski, and Alltop on the subject. Thanks are also due t o the referee for his suggestions on improving the readability of the paper. References [l] W. 0. Alltop, “Extending tdesigns”, J . Combinatorial Theory, Ser. A 18 (1975)
177186. Z . Baranyai, “On the factorization of the complete uniform hypergraph”, Finite [2] and Infinite Sets, Colloq. Math. SOC.Jhnos Bolyai 10 (1975) 91108. NorthHolland, Amsterdam. [3] P. J. Cameron, Parallelisms of Complete Designs London Math. SOC.Lecture Note Series 23 (1976). Cambridge University Press, Cambridge UK. [4] L. G. Chouinard 11, “Partitions of the 4subsets of a 13set into disjoint projective planes”, Discrete Math. 45 (1983) 297300. [5] M. Dehon, “On the existence of 2designs Sx(2,3,v) without repeated blocks”, Discrete Math. 43 (1983) 155171. [6] R. H. F. Denniston, “A small 4design”, Annals of Discrete Math. 291294.
18 (1983)
[7] D. L. Kreher and S. P. Radziszowski, “The existence of simple 6(14,7,4) designs”, J . Combinatorial Theory, Ser. A 43 (1986) 237243. (81 J. X. Lu, “On large sets of disjoint Steiner triple systems, I, 11, 111”, J . Combinatorial Theory, Ser. A 34 (1983) 140146, 147155, 156182. [9] J. X. Lu, “On large sets of disjoint Steiner triple systems, W ,V, VI”, J . Combinatorial Theory, Ser. A 37 (1984) 136163, 164188, 189192.
[lo] L.
Teirlinck, “On the maximum number of disjoint triple systems”, J . Geometry 6 (1975) 9396.
224
A. Hartman
[Ill L. Teirlinck, “On large sets of disjoint quadruple systems”, Am Combinatoria 17 (1984) 173176. I121 L. Teirlinck, “Nontrivial tdesigns without repeated blocks exist for all t”, t o appear.
Annals of Discrete Mathematics 34 (1987) 225242 0 Elsevier Science Publishers B.V. (NorthHolland)
225
Outlines of Latin Squares A.J.W. Hilton Department of Mathematics University of Reading Whiteknights Reading RG6 2 A X UNITED KINGDOM TO A L E X R O S A O N U I S 3 I 3 7 I C T U B I R T U D A Y
ABSTRACT A unified treatment of outline latin squares and of (p,q,r)latin rectangles is presented, together with applications to embedding partial latin squares. 1. Introduction
The generalized latin square A of Figure 1 contains four symbols in each square, and each symbol is contained four times in each row and column. It can actually be obtained from the latin square B of Figure 1 by erasing the dotted cell boundaries and reducing the numbers modulo 3.
B= A =
2 3
2 3
2 3 3 3
2 2
Figure 1
226
A.J. W. Hilton
We show in this paper t h a t a generalized latin square such as A can always be obtained from a latin square B by the kind of process described. This interesting phenomenon has quite a number of applications, ,some of which are considered in this paper. Some of t h e results described here are new, some of t h e proofs of known results are new, and the unified presentation is new. Most of the known results are contained in [l] and [2] by Andersen and Hilton, and the principal result, Theorem 1, is contained (in a slightly different form) in [lo],by Hilton. 2. Outline latin squares * q t ) and U = ( r l , r 2 ,* . . r,) be three Let S = ( p l , p 2 , * * . p s ) , T = (q1,q2, compositions of n. (A composition of n is a vector with positive integer components which add up to n.) If L is a n n X n J a t i n square on symbols 1, * * n then we form the (S,T,U)amalgamated latin square L of size s X t on u symbols al,..., a, by placing ak in cell ( i , j ) of Lf each time t h a t one of r l + * * +rkl+l, . . r l + . . + r k occurs in one of the cells (a,@), where


“E{Pl+
* . *
PE{q1+
+~i~+l, + qj1+1,
. . *
* *
 ,p , + , q1 +
* 
* * *
* *
+pi},
.
and
+ q’i}.
If L is on some other set of symbols, say T ~ ..., , T , , t h e n we may assume that T ~ ..., , 7, are replaced by 1,2, ..., n respectively, and apply the definition above. To illustrate this definition, let n = 6, S = (1,3,2), T = (2,2,2) and U = (1,1,2,2), and let L be the Latin square C of Figure 2.
C=
Figure 2 Then the amalgamated latin square Ls on symbols {1,2,3,5} is obtained by erasing the dotted lines, and replacing th: 4’s by 3’s and the 6’s by 5’s (and keeping the existing l’s, 2’s, 3’s, and 5’s). Then L is depicted as the generalized Latin square D of Figure 3. No significance is attached to the positions of symbols within cells. I t is clear t h a t the number of times each symbol appears in each row, colymn, or cell of D is determined by C and the compositions s, T , and U . Clearly in L
Outlines of Latin Squares
227
D
Figure 3 (i)
row i contains symbol ok pirk times,
(ii) column j contains symbol ok q,rk times, and (iii) cell ( i , j )contains (counting repetitions) p i q i symbols.

Let us call a generalized latin square L** on symbols bl, . ,a,,in which n 2 symbols occur altogether (counting repetitions) and for which there are compositions s = (Pl,* * * , p a ) , T ( q l , + * * ,qt) and U = (rl, * * * ,ru) of n such that (i), (ii), a$ (iii) above hold for L , an (S,T,U)outline latin square. Thus we as%yme here that L satisfies various numerical constraints, but we do not assume that L is obtained from a !$tin square. Our main result is that, nonetheless, there is a latin square from which L can be obtained. Thus if we were presented with the outline latin square D of Figure 3 we would know that there is a latin square from which it could be obtained. Theorem 1: Each outline latin square is an amalgamated latin square. This theorem was originally proved in [lo] (in a slightly different format). The special case of (S,T,U)outline latin squares in which s = t = n , so that S = T = I (the vector of all l’s), has been studied earlier under the name Fsquares. The interest has mainly been in connection with orthogonality. Before proceeding t o prove Theorem 1 we point out that conditions (i), (ii) and (iii) are symmetrical. This symmetry can be seen more vividly if the n X n latin square is thought of as a triangulation (that is, a decomposition into K,’s) of the complete tripartite graph Kn,n,n; this triangulation may be written down in tabular form as an n 2 X 3 orthogonal array. An amalgamated latin square then corresponds to the amalgamation (or identification) of various vertices of the Kn,n,n (where vertices may be amalgamated only if they are in the same part); amalgamation can therefore be represented as an operation on an orthogonal array. In order t o prove Theorem 1 we need to give some graph theoretical definitions and t o state a lemma. An edgecolouring of a multigraph G is a map d:E(G) u. Then a TSQG(v)can
The problem naturally divides itself into the 36 cases for w and w (mod 6). The following table summarizes our approach:
1
D I D I S I D D D D V D
2
D
3
D D D D V S S I D I D I V D V S V D
0 w(mod6)
4 5
I D
I D
I
The cases corresponding t o the nine I’s in Table 1 are impossible by condition ii) of the theorem. The eighteen D’s are covered in section 3; the main tool there is the DoyenWilson theorem on embedding Steiner triple systems, (quasigroups for which S consists of all three identities). The four S’s in Table 1 are done by difference methods in section 4; here the main tool is a theorem of Stern and Lenz [17]. The five V‘s in the table are handled in section 5 by an extension of Vizing’s theorem. But first, we rephrase the problem in terms of graphs.
Embedding Totally Symmetric Quasigroups
251
2. Graph Theoretic Considerations
For terms and notation not defined here, the reader is referred to our bible [3]. (none of our graphs here have multiple edges.) An edge whose two ends are the same is a loop, an edge whose two ends are distinct is a link. A graph on two vertices consisting of exactly one loop and exactly one link is called a lollipop. As usual, K , denotes the complete graph on n vertices and
kl
links, while K z
denotes K , together with a loop at each vertex. K , is called a triangle.
Lemma 1 A TSQGfv) is equivalent to a partition of the edges of K t into triangles, lollipops, and loops. Sketch of Proof Consider the six equations in (*). If a # b # c # a , they correspond to a triangle in K: on the vertices a , b , c . If a # b = c , they reduce t o three equations, and correspond to a lollipop on a,b with the loop at b . If a = b = c , they reduce t o one equation, and correspond to a loop at a . 0 A set M of edges of a graph is a meeting if every vertex of the graph is an end of exactly one edge in M . Thus a loopless meeting is what is usually called a perfect matching, or a 1 factor. Throughout the sequel, w,w, and d are positive integers with d = ww. Lemma 2. An embedding of a TSQG(w) into a TSQG(w) corresponds t o a partition P of the edges of Kd+ into triangles, lollipops, loops, and w meetings. Sketch of proof. Given a 11 correspondence between the meetings and the elements of the TSQG(v), if meeting M corresponds t o element a , then a loop in M at vertex b corresponds to a lollipop on a,b with its loop at b , while a link in M on vertices b,c corresponds t o a triangle on vertices a , b , c . 0 3. The DoyenWilson Theorem.
A TSQG(w) is idempotent (i.e. satisfies x o x = z) if and only if there are no lollipops in the corresponding partition of Lemma 1. For the embedding problem in Lemma 2, again there can be no lollipops, and further each of the meetings must be 1factors. This problem was solved by Doyen and Wilson [ 5 ] ; the necessary and sufficient conditions are that w and w must be congruent to 1 or 3 (mod 6), and w 2 2w+l. This accounts for four of the eighteen D’s in Table 1. For the remainder of this section, let w and w satisfy the DoyenWilson conditions above. Thus we have a partition P of the links of Kd into triangles and w 1factors.
Lemma 3. A TSQG(w1) can be embedded in a TSQG(w1). Proof. Replace a 1factor of P by d / 2 loops and d / 2 lollipops. 0 Lemma 4. If w 2 3, a TSQG(w2) can be embedded in a TSQG(w2). Proof. Replace two of the 1factors of P by d lollipops. 0
D.G. Hoffman and C.A. Rodger
252
Lemma 5. A TSQG(w) can be embedded in a TSQG(w1). Proof. Delete a point z from Kd. For every triangle in P on z,b,c, place a lollipop and a loop on b,c. In every 1factor M of P,replace the link z , a by a loop on a to form a new meeting. 0 Lemma 6. A TSQG(w+l) can be embedded in a TSQG(w+l). Proof. Adjoin to P a ( ~ + l meeting ) ~ ~ consisting of d loops. 0 The four lemmas above give the other fourteen D's in Table 1. 4. The theorem of Stern and Lenr
Stern and Lenz [17] proved a beautiful theorem in order t o give a shorter proof of the DoyenWilson theorem. We will need some definitions before stating this theorem. If z is an integer, we define
y
1 1 .
as follows: find the unique integer to be the absolute value of
= z (mod d) with d/2 < y 5 d / 2 ; then Iz 1 is defined
Y
For the remainder of this section, we assume that the vertices of Kd+ are the elements of the ring z d of integers modulo d. If the edge e has ends and y , we define 5 d/2. So e is a the difference of the edge e t o be the nonnegative integer lacy loop if and only if its difference is 0. (By the way, the function t i  y Id makes z d into a metric space.)
1
Let D = {O,l,

,[d/2]} be the set of differences of edges of Kd+. For any by Gd(S) (or just G ( S ) )the graph on vertices z d with just the edges of Kd+ whose differences lie in S. Thus Kd+ = G ( D ) and Kd = G(D\{O}). If d is even, G({d/2}) consists of a single 1factor. If z ED\{d/2}, then G({z}) is a disjoint union of g cycles, each of length d/g, where g = gcd(z,d).
S
s D , we denote
* *
If z E D , we say z is a good difference (mod d ) if d/g is even, where again g = gcd(z,d). Thus no difference is good if d is odd. If d is even, d / 2 is good, as is any odd difference. Finally, the theorem of Stern and Lenz:
Theorem 1. Let 0# S D\{O}. Then the edges of G ( S ) can be partitioned into 1factors if and only if S contains at least one good difference. 0 Before we can put this theorem to use, we need a preliminary lemma, and one more definition. Lemma 7. Let d be even, let a,b ED\{O,d/2}, with a good, and either b = 2a, or 2a+b = d. Then i) the edges of Gd({O,a,b}) can be partitioned into triangles, lollipops and loops, and ii) the edges of Gd({O,a,b}) can be partitioned into triangles, lollipops, loops, and one meeting. Proof. Note first that a # b. Let k = d/(2 gcd(a,d)), a positive integer since a is good. Let G = G2k({0,1,2}). Since each component of Gd({O,a,b}) is isomorphic t o G , we need only prove the lemma for G . Note that G2k({2}) consists of 2 disjoint kcycles C, and Cp. First partition the edges of G into G, and G2, where G, consists of the
Em bedding Totally Symmetric Quasigroups
253
edges of difference 1 together with C , , and G, the loops together with C,. Obviously, G , is an edgedisjoint union of triangles, and the edges of G , can be partitioned into k lollipops and k loops, proving i). We leave to the reader the simple task of partitioning the edges of G, into lollipops, loops, and one meeting. (Hint  consider separately the two cases k (mod 2)). 0
A three element set of integers is said to be a sumset if one of the elements is the sum of the other two.
Lemma 8. Let k and l be nonnegative integers with l >2k+1 and let E be 0 or 1. Then a TSQG(6k+3+€) can be embedded in a TSQG(61+5+€). Proof. We consider the two cases Ik(mod 2). Case 1. l  k = 2 s 2 k + l . Here d = 12s+2. We first partition D\{0,4s,4~+1,5~+1,6~+1}into 2s1 sum sets as follows: for 1 5 i 5 s, { 2 i  l , 3 ~  ~ , 3 s + i  l } is a sum set, and for 1 5 i 5 s1, { 2 i , 5 s i + 1 , 5 s + i + l } is a sum set. Select 2sk1 of these sum sets. If {z,y,z} is one of these, with say z < y < z , take the d triangles {i,z+i,z+y+i}, i E Z d , for a total of d(2sk1) triangles. Let S be the set of 3k differences, not in the selected sum sets, together with the differences 5s+1 and 6 s + l . Since 6 s + l is a good difference, G ( S ) can be partitioned into 6k+3 1factors by the theorem of Stern and Lenz. Now apply Lemma 7 with a = 4s+l,b = 4s, using i) if E = 0 and ii) if E = 1. Case 2. l  k = 2s+1 2 k + l . Here d = 12s+8. Now we partition D\{0,3s$1,4s+2,4~+3,6s+4} into 2s sum sets as follows: is a sum set, and for 1 5 i 5 s, {2i1,5si+4,5s+i+3} for 1 5 i 5 8 , {2i,3si+1,3s+i+l} is a sum set. The rest of the proof is like Case 1 , except we use Lemma 7 with a = 49+3, b = 4s+2. 0 Lemma 9. Let k,C and E be as in Lemma 8, except now C >2k+2. Then a TSQG(6k+5+~)can be embedded in a T S Q G ( ~ ~ + ~ + E ) . Proof. This is just like the proof of Lemma 8, so we will only specify s,a,b and the sum sets in the two cases. Case 1. l  k = 2s (so d = 1292). Here a = 491, b = 4 s , and the sum sets are { 2 i  l , 5 s  z ~ , 5 s + ~  l }for 1 5 i 5 s1, and {2i,3sa.2,3s+i2} for 1 5 i 5 81. Case 2. l  k = 2s+1 (so d = 128+4). Here a = 4 s + l , 6 = 4s+2, and the sum sets are {2il,3si,3s+il} for 1 5 i 5 s , and {2i,5sa’+2,5s+i+2} for 1 5 i 5 81.0
254
D.G. Hoffman and C.A. Rodger
Lemmas 8 and 9 together account for the four S’s in Table 1. 5 . An Extension of Vizing’s Theorem
A proper k edgecolouring of a simple graph G is an assignment of one of k colours t o each of the edges of G so that no vertex is incident with more than one edge of any given colour. We will abbreviate “proper k edgecolouring” to “kcolouring”. Vizing’s theorem [18], also proved by Gupta [9], states that if G has maximum degree A, then G has a (A+I)colouring. (By the way, Stern and Lenz used Vizing’s theorem t o prove their theorem.) The following extension of this theorem is implicit in the work of Vizing [IS], was proved again by Fournier [7,8] and is also a corollary of a recent paper by the present authors [El.So it surely must be true! Theorem 2. Let G be a simple graph with maximum degree A. Suppose the subgraph of G induced by the vertices of degree A is a forest. Then G can be Acoloured. We need a few more preliminary results before we get down to work. A PSTS(d) (partial Steiner triple s y s t e m of order d ) is a collection of edgedisjoint triangles in &. We denote by s ( d ) the maximum number of triangles in a PSTS(d). Schonheim [16] determined s(d). He showed that s ( d ) = [(d/S)[(dl)/?2]]~, where E = 0 unless d G 5 (mod 6), in which case E = 1. A PSTS(d) is said to be equitable if for all vertices z,y, I t ( z )  t ( y ) I 5 1, where t ( z ) denotes the number of triangles containing the vertex z. The following theorem was established by L.D. Andersen, A.J.W. Hilton, and E. Mendelsohn [l]: Theorem 3. Let d and k be positive integers. Then there exists an equitable PSTS(d) with k triangles if and only if k 5 s ( d ) . 0 We must now examine more closely the structure of the partition of Kd+ described in Lemma 2. So let P be such a partition. We define three simple graphs G,, G, and G3 on the d vertices as follows. The edges of G, are the links in any of the v meetings of P. The edges of G , are the links in the lollipops of P. The edges of G 3 are the links in the triangles of P. Obviously, the graphs G,, G,, and G, enjoy the following properties: i) the edges of G,, G,, and G, partition the edges of K,; ii) the degree of each vertex in G, is either v or v1, and G I can be vcoloured; iii) if G i denotes the graph obtained by adding t o G, a loop at each vertex which has degree v in G,, then the edges of G i can be partitioned into lollipops and loops; and iv) the edges of G3 can be partitioned into triangles. In particular, every vertex has even degree in G3, and the number of edges of G, is a multiple of three. We leave t o the reader the simple proof of the converse: if G,, G 2 , and G3 are graphs on the same d vertices satisfying i)  iv) above, then there is a partition of Kd+ satisfying the conditions of Lemma 2.
Embedding Totally Symmetric Quasigroups
255
Now to work.
Lemma 10. Let v and w satisfy the conditions of the main theorem, with
(v,w) = (1,4)(mod 6 ) , or ( v , ~= ) (5,2)(mod 6), or
(v,w) = (5,4)(mod 6). Then any TSQG(u) can be embedded in a TSQG(w).
Proof. Let T be an equitable P S T S ( d ) with (1/6)d(dv) triangles and let G, be the graph whose edges are the edges of triangles of T. Then every vertex is in (du)/6 triangles of T , and hence has degree dv in G3. Let G2 be the graph with no edges, let G l be the graph consisting of the remaining edges of K d . Then G, has every vertex of degree u1, so it can be vcoloured by Vizing's theorem.0 Lemma 11. Let v and w satisfy the conditions of the main theorem, with (u,w) = (3,4)(mod 6). Then any TSQG(v) van be embedded in a TSQG(w). Proof. Let T' be an equitable P S T S ( d ) with (1/6)(d(dv)+2) triangles and let G i be the graph whose edges are the edges in triangles of T'. Then one vertex, say x, has degree dv+2 in G i , while the remaining vertices have degree du in G i . Then P must contain at least one triangle containing x; let a and b be the other two vertices in such a triangle t . Let T be the PSTS(d) obtained from T" by removing triangle t , and let G, be the resulting graph. Let G2 be the graph with one link on a,b; let G , be the graph consisting of the remaining edges of K d . Then a and b have degree v in G1, and the other vertices have degree v1 in G,, so G, can be vcoloured by Theorem 2. 0
Let v and w satisfy the conditions of the main theorem, with ( v , w )= (5,0)(mod 6). Then any TSQG(u) can be embedded in a TSQG(w).
Lemma 12.
Proof. Let T' be an equitable P S T S ( d ) with (1/6)(d(dv)2) triangles, and let G; be the associated graph. Then one vertex, say x, has degree du2 2 6 in G i , the rest have degree du. Let a be any vertex adjacent to 2. Since a has degree larger than x, there must be a vertex e # x adjacent to a but not to x. Let t be the triangle containing a and c , let b be the third vertex of t . Let T be the PSTS(d) obtained from 2" by removing t , and let G, be the associated graph. Let G2 be the graph with exactly two links, one on a,b and the other on 2 , c . Let G, be the graph consisting of the remaining edges of K , . Then a,b,c and x each have degree v in G,, while the other vertices have degree v1 in G,. Moreover, the subgraph of G1 induced by the vertices a,b,c and x is either a path of length three from a t o x, or a path of length two from a t o b together with the isolated vertex x. In any case, G , can be vcoloured by Theorem 2. 0 We have accounted for the five V's in the table.
Note We would like t o thank the referee for simplifying the proof of Lemma 12.
D.G. Hoffman and C.A. Rodger
256 6. The Exception
Only one part of the main theorem remains.
Lemma 13. If k is a nonnegative integer, then no TSQG(6k+5) can be embedded in any TSQG(12kS12). Proof. Suppose there is such an embedding, let P be the associated partition of the edges of Kd+ given by Lemma 2. Since d = 6k+7 is odd, each of the w meetings must contain exactly 2i+l loops for some i, 0 5 i 5 3k+3. For each such i, let xi be the number of meetings in P with exactly 2i+l loops. Let y be the number of lollipops in P , let z be the number of loops in P and let t be the number of triangles in P. Then 3k+3 i) C 3; = 6k+5 (counting meetings), i0
ii)
3k+3 C (2;
+1)zi +y +z = 6k +7
(counting loops) and
i0
iii)
3k+3 C ( 3 k S  3  i ) ~ ;+ y +3t = (3k +3)(6k +7) (counting links). i0
Subtracting i) from ii), and subtracting (3k+3) times i) from iii) shows that the only nonnegative integer solution is xo = 6k +5, xi = 0 for 1 5 i 5 3k +3, y = 0, z = 2 and t = 2k+2. Referring now to the partition of the edges of K d into G I , G,, and G3 given in section 5, we see that, since z = 2 and y = 0, there are two vertices of degree one in G3, contradicting the fact that the edges of G3 can be partitioned into triangles. 0
References L.D. Andersen, A.J.W. Hilton, and E. Mendelsohn, “Embedding partial Steiner triple systems”, Proc. Lond. Math. SOC.,41 (1980), 557  576. F.E. Bennett and N.S. Mendelsohn, “On the existence of extended triple systems”, Utilitas Math. 14 (1978), 249  267. A. Bondy and U.S.R. Murty, Graph Theory with Applications, American Elsevier, 1976. A. Cruse, “On embedding incomplete symmetric latin squares”, J . Combin. Theory Ser. A, 16 (1974), 18  27. J. Doyen and R. Wilson, “Embeddings of Steiner triple systems”, Discrete Math. 5 (1973), 409  416. T. Evans, “Embedding incomplete latin squares”, Amer. M a t h . Monthly, 67 (1960), 958 961. J.C. Fournier, “Colorations des ar6tes d’un graphe”, Cahiers du C.E.R.O. (Bruxelles), 15 (1973), 311  314. J.C. Fournier, “Me‘thode et thdor4me ge‘ne‘ral de coloration des ar6tes d’un multigraph”, J. Math. Pures Appl., 56 (1977), 437  453.
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[Q] R.P. Gupta, “The chromatic index and the degree of a graph”, Notices Amer. Math. SOC. 13, abstract 66T429.
[lo] D.G.
Hoffman, “Cyclic embeddings of semisymmetric quasigroups and Mendelsohn triple systems”, Ars Combinatoria, to appear. Ill] D.G. Hoffman and C.C. Lindner, “Embeddings of Mendelsohn triple systems”, Ars. Combin. 2 (1981). [12] D.G. Hoffman and C.A. Rodger, “Class one graphs”, J . Gombin. Theory (B), t o appear. [13] C.C. Lindner and T. Evans, Finite embedding theorems l o r partial designs and algebras, Les presses de I’Universite‘ de Montreal, 1977. [14] N. Mendelsohn, “A natural generalization of Steiner triple systems”, Computers in Number Theory Academic Press, New York, 1971, pp. 332  338. 115) H.J. Ryser, “A combinatorial theorem with an application to latin rectangles”, Proc. Amer. Math. SOC.,2 (1951), 550  552. [16] J. Schonheim, “On maximal systems of ktuples”, Studia Sci. Math. Hungar. 1 (1966), 363  368. [17] G. Stern and H. Lenz, “Steiner triple systems with given subspaces : Another proof of the DoyenWilson theorem”, Boll. Un. Math. Ital. A(5) 17 (1980), 109 114. (181 V.G. Vizing, “On an estimate of the chromatic class of a pgraph”(Russian), Diskret. Analiz. 3 (1964), 25  30. [IS] V.G. Vizing, “The chromatic class of a multigraph”, Kibernetilca (Kiev) 3 (1965), 29  39; Cybernetics 3 (1965), 9  17. MR 3417.
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Annals of Discrete Mathematics 34 (1987) 259272 0 Elsevier Science Publishers B.V. (NorthHolland)
259
Cyclic Perfect One Factorizations of K,, E d w i n C. Ihrig
Department of Mathematics Arizona State University Tempe, Arizona 85287 TO A L E X R O S A
O N MIS 3I3TIIcTu BIRTKDAy
ABSTRACT Recently Hartman and Rosa characterized those n for which K,, admits a cyclic one factorization. We show that K,, admits a cyclic perfect one factorization if and only if n is prime. We also show that if n is prime there is a one to one, onto correspondence between cyclic perfect one factorizations on K 2 , and starter induced perfect one factorizations on K,+l. Moreover the full symmetry group of the cyclic perfect one factorization is that of the corresponding starter induced perfect one factorization direct sum 2,.
1. Introduction In [6] Hartman and Rosa define a cyclic one factorization of K,, as a one factorization of K,, (the complete graph on 2 n vertices) which has a cycle of length 2 n in its symmetry group. They then show that a necessary and sufficient condition for K,, to have a cyclic one factorization is that n # 2' for all t 2 2. In this paper we study those cyclic one factorizations that are perfect. The condition of being cyclic is a very natural one for perfect one factorizations. There are exactly three kinds of perfect one factorizations of K,, which have a permutation consisting of of a single non trivial cycle in their symmetry groups. If this permutation has one fixed point then the perfect one factorization is called factor cyclic or vertex1rotational (see [I11 p.46). The GK,,, for p prime are examples of such perfect one factorizations. If the permutation has two fixed points then the perfect one factorization is called factor1rotational [ll],bipyramidal [9], cyclic of type 2 [lo] and generated by an even starter [I]. Again the GK,+I are examples of this type of perfect one factorization. Both of these types of perfect one factorizations have been well studied, and a number of examples other than the GKp+lare given in [4] and [13]. The only remaining possibility is that the permutation has no fixed points. These are the cyclic perfect one factorizations. GA,, are examples of cyclic perfect one
260
E.C.Zhrig
factorizations (see 1111 p.49). However, despite t h e prominent role cyclic perfect one factorizations play, they do not seem t o have been studied in their own right. There are even more subtle reasons why cyclic perfect one factorizations are special. In [8] we have shown t h a t cyclic perfect one factorizations constitute some of the few perfect one factorizations with symmetry groups t h a t are not small. A perfect one factorization is said to have a small symmetry group if there is a vertex such t h a t the only symmetry t h a t fixes it is the identity. This means in particular t h a t the symmetry group is in one to one correspondence with a subset of the vertex set. Then 4.1 of [8] says t h a t any perfect one factorization which does not have a small symmetry group is either cyclic or starter induced. Starter induced perfect one factorizations are those perfect one factorizations t h a t are generated by a starter on a not necessarily abelian group, and as such they constitute natural generalizations of 1factor cyclic perfect one factorizations. With the help of [8] one can show t h a t the condition of being starter induced is equivalent to being factor transitive. These one factorizations are described in more detail in 3.1 and 3.2. Since it appears t h a t the new perfect one factorizations of K,, which will be easiest to find will be the ones whose symmetry groups are not small, it is of interest to discover what perfect one factorizations are starter induced or cyclic. However not only have starter induced perfect one factorizations with starter group Z2n1 been studied extensively, but also those with starter group (Z,)m have been considered in some detail (see [2] and [12]). Thus the natural candidates for new perfect one factorizations are either starter induced perfect one factorizations with a more complicated starter group or cyclic perfect one factorizations. Unfortunately our first result shows t h a t no new orders have cyclic perfect one factorizations. W e show t h a t K,, admits a cyclic perfect one factorization if and only if n is prime. This result is a modification of 4.2 of [8] to include the case when the cyclic perfect one factorization has a small symmetry group. Our main result is slightly more positive in t h a t it shows t h a t there is a one t o one onto correspondence between cyclic perfect one factorizations of K,, and starter induced perfect one factorizations of Kn+l when n is prime. W e state this more explicitly at the end of this section in 1.1. Since there are known starter induced perfect one factorizations on K,,, which are different from GK,,,, we now know of the existence of several new perfect one factorizations of K,, which are different from GA,,. In particular, using [2] (and 2.1 (m) for notation), we find there is a cyclic perfect one factorization on K,, with symmetry [Z2,]Z5,one on K,, with symmetry z34,two on K 3 8 with symmetry [Z3,]z3 and three on K38 with symmetry z38.Using [13]we find there are also new cyclic perfect one factorizations on K,, and K58. The main message of this paper is t h a t the study of all perfect one factorizations with a non small symmetry group reduces to the study of starter induced perfect one factorizations. The part of this area which is as yet unexplored consists of starter induced perfect one factorizations with non abelian starter groups. The known constraints on such a starter group are given in 3.2.
If we wish to consider perfect one factorizations with small symmetry groups, the largest small symmetry groups consistent with our current knowledge of symmetry may also be of interest. Such a group would be strictly vertex transitive, and as such their
Cyclic Perfect One Factorizations of K2,,
261
perfect one factorizations would be a natural generalization of the cyclic perfect one factorizations. The group must be a semidirect product of Z,, with Z,, for n 1 and n, relatively prime with n1n2 = 2n (see [8] 3.13 and [5] p. 146, 9.4.3) and, of course, must not be Z2,. Perhaps this symmetry group might prove a basis for generating new perfect one factorizations. We now summarize t h e contents of the paper. In section 2 we give our notation and show that Kzn admits a cyclic perfect one factorization then n is prime. In section 3 we relate the cyclic perfect one factorizations of K,, t o the starter induced perfect one factorizations of Knl. Also in section 3 we summarize some basic facts about starter induced perfect one factorizations which are needed for the results about cyclic perfect one factorizations. These facts may have interest in their own right and are given in 3.2. For ease of reference we state the main result of the paper below.
Theorem 1.1 The following are true. (a) There exists a cyclic perfect one factorization of K,, if and only if n is prime. (b) For each cyclic perfect one factorization P of K,, one can construct a starter induced perfect one factorization r ( P )of K,+, (use 3.8 and 3.4). (c) Let n be prime. r defines a one t o one onto correspondence between the isomorphism classes of cyclic perfect one factorizations of K,, and the isomorphism classes of starter induced perfect one factorizations on K,+l. (d) We have (see 2.1 (g))
1;
=
22 @ IZq.
Proof See 2.4, 3.15 and 3.16. 0 2. Orders That Admit a Cyclic Perfect One Factorization In this section we will show t h a t K,, admits a cyclic perfect one factorization if and only if n is prime. We will use the same notation as [7] which we briefly summarize in 2.1 and 2.2 below. 2.1 Notation (a)
X
= {1,2,
. . ,2n}
(b) If Y is a set then S, = {T IT is a one to one onto function from Y t o Y}. i is used to denote the identity function. (c) Define L:
s, scJx
by L(7)O
= 7ml.
(d) We use o to denote a one factor  an element of S x t h a t is conjugate to (1,2) (3,4) ... (2n1,2n).
E.C. Ihrig
Let o1 and a, be two one factors. (ol,02) E H if and only if o1 and a, form a Hamiltonian pair; that is, o1together with a, form a Hamiltonian circuit (see [7] 2.3 for equivalent conditions). We use P t o denote a perfect one factorization of Kzn P is a collection of 2n1 one factors any two of which form a Hamiltonian pair. 1; is the full symmetry group of P (I; = {T ESx l ~ ( r ) = P P}). Ip is any subgroup of 1;. T will be used to denote an element of Ip. Let T E Sp Then F , = {y EY I r ( y ) = y} is the set of fixed points of r . If y EY and G is a subgroup of Sy then G,={TEGIr(Y)=y} is the stabilizer (or isotropy) subgroup of y. G = G ' if G is isomorphic to G'. However G C G ' will mean either G is a subgroup of G' or G is isomorphic t o a subgroup of G' depending on the context. If S C G a group then <S> denotes the subgroup generated by S. is the centralizer of 7. (m) [G]+G' denotes the semidirect product of G with G'. G is the normal subgroup and $J EHom(G',Aut(G)). [GIG' is used if $is clear from the context. C ( T )= {r'l
77' = 7%)
2.2 Definition
nor GAPp, (a) A perfect one factorization is called non classical if it is neither the two known classes of perfect one factorizations on K,,, and K2, respectively where p is prime. (b) P is called a cyclic perfect one factorization if 1; contains an element of order 2 n . (c) An element T of ; Z is called reductive if o ( r ) = 2, F, is empty and > 1 (see 3.4 or [7] 3.12 for more information). Note that 2.2 is apparently weaker than the Hartman Rosa definition. We will see in 2.6 that they are in fact equivalent. We start with some lemmas. Lemma 2.3 Let 7 E I p and O ( T ) = 2 n . Then n is odd, IFb(,?I 2 n1 and T " is a reductive element. Proof First we show that n is odd. r"is an element of 1; which has order 2. If r " had fixed points then let I p = and use [7] 5.9 t o find that kpl divides 2n2 which contradicts IZp I = 2 n . Thus T" has no fixed points, and [7] 3.9 shows n is odd. Next we use [7] 3.15 t o conclude that
pb(rlI
47,)= YlY2
* * *
Yrn
where yi are disjoint ni cycles and all the ni are distinct. Since T" commutes with we have that L ( ~ " )  ' ~ ~ L ( T " = ) y j , for some j i . Because the ni are distinct ji must be i. This means that L(T"){~~,C,, * ,on,} = {01,02, * * * ,on,} where b(yi) = (ol,02, . . . ,on,) and 01,02,* * * ,an,EP. Now the order of is n which is odd so that each ni is odd. Since the order of T " is two there must be a o E{al,02, * * ,on,}such that L ( T " ) ~= o. This together with L ( T " )  ' ~ ~ L ( T " )= yi


