Latin Squares with No Small Odd Plexes

Transcription

1 Latin Squares with No Small Odd Plexes Judith Egan, Ian M. Wanless School of Mathematical Sciences, Monash University, Victoria 3800, Australia, Received May 14, 2007; revised September 30, 2007 Published online 7 January 2008 in Wiley InterScience ( DOI /jcd Abstract: A k-plex in a Latin square of order n is a selection of kn entries in which each row, column, and symbol is represented precisely k times. A transversal of a Latin square corresponds to the case k = 1. We show that for all even n>2there exists a Latin square of order n which has no k-plex for any odd k< n but does have a k-plex for every other k 1 n Wiley 4 2 Periodicals, Inc. J Combin Designs 16: , 2008 Keywords: Latin square; transversal; plex; orthogonal partition 1. INTRODUCTION A Latin square of order n is an n n matrix in which n distinct symbols are arranged so that each symbol occurs once in each row and column. A Latin square may be represented by a set of n 2 elements, each of which is an ordered triple (i, j, k) such that symbol k appears at the intersection of row i and column j. In particular, defining a Latin square as a set allows us to use set terminology and notation. A subset of a Latin square which is itself a Latin square is called a subsquare. A k-plex in a Latin square of order n is a subset of kn elements in which each row, column, and symbol is represented exactly k times. A transversal of a Latin square corresponds to the case k = 1. Our main aim is to prove the next statement. Theorem 1.1. For all even n>2, there exists a Latin square of order n which has no k-plex for any odd k< n 4 but does have a k-plex for every other k 1 2 n. Theorem 1.1 will follow immediately from Theorems 5.8 and 6.3. In fact, we will analyze the Latin squares used to prove Theorem 1.1 in sufficient detail to describe all their possible types of orthogonal partitions (in the sense defined in the next section). We define the plex range of a Latin square L, to be the set of values of k for which a k-plex is contained in L. Let R be the plex range of a Latin square L of order n. Then L is said to have a complete plex range if R ={k 0 k n}, aneven plex range if R ={0, 2, 4,...,n}, Journal of Combinatorial Designs 2008 Wiley Periodicals, Inc. 477

2 478 EGAN AND WANLESS and a mixed plex range if R is neither complete nor even. Only squares of even order can have an even plex range. For reasons that will be detailed in the next section, we are interested in studying which plex ranges occur among the Latin squares of order n. Theorem 1.1 represents the first general existence result for Latin squares with a mixed plex range. It shows there is greater variety than had previously been observed. For example, we have a Latin square of order 16 possessing a 5-plex but no transversal or 3-plex. 2. BACKGROUND AND MOTIVATION Although the concept is much older, the name k-plex was first used in [21]. The definition given in Section 1 is a restriction of the original one. In defining a k-plex as a partial Latin square, Wanless [21] allowed that a k-plex be not necessarily completable to a Latin square of the same order. Herein, we are solely concerned with k-plexes which are contained in a Latin square. In statements of a general nature, we sometimes use plex for k-plex. We call a k-plex an odd plex if k is odd. An even plex is a k-plex with k even. Various names for specific cases of k-plexes have been used. Statistical literature sometimes refers to a transversal as a directrix and uses the terms duplex, triplex, and quadruplex for a 2-plex, 3-plex, and 4-plex, respectively. Other names are mentioned in [21]. We call a set of plexes parallel if no two of them share a common element. The union of an a-plex and a parallel b-plex in a Latin square L yields an (a + b)-plex of L. We will frequently use this fact (and its obvious generalization to sets of more than two parallel plexes). The reverse process of dividing an (a + b)-plex into an a-plex parallel to a b-plex is not always possible. If a k-plex K contains no c-plex for 0 <c<k, then it is said that K is indivisible. Otherwise K is said to be divisible. This article sheds some light on the existence of indivisible plexes of certain sizes and our results will be used in a subsequent article to develop the theory further. It is obvious that a Latin square L of order n is an n-plex. Trivially, every Latin square also contains a 0-plex. If L has a k-plex K, then the complement L \ K, isan(n k)-plex. Together a k-plex and its complementary (n k)-plex are an example of an orthogonal partition of L. More generally, if L is decomposed into disjoint parts K 1,K 2,...,K m where K i is a k i -plex, then K 1,K 2,...,K m is called a (k 1,k 2,...,k m )-partition of L. The notation (k α 1 1,kα 2 2,...) can be used as shorthand for a partition with α 1 parts which are k 1 -plexes, α 2 parts which are k 2 -plexes, etc. It is of particular interest when all parts are of the same size. We call a (k n/k )-partition a k-partition. IfLhasa(k α 1 1,kα 2 2,...)-partition then we call the tuple (k α 1 1,kα 2 2,...)apartition type of L. Much interest in orthogonal partitions stems from their application in the design and analysis of experiments. The most important case is a 1-partition but more general orthogonal partitions of Latin squares have been studied from at least as early as 1945 by Finney [6]. Finding an orthogonal partition of a Latin square L is equivalent to finding a frequency square orthogonal to L in the sense, for example, of [3]. Bailey [1] and Gilliland [13] both consider orthogonal partitions in a general setting of which Latin squares are a particular example. In the design of experiments, it is important to be able to find an orthogonal partition of a given type in a Latin square. One approach to finding a partition is to begin with the trivial partition with one part and make successive refinements until a partition of the desired type is obtained. To know when such an approach may get stuck, it is important to understand

