Orthogonal one-factorizations of complete multipartite graphs

The paper provides a complete solution to the existence problem of two orthogonal one-factorizations of a complete balanced multipartite graph Kp×q\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$K_{p\times q}$$\end{document}. In particular, new classes of Howell designs are constructed.


Introduction
We use standard notation K p×q for a complete balanced p-partite graph with each part of cardinality q. Let V (K p×q ) = V 1 ∪ V 2 ∪ . . . ∪ V p , where V i ∩ V j = ∅ whenever i = j. Moreover, we also use the standard symbol K q,q to denote K 2×q , a complete balanced bipartite graph on 2q vertices.
A one-factor in a graph G is a regular spanning subgraph of degree one. A one-factorization of G is a set F = {F 1 , F 2 , . . . , F r } of edge-disjoint one-factors such that E(G) = r i=1 E(F i ). Two one-factorizations F = {F 1 , F 2 , . . . , F r } and F = {F 1 , F 2 , . . . , F r } are orthogonal if |F i ∩ F j | ≤ 1 for all 1 ≤ i, j ≤ r .
Orthogonal one-factorizations of complete graphs are well-studied, mostly in terms of Rooms squares, cf. [7,12]. Let m be an odd integer and let S be a set of m + 1 elements (symbols). A Room square R of side m is an m × m array which satisfies the following properties: (1) every cell of R is either empty or contains an unordered pair of symbols from S, (2) every symbol of S occurs exactly once in each row and exactly once in each column of R, (3) every unordered pair of symbols occurs in precisely one cell in R. Thus each row and each column of R contain m−1 2 empty cells. The existence of two orthogonal one-factorizations, F and F , of a complete graph K 2n is equivalent to the existence of a Room square of side 2n − 1: each row corresponds to a one-factor in F whilst each column represents a one-factor in F .
The existence problem for Room squares is completely settled.
Theorem 1 [14] A Room square of side m exists if and only if m is odd and m = 3 and m = 5.
Two orthogonal one-factorizations of a complete bipartite graph K n,n are equivalent to two orthogonal latin squares of side n. A latin square of side n is an n × n array in which each cell contains a single symbol from an n-element set S, such that each symbol occurs exactly once in each row and exactly once in each column. Two latin squares, L and L , of side n are orthogonal if the n 2 ordered pairs (L(i, j), L (i, j)) are all distinct. Bose, Shrikhande and Parker [3] completely solved the famous Euler's conjecture.

Theorem 2 [3]
A pair of orthogonal latin squares of side n exists whenever n = 2 and n = 6.
The above equivalences can be extended to other classes of regular graphs. Namely, a pair of orthogonal one-factorizations of an s-regular graph G on 2n vertices corresponds to the existence of a Howell design of type (s, 2n), for which a graph G is called an underlying graph, cf. [15]. Let S be a set of 2n symbols. A Howell design H (s, 2n) on the symbol set S is an s × s array that satisfies the following conditions: (1) every cell is either empty or contains an unordered pair of symbols from S, (2) every symbol of S occurs exactly once in each row and exactly once in each column of H , (3) every unordered pair of symbols occurs in at most one cell of H .
Necessary condition for the existence of Howell designs H (s, 2n) is n ≤ s ≤ 2n − 1. The existence of an H (n, 2n) comes from two orthogonal one-factorizations of a complete bipartite graph K n,n if n = 2, 6 and some 6-regular graph if n = 6 [13]. There in no H (2,4). In the other extreme case, an H (2n − 1, 2n) is a Room square of side 2n − 1. The existence of Howell designs has been completely determined for all remaining values of s. An important question related to Howell designs concerns properties of graphs which are underlying graphs of Howell designs. While for s = 2n − 1 and s = 2n − 2 these graphs are unique (the complete graph K 2n and the cocktail party graph K 2n \ F, respectively, where F is a one-factor), determining these graphs in general seems to be hopeless [15,16]. We have to notice that some known constructions may provide Howell designs for certain classes of underlying graphs; in particular, in the case of a powerful recursive "PBD-construction" (cf. [2,17]), the structure of an underlying graph strongly depends on the choice of parameters, parallel classes in a PBD as well as Howell subdesigns used in the recursion.
It is known that a necessary and sufficient condition for the existence of a one-factorization of a complete balanced multipartite graph K p×q is that pq is even [11]. The goal of this paper is to show that balanced complete multipartite graphs are underlying graphs of Howell designs; the main result provides a complete solution to the existence problem of two orthogonal one-factorizations of K p×q .