263
Cyclic Perfect One Factorizations o f K z n
means t h a t L(r")a = a for all a E {a,,a,,...,a,,}.Thus
P
''(rz)
*
C
It is now clear that r" is reductiye since if not (Ft!,?I= 1 (see [7] 3.12). * would then say that P = FL(.? giving r 2 = 1 (use 17) 3.2) which is not possible since o ( r ) = 2n. Since r" is reductive [7] 3.13 gives us t h a t = n so t h a t dF,(,?I 2 n1. This finishes the proof. 0
p8(rm)l
W e are now ready to prove the main result of this section.
Theorem 2.4 K,, admits a cyclic perfect one factorization if and only if n is prime. Proof If n is prime the standard perfect one factorization on K,, is cyclic. This result is stated in [ll] p. 49. It may be seen from the results of [3] which s t a t e t h a t 1; has an element of order n and a central element of order 2. The product of these two elements is t h e desired element of order 2n. To finish we assume t h a t K,, admits a cyclic perfect one factorization. Let r be an element of 1; which has order 2n. 2.3 tells us t h a t L ( r 2 ) has at least n1 fixed points. Since n > 2 we have t h a t r 2 E C ( a ) C(d)for some a,d E P . [7] 2.9 says t h a t all the disjoint cycles of r 2 have the same order. This order must be o ( r 2 )= n . Thus there are two n cycles in the disjoint cycle decomposition of 7'. After a relabelling of vertices we may assume t h a t
n
r 2 = (1,3,5,...,2n1)(2n,2n2,2n4,
* *
,2).
Define r j by r j ( i ) = i + 2 j + l mod 2n. One can verify t h a t r j are t h e only elements of order two in S, which commute with r 2 . 2.3 says t h a t there are a t least n1 elements of P t h a t commute with r 2 . Thus all of the r j except perhaps one must be in P. Observe t h a t r k r j ( i )= i+2(kj) mod 2n, and thus t h e order of r k r j is t h e order of kj mod n. If r k and r j are both in P and k#j mod n then [7] 2.3 says t h a t kj must have order n in 2,. If we fix a j so t h a t r j is i n P we find ( k  j I r k E P and k # j mod n } has n2 distinct elements mod n . T h u s n must be prime since if n is composite there can not be n2 distinct units mod n .
Corollary 2.5 If r €1; and o ( r ) = 2n then r is a 2n cycle. Proof Use the same argument given in 2.4 t o express r 2 as (1,3,5,...,2n1)(2,4,6,
..
*
,2n).
Since r" is a n element of order two t h a t commutes with r 2 we have t h a t r"(1) must be even ( if ~ " ( 1 )were odd then r"{1,3, ...,2721) = {1,3, ...,2n1} and thus 7" would have to have a fixed point). Thus by cyclic relabelling of the even vertices we may ) 2. Since r" commute with r 2 we then find t h a t insure t h a t ~ " ( 1=
r" = (1,2)(3,4) . . . (2nl,2n). We then find t h a t (using t h a t n is odd) 7"''
= (1,4,5,8,9,12,
* *
,2(n1),2n1,2,3,6,7,
*
. . ,2n).
Thus r"+' is a 2n cycle which means t h a t r itself must be a 2n cycle since raising a cycle t o a power can not increase its length. 0
E.C. Ihrig
264
We have now established that cyclic perfect one factorizations can only arise when n is prime. In the next section we set up the correspondence with starter induced perfect one factorizations. 3. Cyclic Perfect One Factorizations of K,, And Starter Perfect One
Factorizations of K,+l In this section we will show that there is a one to one, onto correspondence between cyclic perfect one factorizations of K,, and starter induced perfect one factorizations of K,,,. We will also show that there is a simple relation betweeii the symmetry groups of the two one factorizations. First we review the definition and basic properties of starter induced perfect one factorizations. 3.1 Definition
P is called a starter induced perfect one factorization of K,, subgroup G of I$ such that the following are true. (a) IG I = 2n1. (b) There is an i E X such that Gi = {i}.
if there is a
To reIate this definition t o a perfect one factorization generated by a starter as described in we observe that condition 3.1 means that ] { g ( i ) I g E G } l = ]G [ l 4= 2n1. Thus X may be identified with GU{+ where mis g E G}. Then the starter that generates P identified with the only point not in @ ( iI ) is the one factor that connects 00 t o 1. The following is the basic result concerning starter induced perfect one factorizations.
Theorem 3.2 If P is starter induced then the following are true. There is a unique i, E X such that T ( i , ) = i, for all T €1;. i, is called the ideal point. There is only one subgroup G of I$ that satisfies 3.l(a) and (b). This subgroup is called the starter group. G is a normal subgroup of IF, and for any i E X with i#i, we have
I$ = IGlg(I9i where G is the starter group. 11(T)g = 7gTl (see 2.1 (m)). ( I g i acts on G by fixed point free automorphisms, and thus divides 2(n1).
I(& I
If P is non classical (see 2.2 (a)) then
I is odd.
(Igiis isomorphic to a subgroup of Aut(G) so that 16 c [GII(AUt(G)) where I : Aut(G) +Aut(G) is the identity map.
dl
Proof The proof of this result is essentially contained in 8 . If I$ is G or if P is use 131) all of the statements in 3.2 are trivial. If Ip # G then I$ is not small 2 2 IG I > 2n. Then 4.1 of [S] gives the whole result except the uniqueness of G . The uniqueness of G comes from the following facts. IS] 3.1 says that 1; is solvable. Since 2n1 and lIFp(2n1) are relatively prime ([8]3.4), the generalized Sylow
Cyclic Perfect One Factorizations of K2,,
265
theorem for solvable groups (see [5] 9.3.1 p.141) says that any subgroup of 1; of order 2n1 is conjugate t o G. Since G is normal ([8] 4.1) this means that G is the unique subgroup of order 2n1 of 1;. 0 We are now prepared to consider the relationship between cyclic perfect one factorizations and starter induced perfect one factorizations. Since it will be convenient to deal with vertices other than the integers 1, * . . ,272, we start by expanding our notation to accomodate this need. We also make explicit the distinction between a perfect one factorization and an equivalence class of isomorphic perfect one factorizations. We need to do this since our results concern isomorphism classes while our constructions yield members of these classes. 3.3 Notation (a) KV is the complete graph on the vertex set V. (b) P = (V,P)is a perfect one factorization of Kv if (1) P C Sv and PI= (2) o € P implies that o(o)= 2 and F, = (zi (3) 01,02E H ; that is, oIo2is the product of two disjoint n cycles.
PI1
(c)
Let P = (V,P) and P'= (V',P') be perfect one factorizations. p isomorphism from P t o P' if p is a set isomorphism from V t o pPp' = {pup' I 0 E P } = P'. P 2: P' means that P i s isomorphic t o P.
(d)
[PI = {P'lP'
is isomorphic t o P}.
[ A is
is
V'
an and
said to be a perfect one factorization of
IVI.
(e)
I$ denotes the full symmetry group of P. I p denotes any subgroup of Z$.
We phrase the reduction construction given in [7] in terms of this notation. The proof that the objects defined below are well defined and that r r ( P ) is a perfect one factorization is contained in 171 3.13(c).
DefinitionLemma 3.4 Let P = (V,P) be a perfect one factorization, and let
T be a reductive element of IF (see 2.2 (c)). Define a new perfect one factorization r , ( P ) = (r,(V),r,(P)) as follows: (a) rr(V)= (4 {xi where 7 = zlzz* . zn is the disjoint cycle decomposition of T}. Here corepresents an 'ideal' point which in this case may be any point which is not xi for any i. (b) r,(P)= {r,(o)>Io € C ( T ) } where r,(u)(zi)= ~ q 6 when ' ~ x , o  ' # xi and r,(u)(xi)= mas well as r,(a)(oo) = zi when oxia'= 2;.
u I
DefinitionLemma 3.6 Let P be a perfect one factorization and let reductive element of 1;. Define Tr :
19

I[(P)
by nr(7')(q)= 7'Zi(T')1
T
be a central
E.C. Ihrig
266
nr (7‘)(+ = 00 nr is well defined and a group homomorphism. Proof First observe t h a t since 7 is in the center of Z;, 7’zi(7’)l is again a cycle in the disjoint cycle decomposition of 7. Thus ~ ~ ( 7is’ a) permutation on the set T,(V).It follows t h a t ~ ~ ( 7 is’ )a symmetry of T,(P) from the relation 7rr(T’)1T&)7rr(T‘)
= Tr((7’)1m’).
can be verified to be a group homomorphism by a straightforward calculation. 0
7rr
3.6 Definition
(a) Let
CYC = {PI P
is
cyclic
a
perfect
one
factorization
},
let
[CYC] = {[PI I P ECYC}, and let [CYC],, = {[PI I P ECYC and P = (V,P) with IVl= z n } . (b) Let START = {PI P is a starter induced perfect one factorization }, let [START] = {[PI IP E START}, and let [START],, = {[PI I P E START and P = (V,P)with 2n}. W e need the following lemma before we give the reduction construction t h a t connects the cyclic perfect one factorizations to the starter induced perfect one factorizations.
Lemma 3.7 Let P he a non classical cyclic perfect one factorization. Then Z$ has a unique element of order two which is thus central. Proof If Z$ = Z,, then the result is clear. Otherwise Z ; is not small and we may apply [7] 4.l(a) which says 1; = [Z,,]H where H has odd order. Since n is also odd the Sylow 2 subgroup of Z$ is isomorphic to 2,.Let 2, be t h e unique subgroup of order two in 22,. Since Z2, is normal in Z$ any conjugate of 2, must be in Z ,, and so must be 2,. This means 2, is normal and so the Sylow theorem says t h a t it is the unique subgroup of order 2. DefinitionLemma 3.8 Let IV(/z = p be prime. If Pis a perfect one factorization on Kv let 7 El$ be a reductive element of order two in Z;. Define [CYC] + [START]
T:
by
~ ( [ p l= ) I~r(p)l. T
is well defined.
Proof First consider the case when P = GA,,. Then Z$ has 2 p elements of order two, p of which are reductive (see [3]). One can verify by direct computation t h a t [r,(P)] = GKp+l when 7 is any of the p reductive elements. Thus we need only consider the case when P is non classical. Then 2.3 and 3.7 gives us t h a t Z$ has a unique reductive element 7. Thus we must show two things to verify t h a t T is well
defined. We must show t h a t r,(P) €START and t h a t T,(P) 21 v,(P’) if P 2: P’. Let ? EZ$ and o ( ? ) = 2n. First we claim t h a t G = .lr, is the starter group for T,(P). 00 plays the role of i, in 4.4(b). To verify t h a t nr(F2)is an n cycle (and thus G satisfies the requirements of 3.1) one need only compute (?)2 and z; explicitly after
Cyclic Perfect One Factorizations of K2n
26 7
performing a change of vertices so that .i is in the form (1,2,3,...,2n). Next we let p be t o P' = (V,P'). If T and T' are the elements of order an isomorphism from P = (V,P) two of I$ respectively, then p 7 p  l = T' since p 7 p  l is an element of order two of $1 and $1 has only one element of order two. Thus p z i p  ' will be the disjoint cycles in the disjoint cycle decomposition of T'. This enables us to define a map
b
: r,(W
r+(V
by
b(zi) = pzjp'
and j(0C)= 00
It can then be verified that jr,(a)p' = r ? ~ ( p o p  ' ) ,and thus pr,(P)p' = P' as desired. 0 Our next objective is t o produce an inverse for r . We will call this function s, and we will use the function s^ defined below t o construct s.
3.9 Definition Let n be prime and let P €START with P = (V,P). Let G be the starter group for P and i, be the ideal point. Let i be any element of V with i # i,. Define I C ~: G V{i,} by n i ( g ) = g ( i ) . K . ~is one t o one and onto. Now define where (here we use 2, = {l,l}. with multiplication as the group (a) .^(V) = G @ 2, operation) and {&la EP}. (b) i i ( P ) = {agI g E G and g # i} Here ag is defined by
u
~ g ( ( 9 ' , E ) )= ( g ' g C ,   4
for g € G and E Ez,. 8 is defined by *((g,f)) = (Kr'aKi(g),E) q g , 4 ) = (g,E)
if
if
a K i ( g ) # i,
a K i ( g ) = i,
The subscript i may be omitted if there is no ambiguity concerning which i is used t o define K;. Notice that the one factorization given in 3.9 does not depend on G or i, because of 3.2. We know give a number of lemmas which will gather together enough properties of ii to show that it may be used to define an inverse for r . The next lemma shows that i plays no essential role in the definition of s^i(P).
Lemma 3.10
ii(P)u Sj( P) for all i and j. Proof Since i,jEX{i,} there is a g E G so that g ( i ) = j . Define
268
E.C. Ihrig P:
i(T3
+
ifv)
by
P((g',f)) = (g'gl,c).
Then a calculation yields paglp'
=
agglgl.
Also if a E P , a, = b E i i ( P )and a, = 8 E ij(P)then pa1pl = 0 2 .
This shows t h a t p ( i i ( P ) ) p  ' = ii(P) as desired. 0 Lemma 3.11
i(P)is a perfect
one factorization.
Proof It is a straightforward computation to verify t h a t ag and b are one factors (that they are elements of order two with no fixed points). We must only consider which pairs of one factors are Hamiltonian since Ii(P)I = Ii(V) 11. First observe that G N 2, since n is prime. Because G is abelian we have agag'(g'',€) = (g"(q'g1)'
,€)*
Thus every cycle in the disjoint cycle decomposition of agag'has order o((g'g')'). If g # g', this order is n since n is prime. Thus agag~ is a product of two disjoint n cycles and so (ag:ag,) E H for g # 9'. Next we consider bl and &* for al,a2 E P and crl # a,. We have c ~ & ~ (= ~, (/c'ala2/c(g),E) E) if a2/c(g)# i and ~ , ~ ~ 2 "#( 9 i .)Using (71 2.7 we may after a change of vertex labels assume t h a t i = 1 and t h a t 6 1 = (1,2)(3,4) . * * (2nl,2n) and U, =.(2,3)(4,5) . * * (2n,1). Thus alaz= (2n1,2n3,2n5, . . . ,3,1)(2,4,6,. . . ,2n). If we label (g,l) by rc(g) and (9,1) by n(g) then we have 51&2 =
(2nl,2n 3
,...,1,2,4 ,...,2n)(2n 1,2n
3
,...,1,2,4 ,...,272).
Thus (bl,bz) E H since b1b2is the product of two disjoint cycles. To finish we consider as and 5. We find t h a t (ag8)2(g',€) = (rc'(g'ag')arc(g'),€)
if arc(g') # i and ag'arc(g') # i (use tc(glg2) = gl(lc(gz))). Now g f # i so that gfagf # a. Thus (a,gfrgf) E H and a similar argument to the one given above with a, replaced by a and a1 replaced by g'ag' can be used t o show t h a t (ag&)' is the product of two disjoint cycles. Thus agb is a product of'two or fewer disjoint cycles. The [7] 2.3(c) shows t h a t (ag,&) E H as desired. This completes the proof t h a t &(P)is a perfect one factorization. Lemma 3.12
i ( P ) E CYC. Proof Observe t h a t G @ 2, N Z, since n is prime. Now G @ 2, acts on i(V) = G @ 2, by left translation (Thh' = hh' for all h,h' E G @ Z 2 ) . W e claim t h a t Th E 1; for all h E G @ 2,. This follows from the relations below: = agzfor all ~ 1 ~ E 9G 2 (2) Tg,ag,(Tg,)l Tgbl(Tg)supl = 8 , where a , = galg' for all g € G and a1E P (b)
Cyclic Perfect One Factorizations o f KZn
269
(c) Tl~g(Tl)' = ug1for all g E G where 1 means 0 ,  1 ) E H = b for all CT EP. (d) Tlu(Tl)' Lemma 3.13
If [PI= [P'] then
[&(P)]= [^s(P')].
Proof Let p be an isomorphism from P = (V,P)t o P' = (V',P'). Let P (respectively P') have ideal point i, (respectively io') and starter group G (respectively G'). Since p is an isomorphism p(i,) will be an ideal point for P', and pG(p')' will be a starter group for P'. Thus p ( i , ) = i,~ and p G ( p )  ' = G' by 3.2. We may assume t h a t p ( i ) = i' by 3.10. Define
b : &(V)&(V) by N g , 4 = (PgP%.
Observe by direct computation t h a t $ugj?= upgpl,
and t h a t
fie;' This shows t h a t
= 3,where
0'
= pup'
is the desired isomorphism.
Now we are in the position to define the inverse for r . Definition 3.14 Let P be a non classical perfect one factorization on KV where IVl1 is prime. Let P E START. Define

.!?:[START] [CYC] by
4 P I ) = [&(P)I* Now we come to the main result of the section.
Theorem 3.15 Let n be prime. r : ICYCl,, is a set isomorphism with s as its inverse.

[STARq,.')
Proof We must only verify that 75 = id and sr = id. First we show t h a t r s ( P ) is isomorphic to P. Let P = (V,P), let i, be the ideal point in V and let i # i, be in V. Define p : r,&(V) +V
by P(((!JJ)k771))) = g ( i )
and
270
E.C. Brig
404 = i,. Here T is defined by T ( ( g , E ) ) = (g,E). First observe t h a t the only elements of s^i(P) that commute with T are b for CT EP. One can verify by direct computation t h a t pr,&pp' = a for every a E P .
Thus p gives rise to the required isomorphism from r r g i ( P ) to P. Next we must show that i i r r ( P )is isomorphic to P = (V,P).Let j E V and let T' €15 with T'  T (note Z}. We t h a t o ( T ' ) = 2 p ) . Observe t h a t the starter group G of r r ( P ) is { 7 r r ( ~) I k E let i = ( j , ~ ( j )Define ).
!,I,
p : s.jrr(v)+v by I
p ( ( T r ( T ' Y k , E ) ) = (7
12k+(r+l)p/2
(j).
One can show t h a t pogp' commutes with ( T ' ) ~and is not (7')" so t h a t pogsp' is in P for all g E G . One can also verify that if a ' = r r ( o )then '6 = p a p  ' . This shows that pPp' = i i r r ( P )and the theorem is now complete. 0 We finish this section with a theorem t h a t tells us the relation between the symmetry groups of the cyclic perfect one factorizations and those of the starter induced perfect one factorizations.
Theorem 3.16 Let
[PI E [CrC].Then $1
= I$[P])
22.
Proof First we establish some no_tation. Let F 0 be an element of order 2n of 1;. Let 7 = (?,)_", and = Let G be the subgroup of 1; generated by fl, and let G = x ( G ) where T = xi : I$  + I [ ( p ) is the homomorphism defined in 3.5. Also use r to denote r i . If w, v' E V we say v  d if v and v' are in the same G orbit. Finally define f by
f :V+r(v) by
f ( v ) = (.,?(.)). We now start the proof. First we show that ker(7r) = {i,?}2:Z,. Suppose T ( T ) = i. Then ~ z i 7  l= xi for each i . Since zi is a two cycle we have t h a t r 2 fixes t h e two points not fixed by zi. As this is true for all i, we have t h a t T~ fixes every point in V, and so T is either an element of order two or 1. 3.7 says 7" is the only element of order two in I;, and so ker(7r) = { I , ? } as claimed. To finish we need only produce a splitting a for 7r. Such a splitting is a homomorphism a from I $ p ) to 15 in which x a is the identity on I$p,. Let T By 2.2 we have ~ ( 0 0 = ) 00. Thus 7 ( x i ) = zj for some j. We now define as follows:
47)
o(T)(v) =
v'
if
~ ( f ( v ) )= f(v') and v w ' .
Using the specific representation of ?, as the 2 n cycle (1,2,3,...,2 n ) we find that f  ' ( ~ ( f ( v ) ) ) is a set consisting of one odd number and one even number. Also one can
271
Cyclic Perfect One Factorizations o f K 2 ,
verify that wu' if and only if w and w' have the same parity. Thus 47)is well defined. Observe that 7ra(~)(zi)= 4 ~ ) x i 4 7 )  ~Let . xi = f ( v ) for some w EV. Then xi = (TI,?(.))so that T47)(Zi)=
(47)(~),47)(?(~))) = f ( 4 7 ) (= ~7)()f ( ~ )=) .(xi).
This means that TCY is the identity on Zqp). Next we show that CY is a group ) ) f(w') with 112~~. Also homomorphism. We have a(r2)(v) = w' if ~ ~ ( f ( v = 471)(w') = w" if ~ ~ ( f ( w ' ) ) = /(w") with w"VU". Combining these two statements we ) ) f(w") and wv" so that a ( ~ , 7 ~ ) ( 2= ) ) w", and find that ~ , T ~ ( / ( w = ~ ( 7 ~) 7  Q(7&O2) as desired. To finish we must show that 47)EZ; for each 7 EZ$p)rLet C2(?,) = { a l o ( a ) = 2 and = ?,}. We have IC,(?,) I = n from 2.3 and P C2(?,) = C2(f1){?}.This second statement follows from the fact that L(?,) has order n (which is prime) and acts on P which has 2731 elements so that L ( ? ~ ) has n1 fixed points. Thus p C2(?,) = n1. Now ? E C2(?,) and ? @ P showing that the second statement is correct. Let P, = P C2(7,) and let p2 = P {a1 &?o' = 7). Since 7 is a reductive element of 1; we have that p2I = n . Thus there must be an element of P2 which is not in Pl. Since P2 is invariant under conjugation by 7 , we have P, P2 = 0and so P = Pl P2. First let a EP,. We will show that ~(47))~ E P . Notice that ~ ( 4 7 ) )isanot ? for if so then
n
n
I
n
n
n
c7
Thus we need only show that
CY
p P.
~ ( 4 ~ E) C2(f1). ) a First we observe that
?1Q(7)?;1
Since
=7
= L(Q17))1?
u
= a(7r(?1)77r(?1)1)
is a group homomorphism and Q(T(? ?1a(7)OCk(7)1?~1
1)) =
?,. Let ~ ( 7 , = ) g. We have
= a(gTg')Oa(gTlg')
since ?, commutes with a. Now since G is normal in for some integer I c . This gives us g7gl = rgk and
ZJp)
we have that
Tlg7 = gk+'
+&(47))(4 = 47)4g)ka4g)ka(7). Again we use that a(g) = ?, and that ?! commutes with a so that the expression above becomes a ( ~ ) o a ( ~ )and  ~ so ~ ( 4 7 ) ) (iso )invariant under conjugation by ?,. This means L ( ~ ( T ) )=PP~, C P. To finish we must only consider the case in which a EP,. In this case .(a) and so T T ( O ) T  ~ = .(a') for some d EP,. Notice that ( U ' )  ~ Q ( T ) O ~ T )  ' commutes with . i because each element in this product does. Thus we may apply 7r to this element to find that 7r((d)1C+)aa(7)1) = 7T(.')'7r.i7)7r(
= T ($)IT =
1.
+lo)I
T (C7)T 1
E.C. Ihrig
272
This means that in ocder to show that (c7‘)’47)a47)’ = i we must only show that d ( v ) is in the same G orbit as 47)~747)~(21) for each v EV. We claim that o‘(v)v unless $(TI)= ?(TI). This is true since if v and d ( v ) are in different orbits of G then such that d’(v) = .‘(TI) from 2.3. Thus (d’p’) p H and so d‘ = ? there is a d‘ E C,(?,) as desired. First assume that d(v)w. Since preserves the G orbits, our conclusion as long as a47)‘(~1)47)’(~1).If not then is correct a47)’(~1) = ?47)’(v) = 47)’(?(~1)). This gives us that 7r(a)7’ fixes f (TI). Thus r ( d ) fixes f ( v ) which implies that d ( v )= ?(v) since d has no fixed points, This contradicts our assumption that ~ ( T I ) N Z Jsince ? interchanges the two orbits of G . The only other case we must consider is when a ‘(.) = ?(v). In this case the same calculation as given above shows that 47)a47)’(v) = ?(w), and an anabgous = 1 and argument shows that 47)a47)’(~).7‘(v).This shows that (d)’a(~)a47)’ E I $ , and thus that a is a thus L ( 4 7 ) ) p 2 = P2 C P. This completes the proof that 47) splitting. The theorem is now complete because this splitting shows that 1; is a semidirect product of 2, with I f ; p ) . Since 2, is central in 1; this semidirect product must be a direct product. 0
47)
References [l] B. Anderson. “Finite Topologies and Hamiltonian Paths”, J. Combin. Theory Ser.
B, 14 (1973), 87  93. [2] B. Anderson. “Some Perfect 1Factorizations”, Proc. of the 7th S.E. Conf. Combinatorics, Graph Theory, and Computing, (1976) 79 91. 131 B. Anderson.“Symmetry Groups of Some Perfect 1Factorizations of Complete Graphs”, Discrete Math., 18, (1977), 227  234. 141 B. Anderson and D. Morse. “Some Observations on Starters”, Proc. 5th S.E. Conf. Combinatorics, Graph Theory, and Computing, (1974) 91  108. (51 M. Hail. The Theory of Groups, Macmillan, New York, (1959). [6] A. Hartman and A. Rosa. “Cyclic One Factorizations of the Complete Graph”, Europ. J . Combinat., 6, (1985) , 45  48. [7] E. Ihrig. “Symmetry Groups Related to the Construction of Perfect One Factorizations of K2n”,J . Combin. Theory Ser. B, 40, (1986), 121  151. [8] E. Ihrig. “The Structure of Symmetry Groups of Perfect One Factorizations of K,, ”, t o appear.
[9] A. Kotzig. “Hamiltonian Graphs and Hamiltonian Circuits”, in Proc. Sympos. Smolenice lQ63, Nakl. CSAV, Praha, (1964), 63  82. [lo] N. Korovina. “Sistemy par cikliceskogo tipa”, Mat. Zametki 28, (1980), 271 278. [ll] E. Mendelsohn and A. Rosa. “One Factorizations of the Complete Graph  A
Survey”, J . Graph Theory, 9, (1985), 43  65. [12] R. Mullin and E. Nemeth. L‘AnExistence Theorem for Room Squares”, Canad. Math. Bull. 12, (1969), 493  497. [13] E. Seah and D. Stinson. “A Perfect One Factorization of K,,
”, to appear.
Annds of Discrete Mathematics 34 (1987)273286 Elsevier Science Publishers B.V. (NorthHolland)
0
273
On edge but not vertex transitive regular graphs A.V. Ivanou Institute for System Studies Academy of Sciences of the USSR 9 Prospect 60 Let Oktyabrya 117312 Moscow, USSR TO
ALfX R O S A O N MIS 313TIfTu BIRTUDAY
1. Main definitions and survey of results
Only nondirected graphs without loops and multiple edges are considered in this paper. A graph is called vertex (resp. edge) transitive if its automorphism group acts transitively o n the vertices (resp. edges). A graph is called regular of valency d , if the valency of each vertex is d . Clearly, every vertex transitive graph is regular. Let us consider at greater length the relationship between these two sorts of symmetry. It is easy to give many examples of vertex but not edge transitive graphs. The simplest infinite family of such graphs is described in [12]. Also it is not difficult to construct examples of edge but not vertex transitive graphs. For t h a t we may take any complete bipartite graph I(,,,(m#n). B u t these graphs are not regular. If requires much more effort for the construction of analogous examples of regular graphs. Examples of such graphs were first constructed by J. Folkman[7]. Using the terminology from [ll],we shall call the graph G semisymmetric if G is regular and its automorphism group acts transitively on edges but intransitively on vertices. Every semisymmetric graph is bipartite. This fact follows from a theorem proved by E. Dauber (see [8]). A number of other properties and some infinite families of semisymmetric graphs are described in 171. We shall formulate sufficient conditions for the existence or nonexistence of semisymmetric graphs in the following. Theorem 1 [7]: Let v be a positive integer. No semisymmetric graph with v vertices exists if v satisfies any of the following conditions: l a ) v E I(mod 2); 2a) v = 2p or 2p2, where p is a prime number; 3a) u< 30 and v = O(mod 4); 4a) u< 20. A semisymmetric graph with v vertices exists if u satisfies any of the following conditions: I b ) u = O(mod 2p3), where p is an odd prime number; 2b) v G O(mod 2pq), where p and q are odd prime numbers and q l(mod P 1; 3b) v z O ( m o d 2pq2), where p and q are prime numbers, q is odd and
A.V. Ivanov
274 q = l(mod p); 4b) w> 20 and w
= O(mod 4).
J. Folkman[7] proposed a number of problems concerning semisymmetric graphs, some of which have been solved.
A semisymmetric graph with 54 vertices of valency 3 has been constructed in [3]. It provides an affirmative answer t o problem (4.5) of 17) on the existence of semisymmetric graphs of valency d , where d is a prime number. A semisymmetric graph with 70 vertices of valency 15 has been constructed in 141. So problem (4.3) of [7] on the existence of semisymmetric graphs with w = 2pq vertices, l(mod p ) , also has an where p and q are odd prime numbers, p < q and q affirmative answer. The answer to problem (4.4) of (71 on the existence of semisymmetric graphs of valency d
2 2L
is contained in the papers [4] and [12]. The parameters of the such 4 simplest graph are (v,d)= (20,6). The answer t o problem (4.6) of [7] on the existence of semisymmetric graphs with
w vertices of valency d , where d and 2L are coprime, is contained in Ill]. The first 2 graph of the infinite family described in [ll]has the parameters (v,d)= (112,15). Let us also note the procedure for construction of coverings (see [Z]), allowing for every semisymmetric graph G of valency d t o generate an infinite family of semisymmetric graphs of the same valency. This is the reason for the investigation of semisymmetric graphs with some supplementary properties. One of the results in this field is the complete classification of biprimitive cubic graphs [9].
All these results are very interesting, but for practical use complete lists of semisymmetric graphs are necessary. This work seems t o be the first attempt to construct such lists. The semisymmetric graphs with w vertices (w 5 2 8 ) are enumerated by computer. The nonexistence of semisymmetric graphs with 30 vertices is also proved by an exhaustive search. So a negative answer to Folkman’s problem (4.2) of 171 is obtained. Let us note that 66 is now the smallest value of 21 for which the question of the existence of semisymmetric graphs is not settled. 2. Semisymmetric graphs as incidence systems
As was noted, every semisymmetric graph is bipartite with the same number of vertices in both parts. The bipartite graph G can be represented by (0,l)matrices in a variety of ways. We shall interpret the bipartite graph G as the incidence system S=S(E,B,I) with the set of elements E and the set of blocks B , corresponding t o the vertices from different parts of the graph. Then the incidence ( e i , b j ) E I of an element eiE E and a block bjE B will correspond to the adjacency of the corresponding vertices of the graph. The automorphism group of the bipartite graph G is isomorphic to the automorphism group of the incidence system S . So, the construction of a semisymmetric graph G with w vertices of valency d is equivalent t o the construction of some incidence system S=S(E,B,I) with following properties:
Edge But Not Vertex Transitive Regular Graphs 1.
I=
275
2,
I==v, and every element (resp. block) is incident t o d blocks (resp. 2 elements). 2. The automorphism group y=Aut S satisfies the following conditions : a) y acts transitively on elements and blocks; b) if ye (resp. yb) is the stabilizer of an element e E E (resp. block 6 E B ) in group y , then the blocks 61, . * ,bd (resp. the elements e l , * . , e d ) incident to this element (resp. block) form the orbit of action of group ye (resp. Yb). 3. The dual to S , the incidence system S*=S*(B,E,I),is not isomorphic to S . The construction of all mutually nonisomorphic semisymmetric graphs with v vertices was realized by the use of the technique of constructive enumeration of incidence systems [lo]. The main difficulty was to choose the set of conditions corresponding t o properties 13 which can be tested on the partially constructed incidence matrices. In the beginning we intended to consider the incidence systems having the property I only. In this case it was possible to use the algorithm of constructive enumeration of regular bipartite graphs [I]. We intendec! t o construct all such graphs and then t o choose all graphs with automorphism groups acting transitively on both parts. And in the last stage of investigation we would select semisymmetric graphs. But this method would require very much processor time. For example, during the first two hours of the computer calculations more than 1500 graphs with 30 vertices of valency 3 were constructed and only one of them was transitive on the vertices of each part. Therefore it is desirable to construct solely the graphs which are transitive on the vertices of every part (i.e. incidence systems with properties 1 and 2a). One can not solve this problem in full measure. Instead, a set of the invariants, which approximate , used. The verification of property 3 properties 1 and 2 for the graphs with ~ 5 3 0was was realized by some heuristic methods on the completely constructed incidence systems.

a
3. Invariants of the parts of semisymmetric graphs
Let d be the valency of every vertex of a semisymmetric graph G. The transitivity of the action of Aut S on the set of elements E implies the equality of every characteristic of every two vertices from the same part of G. Let us consider one of these invariants. Recall that the incidence of some element and some block corresponds to some edge of the semisymmetric graph. Taking this fact into account we shall use graph theoretic terminology from now on. Let G be a bipartite graph with vertex set V = V l u V 2 and u€Vl.Note by a? the number of vertices belonging t o V, and connecting with u exactly i paths of length 2. Clearly, the vector au = (uy) 05 d is an invariant of the vertices in Vl. We shall denote this vector by a . (lsisd)the Let u and w be adjacent vertices, uEVl,w€V2. Denote by number of vertices adjacent to w and connecting with u by exactly i paths of length 2. From property 2 it follows that the vector bu''=(6Y) 15 d does not depend on the choice of u or adjacent vertex u. Therefore later we shall omit the superscripts.
is
is
276
A. V. Ivanov
We shall prove the following relations in terms of incidence systems d C U i = i0 d
w,1
Cia; = d(d1) i1
d
C bi
=
d1
i1
dbi = iai
i
=
1,2,
* * *
,d
ai,bi are nonnegative integers for i=1,2, ...,d. The first and third equations follow immediately from the definitions of ai and b i .
To prove the second equation, count the incidences of blocks containing the element u with the other elements in two ways. To prove the last group of equations, count the incidences of blocks containing the element u with the elements corresponding to vertices of G connected to u by i paths of length 2, in two ways. Let us rewrite the system (1) in the following form. d
cai
=
v,1
i0
d
C i a i = d(d1) i1
iai
= O(mod d )
i=1,2, 4  1
ai is nonnegative integer for i = 0,1, ...,d. ai,
Clearly, the systems (1) and ( 2 ) have the same solutions with respect to = 0,1, . . * ,d.
i
Now let us consider some other conditions for the solutions of ( 2 ) to be the invariants of a semisymmetric graph G. Let i* = max(0,2dwl).
It is easy t o see that ai = 0 for i< i* when 2d>v,.
The condition ad# 0 leads to the existence of r = ad+l vertices having the same set of neighbours. These vertices form a block of imprimitivity of size r for the transitive action of the group y = Aut S on the corresponding part of the graph G. Hence it is easy to show t h a t
w =:(mod d
r ) , ai =O(mod r ) for i = 0,1,...,d
= O(mod r ) ,
bi sz O(mod r ) for i = 1,2,...,d
(3) One can show t h a t 3 5 d< w,3 for semisymmetric graph G . From ( 2 ) we have ad1 = d or 0. If ad1 = d , then a. = wld1, ai = 0 for I < d2, ad = 0. In this case it is easy t o prove the transitivity of the action of the automorphism group y on the vertex set of the graph G from the transitivity of the action of y on one part of the graph. Therefore
is
Edge But Not Vertex Transitive Regular Graphs ad1 = 0
217
(4)
The following inequalities can be proved in the same way: ad # d1,
ad # vldl
when 2d >ul
(5) The number of complete bipartite subgraphs K2,i ( 1 5 i< d ) of the semisymmetric graph G , with two vertices from the first part having exactly i common neighbours in graph G , is equal to vl.
ai
Therefore 2 aiwl = O(mod 2 ) for i = O , l ,

*
. ,d
(6)
bi Counting the number of such groups with some fixed vertex gives d. 2 bid
= O(mod 2 )
fori = 1,2,
   ,d
Therefore (7)
Using ( 1 ) to rewrite ( 2 ) with restrictions ( 3 )  ( 7 ) we have:
c a1 d
= vl1
i0
d
C i a i = d(d1) i1 V1
d a,
O(mod
(Ud+l))
= O(mod ( a d + l ) )
= O(mod (ad41)) i = O,l,...,dl
iai = O(mod ( d ( a d + l ) ) ) i = O , l , ...,d1 wlai = O(mod 2 ) i = O , l , ...,d
iai =O(mod 2 ) i a d  1 = 0, ad
= 0,1,
# d1,
ad
*
.
*
,d
# v1d1
if 2d
> Wl
a i 2 0 , i = 0,1, ...,d Denote by N t the set of integer solutions of (8). Let ai be the invariant described (i=1,2) of the semisymmetric graph G . By the above of the vertices from part reasoning given above we have proved
Theorem 2: a i E N t for i = 1,2. T h e nonexistence of some semisymmetric graphs with parameters (w,d) immediately follows from the absence of solutions of (8). Examples of such parameters are (20,5), (20,7), (24, i ) for iE{7,8,9}, (28,i) for i€{7,9,10,11}, (30,9),and ( 3 0 , l l ) . Let a s be the solution of system ( 8 ) and a,*# 0. Fix a pair of vertices from the same part V# exactly i (resp. j) common neighbours with the first (resp. second) vertex of the fixed pair.
278
A. V. Ivanov
Theorem 3: For the vector u*EA$ to be the above described invariant of the vertices of a bipartite graph transitive on each part, the system of relations (9) must have the solution in nonnegative integers. d
C c 13 . . = a '*.  6 .
i = O , 1 , ...,d
j 1 d
C c'.3. = u;
6,,
j = 011,...,d
i1
0
Cij>
(9)
c i j = 0 if (i+j
where
6;
=
t
> d ) or (3d > w 1 and i + j < 3dwl)
1 ifi=O 0 ifi#O
Proof: The set of nonfixed vertices in part V+ of the bipartite graph G is divided into I=(d+l)(d+l) classed by the possible numbers of common neighbours with the fixed vertices. The numbers of vertices c i j in some of these classes must be equal to 0. Let s2=
[A:
::: 01 01 10 01
... *..
1100
00 0°1 be the submatrix of the incidence matrix S corresponding to the fixed pair of vertices. Let us examine some other row ,s = ( s m k ) of the matrix S corresponding t o a vertex having exactly i (resp. j) of common neighbours with the first (resp. second) vertex of the fixed pair. Obviously, fd+l
O0O0
. * * ..*
f2d+l
We have 2d
d
a)
CS,~ + C
f
smf = i + j = d
fd+l
1
v1
C
s!,
f2d+l d
v1

