The recognition of symmetric latin squares

A latin square S is isotopic to another latin square S′ if S′ can be obtained from S by permuting the row indices, the column indices and the symbols in S. Because the three permutations used above may all be different, a latin square which is isotopic to a symmetric latin square need not be symmetric. We call the problem of determining whether a latin square is isotopic to a symmetric latin square the symmetry recognition problem. It is the purpose of this article to give a solution to this problem. For this purpose we will introduce a cocycle corresponding to a latin square which transforms very simply under isotopy, and we show this cocycle contains all the information needed to determine whether a latin square is isotopic to a symmetric latin square. Our results relate to 1‐factorizations of the complete graph on n + 1 vertices, K_{n + 1}. There is a well known construction which can be used to make an n × n latin square from a 1‐factorization on n + 1 vertices. The symmetric idempotent latin squares are exactly the latin squares that result from this construction. The idempotent recognition problem is simple for symmetric latin squares, so our results enable us to recognize exactly which latin squares arise from 1‐factorizations of K_{n + 1}. As an example we show that the patterned starter 1‐factorization for the group G gives rise to a latin square which is in the main class of the Cayley latin square for G if and only if G is abelian. Hence, every non‐abelian group gives rise to two latin squares in different main classes. © 2007 Wiley Periodicals, Inc. J Combin Designs 16: 291–300, 2008     

Symmetries that latin squares inherit from 1‐factorizations

A 1‐factorization of a graph is a decomposition of the graph into edge disjoint perfect matchings. There is a well‐known method, which we call the ??‐construction, for building a 1‐factorization of K_n,n from a 1‐factorization of K_{n + 1}. The 1‐factorization of K_n,n can be written as a latin square of order n. The ??‐construction has been used, among other things, to make perfect 1‐factorizations, subsquare‐free latin squares, and atomic latin squares. This paper studies the relationship between the factorizations involved in the ??‐construction. In particular, we show how symmetries (automorphisms) of the starting factorization are inherited as symmetries by the end product, either as automorphisms of the factorization or as autotopies of the latin square. Suppose that the ??‐construction produces a latin square L from a 1‐factorization F of K_{n + 1}. We show that the main class of L determines the isomorphism class of F, although the converse is false. We also prove a number of restrictions on the symmetries (autotopies and paratopies) which L may possess, many of which are simple consequences of the fact that L must be symmetric (in the usual matrix sense) and idempotent. In some circumstances, these restrictions are tight enough to ensure that L has trivial autotopy group. Finally, we give a cubic time algorithm for deciding whether a main class of latin squares contains any square derived from the ??‐construction. The algorithm also detects symmetric squares and totally symmetric squares (latin squares that equal their six conjugates). © 2005 Wiley Periodicals, Inc. J Combin Designs 13: 157–172, 2005.     

On the spectrum of critical sets in latin squares of order 2n

Suppose that L is a latin square of order m and P ? L is a partial latin square. If L is the only latin square of order m which contains P, and no proper subset of P has this property, then P is a critical set of L. The critical set spectrum problem is to determine, for a given m, the set of integers t for which there exists a latin square of order m with a critical set of size t. We outline a partial solution to the critical set spectrum problem for latin squares of order 2ⁿ. The back circulant latin square of even order m has a well‐known critical set of size m²/4, and this is the smallest known critical set for a latin square of order m. The abelian 2‐group of order 2ⁿ has a critical set of size 4ⁿ‐3ⁿ, and this is the largest known critical set for a latin square of order 2ⁿ. We construct a set of latin squares with associated critical sets which are intermediate between the back circulant latin square of order 2ⁿ and the abelian 2‐group of order 2ⁿ. © 2007 Wiley Periodicals, Inc. J Combin Designs 16: 25–43, 2008     

On the completion of partial latin squares

In this paper a certain condition on partial latin squares is shown to be sufficient to guarantee that the partial square can be completed, namely, that it have fewer than n entries, and that at most $\text{[}\text{(n + 1)}\text{2}\text{]}$ of these lie off some line, where n is the order of the square. This is applied to establish that the Evans conjecture is true for n ? 8; i.e., that given a partial latin square of order n with fewer than n entries, n ? 8, the square can be completed. Finally, the results are viewed in a conjugate way to establish different conditions sufficient for the completion of a partial latin square.     