3 LATIN SQUARES WITH NO SMALL ODD PLEXES 479 TABLE I. Plex Ranges Possible for All Latin Squares of Order n 8 Order n Plex range Number of species Complete Even Mixed indivisible plexes and to know what types of orthogonal partitions the square has. Our main results are stronger than stated in Theorem 1.1 in that we identify all possible types of orthogonal partitions in the squares that we construct. The questions considered in this article are related to a number of problems that have been studied in a variety of combinatorial contexts, which we now mention briefly. From any Latin square, a transversal design can be obtained in the usual manner [3, p. 161] and a k-plex in the Latin square corresponds to a k-parallel class of the transversal design. Non-separable k-parallel classes in Steiner triple systems have been investigated [4, p. 418]. An indivisible k-plex in a Latin square corresponds to a non-separable k-parallel class in the associated transversal design. The notion of a k-plex is very closely related to that of a k-homogeneous latin trade. The latter have recently been shown to have interesting connections with topology and geometry. While most results on k-homogeneous latin trades have concentrated on small values of k, there have also been some results for general k. See [2] and the references therein for details. It is well known that a Latin square of order n is equivalent to a proper edge coloring of the complete bipartite graph K n,n where each of n distinct edge colors represents a symbol in the Latin square. A transversal is represented in the graph by a 1-factor (or perfect matching) in which each edge has a distinct color. Such factors have been called rainbow factors. A k-plex is a k-factor of K n,n in which each of the n colors occurs on k edges and a k-partition is a k-factorization into factors of this type. Zaker [23] uses the connection with k-factors to show that the computational complexity of finding a partial transversal of maximum length in a 2-plex is an NP-complete problem. Graph factorization problems have been well studied [3,20]. Indeed, one of the earliest results of graph theory by Petersen in 1891 proves that every regular graph of even degree has a 2-factorization. The corresponding result for k-plexes in Latin squares remains open, as we shall see below. The number of transversals in a Latin square is well known to be a species invariant and it follows from the same reasoning that the plex range of a Latin square is also a species invariant. We refer the reader to [5] for a definition of species, also known as main class. Drawing upon the works of Finney [6 8], Freeman [10 12], Johnston and Fullerton [15], Killgrove et al. [17], and Wanless [21], we are able to summarize in Table I the Latin squares of order n 8 by the named plex ranges. The 34 species of order n 8 which have a mixed plex range all contain squares with a 3-plex but no transversal. It is conjectured [21] that a Latin square with this property exists for all even orders n>4.