2d
sm/ = d  i  j 5 v12d f=2d+l f 1 fd+l Clearly, the sum of the elements in roy i, i=1,2 ,...,d(resp. column j , j=1,2 ,...,d ) of the matrix C = ( c i j ) must be equal to ai (resp. a>; the total number of vertices in V+having exactly i (resp. j) common neighbours with the first (resp. second) vertex of the fix;d pair. In the analogous equation for i = 0 (resp. j = 0) we must subtract 1 from a . to take into account the second (resp. first) vertex of the fixed pair.
b)
C
Sm/
= d  x s r n~
C
The absence of integer solutions of (9) proves the nonexistence of semisymmetric graphs for some invariants obtained from ( 8 ) for w = 24,28,30. Until now we considered the relations for the invariants corresponding t o one part of the graph G . From the semisymmetric graph G we can define two vectors ( u 1 , a 2 ) corresponding to the vertices from the different parts V,,V2 of the graph.
279
Edge But Not Vertex Transitive Regular Graphs
Theorem 4: Let # be the set of nonnegative solutions of (8). If (not obligatory different) vectors a1,a2E# are the invariants of the vertices of the two parts of the semisymmetric graph G , then d
a)
C i 2 ( a f  a,?) = o i1
1E {1,2} then a;'(1Zi)+j
is
2 0 for 15 d. d Proof: By counting the number of subgraphs K2,2of the graph G in two different ways, we have: b) if af # 0 for some 05 j < d and
211
()C
d
af
i(i1)
v1
=
()C
d
i(i1)
a,?
2 2 i2 Now with regard for the second equality from (1) 2
2
i2
So the relation (10a) has been proved. Let us fix some edge q, of the semisymmetric graph G corresponding to the element ,,s = 1 of the incidence system S. Denote by J i ( q l ) = s f the set of rows of the matrix S satisfying the following conditions VI
sf€Ji(ql)
CSfjSmj j 1
= i,
Sf,
=1
Cl:arly, J i ( q l ) = b f . Let us fix some edge qz, corresponding to the element smn*( n # n ) of the matrix S . For some j 01
aj2# 0
and
C sfn*sfn
=j
f1
From the transitivity of the action of Aut S on the edges we have g ( q 2 ) = q1 for some automorphism gE Aut S. Using the following notations
I,
ri = IJi(qJn ~i(q2)
pi = I J i ( q A / ~ i ( q lI)
from the equality pi
+ ri = IJi(q2)I = bf
and the obvious inequalities pi< a/
 bf ,
ri5j  1,
we have ail
 2bi1 + j 2 1
Now from (1) we have that the proposition (lob) is true for I = 2. Reasoning by the same way for S T ,we obtain (lob) for 1 = 1.
280
A. V. Ivanov
Apparently, considering more precisely the structures of the incidence systems corresponding t o semisymmetric graphs one can add some other relations to the ones described above. In this case one can try to enumerate the semisymmetric graphs with u > 30. Our experience of the construction of regular bipartite transitive on one part graphs show that including each new relation into consideration decreases the number of complete constructed graphs and the computing time of the enumeration. Sometimes if is not necessary t o use the computer at all. For example, the nonexistence of semisymmetric graph G with u = 30, d = 8,a' = (0,0,4,0,2,8,0,0,0)Tfollows from (10). And the search for these graphs required more than four hours of processor time. 4. Algorithm of constructive enumeration of semisymmetric graphs
In [lo] an algorithm for constructive enumeration of incidence systems is presented. By this algorithm all possibilities for the n + l s t row of the incidence matrix when n rows are already constructed are generated and investigated. The computer program for the ICL4/70 realizing this algorithm was used for an exhaustive search of semisymmetric graphs in the following manner. In the beginning the system of relations (8) for u = 20,24,28,30 and 3
21 5 d 5 3 2
was solved. Thus all possible values of the invariants described above for the vertices of semisymmetric graphs with parameters (v,d) were obtained. Some of these values were excluded by (9) and (10). Then all incidence systems which were transitive on the elements for every remaining value a' were constructed by the computer. In the process of construction the relations (9) and (10) were taken into account and verification of' the requirement on subgroup ge (property 2) was made. As a result, about 65% for u 5 28 and and 10% for w = 30 of all constructed incidence systems actually satisfied property 2. From the set of constructed incidence systems corresponding to edge transitive bipartite graphs we selected those satisfying property 3. For this the algorithm described in [6] would be used except for cases where a' # a2. 5. Results of constructive enumeration of semisymmetric graphs with u
5 30
The present paper is one of the first attempts t o constructively enumerate combinatorial objects largely defined by the transitivity of their automorphism groups. Therefore information about all completely constructed incidence systems is presented later in the tables. This information allows one to estimate the degree of approximation of the transitivity properties by the described methods. The invariants of the parts of semisymmetric graphs which are not excluded by (9) and (10) were obtained from (8) for different sets of parameters (v,d) and are presented in Tables l a  Id. In the same tables information on the numbers of constructed incidence systems is given. is the number of vectors ( a i ) for Notations : d is the valency of the graph; given d ; N is the number of complete constructed incidence systems (i.s); N , is the number of i s . which are transitive on the elements; N , is the number of i s . which are transitive on the elements and on the blocks; N , is the number of i s . corresponding to
Edge But Not Vertex Transitive Regular Graphs
281
the edge transitive graphs; T is the processor time.
d N j a. a l a2 a3 a 4 a5 3 1 3 4 1 4 2 0 3 3 4 1 6 1 0 2 0 3 0
I
6 0 8 0 4 0 0 0
0 4 0 6 4 0 3 0
0 0 0 0 0 8 0 6
a6
N N , N 2 N, T(sec.)
3 1 1 1 1 1 1 1 0 1 1 1 1 0 0 0 1 1 1 6 0 0 1 1 3 0 0 1 1 Total: 10 8
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
6.7 0.5 0.5 1.2 4.2 0.5 1.8 5.4
8
8
20.8
Table la. v=20.
d Nj 3 1 4 1 2 3 4 5 5 1 6 1 2 3 4 5 6 7 8
a.
5 6 2 5 3 1 1 3 0 2 0 1 0 1 0
a l a2 a3 a4 a5 a6 N N , N 2 6 0 0 1 1 1 0 4 0 1 2 2 2 8 0 0 1 1 1 1 1 1 1 0 6 0 0 5 3 3 4 4 0 0 2 1 1 8 2 0 0 010 0 0 0 1 1 1 0 0 6 0 0 2 1 1 1 0 9 0 0 0 2 1 1 1 0 0 8 0 0 1 1 1 1 0 6 4 0 0 1 0 0 0 0 3 4 3 0 0 0 0 0 0 6 2 3 0 0 6 3 3 0 0 1 0 0 0 0 0 0 0 0 3 8 0 0 0 2 2 2 Total: 24 18 18 Table Ib. u=24.
N, 1 2 1 1 2 1 1 1 1 1 0 0 1 0 2
T(sec.) 30.5 1.1 0.8 3.3 32.4 6.0 4.7 0.5 0.6 0.7 0.3 83.6 44.5 4.3 22.4
15
3.3min.
A.V. Ivanov
282
d N j a,, a l a2 a3 a4 a5 a6 a7 as N N , N 2 N, 3 1 4 1 2
7 8 4 3 7 4 5 5 3 6 1 5 1 3 6 1 2 2 0 3 3 4 2 5 1 6 1 7 0 8 1 0
6 0 0 0 4 0 1 8 0 0 1 0 6 0 0 4 4 0 0 8 2 0 0 1 2 0 0 0 0100 0 0 0 6 4 0 0 0 1 2 0 0 0 0 3 4 3 0 0 6 2 3 0 0 9 0 3 0 0 6 6 0 0 0 9 4 0 0 0 0 0 1 2 0
1 1 0 0 0 0 0 0 0
1
1 1 0 1 0 0 1 1 0 2 0 0 1 0 0 1
T(sec.) l0min. 1.2 1.6 13.4 2min. lmin. 32.5 10.8 0.2 2.1 1.7min. 1.Omin. 26.2 3.8min. 1.3min. 1.5
20 11 10 9
22.5min.
4 2 2 3 0 0 1 1 0 3 0 0 1 0 0 3
2 1 1 1 0 0 1 1 0
2 0 0 1 0 0 1
2 1 0 1 0 0 1 1 0
2 0 0 1 0 0 1
Total:
Table lc. v=28.
Edge But Not Vertex Transitive Regular Graphs
d Ni 3 1 4 1 2 3 4 5 1 6 1 2 3 4 5 7 1 8 1 2 10 1 12 1
a0 a1 a 2 . a 3 a4 a5
8 6 0 0 8 0 6 0 0 6 4 4 0 0 4 8 2 0 0 212 0 0 0 4 0 1 0 0 0 6 0 0 6 0 4 0 010 0 0 6 0 8 0 2 0 6 6 0 0 012 2 0 0 0 014 0 0 0 0 8 2 0 0 0 0 1 4 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 4 0
as a7 a8 ag a10 a l i . a l 2
2 0 0 0 0 0 0 4 0 0 0 0 1 0 0 0
0 0 0 0 0 8
0 6 Total:
0
283
N N1 N2 Ne
57 1 4 0 4 0 1 0 1 0 0 1 7 1 0 0 1
2 0 2 0 3 0 1 0 1 0 0 1 2 1 0 1
2 0 2 0 3 0 1 0 1 0 0 1 2 1 0 1
1 0 1 0 1 0 1 0 1 0 0 1 1 1 0 1
78 14 14 9
T(SeC.)
2h.10m. 11.5 18.1 16.8min. 13.8min. 5.0 0.9 1.8 41.0 3.5min. 1.2min. 25.2 3.0min. 29.3 47.2 4.5min. 3h.llm.
Table Id. w=30.
Table 2 contains the orders of automorphism groups of constructed incidence systems corresponding to the edge transitive grapis. Here NO is the number of i s . S in succession; Ng is the number of dual to S i.s. S . Clearly, the i s . S corresponds to a semisymmetric graph iff NO # Ng.
A. V. Ivanov
284
No 1 2 3 4 5 6 7 8 9
10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41
u 20
24
24 28
30
d 3 4 4 4 4 6 6 6 3 4 4 4 4 4 4 4 5 6 6 6 6 6 6 3 4 4 4
Ndo
AutS
N9
1 1 2 3 4 1 2 3 1 1 1 2 3 4 4 5 1 1 2 3 6
120 21O.10 25.120 25.120 40 25120 25.120 120 72 212*72 21272 26.48 26*48 48 48 384 240
5
1 2 2 5 1 1 2 4 1 3 1 1 2 1
1 2 4 3 5 7 6 8 9 10 11 13 12 14 15 16 17 18 20 19 21 22 23 24 25 26 27 28 31 30 29 32 33 34 35 36 37 38 39 40 41
6 6 6 8 3 4 4 6 6 7 8 8 12
8
8 1 1 3 6
Table 2
(3!)4*8
26(3!)424 26(3!)424 144 576 576 56448 214*14 56448 336 336 27*42 214168 27.42 214*168 720 60 360 #(3!)"10 720 20160 720 20160 720
Edge But Not Vertex Transitive Regular Graphs
285
In summary, we constructed 5 pairs of nonselfdual incidence systems which were transitive on the elements and on the blocks. They correspond t o 5 semisymmetric graphs with 201u @
[email protected] for Ge's isomorphic t o a graph G , then H is said t o have a cyclic Gdecomposition. In this paper, we consider only the case when any orbits corresponding t o GL's are full. For each graph G,, let
C
AG,(z)=
(zij
G, I'lEE(Gt1
+ z J  1.~
Then ~ ( z ) = A G , ( z ) + * . . + A G , ( z ) holds. A cyclic Gdecomposition of XK, is called a cyclic (v,k,X) Gdesign, where k is the order of G . Note that a cyclic Gdesign is always balanced.
Lemma 2. For an even integer v, if there exists a cyclic (v,k,X) Gdesign with no short orbit, then X must be even. Proof: be a cyclic Gdecomposition of XK,.
Let
[email protected]*
[email protected] i>
(0,
is contained in a graph G t + j for a certain V
.
a graph G e + j +
2
Ge + j and Ge +j+
Since the
e
If the edge
and j , then it is also contained in
cyclic Gdecomposition
has no short orbit,
21
 must
be distinct. Thus the set of all graphs having the edge 2 (0, f }can be partitioned into pairs of graphs (Ge+j,Ge + j + " ) I s . Hence X must 2 be even. 2. Difference arrays and Graph arrays.
Let F be u1
F(y)=
C
a cyclic graph of order w with a characteristic polynomial
fpyp, then f=F(l)=Cf,
is the degree of each vertex. Let G be a simple
P0
graph, which is defined t o be a graph with no loops and no multiple edges, with vertex A k X f array D = ( d i j ) is called an (F,G)difference set V(G)={O,l, . . . ,kl}. array if dj . €2, and every vertex p of F occurs exactly f p times among the differences {dijdpj (mod v ) I j=1,2,..,,f} for any two rows i and i' such that {i,i'}€E(G), that is,
6
ydiidi*j = F ( y )
(y"=l)
(2)
j0
holds for any {i,i'}€E(G). Example 2. (i) Let F be a cyclic graph of order 5 with a characteristic polynomial F ( y ) = l + y + y 4 and G be a path of length 4. Then the following array is an (F,G)difference array:
A Product Theorem for Cyclic Graph Designs 0 0 0 0
0 1 0 1
291
0 4 0 4
(ii) Let F be K , with a loop on each vertex and G be K,, then the following array is an (FIG)difference array: 0 0 0 0
0 1 2 3
0 2 4 1
0 3 1 4
0 4 3 2
In the case when a cyclic graph F is the complete graph K , with a loop on each vertex, a (XF,Kk)difference array is called a ( v , k , X ) row difference scheme, which is useful t o construct cyclic BIB designs (see Jimbo and Kuriki IS]). If there is a mapping 4 from V(G) onto a set of n colors such that $(a)#+(b) for any {a,b}EE(G), then a graph G is said t o be ncolorable. Lemma 3. If G is 2colorable then an (FIG)difference array exists for any cyclic graph F . Proof: We have the lemma by constructing an array such that the rows corresponding t o the first color are all zero and in the rows corresponding t o the second color, every vertex p of F adjacent to the vertex 0 occurs f p times, where F ( y ) = C fpyP is a characteristic polynomial of F. 0 Let G be a simple graph of order k = p ( G ) with e = e ( G ) edges. Let G * be a directed graph obtained from G by replacing each edge by two arcs, one directed each way, between the same vertices. A kX2e matrix A=(aij) with elements from V(G)={O,l, . . . , k  l } is called a graph balanced array for G I if every arc of G f occurs exactly once in the set of ordered pairs {(a;j,ai,j) lj=1,2, . . . ,2e}, for any {i,i')€E(G). Lemma 4. A graph balanced array exists for any 2colorable graph G . An m X N matrix A with entries from a set of s( 2 2 ) elements is called an orthogonal array of size N , m constraints, s levels, strength t and index XI if any t X N submatrix of A contains all possible t X I column vectors with the same frequency A. Lemma 5. If there exists an orthogonal array of size k2, k + l constraints, k levels, strength 2 and index I, then a Kkarray exists.
Proof: Without loss of generality, we can assume that a (k+1)Xk2 orthogonal array A=(ai j ) is standardized as follows: ak+l,pk+r=qfor any O s q s k  1 and 1 < r s k ,
M.Jimbo and S. Kuriki
292
and ai,k(k,)+j=jl
for any l < i < k
and 1 s j s k .
Then the submatrix A = ( a i j ) ( i = l , . . . , k ; j = l , . . . , k ( k  l ) ) is a Kkarray. 0 Example 3. The following are graph balanced arrays for graph G’s which are not 2colorable and k 5 5 except K , and K 5 :
1
1
(ii)
012012 120201 201120 0123401234 1234040123 2340134012 3401223401 4012312340 011202033404 102120304340 220011440033 102120304340 220011440033
(iii)
But the existence problem of graph balanced arrays is not solved, in general. The following theorem shows a way t o construct an (F,G)difference array by using a graph balanced array.
Theorem 1: Let G be a simple graph with a graph balanced array. If a cyclic graph F has a cyclic Gdecomposition with no short orbit, then there exists an (FIG)difference array. Proof: Let F = < G , > @ *  . @ < G , > be a cyclic Gdecomposition, where GL’S are isomorphic to a graph G with k vertices and e edges. Then we have F(y)=AG,(y)+.*.+AG,(y). Let 4t:V(G)+V(Gl) be isomorphisms. For k X 2 e Garray A = ( a i j ) , let D L = ( b l ( a i j ) for ) e = 1 , 2 , . . . ,a. Then every arc of GL occurs exactly once in the*set of ordered pairs {(#J(( a i j ) , # (~a i j j ) ) I j = 1 , 2 , ...,2e) for any {i,i’}W(G), where Gl is a directed graph corresponding to a graph G a s before. Hence we have 2e 2
y+das,)+t
(as,j)
=
A G~ (y).
j 1
Thus a k X 2 e a array D = [ D , D , *  * D , ] is an (FIG)difference array. 0
A Product Theorem for Cyclic Graph Designs
293
3. A product theorem for cyclic decompositions.
In this section we consider a composition method (or a product method) of two cyclic Gdecompositions. The following lemma is obvious:
Lemma 6. If cyclic graphs H and F with V ( H ) = V ( F )have cyclic Gdecompositions, then H @ F , X H and I , *H have cyclic Gdecompositions, where the graph I , is a cyclic graph of order w with a loop for each vertex and no other edges. Theorem 2. Let G be a simple graph. Let H and F be cyclic graphs. If (i) H has a cyclic Gdecomposition with no short orbit, and if (ii) there is an (F,G)difference array, then the cyclic product H*F has a cyclic Gdecomposition with no short orbit. Proof: Let a k X f array D = ( d i j ) be an (F,G)difference array. Let
[email protected]@ b e a cyclic Gdecomposition, where GL'S are isomorphic to G . Then we have H ( z ) = A G 1 ( z ) +  .  + A G , ( z ) . Let +L:V(G) = {0,1, . . . ,kl}+V(G!) be an isomorphism. For each base graph GL of H , construct base graphs G t (j=1,2, . . . ,f ) of H*F which have vertex sets V(GLj)={4e((i)+v1d,j
I iW(G))
(mod
~ 1 ~ 2 )
and which is isomorphic to G , where w1 and w 2 are the orders of H and F , respectively. Then we obtain by (2)
= c
zh ( i )  h ( i ' )
( i ,i')EE(G*)
zul(d,,d,~,) j 1
=AG,(z)F(zU1), where zu1u2=1. Hence
H(z).F(z")=
5
AGL (z)*F(z")
e1
=$
AGtj(z),
l = 1 j1
which implies that H*F can be decomposed cyclically by base graphs G l j ' s . Hence the theorem is proved. 0 By combining Theorems 1 and 2, we obtain the following corollary:
Corollary 1. If cyclic graphs H and F have cyclic Gdecompositions with no short orbit, and if there is a graph balanced array for G , then a cyclic Gdecomposition of H*F exists.
M.Jimbo and S. Kuriki
294
The following corollary is the direct consequence of Theorem 2 and Lemma 3. CoroIIary 2. Let a simple graph G be 2colorable. If H has a cyclic Gdecomposition with no short orbit, then for any cyclic graph F , H*F has a cyclic Gdecomposition with no short orbit. Corollary 3. Under the same assumption as Theorem 2 H * ( F @ X I ) has a cyclic Gdecomposition with no short orbit, where V ( F ) = V ( I ) .
Proof: Note that H * ( F @ X Z ) = ( H * ~ ) $ ( X H * Z )and that p ( G ) X l zero vector is an (Z,G)difference array. Hence H*Z has a cyclic Gdecomposition with no short orbit 0 by Theorem 2. Thus the corolla.ry is proved by Lemma 6. Theorem 3. Let G be an ncolorable simple graph of order k. If there are (i) a cyclic (wl,k,Xl) Gdesign with no short orbit, (ii) a cyclic ( w 2 , k , X 1 X 2 ) Gdesign with no short orbit and (iii) a (w2,n,X2)row difference scheme, then there exists a cyclic (v1v2,k,X,X2) Gdesign with no short orbit. Proof: We have only t o show that the graph X1X2Kulu, has a cyclic Gdecomposition. Let A , , . . . ,& be n color classes of G . Let D = ( d i i ) be a (v2,n,X2) row difference scheme. Construct kXX2v2 array D as follows. Let every vertex of G correspond to a row of an array D arbitrarily. Make copies of the ith row of D to the rows which correspond t o vertices of Ai for every i . Then it is easily shown that the array D is a (X2(
[email protected]),G~differencearray. From the assumption (i), X,K,, has a cyclic Gdecomposition with no short orbit. Hence by Theorem 2, the cyclic graph X1Ku1*X2(Ku2$Zu~ has a cyclic Gdecomposition with no short orbit. And Zul *A1 ASKu,has a cyclic Gdecomposition with no short orbit by Lemma 6. Finally, by (1)of Lemma 1 X1 X 2 Ku 1 u ,
={X,K,,
* X2(Kup
@LJ2))@(~VJ,
*X1X2Kuph
we have the theorem. 0 The following theorem can be shown from Corollary 1 by the similar argument to Theorem 3: Theorem 4. Let G be a graph of order k with a graph balanced array. If there are
(i) a cyclic (vl,k,Xl) Gdesign with no short orbit and (ii) a cyclic (v2,k,X2) Gdesign with no short orbit, then there exists a cyclic (vlv2,k,XlX2) Gdesign with no short orbit.
A Product Theorem for Cyclic Graph Designs
295
Acknowledgements This research was supported in part by a GrantinAid for Scientific Research of the Ministry of Education, Science, and Culture under Contract Number 321600961530017, and by a Research Grant of Science University of Tokyo under Contract Number 861001. References
J.C. Bermond and D. Sotteau, “Graph decomposilions and Gdesigns”, Proc. 5 t h British Combinatorial Conf . (1975) 5372.
M.J. Colbourn and C.J. Colbourn, “Recursive constructions for cyclic block designs”, J . Statist. Planning and Inference 10 (1984) 97103. H. Enomoto and K. Ushio, “Ckfactorizations of the complete bipartite graphs” (in Japanese), Kyoto Uniw. Inst. Math. Kokyuuroku 587 (1986) 5257.
P. Hell and A. Rosa, “Graph decompositions, handcuffed prisoners and balanced Pdesigns”, Discrete Math. 2 (1972) 229252. J.D. Horton, “Resolvable path designs”, J. Combin. Tiieory (A) 39 (1985) 1 17 131.
C. Huang and A. Rosa, “On the existence of balanced bipartite designs”, Utilitas Math. 4 (1973) 5575.
C. Huang, “Resolvable balanced bipartite designs”, Discrete Math. 14 (1976) 319335. M. Jimbo and S. Kuriki, “On a composition of cyclic 2designs”, Discrete Math. 46 (1983) 249255. A. Rosa and C. Huang, “Another class of balanced graph designs: Balanced circuit designs”, Discrete Math. 12 (1975) 269293. S. Yamamoto, H. Ikeda, S. Shigeeda, K. Ushio and N. Hamada, “On clawdecomposition of complete graphs and complete bigraphs”, Hiroshima Math. J . 5 (1975) 3342.
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Annals of Discrete Mathematics 34 (1987) 297300
297
0 Elsevier Science Publishers B.V. (NorthHolland)
A New Class of Symmetric Divisible Designs Dieter Jungnickel
Mathematisches Institut JustusLiebigUniversitzt Giessen Arndtstrasse 2 D6300 Giessen Federal Republic of Germany TO A L C X R O S A
O N MIS 3I3?IEm BIRT)IDAY
ABSTRACT Using affine designs, we give a new construction for symmetric divisible designs. This yields the existence of the class of such designs with parameters m=qd++q+2, n=qd+l, k=q q +.+q2+q+l)+qd+', Xl=qd(qd'+.+q+l)+qd+', and &=i'(id'+..'+q+3) for all prime powers q and all positive integers d . We also construct relative difference sets with these parameters. 1. Introduction
Let A be an affine design S x ( 2 , k ; w )(see Beth, Jungnickel, and Lenz [ I ] for background from design theory). In 1971, Wallis [6] proved t h a t one can construct a symmetric design with parameters ( 1 ) w* = ( r + l ) w , k * = kr and A * = kX ( = ( r  1 ) ~ ) ; In particular, using AGd,(d,q)one gets the series of symmetric designs for ( 2 ) w* = qd+'(qd++q+2), k* = q d ( q d +  + q + l )
and A * = qd(qd'++q+l);
and using a Hadamard 3design Sn1(3,2n;4n)one gets ( 3 ) W * = 16n2, k* = 2 n ( 4 n  l ) , and A' = 2 n ( 2 n  l ) .
Wallis used strongly regular graphs to prove this result; a considerably simpler construction using "auxiliary matrices" was given by Lenz and Jungnickel [ 4 ] . These authors also observed that their construction could be used to establish the existence of difference sets with parameters (2), a result due t o McFarland [ 5 ] . In the present note, we use a simple modification of the LenzJungnickel construction t o produce a new series of symmetric divisible designs (from affine designs) and also a new series of relative difference sets.
D.Jungnickel
298
2. New Symmetric Divisible Designs In this section, we prove the analogue of Wallis’s theorem (see e.g. Jungnickel [3] for divisible designs and the notation used here): Theorem 1: Assume the existence of an affine design S i ( 2 , k ; v ) . Then there also exists a symmetric divisible design with parameters (4) m = r + l , n=v, k’=kr+v, X,=kX+v,
and X2=k(X+2).
Proof: Let A be the given affine Sh(2,k;v) and label the points and parallel classes of A as pl,...,pv and Bl,...,Br, respectively. Drake and Jungnickel [2] have constructed “auxiliary matrices” M ~...,,M,. from A as follows. P u t Mh = (mfj), ( h = l , ...,r ; i,j=l,...,v), where
t
1 if pi and pi are on a block from Bh
mfi =
0 otherwise.
tih
By the definition of an affine design, one then gets the following equations: (5)
MhM&
=
if h#h‘ ifh=h’
(h,h‘=l, ...,r)
and thus (6) MIMT
+  + MrMF=
k(rX)f+kXJ.
Now let Mn=J and replace in the group table of residues mod r+l each entry i by Mi; that is, put Mo M i ... M,. LO Ml M 2 ... Mo Ll
L
=
L,. From (5) and (6), one obtains
if h#h’ rX)I+(kX+v)J if h=h‘. Thus L is the incidence matrix of the desired symmetric divisible design, since ( r  1 ) p = k X for an affine design. 0 We remark that replacing Mo by the zero matrix in the above proof gives the LenzJungnickel proof of Wallis’s theorem (see also [l, 11.8.161). Using AG,,(d,q) in Theorem 1 gives the following result: Theorem 2: Let q be a prime power and d a positive integer. Then there exists a symmetric divisible design with parameters
A New Class of Symmetric Divisible Designs
299
Note that using Hadamard 3designs in Theorem 1 does not yield anything interesting. One obtains a symmetric design the complement of which has parameters (3) above. Similarly, the case g=2 of Theorem 2 results in the complement of a symmetric design with parameters (2). Thus Theorem 2 is interesting only for 9#2. 3. New relative difference sets
As in [4], one easily sees that the group H = 2, @ EA(gd+') (where E A ( z ) denotes the elementary abelian group of order z) acts regularly as an automorphism group of the symmetric divisible design D constructed in Theorem 2. Thus D corresponds to a relative difference set D with parameters (7) in G . (See [3] for relative difference sets and their relation to divisible designs). Using a modification of McFarland's construction [5] for difference sets with parameters (2), we can obtain relative difference sets with parameters (7) in other groups: Theorem 3: Let q be a prime power and d a positive integer. Moreover, let G be Then there exists a relative difference set with any group of order qd+.+9+2. parameters (7) in H = G @ EA(gd+'). Proof: Let U2,...,Urn be the subgroups of EA(qd+') given by the hyperplanes of this group considered as the (d+l)dimensional vector space over GF(g). (Note that the number of hyperplanes is indeed m1). Also, label the elements of G as 91, ...,gm. Now put D = D,UD2 where Dl={(gi,ui):i=2, ...,m;uiEUi} and D2={(gl,z):zEEA(qd")). It is not difficult t o check that D is the desired relative difference set. 0 We remark that D, is a difference set with parameters (2) in H as constructed by McFarland [5] (see also 11, VI.7.31). References Th. Beth, D. Jungnickel, and H. Lenz, Design Theory, Bibliographisches Institut, MannheimWienZiirich, and Cambridge University Press, 1985. D.A. Drake and D. Jungnickel, "Klingenberg structures and partial designs 11", Pacific J . Math. 77 (1978) 389415. D. Jungnickel, "On automorphism groups of divisible designs", Canadian J. Math. 34 (1982) 257297. H. Lenz and D. Jungnickel, "On a class of symmetric designs", Arch. Math. 33 (1979) 590592.
R.L. McFarland, "A family of difference sets in noncyclic groups", J. Comb. Th. A15 (1973) 110. W.D. Wallis, "Construction of strongly regular graphs using affine designs", Bull. Austral. Math. SOC.4 (1971) 4149.
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Annals of Discrete Mathematics 34 (1987) 301306 0 Elsevier Science Publishers B.V. (NorthHolland)
301
2(25,10,6)Designs Invariant under the Dihedral Group of Order Ten Stojan Kapralov, I v a n Landgev, and Vladimir Tonchev
Institute of Mathematics Sofia 1090 P.O. Box 373
BULGARIA TO A L E X ROSA O N MIS 3I37TE73l BIRTUDAY
In this paper we enumerate all 2(25,10,6) designs which are invariant under the dihedral group of order 10, which splits the point set into 5 orbits of length 5. A few designs with these parameters were previously known; one design is mentioned in the survey paper of Mathon and Rosa [5],and others are constructed by Hanani [2] and Tran Van Trung (111. The concepts and notations in this paper are in accordance with those in [7]. As a first step we construct all possible orbit matrices of a 2(25,10,6) design with ) respect to an automorphism of. order 5 without fixed points or blocks. If M = ( m j j is such a n orbit matrix, the following equalities hold:
c i1 5
mij
= 10
( 1 9 2 ) ;
C mij
= 16
(lsig);
= 24
(lsig);
8
j= 1 8
C rn;j(mijl)
j= 1
8
Cm,jmpj
=
30
(@#P).
j 1
Using an algorithm described in [3], we found 50 integer matrices with nonnegative entries satisfying (1). Every such matrix is extended (if possible) to an incidence matrix of a 2(25,10,6) design, by use of an algorithm from [4]. As a result we found 106 designs. To select t h e nonisomorphic ones we use some invariants introduced in [9]. Let D be a design. Let C; be the number of pairs occurring together with a given point x in exactly j blocks. For every point x, we compute the vector
cx= (Ci, c;.,..., C ? ) .
302
S. Kapralou, I. Landgev and V. Toncheu
The set of 106 designs is divided into 43 classes by these invariants. A more detailed investigation showed that designs belonging to the same class are isomorphic, i.e. 43 is the exact number of nonisomorphic 2(25,10,6) designs invariant under Dl0, dividing the point set into 5 orbits of length 5. The 43 nonisomorphic 2(25,10,6) designs are listed in Table 1, where we use the following notations:
A = U + U 4 ; B=U2+U3; A ' = I + A ; B'=I+B; H = A + B , where U is the circulant with first row 0 1 0 0 0, and I is the 5x5 identity matrix; a zero matrix will be denoted simply by 0. Table 1
I
A' 0 I A'A B'A A'B
A' B' B B 0 A B I BIB
I
A' B' B' B B A 0 A A ' B A'O B'O B
A' 0 I A'A B'A A'B
B' A A' A 0
No. 1 I I
I B'B'A' I B B' A'A'O A I B'
No. 2 I A' A' B' B' A ' A B I A A'B B'B 0 0 I B A B B'A 0 A A'
I I H I I I B'A A'B'B' I B A'
No. 3 I I I . B'B'A' I B B' A'I A A B'I
No. 4 A B I B ' H A B H A ' I A'A A I B B' A' I I I 0 I B A B
I I B I A I H I I I A' B' A'B'B'
No. 5 0 I B B B B'B'B' A B A A H O B B A A A ' B ' B A ' O I A ' A B A'O I B ' A A'A'I 0 A B'A B
0
B A A' A'
I B A B B'
A B €3 A A H A'A'B A B'O 0 A I
No. 7 A'B'B' O A B B'O I I A'B BIB B
0
A A A ' A'
I B A A A'
I

