Group divisible designs in MOLS of order ten

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1 Des. Codes Cryptogr. (014) 71:83 91 DOI /s Group divisible designs in MOLS of order ten Peter Dukes Leah Howard Received: 10 March 011 / Revised: June 01 / Accepted: 10 July 01 / Published online: 5 July 01 Springer Science+Business Media, LLC 01 Abstract The maximum number of mutually orthogonal latin squares (MOLS) of order 10 is known to be between and 6. A hypothetical set of four MOLS must contain at least one of the types of group divisible designs (GDDs) classified here. The proof is based on a dimension argument modified from work by Dougherty. The argument has recently led to the discovery of a counterexample to Moorhouse s conjecture on the rank of nets, found by Howard and Myrvold. Although it is known that even three MOLS can admit no nontrivial symmetry group, we are hopeful this classification via GDDs and dimension can offer some structure to aid the eventual goal of exhausting the search for four MOLS of order 10. Keywords Code Dimension Net MOLS GDD Mathematics Subject Classification (000) Primary 05B15 Secondary 51E14 1 Introduction A latin square of order n is an n n array with n symbols such that every row and column is a permutation of the symbols. Two latin squares A and B are orthogonal if the ordered pairs (A ij, B ij ) exhaust all n possible ordered pairs of the symbols. For example, it is easy to see that there do not exist orthogonal latin squares of order. On the other hand, the addition and subtraction tables for finite fields of odd characteristic furnish orthogonal latin squares of all odd prime power orders. Communicated by C. Mitchell. P. Dukes (B) Department of Mathematics and Statistics, University of Victoria, Victoria, BC V8W 3R4, Canada dukes@uvic.ca L. Howard Department of Mathematics, Camosun College, Victoria, BC V8P 5J, Canada howardl@camosun.bc.ca

2 84 P. Dukes, L. Howard A straightforward argument shows that there can exist at most n 1 pairwise orthogonal, or mutually orthogonal latin squares of order n, with equality if and only if there exists a projective plane of that order. In general, a family of mutually orthogonal latin squares (MOLS) of order n is abbreviated MOLS(n) and the maximum size of such a family is denoted N(n). A wide body of literature on MOLS(n) is summarized in [4], chapter III.3; in particular a table of values of N(n) is given. As many lower bounds on N(n) come from a product construction, the most challenging case is generally n = 4t + for some integer t. In 1959 Bose and Shrikhande [1] disproved Euler s 178 conjecture that there do not exist orthogonal latin squares of order 4t + by constructing a pair of order. The same year Parker [1] established the existence of a pair of order 10. In 1960 Bose, Shrikhande and Parker [] showedthatthere areatleasttwo MOLS(4t + ), exceptfororders and6. Since then, much research has been conducted on MOLS of order 4t +. In 005 McKay, Meynert and Myrvold [10] showed that if three MOLS(10) exist then each latin square in the triple has trivial symmetry. A result of Bruck [3] gives one of the few known general upper bounds on MOLS. Theorem 1.1 If N(n) <n 1 then N(n) <n 1 (n) 1 4. In 1989 Lam, Thiel and Swiercz s computer search [9] revealed no projective plane of order ten; hence N(10) <9. Taken together with Theorem 1.1, this yields the current best upper bound of N(10) 6. That is, there are at most six MOLS(10). MOLS are equivalent to two other combinatorial objects: nets and transversal designs. First, a k-net of order n is an incidence structure of n points and kn lines, so that every pair of points is on at most one line and every line contains n points; furthermore, the lines partition into k parallel classes,each containing n disjoint lines. Dually, a transversal design TD(k, n) is a collection of kn vertices and n blocks, each containing k vertices; the kn vertices are partitioned into k groups, each of size n, and every pair of vertices from different groups belongs to exactly one block. By interchanging points with blocks, lines with vertices, and parallel classes with groups, it is clear that k-nets of order n and transversal designs TD(k, n) are equivalent objects. Moreover, a set of k MOLS(n) is equivalent to either of these. In the net context, the n points correspond to the n cells of the latin squares. The parallel classes of lines are formed by the rows, columns, and, for