4 480 EGAN AND WANLESS There is no such complete categorization of the plex range of the Latin squares of orders n>8. The existence of a transversal in every Latin square of order 9 has been shown [19] via computer enumeration, thereby proving a specific case of Ryser s Conjecture that every Latin square of odd order has a transversal (see [5, pp. 32, 486]). A conjecture by Rodney (see [3, p. 143]) that every Latin square has a duplex remains open. It might be hoped that some progress on these and other longstanding problems may be made by developing a general theory of k-plexes and how Latin squares may fail to contain k-plexes for certain values of k. General existence results with respect to the complete and even plex ranges have previously been shown. We next complete our survey of these earlier findings and shall see that a result like Theorem 1.1 is not expected for the Latin squares of odd order. It is widely known that a Latin square has an orthogonal mate if and only if it has a 1-partition and that such squares exist for all orders n>2 except n = 6 (see [5] or [18]). A square with a 1-partition necessarily has a complete plex range although the converse is false. Indeed, Table I records that there are examples of Latin squares of order 6 which possess a complete plex range, thus: Theorem 2.1. plex range. For all n>2, there exist Latin squares of order n which have a complete The next theorem, first stated in [21], is also relevant. Theorem 2.2. Let G be a group of finite order n with a non-trivial cyclic Sylow 2-subgroup. The Cayley table of G contains no k-plex for any odd k but has a 2-partition and hence contains a k-plex for every even k in the interval 0 k n. The Cayley table of Z n thus provides an example of a Latin square with an even plex range for every even order n. Corollary 2.3. plex range. For all even n, there exists a Latin square of order n which has an even Theorem 2.2 and well-known results by Hall and Paige [14] provide a complete categorization of finite soluble groups by plex range. The Cayley table of any such group has an even plex range if the group has a non-trivial cyclic Sylow 2-subgroup, and otherwise it has a complete plex range. If the Hall Paige Conjecture is true then this categorization extends to all finite groups. It would follow that the Cayley table of a non-soluble finite group has a complete plex range and that there is no group whose Cayley table has a mixed plex range. Conjectures in [21] say that every Latin square contains n 2 parallel 2-plexes. This strengthens both Ryser s Conjecture and Rodney s Conjecture. The conjectures in [21] imply that every Latin square of odd order has a complete plex range and that if R is a mixed plex range then the only possibility for k/ R is that k is odd. These conjectures are known [21] to be true for all Latin squares of order n DEFINITIONS AND NOTATION Since it may be that all Latin squares of odd order have a complete plex range, we focus for the rest of the article on squares of even order n = 2h. We will use Z n ={0, 1,...,n 1}

5 LATIN SQUARES WITH NO SMALL ODD PLEXES 481 to index our rows, columns, and symbols and all calculations of such indices will be modulo n. We partition Z n into E ={0, 2, 4,...,n 2} and D ={1, 3, 5,...,n 1} and adopt the convention of ordering the indices 0, 2, 4,...,n 2, 1, 3, 5,...,n 1. We stress that when we write a Latin square, the rows and columns appear in this order. We adopt the notation L[i, j] = k to define L. For example, Z n [i, j] = (i + j) mod n defines the Cayley table of the group Z n, the integers under addition mod n. Then if n = 2h, the Latin square Z n, exhibits the unique subgroup of index 2 in Z n : n 2 ( ) A00 A 01 Z n = where A 00 = , (1) A 10 A n n n A 01 = A 10 = and A = n n 4 n n n 2 A reader familiar with the definition of a step type Latin square (see [5, p. 445] or [21]) will recognize that Z 2h is an h-step type Latin square. That is, each block A ij, i, j {0, 1}, is an h h subsquare such that A ij and A i j contain the same symbols if and only if i + j i + j mod 2. Definition 3.1 (Latin square P n ). When n = 2h = 4q for an integer q 1, define (i + j + 2) mod n if i = n 3 and j D, P n [i, j] = (i + j 2) mod n if i = n 1 and j D, (i + j) mod n otherwise. Alternatively, P n may be thought of as being obtained from Z n, as shown in (1), by interchanging the last two rows in the block A 11. An illustration of P 16 can be found in section 5. Definition 3.2. (Latin square Q n ). When n = 2h for an odd integer h 3, define n 1 if i = j = 0 or i = j = n 1, 0 if {i, j} ={0,n 1}, Q n [i, j] = (i + j 2) mod n if i = 1 and j D, (i + j + 2) mod n if i = n 1 and j D \{n 1}, (i + j) mod n otherwise. Alternatively, Q n may be considered to be the Latin square obtained from Z n, as shown in (1), by interchanging the first and last rows in the block A 11, and then turning the following (2) (3)

6 482 EGAN AND WANLESS ( ) (0, 0, 0) (0, n 1, n 1) resultant intercalate (subsquare of order 2):.An (n 1, 0,n 1) (n 1,n 1, 0) illustration of Q 14 is at the end of this section. The notation which we describe in the remaining part of this section relies on context to make clear which Latin square L of order n we are considering. Define row i of L, denoted r i, to be the set of elements in the ith indexed row of L. That is; r i = { (i, j, k) L j, k Z n }. Also any subset X L may be specified by simply indicating the columns represented in X for each row of L.IfX is clear from context, we define {j (i, j, k) r i X for some k Z n }. The function : L Z n is given by (e) = (z x y) mod n for each e = (x, y, z) L. Define L and a L by ={e L (e) 0}, a ={e L (e) = a}. In the case when L = P n, it is immediate from (2) that = 2 2 where, 2 = { (x, y, z) r n 3 y D }, 2 = { (x, y, z) r n 1 y D }. Similarly, when L = Q n, = 2 2 I where I = 1 1, (4) 2 = { (x, y, z) r 1 y D }, 2 = { (x, y, z) r n 1 y D \{n 1} }, 1 = { (0, 0, n 1) }, 1 = { (0,n 1, 0), (n 1, 0, 0), (n 1,n 1,n 1) }. We next illustrate with Q 14 our adopted notation and convention for writing Latin squares. The borders of the square contain the row and column indices. Marks indicate elements of and circled entries identify I. The darker shaded cells identify a 3-plex K, and lighter shaded cells identify an 8-plex named J. It should be apparent that Q 14 \ (K J) is another 3-plex and we have at hand a (3 2, 8)-partition of this square.