Cropper's question and Cruse's theorem about partial Latin squares

In 1974 Cruse gave necessary and sufficient conditions for an r × s partial latin square P on symbols σ₁,σ₂,…,σ_t, which may have some unfilled cells, to be completable to an n × n latin square on symbols σ₁,σ₂,…,σ_n, subject to the condition that the unfilled cells of P must be filled with symbols chosen from {σ_{t + 1},σ_{t + 2},…,σ_n}. These conditions consisted of r + s + t + 1 inequalities. Hall's condition applied to partial latin squares is a necessary condition for their completion, and is a generalization of, and in the spirit of Hall's Condition for a system of distinct representatives. Cropper asked whether Hall's Condition might also be sufficient for the completion of partial latin squares, but we give here a counterexample to Cropper's speculation. We also show that the r + s + t + 1 inequalities of Cruse's Theorem may be replaced by just four inequalities, two of which are Hall inequalities for P (i.e. two of the inequalities which constitute Hall's Condition for P), and the other two are Hall inequalities for the conjugates of P. © 2011 Wiley Periodicals, Inc. J Combin Designs 19:268‐279, 2011     

Latin squares with restricted transversals

We prove that for all odd m ≥ 3 there exists a latin square of order 3 m that contains an ( m ? 1 ) × m latin subrectangle consisting of entries not in any transversal. We prove that for all even n ≥ 10 there exists a latin square of order n in which there is at least one transversal, but all transversals coincide on a single entry. A corollary is a new proof of the existence of a latin square without an orthogonal mate, for all odd orders n ≥ 11 . Finally, we report on an extensive computational study of transversal‐free entries and sets of disjoint transversals in the latin squares of order n ? 9 . In particular, we count the number of species of each order that possess an orthogonal mate. © 2011 Wiley Periodicals, Inc. J Combin Designs 20:124‐141, 2012     

Transversals and multicolored matchings

Ryser conjectured that the number of transversals of a latin square of order n is congruent to n modulo 2. Balasubramanian has shown that the number of transversals of a latin square of even order is even. A 1‐factor of a latin square of order n is a set of n cells no two from the same row or the same column. We prove that for any latin square of order n, the number of 1‐factors with exactly n ? 1 distinct symbols is even. Also we prove that if the complete graph K_2n, n ≥ 8, is edge colored such that each color appears on at most edges, then there exists a multicolored perfect matching. © 2004 Wiley Periodicals, Inc.     

On the separation power and the completion of partial latin squares

Let n and m be natural numbers, n ? m. The separation power of order n and degree m is the largest integer k = k(n, m) such that for every (0, 1)-matrix A of order n with constant linesums equal to m and any set of k 1's in A there exist (disjoint) permutation matrices P₁,…, P_m such that A = P₁ + … + P_m and each of the k 1's lies in a different P_i. Almost immediately we have 1 ? k(n, m) ? m ? 1, yet in all cases where the value of k(n, m) is actually known it equals m ? 1 (except under the somewhat trivial circumstances of k(n, m) = 1). This leads to a conjecture about the separation power, namely that k(n, m) = m ? 1 if $\text{m ? [}\text{n}\text{2}\text{] + 1}$. We obtain the bound $\text{k(n, m) ? m ? [}\text{n}\text{2}\text{] + 2}$, so that this conjecture holds for n ? 7. We then move on to latin squares, describing several equivalent formulations of the concept. After establishing a sufficient condition for the completion of a partial latin square in terms of the separation power, we can show that the Evans conjecture follows from this conjecture about the separation power. Finally the lower bound on k(n, m) allows us to show, after some calculations, that the Evans conjecture is true for orders n ? 11.     

Bachelor latin squares with large indivisible plexes

In a latin square of order n , a k ‐plex is a selection of kn entries in which each row, column, and symbol occurs k times. A 1 ‐plex is also called a transversal. A k ‐plex is indivisible if it contains no c ‐plex for 0 < c < k . We prove that, for all n ≥ 4 , there exists a latin square of order n that can be partitioned into an indivisible ? n / 2 ?‐plex and a disjoint indivisible ? n / 2 ?‐plex. For all n ≥ 3 , we prove that there exists a latin square of order n with two disjoint indivisible ? n / 2 ?‐plexes. We also give a short new proof that, for all odd n ≥ 5 , there exists a latin square of order n with at least one entry not in any transversal. Such latin squares have no orthogonal mate. Copyright © 2011 Wiley Periodicals, Inc. J Combin Designs 19:304‐312, 2011     