Constructions
We first discuss a general recursive construction which in fact is an application of a standard "expansion by latin squares" method. Proof Let X be the vertex set of K p×q and let (Y, Y ) be the vertex set of K m,m . Let We replace each edge of K p×q with one-factorization E z as follows: the edge {(i, j), (k, l)} belongs to one-factor D z s,t if {i, k} is an edge of F z s and { j, l} is an edge of E z t . To prove orthogonality of D 1 and D 2 we suppose to the contrary that there are two distinct edges, {(i, j), (k, l)} and {(i , j ), (k , l )} of K p×qm that belong together to the same two one-factors, D 1 s,t and D 2 s ,t . We consider two cases: (1) i = i and k = k . Then j = j and l = l . Moreover, { j, l} and { j , l } are both in the same two one-factors E 1 t and E 2 t , a contradiction to the orthogonality of E 1 and E 2 . (2) i = i or k = k . Then {i, k} and {i , k } are two distinct edges of both F 1 s and F 2 s , a contradiction to the orthogonality of F 1 and F 2 .
When q = 1 we immediately get the following.

Corollary 6
Let p and q be integers such that p is even, p ≥ 8, m ≥ 3 and m = 6. There exists a pair of orthogonal one-factorizations of a complete multipartite graph K p×m .
The second construction is based on Room frames. Let {S 1 , S 2 , . . . , S k } be a partition of the set S. An {S 1 , S 2 , . . . , S k }-Room frame is an |S| × |S| array, F, indexed by S, which satisfies the following properties: (1) every cell of F is either empty or contains an unordered pair of symbols from S, (2) the subarrays S i × S i are empty, for 1 ≤ i ≤ k (these subarrays are called holes), (3) every symbol x / ∈ S i occurs exactly once in each row s and exactly once in each column t, for any s, t ∈ S i , (4) pairs occurring in F are those {s, t}, where (s, t) The type of a Room frame F is a multiset {|S i | : 1 ≤ i ≤ k}. An "exponential" notation is used to describe types; a Room frame has type t u 1 1 t u 2 2 . . . t u l l if there are u i subsets of cardinality t i , 1 ≤ i ≤ l. A Room frame of type t u (one hole size) is called uniform. In particular, a Room square of side m is equivalent to a Room frame of type 1 m .
The existence problem for uniform Room frames is completely solved. Room frames are key structures in the "filling in holes" construction for Howell designs, cf. [4]. In particular, applying this construction for uniform Room frames yields Howell designs with complete balanced multipartite graphs as underlying graphs.

Lemma 8
Let t and u be integers such that t ≥ 3, t = 6, u ≥ 4 and t (u − 1) is even. Then there exists a Howell design H (ut, ut + t) whose underlying graph is K (u+1)×t .
Proof By Theorem 7, there exists a Room frame F of type t u on a set S of cardinality tu. Let S 1 , S 2 , . . . , S u be sets corresponding to holes of F, S i ⊂ S and |S i | = t for each i = 1, 2, . . . u. Let S u+1 be a set containing t elements, none of them in the set S.
For each pair of sets (S i , S u+1 ), i = 1, 2, . . . u, by Theorem 2, there exists a pair of orthogonal latin squares of side t which correspond to two orthogonal one factorizations of complete bipartite graph K t,t with bipartition (S i , S u+1 ), and moreover which are equivalent to a Howell design H i of type (t, 2t) on the set S i ∪ S u+1 . It is easy to see that each hole S i × S i of F can be filled with H i . In this way we obtain a Howell design H on the set S ∪ S u+1 . Notice that none of unordered pairs with both elements in the same S i , i = 1, 2, . . . , u + 1, occurs in H . Thus K (u+1)×t is an underlying graph of H .
The well-known starter-adder construction, as a basic method to obtain Room squares, can be generalized for Howell designs, cf. [1]. Let G be an abelian group of order s. A Howell starter in G, where s + 1 ≤ 2n ≤ 2s, is a set S s,n = {{x i , y i } : If S s,n is a Howell starter, then an ordered set A s,n = {{a i } : 1 ≤ i ≤ n} is an adder for S s,n if elements in A s,n are distinct and {x i +a i : In what follows, we use notation S A s, for a Howell starter S s,n together with an adder A s,n . Moreover, we take the cyclic group Z s as G. (s, 2n), generated by S s,n and A s,n , has an underlying graph K ( p+1)×q .