7 LATIN SQUARES WITH NO SMALL ODD PLEXES 483 (5) The illustrated 3-plex K is defined by the columns c(i) inr i K as follows: c(0) ={0, 2, 9}, c(1) ={1, 3, 5}, c(2) ={2, 6, 11}, c(3) ={0, 12, 5}, c(4) ={4, 8, 13}, c(5) ={0, 2, 7}, c(6) ={6, 8, 1}, c(7) ={4, 9, 13}, c(8) ={10, 3, 7}, c(9) ={4, 1, 11}, c(10) ={10, 5, 9}, c(11) ={6, 3, 13}, c(12) ={12, 7, 11}, c(13) ={8, 10, 12}. (6) Later, in Section 6, we give a generalized definition of J. 4. NECESSARY CONDITIONS The next lemma is pivotal to the results which follow. The case for k = 1 was used in [22] to show the existence of Latin squares without orthogonal mates. Lemma 4.1. Let K be a k-plex in a Latin square L of order n. If n is even and k is odd, then e K (e) 2 1 n mod n. Otherwise e K (e) 0 mod n. Proof. By definition, K contains k representatives of each row, column, and symbol so, e K n 1 (e) = k z=0 n 1 z k x=0 n 1 x k y=0 y = 1 kn(n 1). 2 If n is even and k is odd, then 2 1 kn(n 1) 2 1 n mod n since k(n 1) is odd. Otherwise k(n 1) is even so 2 1 kn(n 1) 0 mod n. We now use Lemma 4.1 to show that P n and Q n have no small odd plexes.

8 484 EGAN AND WANLESS Lemma 4.2. Let k be an odd integer. If k< 4 1 n or k> 4 3 n, then the Latin square P n has no k-plex. Proof. Assume that K P n is a k-plex for some odd k. By Lemma 4.1, (e) 1 n mod n. (7) 2 e K Since K cannot satisfy (7) without inclusion of either 4 1 n elements of 2 or 4 1 n elements of 2, there must be a row in K with at least 4 1 n elements. As n is even, the complement of K is an (n k)-plex with n k odd. Hence n k 4 1 n,sok 4 3 n. Corollary 4.3. If q is even, then a partition of the Latin square P 4q includes at most two odd plexes. Proof. Any partition has an even number of odd plexes, as 4q is even. Since q is even, an odd k-plex has k q + 1 (by Lemma 4.2) so four or more odd plexes in a partition is impossible. Lemma 4.4. Let k be an odd integer. If k< 4 1 (n 2) or k> 4 1 (3n + 2), then the Latin square Q n has no k-plex. Proof. Assume that K Q n is a k-plex for some odd k. By Lemma 4.1, (e) 1 n mod n. (8) 2 e K Since h = 2 1 n is odd, it is immediate that K contains an odd number of elements of I as defined by (4). We consider two cases. Case 1. e K I (e) =±1. Then (8) requires that 2 K 2 1 (h 1) or 2 K 2 1 (h 1). Case 2. e K I (e) = 3. Then (8) requires that either (i) 2 K 2 1 (h + 3), or (ii) 2 K 2 1 (h 3). However, in subcase (ii), K also has two elements of I in r n 1. Hence, each possibility implies the existence of at least 2 1 (h 1) = 4 1 (n 2) elements in the same row of K.Asnis even, the complement of K is an (n k)-plex with n k odd. Thus n k 4 1 (n 2), so k 4 1 (3n + 2). Plexes in squares of order six have been studied in detail. Our square Q 6 belongs to species I in Fisher and Yates catalogue [9]. Hence from Finney [8], we know that Q 6 hasa(1 2, 2 2 )-partition but no (1 4, 2)-partition. Also, none of its eight distinct transversals intersect I r 5. More generally, Lemma 4.5. Let h 3 mod 4 and K Q 2h be a k-plex where k = 2 1 (h 1). Then K contains precisely one element of I and that element must be in r 0. Proof. It is clear from Lemma 4.4 that k is the minimal odd value for which a k-plex is possible. Assume that I r n 1 K > 0. By Lemma 4.1, we have two cases. Case 1. e K I (e) = 3.

