Orthogonal one-factorizations of complete multipartite graphs

Abstract

The paper provides a complete solution to the existence problem of two orthogonal one-factorizations of a complete balanced multipartite graph $K_{p\times q}$. In particular, new classes of Howell designs are constructed.

1 Introduction

We use standard notation $K_{p\times q}$ for a complete balanced p-partite graph with each part of cardinality q. Let $V(K_{p\times q})=V_1\cup V_2 \cup \ldots \cup V_p$, where $V_i\cap V_j=\emptyset$ whenever $i\ne j$. Moreover, we also use the standard symbol $K_{q,q}$ to denote $K_{2\times q}$, 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 $\mathcal{F}=\{F_1,\,F_2,\ldots ,F_r\}$ of edge-disjoint one-factors such that $E(G)=\bigcup _{i=1}^r E(F_i)$. Two one-factorizations $\mathcal{F}=\{F_1,\,F_2,\ldots ,F_r\}$ and $\mathcal{F'}=\{F'_1,\,F'_2,\ldots ,F'_r\}$ are orthogonal if $|F_i\cap F'_j|\le 1$ for all $1\le i,j\le r$.

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 squareR of side m is an $m\times m$ 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 $\frac{m-1}{2}$ empty cells.

The existence of two orthogonal one-factorizations, $\mathcal{F}$ and $\mathcal{F'}$, 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 $\mathcal{F}$ whilst each column represents a one-factor in $\mathcal{F'}$.

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 $m\ne 3$ and $m\ne 5$.

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 siden is an $n\times n$ 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 $n\ne 2$ and $n\ne 6$.

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 designH(s, 2n) on the symbol set S is an $s\times s$ 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 $n\le s\le 2n-1$. The existence of an H(n, 2n) comes from two orthogonal one-factorizations of a complete bipartite graph $K_{n,n}$ if $n\ne 2,6$ 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.

Theorem 3

[17] If s is odd and $n<s<2n-1$ then there exists an H(s, 2n), except that H(5, 8) does not exist.

Theorem 4

[2] If s is even and $n<s<2n-1$ then there exists an H(s, 2n).

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 $K_{2n}\setminus F$, 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 “$\mathrm{PBD}$-construction” (cf. [2, 17]), the structure of an underlying graph strongly depends on the choice of parameters, parallel classes in a $\mathrm{PBD}$ 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 $K_{p\times q}$ 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 $K_{p\times q}$.

2 Constructions

We first discuss a general recursive construction which in fact is an application of a standard “expansion by latin squares” method.

Lemma 5

Let p, q and m be integers such that $p\ge 2$, $q\ge 1$, $m\ge 3$ and $m\ne 6$. Suppose there exist two orthogonal one-factorizations of the complete multipartite graph $K_{p\times q}$ and moreover two orthogonal one-factorizations of the complete bipartite graph $K_{m,m}$. Then there exists a pair of orthogonal one-factorizations of the complete multipartite graph $K_{p\times qm}$.

Proof

Let X be the vertex set of $K_{p\times q}$ and let (Y, Y) be the vertex set of $K_{m,m}$. Let $\mathcal{F}^1$, $\mathcal{F}^2$ be two orthogonal one-factorizations of $K_{p\times q}$ on the set X such that $\mathcal{F}^z=\{F_1^z,F_2^z,\ldots ,F_{q(p-1)}^z\}$, $z=1,2$. Moreover, let $\mathcal{E}^1$, $\mathcal{E}^2$ be a pair of orthogonal one-factorizations of $K_{m,m}$ on (Y, Y) and $\mathcal{E}^z=\{E_1^z,E_2^z,\ldots ,E_{m}^z\}$, $z=1,2$.

For each $z=1,2$ we construct a one-factorization $\mathcal{D}^z=\{D_{s,t}^z:\; s=1,2,\dots ,q(p-1),\, t=1,2,\dots ,m\}$ of $K_{p\times qm}$ on vertex set $X\times Y$. We replace each edge of $K_{p\times q}$ with one-factorization $\mathcal{E}^z$ as follows: the edge $\{(i,j),(k,l)\}$ belongs to one-factor $D_{s,t}^z$ if $\{i,k\}$ is an edge of $F_s^z$ and $\{j,l\}$ is an edge of $E_t^z$.

To prove orthogonality of $\mathcal{D}^1$ and $\mathcal{D}^2$ we suppose to the contrary that there are two distinct edges, $\{(i,j),(k,l)\}$ and $\{(i',j'),(k',l')\}$ of $K_{p\times qm}$ that belong together to the same two one-factors, $D_{s,t}^1$ and $D_{s',t'}^2$. We consider two cases:

• (1) $i=i'$ and $k=k'$. Then $j\ne j'$ and $l\ne l'$. 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 $\mathcal{E} ^1$ and $\mathcal{E} ^2$.

• (2) $i\ne i'$ or $k\ne k'$. 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 $\mathcal{F} ^1$ and $\mathcal{F} ^2$.

$\square$

When $q=1$ we immediately get the following.