No. 6 B B B B'B'B' A A H O B B B ' A ' B A'O I A B'O I A ' B I 0 A B'B A No. 8
0
A B A A
I A B B B'A'B' B B A H O A B A B ' A ' B B'O I ' A A A ' O I A ' A 'B'I 0 B BIB A
303
2(25,10,6) Designs Invariant
No. 9 A B B B'A'B' A B H O B A A ' B ' A A'O I B B'O I A ' A 0 A A ' A B I
No. 10 0 I B A B B'A'B' A B B A H O B A B A B'A'A A ' O I A ' A A A ' O I B'B B'A'I 0 B BIB A
No. 11 0 I A B A A'A'B' B A A B H O B A B A B'A'A A ' O I B ' B B B'O I B ' B A'B'I 0 A A'B A
No. 12 0 I B B B B ' B ' B ' A B A A H O B B A A'I B'A A ' O B A ' A A I B B'A'O A ' A B ' A 0 I B A'
0 I A B B A BIB A'B'
0
A A A' A'
I B A A A
No. 13 B B B B'B'B' A A H O B B ' I A ' B B'O A B I A A'B'O A ' B 0 I A B'
0
B A A' B'
I B A B A
B A 'I A A
No. 15 I B B A B A A A H A A'I A ' A A'A B I B BIB B ' A 0 0
0 I B A B A A B'I A'B B A ' A B'
0
A A A A
I B A ' 'A ' A
A'B'B' O B B A'O A B'B'O I B A'
No. 17 A B B'A'B' B H O A B A ' B A'O A I A B'B'O B 0 I A A'
B B B A A H I B ' A B A'O A'O B
No. 19 B'B'B' O B B A ' O B I B'A B'A I
0
A B A A
I B B ' ' A 'A
No. 14 A B A'B'B' A H O A B A ' A B'O B I A A'B'O ' B 0 I B B'
No. 16 A B B B ' A ' B ' A B H O B A I A ' B B ' O A B I B B'B'O A ' A 0 I A A'
0 I A B B A B B'I A'B A B ' A A '
0
A A A A
I B A' 'A ' A
No. 18 A B A'A'B' B H O A A A ' A B'O B I B B'B'O A O I B A '
No. 20 B B B B'B'B' A A H O B B I A ' B B'O A A B'O I A ' B B'O A A ' B I
S.Kapralov, 1.Landgev and V. Tonchef.
04
0
B A A A 0
A B A B
I A B B A A B'I A ' B B B ' ' A A ' O
No. 21 B A'B'B' H O B A ' A A'O B O I B ' A A B'B I
I B A B A B A'I B ' A A A ' 'A A ' O
No. 23 B B'A'B' H O A B ' A B'O B O I B'B B A ' A I
I A B A A B B'I A ' B B B ' ' A B'O
A H ' A O A
0
A B A A 0
B B B A
No. 22 I A B B B'A'B' B B A H O A B B ' I A ' B B'O A ' A A A ' O I A ' A ' A B ' O B B ' B I No. 24 I B B A B'A'B' B B A H O A A A ' I B ' B B'O A ' A A A ' O I A'B ' A B'O B A ' A I No. 26
No. 25 A'A'B' O B A A'O A I B'B A ' B I
A A A A
I B B B B B A A H A'A'B'O 'A I A B 'A B I A
B A A' B I
No. 27 B A'B'B' H O A B O I B A A B ' O B' B B'A'O
0 B A B A
I A B B B'A'B' A A A H O B B A'A'A'O I A A ' B I A B B ' O A' ' A B I B B'B'O
B A A ' I A
A A A ' B I
No. 29 B B'A'B' H O B B O I A A B B ' O A' B B'B'O
No. 30 B H I A I B I B ' A ' I H A I B I A A ' B I B A B'B'O 0 I A B B B'A'B' A B B A H O A B
A' I A' B'A'O 0 I A A A B B A ' B '
B' B A A' O
No. 31 I B' I I I A B'B B A'A'B' A' 0 I B' B'B A I
0
B B B A 0
B A A' A' 0
B A B A
I B B' B A
I A A ' ' A 'B
A A A' I A
0
' B ' B ' O B B I A B B ' O A' A'B'O
No. 28
A' I B'B
B' B 0 I A B A ' O A B'A'
No. 32 I A' I B' I B'O A ' A I B A B'A'A' B'B'I B A I B A 0 A'
305
2(25,10,6) Designs Invariant
B B ' I B B ' B B ' H H O B 0 f A
No. 33 I A B I H I B I A I I A B I B B A A A A A A'A'A'
I
H
No. 35 A I I B B B ' H A I H B A'A I B H A B ' I I B H A I I A B I A O B A B A B A
I I I ' H
I I B B H
No. 37 I I I A' B' A' B' B A H I I A A ' I H A A A ' I I H I A B ' B I I B A I I I B ' B
I I B B H
I I B A I H H A B I
No. 39 I A' B' A' B' H I I A A ' A A A ' I I I I B B ' I A B ' I I B
No. 41 I I I I A' B' A' I A B H I I A A B B B ' A I H I B A ' A A H I I H A B I I I B '
I A B A A
I B A A '
I
B B B ' ' B A A
I
B' ' I
I B
B B B ' B ' 0 I B H
No. 34 I I H I A B I I H B B I A I H I B A B I A A A A'A'A' B B O A A A
'
No. 36 A I I B B B ' A B A ' I H A B H A B ' I I H I A A I B I O B A B A B
I I A B H
I I I ' H
No. 38 I A' A' B' H I I A B A B A ' I I I A A ' B B ' I I
B' ' I I B
No. 40 I A' A' H I I A A B A ' I B ' I B I A
B' B I I A '
B' ' I B I
No. 42 I I I A' A' B B H I I A B A ' A I H H A I B ' I I A B A I A
B' B I I '
B' ' I B I
I I B B I H H A A I
I I I B A I B H H A I I A B H
H I B A A
I B H A I
No. 43 A' A' B' B'
' I B I A I B H
I I H I H I H I I I I I
It is of interest to find out which of the above designs are embeddable into symmetric 2(41,16,6) designs. The 2(25,10,6) design given in [2] is nonembeddable, since it contains a pair of blocks intersecting in 7 points. Recently, Tran Van Trung Ill] constructed a nonembeddable 2(25,10,6) design invariant under Dl0 (isomorphic to number 20 in Table 1).
S. Kapralov, I. hndgev and V. Tonchev
306
We checked the embeddability of our 43 quasiresidual 2(25,10,6) designs with the help of an algorithm described in [6]. It turns out that the only embeddable designs are numbers 33 and 28. The first one extends t o the symmetric 2(41,16,6) design found by Bridges, Hall, and Hayden [I], and Tran Van Trung [ll];the second one extends to its dual 18). The remaining designs are quasiresidual but not residual.
References (1) W.G. Bridges, M. Hall Jr., and J.L. Hayden, “Codes and designs”, J . Comb. Theory A31 (1981) 155174. H. Hanani, “Balanced incomplete block designs and related designs”, Discrete [2] Math. 11 (1975) 255369. (31 S. Kapralov, “Algorithms for generating of orbit matrices”, Proceedings o f the Conference; 100th anniversary of the birth of academician L. Tchakalov, Samecov, 1986, proceedings t o appear. [4] I. Landgev, “On symmetric 2(41,16,6) designs invariant under the Frobenius group of order lo”, Compt. Rend. de I’Acad. Bulg. des Sci., to appear. [5] R. Mathon and A. Rosa, “Tables of parameters of BIBDs with 9241 including existence, enumeration, and resolvability results”, Ann. Disc. Math. 26 (1985) 275308. 161 V. Tonchev, “Quasiresidual designs, codes, and graphs”, Collog. Math. SOC. Jan& Bolyai 37 (1981) 685695. [7] V. Tonchev, Combinatori’al Configurations: Designs, Codes, Graphs Nauka i Iskustvo, Sofia, 1984. [8] V. Tonchev, “The isomorphism of certain symmetric block designs”, Compt. Rend. de 1’Acad. Bulg. des Sci. 2 (1985) 161164. [9] V. Tonchev, “Hadamard matrices of order 28 with automorphisms of order 7”, J . Comb. Theory A40 (1985) 6281.
[lo] Tran
Van Trung, “The existence of symmetric block designs with parameters (41,16,6) and (66,26,10)”, J . Comb. Theory A33 (1982) 201204.
[11] Tran Van Trung, “Nonembeddable quasiresidual designs with k
n I
6bo
+ 3bI (2‘FZ], and bo + b , = 50  b4 =
We easily conclude that z = 4, and we are done. 0
Theorem 2.2 Let p be a prime dividing the order of the automorphism group G of an S(2,4,25). Then, p = 2,3,5 or 7. Further, if a E G has order p and: (i) p = 3, then a fixes 1 or 4 points; (ii) p = 5, then a fixes no points; (iii)p = 7, then a fixes 4 points. Proof: Let a be an automorphism of order p of an S(2,4,25). If p E{11,13,17,19,23} then cz fixes z points where z €{2,3,6,8,12,14}. Our result then follows from the previous lemma n Theorem 2.3 If 3k IG then k 5 2 and if H is a subgroup of G , with = 9, then H is elementary abelian, i.e. H is isomorphic t o Z3XZ3.
I
I
I
Proof: Assume that k > 2 so there is a subgroup H of order 9. If there were an automorphism of order 9 of the S(2,4,25) then its cube would fix 7 or 16 points, contrary to Theorem 2 . 2 , so H cannot be cyclic. Using the CauchyFrobeniusBurnside
Steiner Systems S(2,4,25) Invariant
309
lemma, one may compute the number m of orbits of H on X by:
where ~ ( hdenotes ) the number of points fixed by h . Since the set of points fixed by an automorphism of order 3 forms a subsystem S(2,4,r), any such automorphism can fix either 1 or 4 points of X . Moreover, the nonidentity elements of H split into four subgroups of order 3 and any pair of elements of order 3 belonging to t h e same subgroup fix the same number of points. Thus, equation (2) can be rewritten as follows: 1 (3) 9 where a b = 8 and a,b are even. Under these conditions the only solution to (3) is a = 6 = 4. Hence, H partitions t h e point set X into 5 orbits, and t h e orbit lengths are clearly 1,3,3,Q,9. Now, H cannot be contained in a subgroup I( < G of order 27, for otherwise t h e stabilizer in K of a point from an Horbit of length 3 would be a subgroup of order 9 fixing at, least 3 points, contradicting t h e above conclusion. This completes our proof 0
m = .(25+a.l+b.4)
+
3. The S(2,4,25)’s invariant under G = Z3XZ3
Having established the structure of a group of automorphisms of order 9, as a permutation group, we proceed to show t h a t there are precisely five isomorphism classes of Steiner systems S(2,4,25) admitting such a group H as a group of automorphisms. From the orbit structure of H on the 25 points, we see t h a t it is represented regularly on t h e two pointorbits of length 9, has two pointorbits of length 3, and fixes a point. The only question is whether the stabilizer in H of a point out of one orbit of length 3 is the same subgroup as the stabilizer of a point out of the other orbit of length 3. We quickly see, however, t h a t this cannot happen because otherwise there would exist a n element of order 3 fixing 7 points , contrary to Theorem 2.2. Without loss of generality we choose H = 1 then obviously it can not be selected, so y e delete all columns of A which have entries greater than 1 to arrive at a submatrix A of size 36 X 739. We can easily see t h a t an S(2,4,25) must be comprised of 2 orbits of length 1, and at least 4 orbits of length 3 on 4sets. A n analysis of the orbits of length 3 shows, however, t h a t no more t h a n 4 can be chosen. Therefore, a design consists of exactly 2
E.S. Kramer, S.S. Magliveras and V.D. Tonchev
310
orbits of length 1, 4 orbits of length 3 and 4 orbits of length 9 of 4sets. There are exactly 2 Horbits of length 1 on 4sets, hence both must be chosen for an S(2,4,25). There are exactly 20 orbits of length 3 on 4subsets and exactly 4 must be chosen in a design. Using this information, we can further restrict the scope of the search for solutions. With the above restrictions, and using a new, recursive algorithm we find all binary solutions t o (4)
A X = l
with x of dimension 739. There are a total of 1944 solutions t o (4) which form 6 orbits, each of length 324, under the action of N = Ns,(H), the normalizer in S,, of H . Here the order of N is g3*4.Four of the Norbits are distinguished among themselves and from the union of the remaining two by means of the blockgraph invariant. The remaining two orbits have the same blockgraph invariant and fuse under 7r = (1)(2
3)(4 7115 9)(6 8)(10 lS)(ll 17)(12 16)(13 15)(14)(19)(20 21)(22)(23 24)(25).
Thus we have established:
Theorem 3.1 There are exactly 5 isomorphism classes of S(2,4,25)’s admitting a group of automorphisms of order 9. The orders of G are 504, 63, 9, 9 and 9 respectively. As a consequence, the only remaining S(2,4,25)’s that can exist are those with automorphism group G , of order IG I = 2’3j, j 5 1. 4.
The 5(2,4,25)’s
In this section we present the S(2,4,25)’s known t o us. For each design we give generators of the automorphism group G , representatives for each blockorbit, the orbit sizes, and the blockgraph invariants. The permutations a and /3 appearing below have been given in section 3.
Design 1.
H
5G
=
I
= , (G = 78.9 = 504, where:
rl = (1 2 24 12 8 18 19 9 7 23 17 4 14 21 5 6 22 13 3 10 20)(11 16 15)(25) I
I ORBSIZE 123191 42 8 15925
ORBREP.
I
BLOCKGRAPH INVARIANT
I n,, n5 7t6
n7
n8 ng
7t10
rill
I 1 0 30 204 411 372 158 0 0 0 63 126 462 378 147 0
7112
0 0
311
Steiner Systems S(2,4,25)Invariant
H = 5 G = , IG I = 7.9 = 63, where:
Design 2. 72 =
(1 21 13 18 4 15 2 5 19 17 10 8 16 6 9 20 12 14 3 11 7)(22 23 24)(25)
BLOCKGRAPH INVARIANT
Design 3.
0RB.REP. 1 2 319 1 4 722 19 20 21 25 22 23 24 25
ORBSIZE 21 21 7 1
G = ,
IG I = 9,
n4 n5 n6 n7 0 30 208 396 0 33 202 399 3 27 207 393 0 63 126 462
n8 414 411 414 378
ng nlo n l l n12
110 18 0 116 15 0 120 12 0 147 0 0
0 0 0 0
1
BLOCKGRAPH INVARIANT
0RB.REP.
1 15 21 24 1 6 10 11 1 12 13 23 1 14 16 20 10 13 16 25 1 5 925 1 4 722 1 2 319 19 20 21 25 22 23 24 25
Design 4.
G
=
,
ORBSIZE 9 9 9 9 3 3 3 3 1 1
n 4 725 n 6
0 0 0 0 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0
54 45 45 44 49 48 46 36 63 48
%9 7E10
n7
152 429 179 399 170 429 171 427 158 428 165 422 174 408 196 405 141 399 153 453
411 429 393 403 412 414 421 399 477 390
121 112 130 117 118 114 117 131 78 120
rill 9 12 9 14 9 13 9 9 18 9
n12
0 0 0 0 1 0 1 0 0 3
IG I = 9,
BLOCKGRAPH INVARIANT
n4 n 5 n6
1 11 13 21 1 16 18 23 1 6 10 14 1 15 20 24 10 13 16 25 1 2 319 1 4 722 1 5 925 19 20 21 25 22 23 24 25
9 9 9 9 3 3 3 3 1 1
0 0 0 0 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0
41 39 39 39 31 54 43 36 63 39
n7
n8
n9
185 191 187 179 203 142
409 405 415 432 401 462 405 419 399 372
403 399 393 390 403 372 406 381 477 453
183 192 141 198
n10 lZll n12
130 8 136 6 134 8 121 15 127 9 140 6 135 3 141 7 78 18 102 9
0 0 0 0
1 0 1 0 0 3
E.S. Kramer, S.S. Magliveras and V.D. Tonchev
312
Design 5.
G = , IG I = 9, BLOCKGRAPH INVARIAVT
n4 n5
1 1 1 1 1 10 10 1 22 19
Design 6.
9 9 9 9 3 3 3 3 1 1
14 21 24 13 17 20 6 11 22 4 10 15 2 319 13 16 22 11 12 25 5 925 23 24 25 20 21 25
0 0 0 0 1 1 0 0 0 0
0 0 0 0 0 0 0 0 0 0
n6
n7
59 58 54 51 61 54 76 48 63 48
153 400 144 432 162 396 162 414 130 435 150 426 111 420 168 417 132 444 153 453
n8
n9
442 410 450 426 433 411 463 411 396 390
n10 lZll n12
117 126 102 111 106 131 99 123 141 120
5 6 12 12 9 3 6 9 0 9
0 0 0 0 1 0 1 0 0 3
G = ,IG I = 2.3.25 = 150, where:
r3= (1 23)(2 24)(3 25)(4 21)(5 22)(6 17)(7 18)(8 19)(9 20)(10 16)(11)(12)(13)(14)(15) r4 = (1 25 5)(2 19 10)(3 13 15)(4 7 20)(6 21 24)(8 14 9)(11 22 18)(12 16 23)(17) BLOCKGRAPH INVARIANT
0RB.REP. 1 2 625 1 3 11 19
Design 7.
G
ORBSIZE 25 25
n8 ng nlo nll n12 0 33 198 423 363 156 3 0 0 45 171 435 369 156 0 0
n4 n5 n6 n,
0 0
(G I = 21, where:
= ,
rs= (1 2 3 4 5 6 7)(8 9
10 11 1 2 13 14)(15 16 17 18 19 20 21)(22)(23)(24)(25)
rs = (1 2 4)(3 6 5)(7)(8 9 11)(10 13 12)(14)(15 16 18)(17 20 19)(21)(22 23 24)(25) BLOCKGRAPH INVARIANT
0RB.REP. 1 9 17 22 1 2 414 1 8 15 25 1 18 20 21 8 9 11 21 22 23 24 25
ORELSIZE
21 7 7 7 7 1
nq n5 n6 n7 n8
0 0 0 0 0 0
0 0 0 0 0 0
43 46 38 40 42 35
174 176 198 178 175 203
427 402 393 432 438 399
n9 nlo n l l n12 394 425 403 383 366 385
129 121 141 137 152 154
8 3 3 3 3 0
1 3 0 3 0 0
Steiner Systems S(2,4,25) Invariant
Design 8.
G = ,
313
IG I = 6, where:
y7 = (1 20 16)(2 10 3)(5 17 18)(4 19 22)(6 9 25)(11 15 14)(7 21 12)(8 13 24)(23) 78
= (1)(2 18)(3 5)(4 22)(6 15)(7)(8 24)(9 11)(10 17)(12 21)(13)(14 25)(16 20)(19)(23)
BLOCKGRAPH INVARIANT
0RB.REP. 1 2 425 1 3 811 2 3 14 19 2 6 912 1 6 715 1 10 13 17 1 12 19 21 2 7 18 23 2 17 21 24 4 6 814 4 12 13 24 8 9 15 23 1 16 20 23 4 19 22 23
ORBSIZE 6 6 6 6 3 3 3 3 3 3 3 3 1 1
n4 n5 n s n7 n8 n9 nl0 n I 1nI 2 0 0 45 175 415 405 128 8 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 39 195 1 40 188 0 43 188 0 52 167 0 49 161 0 51 156 0 46 174 0 36 199 0 63 136 0 46 177 2 60 137 0 33 192 0 57 147
384 400 392 402 433 431 416 399 421 407 420 441 435
423 407 416 422 395 409 398 399 440 407 438 345 393
129 136 133 130 130 120 138 137 107 135 112 162 144
5 4 4 3 8 7 4 6 8 4 7 3 0
0 0 0 0 0 2 0 0 1 0 0 0 0
Added in Proof: R. Mathon, E. Kramer, and S. Magliveras have found eight new S(2,4,25)’s admitting an automorphism of order 3. They also establish that there are no new S(2,4,25)’s with an automorphism of order 2. In view of the above, all S(2,4,25)’s with nontrivial automorphism group are now known.
References T. Beth, D. Jungnickel, H. Lenz, Design Theory, Bibliographisches Institut, MannheimWienZurich, and Cambridge University Press, Cambridge, 1985. A. E. Brouwer , (personal communication) F. C. Bussemaker, R. Mathon, J.J. Seidel, “Tables of twegraphs”, Combinatom’cs and Graph Theory, Proc. Conf. Calcutta ISSO, Lecture Notes Math. 885, Springer 1981, pp. 70  112. H. Gropp (personal communication). M. Hall, Combinatorial Theory, 2nd Edition, Wiley, N.Y. 1986. E.S. Kramer, D.M. Mesner, “tdesigns on hypergraphs”, Discrete Math. 15 (1976) 263  296.
E.S. Kramer, D. Leavitt, S.S. Magliveras, “Construction Procedures for tdesigns and the existence of New Simple 6designs”, Annals of Discrete Math. 26 (1985) 247  274.
314
E.S. Kramer, S.S.Magliveras and V.D. Tonchev
[8] R. Mathon, and A. Rosa, “Tables of Parameters of BIBDs with r 5 41 including Existence, Enumeration, and Resolvability Results”, Annals of Discrete Math. 26 (1985) 275  308. [9] L.P. Petrenjuk and A.J. Petrenjuk, “List of 8 S(2,4,25)’si’, (preprint). (lo] V.D. Tonchev, “TWOnew Steiner systems S(2,4,25)”, Compt. Rend. Acad. Bdg. S C ~39 . (1986) 47  48.
Annals of Discrete Mathematics 34 (1987) 315318 0 Elsevier Science Publishers B.V. (NorthHolland)
315
Simple 5(28,6,X) Designs from PSL2(27) Donald L. Kreher and Stanisyaw P. Radziszowski School of Computer Science and Technology Rochester Institute of Technology Rochester, NY 14623 U.S.A. TO A L C X ROSA O N MIS 3IITTIETU BIRTUDAY
ABSTRACT Simple 5(28,6,X) designs with PSL2(27) as an automorphism group are constructed for each X, 2 5 X 5 2 1 . 1. Introduction
A tdesign, or t(v,k,X) design is a pair ( X , B ) with a vset X of points and a family B of ksubsets of X called blocks such t h a t any t points are contained in exactly blocks. A t(v,k,X) design ( X , B ) is simple if no block in B is repeated and is said to have G g y m ( X ) as an automdrphism group if whenever K is a block Ke={xa:xU(J is also a block for all a E G . In this paper we show t h a t for each A, 2 5 x 1 2 1 , a 5(28,6,X) design exists with G=PSL2(27) as an automorphism group, leaving open only the existence of a 5(28,6,1) design with some group other than PSL2(27). Incidently, Denniston showed that a 5(28,7,1) design does indeed exist with PSL2(27), see [2]. 2. Preliminaries As with our construction of two disjoint nonisomorphic simple 6(14,7,4) designs [6] a s well as with the only other known examples of 6designs with small X [4,9], we start with the following observation of Kramer and Mesner [3]: A t(v,k,X) design exists with G 5 Sym(X) as an automorphism group if and only if there is a (0,l)solution vector U to the matrix equation
AtkU = XJ, where a. The rows of Atk are indexed by the Gorbits of tsubsets of X; b. The columns of Atk are indexed by the G m b i t s of ksubsets of X c.
Atk[A,rj = I{
[email protected]:K>T,)I
d.
J=[l,l,l,
...,lIT.
where TOEAis any representative;
(1)
316
D.L.Kreher and S.P.Radziszowski
If we choose t.he group G to be PSL2(27) acting on the projective line X = G F ( 3 3 ) u 00 then G has order 9828 and an isomorphic copy is generated by the permutations a,,6, and 7 on {0,1,2,...27) given below:
= (0,26,27)( 1,2,10)(3,6,4)(5,15,19)(7,11,21)(8,17,14)(9,18,12)(13)(16,24,25)(20,23,22);
p = (0,13)(1,12)(2,11)(3,10)(4,9)(5,8)(6,7)(14,25)(15,24)(16,23)(17,22)(18,21)(19,20)(26,27); 7 = (0,23,22,10,11 ,I ,27,26,25,15,16,4,3)(2,9,13,17,24,19,8,20,14,12,6,18,7)(5)(21).
In this action G has exactly 10 orbits of 5subsets and exactly 54 orbits of 6subsets, see figure 1 and 2. Hence, the matrix belonging to G has 10 rows and 54 columns and is displayed in figure 3.
01246 01236 012411
012411 01248 01238 01258
Figure 1. Orbit representatives of 5subsets
Figure 2. Orbit representatives of 6subsets
Simple 5(28,6,X)Designs
317
111100110200120000201000000101111000001001000010000001 002121101110000000000011000100000001001010101011001101
100021110011200000111001100110100000010000010001100010 000000000100011020110112111020000000000000200100111100 011001001011100010020100100000000000010111011000021200
102000000100000012000001000000011101100100112121000100 200000000011010210001000010110010101100002120000010010
000000010100011000110010011001020002000000000102110121
000000001020200010110001001210010120101000001011000010 100120000000011000000100000011101001010112111010010010
Figure 3. The
matrix belonging to PSLz(27).
The task of solving the resulting system (1) of 10 linear Diophantine equations in 54 unknowns for a (0,l)solution vector U was accomplished by our basis reduction algorithm presented at the 17th Southeastern International Conference on Combinatorics Graph Theory and Computing, [5]. We remark t h a t this algorithm is based in part on the L3 algorithm of A.K. Lenstra, H.W. Lenstra and L. Lov6sz [8], and was inspired by the work of J.C. Lagarias and A.M. Odlyzko [7]. 3. Results In figure 4 for each X, 2 5 x 5 1 1 we give a list of orbits of 6subsets by number whose union forms a 5(28,6,X) design. Noting t h a t the complement of a 5(28,6,X) design is a 5(28,6,23X) design, orbit number lists for X 2 11 were not given, and in the interest of brevity we give only one solution for each parameter situation. It is easy to see that these solutions satisfy equation (1). Furthermore, the industrious reader can check using a n algorithm similar to the one found in [l] that these are indeed 5(28,6,X) designs.
X
1
2 3 4 5 6 7 8 9 10 11
I 12 15 18 28 3051
Orbit numbers 7 8 9 22 26 27 44 47 2 21 22 35 39 40 43 44 48 49 54 2 8 22 28 39 43 44 47 49 52 53 1 10 11 13 25 30 36 39 43 45 49 50 1 4 7 8 11 12 15 18 22 23 25 26 30 34 35 37 38 39 40 46 51 54 1 2 8 9 10 11 14 24 29 31 36 37 38 45 48 50 51 3 8 9 13 14 19 22 24 28 36 42 43 44 45 49 52 53 1 2 3 5 6 13 16 17 19 22 24 29 30 32 36 39 47 50 52 53 2 3 5 9 11 13 14 17 19 24 25 30 32 36 42 43 44 47 49 52 53
Figure 4. Designs from PSLz(27).
D.L. Kreher and S.P. Radziszowski
318
Finally, we state the following theorem. Theorem 2.1 PSL2(27) is an automorphism group of a simple 5(28,6,x) design for all A, 2 9 5 2 1 and cannot be an automorphism group of a 5(28,6,x) design when X=1.
Proof: The first part of the theorem is established by figure 4. To show that G=PSL2(27) cannot be the automorphism of a 5(28,6,1) design we consider the Gorbit lengths of 6subsets. These are: 3276, 4914 and 9828. Thus if a 5(28,6,1) design is t o be constructed as a union of Gorbits then the number of blocks, 16380, must be written as a sum of these numbers. Congruence modulo 3 and the fact that there are only 4 orbits of length 3276, namely numbers 2, 16, 18, and 35, says that exactly two of them must be used. But, this is impossible since the row sum of any two of the corresponding columns of the matrix contains an entry greater than 1, contradicting X=1. Acknowledgements Both authors’ research was supported under NSF Grant number DCR8606378. References 111 R.H.F. Denniston, “Some New 5designs”, Bull. London Math. SOC.8 (1976) 263267.
121 131 141
151
161 171 181 191
R.H.F. Denniston, “On the problem of the Higher Values of t”, Annals of Discrete Math. 7 (1980) 6570. E.S. Kramer and D.M. Mesner, “tDesigns on Hypergraphs”, Discrete Mathematics 15 (1976) 263296. E.S. Kramer, D.W. Leavitt and S.S. Magliveras, “Construction Procedures for tDesigns and the existence of New Simple &Designs”, Annals of Discrete Mathematics 26 (1985) 247274. D.L. Kreher and S.P. Radziszowski, “Finding Simple tDesigns by Basis Reduction”, Proceedings of the 17th Southeastern Conference on Combinatorics, Graph Theory and Computing, Congressus Numerantium 55 (1986) 235244. D.L. Kreher and S.P. Radziszowski, “The Existence of Simple 6(14,7,4) designs”, Journal of Combinatorial Theory (A)43 (1986) 237243. J.C. Lagarias and A.M. Odlyzko, “Solving LowDensity Subset Sum Problem”, Journal of the ACM 32 (1985) 229246. A.K. Lenstra, H.W. Lenstra and L. Lov&z, “Factoring Polynomials with Rational Coefficients”, Mathematische Annalen 261 (1982) 515534. S.S. Magliveras and D.W. Leavitt, Simple 6(33,8,36) Designs from PT‘L2(32), in Computational Group Theory, Proceedings of the London Mathematical Society Symposium on Computational Group theory, Academic Press (1984) 337352.
Annals of Discrete Mathematics 34 (1987)319338 0 Elsevier Science Publishers B.V. (NorthHolland)
The existence of partitioned balanced tournament designs of side 4n + 3 E.R. Lamken School of Mathematics Georgia Institute of Technology Atlanta, Georgia U.S.A.
S.A. Vanstone Department of Combinatorics and Optimization University of Waterloo Waterloo, Ontario, N2L 3G1 CANADA TO ALCX ROSA O M MIS 3I77IC7U BIRTUDAY
ABSTRACT A balanced tournament design, BTD(n), defined on a 2nset V is 2n an arrangement of the ( ) distinct unordered pairs of the elements of V into an n X 2n  1 array such t h a t (1) every element of V is contained in precisely one cell of each column and ( 2 ) every element of V is contained in at most two cells of each row. If we can partition the columns of a BTD(n) into three sets C,,C2,C3of sizes 1, n1, n1 respectively so t h a t the columns in C , u C , form an H(n,2n) and the columns in C , U C3 form an H(n,2n), then the BTD(n) is called partitionable. W e denote a partitioned balanced tournament design of side n by PBTD(n). In this paper, we prove the existence of PBTD(n) for n ~3 (mod 4), n 2 7 with three possible exceptions.
319
E.R. h m k e n and S.A. Vanstone
320
1. Introduction. A balanced tournament design, BTD(n),defined on a 2nset V is an arrangement 2n of the ( ) distinct unordered pairs of the elements of V into an n X 2n  1 array such that ( 1 ) every element of V is contained in precisely one cell of each column and ( 2 ) every element of V is contained in at most two cells of each row.
i of a BTD(n) is called a deficient element of row i. The two deficient elements of row i are referred to as the deficient pair of row i. The existence of BTD(n)s was established in [ l l ] . (A simpler proof of this result appears in [7].)
An element which is contained only once in row
Theorem 1.1. FOTn a positive integer, n # 2, there exists a BTD(n). For the tournament scheduling aspects of these designs the reader is referred to [Ill. Let V be a set of 2n elements. A Howell design of side s and order 2n, or more briefly an H(s,2n),is an s X s array in which each cell is either empty or contains an unordered pair of elements from V such that (1) each row and each column is Latin (that is, every element of V is in precisely one cell of each row and column) and
( 2 ) every unordered pair of elements of V is in at most one cell of the array. It follows immediately from the definition of an H(s,2n)that n 5 s 5 2n1. If we can partition the columns of a BTD(n) defined on V into three sets C,,C,, C3 of sizes 1, n1, n1 respectively so that the columns in C , U C , form an H(n,2n) and the columns in C, U C, form an H(n,2n), then the BTD(n) is called partitionable. We denote the design by PBTD(n). An example of a PBTD(5) is displayed in Figure 1. One motivation for studying partitioned designs is for the tournament scheduling problem itself. The reader is referred to [Q] for details. Applications t o other areas of design theory are described in the next paragraph. Partitioned balanced tournament designs are related to Room squares. A Room 1, RS(2n+1), is an H(2n+1,2n+2). Let R be a RS(2n+1). Since square of side 2n each row or column of R contains n empty cells, a t X t subarray of empty cells in R must have t 5 n . If R contains an n X n subarray of empty cells, we say R contains a maximum empty subarray and denote R by MESRS(2n+1). The connection between MESRSs and PBTDs was provided by D.R. Stinson in [13].
+
Tournament Designs of Side 4n+3
321
Cl Figure 1 A PBTD(5)
Theorem 1.2. There exists a MESRS(2n+1) i f and only i f there exists a PBTD ( n 1>. There is an extensive literature o n Room squares and it is known that Room 1 exist if and only if 2n 1 # 3 or 5 (Mullin a n d Wallis [lo]).In squares of side 2n (131, D.R. Stinson established the nonexistence of a MESRS(7) and the existence of M E S R S ( m ) for m = 9 and m = 11. He showed t h a t two recursive constructions for Room squares could be used to provide infinite classes of MESRSs and conjectured t h a t MESRS(2n+l)s exist for all n 2 4. New constructions for balanced tournament designs have led us to investigate the existence question for MESRS(2n+l)s by considering the existence of PBTD(n+l)s. We can now show t h a t PBTD(n) exist for n a positive integer, n 2 5, with 12 possible exceptions. I n [8],we showed t h a t PBTD(n) exist for n = 1 (mod 4), n 25,with the
+
+
+
possible exception of n = 9. In this paper, we will use this result and a frame construction to construct PBTDCn) for n ~3 (mod 4). In [9], we use the existence of PBTD(n) for n = 1 (mod 2) and a new recursive construction to construct PBTD(n) for n = 0 (mod 2). In order to describe our constructions, we will need several definitions and preliminary results. These are contained in the next section. W e include a summary of previous results on PBTDs in section 2. In section 3, we describe some new frame constructions. In order to use many of the constructions described in sections 2 and 3, we need the existence of PBTD(n) for some small values of n . These are provided in section 4. The frame constructions are used in section 5 to find PBTD(n) for n = 3 (mod 4) with 3 possible exceptions. 2. Preliminary definitions and results.
In order to describe our constructions, we will need several definitions and results. These are collected in this section. In (81 we described a direct construction for PBTD(n+l)s. This construction uses a starter for a Howell design of side 2n and order 2n 2, H(2n,2n+2). (For definitions and results on starters and adders for H(2n,2n+2), see [ I ] . )
+
Theorem 2.1. [8]Let n
= 1 (mod 2).
Let S =
. ,{znI,V,I},{CYlzl},{~~2}} fJe a starter for an H(2n,2n+2) d e f i n e d o n Z 2 , U {cr,oo}. Let S’ = S  {zi,yi} for some i where Iziy; I 1 (mod 2), 1 5 i 5 n1. Suppose there is a n adder A = ( a l , a z..., , a n ) o f n distinct elements o f {{z~,Y1},{3c2,Y2},
* *
E.R. Lamken and S.A. Vanstone
322
z,
such that
(i)
ai E {0,2,4
+
,...,2n2)
f o r i = 1,2 ,...,12 and
I
(ii) S' A = z,,  {ul,wl} where lulwl = n . Then there is a PBTD(n+l). We use this result t o find several small PBTDs of even side.
Lemma 2.2. [8,9] There emkt PBTD(n+l) for n E {5,7,9,11,13,15,17,19,21}. The main construction for finding PBTD(n)where n 1 (mod 4) is a frame construction. We will also use frames t o construct PBTD(n)for n = 3 (mod 4). Let V be a set of w elements. Let G1,G2,...,G, be a partition of V into m sets. A {G1,G2,...,G,)frame F with block size k , index A and latinicity p is a square array of side 21 which satisfies the properties listed below. We index the rows and columns of F by the elements of V. (1) Each cell is either empty or contains a ksubset of V. (2)
Let Fi be the subsquare of F indexed by the elements of Gi. Fi is empty for i = 1,2 ,...,m .
(3) Let j E Gi. Row j of F contains each element of V  Gi p times and column j of F contains each element of V  Gi p times. (4)
The collection of blocks obtained from the nonempty cells of F is a GDD(v;k;Gl,G2,...,G,;O,X). (See [17] for the notation for group divisible designs
(GDD1.1
If there is a {GI,G2,...,Gm)frame H with block size k , index A and latinicity p such that (1) Hi = Fi for
(2)
i
= 1,2,
...,m and
H can be written in the empty cells of F 
m u Fi, i1
then H is called a complement of F and denoted F. If a complement of F exists, we call F a complementary {G1,G2,...,G,)frame. A complementary {G1,G2,...,G,)frame F is said to be skew if at most one of the cells (i,j)and ( j , i ) (i # j) is nonempty.
I
We will use the following notation for frames. If IGi = h for i = 1,2, ...,m , we call F a (p,X;k,m,h)frame. The type of a {G1,G2,...,Gm&frame is the multiset { IG, 1, IG2 1, . . . ,IG, We will say that a frame has type ty't;' * * tp if there are ui Gj's of cardinality ti, 1 5 i 5 k. For notational convenience, if p = = 1 and k = 2, we will denote a frame simply by its type.
I}.
The constructions which use frames also use sets of mutually orthogonal parti...,Sm}be a partition of a tioned incomplete Latin squares (OPILS). Let P = {S1,S2, set S ( m 2 2). A partitioned incomplete Latin square, having partition P , is an I SI X I S array L , indexed by the elements of S , satisfying the following properties.
I
(1) A cell of L either contains an element of S or is empty.
Tournament Designs of Side 4n+3
323
( 2 ) The subarrays indexed by Si X Si are empty for 1 5 i 5 m. (3) Let j ESi. Row j of L contains each element of S  Si precisely once and column j of L contains each element of S  Si precisely once. The type of L is the multiset { ISl I, IS2I,...,ISmI}. If there are ui Sj’s of cardinality t i , 1 5 i 5 k , we say L has type t ~ ’ t *~* ** t;”. Suppose L and M are a pair of partitioned incomplete Latin squares with partition P . L and M are called orthogonal if the array formed by the superposition of L and M , L o M , contains every ordered pair in S X S 
um (Si X S i ) precisely once.
A
i1
set of n partitioned incomplete Latin squares with partition P is called a set of n mutually orthogonal partitioned Latin squares of type {IS, 1, IS, I,...,ISmI} if each pair of distinct squares is orthogonal. The following frame construction for PBTDs was proved in 181.
Theorem 2.3. If there exist a complementary {G1,G2,...,Gm}frame ( m 2 2), a pair o f orthogonal partitioned incomplete Latin squares o f type { IG, 1, IG, 1, ...,IG, I} and m
PBTD( IGi 1+1) for i = 1,2,...,m , then there i s a P B T D ( ( C IGi 1)+1). i1
This construction combined with a direct product gives the next result.
Theorem 2.4. 181 I f there exist a complementary {GI,G2,...,G,)frame ( m 2 2 ) , a pair o f orthogonal partitioned incomplete Latin squares o f type { IG, 1, IG, 1, ...,IG, I}, a pair o f orthogonal Latin sqztares o f side n and PBTD(n IG; 1+1) for i = 1,2,...,m , m
then there i s a PBTD(nC IG, 1+1). i‘1
W e used Theorem 2.3 and the existence of complementary 4” frames and OPILs of type 4” for n 2 4, n # 6, t o prove the following.
Theorem 2.5. [8]Let n be a positive integer. There exists a PBTD(4n+l) except possibly for n = 2. There are several other applications of the frame construction. One example is: Theorem 2.6. (81 Let n = 1 (mod 2), n 2 7. If there is a PBTD(m+l) ( m # 6 ) , then there i s a PBTD(mn+l). Other results using Theorems 2.3 and 2.4 can be found in [ 8 ] . Finally, we will also use two product constructions from [13]. Theorem 2.7. Suppose there exist PBTD(m) and PBTD(n) and a pair o f orthogonal Latin squares of order m . Then there exists a PBTD(mn). Theorem 2.8. Suppose there exists a PBTD(2m)and a pair of orthogonal Latin squares of side m. Then there exists a PBTD(6m).
E.R. Lamken and S.A. Vanstone
324 3. Frame Constructions.
In this section, we describe some new frame constructions.
Theorem 3.1. Let n = 0 (mod 2). I f there exists a complementary 2" f r a m e and a pair of orthogonal partitioned incomplete L a t i n squares o f tvpe 2", t h e n there i s a PBTD ( 2 n 1).
+

= 1,2,...,n}. Let
Proof. Let V = {ui,%b = 1,2,...,n } and let V = {Ci,C; Gi = {ui,vi} and let Gi = {Ci,C;}for i = 1,2 ,...,n.
Let F , be a complementary {G1,G2,...,G,) frame definedonV. F , isof type 2". Let F , be a complement of F , defined on V. F, will be a { G 1 , G P , . . ,G,&frame of type 2". F will denote the array of pairs formed by the superposition of Fl and F,, F = F , o F2. Let L , and L , be a pair of orthogonal partitioned incomplete Latin squares of type Suppose L, is defined on V with  partition {{u1,uz},{u1,u2}, . . . ,{~nl,un},{w,~,un}} and suppose L , is defined on V with partition {{C1,C2},{C1,C2}, . . . ,{Cnl,Cn},{~nl,Cn}}. L will be the array of pairs formed by the superposition of Ll and L,, L = Ll o L,.
2".
Fl, LEI Frl,r [I l El,
Define the following arrays.
4=
Ci =
m;
and
Qu;
Ri = u; ui
ui
vi
for i = 1,2 ,...,n.
For i = 1,3 ,...,n1 (i = 1 mod 2),
Ui+l Ui+l