9 LATIN SQUARES WITH NO SMALL ODD PLEXES 485 Then I r n 1 K =2so that Lemma 4.1 implies that either 2 K 2 1 (h + 3) or 2 K + I r n 1 K 2 1 (h 3) + 2 = 2 1 (h + 1). Case 2. e K I (e) = 1. Then I r n 1 K 1 and Lemma 4.1 implies that either 2 K 2 1 (h + 1), or, 2 K + I r n 1 K 2 1 (h 1) + 1 = 2 1 (h + 1). Hence in all cases K has at least 2 1 (h + 1) >kelements in either r 1 or r n 1 which is a contradiction. Since I K is odd, K contains precisely one element of I in r 0. Lemma 4.6. A partition of the Latin square Q 2h includes at most two odd plexes. Proof. The square is of even order so the number of odd plexes in the partition is even. By Lemma 4.4, the number is not more than four. We now show that four is impossible. Assume there is a (k 1,k 2,...,k m )-partition where k i is odd for i = 1, 2, 3, 4. We have two cases. Case 1. h 1 mod 4. Since 4 1 (n 2) is even, Lemma 4.4 implies that k i 4 1 (n + 2) > 4 1 n for i = 1, 2, 3, 4 which is impossible. Case 2. h 3 mod 4. It follows from Lemma 4.5 that at most two of the k i are minimal, say k 1 and k 2. Therefore k 3,k 4 k = 2 1 (h + 3). However, then we have 4 i=1 k i n + 2 which is a contradiction. 5. PARTITIONS OF P n In this section, we show partitions of P n and determine its plex range. Throughout this section, we assume that n = 2h = 4q for an integer q 1. We first show the even values in the plex range of P n. Lemma 5.1. The Latin square P n has a 2-partition and hence has k-plexes for all even k in the interval 0 k n. Proof. For each a E, we construct a 2-plex J a. The columns c(i)inr i J a are as follows: {i + a, i a} if i E, {i 3 + a, i + a} if i D and i<n 3, {i 3 + a, i a} if i = n 3, {i 3 + a, i 2 + a} if i = n 1. Next is a partition with odd plexes. Lemma 5.2. The Latin square P n has an (h 1,h+ 1)-partition. Proof. We build an (h 1)-plex K by specifying the columns c(i) to use in r i K. c(n 2) = E \{h}, c(h 2) = D \{h + 1},

10 486 EGAN AND WANLESS For i {0, 2, 4,...,h 4}, c(1) ={j E j h} {j D j<h 2}, c(n 1) ={j E j<h 2} {j D j>h}. c(n 4 i) = E \{n 2 i}, c(h 4 i) = D \{n 1 i}, c(h i) = E \{i}, c(3 + i) = D \{1 + i}. It is routine to check that K is an (h 1)-plex. The next corollary follows from Lemma 4.2. Corollary 5.3. For all n 8, the Latin square P n has a mixed plex range. We point out that the Latin square P 4 is of the same species as the Cayley table of Z 2 Z 2. It has a complete plex range. Our next two lemmas use partitions of the plexes constructed in Lemma 5.2 to create smaller plexes. Lemma 5.4. The (h + 1)-plex in P n defined in the proof of Lemma 5.2 contains a set of n8 parallel 2-plexes. Proof. Let m = 4 ( ) n 8 1. For each a {0, 4, 8,...,m}, we find a 2-plex Ja. The columns c(i) inr i J a are as follows: {n 4 + i a, n 2 + i a} if i E and i<h, {1 + i a, 3 + i a} if i E and i h, {h 5 + i a, h 3 + i a} if i D and i<h, {i h a, 2 + i h a} if i D and h<i<n 3, {2 + i h a, 4 + i h a} if i = n 3, {i 2 h a, i h a} if i = n 1. Lemma 5.5. If q is even, then the (h 1)-plex in P n defined in the proof of Lemma 5.2 contains a set of n 8 1 parallel 2-plexes. Proof. Let m = 4 ( n 8 2 ) = h 8. For each a {0, 4, 8,...,m}, we find a 2-plex J a. For i {0, 4, 8,...,h 4}, {1 + i a, 3 + i a}, c(2 + i) ={h i a, h i a}, c(h + i) ={i a, 2 + i a}, c(h i) ={h + i a, h i a},