Construction of doubly diagonalized orthogonal latin squares

A transversal T of a latin square is a collection of n cells no two in the same row or column and such that each of the integers 1, 2, …, n appears in exactly one of the cells of T. A latin square is doubly diagonalized provided that both its main diagonal and off-diagonal are transversals. Although it is known that a doubly diagonalized latin square of every order n ≥ 4 exists and that a pair of orthogonal latin squares of order n exists for every n ≠ 2 or 6, it is still an open question as to what the spectrum is for pairs of doubly diagonalized orthogonal latin squares. The best general result seems to be that pairs of orthogonal doubly diagonalized latin squares of order n exist whenever n is odd or a multiple of 4, except possibly when n is a multiple of 3 but not of 9. In this paper we give a new construction for doubly diagonalized latin squares which is used to enlarge the known class for doubly diagonalized orthogonal squares. The construction is based on Sade's singular direct product of quasigroups.     

An inequality for pseudo-subplanes of sets of orthogonal latin squares

Let L₁, L₂,…, L_t be a given set of t mutually orthogonal order-n latin squares defined on a symbol set S, |S| = n. The squares are equivalent to a (t + 2)-netN of order n which has n² points corresponding to the n² cells of the squares. A line of the net N defined by the latin square L_i comprises the n points of the net which are specified by a set of n cells of L_i all of which contain the same symbol x of S. If we pick out a particular r × r block B of cells, a line which contains points corresponding to r of the cells of B will be called an r-cell line. If there exist r(r ? 1) such lines among the tn lines of N, we shall say that they form a pseudo-subplane of order r-the “pseudo” means that these lines need not belong to only r ? 1 of the latin squares. The purpose of the present note is to prove that the hypothesis that such a pseudo-plane exists in N implies that $\text{r}{}^3\text{? (t + 2)r}{}^2\text{+ r + nt ?}{}^1\text{0}$.     

Doubly diagonal orthogonal latin squares

A pair of doubly diagonal orthogonal latin squares of order n, DDOLS(n), is a pair of orthogonal latin squares of order n with the property that each square has a transversal on both the front diagonal (the cells {(i, i):1?i?n}) and the back diagonal (the cells {(i, n + 1?i): 1?i?n}). We show that for all n except n = 2, 3, 6, 10, 12, 14, 15, 18 and 26, there exists a pair of DDOLS(n). Obbviously these do not exist when n = 2, 3 and 6.     

Multi-latin squares

A multi-latin square of order n and index k is an n×n array of multisets, each of cardinality k, such that each symbol from a fixed set of size n occurs k times in each row and k times in each column. A multi-latin square of index k is also referred to as a k-latin square. A 1-latin square is equivalent to a latin square, so a multi-latin square can be thought of as a generalization of a latin square.In this note we show that any partially filled-in k-latin square of order m embeds in a k-latin square of order n, for each n≥2m, thus generalizing Evans’ Theorem. Exploiting this result, we show that there exist non-separable k-latin squares of order n for each n≥k+2. We also show that for each n≥1, there exists some finite value g(n) such that for all k≥g(n), every k-latin square of order n is separable.We discuss the connection between k-latin squares and related combinatorial objects such as orthogonal arrays, latin parallelepipeds, semi-latin squares and k-latin trades. We also enumerate and classify k-latin squares of small orders.     

Atomic Latin squares of order eleven

A Latin square is pan‐Hamiltonian if the permutation which defines row i relative to row j consists of a single cycle for every i ≠ j. A Latin square is atomic if all of its conjugates are pan‐Hamiltonian. We give a complete enumeration of atomic squares for order 11, the smallest order for which there are examples distinct from the cyclic group. We find that there are seven main classes, including the three that were previously known. A perfect 1‐factorization of a graph is a decomposition of that graph into matchings such that the union of any two matchings is a Hamiltonian cycle. Each pan‐Hamiltonian Latin square of order n describes a perfect 1‐factorization of K_n,n, and vice versa. Perfect 1‐factorizations of K_n,n can be constructed from a perfect 1‐factorization of K_n+1. Six of the seven main classes of atomic squares of order 11 can be obtained in this way. For each atomic square of order 11, we find the largest set of Mutually Orthogonal Latin Squares (MOLS) involving that square. We discuss algorithms for counting orthogonal mates, and discover the number of orthogonal mates possessed by the cyclic squares of orders up to 11 and by Parker's famous turn‐square. We find that the number of atomic orthogonal mates possessed by a Latin square is not a main class invariant. We also define a new sort of Latin square, called a pairing square, which is mapped to its transpose by an involution acting on the symbols. We show that pairing squares are often orthogonal mates for symmetric Latin squares. Finally, we discover connections between our atomic squares and Franklin's diagonally cyclic self‐orthogonal squares, and we correct a theorem of Longyear which uses tactical representations to identify self‐orthogonal Latin squares in the same main class as a given Latin square. © 2003 Wiley Periodicals, Inc.     