each latin square, the set of cells containing a fixed entry. In the TD context, the vertices correspond with row labels, column labels, and symbols in each of the k squares. A block is necessarily transverse to the groups and indicates which symbols appear in each square in a given row and column. Proposition 1. The existence of the following are equivalent: (1) k MOLS(n), () a k-net of order n, and (3) a transversal design TD(k, n). The definition of a TD(k, n) can be weakened so that the k groups have different sizes and blocks do not necessarily intersect each group. We call such an object a group divisible design, or GDD. The GDDs under consideration here have block sizes of, 4, or 6 and each vertex lies in exactly 10 blocks of the GDD. Given a GDD, the groups can be added to the set of blocks to produce a pairwise balanced design, or PBD. Every pair of distinct vertices now belongs to exactly one block. The type of a GDD specifies the group sizes in exponential notation. For example, a GDD with a group of size, three groups of size 4 and a group of size 6 is of type

3 GDDs in MOLS of order Specializing to our context, a GDD embeds in a TD(6, 10) if it is possible to extend each group to 10 vertices, extend each block to six vertices and enlarge the set of blocks in such a way that each pair of vertices appear together in one group or one block, but not both. Here the possible existence of four MOLS(10) is studied in terms of the corresponding TD(6, 10), and the possible GDDs which are subdesigns of the TD. The two main results in this paper are a classification of the possible types of GDDs that occur in this way, and a proof that one of these possible GDDs does not embed in a TD(6, 10). It is our hope that the classification of GDDs could provide a preliminary framework for a computer search for four MOLS(10). The feasibility of at least a partial search was demonstratedbyhowardandmyrvold[8] who found a GDD embedded in a TD(4, 10), providing a counterexample to Moorhouse s conjecture [11] on the rank of nets. If it could be shown that none of the GDDs in our classification extend to a TD(6, 10), then an upper bound of N(10) 3 could be established. Main results The general approach of examining codes arising from designs was pioneered by Hall [7]. Stinson [13] gave a new proof of the nonexistence of two MOLS(6) using design-theoretic techniques and a concise dimension argument. Some of the techniques in this paper, particularly those in the proof of Proposition.7, follow Stinson s treatment of sub-pbds in a hypothetical transversal design. Dougherty [6] gavean expandeddimension argumentbased on that of Stinson, but now phrased in the language of nets, to show the nonexistence of two MOLS(6). Dougherty s dimension argument is valid for finite fields of prime order p and for nets whose order is divisible by p. This generality allows Dougherty s techniques to be applied to nets of order 10. A k-net will be denoted N k for brevity. The number k is the net s degree. Somevector spaces associated with nets F generated by the characteristic functions of lines in N k.these characteristic functions indicate which points lie on each line; in a net of order n the vectors will have n components. Let D (N k ) denote the vector space over F generated by the characteristic functions of vectors of the form l + m where l and m lie in the same parallel class. The dimensions of C and D over F for nets of degree one are as follows: dim C (N 1 ) = n and dim D (N 1 ) = n 1. Any linear dependency in the vector space C (N k ) is called a relation. For instance, a 6-net of order 10 has trivial relations generated by the following five relations: 1, 1 3, 1 4, 1 5,and 1 6,wherethe i are the parallel classes of the net. Since such a net has 60 lines, its dimension is at most 55. Any relation consists of a set of lines of the net that intersects each point of the net an even number of times (possibly zero). The weight of a parallel class with respect to a relation is the number of lines it contributes to the relation. The following four results are due to Dougherty [6]: Theorem.1 If k is even or if N k has a transversal then, for even orders, dim C (N k ) dim D (N k ) = k. Lemma. If any parallel class has odd weight with respect to a relation in a net N k of even order then dim C (N k ) dim D (N k )<k. Lemma.3 The weight of each parallel class is 0, n, or n for any relation in a net N 3.