11 LATIN SQUARES WITH NO SMALL ODD PLEXES 487 c(1 + i) ={h 5 + i a, h 3 + i a}, c(3 + i) ={n 5 + i a, n 3 + i a}, c(h i) ={h 4 + i a, h 2 + i a}, c(h i) ={n 4 + i a, n 2 + i a}. We can now say which odd plexes exist in P 4q when q is even. Lemma 5.6. If q is even, then the Latin square P 4q has a (2 q 1, (q + 1) 2 )-partition and hence has k-plexes for all odd k in the interval 4 1 n + 1 k 4 3 n 1. Proof. We have the following refinements of the (h 1,h + 1)-partition of Lemma 5.2. Lemma 5.4 gives a division of the (h + 1)-plex into n 8 = 2 1 q parallel 2-plexes and a parallel (odd) (q + 1)-plex. Lemma 5.5 gives a division of the (h 1)-plex into n 8 1 = 2 1 q 1 parallel 2-plexes and a parallel (q + 1)-plex. Thus, we have the claimed partition. The union ofa(q + 1)-plex with as many of the 2-plexes as required forms a k-plex for each odd k in the interval 4 1 n + 1 k 4 3 n 1. The following illustrates the (2 3, 5 2 )-partition of P 16 in Lemma 5.6. The circled elements mark the 2-plex specified by Lemma 5.5 as contained within the (dark) 7-plex given in the proof of Lemma 5.2. The two 2-plexes from Lemma 5.4 are identified by paler shading. Marks indicate elements of. We have seen that when q is even a partition of P 4q has at most two odd plexes. In contrast, when q is odd P 4q has a partition containing four odd plexes. Lemma 5.7. If q is odd, then the Latin square P n has an h-partition in which each h-plex can be divided in both of the following ways: (9)

12 488 EGAN AND WANLESS 1. into q parallel 2-plexes, or; 2. into two parallel q-plexes. Thus when q is odd, P 4q has a q-partition and a (2 q,q 2 )-partition. Hence, it has k-plexes for all odd k in the interval 1 4 n k 3 4 n. Proof. We construct two parallel h-plexes H and H by taking the union of q parallel 2-plexes and then divide each of H and H into two parallel q-plexes. We first partition D and E as follows: E 0 ={i E i 0 mod 4}, E 2 ={i E i 2 mod 4}, D 1 ={i D i 1 mod 4}, D 3 ={i D i 3 mod 4}. We use the parallel 2-plexes J a as defined in the proof of Lemma 5.1. Let H = a E0 J a and H = a E2 J a.nowh and H each contain q parallel 2-plexes and together represent an h-partition which by its definition can be refined into a 2-partition. We now need only show that each of H and H contain a q-plex. In H, we define a q-plex K 0. The columns c(i)inr i K 0 are as follows: E 0 if (i E 0 and i h 2) or (i D 3 and i<h), E 2 if (i E 2 and i<h)or(i D 1 and i h 1), D 1 if (i E 0 D 1 and i<h 2) or i = n 1, D 3 if (i E 2 and i h) or(i D 3 and h<i<n 1). In H, we define a q-plex K 2. The columns c(i) inr i K 2 are as follows: E 2 if (i E 0 and i h 2) or (i D 3 and i<h), E 0 if (i E 2 and i<h)or(i D 1 and i h 1), D 3 if (i E 0 D 1 and i<h 2) or i = n 1, D 1 if (i E 2 and i h) or(i D 3 and h<i<n 1). It is routine to check that each of the above q-plexes is contained in its respective h-plex. It follows that K 1 = H \ K 0 and K 3 = H \ K 2 are also q-plexes so we have divided the h-partition into a q-partition. Dividing H in one way and H in the other provides a (2 q,q 2 )-partition. Thus, taking plexes from this partition we can, by the union of one q-plex with as many of the 2-plexes as required, form a k-plex for each odd k in the interval 4 1 n k 4 3 n. Combining the above results, we can now establish exactly which partitions of P n are possible. Theorem 5.8. The plex range of the Latin square P n is the set { 14 n k 4 3 n k is odd} {0, 2, 4,...,n}. Moreover P n possesses a partition of every type consistent with its plex range.