A product theorem for row-complete Latin squares

In this article it is shown how to construct a row-complete latin square of order mn, given one of order m and given a sequencing of a group of ordern. This yields infinitely many new orders for which row-complete latin squares can be constructed. © 1997 John Wiley & Sons, Inc. J Combin Designs 5: 311–318, 1997     

Some arrowing results for trees versus complete graphs

For an arbitrary tree T of order m and an arbitrary positive integer n, Chvátal proved that the Ramsey number r(T, K_n) = 1 + (m ? 1) (n ? 1). for graphs G, G₁, and G₂, we say that G arrows G₁ and G₂, written G → (G₁, G₂), if for every factorization G = R ⊕ B, either G₁ is a subgraph of R or G₂ is a subgraph of B. it is shown that (i) for each l ≥ 2, K₁+ (m?1)(n?1) ?E(K₁) → (T, K_n) for m ≥ 2/ ? 1 and n ≥ 2; (ii) K₁ +,(m ?1)(n ?1) ? E(H) → (T, K_n), where H is any tree of order m ? 1, m ≥ 3 and n ≥ 2. It is further shown that result (i) is sharp with respect to the inequality m ≥ 2/? 1; in particular, examples are given to show that K1 + (2l?3)(n?1) E(K₁) ? (P_21?2, K_n) for all n ≥ 2, where P_21?2 denotes the path of order 21 ? 2. Also result (ii) is sharp with respect to the order of H; examples aregiven to show that K₁ + _(m?1)(n?1)? E(K(1, m ? 1)) ?(T, K_n)for any tree T of order m and any n ≥ 2.     

Partitioning complete multipartite graphs by monochromatic trees

The tree partition number of an r‐edge‐colored graph G, denoted by t_r(G), is the minimum number k such that whenever the edges of G are colored with r colors, the vertices of G can be covered by at most k vertex‐disjoint monochromatic trees. We determine t₂(K(n₁, n₂,…, n_k)) of the complete k‐partite graph K(n₁, n₂,…, n_k). In particular, we prove that t₂(K(n, m)) = ? (m‐2)/2ⁿ? + 2, where 1 ≤ n ≤ m. © 2004 Wiley Periodicals, Inc. J Graph Theory 48: 133–141, 2005     

All cycle‐complete graph Ramsey numbers r(Cm,K6)

The cycle‐complete graph Ramsey number r(C_m, K_n) is the smallest integer N such that every graph G of order N contains a cycle C_m on m vertices or has independence number α(G) ≥ n. It has been conjectured by Erd?s, Faudree, Rousseau and Schelp that r(C_m, K_n) = (m ? 1) (n ? 1) + 1 for all m ≥ n ≥ 3 (except r(C₃, K₃) = 6). This conjecture holds for 3 ≤ n ≤ 5. In this paper we will present a proof for n = 6 and for all n ≥ 7 with m ≥ n² ? 2n. © 2003 Wiley Periodicals, Inc. J Graph Theory 44: 251–260, 2003     

Orthogonal latin square graphs

An orthogonal latin square graph (OLSG) is one in which the vertices are latin squares of the same order and on the same symbols, and two vertices are adjacent if and only if the latin squares are orthogonal. If G is an arbitrary finite graph, we say that G is realizable as an OLSG if there is an OLSG isomorphic to G. The spectrum of G [Spec(G)] is defined as the set of all integers n that there is a realization of G by latin squares of order n. The two basic theorems proved here are (1) every graph is realizable and (2) for any graph G, Spec G contains all but a finite set of integers. A number of examples are given that point to a number of wide open questions. An example of such a question is how to classify the graphs for which a given n lies in the spectrum.     

Existence of frame sols of type 2nu1

An SOLS (self-orthogonal latin square) of order n with n₁ missing sub-SOLS (holes) of order h_i (1 ? i ? k), which are disjoint and spanning (i.e., Σ _1?i?kn_ih_i = n), is called a frame SOLS and denoted by FSOLS(h₁ⁿ¹ h₂ⁿ² …h_k^nk). In this article, it is shown that for u ? 2, an FSOLS(2ⁿu¹) exists if and only if n ? 1 + u. © 1995 John Wiley & Sons, Inc.