Bi
i+l
Cli =
R'i =
For i = 2,4,...,n (i
and
ui Ci
Qui+l
~i ~ i + 1 ~i ui+l
.
0 mod 2),
Bi =
and
=
R'; We construct two 2 n
. I[
vi1 vi1
=
+ 1 x 2 n + 1 arrays using these arrays and L and F.
Tournament Designs of Side 4n+3
325
2n
HI
=
I
1
P
1
I
IB ,
H', =
B 2 :
R', R',
uvu
R',
?I
'n
uoo
Every element in V {a,.} occuls once in each row and once in each column of H , and HI,. Let C denote the 2n x 1 array [C,C,* . . CnlT. We permute the rows of H', so that the last column of.the resulting array H 2 is [C,C, . . * C,{a,co}]*. We can now write H , and H , in the following forms.
H , and H , are both H(2n+1,4n+2)s defined on V U
7U
{a,co}.
The array
is a PBTD(2n+1) defined gn V U V U {a,co}.It is straightforward t o vel..; that a , . } occurs precisely once in B . Every element in everydistinct pair in V U V u { V u V U {a,co}occurs at most twice in each row of B and precisely once in each column of B . The partitioning of B is given by H , and H,. 0
Corollary 3.2. There exists a PBTD(13).
326
E.R. Lumken and S.A. Vanstone
Proof. There exists a complementary 26 frame [8] and a pair of orthogonal partitioned incomplete Latin squares of type 26 151. This corollary provides a proof for the note added to the end of [8]. A PBTD(n) is said to be missing a PBTD(m) if it contains an m X (2m  1 ) empty subarray which could be filled in with a PBTD(m). The PBTD(m) need not exist. We state without proof a result similar to Theorem 3.1 which provides PBTD(n)s which are missing a PBTD(3).
Theorem 3.3. Let n = 1 (mod 2). If there exists a complementary 2" frame and a pair of orthogonal partitioned incomplete Latin squares of type 2", then there i s a PBTD(2n+l) which i s missing a PBTD(3). Corollary 3.4. Let n a PBTD(3).
G
1 (mod 2), n
2 5.
There i s a PBTD(2n+1) which i s missing
Proof. There exist complementary 2n frames IS] and a pair of OPILs of type 2n 15) for n 1 (mod 2 ) , n > 5. 0 The idea of using different groups for the complementary frame and the pair of orthogonal partitioned incomplete Latin squares is also the basis of the following result.
Theorem 3.5. Let m = 1 (mod 2), m 2 3. I f there exists a complementary 2mn frame, a pair of orthogonal partitioned incomplete Latin squares of type m2" and a pair of orthogonal Latin squares of side m 1, then there i s a PBTD(2mn+I).
+
v
Proof. Let V = {uj,w/li =,1,2 ,...,m , j ~ 1 ,,..., 2 v} and let = {ii{,$b = 1,2,...,m , j = 1,2 ,...,n}. Let G/ = {u!,v/} and let G/ = {E/,G/} for i = 1,2,...,m and j = 1,2 ,...,n . Let F , be a complementary {G: ,...,GA,GT ,...,G i ,...,G: ,...,Gk)fr_ame_defined on V. F , is of type 2mn. Let F2 be a complement of F , defined on V. F, will be a 7 ... . . . ?GmI . . . , Gn 1 ? . . . , g l > f r a m e of type 2mn. F will denote the array of pairs formed by the superposition of F , and F,, F , o F2. Let L , and L , be a pair of orthogonal partitioned incomplete Latin squares of type m2". Suppose L l . is defined on V with partition {U',...,V , V 1,...,V"} where U i= y , , u i l . 5' ?tif} and V' = {u:, . I: ,uX}.. Suppose L , is de_f!nedon with partition { V ?. . . , V , $ . . . ,p} where U' = {qJ . . . ,gm}and v' = . . . ,$,}. L will be the array of pairs formed by the superposition of L , and L,, L = L , 0 L,. Let M , and M2 be a pair of orthogonal Latin squares of side m 1. Let Mi be the array,of pairs formed by the superpositionzf M I and M , where M1 is defined on the set u" u {a}and M , is defined on the set U' U {m}. Let Ni be the array of pairs formed by the superposition ofM, and M2 where M1 is defined on the set V" U {a} and M2 is defined on the set U {a}. Furthermore, suppose Mi and Ni are parti
7gk,el
{G,
+
tioned in the following way.
327
Tournament Designs of Side 4n+3
M.
=
Nj
Let J , =
[{ui,G}
* * *
{ u & , & } ] ~ and let K i =
[{ui,4}. . . { u ~ , & } ] ~
Define the following arrays for i = 1,2 ,...,m and j = 1,2 ,...,n.
r' l , 1 &J
F? =
C J=
CYij!
We construct two 2mn
1,
and
i;J 1 1
+ 1 x 2 m n + 1 arrays as follows.
E.R. Lamken and S.A. Vanstone
328
v
Every element in V U U {a;oo} occurs once in each row and once in each column of H , and H f 2 . Let C denote the 2mn X 1 array [C: * Ch * * C;  * C;]'. We per
 


mute the rows of HI2 so that the last column of the resulting array H , is [c;. . . c; . . . c; . . . C~(CY,CO}]~. We can now write H I and H 2 in the following forms.
H , and H , are H(2mn+1,4mn+2)s defined on V u
v U {a,..}.
Tournament Designs of Side 4n+3
329
It is straightforward t o verify that the array
is a PBTD(2mn+l) defined on V U 7 U {a,.}. Every distinct pair in V U U { a , ~ } occurs at most twice in each row of B and every element occurs precisely once in each column of B . The partitioning of B is given by H , and H2. 0
Corollary 3.6. Let n
E
1 (mod 2), n
2 3.
There i s a PBTD(Gn+l).
Proof. We use Theorem 3.3 with m = 3 and n G 1 (mod 2), n 2 3. Since there exists a complementary 23n frame [8], a pair of orthogonal partitioned incomplete Latin squares of type 32n [8,19] and a pair of orthogonal Latin squares of side 4, we can construct a PBTD(Gn+l). 0
Corollary 3.7. Let N for x E N .
=
{71,99,155,183,111,131,171,191}.
There e z i s t s a PBTD(z)
Proof. We use Theorem 3.3 with the values of m and n listed in the table below. The existence results for the complementary frame and pair of orthogonal partitioned Latin squares (OPILS) can be found in [8]. m n Complementaryframe OPILS PBTD 7 7 7 7 11 13 17 19
5 7 11 13 5 5 5 5
2a3 249 277 291 255 265 285 295
7'" 7 l4 722 726 11'O 131° 171° 19'0
71 99 155 183 111 131 171 191
0
The next results are generalizations of Theorem 3.5. We will use them to produce some small designs. Since the proofs are quite similar to the proof of Theorem 3.5, we omit them.
Theorem 3.8. If there e x i s t s a complementary 2n f r a m e , a p a i r o f orthogonal k partitioned incomplete L a t i n squares of type tyLti2 . . t p where C u i t i = 2 n and

i1
0 (mod 2) for all i , a n d a p a i r o f orthogonal L a t i n squares o f order ti i = 1,2,...,k , t h e n there i s a PBTD(2n+l).
ui
G
Corollary 3.9. There exist PBTD(e) for
e
= 75,95 and 167.
+ 1 for
E.R. Lomken and S.A. Vanstone
330
Proof. (i) e = 75. There is a 237 complementary frame [8]. We use a GDD(74;{7,8,9);{7,8};0,1) to construct a pair of OPES of type 7%'. (This GDD is constructed in [3].) (ii) e = 95. There is a 247 complementary frame [8\ We use CDD(94;{8,9,10},{3,11};0,1) t o construct a pair of OPILs of type 11 3 . (iii) e = 167. There is a 283 complementary frame. We use GDD(166;{8,9,10};{19,7};0,1) to construct a pair of O P E S of type 19872. 0
a a
+
0 (mod 2). If there exists a Theorem 3.10. Let 2 n = tu 6 where t complementary 2" frame, a pair of orthogonal partitioned incomplete Latin squares 1, then there i s a of type t"6 and a pair of orthogonal Latin squares of order t PBTD ( 2 n 1) . We note that this construction uses the special H(7,14) constructed for a PBTD(7) in [12]in place of a pair of orthogonal Latin squares of order 7. (Both of the Howell designs D , and D , of Table 3 in [12]have the required property.)
+
+
Corollary 3.11. There exist P B T D ( e ) lor
e
= 39,47 and 51.
Proof. If there is a set of three mutually orthogonal Latin squares of side m where m 2 7 , then there is a GDD(4m+6,{4,5},{4,6*};0,1). This group divisible design can be used to construct a pair of OPILs of type 4m6. (i)
e
= 39. Since there is a complementary 219 frame [8]and a pair of OPILs of type 4'6, we can apply Theorem 3.10 with t = 4 and u = 8.
e
(ii)
= 47. There is a complementary 223 frame [8]and a pair of OPILs of type 856. To construct the OPILs of type 856, we use the GDD(23,{4,5},{4,3*};0,1) from [19] as follows. Expand by a factor of 2, replacing each block with a pair of OPILs of type 24 or 25. The resulting design is a pair of OPILs of type 8'6. We use Theorem 3.10 with t = 8 and u = 5 to construct a PBTD(47).
(iii)
= 51. Since there is a complementary 225 frame [8] and a pair of OPILs of type 4116, we can apply the theorem with t = 4 and u = 11. 0
e
Let F be a complementary 2n frame. F is called row complementary if we can construct a complement for F by interchanging rows i and i 1 for i = 1,3,...,2n1.
+
Suppose F is a {G1,Gz,...,Gn>frame where Gi = {q,Zi}.
Let V =
5 G i . Let F , be
i1
the n X n subarray of F indexed by rows 1,3,5 ,...,2n1 and columns 1,3,5 ,...,2n1. If F can be written so that row i and column i of F , contain every element of V  Gi precisely once, we call F partitionable. In Figure 2, we display a row complementary partitioned 26 frame 141. Row complementary partitioned 2" frames can also be used t o construct partitioned balanced tournament designs.
Theorem 3.12. I f there i s a row complementary partitioned 2" frame, a P B T D ( m + l ) and a pair of orthogonal Latin squares of order m , t h e n there is a PBTD (mn+ I ) .
Tournament Designs of Side 4 n f 3
331
21 41
21
001
20 MI 30
31
40
001 31 41
10
40
05
00 11
05 I1
20
31
05 10
30 30
20
01
4 1
21
31
01 20
40
Figure 2
A row complementary partitioned 26 frame defined on (2, X Z2) U { c q , q }
Proof. Let V = {q,C = 1,2 ,...,n) and let Gj = {q,C). Let M = {1,2,...,m } . Let F be a row complementary partitioned 2" frame defined on V. F will be a {GI,G2,...,Gn)frame. Let Ri be the 1 X 2 n array formed by superimposing rows i and i 1 of F for i = 1,3,5,...,2n1. Let R be the n X 2 n array constructed as follows:
+
R=
332
E.R. Lamken and S.A. Vanstone
Let N , and N 2 be a pair of orthogonal Latin squares of side m defined on M . Let N be the array of pairs formed by the superposition of N 1 and N 2 , N = N , o N 2 . Nzywill be the m X m array of pairs formed by replacing each pair (a,b) in M with the pair (az,by). Let Pi be a P B T D ( m + l ) defined on (A4X G i ) U {cu,co}. Pi can be written in the following form.
We now use R to construct a P B T D ( m n + l ) defined o n ( M X V) U {&,co}. Replace each pair ( x , y ) in R with the m X m array Nzy. Replace the 1 X 2 empty array in row i and columns 2il,2i (i = 1,2,...,7 2 ) with the m X 2 m array Ai. We add a new row and a new column to the resulting mn X 2 m n array, P'. P will be the following (nm+l) X (2nm+1) array.
P=
It is straightforward t o verify t h a t P is a P B T D ( n m + l ) defined on ( M X V) U {cu,co}. The partitioning of P is given by columns 1,3,...,2nml,2nm+l and columns 2,4 ,...,2nm,2nm+1. 0
Corollary 3.13. I f there i s a P B T D ( m + l ) and a pair of orthogonal L a t i n squares of side m , then there i s a PBTD(6m.+1). Proof. A row complementary partitioned 26 frame is displayed in Figure 2. 0 4. Small designs.
In order to apply our constructions to the case n ~3 (mod 4), we will need the existence of several P B T D ( n ) s where n is small. We will use the following result which is proved in [9].
Theorem 4.1. Let n = 1 (mod 2). I f there exists a P B T D ( n + l ) generated by a starteradder pair on Z 2 , U {a,.}, a P B T D ( m ) , a P B T D ( m + k ) , a pair o f orthogonal Latin squares of order m and a n IA(m+k,k,4), then there i s a P B T D ( ( n + l ) m + k ) .
Tournament Designs of Side 4n+3
333
This theorem requires the existence of IA(m+lc,k,4)s. These are provided by a recent result due to L. Zhu and K. Heinrich.
Theorem 4.2. [6]An I A ( n , k , 4 ) exists i f and only i f n Lemma 4.3. Let n = 3 (mod 4 ) . possibly for n E {11,15,27}.
2 31c and
(n,lc)# (6,l).
There exists a PBTD(7i) for 7 5 n
Proof. Table 4.1 contains a list of constructions for PBTD(n) for n 7 5 n 5 203,and n $? {11,15,27}.
5 203 except
= 3 (mod 4),
Table 4.1 Constructions for PBTD(n) for n
n 7 19 23 31 35 39 43 47 51 55 59 63 67 71 75 79 83 87 91 95 99 103 107 111 115 119 123 127 131
ZE
3 (mod 4), 7 5 n
Construction
1121 6.3+1 4.4+7 6.5+1 5.7 4.8+7 6.7+1 8.5+7 4.11+7 6.9+1 8.7+3 6.10+3 6.11+1 14.5+1 6.13+1 8.10+3 6.14+3 6.15+1 14.7+1 6.17+1 6.17+5 22.5+1 6.19+1 8.14+7 12.1013 6.21+1 10.13+1
3.6 2.3 3.6 2.7 3.11 3.6 3.11 3.11 3.6 4.1 4.1 3.6 3.7 3.9 3.6 4.1 4.1 3.6 3.9 3.7 3.6 4.1 3.7 3.6
4.1 4.1 3.6 3.7
n
5 203 and n
ft? {11,15,27}.
Construction
135
6.21+9
4.1
139 143 147
6.23+1 8.17+7 8.18+3
3.6 4.1 4.1
151
6.25+1
3.6
155 159 163 167 171 175 179 183 187 191 195 199 203
14.11+1 12.13+3 6.27+1
3.7 4.1 3.6 3.9 3.7 3.6 4.1 3.7 3.6 3.7 4.1 [9] 3.6 4.1
34.5+1 6.29+1 8.22+3 14.13+1 6.31+1 38.5+1 6.31+9 6.33+1 20.10+3
E.R. Lamken and S.A. Vanstone
334 5. P B T D ( n ) for n
= 3 (mod 4).
We will use the frame construction, Theorem 2.3, and the existence of PBTD(n) for n = 1 (mod 4 ) to construct P B T D ( n ) for the case n G 3 (mod 4). We require a complementary frame and a pair of orthogonal partitioned incomplete Latin squares of type (4m)4(2t)'. The complementary frame was constructed in [14].
Theorem 5.1. Let m be a positive integer, m >_ 4, m # 6,lO and let t be a nonnegative integer such that 0 5 t 5 3 m . Then there is a skew f r a m e of type (4m)4(2t)'. We will use the same construction which was used t o find the skew frame t o provide the OPILs of type ( 4 ~ n ) ~ ( 2 t )We l . need the existence of OPILs of types 44, 45, 442 and 446. Lemma 5.2. [8] For n = 4,5 there is a pair of orthogonal partitioned incomplete Latin squares o f type 4". Lemma 5.3. There is a pair of orthogonal partitioned incomplete orthogonal Latin squares o f type 442.
+
+
Proof. Let G' = {(0,0),(0,2),(2,0),(2,2)}, G 2 = G' ( O , l ) , G 3 = G' (1,O) and 00 G 4 = G' (1,l). Let V = G' U {cq,q}. Let G? = GI X {i} for i = 1,2 and
u
+
11
j = 1,2,3,4. Define Gf
=
{(cq,i),(q,i)} for i
=
1,2. Let V; =
u5 GI. for i = 1,2. j 1
Let F be a Ekew frame of type 442 generated by an intransitivcframe starteradder (see [14]). F will denote the complement of F. We write F and F on the symbol set V, U V, so that F o
F
contains every pair in V,X V,
5
 (UG i j 1
once. The starters and adders for the frames are listed below. F: Starter S 11,1 q , 2 Adders: cq,l 3072 12,l 01,2 03,l 13,2 10,l 23,2 21,l 33,2 C 31,l 32,2
R
10,l
33,2
01 23 31 32 11 30
X G i ) precisely
Tournament Designs of Side 4ni3