13 LATIN SQUARES WITH NO SMALL ODD PLEXES 489 Proof. A partition with no odd plexes is of a type which can be obtained by the union of plexes from the 2-partition of Lemma 5.1. If n 0 mod 8 then a (q + 1)-plex is the smallest odd plex permitted by Lemma 4.2. A partition with odd plexes has exactly two odd plexes by Lemma 4.3 so it is of a type which can be generated by the (2 q 1, (q + 1) 2 )-partition of Lemma 5.6. In the case that n 4 mod 8, a partition with four odd plexes is a q-partition. A partition with precisely two odd plexes can, for all possible partition types, be formed by the union of plexes from a (2 q,q 2 )-partition. Thus, the partitions of Lemma 5.7 generate every possible partition type for odd plexes in this case. 6. PARTITIONS OF Q n In this section, we investigate partitions of Q n and determine its plex range. We assume throughout this section that n = 2h for an odd integer h 3. Lemma 6.1. The Latin square Q n has a 2-partition and hence has k-plexes for all even k in the interval 0 k n. Proof. For each a E, we construct a 2-plex J a. The columns c(i)inr i J a are as follows: {i + a, i 1 + a} if i E, {i a, i 2 + a} if i D \{1,n 1}, {i a, i 4 + a} if i = 1, {i a, i + a} if i = n 1. Notice that the 2-plex J 0 defined in the proof of Lemma 6.1 contains all of I so it follows from Lemma 4.1 that Q n \ J 0 is an (n 2)-plex which contains no odd plexes. We now show that Q n itself contains many odd plexes. In the next lemma, the specified value of k is the smallest odd value that is not prohibited by Lemma 4.4. Lemma 6.2. The Latin square Q 2h has a (k 2, 2 h k )-partition where k = 2 h Proof. First, we define an even plex J which is the union of h k parallel 2-plexes. Then we find an odd k-plex K in the 2k-plex Q 2h \ J, thus collectively showing a (k 2, 2 h k )- partition. Let m = 2(h k 1). For each a {0, 2, 4,...,m}, we construct a 2-plex J a. The columns c(i) inr i J a are as follows: {i a, i 2 a} if i E or (i D \{1,n 1} and a<m), {i a, i 4 a} if i = 1 and a<m, {i a, i a} if i = n 1 and a<m, {i a, n 1} if i = 1 and a = m, {i a, i} if i D \{1,n 1} and a = m, {i a, 1} if i = n 1 and a = m. Then J = a {0,2,4,...,m} J a. (An illustration of J Q 14 was given in (5).)

14 490 EGAN AND WANLESS Next we construct the odd k-plex K. Let t = h 4, and let X i ={i + 2,i+ 4,i+ 6,...,i+ 2t}, Y i ={n 1 4t + i, n + 1 4t + i, n + 3 4t + i,...,n 3 2t + i}, V i ={h + i + 4,h+ i + 8,h+ i + 12,...,h+ i + 4t}, W i ={i + 2,i+ 6,i+ 10,...,i 2 + 4t}. The columns c(i)inr i K are given in two cases. Case 1. h 1 mod 4. For i E, and for i D, { Xi {i} Y i if i 2t or i h + 2t + 1, X i {i 1} Y i otherwise, {0} {1, 3, 5,...,h 2} if i = 1, {h 1,h+ 1,h+ 3,...,n 2} if i = n 1, V i W i {i + 4t} if h 2t i h + 2t, {h + i} V i W i otherwise. Case 2. h 3 mod 4. In the case h = 7, we refer to (6). For h 7, we define K as follows. For i E, and for i D, { Xi {i} Y i if i h 1, X i {i 1} Y i otherwise, {1, 3, 5,...,h 2} if i = 1, {h + i + 2} V i W i if 3 i h 4, V i {h + i + 10} W i if i = h 2, ( Vi \{h + i + 8} ) {h + i + 10,i t} W i if h i n 7, V i W i {i t} if n 5 i n 3, {h + 1,h+ 3,h+ 5,...,n 2} if i = n 1. Combining the above results, we can now establish exactly which partitions of Q n are possible.