4 86 P. Dukes, L. Howard Lemma.4 If a net N 3 of order two modulo four has a transversal then dim C (N 3 ) dim C (N ) = n 1. These can be used to give an upper bound on the dimension of nets of even order: Proposition.5 Let N k be a net of even order n such that k is even or N k has a transversal. Then dim C (N k ) n +k. Proof When n is even C (N k ) D (N k), implying that dim C (N k ) dim D (N k) = n dim D (N k ). When k is even or N k has a transversal, this gives dim C (N k ) n + k dim C (N k ) or dim C (N k ) n +k. Corollary.6 A 6-net of order ten contains at least two linearly independent nontrivial relations. Proof The net has 60 lines and five linearly independent trivial dependencies, yielding a dimension of 55 minus the number of linearly independent nontrivial dependencies. By Proposition.5,dimC (N 6 ) = 53 for a 6-net of order ten. Since a 6-net of order 10 has even degree, Dougherty s Theorem.1 and Lemma. imply that each parallel class has even weight with respect to any such relation. Furthermore, Lemma.4 implies that there are no nontrivial relations in fewer than four parallel classes. It is convenient to view each relation in a 6-net of order 10 as a GDD with blocks of size, 4 or 6 embedded in a TD(6, 10). The vertices of the GDD correspond to the lines of the relation, with groups corresponding to the parallel classes of the net. Each block of the GDD arises from a point of the net, indicating which lines of the net intersect at that point. In any such GDD, each vertex lies in exactly 10 blocks of the GDD. Note that some parallel classes may not contribute any lines to the relation; therefore the GDD may have some groups of size 0. Proposition.7 The number of lines in a relation in a net N k of order ten is divisible by four, provided that k is even or N k has a transversal. Proof Consider the GDD resulting from the dual of the relation. Construct a PBD by adding the groups to the block set. Let m be the number of lines in the relation, equal to the number of vertices in the PBD. All block sizes are necessarily even by the definition of a relation and because all groups of the GDD are of even size. Let b, b 4, b 6, b 8, b 10 be the number of blocks of appropriate size in the PBD. Each vertex of the PBD appears in precisely eleven blocks: b + 4b 4 + 6b 6 + 8b b 10 = 11m. Each pair of distinct vertices in the PBD appear together in precisely one block: b + 6b b 6 + 8b b 10 = m(m 1). Subtracting the first equation from twice the second yields 8b 4 + 4b b b 10 = m(m 1) so that m(m 1) is divisible by eight. This implies that four divides m. Theorem.8 The following types of GDDs in a TD(6, 10) admit the possibility of a relation: 4 4, , 4 3, , 4 5, 6, 5 6 1, 4 4, , 4 4, , 4 6. Any other admissible type of GDD can be achieved from one of the above by complementing the vertex set in an even number of groups.

5 GDDs in MOLS of order Proof Consider the GDDs embedding in a TD(6, 10) that represent relations. By Proposition.7 the number of vertices in any such GDD is divisible by four. The total number of vertices when any group is deleted must be at least ten since any deleted vertex appears in exactly ten blocks. The vertices in any pair of groups whose combined size exceeds ten can be complemented to get a combined size below 10. This reduces the total number of vertices under consideration. Together with Proposition.7 the list is reduced to the types: , 4 4, , 4 3, , 4 5, 6, 5 6 1, 4 4, , 4 4, , 4 6.Now suppose that a GDD of type embedded in a TD(6, 10) admits a relation. Since every vertex of the GDD appears in exactly ten blocks of the GDD, and recalling that all block sizes are even, all blocks through the largest group must be blocks of size two. Such GDDs therefore consist of 8 blocks of size zero and 9 blocks of size two. Such GDDs cannot embedinatd(4, 10), for a contradiction. GDDs of type in a TD(6, 10) therefore do not admit a relation. Since a GDD is embedded in a TD(6, 10) if and only if one complementable to it is, at least two GDDs chosen from the types above are forced to occur in any TD(6, 10). The goal is now to describe the structure that is forced on each of the GDDs above. Consider the structure produced when all the vertices from one group of the GDD are deleted. This structure will be called a conformation. The blocks now have size 1 through 5. Let b 1, b, b 3, b 4, b 5 represent the number of blocks of each size. It will be convenient to let b 0 represent the number of blocks of size 0. Suppose a group of size g is deleted from the GDD. Let L be the number of vertices remaining in the GDD and P be the number of pairs of vertices remaining that lie in different groups. Given a GDD of type 4 4 and a resulting conformation of type 4 3, P = 3 4 = 48. The quantity C comes from the same calculation for 10 minus each group size. In this example C = 3 6 = 108. The quantity copairs is 3b 0 +b 1, 6b 0 +3b 1 +b or 10b 0 +6b 1 +3b +b 3 depending on whether the original GDD had four, five or six groups respectively. A conformation then produces blocks of various sizes according to solutions of the following system: b 0 + b 1 + b + b 3 + b 4 + b 5 = 100 b + 3b 3 + 6b b 5 = P copairs = C b 1 + b 3 + b 5 = 10g b 1 + 3b 3 + 5b 5 = Lg The last two restrictions insist that each deleted vertex contributes to 10 odd-sized blocks and that each deleted vertex appears in a block with each vertex in another group precisely once. When the GDD has four or five groups of nonzero size, b 5 must equal zero and the above system has a unique solution. When the GDD has six groups of nonzero size, given an integral solution (b 0, b 1, b, b 3, b 4, b 5 ) to the above system with b 5 = 0, observe that (b 0 r, b 1 + r, b + r, b 3 r, b 4 r, r) is also an integral solution for any positive integer r. Combinatorial feasibility of the GDD requires that all solutions be nonnegative. An upper bound on r, the number of blocks of size six in the GDD, helps to restrict the structure of the GDD. Here are some upper bounds on this number: the product of the two smallest group sizes in the GDD, b 0, 1 b 3, b 4.Thelast three follow from the nonnegativity condition above.