F
Starter
S
oq,l
C
1391 23,l 01,l 30,l 32,l 33,l
12,2 oq,2 11,2 31,2 21,2 03,2 10,2
R
32,l
31,2
335
Adders:
03 21 10 12 33 13
We can construct a pair_ of orthogonal partitioned incomplete orthogonal Latin squares of type 442 from F o F. 0
Lemma 5.4. There i s a pair of orthogonal partitioned incomplete L a t i n squares of type 446. Proof. Let G? be defined as in the proof of Lemma 4.16 for i = 1,2 and j = 1,2,3,4. Define G: = {(q,i),(oq,i) ,...,(.g,i)}for i = 1,2. let
=
u5 G?. for i = 1,2.
j 1
Let F be a Ekew frame of type 446 generated by an intransitivcframe starteradder (see [14]). F will denote the complement of F . We write F and F on the symbol set V , U V, so that F o
F contains
every pair in V , X V, 
once. We list starters and adders for F and F below. F S 30,l oq,2 A: 0q,1 03,2 og,l 1272 W,,l 11,2 3371 og,2 2371 %,2
c
01,l 13,l 32,l
10,2 21,2 31,2
R
33,l 01,l 32,l
10,2 11,2 21,2
5
(uG{ X G i ) precisely j 1
23 32 31 01 30 11
E.R. Lumken and S.A. Vanstone
336
C
10,l 11,l 21,l
33,2 01,2 32,2
R
31,l 21,l 10,l
32,2 13,2 01,2
We can constLuct a pair of orthogonal partitioned incomplete Latin squares of type 446 from F o F . 0
Theorem 5.5. Let 0 5 t 5 3 m . If there i s a T D ( 5 , m ) , t h e n there i s a pair o f orthogonal partitioned incomplete Latin squares o f type ( 4 m ) 4 ( 2 t ) 1 . Proof. The construction for this proof is the same construction that was used t o construct skew frames of type ( 4 m ) 4 ( 2 t ) 1in [14]. For completeness, we include the construction. Let ( X , G , A ) be a T D ( 5 , m ) with G = {Gl,G2,...,G5}. Define a weighting 20: X + {0,2,4,6} by setting W ( X ) = 4 if 2 EX  G 5 and by defining W ( X ) for x E G, so that C W ( Z ) = 2t. z G 5
Apply Wilson's Fundamental Construction [18]. A block A € A requires a pair of orthogonal partitioned incomplete Latin squares of one of the types 44, 45, 442 or 446. These designs are provided by Lemmas 5.2, 5.3 and 5.4. The result of this construction . is a pair of orthogonal partitioned incomplete Latin squares of type ( 4 m ) 4 ( 2 t ) 1 0 We can now apply Theorem 2.3 in the following form.
Theorem 5.6. Let m be a positive integer, m 2 5, m # 6,lO and let t be a nonnegative integer such that 0 5 t 5 3 m . If there is a PBTD(St+l), t h e n there i s a PBTD (16 m +2t 1) .
+
We consider four cases n
= 3,7,11 and 15 (mod 16).
Lemma 5.7. (i)
If n
(ii) I f n (iii) If n
= 3 (mod 16) and n 2 195, then there exists a P B T D ( n ) . = 7 (mod 16) and n 2 183, then there exists a P B T D ( n ) . G
11 (mod 16) and n
2 219, t h e n there exists
a PBTD(n).
Tournament Designs of Side 4n+3
(iv) I f n
337
= 15 (mod 16) and n 2 191, then there exists a PBTD(n).
Proof. There exist PBTD(n) for n = 19,7,43 and 31 (Lemma 4.3). We apply Theorem 5.6 with (i) t = 9 and m 2 11, (ii) t = 3 and m 2 11, (iii) t = 21 and m 2 11 and (iv) t = 15 and m 2 11. 0 Combining the results of Lemma 4.3 and Lemma 5.7, we have the following.
Theorem 5.8. Let n n
E {11,15,27}.
= 3 (mod 4).
There exists a PBTD(n) except possibly for
Added In Proof There exists a PBTD(27). Apply Lemma 4.2 in I191 t o construct a pair of OPILS of type 456. Use Theorem 3.10 of this paper to construct a PBTD(27). References. PI Anderson, B.A., Howell designs of type H ( p  l , p + l ) , J. Combinatorial Theory (A) 24 (1978), 131140.
Bose, R.C., Parker, E.T. and Shrikhande, S., Further results on the construction o f mutually orthogonal Latin squares and the falsity of EuEer’s conjecture, Canadian J. Math. 12 (1960), 189203. [31 Colbourn, C.J., Manson, K.E. and Wallis, W.D., Frames for twofold triple systems, Ars Combinatoria 17 (1984), 6978. 141 Dinitz, J.H. and Stinson, D.R., Further results on frames, Ars Combinatoria 11
I21
(1981), 275288. 151 Dinitz, J.H. and Stinson, D.R., MOLS with holes, Discrete Math. (1983), 145154.
Heinrich, K. and Zhu, L., Existence of orthogonal Latin squares with aligned subsquares, Discrete Math. 59 (1986) 6978. Lamken, E.R. and Vanstone, S.A., The existence o f factored balanced tournament 171 designs, Ars Combinatoria 19 (1985) 157160. PI Lamken, E.R. and Vanstone, S.A., Partitioned balanced tournament designs o f side 4 n + l , Ars Combinatoria 20 (1985) 2944. 191 Lamken, E.R. and Vanstone, S.A., The existence o f partitioned balanced tournament designs, Annals of Discrete Math. (elsewhere in this volume). [lo] Mullin, R.C. and Wallis, W.D., The existence of Room squares, Aequationes Math. 13 (1975) 17. [ l l ] Schellenberg, P.J., van Rees, G.H.J. and Vanstone, S.A., The existence of balanced tournament designs, Ars Combinatoria 3 (1977) 303318. [12] Seah, E. and Stinson, D.R., An assortment of new Howell designs, Utilitas Math. (to appear). I61
338
E.R. Lamken and S.A. Vanstone
[13] Stinson, D.R., Room squares m'th maximum empty subarrays, Ars Combinatoria 20 (1985) 159166. [14] Stinson, D.R., Some Classes of Frames and the Spectrum of Skew Room Squares and Howell Designs, Ph.D. Thesis, University of Waterloo, 1981. [15] Stinson, D.R. and Wallis, W.D., A n even side analogue of Room squares, Aequationes Math. 27 (1984) 201213. [16] Todorov, D.T., Three mutually orthogonal Latin squares of order 14, Ars Combinatoria 20 (1985) 4548. [17] Vanstone, S.A., Doubly resolvable designs, Discrete Math. 29 (1980) 7786. [18] Wilson, R.M., Constructions and uses of pairm'se balanced designs, Math. Centre Tracts 55 (1974) 1841. (191 Stinson, D.R., and Zhu, L. On the existence of MOLS m'th equalsized holes, Aequationes Math. (to appear).
Annals of Discrete Mathematics 34 (1987) 339352 0 Elsevier Science Publishers B.V. (NorthHolland)
The existence of partitioned balanced tournament designs E.R. Lamken School of Mat hematics Georgia Institute of Technology Atlanta, Georgia U.S.A.
S.A. Vanstone Department of Combinatorics and Optimization University of Waterloo Waterloo, Ontario, N2L 3 G l CANADA TO A L E X R O S A O N X I S 3I3TIETU BIRTUDAY
ABSTRACT
A balanced tournament design, BTD(n), defined on a 2nset V is an arrangement of the distinct unordered pairs of the elements of V into a n n X 2n1 array such t h a t (1) every element of V is contained in precisely one cell of each column and (2) every element of V is contained in at most two cells of each row. If we can partition t h e columns of a BTD(n) into three sets C,,C,,C, of sizes l,nl,nl respectively so t h a t t h e columns in C,U C, form a n H ( n , 2 n ) and t h e columns in C , U C, form an H ( n , 2 n ) , then the BTD(n) is called partitionable. We denote a partitioned talanced tournament design of side n by PBTD(n). In this paper, we prove the existence of P B T D ( n ) for n 2 5 with 12 possible exceptions.
(y)
339
340
E.R. Lamken and S.A. Vanstone
1. Introduction. A balanced tournament design, BTD(n), defined on a 2nset V is a n arrangement 2n of the ( ) distinct unordered pairs of the elements of V into an n X 2nI array such that (1) every element of V is contained in precisely one cell of each column and (2) every element of V is contained in at most two cells of each row. An element which is contained only once in row i of a BTI>(n) is called a deficient element of row i. The two deficient elements of row i are referred to as the deficient pair of row i. The existence of BTD(n)s was established in [S]. (A simpler proof of this result appears in [4].) Theorem 1.1. For n a positive integer, n # 2, there exists a BTD(n). Balanced tournament designs can be used to represent round robin tennis tournaments. In terms of scheduling a tennis tournament we think of the rows of the design as representing courts, the columns as rounds of play and the elements as players or teams. The properties of t h e array give us a schedule of play which has each player playing each other player precisely once, each player plays each round and no player plays more t h a n twice on any court. We would now like to consider BTD(n)s with additional properties. In terms of tournaments, suppose t h a t we also require t h a t each player plays precisely once on each court in the first n rounds of the tournament. A B T D ( n ) with this property is called a factor balanced BTD(n) and is denoted by FBBTD(n). In graphical terminology, the first n pairs of any row in the design form a Ifactor of K2n. In addition to this, if each player plays precisely once on each court during the final n rounds, the design is said to be partitioned and is denoted by PBTD(n). Clearly, the existence of a P B T D ( n ) implies (by definition) t h a t there exists a FBBTD(n). The converse is false as the example in Figure 1 illustrates a FBBTD(6) which is not a PBTD(6). I t is a relatively easy task to determine the spectrum of FBBTD(n,)s. "71) In the sequel we study the more difficult problem of the existence of PBTD(n). W h a t make the problem difficult is the fact t h a t these designs are not pariwise balanced design closed (PBDclosed). We begin by giving a more formal definition.
Figure 1
FBBTD (6)
Tournament Designs
341
Let V be a set of 2n elements. A Howell design of side s and order 2n, or more briefly a n H(s,2n), is an s X s array in which each cell is either empty or contains an unordered pair of elements from V such t h a t ( 1 ) each row and each column is Latin (that is, every element of V is in precisely one
cell of each row and column) and ( 2 ) every unordered pair of elements of V is in at most one cell of the array. It follows immediately from the definition of an H(s,2n)t h a t n 5 s 5 2721.
If we can partition the columns of a BTD(n) defined on V into three sets C,,C2,C3 of sizes 1,n1,n1 respectively so t h a t the columns in C , U C , form an H ( n , 2 n ) and the columns in C, U C , form an H(n,2n),then t h e BTD(n)is called partitionable. W e denote the design by PBTD(n). A n example of a PBTD(5) is displayed in Figure 2. Partitioned balanced tournament designs are related to Room squares. A Room square of side 2n+l, RS(2n+1), is an H(2n+1,2n+2). Let R be a RS(2n+l). Since each row or column of R contains n empty cells, a t X t subarray of empty cells in R must have t 5 n . If R contains a n n X n subarray of empty cells, we say R contains a maximum empty subarray and denote R by MESRS(2n+l). The connection between MESRSs and PBTDs was provided by D.R. Stinson in [lo].
Figure 2 A PBTD ( 5 )
Theorem 1.2. There exists a MESRS(2n+1) i f and only i f there exists a PBTD (72 1 ) .
+
There is an extensive literature on Room squares and it is known t h a t Room squares of side 2n+l exist if and only if 2n+1 # 3 or 5 (Mullin and Wallis [S]). In [ l o ] , D.R. Stinson established the nonexistence of a MESRS(7) and t h e existence of MESRS(m) for m = 9 and m = 11. He showed t h a t two recursive constructions for Room squares could be used t o provide infinite classes of MESRSs and conjectured t h a t MESRS(2n+l)s exist for all n 2 4. This paper settles the conjecture in the affirmative with the possible exception of 12 values of n . W e have investigated the existence of MESRS(2n+l)s by considering the existence of PBTD(n+l)s. In [5] and [6), we showed t h a t PBTD(n) exist for n F 1 (mod 2), n 2 5 , with four possible exceptions. In this paper, we describe a new recursive construction for PBTD(n)s. This construction along with t h e existence of PBTD(n)sfor n = 1 (mod 2 ) is used to establish the existence of PBTD(n)sfor n G O (mod 2). W e describe this construction in section 3. In the next section, we collect several results which we will need from previous papers. In section 4, we apply the
E.R. Lamken and S.A. Vanstone
342
construction and provide the small designs needed t o complete the case n with eight possible exceptions.
E0
(mod 2 )
2. Preliminary results.
In this section, we collect for future reference some of the previous constructions and results on partitioned balanced tournament designs. In [5],we described a direct construction for PBTD(n+l)s. This construction uses a starter for a Howell design of side 2 n and order 2n+2, H(2n,2n+2). (For definitions and results on starters and adders for H(2n,2n+2), see [l].)
Theorem 2.1. [5]Let n
= 1 (mod 2).
Let
s = { { x l , ~ l } , { x 2 , ~ 2.}., . ,{x,lty,,l},{cu,zl},{~~2}}
be a starter f o r an H(2n92n+2) defined on Z,,, u { c Y , ~ }Let . S’ = S  {xi,yi} for some i where Ixiyi I = 1 (mod 2), 1 5 i 5 n1. Suppose there i s an adder A = (a1,a2,...,a,,) of n distinct elements of Z,, such that (i) ai E {0,2,4 ,...,2n2) for i = 1,2 ,...,n and A = Z,,,  {ul,vl} where lulvl I = n (ii) S’ Then there i s a PBTD(n+l).
+
We use this result to find several small PBTDs of even side.
Lemma 2.2. There e d s t PBTD(n+l) for n E (5,7,9,11,13,15,17,19,21}4 Proof. Starters and adders for P B T D ( n + l ) for n = 5,7,9 and 11 appear in [5]. We list starters and adders for PBTD(n+l) for n = 13,15,17,19 and 21 in Table 2.1. 0 The main constructions for finding P B T D ( n ) when n = 1 (mod 2 ) are frame constructions. The details of these constructions can be found in [5]and [6]. An example of this type of construction is the following.
Theorem 2.3. [5] If there emsts a complementary {GI,G2,...,G,)frame ( m 2 2), a pair of orthogonal partitioned incomplete Latin squares with partition {G1,G2,...,G,,,} m
and PBTD( IGi 1+1) f o r i = 1,2, ...,m, then there i s a PBTD((C IGi 1)+1). i1
We used Theorem 2.3 together with existence results for frames and orthogonal partitioned incomplete Latin squares t o prove the existence of P B T D ( n ) for n = 1 (mod 2).
Theorem 2.4. [5]Let n for n = 9. Theorem 2.5. [6]Let n for n E {11,15,27}.
G
1 (mod 4), n
2 5.
There e d s t s a P B T D ( n ) ezcept possibly
3 (mod 4), n
2 7.
There exists a P B T D ( n ) except possibly
Several other applications of the frame constructions can be found in [5]. We will use the following result in this paper.
Tournament Designs
Theorem 2.6. [5]Let n
= 1 (mod 2),
n
2 7.
343
If there is a PBTD(m+l), m # 6, then
there i s a PBTD(mn+l).
Table 2.1 Starteradder pairs for PBTD(n+l) for n = 13,14,17,19,21. n = 13 S 2,4 A 0 11,22 10 {xi
3,6 12
5,9 18
7,12 4
8,20 16
418 20
0419 6
10,16 24
14,21 8
17,25 14
15,24 2
13,23 22
YVi}: 071 7720
{UI7f4}:
n = 15 S 2,4 A 0 16,27 29 {xi , Y i } :
3,6 2
5,9 10
7,12 16
8,14 12
15,22 26
11,23 20
13,26 14
10,24 6
a,l9 18
0421 8
17,25 22
20,29 4
18,28 24
071
{wJ: 6721 n = 17 S 2,4 A
0
3,6 2
5,9 16
7,12 22
8,14 12
11,18 6
17,2 32
19,28 18
21,31 14
22,33 10
20,32 30
16,29 24
13,27 20
15,20 26
10,26 4
423 8
0424 28
3,6 2
5,9 4
7,12 14
8,14 22
10,17 12
15,23 10
18,27 26
21,31 6
25,37 24
20,33 8
22,36 16
19,34 36
16,32 18
13,30 32
11,29 28
q26 30
{xi,Yi}: 071 {ul,vl}: 10,27 n = 19 S 2,4 A 0 24,35 34
28 20
E.R. Lumken and S.A. Vanstone
344
n S
= 21
A
2,4 0
3,6 2
5,9 4
7,12 14
8,14 24
10,17 10
11,19 20
20,29 8
21,31 12
25,37 40
28,41 26
26,40 16
24,39 6
22,38 38
15,32 34
18,36 22
16,35 36
13,33 28
427 32
q30 18
{xi ,Yi}: 071 {ul,vl}: 15,36 In section 4, we will also use two product constructions from [lo].
Theorem 2.6. Suppose there exist PBTD(m) and P B T D ( n ) and a pair o f orthogonal Latin squares of order m . Then there exists a P B T D ( m n ) . Theorem 2.7. Suppose there exists a P B T D ( 2 m ) and a pair o j orthogonal L atin squares o f side m . Then there exists a P B T D ( 6 m ) . 3. Recursive Constructions. In this section, we prove the main recursive construction of the paper. This construction uses incomplete orthogonal arrays, I A ( n ,k,s)’s. Let V be a finite set of size n. Let K be a subset of V of size k. An incomplete orthogonal array I A ( n , k , s ) is an (n2 k2) X s array written on the symbol set V such that every ordered pair of V X V  ( K X K ) occurs in any ordered pair of columns from the array. An I A ( n , k , s ) is equivalent to a set of s  2 mutually orthogonal Latin squares of order n which are missing a subsquare of order k. We need not be able to fill in the k X k missing subsquares with Latin squares of order k.
Theorem 3.1. Let n = 1 (mod 2). If there exists a P B T D ( n + l ) generated b y a a,..}, a P B T D ( m ) , a P B T D (m + k ), a pair of orthogonal starteradder pair on Z,, U { Latin squares of order m and an IA(m+k,k,4), then there i s a PBTD((n+l)rn+k). Proof. Let B be a P B T D ( n + l ) generated by a starteradder pair on Z,, U { a , . . } . Let Cj denote column i of B for i = 1,2,...,2n+1. Suppose row i of C2n+lcontains the pair {ui,wi}where Iuiwi = n for i = 1,2,...,n and row n 1 of C2n+lcontains the pair {a,oo}.The partitioning of B into H(n+1,2n+2)s will be C1 U C, U * * U C2n1 U C 2 n + l and C, U C4 U * * . U C,, U C 2 n + l . Let {sl,t,} be a pair in column 1, row j , 1 5 j 5 n , of B such that lsltl 1 (mod 2). Let si = s1 i  1 (mod 2 n ) and let ti = t , + i  1 (mod 2 n ) for i = I,&...,%. Then {si,ti} occurs in column i of B . Since Is,tl 1 (mod 2),
I
+
7
I
+
u {s2j+l,t2j+1}= Z,,
n1
j0
{s1,s2 ,...,s2,}
and
(j{ ~ ~ ~ =, tZ,,., ~ }Every element j 1
I
of Z ,,
occurs once in
and once in {t1,t2,...,tzn}. I f {si,ti} occurs in row j of B , then {s,+i,tn+i}
Tournament Designs also occurs in row j. Note that if i is odd, then n
345
+ i is even.
Let M = {1,2,...,m } . Let Ll and L2 be a pair of orthogonal Latin squares of side m defined on M . L will be the array of pairs formed by the superposition of Ll and L,. L will denote the array of pairs formed by replacing each pair ( a , b ) in L with IY the pair (az,by). We use an I A ( m + k , k , 4 ) t o construct a pair of orthogonal Latin squares of order m k , Ngi and Nti, which are missing subsquares of order k . Let p = {pl,P2,. . . ,pk} and let 7 = {71,72, . . . ,rk}.Suppose NSi is defined on ( M X {si}) U p and is missing a subsquare of order k defined on P. Suppose Nti is defined on ( M X { t i } ) U y and is missing a subsquare of order k defined on 7. Let Naifi be the array of pairs formed by the superposition of Nai and Nti. Nsitican be written in the following form.
+
Kl
Kz
I
K3
1
K4
I
K5
E.R. Lamken and S.A. Vanstone
346
row j row j + l row j+2
nm
k
k
where 02, is an empty array of size ( m + k ) X 2 k . The resulting array is of size ( n + l ) m k X (2mn+2ml+2k) and is missing a subarray of size ( m + k ) x (2(m+k)1) in the lower right hand corner. Fill this array in with the array Bn+l. Let P denote the resulting array. The form of P is indicated below. The rows and columns of B have been permuted so that {sl,tl} occurs in cell ( 1 , l )of B .
+
P is a PBTD((n+l)m+k) defined on V = (A4X V) U W. Every element of V occurs once in each column and at most twice in each row of P. Every distinct unordered pair in V occurs once in P. We use the collections of columns indicated in the diagram above t o construct a partitioning of P. C, U C, U is an H ( ( n + l ) m
U C2n1 U D1 U D4 U D3
+ k, 2(n+l)m+2k) and C2 U C4 U
is an H ( ( n + l ) m
* *
* * *
U C2, U D2 U D, U D3
+ k,2(n+l)m + 2k). 0
Tournament Designs
347
This construction can be used with any PBTD(n+l) which has a partitionable transversal. Let B be a PBTD(n+l) defined on V U {a;oo}. W e write B in the following form.
The partitioning of B is given by C, U C, and C, U C,. W e consider the pairs in B as ordered pairs. Suppose there is a collection of 2n pairs T in A , one in each column and two in each row such t h a t each element of V occurs once as a first coordinate and once as a second coordinate in T , (i.e. a transversal of A ) . Let I;. denote the pairs of T in C ifor i = 1,2. If the pairs in I;. occur in n distinct rows a n d n distinct columns and every element of V occurs precisely once in for i = 1 and i = 2, then we say B has a partitionable transversal T .
Theorem 3.2. I f there exists a PBTD(n+l) un'th a partitionable transversal, a PBTD(m), a PBTD(m+k), a pair of orthogonal Latin squares of order m and a n IA(m+k,k,4), then there is a PBTD((n+l)m+k). We omit the proof of Theorem 3.2 since it is quite similar to t h a t of Theorem 3.1. Theorem 3.1 is the main construction used to find PBTD(n) for n ~0 (mod 2). In order to use this construction, we need t h e following recent result on IA(n,k,4)s. Theorem 3.3. [3]An IA(n,k,4) exists i f and only i f n 2 3k. The last construction in this section is a special case of t h e singular indirect product. The proof is similar to the proof of the singular direct product for MESRSs [lo] and is omitted for t h a t reason. Theorem 3.4. If there exists a PBTD(n),a PBTD(m) which i s missing as a subarray a PBTD(k) (the PBTD(k) need not exist), a PBTD(kn) and a n IA(m,k,4), then there is a PBTD(nm). Corollary 3.5. There is a PBTD(152). Proof. Let n = 8 , m = 19 and k = 3 in Theorem 4.3. There exists a PBTD(8),a PBTD(19)which is missing a PBTD(3) (Corollary 3.4), a PBTD(24) and a n IA(19,3,4) 131.
E.R. Lamken andS.A. Vanstone
348 4. P B T D ( n ) for n
= 0 (mod 2).
In order to apply our construction to the case n = O (mod 2), we will need the existence of P B T D ( n ) for n = 0 (mod 2 ) and 6 5 n 5 364.
Lemma 4.1. Let n = 0 (mod 2). There exists a P B T D ( n ) f o r 6 possibly f o r n E {26,28,32,34,38,44,58,94}. Proof. In Table 4.1, we list constructions for P B T D ( n ) for n 6 5 n 5 230, and n !$ {26,28,32,34,38,44,58,94}. 0 Lemma 4.2. Let n
E0
5 n 5 230
= 0 (mod 2 ) ,
(mod 2). There exists a P B T D ( n ) for 232
5 n 5 364.
Proof. The constructions for P B T D ( n ) where n = 0 (mod 2 ) and 232 5 n listed in Table 4.2. 0 W e can now prove the existence of P B T D ( n ) where n E 0 (mod 2), n possible exceptions.
Theorem 4.3. Let n for n
= 0 (mod 2), n 2 6.
E {26,28,32,34,38,44,58,94}.
except
5 364 are 2 5, with
8
There exists a P B T D ( n ) except possibly
Proof. There exists a P B T D ( n ) for n = 0 (mod 2 ) , 6 5 n 5 364, except possibly for n E {26,28,32,34,38,44,58,94} (Lemmas 4.1 and 4.2). (mod 2 ) , n 2 3 6 6 . W e write n = 6 ( 4 m + l ) k where Let n = O k E {0,2,4, ...,20,22}. For ?a 2 15 and k E {0,2,...,22) there exist PBTD(4m+1) (Theorem 2.4) and PBTD(4m+l+k) (Theorem 2.5). Since there exists a pair of orthogonal Latin squares of order 4 m 1 ( [ 2 ] and ) an IA(4m+l+k1k,4) (Theorem 3.3), we can apply Theorem 3.1 to construct P B T D ( n ) for n = 0 (mod 2 ) and n 2 366. 0
+
+
5. Conclusions.
We have settled the existence question for P B T D ( n ) s with twelve possible exceptions. Theorems 2.4, 2.5 and 4.3 can be summarized as follows.
Theorem 5.1. Let n be a positive integer, n possibly for n
2 5.
There exists a P B T D ( n ) except
E {9,11,15,26,27,28,32,34,38,44,58,94}.
It should be noted t h a t the recursive constructions together with t h e existence of PBT’(n) for small n could be used to reduce t h e number of exceptional cases. The next result illustrates this.
Tournament Designs
Constructions for PBTD(n) n 6 {26,28,32,35,38,44,58,94}. n Construction 6 StarterAdder 2.2 8 StarterAdder 2.2 10 St arterAdder 2.2 12 StarterAdder 2.2 14 StarterAdder 2.2 16 St arterAdder 2.2 18 StarterAdder 2.2 20 Start erAdder 2.2 22 StarterAdder 2.2 24 4.6 2.7 30 5.6 2.6 36 2.5 7.5+1 40 5.8 2.6 42 6.7 2.7 46 2.5 9.5+1 48 2.6 6.8 7.7+1 50 2.5 6.8+4 52 3.1 54 6.9 2.7 56 11.5+1 2.5 6.10 60 2.6 6.10+2 62 3.1 64 8.8 2.6 66 13.5+1 2.5 68 8.8+4 3.1 70 5.14 2.6 72 6.12 2.6 74 6.12+2 3.1 76 15.5+1 2.5 78 7.11+1 2.5 10.8 2.6 80 9.9+1 2.5 82 3.1 10.8+4 84 17.5+1 2.5 86 3.1 6.14+4 88
Table 4.1. for n GO n
90 92 96 98 100 102 104 106 108 110 112 114 116 118 120 122 124 126 128 130 132 134 136 138 140 142 144 146 148 150 152 154 156 158 160
349
(mod
Construction 6.15 2.7 7.13+1 2.5 5.19+1 2.5 8.12+2 3.1 11.9+1 2.5 6.17 2.6 10.10+4 3.1 21.5+1 2.5 6.18 2.6 6.17+8 3.1 8.14 2.6 8.14+2 3.1 23.5+1 2.5 9.13+1 2.5 6.20 2.6 10.12+2 3.1 10.12+4 3.1 3.1 10.12+6 8.16 2.6 13.10 2.6 3.1 16.8+4 8.16+6 3.1 8.16+8 3.1 2.7 6.23 10.14 2.6 3.1 10.14+2 3.1 10.14+4 3.1 10.14+6 3.1 12.12+4 3.1 12.12+6 19.8 3.5 3.1 6.25+4 3.1 6.24+12 3.1 6.25+8 2.6 10.16
2), n
162 164 166 168 170 172 174 176 178 180 182 184 186 188 190 192 194 196 198 200 202 204 206 208 210 212 214 216 218 220 222 224 226 228 230
6
sn 2 2 3 0
Construction 3.1 10.16+2 10.16+4 3.1 10.16+6 3.1 10.16+8 3.1 17.10 2.6 8.21+4 3.1 10.17+4 3.1 8.21+8 3.1 8.22+2 3.1 10.18 2.6 10.18+2 3.1 10.18+4 3.1 10.18+6 3.1 11.17+1 2.5 9.21+1 2.5 8.24 2.6 8.24+2 3.1 12.16+4 3.1 16.12+6 3.1 10.20 2.6 20.10+2 3.1 20.10+4 3.1 8.25+6 3.1 8.25+8 3.1 10.21 2.6 8.25+12 3.1 10.21+4 3.1 6.36 2.6 3.1 10.21+8 10.22 2.6 22.10+2 3.1 22.10+4 3.1 14.16+2 3.1 16.14+4 3.1 16.14+6 3.1
and
350
E.R. Lamken and S.A. Vanstone
Table 4.2. Constructions for PBTD(n) for n = 0 (mod 2), 232 n Construction n 3.1 300 232 14.16+8 302 6.39 234 2.7 304 8.29+4 236 3.1 306 14.17 238 2.6 308 10.24 240 2.6 242 12.20+2 3.1 310 244 312 12.20+4 3.1 314 246 10.24+6 3.1 316 248 2.6 8.31 250 318 10.25 2.6 252 320 10.24+12 3.1 322 10.25+4 254 3.1 324 256 3.1 12.21+4 326 258 10.25+8 3.1 328 20.13 260 2.6 330 262 12.21+10 3.1 332 16.16+8 264 3.1 266 334 12.22+2 3.1 336 6.43+10 268 3.1 270 338 6.45 2.7 272 340 12.22+8 3.1 274 342 7.39+1 2.5 276 344 6.46 2.7 278 346 6.46+2 3.1 280 348 6.46+4 3.1 350 6.46+6 282 3.1 6.46+8 3.1 284 352 286 354 6.46+10 3.1 288 356 6.48 2.6 290 6.48+2 3.1 358 292 360 6.48+4 3.1 294 362 6.48+6 3.1 364 296 3.1 6.48+8 298 3.1 14.21+4
5 n 5 364. Construction 10.30 2.6 3.1 6.50+2 3.1 6.50+4 3.1 6.50+6 2.6 14.22 3.1 22.14+2 3.1 22.14+4 22.14+6 3.1 14.22+8 3.1 2.6 6.53 6.53+2 3.1 3.1 6.53+4 2.6 8.41 3.1 8.41+2 3.1 8.41+4 3.1 6.54+6 3.1 6.54+8 3.1 6.54+10 3.1 6.54+12 3.1 6.54+14 3.1 6.54+16 3.1 6.54+18 3.1 6.54+20 3.1 6.54+22 3.1 6.54+24 3.1 6.54+26 3.1 6.56+16 3.1 6.56+18 3.1 6.56+20 3.1 6.56+22 3.1 6.56+24 3.1 6.60+2 3.1 6.60+4
Tournument Designs
351
Lemma 5.2. If there exists a PBTD(9), then there exist PBTD(n) for n E {44,58,94}. Proof. The following table lists the constructions to be used. n Construction 44 6.7+2 3.1 58 6.9+4 3.1 94 10.9+4 3.1 It is interesting t h a t the existence of partitioned balanced tournament designs implies the existence of several other types of designs. We have already noted the equivalence between MESRSs and PBTDs. PBTDs can also be used to prove the existence of a n even sided analogue for Room Squares.
+
2. Let F be a Let n be a nonnegative integer and let S be a set of size 2n partition of S into unordered pairs. A house of order n 1 is a 2n+2 X 2n+2 array H which satisfies the following:
+
( 1 ) every cell of H either is empty or contains an unordered pair of elements of S ,
( 2 ) every element of S occurs in precisely one cell of each row and column of H , ( 3 ) the pairs in F each occur in precisely two cells of H , whereas every other pair of elements occurs in exactly one cell of H , ( 4 ) the pairs in t h e first and second rows of H are precisely those in F , and (5) every column of H contains one pair from F. A house of order n 1 is a n approximation to a Room square of side 2n is known t h a t houses of order n exist for n a positive integer and n # 2, [ l l ] .
+
+ 2.
It
If we can permute the columns 1 ,...,2n+2 and rows 3,...,2n+2 of a house of order so t h a t t h e n+1 X n+l subarray H , indexed by rows and columns 2 j 1 for j = 0,1,...,n is a n H(n+1,2n+2) and the n+l X n+l subarray H , indexed by rows and columns 2 j for j = 1,2,...,n+l is an H(n+1,2n+2), then we call H a partitionable 1. We note t h a t H , and H , will form a pair of almost disjoint house of order n H(n+1,2n+2)s. (They will have the same first row.) n
+
+1 H
+
+
Therefore, a partitionable house of order n 1 can be used to construct a PBTD(n+l). Conversely, it is easy to construct a partitionable house of order n 1 from a PBTD(n+l). The following theorem lists the designs which are equivalent to PBTDs.
Theorem 5.3. For n a positive integer, the follouring designs are equivalent. (i) a PBTD(n+l) (ii) a MESRS(2n+l) [lo] (iii) a pair of almost disjoint H(n+1,2n+2)s and [lo] (iv) a partitionable house of order n 1.
+
+
E.R. Lamken and S.A. Vanstone
352
Added In Proof Recently, PBTD(n)’s for n = 27, 32, 38, 58, and 94 have been constructed. This leaves seven possible exceptions to the existence of PBTDs. For the case n = 27, the reader is referred t o [6], and for n = 32, 38, 58 and 94 t o E.R. Lamken, “A note on partitioned balanced tournament designs”, Ars Combinatoria 24 (1987). References Anderson, B.A., Howell designs of type H(pl,p+l), J. Combinatorial Theory (A) 24 (1978), 131140.
Bose, R.C., Parker, E.T. and Shrikhande, S., Further results on the construction of mutually orthogonal Latin squares and the falsity of Euler’s conjecture, Canadian J. Math. 12 (1960), 189203. Heinrich, K. and Zhu, L., Existence of orthogonal Latin squares with aligned subsquares, Discrete Math. 59 (1986) 6978. Lamken, E.R. and Vanstone, S.A., The existence of factored balanced tournament designs, Ars Combinatoria 19 (1985) 157160. Lamken, E.R. and Vanstone, S.A., Partitioned balanced tournament designs of side 4 n + l , Ars Combinatoria 20 (1985) 2944. Lamken, E.R. and Vanstone, S.A., The existence o f partitioned balanced tournament designs of side 4n+3, Annals of Discrete Math. (elsewhere in this volume). Lamken, E.R. and Vanstone, S.A., Balanced tournament designs and related topics, Preprint. Mullin, R.C. and Wallis, W.D., The existence of Room squares, Aequationes Math. 13 (1975) 17. Schellenberg, P.J., van Rees, G.H.J. and Vanstone, S.A., The existence o f balanced tournament designs, Ars Combinatoria 3 (1977) 303318. [lo] Stinson, D.R., Room squares with maximum empty subarrays, Ars Combinatoria 20 (1985) 159166. [ l l ] Stinson, D.R. and Wallis, W.D., An even side analogue o f Room squares, Aequationes Math. 27 (1984) 201213.
[12] Todorov, D.T., Three mutually orthogonal Latin squares of order 14, Ars Combinatoria 20 (1985) 4548.
Annals of Discrete Mathematics 34 (1987) 353362 0 Elsevier Science Publishers B.V. (NorthHolland)
353
Constructions for Cyclic Steiner %designs Rudolf Mathon Department of Computer Science University of Toronto Toronto, Ontario, Canada M5S 1A4 TO A L E X ROSA
O N MIS 3I3TIETU BIRTUDAY
ABSTRACT This paper surveys direct and recursive constructions for cyclic Steiner 2designs. A new method is presented for cyclic designs with blocks having a prime number of elements. Several new constructions are given for designs with block size 4 which are based on perfect systems of difference sets and additive sequences of permutations.
1. Introduction A balanced i n c o m p l e t e block design (briefly BIBD) with parameters (w,k,X) is a pair (V,B) where V is a wset and B is a collection of ksubsets of V (called blocks) such t h a t every 2subset of V is contained in exactly X blocks. A S t e i n e r %design is a (w,k,X)BIBD with X = 1. An a u t o m o r p h i s m of a BIBD (V,B)is a bijection 4: V +V such t h a t the induced mapping a: B + B is also a bijection. The set of all such mappings forms a group under composition called the automorphism group of the design. A (w,k,X)BIBD is cyclic if it has a n automorphism consisting of a single cycle of length w. Cyclic (v,k,X)BIBD’s will be denoted by C(v,k,X). A (v,k,X) d i f f e r e n c e f a m i l y (briefly DF) is a collection of ksubsets D,, . . . ,Dt of t h e integers 2, modulo w such t h a t for each nonzero z €2, the congruence di  d j = z ( m o d w ) has exactly solution pairs ( d i , d j ) with d i , d j E DI, for some 1. A (w,k,X)D F is called s i m p l e if X = 1. It is easily verified t h a t a necessary condition for the existence of a (v,k,X)D F is X(w  1) = 0 m o d k(k  1). In particular, if a simple D F exists then w G 1 m o d k(k  1). A (v,k,X) D F generates a cyclic BIBD C(w,k,X) with V = 2, and B = { d D I I O s i < w , I l l s t } , where a : V + V , a ( z ) = z + l m o d v and t = X(w  l ) / ( k ( k  1)). The t blocks D,, . . . ,D, are called starter or base blocks of the design (V,B) (they are representatives of the orbits of B under a). An orbit analysis of a cyclic Steiner 2design C(v,k,l) yields the following necessary existence condition:
354
R. Mathon u
= 1, k
mod
k(k  1).
+
+
(1)
The case u = k(k  1)t 1 corresponds to a simple DF. If u = k(k  1)t k then 1 starter blocks Do,Dl, . . . ,Dt, where Do = {O,m,2m, . . . ,(k  l)m}, there are t m = (k  1)t 1 generates an morbit and D,, . . ,Dtgenerate t uorbits under 6. It is clear, that the differences in D1, . ,Dt cover the elements Zw\poexactly once.
+
+
..
.
Two difference families D = {D,, . . . ,Dt} and D ' = {D;, equivalent if for some integers r,sl, . . . ,st
{Di,
. . . ,Dt'} = {rDl + a,, . . . ,rDt + s t } mod u.
. . . ,Dt'} are said to be (2)
If D is equivalent with itself, then the corresponding r is called a multiplier of D and 7 : z + rz, z E 2, is an automorphism of the cyclic design. Cyclic designs have a nice structure and interesting algebraic properties. Their concise representation makes them attractive in applications and for testing purposes. Cyclic BIBD's and difference systems have been studied by many authors [3], [7], (lo], [13]. Results concerning cyclic Steiner 2designs are surveyed in [5]which also contains a fairly extensive bibliography. The present paper addresses the problem of existence of cyclic Steiner 2designs C(u,k,l). In the next two sections we discuss direct and recursive constructions for general block sizes k. In addition to known techniques, several new constructions are presented for k = 4 and 5. We conclude with a list of open problems. The paper significantly extends the existence results given in [5\ for cyclic Steiner %designs with block sizes k > 3. 2. Direct Constructions
The majority of direct methods for constructing cyclic designs are based on finite fields. In this section we survey those constructions which apply to Steiner 2designs and apply them to generate some new designs with blocks of prime size. We begin with two general constructions of Wilson for (u,k,l) difference families 1131.
+
Theorem 1 Let p = k(k  1)t 1 be a prime and CY a primitive root of Zp. Let Hm be the multiplicative subgroup of Z,\{O} generated by d" and let w = dmf. (i)
+
If k = 2m 1 is odd and {w1,w21, . . . ,wml} is a system of representatives for the cosets d H m , i = O,l, . . . ,m1, then the blocks Di+, = {d"',wd"' ,..., 2m # i }, i = O , l , . . . ,t  1 form a ( p , k , l ) DF.
(ii) If k = 2m is even and {l,w1, . . . ,wrnl  1) is a system of representatives for the ,cosets d H m , i = O , l , . . . ,m  1, then the blocks Di+, = { O , d " ' , wd"', . . . , W2m20(ni }, i = 0,1, . . . ,t  1 form a (p,k,l) DF in Zp.
+
Theorem 2 Let p = k(k  l)t 1 be a prime and CY a primitive root of Zp. If there exists a set B = {b,, . . . , b k } subset Zp such that {b  bi I 1 5 i < j 5 k} is a system of representatives for the cosets dHm', i = 0,1,. . . ,m  1, where m = k(k  1)/2 and H m is the subgroup of Z,\{O} generated by d",then D,+l = g m ii~=,0,1, . . . ,t  1 is a ( p , k , l ) DF in zP.
Constructions for Cyclic Steiner 2Designs Our next result concerns the case v
+
355
= k mod k(k1).
+
Theorem 3 Let k = 2m 1 and p = 2mt 1, t 2 2 be two odd primes and let a be a primitive root of Zp. Define m  1 numbers r, by the equations ar8= d  1, i = 1, . . . ,m  1. If there exists a p EZk such that the 2m elements f l , k(pi = 1, . . . ,m  1 are all distinct in zk, then the blocks
 102m}
Do = {00,01,
D,+1 = {oo,aia',ay:~, . . . ,a"""&L}, form a (kp,k,l) DF in
i
= 0,1,
. . . ,t
(3)
1
zkp.
Proof We note, that since in the family of blocks B = {B1,. . . ,Bt}, Bi+l = {O,d,. . . , dmtt+i} each nonzero difference appears exactly k  2m 1 times, B forms a (p,lc,k) DF in Zp. To complete the proof, it suffices to show that for any fixed difference in B the corresponding subscript differences cover every nonzero element of 2, exactly once. Since for each i , (2' 1) = 1 this is equivalent to the assumption that f l , $(p"'  l)prs,i = 1, . . . , m  1 are distinct in zk. Finally, since k and 0 p are distinct primes the design is cyclic in Zkp.
+
We will apply Theorem 3 t o blocksize k = 7. Then m = 3 and p is a prime of the 1, t 2 2. If cr is a primitive root of Zp, then dt = 1 and since form p = 6t
+
(02
+ 1 ) d t = d t  1 = (02 + 1)(02  1)
we have 02  1 = #. Let r . be the solution of some p E 2, the 6 numbers
et,
[email protected] d
=
 11, et'(p2'  1) elements of 2,. Since Pt cannot be
d t  1.
We require that for (41
cover the nonzero congruent to 1 modulo 7, we see that t = 1 or 2 mod 3. If t = 1 mod 3, then (4) are distinct if either p = 2 and r = 0 mod 3, or /3 = 4 and r = 2 mod 3. If t G 2 mod 3, then we need either p = 2 and r = 1 mod 3, or p = 4 and r E 0 mod 3. Combining all these conditions we obtain the following result.
+
Corollary 4 Let p = 6t 1 be a prime, t 2 2 , t f 0 mod 3, and let a be a primitive root in Zp. Then the blocks (3) form a (7p,7,1) DF for some p E 2, if and only if t f r mod 3, where r satisfies d = d t  1. We note, that for some values of t we obtain two nonisomorphic cyclic designs. If t = 4 mod 6, then (4) are distinct also if either /3 = 3 and r = 2 mod 3, or p = 5 and T = 0 mod 3. If t = 2 mod 6, then (4) are distinct also if either p = 3 and r = 0 mod 3, or p = 5 and T = 1 mod 3.
For k = 7 solutions exist when t = 2 * , 5, 7, 13, 16*, 26*, 35, 37, 38*, 40*,46*, 47, etc. The base blocks for t = 2 * , 5 and 7 are
R, Mathon
356
The solutions for t = 5 and 7 are first examples of BIBD's with the parameters (217,7,1) and (301,7,1), respectively. For k = 11 solutions exist when t = 33, 54*, 57, 91, 94*, etc. and for k = 13, t = 13, 19, 59, etc. (* indicates 2 solutions). We conclude this section with a wellknown result in finite geometries [6].
Theorem 5 Let q be a prime power. Then the lines in the projective geometry PG(n,q),n 2 2 form a cyclic design with parameters ((q"+'  l)/(q  l), q 1, 1).
+
3. Recursive Constructions
Given two difference families it is sometimes possible to combine them to construct a new one. Several such constructions are known for general cyclic BIBD's [4] [8] [14]. To apply them, various conditions on the block sizes are usually required. We begin with a constructlon by C.J. Colbourn and M.J. Colbourn [4].
Theorem.6 Let Afi = {O,ail, . . . ,aik1}, i = 1, . . . ,t be a ( v , k , l ) D F in 2, and let Bsj = {O,bJ1, . . . ,bJkl}, j = 1, . . . ,s be a ( w , k , l ) D F in 2,. (i)
+
If v = k(k  1)t 1 and w is relatively prime t o (k  I)!, then for i = 1, . . . , t , j = I , , . . ,s and 1 = 0,1,. . . , w  1
is a (vw,k,l) DF in Z,,.
(ii) If u = ka, w = k p and 6 , is relatively prime to (k  I)!, then for i = 1, . . . , t , j = 1 , . . . ,s and 1 = 0,1,. . . ,w  1 {O,ail
+ 1v7ai2 + 21Uj . . , a i k  l + (kl)lv}
{ O , a b l ,abj2, {o,cYp,2&,
. . . ,cYbjkl
* * *
}
,(kl)olp}
is a ( k a p , k , l ) D F in Zk,p. Here CY = (k  l ) t full orbit base blocks A t i , Bsj are considered.
I
(6)
+ 1, p = (k  1)s 4 1, and only
Constructions for CycIic Steiner 2Designs
357
We note t h a t the construction can be used if either w or p are prime. Then the existence of a ( w , k , l ) D F implies the existence of a (wn,k,l) D F for every n 2 1. Similarly, from a (kp,k,l) D F we obtain a (kp",k,l) DF. Also, if a ( v , k , l ) D F exists with v G 1 mod k(k  1) and prime k then there exists a (vk,k,l) DF. In [8] M. Jimbo and S. Kuriki have introduced a more general construction for cyclic BTSD's which is based on orthogonal arrays. Applying i t to Steiner 2designs we obtain the following typical result.
Theorem 7 Suppose there exists a C ( v , k , l ) and a C ( w , k , l ) , where w = 1 mod k(k  1) and k is a n odd prime. Then there exists a C(vw,k,l). If, in addition, w = 1 mod k(k  I), then the conclusion holds for k a prime power. So, for example, if k is an odd prime not dividing v , then t h e existence of a C ( v , k , l ) implies t h e existence of both C(vn,k,l) and C(ku",k,l) for any n 2 1.
The next construction employs cyclic pairwise balanced designs. A p a i r w i s e balanced d e s i g n (briefly PBD) is a pair (V,B)where V is a vset and B is a collection of subsets of V (blocks) such t h a t every 2subset of V is contained in exactly one block. ,kn} is t h e set of block sizes. A PBD will be denoted by ( v , K , l ) , where K = {kl,
. ..
Theorem 8 Suppose there exists a cyclic ( v , K , l ) PBD with K = {kl, . . . ,kn} and t h a t for each ki there exists a (ki,k,l) Steiner 2design. Then there exists a C ( v , k , l ) . Proof Replace each base block in the PBD by the blocks of the corresponding Steiner 0 %design to obtain the base blo.cks of the final C ( v , k , l ) . In the next section we shall give some other recursive constructions for cyclic designs with blocks of size 4 and 5 which are based on t h e concepts of perfect systems of difference sets and additive sequences of permutations. 4. Special Constructions
The existence question for cyclic Steiner triple systems has been completely settled by Peltesohn [lo],who constructed C(v,3,1) for all = 1,3 mod 6, v # 9. For block sizes k > 3 the existence problem for C ( v , k , l ) remains unsolved. The state of affairs is most promising for the cases k = 4 and 5 . In order to present additional recursive constructions we require a few more definitions. of t ksubsets Di = (dio,di,,. . ,dikPl}, 0 = di0 < d', 1, . . . , t is said t o be a p e r f e c t d i f f e r e n c e f a m i l y (PDF) in Z,, v = k(k  1)t 1, if the t k ( k  1)/2 differences d t l  dIi, 0 5 j < 1 < k cover the set {1,2, . . . , t k ( k  1)/2}. PDF's are equivalent t o regular perfect systems of difference sets starting with 1, which have been studied by many authors (see [I] for a recent survey). It has been shown [2] t h a t PDF's can exist only when k is 3,4 or 5. For k = 3 the existence of a P D F is related t o Skolem's partitioning problem [l]. A
collection
< . . < d Z k w l ,i *
+
~
=
358
R. Mathon
+
Let X' be the mvector (r,r+l, . . . ,1,0,1,...,r  l , r ) , m = 2r 1 and let X 2 , . . . ,X"be permutations of X ' . Then X', . . . ,X" is an additive sequence o j p e r m u t a t i o n s (ASP) of order m and length n if the vector sum of every subsequence of consecutive permutations is again a permutation of X ' . ASP play an important role in recursive constructions for PDF's and vice versa [I] [ll][12].
Block size 4 We begin with two direct constructions.
+ +
Theorem 9 let p = 12t 1, t 2 1 be a prime and let a be a primitive root of Zp. (i) (131 [13]) If p # z2 36y2 for any integers z and y then
{ O , @ , ~ ~ * + ~ ~ , O H ~ +i~ = ~ } 0,1, . . . ,t  1
(7)
is a (p ,4,1) DF in Zp. (ii) ([5]) If a 3 mod 4 (and such an a always exists in Zp) then {O a4i a4i+3 4i+6 } i = 0 , . . . ,3t  1) 9 , ,a {0,~4i+1,,4*+4j+l #+4j+l} j = 0, . . . ,t  1) {O,P ,2P,3P
I
1
form a (4p,4,1) DF in ZSp.
The next two constructions will exhibit the relationship between PDF's and ASP
Theorem 10 ([3] [13]) Let Di = {O,ai,bi,ci}, i = 1, . . . ,t be a PDF in Z12t+land let X ' , X 2 , X 3 be an ASP of order m = 2r 1, r 2 2 and length 3. Then
+
(i)
For i = 1, . . . ,t and j = 1, = {O,ma,
cover the set {r+l,r+2,
X'
. . . ,m the 6 t m positive
differences in the family
+ a;,mbi + P,,mci + 7,)
. . . ,r+6tm}.
+ P,X' + X* + x3,respectively.
Here
a,P
(9) and 7 are the mvectors X ' ,
(ii) For i = 1, . . . ,t \
Xi' = (c ,a c ,b ,b c ,a b ,a ,a ,b a ,c b ,b ,c a ,c)i
Xi'
= ( c b ,c ,b a ,c a
,b c ,a c ,b ,a ,b ,E ,a b ,a
)i
Xi" = ( b a ,b ,ac ,a ,c ,c b ,b c ,c ,a ,c a ,b ,a b)i the (12t+l)vectors X j = (O,Xl, . 1% 1 and length 3.
+
. . ,Xjt), j = 1,2,3
form an ASP of order
In order to utilize products of the form (9) for constructing new difference families we need to find additional base blocks with differences covering the set (1, . . . ,+} and possibly {r 6tm 1, . . . ,6z} for some z 2 1.
+
+
359
Constructions for Cyclic Steiner 2Designs
We list now the known recursive constructions for 1 5 m
5 25.
Theorem 11 Let D ( t ) = {Dl,. . . ,Of} be a PDF and let A(mt) = {Al, . . . ,bt} be , l,O,l ,..., r1,r). Then 1 and a = (r,r+l, * defined by (9), where m = 2r 1. For r = 2
+
p=
(2,0,2,l,l),

7 = (0,2,1,1,2)
D(5t+1) = A(5t) U {0,1,30t+4,30t+6}
is a PDF in Z60f+13. 2. For r = 3
p=
(1,2,3,3,2,1,0),
7 = (2,1,3,0,3,1,2)
D(7t+1) = A(7t) U {0,2,3,42t+7}
is a DF in
Z8,,+,3.
3. For r = 6
,B = (4,5,1,2,3,6,6,5,1,3,0,2,4) 7 = (1,5,6,3,3,4,4,2,5,2,6,1,0)
D(13t+1) = A(l3t) U {0,1,4,6}
is a PDF in 4. For r = 9
p 7
Z156t+13.
= (6,7,1,2,3,5,3,9,4,7,8,0,5,9,1,6,2,4,8) (2,1,8,9,3,1,5,6,5,2,7,7,3,8,4,6,0,9,4)
D(l9t+4) = A(19t) U {0,1,7,~+23}U { 0 , 2 , ~ + 1 4 ~ + 1 9 } U {0,3,z+13,z+21} U {0,4,z+15,z+24},
is a PDF in
z = 114t
2228t+49.
5. For r = 11
p ~(0,2,1,7,2,6,1,5,4,10,11,9,6,4,8,3,11,8,10,3,5,7,9) 7 = (9,5,4,10,0,7,10,9,8,4,3,1,5,11,8,2,6,2,1~~6,~~~~~)
D(23t+5) = A(23t) U {0,1,8,~+28}U {O72,z+14,z+24} U {0,3,z+18,z+29} U {0,4,z+17,~+23} U {0,5,z+21,z+30},
is a PDF in Z2T6t+61. 6. For r = 12
using a,p,7 and A(5t) from 1 t o obtain A(25t) D(25t+5) = A(25t) U {0,1,~+18,~+29} U {0,4,z+20,z+26}
z = 138t
R. Mathon
360
U {0,3,8,2+27} U { 0 , 7 , ~ + 2 1 , ~ + 3 0U } {0,10,12,2+25},
is a PDF in
2 =
150t
Z300t+61, and
D(25t +6) = A(25t) U {O,l,z+34,2+36} U {0,3,2+18,~+29} U {0,4,~+20,s+28} U {0,5,~+22,2+32}
U {0,6,~+19,~+31U } {0,7,~+21,2+30}, x = 150t
is a PDF in Zmot,,,. Proof Use (9) t o check that the required sets are covered by all differences from the base blocks. 0
We note that the constructions 2,5,6b are new and that 1,3,4,6a have been known [Ill. The ASP with r = 11 has been found by P.J. Laufer.
If we apply all methods listed in Sections 2, 3 and 4 and add the computer generated DF’s from [5] we obtain the following results for 1 5 t 5 50: (12t+1,4,1) PDF
t
=
1,48,14,21,23,26,28,3031,36,41
(12t+1,4,1) DF
t
= 1,310,1415,1921,23,26,2831,3436,38,4~41,43,45,50
(12t+4,4,1) DF
t
= 36,12,20,24,30,32,36,43.
Block size 5 As before, two direct constructions are known.
+
Theorem 12 Let p = 20t 1, t 2 1 be a prime and let cy be a primitive root of Zp. ([3] [13]) If p # x 2 100y2 for any integers x and y then (i)
+
{giIa4t+2i
d 3 f + 2 i $2t+2i 7
7
is a (p,5,1) DF in Zp. (ii) ([5]) If d 1 = d(d
+
$0t+2i} 7
i
= O,J,. . . , t