Lemma 9 Suppose that there exist a Howell starter S s,n together with an adder A s,n in Z s such that q = 2n − s is a divisor of s and moreover none of the pairs in S s,n has the difference of its elements divisible by p = s/q. Then a Howell design of type
Proof An H (s, 2n) is constructed on the symbol set V of cardinality 2n. Let (V 0 , V 1 , . . . V p ) be a partition of V such that V j = { j, j + p, j + 2 p, . . . , j + (q − 1) p}, where j = 0, 1, . . . , p − 1, and V p = {∞ 1 , ∞ 2 , . . . , ∞ q }.
Let us label rows and columns of H (s, 2n) by elements of Z s . The first row consists of pairs {x i , y i }, for i = 1, 2, . . . s −n, and pairs {x i , ∞ i−s+n }, for i = s −n +1, s −n +2, . . . n, each of them in column −a i . It is easy to see that all these pairs form a 1-factor of K ( p+1)×q on V . The remaining cells of the Howell design are filled out by developing the square via the group Z s ; that is, the pair {x i + k, y i + k} is placed in row k and column −a i + k, and also the pair {x i + k, ∞ i−s+n } in a cell in row k and column −a i + k, where all arithmetic is modulo s. Thus, in particular, the first column consists of pairs {x i + a i , y i + a i }, for i = 1, 2, . . . s − n, and pairs {x i + a i , ∞ i−s+n }, for i = s − n + 1, s − n + 2, . . . n, which obviously constitute a 1-factor of K ( p+1)×q . Due to cyclic rotation of rows and columns we obtain two orthogonal one factorizations of K ( p+1)×q .
An elementary verification shows that the following sets S A s,n are Howell starters and adders for Howell designs of type (s, 2n) whose underlying graphs are K 4×q , where q = n/2. Notice that none of the pairs in starters has the difference of elements divisible by 3. ≡ 3 (mod 24), s ≥ 27 Let s = 24m + 3 and n = 16m + 2, m ≥ 1.

Construction 1 s
Construction 2 s ≡ 9 (mod 24), s ≥ 33 Let s = 24m + 9 and n = 16m Some examples of small order have to be constructed separately.

Main results
Lemma 10 For every even positive integer q there exist two orthogonal one-factorizations of K 3×q .
Proof We consider separately the following cases. If q = 2 then K 3×2 is the cocktail-party graph and the assertion immediately holds by Theorem 4. For q = 4 we use two orthogonal one-factorizations of K 3×4 from Example 1. For q ≥ 6 and q = 12 we apply the general recursive construction given in Lemma 5 taking as initial graphs K 3×2 and K q 2 , q 2 . If q = 12 we apply the same construction but we use orthogonal one-factorizations of K 3×4 and K 3,3 .

Lemma 11
For every integer q ≥ 2 there exist two orthogonal one-factorizations of K 4×q .
Proof If q = 2 then two orthogonal one-factorizations of the cocktail party graph K 4×2 exist by Theorem 4. If q = 3 or q = 4, two orthogonal one-factorizations of K 4×3 and K 4×4 are given in Examples 2 and 3, respectively. For odd q ≥ 5 the existence is satisfied by Constructions 1-4 and Lemma 9. For even q ≥ 6 and q = 12, the general recursive construction given in Lemma 5 can be used taking as initial graphs K 4×2 and K q 2 , q 2 . If q = 12 we apply the same construction but we use orthogonal one-factorizations of K 4×4 and K 3,3 .

Lemma 12
Let p, q be integers such that p is odd and p ≥ 5, q is even and q ≥ 2. Then there exist two orthogonal one-factorizations of K p×q .
Proof If q = 2 then two orthogonal one-factorizations of K p×2 exist by Theorem 4. For q ≥ 4 and q = 6, the existence of two orthogonal one-factorizations of K p×q follows directly from Lemma 8. If q = 6, a construction in Lemma 5 can be applied for initial graphs K p×2 and K 3,3 .

Lemma 13
Let p, q be integers such that p is even, p ≥ 6 and q ≥ 2. Then there exist two orthogonal one-factorizations of K p×q .
Proof If q = 2 then K p×2 is the cocktail-party graph and the assertion holds by Theorem 4. For q ≥ 3 and q = 6, the existence of two orthogonal one-factorizations of K p×q follows from Lemma 8. If q = 6 then we apply the general recursive construction given in Lemma 5 taking K p×2 and K 3,3 as initial graphs.
Combining Lemmas 10-13 together with Theorems 1 and 2 gives the main result.