6 88 P. Dukes, L. Howard Consider any GDD corresponding to a relation, and consider any group of size g in the GDD (not necessarily the largest group). Let c, c 4, c 6 represent the number of blocks of each size through a given vertex in this group. Let L be the number of vertices in the GDD not in the group. Then c + c 4 + c 6 = 10 because each vertex appears in exactly 10 blocks of the GDD and c + 3c 4 + 5c 6 = L because each vertex must meet each of the vertices from the other groups. These two conditions are called regularity conditions. Subtracting the first regularity condition from the second we have c 6 L This gives another upper bound on the number of blocks of size six in the GDD: g L When there are no blocks of size six in the GDD, the solution to the system of regularity conditions is unique: Otherwise (c, c 4, c 6 ) = (c, c 4, c 6 ) = ( 1 (30 L), 1 ) (L 10), 0. ( 1 (30 L) + h, 1 ) (L 10) h, h for some h between 0 and the maximum number of blocks of size six in the GDD. The regularity conditions will now be used to exclude a GDD of type from embedding in a TD(6, 10). Proposition.9 A GDD of type does not embed in a TD (6, 10). Proof Complement the vertex sets of two groups to produce a GDD of type Any vertex in the largest group has to occur in blocks with 1 other vertices and must occur in exactly ten blocks. Thus there are no blocks of size six in the GDD. The conformation produced when the group of size eight is deleted has (b 0, b 1, b, b 3, b 4, b 5 ) = (13, 7,, 8, 5, 0). Let d 3 and d 4 be the numbers of blocks of sizes 3 and 4 that intersect the fifth group of the conformation. By counting pairs of vertices where one lies in the fifth group and one lies in one of the first four groups, 3d 4 + d 3 3. There are 5 d 4 blocks of size four and 8 d 3 blocks of size three containing only vertices from the first four groups. Counting pairs of vertices lying in two of the first four groups, 3d 4 + d 3 + 6(5 d 4 ) + 3(8 d 3 ) 4. This simplifies to 30 3d 4 + d 3. Thus 30 3d 4 + d 3 3. Since d 4 5andd 3 8, the only solution is d 4 = 5, d 3 = 8. Altogether this gives 13 blocks of size four through the fifth group of the GDD of type On the other hand the regularity conditions are c + c 4 = 10 and c + 3c 4 = 16 for a vertex in the fifth group of the GDD. This means that there are three blocks of size four through each vertex in the fifth group, a contradiction. A GDD of type does not embed in a TD(6, 10). Since a GDD embeds in a TD(6, 10) if and only if one complementable to it does, GDDs of type do not embed either. The Eleven Types of GDDs Possible in a TD(6, 10): The largest group is chosen for deletion. The maximum r value, that is the maximum number of blocks of size six in the GDD, is given where applicable. When r > 0 is possible, additional solutions are given by (b 0 r, b 1 + r, b + r, b 3 r, b 4 r, r). 1. GDD of type 4 4 with a group of size four deleted: (b 0, b 1, b, b 3, b 4, b 5 ) = (4, 36, 36, 4, 0, 0)

7 GDDs in MOLS of order Since the number of blocks through a vertex in the deleted group is always ten, and each deleted vertex has to pair with the 1 vertices from the other groups, each deleted vertex must appear in one block of size four and nine blocks of size two in the GDD. This applies no matter which group is chosen for deletion. This is just a restatement of the regularity conditions. Using just this structure, Howard and Myrvold [8] generated some examples of GDDs by computer and tried to extend them to a TD(4, 10). Several examples of GDDs embedded in a TD(4, 10) were found. Some