 1) for some odd integers r and 8 then = 0,1, . . . ,t  1
{O,QZi,QZi+r,gt+2i,QZt+2i+r} i
(11)
(12)
{O,P7 % ,3P94P 1 form a (5p,5,1) DF in Z S p . Concerning PDF’s with blocks of size 5 and ASP of length 4, results can be proved which are similar t o those stated in Theorem 10 [l]. They can be used t o derive the following construction.
Constructions for Cyclic Steiner 2Designs
361
Theorem 13 Let D ( t ) = {D1,. . . J t } be a PDF in C,,,+, and let D ( s ) = {D1, . . . ,D8}be a DF in C208+1.Then a DF D ( r ) exists in C20r+l,r = 2Ost+s+t and D ( r ) is perfect whenever D ( s ) is perfect. Proof Use D ( t ) to construct an ASP of length 4 and order m = 20t+1 [l]. With help of this ASP construct the blocks A(ms) in a similar way as in (9). Then D ( r ) = A(ms)u D ( s ) .
PDF’s with k = 5 can exist only if t is even and t 2 6 [I]. They have been enumerated for t = 6 [9] and examples are known for t = 8, 10, 732, 974, etc. Difference families are known for the following values of t , 1 5 t 5 50: (20t +1) PDF
t
= 6, 8, 10
(20t +1) DF
t = 13,6,8,10,12,14,2122,30,3233,35,41,4344 (20t +5) DF
t = 35,7,910,13,15,18,22,2425,2728,30,34,37,39~40,4243,45~48~50~ Open Problems 1. Does there exist a (12t+1,4,1) DF for every t 2 31 Can all of these DF’s be perfect if t 2 4 1 2. Does there exist a C(w,4,l)‘for every w # 16, 25 and 28? 3. Does there exist an ASP of length 3 for every order m 2 5, m # 9,10? 4. Do there exist C(v,5,1) for w = 81 and 85? 5. Construct examples of PDF’s D ( t ) for k = 5 and even t 2 12. 6. Construct examples of ASP of length 4 for orders m 2 7 . Acknowledgement Research supported by NSERC Grant A8651. References [l] J. Abrham, “Perfect systems of difference sets
 a survey”, Ars Combinatoria 17A
(1984) 536.
[2] J.C. Bermond, A. Kotzig, J. Turgeon, “On a combinatorial problem of antennas in radioastronomy”, Proe. 18th Hungarian Combinatorial Colloquium, North Holland, 1976, 135149. [3] R.C. Bose, “On the construction of balanced incomplete block designs”, Ann. Eugenics 9 (1939) 353399.
362
R. Mathon
[4] M.J. Colbourn, C.J. Colbourn, “On cyclic block designs”, Math. Report of Canadian Academy of Science 2 (1980) 2126. [5] M.J. Colbourn, R.A. Mathon, “On cyclic Steiner 2designs”, Ann. Discrete Math. 7 (1980) 215253. (61 M. Hall, Jr., Combinatom’al Theory, Blaisdell, Waldham, Mass. (1967). [7] H. Hanani, “Balanced incomplete block designs and related designs”, Discrete Math. 11 (1975) 255369. 181 M. Jimbo, S. Kuriki, “On a composition of cyclic 2designs”, Discrete Math. 46 (1983) 249255. [9] P.J. Laufer, “Regular perfect systems of difference sets of size 4 and extremal systems of size 3”, Ann. Discrete Math. 12 (1982) 193201.
[lo] R.
Peltesohn, “Eine L6sung der beiden Heffterschen Differenzenprobleme”, Compositio Math. 6 (1939) 251257. [Ill D.G. Rogers, “Addition theorems for perfect systems of difference sets”, J. Lond. Math. SOC.(2) 23 (1981) 385395. [12] J.M. Turgeon, “Construction of additive sequences of permutations of arbitrary lengths”, Ann. Discrete Math. 12 (1982) 239242. [13] R.M. Wilson, “Cyclotomy and difference families in elementary abelian groups”, J . Number Theory 4 (1972) 1747. [14] R.M. Wilson, “Constructions and uses of pairwise balanced designs”, Combinatorics (eds. M. Hall, Jr. and J.H. van Lint), Mathematical Centre, Amsterdam, 1975, 1942.
Annals of Discrete Mathematics 34 (1987) 363370
363
0 Elsevier Science Publishers B.V. (NorthHolland)
On the Spectrum of Imbrical Designs E. Mendelsohn and A. Assaf Department of Mathematics University of Toronto Toronto, Ontario, M5S 1Al CANADA TO A L E X R O S A O N T U f C L E U E N T U A N N I V E R S A R Y 0 3 U I S TUIR7YNINTU BIRTUDAY
ABSTRACT An imbrical design is a minimal, but not necessarily minimum, covering. Thus, a pair ( V , D ) is an ID(v,k,X,b) if and only if
(4 PI= v; (b) D is a collection of 6 Icsubsets of V called blocks; (c) every pair of elements of V is in at least X blocks; (d) for every 6 € D there is a pair {z,y}C6 so that z y is in exactly X blocks ({z,y) is called essential for 6); (d') equivalently, if D ' C D , (V,D') is not an imbrical design. Given w, k , and A, the spectrum Spec(v,k,X) = (6:there exists In this paper we determine Spec(v,3,1) for all w and Spec(v,4,1) except for ( I J , ~ ) , where w €{25,28,37,40,85), 6 = 1 21 u1 v(vl) 1. The results are that min Spec(v,3,1) = 12 3 2
ID(w,k,X,b)}.
+
max Spec(v,3,1) =
I.;'],
[[11,
and that Spec(v,3,1) is an interval except
that IJ3, 6=2 and w=7, 6=8 do not exist. For k=4 we h ve s'milar results: min ~pec(w,4,1)=
I[]], u1 max 4 3 2,
~pec(w,4,1)=
by2] and
Spec(v,4,1) is an interval except possibly for the noted values, and v=4, 6=2, v=13, 6=14 do not exist.
E. Mendelsohn and A. Assaf
364
1. Introduction
The idea of a maximal packing which is not maximum in the case k=3, X=1 is called a maximal partial triple system, and many papers have been written about them (see [1,2,5,7], for example). The concept has not been extended to k = 4, and seems to be quite difficult. In this paper, we dualize the results t o coverings and obtain results both for k=3, X=l, and for k=4, X=1. It is quite likely t h a t if the covering of pairs by quintuples were solved, we could extend our methods to them as well. W e introduce the new term imbrical design for a covering which is not necessarily the smallest, to avoid confusion. This is the same reason t h a t the term maximal partial triple system replaced the term maximal nonmaximum packing. A pair (V,D) is a n ID(v,k,X,b) if and only if (a)
PI= v ;
(b) D is a collection of b ksubsets of V called blocks; (c) every pair of elements of V is in at least X blocks; there is a pair {z,y}Cb so t h a t zy is in exactly (d) for every b called essential for b);
blocks ({z,y} is
(d’) equivalently, if D’CD, (V,D’) is not an imbrical design. Given v, k , and 1,the spectrum Spec(v,k,X) = {b:there exists ID(v,k,X,b)}. In this paper we determine Spec(v,3,1) for all v and Spec(v,4,1) except for ( v , b ) , where 1 vE{25,28,37,40,85}, b = v(v1) 1. The results are t h a t min Spec(v,3,1) = 12 21 v1 rTrTii, max ~ ~ ~ + , 3 , =. 1 ) I;’], and t h a t Spec(v,3,1) is a n interval except t h a t
+
1;j
and v=7, b=8 do not exist. For k= we have similar results: min 21 v1 and Spec(v,4,1) is an interval ~ p e c ( v , 4 , 1 )= max ~ p e c ( v , 4 , 1 )= 4 3 except possibly for the noted values, and v=4, b=2, v=13, b=14 do not, exist. v=3, b=2
[[]I,
2. Augmentations and interval designs
In this section we determine max Spec(v,k,l) for k=3 and k=4, and find a condition on an imbrical design with b blocks so t h a t if such a design exists then [b,max Spec(v,k,l)] Spec(v,k,l). These are called interval designs. The remainder of this section is devoted to showing t h a t (a) If k=3, v$1,3 mod 6, then the minimal covering is an interval design. (b) If k=3, v 3 , 3 mod 6, there exists an interval design with (c)
6 If k=4, v*1,4 mod 12, the minimal covering is an interval design.
(dl
If k=4, v 3 , 4 mod 12, an interval design exists with
+ 2 blocks.
vO + 2 blocks. 12
W e now give a basic augmentation procedure. Let (V,B) be an ID(v,k,l,b). Let a={ q , c q , * * * ql} E B be fixed. Let a be a block of B , a = {ao,ul, . . . ~ k  ~ } with { u ~ , u ~ essential ~} (i.e., in no other block). Thus not both uO,akl E m Assume
,
Spectrum of Im brical Designs
365
without loss of generality t h a t if one of them is in m, it is a,, and a, = qP1. The basic augmentation procedure is (1) Delete from B the block {a,,,
*
..u ~  ~ } .
} oq has the least i such t h a t (2) Add to B t h e block {cq,ao,al, * * ~ k  ~where cq €*a, provided there is an essential edge in this new block.
(3) Add to B t h e block {cq,al,a2, . * . ~ k  ~where } cq has the least i such t h a t q €*a, provided there is an essential edge in this new block. Add t h e block { q , c q , . . . q  3 , a o , a k  l } . This process yields A,(V,B,a). (4)
Lemma 1: A,(V,B,a) is a n ID(v,k,l,b+i) where i€{0,1,2}, and increases the number of blocks of the form { q , c q , q3,~,v}. Proof: As {ao,uk} is an essential edge of a , the block added in (4) must be present if the block a is deleted. 0
Theorem 1: Max Spec(v,k,l) = [v5+2].

Proof: Let V, = {q,* q3} u V", and B, = {{q* ,* q3,21,v}: u,w€V"}. It is clear t h a t every pair is covered at least once, and t h a t {u,v} is essenti 1 [prop r t y (d)] for {q, * * ~ ~ C  ~ , U , T J }Thus . (V,,B,) is an imbrical design with 6 = 1;+2].

Suppose
that
we
have
an
imbrical
design
with
a
fixed
block
{q,cq,. . . ql}, and a block a so t h a t a n 00 # {q,cq, * * * 0 0 ~  ~ } ;then form A,(V,B,a). This does not decrease the number of blocks, a n d increases the ... This process only terminates when we number of blocks containing {q, 0 achieve (V,,B,). 00=
We call an ID an interval design if there exists a block a* so t h a t A,(V,B,a*) is an ID(v,k,l,b+l), and further t h a t no augmentation A,(V,B,a') changes this property of augmenting by one.
Theorem 2: If an ID(w,k,l,b) is an interval design then Further, such designs exist for k = 3 and k = 4 , Spec(v,k,l) 2 for the cases where the BIBD exists).
+
[
6 , 1:+21]
Spec(v,k,l).
6 = min Spec(v,k,l) (or min
Proof: Any augmenta ion t h t increases t h e number of blocks increases it by one or by two. Since b1 ) then a x + b z + c ( m o d p t ) is a ptcycle if and only if b r l mod p and c r O ( m o d p ) for primes p >2. (For p = 2 , the condition becomes b =1 mod 4 , c #=0 mod 2.) Lambossy also established t h a t these regular cyclic subgroups are not conjugate in the affine group  but this is a trivial exercise. The above results seem to have been ignored by most, if not all, researchers and it was not until recently t h a t the BaysLambossy conjecture was disproved by N. Brand 131. 3. Cyclic Designs of Order p t
To construct isomorphic cyclic designs which are not multiplier equivalent one must first be able to construct cyclic designs which is, in general, a difficult and unresolved issue. However, R. M. Wilson [14] established t h a t for primes p , p =l(mod k(kl)), which are sufficiently large and satisfy t h e obvious necessary conditions, there exist cyclic (p, k, l) designs ( and hence cyclic (p,k,X) designs). Lemma 3.1: If there exists a cyclic (p,k,X) then there exists cyclic ( p f , k , X ) designs for all t >O which have multiplier automorphism P:x +bx (mod p t ) for b ~1 mod p. Proof: Let p be an odd prime, (i.e., msume k>2 ). The group of units in the ring of integers (mod p t ) has order p t  p1 and is cyclic. Let a be t h e generator of this multiplicative group. Note t h a t =JP')(mod p f ) is a generator for t h e subgroup of units m , m = 1 mod p . Moreover, a = ~ ( m o dp ) where is a generator in the multiplicative group of integers (mod p).
b
Isomorphism Problems for Cyclic Block Designs
387
Let A, = {z lz=i mod p} and let Bt denote the set of orbit representations in a cyclic (pt,k,X) design. For each block 1 ={i1,i2,. . . ,ik}EBl choose a block l'={zl,z2, . . * , z k } such that zj'ij mod p , for j = 1 , 2 ,...,k. Define
Ut
=
{bil'={bizl,biz2,
* * *
,bizk}b=o,l, *
*
,pt'
1)
and
and
where the computations are all mod pt and b is the generator of the multiplicative subgroup of units m , m =l(mod p). The proof that B, is a set of orbit representatives for a cyclic (pf,k,X) design follows inductively from the existence of B , and is straightforward. Note that Bt will always have some multiplier automorphism m=1 mod p , regardless of the choice of Btl. Also note that there are many different choices of I' for each Z EB, (essentially (p t  1 )k1 choices). From the previous discussion of P. Lambossy's work we can conclude that the designs constructed above will have regular cyclic subgroups and these will have isomorphic mates. N. Brand [4] shows that the isomorphism must be of a certain form and hence we can conclude that the fixed points of the isomorphism are all z EZ,t such that z GO (mod p). This also follows from Lambossy's characterization of the ptcycles, (P:z+bz+c,b=l mod p). We conclude that both the original design and its isomorphic mate will have the same identical cyclic subdesign on the set A,={z I z = O mod p}. Hence any multiplier isomorphism between these designs must be congruent (modulo pf') t o a multiplier automorphism of the cyclic subdesign. If we construct a cyclic design for which this is impossible then we have a design with an isomorphic mate which is not multiplier equivalent.
Lemma 3.2: If m is a multiplier automorphism of a cyclic (p,k,X) design (Z,,B) and m fixes one orbit, then m fixes a block in that orbit. Proof: Assume for I ={z1,z2,. . * ,zk}EB,ml =l+i (i.e. , ml ={mx Iz E l } and Z+i={z+i IzEZ} ) where all computations are reduced modulo p. Then m'l = l +(m'1+m'2+ . . . + l ) i = l for some T . But each orbit under the additive group is full having length p. Hence rn'l+ . * * + l = ( m r  l ) / ( m  l ) = O (mod p). Thus m'E1 mod p or i=O mod p , but r divides p  1 and since every suborbit under multiplication by m has length r or 1 , then k r + n * l = p and thus there is a block ,'Z with ml'=l'. Theorem 3.3: If there exists a cyclic (p,k,X) design with multiplier automorphism m then there exists a cyclic ( p t , k , X ) design with all multiplier automorphisms m'=m (mod p). Proof: We can assume that m generates the multiplicative automorphism group for the cyclic (p,lc,X) design (Z,,B) and that we can choose orbit representatives m ' l a such that l,ml, * ,m'l are in different orbits and m'"l=l whether or not
388 m r + l = 1
K.T. Phelps mod p . Let B , be the set of these orbit representatives and let a be the
generator of the multiplicative group in Z,t,t> l . Then a=% mod p where g is a generator for the multiplicative group in Z p . Let 1 = & ' 1 , ~ r 2 , . . . ,ark}@and 1 m (mod p ) . Let c =aP , and m'=cs (mod p ' ) . Define l'={crl,cr*, * * ,crk}€Ut for each 1 (as in Lemma 3.1). Since c ~g mod p then c s =aS=m mod p and for each c' El',gr El ,c' =ar(mod p). Moreover, c'l' corresponds to ml Proceed as in Lemma 3.1 so t h a t b;l'EUt for each I'EU, and Bt=pBt,UUt. Obviously, m ' r m mod p will always be a multiplier automorphism for Ut since m'=csbi mod p t for some i. The result now follows by induction on t assuming t h a t every m'=m mod p is a multiplier automorphism for B,,(mod pt').
=aS ml
el.
Corollary 3.4: If there exists a cyclic ( p , k , X ) design then there exists isomorphic cyclic (pf,k,X) designs which are not multiplier equivalent, for all t > l , where p r l mod k(k 1) and p is a prime. Corollary 3.5: For all primes p > c k a constant dependent on k , there exists isomorphic cyclic (p',k,X)designs, t >1 which are not multiplier equivalent. There are some composition methods for constructing cyclic designs (nm&,A) from cyclic design (n,k,X)and (m,k,X) (cf., Jimbo, Kuriki [S]). The simplest is the "direct product" which can be defined when Ic is a prime and n and m are relatively prime. Using the standard direct product construction we have:
Theorem 3.6: If there exist cyclic (n,q,X) and cyclic (m,q,X) designs, where q is a prime and n and m are relatively prime and if there are cyclic isomorphic designs (n,q,X) which are not multiplier equivalent. The above construction generalizes the approach of t h e author for t h e case k = 3 [13], which was originally established by N. Brand [3]. 4. Cyclic Designs ( p q , k , X )
Since a (n,2,1) design is just the complete graph we assume throughout t h a t k 2 3 . P. P. P i l f y ([10],[11]) proved t h a t if n and 4 ( n ) are relatively prime (where n is a product of distinct primes), then any isomorphic cyclic ( n , k , X ) designs must be multiplier equivalent. Thus we restrict our attention to the case n=pq where q lp 1,for primes p ,q. Alspach and Parsons [I] established t h a t any cyclic (pq,Ic,X)p >q having a sylow psubgroup of order p must be multiplier equivalent to any isomorphic cyclic ( p q , k , X ) design. Godsil [7] points out t h a t cyclic ( p q , k , X ) designs have automorphism groups t h a t are 2transitive or imprimitive. Combining these observations, we conclude t h a t in order to construct cyclic ( p q , k , X ) designs which have sylow psubgroups of order p t , t > l . The sylow psubgroups of such cyclic designs are equivalent to a cyclic code of length q over GF(p).W e use these to construct cyclic transversal designs. Lemma 4.1: There exists cyclic transversal designs (Z,,,B) with groups of size p and blocks of size q , where the sylow psubgroups of t h e automorphism group has order p 2 , whenever q (p1 and p is a prime (note: q need not be prime). Proof: Let P be a cyclic code of length q , dimension 2 over GFLp).. Since q b1, the qth roots of unity are in G F ( p ) ; then P = { s ( l , l , , . . , l ) + t ( l , ~ , r , ,r4') I s , t E Z p } where r 4 = l mod p . P has distance q1 , since the roots of the generator polynomial
Isomorphism Problems for Cyclic Block Designs
389
are r , r 2 , . . . ,vqP2. Now the groups of the design are G i = Z p X { i } , i = O , l , ...,q 1 and for each (zo,zl, * . * ,zql)Ep, we have a block {(z,O),(z,l), * * * , ( z , q  l ) } E B . This obviously gives us a cyclic transversal design with a sylow psubgroup isomorphic t o P. For each b =(bo,61,...,6q1)€P define a permutation c U * : ( z , i ) ~ ( z f b i , i ) ( m o dp,mod q ) . These permutations along with /3:(z,i)+(z,i+l)generate a cyclic group having the desired sylow psubgroup. Note t h a t for every nonzero m EZp,(m,l) is a multiplier automorphism for the above design.
P. P. P i l f y [lo] claims t h a t the automorphism group of the above design contains a subgroup with has nonconjugate cyclic subgroups. However, they are conjugate in the full group. If (Z,X.Z,,B) is the design constructed above then:
Lemma 4.2: For primes p , q , q b1, any cyclic design isomorphic t o ( Z p X Z q , B )as constructed above, must be multiplier equivalent to it. Proof: If (ZpXZ,,B’) is the isomorphic cyclic design with c~:(a:,i)+(z+l,i+l) (mod p,mod q ) as the pqcycle then the automorphism group contains a sylow psubgroup of order p 2 which is a cyclic code of length q over G F ( p ) having dimension 2 and distance q1. Since q b1 and (1,1,1,.:.,1)is in the code then * p ) then the two the generator polynomial has roots L , L ~ *, . . ,f2. Since r ~ r(mod codes are multiplier equivalent. The blocks in B’either correspond to the words in the code or a coset of it. Being cyclic, the latter is impossible. Hence is L E r ’ (mod p ) then (I$) is a multiplier automorphism for the design. Corollary 4.3: Isomorphic cyclic (pq,q,l) designs are always multiplier equivalent for primes p ,q with p >q. Proof: If q]p1 then p q and 4 ( p q ) are relatively prime; hence the corollary follows from P. P. Pilfy’s result ([lO],[ll]). If q b1, then we may as well assume by our previous discussion t h a t the sylow psubgroup has order p 2 and the automorphism group is imprimitive with Z p X { i } as the imprimitive sets (cf. [14]). This all means t h a t there must exist cyclic (p,q,l) designs on the sets Z p X { i } ,i = O , l , . . . ,q1. The sets Z p X { i } along with the p 2 blocks cutting across these sets forms a cyclic transversal design with a sylow psubgroup of order p 2 which is equivalent to a cyclic code. Hence the transversal design is (multiplier) isomorphic t o the design constructed in Lemma 3.1. The two isomorphic designs will have isomorphic cyclic subdesigns (p,q,l) which are multiplier isomorphic (mod p ) (Bays [2]) and isomorphic cyclic transversal designs which are multiplier isomorphic (mod q ) . Hence the designs are multiplier isomorphic. In general, for a cyclic ( p q , k , l ) design ( p , q primes, p > q ) to have a sylow psubgroup of order p t , t > l there must exist cyclic ( p , k , l ) and (q,k,l) designs. This follows easily from elementary properties of designs along with several remarks by Alspach, Parsons [l]. Lemma 4.4: If a cyclic ( m , k , l ) design does not exist for either m=p or m = q , p , q primes, then any two isomorphic cyclic ( p q , k , l ) designs will be multiplier equivalent. Proof: There must be a subdesign on the set of fixed points of any automorphism (subgroup). Assuming p > q and the sylow psubgroup has order p t , t>l , then each set Z p x { i } will be the set of fixed points for some subgroup of the sylow psubgroup [I]. Hence there must exist a cyclic ( p , k , l ) . The orbits under the sylow psubgroups are the sets Z p X { i } . If one block cuts across k of these sets then exactly p 2 blocks cut
K.T. Phelps
390
across them. The sets of groups which have exactly p 2 blocks cutting across from a cyclic (q,k,l) design. Hence it either of these cyclic subdesigns do not exist then the automorphism group cannot have a sylow psubgroup of order greater than p . The result now follows from a lemma of Alspach, Parsons 111, (cf. [14]). We now turn t o the main theorem of this section. The author has tried valiantly t o prove that isomorphic cyclic (pq,3,1) designs are multiplier equivalent. The basic difficulty was that it simply is not true as we shall see. Theorem 4.5: For primes p , q , q b1, if there exists a cyclic ( p , k , X ) design having multiplier automorphism r , with r q = 1 mod p and there exists a cyclic ( q , k , X ) which does not have l(mod q ) as a multiplier automorphism. Then there exist isomorphic cyclic ( p q , k , X ) designs which are not multiplier equivalent. Proof: Let ( Z p X Z q , B )be the cyclic transversal design as constructed in Lemma 4.1. We construct a cycle ( p q , k , X ) design by placing appropriate copies of the cyclic ( p , k , X ) and the cyclic ( q , k , X ) designs on the groups and blocks respectively. Define an isomorphism f : ( x , i ) + ( r  * z , i ) . (This is essentially the mapping that P. P. Pfilfy uses [lo].) The image of this mapping is an isomorphic cyclic ( p q , k , X ) design. Note that the cyclic subdesigns on Z,X{i} and {O}XZq remain unchanged. Hence any multiplier isomorphism must be congruent (mod q ) to a multiplier automorphism of the cyclic sub (q,k,X). On the other hand the mapping gives isomorphic cyclic transversal designs which are multiplier equivalent where the multiplier is (1,1) (mod p,mod 9). Hence these designs are not multiplier equivalent, since l(mod q ) is not an automorphism of the cyclic subdesign ( q , k , X ) .
Corollary 4.6: For all primes p = 7 mod 12, if q b1, q>3 and, q = l mod 6, there exists isomorphic cyclic (pq,3,1) designs which are not multiplier equivalent. Proof: The wellknown
block
automorphism group of order
transitive
cyclic ( p ,3,1) design
has multiplier
(cf. Phelps [13]) and 1 (mod q ) is never a
2 multiplier automorphism for a cyclic (q,3,1) design, q >3.
Note that the proof of the Theorem does not depend on whether q is a prime.
R. M. Wilson’s [15] constructions ensure that cyclic (p,k,X) designs exist where p ~1 (mod qiE(k1)) (infinitely often). The constructions also ensure that these designs will have the qth roots of unity as multipliers again infinitely often. 5. Conclusions Because R. M. Wilson’s results guarantee the existence of cyclic ( p , k , X ) designs of order p , we have: Theorem 5.1: For any k > 2 , X>O there exists isomorphic cyclic (m,k,X) designs which are not multiplier equivalent. Isomorphic circulant graphs on n vertices, are always multiplier equivalent for n = p q , ([I], [7], [14]) where p , q are distinct primes. the results of this paper establish that this is not true in general for isomorphic circulant (kuniform) hypergraphs on p q vertices when q b1.
Isomorphism Problems for Cyclic Block Designs
391
The most important question remaining is whether isomorphism of cyclic block designs is isomorphismcomplete or not (cf. [5]).
References
B. Alspach and T. Parsons, “Isomorphism of circulant graphs and digraphs”, Discrete Math. 25 (1979) 97  108. S. Bays, “Sur les systemes cyclique de triples de Steiner differents pour N premier (on puissance d u nombre premier) de la forme 6n 1, I1  111”, Comment. Math. Helv. 3 (1931) 22  41.
+
N. Brand, “On the BaysLambossy Theorem”, preprint. N. Brand, “Some combinatorial isomorphism theorems”, preprint. M. J. Colbourn and C. J. Colbourn, “Concerning the complexity of deciding isomorphism of block designs”, Discrete A p p l . Math. 3 (1981) 155  162. M. J. Colbourn and R. Mathon, “On cyclic Steiner Zdesigns”, Annals of Discrete Math. 7 (1980) 215  251. C. Godsil, “On Cayley graph isomorphism”, Ars Combinatoria 15 (1983) 231 246.

M. Jimbo and S. Kuriki, “On a composition of cyclic %designs”, Discrete Math. 43 (1983) 249  255.
P. Lambossy, “Sur une maniere de differencier les fonctions cycliques d’un form donne‘e”, Comment. Math. Helv. 3 (1931) 69  102. P.P. PBlfy, “On regular pronormal subgroups of symmetric groups”, Acta Math. A c a d . Sci. Hungar. 34 (1979) 287  292. P. P. Pilfy, “Isomorphism problems for relational structures with a cyclic automorphism”, preprint.
R. Peltesohn, “Eine Lasung der beiden Heffterschen Differenzen probleme”, Compositio Math. 6 (1939) 251  257. K. T. Phelps, “A construction of Steiner triple systems of order p””, Discrete Math., to appear. K. T. Phelps, “Isomorphism of circulant combinatorial structures”, preprint. R. M. Wilson, “Cyclotomy and difference families in elementary abelian groups”, J . Number Theory 4 (1972) 17  47.
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Annals of Discrete Mathematics 34 (1987)393408 0 Elsevier Science Publishers B.V. (NorthHolland)
393
Multiply Perfect Systems of Difference Sets D.G. Rogers Fernley House, The Green Croxley Green, WD3 3HT UNITED KINGDOM TO A L E X R O S A O N X I S 3I3TIETU BIRTUDAY
ABSTRACT
A family of difference sets of valency at least 2 is multiply perfect for c if every integer in a run of consecutive integers starting with c occurs a prescribed constant number of times in the difference sets of the family. The theory of perfect systems extends to multiply perfect systems with the advantages t h a t in practical situations we have a n economical check on our d a t a and t h a t theoretically we have a possible way of circumventing certain parity constraints. Associations with block designs and graph labelling also either carry over or generalize in a natural way. 1. A general problem in measurement
Quite often in making measurements a i , O2c1 or S=1, c=2 and m = 1 or 2. Note that since m is in M0,2cl exactly when m is not in Mo,2c1,the (m,3,c)'systems provided by Theorem 2 are examples of minimal systems as described in Section 1. It seems likely that Theorem 2 extends t o other values of 6 (compare 121; p.1041). Since A is always an integer when X is even, the sufficiency of the condition in Theorem 1 when 6=0 follows in general if it is sufficient for = l+f and m EM&1, for E = 0 or 1, by taking appropriate unions of systems for these values of A. But when 6 =O and h=2, the condition in Theorem 1 is just
D.G. Rogers
400
m >2c1; (6) and it is natural to conjecture that, in this case, (6) is indeed sufficient. We confirm this for all but finitely many exceptional values of m for each c z l as a consequence of Theorem 2 and some special constructions.
Theorem 3: There is an (m,3,~)~system at least when m
2 7(2c1).
For small values of c this is readily improved as follows. Theorem 4: For l < c 5 5 , there is an (m,3,~)~system for m22c1 except possibly for m = 6,7,11,14, when c = 3; m = 9,10,13, when c = 4; m = 10,11,14,19,23, when c = 5. We discuss Theorems 1 and 2 in Section 5 and establish Theorems 3 and 4 in Section 6. 5. Necessary conditions: proofs
Proof of Theorem 1: Let the sets A,. = {0,al(r),a2(r)}, l < r < X m , form an (m,3,~)~system.Then summing the elements of the difference sets D ( R , ) , 1 IU, 2 2; or G has the form o f Figure 2, where IU 2 3 and without loss of generality IU,, > lU, 2 3.
I
Figure 1
I
I
I
I
Figure 2
Proof: Let w , z be two distinct fixed points of a focal graph G . Then clearly any edge e must be incident with w or z since S, fixes both w and z . Now suppose that G has three distinct fixed points u,w,z. Since G has no isolated vertex and any edge
Some Remarks on Focal Graphs
413
incident with one of the three fixed points must also be incident with one of the other two, it follows that V ( G )= {a,w,z}. But no graph of order 3 has three fixed points. Again, let w,z be two fixed points. Then [w,z]p E ( G ) , otherwise S[w8zl = Aut G would fix wz # w , z , so that G would have three fixed points. It follows that every edge of G is incident with exactly one of w,z. In other words, if we put U = v,nv,, U, = V,\U, U, = Vz\U, then V,UUUU, = V(G)\{w,z} is an independent set. Thus G has the form given in Figure 1 or Figure 2 according as U is empty or not. 0 Note that in the case where G is connected the two fixed points are also cutpoints of G. Now consider a disconnected focal graph G . As can be seen from Figure 1 the components of G need not be focal. Suppose G has a nonfocal component H,. By definition of focality, Ho has at least one edge, say [zo,yo]. Let Hl (# Ho) be the component of G containing z := xoyo. Since any automorphism of Hl can be extended t o an automorphism of G it follows that z is a fixed point of Hl (and hence of G ) , and that z is the focus of every edge of Ho. Moreover, the stabiliser of any edge e of Ho cannot fix any vertex of Ho except possibly the two endpoints of e , Le., H , is (1,O)displacing. We sum this up in: (3.2) Proposition: A n y disconnected focal graph G is of one of t h e following three types: (1) A l l components of G are focal. (2) There are both focal a n d nonfocal components. I n this case, G h a s a unique f i x e d point z , and z lies in one of t h e focal components; every nonfocal component i s (1,O)displacing. ( 3 ) No component is focal. In this case G h a s two f i x e d p o i n t s and is t h e graph o f Figure 1.
We conclude this section with a characterisation of focal trees. (3.3)Proposition: A n y focal tree h a s one of t h e f o r m s given in Figure 5.
Figure 3
Proof: Let T be a focal tree. If T has a central edge eo = [ z l , z z ] then , eo is involutorial, otherwise z1,zPare two adjacent fixed points, contrary t o (3.1). However, if eo is involutorial, then Se, displaces all vertices of T , again a contradiction. It follows that
G.Sabidussi
414
T has a central vertex z and t h a t all edges of T are noninvolutorial. Hence given any edge e , S, fixes all vertices between e and z , which means that every vertex is at distance at most 2 from z. If a EV,, then any permutation of Vu\{z} belongs to S I ~ , ~ ] . Hence a is of degree at most 2. If z is incident with a pendant edge e l , then it IS incident with exactly two such edges since the stabiliser of el permutes all other pendant edges incident with z . Hence T has one of the forms given in Figure 3 depending on whether there are no vertices at distance 2 from z , no pendant vertices adjacent t o z , or both types of vertices exist. In the last two cases the number of vertices at distance 2 from z is 2 3, otherwise any S, fixes more t h a n one vertex not on e . 0 4.
T h e block structure of focal graphs
In this section we consider the blockcutpoint tree TG of a connected focal graph G . Recall t h a t the blockcutpoint tree of any connected graph is of even diameter and has a central vertex. An extremal block of G is one which is an endpoint of a longest path in T G . An extremal cutpoint is the unique cutpoint of G belonging t o an extremal block. A block is trivial if it consists of a single edge. (4.1) Lemma: I f G h a s a nontrivial extremal block, t h e n d i a m TG = 2 or
4.
Proof: Let B,a,B, . a,B, be a longest path in T G , where each B, is a block, and each ai is a cutpoint of G , and suppose t h a t n 2 2 . By hypothesis we may also assume t h a t B, is not a single edge. Hence there is an edge e EE(B,) which is not incident with a l . Then S, fixes B , and hence also al. This implies t h a t Sa, C SB,. For if there is a o E Su, with aB, # B,, then
Bna,
*

a2B,al(aBl)(oa2) . .  (oa,)(aB,)
is a path in TG of length 2(2n1), whence n 5 1, a contradiction. O n t h e other hand, Su, displaces all vertices of G not in B,. In particular, there is a T E S u l which displaces a2. Hence
Brian . . . B2a2B1(7a2)(7B2). . (Ta,)(rB,) length 2(2n2) and therefore n 5 2 as claimed. 0 *
is a path of
If d i a m T G = 2 , i.e., if TG is a star, then the center of T G is a cutpoint of G , say z . Clearly z is a fixed point of G. If d i a m TG = 4, then G has a central block B and all other blocks meet B. Moreover: (4.2) Lemma: I f G has a nontrivial extremal block, and i f d i a m TG = 4, t h e n the central block B i s likewise nontrivial; and i f G has a f i x e d point z , t h e n z i s a cutpoint belonging t o a nontrivial extremal block. Proof: If B is an edge, say [a,a’], then the stabiliser of any edge not incident with a or a’ fixes both a and a’, a contradiction. Now assume t h a t G has a fixed point z. Let B,, * * * ,B, be the extremal blocks of G , ai the unique cutpoint in Bi,i = 1, * * * ,n. Suppose t h a t z is not a cutpoint. If z belongs to some B;, take any edge e E E ( G ) \ F ( B i )not incident with a, (such an edge exists because d i a m TG = 4). Since z is a fixed point, S, must fix both z and a; which is impossibie. If z does not belong to any Bi, choose e in some nontrivial block B j such t h a t e is not incident with a j . Then S, fixes the two distinct vertices a j and
Some Remarks on Focal Graphs
415
z , the same contradiction as before.
It follows t h a t z = ai for some i . A t least one of the extremal blocks containing z must be nontrivial, for if not, take any e in a nontrivial block B k , e not incident with a k . Then S, fixes both a k and z , and by assumption the two vertices are distinct. 0 An example illustrating the situation described in Lemma (4.2) is the graph of Figure 4.
Figure 4 (4.3)Proposition: d i a m TG
5 8 for a n y connected
focal graph G .
Proof: Let B,a,B, . . . a,B, be a longest path in TG. By Lemma (4.1) we may assume t h a t every extremal block of G is trivial. Let A be the set of pendant edges of G incident with a , (i.e., the extremal blocks containing a,).
Case 1: I.1I = 2, say A = {e,e’}, where e = [ a l , u ] , e’ = [al,u’]. Then u a , = u’ and S, displaces all vertices other than al,u,u’; in particular, S, displaces a i , i 2 2 . Hence by the argument used in the proof of (4.1), TG contains a path of length 2(2n2), i.e., n 5 2.
I
Case 2: # 2. Take any e = [ a l , u ] € A . Then u a , is not incident with any edge in A . If u a l €V(B1)\{a2}, then S, fixes B 1but displaces a 2 . Hence TG contains a path of length 2(2n2), so t h a t n 5 2 as before. If ua, EV(Bi) with i 2 2 , then S, fixes a,, and hence u a l = a 2 . On the other hand, if n 3, then S, displaces a3. Hence TG contains a path of length Z(2n4) or 2(2n3), depending on whether S, does or does not stabilise B,. In either case n 5 4. 0
>
I.1
Note t h a t the case I = 1 can occur only if B , is trivial. For if A = { [ a , , ~ ] } , then Sa, = S, and hence the stabiliser of any edge of B , incident with a l would fix both u and a,, and this is possible only if B , consists of the single edge [al,a2]. We therefore obtain: i f d i a m TG = 6 or 8, t h e n the extremal blocks of G are single edges, and a n y extremal cutpoint a i s either incident w i t h at least three pendant edges or w i t h exactly one. I n the latter case the unique nonextremal block containing a is likewise trivial. If d i a m TG = 8, t h e n t h e central block of G i s nontrivial. The cases t h a t d i a m TG = 6 or 8 can actually occur. A n example for diameter 6 is the focal tree T, of Figure 3. For diameter 8 , take any connected, sharply focal graph H and at each x E V ( H ) attach a copy of T,, using z as the vertex of attach8. It follows t h a t d i a m TG = 8 ment. This yields a focal graph G with d i a m TG and hence t h a t H consists of a single block. We state this separately as:
>
416
G. Sabidussi
(4.4) Corollary: A n y connected, sharply focal graph i s %connected.
5. Structure near a fixed point
In this section we consider connected focal graphs hawing exactly one f i x e d point z. In such a graph all edges not incident with z have the same focus, viz. z. One can
not hope, therefore, t o obtain much information about G except in the vicinity of z. We shall derive some simple structural details about the part of G lying within distance 2 of z , but except in the bipartite case they are far from adding up t o a characterisation of even this small portion of G. The bipartite case, on the other hand, is close t o trivial. We shall need some terminology and notation. For i E N put
A, := {x E V ( G ) : d i s t ( x , z ) = i}, Bi := { x € A i : V,nAi+, #
0).
Given x € A i ,i 2 1 , define the set of upper [lower] neighbours of x t o be U, := V,nAi+, [L, := V,nA,,]. If U, # 0, x is t e r m i n a l .
G has the following simple properties: A, = { x z : x EB,}. M o r e precisely, i f x i s a neighbour of z , t h e n U, i s e m p t y or (1) consists o f t h e single vertex x z . Let y E U , and consider e = [ x , y ] . Then S, C S, = S[,,,] C S,,, i.e., S, fixes z and x z # x , z . The only way this can happen is if x z = y . (2) Let x , y EV,. If x i s t e r m i n a l and y i s not, t h e n x a n d y are nonadjacent. Otherwise [ x , y ] is a noninvolutorial edge and S[,,yl= SznSy would fix both z and YZIt is an immediate consequence of (2) that z i s a cutpoint of G i f and only i f it h a s t e r m i n a l neighbours (i.e., if # 0).
Alp,
Let Go,G, be the subgraphs of G induced by { z } U B , U U { A , : i B } and ( z } U ( A , \ B , ) , respectively. Clearly G o U G l = G and G o n G l = z . Both G o and G , may consist of several blocks. We shall consider G , first, assuming that # 0. Note that if e = [ x , y ] E E ( G , ) , then x y EV(G,). Otherwise x y EV(Go)\{z} and since the action of S, on G o is the same as that of the full automorphism group of G , it would follow that x y is a fixed point of G , a contradiction. P u t D := G,z. Since V ( D ) is the neighbourhood of z in G I , the structure of G , is given by that of D . The vertices of D are invariantly defined, hence ob E A u t D for every o E A u t G . Conversely, every automorphism of D can be extended t o an automorphism of G by taking the identity on V(G)\V(D). This says in S ( x ; D )= {ob:a€S[,,,l} for any x EV(D). Since particular that x z EV(G,)\{z} = V ( D ) and x z # x , it follows that S ( x ; D )fixes exactly one vertex of D different from x , i.e., D is pointfocal. Similarly, one shows that D is (1,O)displacing, using that x y = z for every [ x , y ] E E ( D ) . We sum this up as: (5.1) Lemma: G,z i s (1,O)displacing and pointfocal, t h e focal m a p being x  x z . Conversely, t h e Zykowsum of a n y point focal, (1,O)displacing graph and a onepoint graph c a n play t h e role of G I .
Alp,
417
Some Remarks on Focal Graphs
W e now t u r n to the consideration of Go, continuing the enumeration of simple facts begun above. (3)
b, I = 1 or 2 3 f o r a n y x E A , , i 2 2 ; Iu, 1 2 3 f o r a n y x EB,,i 2 2 . Let x EBi and suppose x has a unique upper neighbour u . Then for any w EL, the stabiliser of [w,x]fixes both u and z. Hence lU, I 2 2. If U, = {u,v}, then S~,,ul would fix t~ and z ; similarly for L,.
I f z EAi, i _> 2, and /L, I = 1, t h e n x is terminal. For if there is a y EU,, then Slz,vlfixes z as well as the unique lower neighbour of x . (5) Partition B , into equivalence classes by setting x y 9 x z = y z , where x , y E B , . By (1) the equivalence class containing x is L,, = V,,nV, = V,,nB,. (4)