15 LATIN SQUARES WITH NO SMALL ODD PLEXES 491 Theorem 6.3. The plex range of the Latin square Q n is the set { 14 (n 2) k 4 1 (3n + 2) k is odd} {0, 2, 4,...,n}. Moreover Q n possesses a partition of every type consistent with its plex range. Proof. The 2-partition of Lemma 6.1 generates an example of a partition of any type that has no odd plex. A partition with odd plexes has exactly two odd plexes by Lemma 4.6. Therefore, we can obtain a partition of any type with odd plexes (except those ruled out by Lemma 4.4) by using the partition described in Lemma 6.2. Corollary 6.4. For n 10, the Latin square Q n has a mixed plex range. 7. CONCLUSION We have established the existence of Latin squares of even order with a plex range which is quite distinct from those which have previously been described. At the heart of the methodology is Lemma 4.1. This simple statement has also proved useful in [22] to answer the existence question for Latin squares without orthogonal mates. The smallest cases arising from Theorems 5.8 and 6.3 which are noticeably different from any previously recorded examples of mixed plex ranges are P 16 and Q 18 since these Latin squares have neither a 3-plex nor a 1-plex, but do have a 5-plex. Questions raised in [21] on the possible absence of a maximum number of parallel 3-plexes, both within and outside of the class of step-type squares, are answered by these two examples. The construction of P n and Q n could conceivably be adapted to produce examples of Latin squares with other plex ranges. We leave open a conjecture in [21] that for all even n>4 there exists a Latin square of order n which has no transversal but does possess a 3-plex. It is a simple observation that in an even order square every k-plex is indivisible, provided k is taken to be the smallest odd value in the plex range. It follows that the Latin squares that we have constructed contain relatively large indivisible k-plexes, in the sense that k grows linearly with the order of the square. This improves substantially on the result in [21], which had k = O( n). In a subsequent article, we intend to investigate the question of how large an indivisible plex in a square of order n can be. It is also not known whether a Latin square can have a plex range substantially different from examples seen thus far. For example, does there exist a Latin square which has an a-plex and a c-plex but does not have a b-plex for odd integers a<b<c? An analogous question concerning k-factors of graphs is answered in the negative by Katerinis [16]. REFERENCES [1] R. A. Bailey, Orthogonal partitions in designed experiments, Des Codes Cryptogr 8 (1996), [2] N. J. Cavenagh, D. M. Donovan, and E. S. Yazici, Minimal homogeneous latin trades, Discrete Math 306 (2006), [3] C. J. Colbourn and J. H. Dinitz (Editors), Handbook of Combinatorial Designs, Second Edition, Chapman & Hall/CRC, Boca Raton, 2007.

16 492 EGAN AND WANLESS [4] C. J. Colbourn and A. Rosa, Triple Systems, Clarendon Press, Oxford, [5] J. Dénes and A. D. Keedwell, Latin Squares and Their Applications, Akadémiai Kiadó, Budapest, [6] D. J. Finney, Some orthogonal properties of the 4 4 and 6 6 latin squares, Ann Eugenics 12 (1945), [7] D. J. Finney, Orthogonal partitions of the 5 5 latin squares, Ann Eugenics 13 (1946), 1 3. [8] D. J. Finney, Orthogonal partitions of the 6 6 latin squares, Ann Eugenics 13 (1946), [9] R. A. Fisher and F. Yates, Statistical Tables for Biological, Agricultural and Medical Research, Second Edition, Oliver and Boyd, London, [10] G. H. Freeman, Duplexes of 4 4, 5 5 and 6 6 latin squares, Utilitas Math 27 (1985), [11] G. H. Freeman, Orthogonal partitions of 6 6 latin squares of set X, Utilitas Math 41 (1992), [12] G. H. Freeman, Triplex properties of 6 6 latin squares, Utilitas Math 41 (1992), [13] D. C. Gilliland, A note on orthogonal partitions and some well known structures in design of experiments, Ann Statist 5 (1977), [14] M. Hall and L. J. Paige, Complete mappings of finite groups, Pacific J Math 5 (1955), [15] B. Johnston and T. Fullerton, Construction of the 3 2 orthogonal partitions of 6 6 latin squares of set X, Congr Numer 42 (1984), [16] P. Katerinis, Some conditions for the existence of f-factors, J Graph Theory 9 (1985), [17] R. Killgrove, C. Roberts, R. Sternfeld, R. Tamez, R. Derby, and D. Kiel, Latin squares and other configurations, Congr Numer 117 (1996), [18] C. F. Laywine and G. L. Mullen, Discrete Mathematics Using Latin Squares, Wiley, New York, [19] B. D. McKay, J. C. McLeod, and I. M. Wanless, The number of transversals in a latin square, Des Codes Cryptogr 40 (2006), [20] M. D. Plummer, Graph factors and factorization: : A survey, Discrete Math 307 (2007), [21] I. M. Wanless, A generalisation of transversals for latin squares, Electron J Combin 9 (2002), R12. [22] I. M. Wanless and B. S. Webb, The existence of latin squares without orthogonal mates, Des Codes Cryptogr 40 (2006), [23] M. Zaker, Maximum transversal in partial latin squares and rainbow matchings, Discrete Appl Math 155 (2007),