of the corresponding nets had dim C (N 4 ) dim C (N 3 ) 6. These nets yielded counterexamples to Moorhouse s conjecture [11]that dim C p (N k ) dim C p (N k 1 ) n k + 1 for p any prime dividing the net s order at most once. Continuing with the classification:. GDD of type with a group of size six deleted: (b 0, b 1, b, b 3, b 4, b 5 ) = (4, 60, 1, 0, 4, 0) 3. GDD of type 4 3 with a group of size four deleted: (b 0, b 1, b, b 3, b 4, b 5 ) = (5, 36, 34, 4, 1, 0) 4. GDD of type with a group of size six deleted: (b 0, b 1, b, b 3, b 4, b 5 ) = (14, 48, 4, 1,, 0) 5. GDD of type 4 5 with a group of size four deleted: (b 0, b 1, b, b 3, b 4, b 5 ) = (15, 8, 4, 1, 3, 0) 6. GDD of type 6 with a group of size two deleted: (b 0, b 1, b, b 3, b 4, b 5 ) = (40, 0, 40, 0, 0, 0); r = 0byb 4 = 0 7. GDD of type 4 4 with a group of size four deleted: (b 0, b 1, b, b 3, b 4, b 5 ) = (6, 36, 3, 4,, 0); r = 0byg L 10 4 =0 8. GDD of type with a group of size six deleted: (b 0, b 1, b, b 3, b 4, b 5 ) = (15, 48,, 1, 3, 0); r 3byb 4 = 3 9. GDD of type 4 4 with a group of size four deleted: (b 0, b 1, b, b 3, b 4, b 5 ) = (16, 8, 40, 1, 4, 0); r 4 by two groups of size two 10. GDD of type with a group of size six deleted: (b 0, b 1, b, b 3, b 4, b 5 ) = (9, 36, 6, 4, 5, 0); r 5byb 4 = GDD of type 4 6 with a group of size four deleted: (b 0, b 1, b, b 3, b 4, b 5 ) = (10, 0, 40, 0, 10, 0); r 8byg L 10 4 =8 Each GDD listed above gives rise to a number of GDDs with a different type by complementing the vertex sets in an even number of groups. When investigating a specific GDD there is a trade-off between the ease of working with a GDD having a small vertex set versus the stringent restrictions imposed on a GDD with a large number of blocks of sizes four and six. For instance, the GDD of type 4 5 corresponds to: a GDD of type with a group of size six deleted: (b 0, b 1, b, b 3, b 4, b 5 ) = (7, 36, 30, 4, 3, 0) a GDD of type with a group of size six deleted: (b 0, b 1, b, b 3, b 4, b 5 ) = (3, 4, 30, 36, 7, 0) Varying the group size chosen for deletion provides the same information as the regularity conditions discussed above. Recently Delisle [5] determined by computer search that any TD(4, 10) containing GDDs corresponding to two or more linearly independent relations of type 4 4 does not extend to a

8 90 P. Dukes, L. Howard TD(5, 10). This work was motivated by the search for 3 MOLS(10). By Theorem.8, relations of type 4 4 are the only possible type of relations in a TD(4, 10), up to complementing vertices in an even number of groups. Thus there can be at most one non-trivial relation in any TD(4, 10) that extends to a TD(5, 10). In other words, any 4-net of order 10 which can be extended to a net of higher degree must have dim C (N 4 ) equal to 36 or Further results on dimension Some additional results on the dimensions of C (N k ) are established for nets of order 10. Implications for sets of MOLS(10) are also discussed. Proposition 3.1 If k is even or N k has a transversal, then dim C (N k )>dim C (N k 1 ) for any subnet N k 1 of N k. Proof Since k is even or N k has a transversal by Dougherty sn Theorem.1 and Lemma. the evenness condition applies to weights of parallel classes in any relation. This means that the codeword corresponding to any transversal of N k 1 lying in the kth parallel class must be linearly independent over C (N k 1 ). Proposition 3. If dim C (N 5 )<46 or dim C (N 6 )<47 for a net of order ten then there is a nontrivial relation in four or five classes of the net. Proof If dim C (N 5 )<46 = then there is a