Hence: a n y t w o distinct vertices in A, have disjoint lower neighbourhoods. (6)
If x EB,, t h e n S, does not stabilise a n y L,, u € A , , u # x z . If it did, then
S[,>,]= S, would fix both u and zz. Properties (3)(6) refer t o edges of G whose endpoints have different distances from z. For a bipartite graph this is the entire set of edges. In this case t h e information obtained above, in particular (5),(3),(6), describes G to within distance 2 from z. (5.2) Proposition: G i v e n a connected, bipartite, focal graph G having a unique f i x e d point z , let G , be t h e subgraph of G induced by t h e vertices of G w i t h d i s t ( x , z ) 5 2. Then G, is t h e sum of a f a m i l y of rooted graphs ( H i 1 z ; ) i E(sum ~ modulo identification o f the roots) such that I;) at m o s t one Hi is a focal tree and zi i t s central vertex; (ii) all other Hi’s are complete bipartite graphs of t h e f o r m K2,n,,n, # 2, and zi belongs t o t h e twoelement colour class; and (iii) every block of t h e form Kz,n,occurs at least in triplicate, i.e., if Hi 1~ KZsn,,t h e n there are at least t w o other indices j , k E I for w h i c h H j N Hk 2: Conversely, a n y s u c h graph is focal and if it i s not a tree, c a n be extended t o a bipartite focal graph of a n y diameter. The extension to arbitrary diameter 2 3 can be carried out as shown in Figure 5.
.....
F3
Figure 5
G.Sabidussi
418
Suppose t h a t for a given n # 2, G 2 has exactly r blocks Hill* * . ,Hi, isomorphic t o K2,n. Take the composition p = P[@,] of the discrete graph K , with a path P of any length and identify the top vertices of Hi,, . . . ,Hi, with the “endpoints” of P (u,v,w in Figure 5). This yields a bipartite focal graph whose diameter depends solely on the length of P. F3 is the simplest such graph. The graph of Figure 5 shows incidentally t h a t the number of foci in a focal graph is not tied to the diameter of the graph (it may be as low as 4). Also, at least for small diameters, there is no dependence on the connectivity. For any k 2 3 one can construct kconnected focal graphs of diameter 3 or 4 with exactly 5 or 7 foci, respectively. O n the other hand, it follows from property (1) earlier in this section t h a t in the presence of a unique fixed point z , any kconnected focal graph of diameter 2 5 must have at least k + l foci (viz. z and A2).
Acknowledgements I wish to thank Alex Rosa for having pointed out to me the existence of focal graphs and to have listened patiently t o numerous false conjectures of mine about them. Support of the Natural Sciences and Engineering Research Council of Canada, Grant A7315, is gratefully acknowledged. References [l] J. Feigenbaum, J. Herschberger, and A. Schaffer, “A polynomial time algorithm for finding the prime factors of Cartesian product graphs”, Discrete A p p l . Math. 12 (1985) 123138. [2] A. Rosa, in: Combinatorics 79 (Part 11) (M. Deza and I.G. Rosenberg, eds.) Annals of Discrete Math. 9 (1980) 307. [3] G. Sabidussi, “The composition of graphs”, Duke Math. J. 26 (1959) 693696. [4] G. Sabidussi, “Graph multiplication”, Math. 2. 72 (1960) 446457. [5] P. Winkler, “Factoring a graph in polynomial time”, European J. Combin., t o appear.
Annals of Discrete Mathematics 34 (1987) 419436 0 Elsevier Science Publishers B.V. (NorthHolland)
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Some Perfect OneFactorizations of K,, E. Seah and D.R. S t i n s o n Department of Computer Science University of Manitoba Winnipeg, Manitoba, R3T 2N2 CANADA TO A L E X R O S A O N X I S 3I3TICTu BIRTUDAY
ABSTRACT We use orderly algorithms to generate 16 new perfect onefactorizations of KI4. This provides a lower bound of 20 nonisomorphic perfect onefactorizations of K,, (the previous bound was 4). 1. Introduction
Let Gr be an rregular graph on n vertices. A onefactorization of G r is a partin tion of the edgeset of Gr into r onefactors, each of which contains  edges that 2 partition the vertex set of G r . A complete graph on n vertices, K,, is (n1)regular.
A perfect onefactorization (or P1F) is a onefactorization in which every pair of distinct onefactors forms a Hamiltonian cycle of the graph. PlFs are known t o exist on K,,, and K,,, where p is a n odd prime. These are called GKp+l and GA,,, respectively; see [7] and [2]. PlFs are also known to exist on K16,K z s ,K244,and K3,, (see 111). Recently, the authors discovered PlFs for K,, (in [lo]) and K5,, (unpublished). No examples of a PlF are known for any other values of n . I t has been conjectured t h a t every K,, has a P l F , but this appears to be a very difficult problem. A summary of known results on PlFs is given in [ 8 ] . Denote by N ( n ) the number of nonisomorphic PlFs of K,. known:
The following is well
Theorem 1.1 [Q]:N ( 4 ) = N ( 6 ) = N ( 8 )= N(10) = 1, and N(12) = 5. 0 In this paper, we investigate the number of nonisomorphic PlFs of K14. Four nonisomorphic PlFs of K,, were shown t o exist in [l].In this paper, we construct 16 new PlFs and hence improve the lower bound to 20. Theorem 1.2: N(14) 2 20.
420
E. Seah and D.R. Stinson
We also compute the automorphism groups of these PlFs. We find examples where the automorphism group has order 2, 3, 4, 6, 12, 84, and 156. It is interesting t o note that of all the 20 PlFs known so far, none is automorphismfree. The existence of an automorphismfree P1F for K,, would lead one to suspect that there might be some of these for K14. The basic methods used to establish the results of this paper are orderly algorithms. We consider only nonisomorphic PlFs, by eliminating isomorphic structures a s they are being constructed. These algorithms are described in the remainder of the paper.
2. Orderly algorithms for generating PlFs of a complete graph In this section, we outline two orderly algorithms that can be used to generate all the (nonisomorphic) PlFs of a complete graph K2,. These are modifications of the orderly algorithm described in [Ill. We first need t o define orderings on edges, onefactors, etc., of Kzn. All orderings are defined lexicographically, as follows. Suppose that the vertices are numbered 1, ..., 2n. An edge e will be written as an ordered pair ( p , p ’ ) with l l p + ’ g n . For any two edges e l = ( p 1 , p f l ) and e2 = ( p , , ~ ’ , ) , we say el < e2 if either of the following is true: (i) pl3, the dicyclic group Q2” can be symmetrically sequenced.

Given a symmetric sequencing of a group G , one can construct a onefactorization (not necessarily perfect) of K ~ I (see + ~ [ 3 ] ) . It thus seems hopeful t h a t symmetric sequencings of Q2” can be used to construct PlFs of K4n+2.However, it remains to be seen whether symmetric sequencings will give us a new class of PlFs.
Added in Proof: The P1F of K,, is presented in “A perfect onefactorization of K,,”, by E.C. Ihrig, E. Seah and D.R. Stinson, J . Comb. Math. Comb. Comput. 1 (1987) 217219. As well, an exhaustive enumeration of PlFs of K , , having nontrivial automorphism groups has recently been done by the authors. This work is contained in “ O n the enumeration of onefactorizations of the complete graph containing prescribed automorphism groups”, Math. Comp., to appear. There are a total of 21 such PlFs, including the 20 presented in this paper. Acknowledgements We would like to thank Bruce Anderson and Edwin Ihrig for their useful comments. As well, we are indebted to the referee for pointing out reference [ 6 ] . References [ I ] B.A. Anderson, “Some perfect onefactorizations”, Proc. Seventh Southeastern C o n f . Combinatorics, Graph Theory, Computing, 1976, pp. 7991. [2] B.A. Anderson, “Symmetry groups of some perfect onefactorizations of complete graphs”, Discrete Math. 18 (1977) 227234. [3] B.A. Anderson, “Sequencings of dicyclic groups”, Ars Combinatoria 23 (1987) 13 1 142.
E. Seah and D.R.Stinson
4 26
[4] E.C. Ihrig, “Symmetry groups related t o the construction of perfect one factorizations of K2n”,J . Comb. Theory B40 (1986) 121151. [5] E.C. Ihrig, “The structure of symmetry groups of perfect one factorizations of K2,,”, to appear. [6] N.P. Korovina, “On the construction of closed triples of pair groups of order 12” (in Russian), Kombinatornyi Anal. 2 (1972) 4245. [7] E. Lucas, Recre‘ations Mathkmatiques, Vol. 2, GauthierVillars, Paris, 1883 (Sixir5me Recre‘ation, Les Jeux des Demiselles, 161197). [8] E. Mendelsohn and A. Rosa, “Onefactorizations of the complete graph vey”, J . Graph Theory 9 (1985) 4365.

a sur
[9] L.P. Petrenyuk and A.Y. Petrenyuk, “Intersection of perfect onefactorizations of complete graphs”, Cybernetics 16 (1980) 6 9 .
[lo] E.
Seah and D.R. Stinson, “A perfect onefactorization of K36”lDiscrete Math., t o appear.
[ll] E. Seah and D.R. Stinson, “An enumeration of nonisomorphic onefactorizations and Howell designs for the graph Klo minus a onefactor”, Ars Combinatoria 21 (1986) 145161.
Appendix Set 1:
1.2 I = 84 (GA14)
A =
91992 91 = (1 12 11 8 3 7)(2 14 13 10 5 9) 92 = (1 13 4 9 8 5 12 2 11 6 7 10 3 14)
91
92
induces induces
( f 2 f 4 f8 f l 2 f l l f7 f31lf.5 f€if(f9 f10)(f13 f14) ( f 2 f 3 f 4 f l l f 7 f 1 2 ) ( f 5 f l 0 f 1 4 f 6 f 9 f13)
1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 110 111 112 113 114
3 4 2 5 2 6 2 4 2 3 2 9 210 2 8 2 7 213 214 212 211
5 6 4 7 3 8 3 9 410 3 5 312 313 3 6 3 7 311 314 310
7 8 6 9 510 6 8 5 7 411 4 6 4 5 414 412 4 8 4 Q 413
910 811 712 713 814 613 ,514 612 511 5 9 513 5 8 512
1112 1013 914 1012 911 812 711 714 813 614 610 611 6 7
1314 1214 1113 1114 1213 1014 913 1011 912 810 7 9 710 8 9
Perfect OneFnctorizations
421
Set 2: bI=4. A 22x2, g1 = (1 2)(3 5)(4 6)(7 9)(8 10)(11 13)(12 14) 92 = (1 2)(3 6)(4 5)(7 10)(8 9)(11 14)(12 13) 9 1 induces (f.5 f 6 ) ( f 9 f l O ) ( f l 3 f 1 4 ) 92 induces ( f 3 f 4 ) ( f 7 f 8 ) ( f l l f12) 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 110 111 112 113 114
3 4 2 5 2 6 2 4 2 3 2 9 210 2 8 2 7 213 214 212 211
5 6 4 7 3 8 3 9 410 313 3 5 312 3 6 314 3 7 311 310
7 8 6 9 510 6 8 5 7 4 6 414 4 5 411 4 8 413 4 9 412
910 811 712 713 814 511 612 613 514 512 5 9 5 8 513
1112 1013 914 1012 911 812 711 714 813 610 611 614 6 7
1314 1214 1113 1114 1213 1014 913 1011 912 7 9 810 710 8 9
Set 3:
I.2 I = 12.
A =Z2XZ6 A = < 91792
>
91 = (3 11 10 4 12 9)(5 13 8 6 14 7) 9 2 = (1 2)(3 6)(4 5)(7 10)(8 9)(11 14)(12 13) 91 induces ( f 3 fii f i o f 4 f 1 2 f g ) ( f s f 1 3 1 8 f 6 f 1 4 f7) 92 induces ( f 3 f d f l l f l 2 ) ( f 9 f l 0 )
1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 110 111 112 113 114
3 4 2 5 2 6 2 4 2 3 210 2 Q 2 7 2 8 213 214 212 211
5 6 4 7 3 8 3 9 410 3 6 312 313 3 5 314 3 7 311 310
7 8 6 9 510 6 8 5 7 411 4 5 4 6 414 4 8 413 4 9 412
910 811 712 713 814 514 613 511 612 512 5 9 5 8 513
1112 1013 914 1012 911 813 714 812 711 610 611 614 6 7
1314 1214 1113 1114 1213 912 1011 1014 913 7 9 810 710 8 9
E. Seah and D.R. Stinson
428
Set 4: b[=6. A 2: Z6 A= g1 = (3 10 7 4 9 8)(5 14 12 6 13 11) 91 induces f f 3 f l 0 f 7 f 4 f 9 f 8 ) ( f S f 1 4 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 1 10 111 112 113 114
3 4 2 5 2 6 2 4 2 3 212 211 213 2 14 2 7 2 8 210 2 9
5 6 4 7 3 8 3 9 410 311 3 6 310 3 '5 314 3 7 312 313
f l 2 f6 f 1 3 f l l )
7 8 6 9 510 6 8 5 7 4 5 412 4 6 4 9 4 8 413 414 411
910 811 712 713 814 613 514 511 6 12 513 5 9 5 8 512
1112 1013 914 1012 911 8 9 710 714 7 11 610 614 611 6 7
1314 1214 1113 1114 1213 1014 913 812 8 13 912 1011 7 9 810
Set 5: I.11=6.
A
N
26
A = < g i > g1 = (3 10 7 4 9 8)(5 13 11 6 14 12) 91 induces ( f 3 110 f7 f 4 f 9 f8)(f5 /13
1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 110 111 112 113 114
3 4 2 5 2 6 2 4 2 3 211 212 214 213 2 8 2 7 2 9 210
5 6 4 7 3 8 3 9 410 312 3 5 310 3 6 3 7 313 314 311
f l l f6 f 1 4 f12)
7 8 6 9 510 6 8 5 7 4 6 411 4 5 4 9 414 4 8 412 413
910 811 712 714 813 513 614 611 512 5 9 514 511 5 8
1112 1013 914 1011 912 8 9 710 713 711 613 610 6 7 612
1314 1214 1113 1213 1114 1014 913 812 814 1012 911 810 7 9
429
Perfect OneFactorizations Set 6: b(=6.
A
2: Z6
A = < g i >
g 1 = (3 14 12 4 13 11)(5 9 8 6 10 7) g1 induces ( f 3 f 1 4 f 1 2 f 4 f 1 3 f d f 5 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 110 111 112 113 114
3 4 2 5 2 6 2 4 2 3 212 211 213 214 2 7 2 8 210 2 9
5 6 4 7 3 8 3 9 410 310 313 3 7 312 314 3 5 311 3 6
fg
f a fs
7 8 6 9 510 6 8 5 7 414 4 9 411 4 8 4 6 413 4 5 412
flo f7)
910 811 712 714 813 511 514 512 5 9 5 8 6 7 614 513
1112 1013 914 1011 912 613 612 610 611 913 911 7 9 711
13 14 12 14 11 13 12 13 11 14 8 9 7 10 8 14 7 13 10 12 10 14 8 12 8 10
Set 7: (A(=2.
A u2 2 A = < g l >
g1 = (3 4)(5 6)(7 8)(9 10)(11 12)(13 14)
g 1 induces
(.f3 f 4 ) ( f 5 f 6 ) ( f 7 f a U 9 flo)(fll
1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 110 111 112 113 114
3 4 2 5 2 6 2 4 2 3 212 211 2 8 2 7 213 214 2 9 210
5 6 4 7 3 8 3 9 410 3 6 314 3 7 312 310 313 311 3 5
7 8 6 9 510 614 513 413 4 5 411 4 8 414 4 9 4 6 412
f l 2 ) ( f l 3 f14)
910 811 712 713 711 511 612 514 5 9 512 5 7 5 8 6 7
1112 1013 914 812 814 8 9 710 610 613 6 8 611 714 813
13 14 12 14 11 13 10 11 9 12 10 14 9 13 12 13 11 14 7 9 8 10 10 12 9 11
E. Seah and D.R.Stinson
430
Set 8: b)=2.
A 21 2, A= g1 = (3 4)(5 6)(7 8)(9 10)(11 12)(13 14)
g1 induces ( f 3 f 4 ) ( f 5 f 6 ) ( f 7 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 110 111 112 113 114
Set 9:
3 4 2 5 2 6 2 4 2 3 2 9 210 211 2 12 214 213 2 8 2 7
f8)(f9
5 6 4 7 3 8 311 412 3 6 314 310 3.7 312 3 5 3 9 313
flO)(flI
7 8 6 9 510 613 514 413 4 5 4 8 4 9 4 6 411 414 410
f12)(f13
910 811 712 7 9 711 5 8 6 7 512 513 5 9 610 5 7 511
f14)
1112 1013 914 812 810 1012 911 614 6 11 710 714 612 6 8
13 14 12 14 11 13 10 14 9 13 11 14 12 13 7 13 8 14 8 13 8 9 10 11 9 12
I
= 12. A 2: Q6 (dicyclic group) A = < 91, 92 > g1 = (3 5 4 6)(7 13 8 14)(9 11 10 12) g2 = (3 13 10 4 14 9)(5 11 7 6 12 8) 91 induces ( f 3 f 5 f 4 f 6 ) ( f 7 f 13 f 8 f l 4 ) ( f 9 f I I 92 induces ( f 3 1 1 3 f l 0 f4 f 14 f 9 ) ( f 5 fll f 7 f 6
1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 110 111 112 113 114
3 4 2 5 2 6 2 4 2 3 2 9 210 2 8 2 7 214 213 211 212
5 6 4 7 3 8 311 412 314 3 7 313 3 5 3 6 310 312 3 9
7 8 6 9 510 613 514 4 8 413 4 6 414 4 9 4 5 410 411
fl0 fl2)
f l 2 f8)
910 811 712 7 9 711 511 5 9 512 611 5 8 6 7 5 7 513
1112 1013 914 812 810 610 612 714 813 713 814 614 6 8
1314 1214 1113 1014 913 1213 1114 1011 912 1012 911 8 9 710
431
Perfect OneFactorizations Set 10:
I = 12. A 2:& 6 (dicyclic group) A = < 911 9 2 > g1 = (3 5 4 6)(7 12 8 11)(9 13 10 14) 92 = (3 7 14 4 8 13)(5 10 11 6 9 12) 91 induces (f3 f 5 f 4 f 6 ) ( / 7 f 1 2 f 8 f l l ) ( f 9 f13 9 2 induces (f3 f 7 f14 f 4 f 8 f 1 3 ) ( f 5 f l 0 f l l f 6 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 110 111 112 113 114
3 4 2 5 2 6 2 4 2 3 210 2 9 2 7 2 8 213 214 212 211
5 6 4 7 3 8 313 414 311 3 ' 6 312 3 7 3 5 310 314 3 9
7 8 6 9 510 612 511 4 5 412 4 8 411 4 9 4 6 410 413
Set 11: bI=6. A 2: Z6 A= g1 = (3 7 11 4 12 8)(5 10 14 6 9 13) 91 induces (f3 f l l f 7 1 4 f 1 2 f 8 ) ( f 5 f l 0 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 110 111 112 113 114
3 4 2 5 2 6 2 8 2 7 214 213 2 4 2 3 210 2 9 212 211
5 6 4 7 3 8 310 314 3 9 312 3 5 4 6 3 7 311 3 6 313
f l 0 f14) f 9 f12)
910 811 712 711 710 613 514 513 5 9 6 8 5 7 5 8 512
1112 1013 Q14 8 9 812 814 713 610 614 714 813 611 6 7
1314 1214 1113 1014 913 912 1011 1114 1213 1012 911 7 9 810
f 1 4 f 6 f 9 f13)
7 8 6 9 510 413 4 9 411 410 611 512 412 4 8 414 4 5
910 811 712 612 511 513 5 9 710 713 514 5 7 5 8 6 7
1112 1013 914 714 813 610 614 814 8 9 6 8 613 7 9 810
1314 1214 1113 911 1012 812 711 1213 1114 913 1014 1011 912
E. Seah and D.R. Stinson
432 Set 12: I.i\=6.
A
~
2
6
A= g1 = (3 8 13 4 7 14)(5 9 12 6 10 11) 91 induces ( f a f 8 f13 f 4 f7 f 1 4 ) ( f 5 f!3
Set 13:
f l 2 f6 f l 0 f l l )
1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 110 111 112 113 114
3 4 2 5 2 6 2 8 2 7 210 2 9 213 214 2 3 2 4 212 211
5 6 4 7 3 8 313 3 5 312 3 6 3.11 3 7 410 3 9 314 310
7 8 6 9 510 4 6 414 4 5 411 4 8 412 5 9 5 8 4 9 413
910 811 712 711 812 614 513 514 511 6 7 610 5 7 512
1112 1013 914 912 913 813 714 612 613 814 713 611 6 8
13 14 12 14 11 13 10 14 10 11 9 11 10 12 7 10 8 9 12 13 11 14 8 10 7 9
1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 1 10 1 11 1 12 1 13 1 14
3 4 2 5 2 6 2 9 210 213 214 212 211 2 8 2 7 2 4 2 3
5 6 4 7 3 8 3 7 312 311 313 314 3 9 310 3 5 3 6 4 5
7 8 6 9 510 411 4 8 414 412 410 413 4 6 4 9 5 8 6 7
910 811 712 613 514 5 9 511 513 5 7 512 611 710 8 9
1112 1013 914 814 713 612 610 6 8 614 714 813 912 1011
1314 1214 1113 1012 911 810 7 9 711 812 913 1014 1114 1213
I = 156.
Perfect OneFactorizations
433
Set 14:
b I = 12.
A
N
212
A= g 1 = (3 5 9 8 14 12 4 6 10 7 13 11) 91 jnduces ( f 3 f 5 f9 f8 f14 f12 f4 f6
1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 110 111 112 113 114
3 4 2 5 2 6 2 9 210 213 214 2 8 2 7 2 3 2 4 211 212
5 6 4 7 3 8 314 311
310 3 7 313 3 ' 5 4 5 3 6 312 3 9
f l 0 f7 f13 f l l )
7 8 6 9 510 412 413 4 8 4 9 4 6 414 6 7 5 8 410 411
910 811 712 613 514 5 9 511 512 611 814 713 5 7 513
1112 1013 914 711 7 9 612 610 714 813 913 911 614 6 8
1314 1214 1113 810 812 1114 1213 1011 912 1012 1014 8 9 710
1112 1013 914 713 7 9 813 714 610 614 8 9 710 711 812
13 12 11 8 8 10 9 11 12 10 9 9 10
Set 15:
b I = 12.
A
212
A= g1 = (3 5 14 9 7 11 4 6 13 10 8 12) 91 induces (f3 f 5 f14 f g f 7 fll f 4 f 6 f 1 3 f l 0 f 8
1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 110 111 112 113 114
3 4 2 5 2 6 214 213 211 212 2 7 2 8 2 4 2 3 210 2 9
5 6 4 7 3 8 3 9 311 3 5 310 312 3 7 313 414 314 3 6
7 8 6 9 510 412 410 4 9 4 6 4 8 411 514 5 8 4 5 413
f12)
910 811 712 611 512 612 511 513 5 9 6 7 613 6 8 5 7
14 14 13 10 14 14 13
14 13 12 11 12 11
E. Seah and D.R.Stinson
434 Set 16: bI3.
A
N
Z3
A= g 1 = (1 5 14)(3 11 13)(4 8 7)(6 12 10) 91 induces (f? f 3 f l O ) ( f 4 f 9 f12)(f5 f13 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 110
I l l 112 113 114 Set 17:
3 4 2 5 2 6 2 4 2 3 211 2 9 212 214 213 210 2 8 2 7
5 6 4 7 3 8 310 412 314 312 313 3 6 3 9 311 3 7 3 5
fl4)(fS
7 8 6 9 510 612 513 4 6 414 410 4 8 4 5 413 4 9 411
f7 f l l )
910 811 712 7 9 710 512 5 7 5 8 511 6 7 5 9 514 610
1112 1013 914 813 814 810 613 614 713 812 6 8 611 8 9
1314 1214 1113 1114 911 913 1011 711 912 1014 714 1012 1213
I = 3.
A = Z3 A=
g1 = (1 8 3)(2 9 14)(4 11 5)(7 13 10) 91 induces ( f 2 fll f10)(f3 f8
1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 110 111 112 113 114
3 4 2 5 2 6 2 4 214 211 212 2 8 213 210 2 9 2 7 2 3
fdf5 5 6 4 7 3 8 3 9 312 310 3 5 311 314 313 3 7 3 6 4 8
f14 f 9 ) ( f 6
7 8 6 9 510 613 4 5 414 4 9 413 412 4 6 411 410 5 7
f 1 2 f13)
910 811 713 712 711 5 9 614 514 5 8 512 513 511 610
1112 1013 912 814 810 612 710 6 7 611 714 6 8 812 911
1314 1214 1114 1011 913 813 1113 1012 7 9 8 9 1014 914 1213
Perfect OneFactorizations
435
Set 18:
bI=X
A
N
23
A = < g 1 >
g1 = (1 5 9)(2 7 14)(4 6 8)(10 13 11) 91 induces
( f a f13 f 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 110 111 112 113 114
d f 3
fs
3 4 2 5 2 6 2 4 213 210 212 2 7 214 2 3 2 9 211 2 8
/14)(fS
5 6 4 7 3 8 313 312 314 3 5 310 311 4 8 3 7 3 6 3 9
f1o f9)(f7 f l l 112)
7 8 6 9 511 612 4 9 411 410 412 413 514 4 6 414 4 5
910 811 710 714 5 8 512 614 513 5 9 610 510 5 7 613
1112 1013 914 810 711 6 8 7 9 611 6 7 713 813 8 9 712
13 14 12 14 12 13 9 11 10 14 9 13 11 13 8 14 8 12 9 12 11 14 10 12 10 11
Set 19: bI=3.
A
=Z3
A= g1 = (1 4 7)(2 11 5)(3 12 10)(6 8 14) 9 1 induces (1'2 f 1 4 f 1 0 ) ( f 3 f 7 f 4 ) ( f 5 1 1 1 1 1 1 1 1
2 3 4 5 6 7 8 9
110 111 112 113 114
3 4 2 5 2 6 213 2 7 211 212 2 4 2 3 214 214 2 9 2 8
5 6 4 7 3 8 314 310 313 3 5 3 7 4 6 312 3 6 311 3 9
f 9 f12)(f6
7 8 6 9 511 410 413 412 414 5 8 5 7 4 8 4 9 4 5 411
f l l f13)
910 811 710 611 512 514 6 7 612 813 5 9 513 6 8 510
1112 1013 914 7 9 814 610 913 1014 912 614 711 714 613
13 14 12 14 12 13 8 12 9 11 8 9 10 11 11 13 11 14 7 13 8 10 10 12 7 12
E. Seah and D.R. Stinson
436
1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 1 10 111 112 113 114
3 4 2 5 2 6 2 7 212 2 9 214 2 8 2 11 213 2 4 2 3 210
5 6 4 7 3 8 3 6 311 314 310 312 3' 9 3 7 313 4 8 3 5
7 8 6 9 511 410 414 4 5 411 413 4 6 4 9 514 512 412
910 811 713 812 5 9 6 8 5 7 510 5 13 5 8 6 7 610 613
1112 1013 912 914 710 1011 612 611 7 12 614 810 7 9 711
1314 1214 1014 1113 813 1213 913 714 8 14 1012 911 1114 8 9
Annals of Discrete Mathematics 34 (1987) 437440 0 Elsevier Science Publishers B.V. (NorthHolland)
437
A Construction for Orthogonal Designs with Three Variables Jennifer Seberry Department of Computer Science University College The University of New South Wales Australian Defence Forces Academy Canberra, A.C.T. 2600 AUSTFtALIA TO A L & X R O S A
O N X I S 3IITTIETU BIRTUDAY
ABSTRACT We show how orthogonal designs OD(48p2t;16p2t, 16p2t ,16p2t) can be constructed from an Hadamard matrix of order 4p and an O D 4 t ; t , t , t , t ) . This allows us t o assert that OD(48p2t; 16p t,16p2t,16p2t) exist for all t , p 5 102 except possibly for t E {67,71,73,77,79,83,86,89,Yl,Q7}.These designs are new.
1
1. Introduction
Let N = (hij) be a matrix of order h with h;j E { l ,  l } . H is called an Hadamard matrix of order n , if H H T = hI,, where I, denotes the identity matrix order of h . An orthogonal design A , of order n , type ( p l , p 2 , .. . , p , ) , denoted O D ( n ; p l , p 2 , .. . , p , ) , on the commuting variables ( * x l , h 2 , . . . ,+rU,O) is a square matrix of order n with entries * x k where each xk occur p k times in each row and column such that the rows are pairwise orthogonal.
In other words A A T = (pis:
+ + p"X;)I,. *.*
It is known that the maximum number of variables is an orthogonal design is p ( n ) , the d , 0 L d 0, and we may take the first two blocks to be 12345, 12678. First, suppose t h a t no pair z'j ever occurs three times; then the blocks can be written in the form 12345, 12678, ISTXX, 29TXX, BXXXX, TXXXX. If B, is 19T34, then the last blocks would have to be 93678 and T4678; this is not possible. Hence there is no loss of generality in taking B3 as 19T36 and B, as 29T47. This forces t h e final blocks to be 95837 and T5846. The excess graph is shown in Figure 5; it is simply t h e ubiquitous Petersen graph. Now suppose t h a t some pair, say 12, can occur three times. W e may then start with t h e blocks 12345, 12678, 129T3. The remaining blocks must be 3678X, Xxxxx, Xxxxx. Since (45) and (9T) are equivalent pairs, we may take B, to be 34678; then the remaining blocks must be 9T546 and 9T578. The excess graph is shown in Figure 6. W e sum u p the results of this section in Theorem 1: If w = 10, then N(v) = 6, and there are two covering designs, namely, 12345, 12678, 1369T, 2479T, 35789, 4568T, and 12345, 12678, 1239T, 34678, 4569T, 5789T. 0
W e note t h a t the first design is embeddable in the standard BIBD with w = 11, k = 5, and X = 2 (add the blocks 1489E, 157TE, 238TE, 25693, 34673).
Small CoveringDesigns with Block Size Five
445
Figure 5
Figure 6 4. The Case v = 11.
If w = 11, N(w) = 7, and / ( i ) > 2 for all i. Suppose, if possible, t h a t f ( E ) = 5; then the blocks are 12345, Xxxxx. Consider now t h e sets A = {1,2,3,4,5} and B = {6,7,8,9,T}; then the number of pairs a i b j ( a i  ; b j a ) is 25. Each of the first five blocks contributes 0, 3, or 4 such pairs; the last block contributes 0, 4, or 6. We at once see t h a t the only possibilities are 25 = 6 4(4) 3, or 26 = 6 t 5(4). However, we also need ten pairs bibj, and (in both cases) there are only four pairs from the four blocks t h a t have a (2,2) split from A and B . I t follows t h a t the last block must have the form aabbb, and t h e fifth block must have the form Eabbb. We thus have a total of 16 elements from A and 14 from B in the cover; this is a contradiction, since j ( i ) > 2 for i selected from B. We thus see t h a t no element can have frequency 5; hence there must be nine elements of frequency 3, and two elements of frequency 4.
+
+
R.G. Stanton
446
We now consider the possibility that 2,X; let the first two blocks be 12345, 12346. At most two of the elements 1,2,3,4, can have frequency 4; so we may let f(1) = 3. Then 1 must appear with 7,8,9,T,E in a single block, and this is certainly impossible. Thus z4 = 0 in all cases. If x o X , we can write the blocks as 12345, 6789T, (xxxxX)2.(It is possible that a symbol E appears in one of the last two blocks.) If we set A = {1,2,3,4,5} and B = {6,7,8,9,T}, then there must be 25 pairs a i b j . But the maximum number of such pairs is 3(4) 2(6) < 25. Hence we conclude that zo = 0.
+
Now let T and E be the symbols of frequency 4, and consider the block TE123. The intersection numbers of this block satisfy the equations 21
+ 5 2 + 2 3 = 6, 2 1 + 2x2 + 323 = 12.
Either z 3 X or x2 = 6. In the latter case, we can write the six other blocks in the form aabbb, where A = {T,E,1,2,3} and B = {4,5,6,7,8}. Thus we have to have six triples that cover all pairs of the form bb; these can be written uniquely as 456, 478, 495, 578, 769, 869. Now we need to add the symbol 3 to two of the blocks so that all of the pairs 34, 35, 36, 37, 38, 39, occur. This is impossible, and so we conclude that 2,X. Consequently, we may assume two initial blocks TE123 and TE145. Since there can be only two symbols of frequency 4, we may take f(1) = 3; hence 16789 is a block. Since x4 = 0, we see that f ( T ) = f ( E ) = 4. Now let A = {T,E,2,3,4,5} and B = {6,7,8,9}. There must be 24 pairs of the form ab, and this requires us to have two elements from B in each of the last four blocks. The distribution 67XXX, 68=, 79XXX, 89=, cannot be completed; so the 89XXX. last four blocks must be taken as 07XXX, 67XXX, 89=, If T and E occur together in these last four blocks, the skeleton is readily completed t o be 67TE2, 67345, 89TE3, 89245. The corresponding excess graph is shown in Figure 7.
On the other hand, if T and E do not occur together, then the last four blocks may be taken as 67T24, 67335, 89T34, 89325, and the corresponding excess graph is shown in Figure 8. (Note that the completion 67T24, 67335, 891‘25, 89334, is an iscmorphic cover under application of the permutation (TE)(25)(34).) Both of these covers are obtainable by the “star” method of Mullin (cf. [S]). Take the geometry 124, 235, 346, 457, 561, 672, 713, and leave the points 4,5,7, fixed. The blocks of Mullin’s “star” cover then are 457, 5616*1*, 6726*2*, 7131*3*, 1241*2*, 2352*3*, 3463*6*. If we inflate 457 to be the block 22*457, we obtain the cover corresponding to Figure 7 (the isomorphism is easily obtained from the excess graph). On the other hand, if we inflate 457 t o be the block 23457, then we have the cover corresponding t o Figure 8 (again, the isomorphism is easily obtained from the excess graph). The two cases come from the fact that the excess graph of a “star” cover is merely made up of four double links. In Figure 7, we use two points from the same double link to inflate the triple, whereas in Figure 8 we use points from different double links. So the Mullin “star” cover is unique, but there are two nonisomorphic standard covers. We state this result as
Small Covering Designs with Block Size Five
Figure 7
Figure 8
Theorem 2: If w = 11, we have two covers given by TE123, TE145, TE267, TE389, 16789, 24589, 34567, and TE123, TE145, 16789, T2467, T3489, E2589, E3567. 0
447
R.G. Stanton
448
5. Conclusion.
All nonisomorphic covers by quintuples have been determined for w