nontrivial relation in the net, which is necessarily a relation in at most five classes. If dim C (N 6 )<47 then any subnet N 5 has a transversal by the existence of N 6 and Proposition 3.1 applies to give dim C (N 5 ) < 47 1 = 46. This is a particularly useful result because only five of the GDD types in Theorem.8 contain vertices in at most five groups, and hence give a relation in at most five parallel classes. The combinatorial structure of these GDDs is well understood. Thus a proof of the nonexistence of 4 MOLS(10) could happen in two ways: by showing that the dimension of C (N 6 ) is too high and applying the dimension argument or by showing that the dimension of C (N 6 ) is low and showing that none of these five GDDs embed in a transversal design. Proposition 3.3 If N 6 is a 6-net satisfying dim C (N 6 ) = 53, and N 6 can be extended to a 7-net, N 7,thenN 7 has no transversal. Proof If dim C (N 6 ) = 53 then by Dougherty s Theorem.1, dimd (N 6 ) = 47 and dim D (N 6) = 53. Since C (N 6 ) D (N 6) this implies that C (N 6 ) = D (N 6). Any transversal of N 6 belongs to D (N 6) and therefore lies in C (N 6 ). A relation in N 7 with odd weight in the seventh parallel class is thus produced. This means that N 7 has no transversal. Corollary 3.4 If N 6 is a 6-net satisfying dim C (N 6 ) = 53, thenn 6 canbeextendedtoat most a 7-net, or equivalently to at most five MOLS of order 10. Proof By Proposition 3.3 the net N 7 has no transversal and certainly the net cannot extend to an eighth parallel class. Thus there are at most five MOLS(10). To conclude, it is worth mentioning that further structural results are possible for some of the GDDs listed here. This has been left to a separate study.

9 GDDs in MOLS of order Acknowledgments This research of the first author is supported by NSERC. Thanks to the referees for careful reading which helped greatly improve the presentation. References 1. Bose R.C., Shrikhande S.S.: On the falsity of Euler s conjecture about the non-existence of two orthogonal Latin squares of order 4t +. Proc. Natl Acad Sci. U.S.A. 45(5), (1959).. Bose R.C., Shrikhande S.S., Parker E.T.: Further results on the construction of mutually orthogonal Latin squares and the falsity of Euler s conjecture. Can. J. Math. 1, (1960). 3. Bruck R.H.: Finite Nets II. Uniqueness and embedding. Pac. J. Math. 13, (1963). 4. Colbourn C.J., Dinitz J.H. (eds.): The CRC Handbook of Combinatorial Designs, nd edn. CRC Press, Boca Raton (006). 5. Delisle E.: The search for a triple of mutually orthogonal latin squares of order ten: looking through pairs of dimension thirty-five and less. M.Sc. Thesis, University of Victoria (010). 6. Dougherty S.T.: A coding theoretic solution to the 36 officer problem. Des. Codes Cryptogr. 4, 18 (1994). 7. Hall M. Jr.: Combinatorial Theory, nd edn. Wiley, New York (1986). 8. Howard L., Myrvold W.: A counterexample to Moorhouse s conjecture on the rank of nets. Bull. Inst. Combin. Appl. 60, (010). 9. Lam C., Thiel L., Swiercz S.: The non-existence of finite projective planes of order 10. Can. J. Math. XLI(6), (1989). 10. McKay B.D., Meynert A., Myrvold W.: Small Latin squares, quasigroups, and loops. J. Combin. Des. 15(), (006). 11. Moorhouse G.E.: Bruck nets, codes and characters of loops. Des. Codes Cryptogr. 1, 7 9 (1991). 1. Parker E.T.: Constructions of some sets of pairwise orthogonal Latin squares. Proc. Natl Acad Sci. U.S.A. 45, (1959). 13. Stinson D.R.: A short proof of the nonexistence of a pair of orthogonal Latin squares of order six. J. Combin. Theory Ser. A 36, (1984).