Corollary 6

Let p and q be integers such that p is even, $p\ge 8$, $m\ge 3$ and $m\ne 6$. There exists a pair of orthogonal one-factorizations of a complete multipartite graph $K_{p\times m}$. $\square$

The second construction is based on Room frames. Let $\{S_1,S_2,\ldots ,S_k\}$ be a partition of the set S. An $\{S_1,S_2,\ldots ,S_k\}$-Room frame is an $|S|\times |S|$ 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 $S_i\times S_i$ are empty, for $1\le i\le k$ (these subarrays are called holes),

• (3) every symbol $x\not \in S_i$ occurs exactly once in each row s and exactly once in each column t, for any $s,t\in S_i$,

• (4) pairs occurring in F are those $\{s,t\}$, where $(s,t)\in (S\times S)\setminus \bigcup _{i=1}^{k}(S_i\times S_i)$.

The type of a Room frame F is a multiset $\{|S_i|:\; 1\le i\le k\}$. An “exponential” notation is used to describe types; a Room frame has type $t_1^{u_1}t_2^{u_2}\ldots t_l^{u_l}$ if there are $u_i$ subsets of cardinality $t_i$, $1\le i\le l$. 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.

Theorem 7

[5, 6, 8,9,10] Suppose t and u are positive integers, $u\ge 4$ and $(t,u) \ne (1,5)$ and (2, 4). Then there exists a uniform Room frame of type $t^u$ if and only if $t(u-1)$ is even.

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 $t\ge 3$, $t\ne 6$, $u\ge 4$ and $t(u-1)$ is even. Then there exists a Howell design $H(ut,ut+t)$ whose underlying graph is $K_{(u+1)\times t}$.

Proof

By Theorem 7, there exists a Room frame F of type $t^u$ on a set S of cardinality tu. Let $S_1,S_2,\ldots ,S_u$ be sets corresponding to holes of F, $S_i\subset S$ and $|S_i|=t$ for each $i=1,2,\ldots u$. Let $S_{u+1}$ be a set containing t elements, none of them in the set S.

For each pair of sets $(S_i,S_{u+1})$, $i=1,2,\ldots u$, 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 $S_i\cup S_{u+1}$. It is easy to see that each hole $S_i\times S_i$ of F can be filled with $H_i$. In this way we obtain a Howell design H on the set $S\cup S_{u+1}$. Notice that none of unordered pairs with both elements in the same $S_i$, $i=1,2,\ldots , u+1$, occurs in H. Thus $K_{(u+1)\times t}$ is an underlying graph of H. $\square$

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 $s+1\le 2n\le 2s$, is a set $S_{s,n}=\{\{x_i,y_i\}:\; 1\le i\le s-n\}\cup \{\{x_i\}:\; s-n+1\le i\le n\}$ that satisfies:

• (1) $\{x_i:\; 1\le i\le n\}\cup \{y_i:\; 1\le i\le s-n\}=G$,

• (2) $(x_i-y_i)\ne \pm (x_j-y_j)$ if $i\ne j$.

If $S_{s,n}$ is a Howell starter, then an ordered set $A_{s,n}=\{\{a_i\}:\; 1\le i\le n\}$ is an adder for $S_{s,n}$ if elements in $A_{s,n}$ are distinct and $\{x_i+a_i:\; 1\le i\le n\}\cup \{y_i+a_i:\; 1\le i\le s-n\}=G$.

In what follows, we use notation $SA_{s,n}=\{\{x_i,y_i\}^{a_i}:\; 1\le i\le s-n\}\cup \{\{x_i\}^{a_i}:\; s-n+1\le i\le n\}$ for a Howell starter $S_{s,n}$ together with an adder $A_{s,n}$. Moreover, we take the cyclic group $\mathbb {Z}_s$ as G.

Lemma 9

Suppose that there exist a Howell starter $S_{s,n}$ together with an adder $A_{s,n}$ in $\mathbb {Z}_s$ 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 (s, 2n), generated by $S_{s,n}$ and $A_{s,n}$, has an underlying graph $K_{(p+1)\times q}$.

Proof

An H(s, 2n) is constructed on the symbol set V of cardinality 2n. Let $(V_0,V_1,\ldots V_p)$ be a partition of V such that $V_j=\{j,j+p,j+2p,\ldots ,j+(q-1)p\}$, where $j=0,1,\ldots ,p-1$, and $V_p=\{\infty _1,\infty _2,\ldots ,\infty _q\}$.

Let us label rows and columns of H(s, 2n) by elements of $\mathbb {Z}_s$. The first row consists of pairs $\{x_i,y_i\}$, for $i=1,2,\dots s-n$, and pairs $\{x_i,\infty _{i-s+n}\}$, for $i=s-n+1,s-n+2,\ldots n$, each of them in column $-a_i$. It is easy to see that all these pairs form a 1-factor of $K_{(p+1)\times q}$ on V. The remaining cells of the Howell design are filled out by developing the square via the group $\mathbb {Z}_s$; that is, the pair $\{x_i+k,y_i+k\}$ is placed in row k and column $-a_i+k$, and also the pair $\{x_i+k,\infty _{i-s+n}\}$ 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 $i=1,2,\dots s-n$, and pairs $\{x_i+a_i,\infty _{i-s+n}\}$, for $i=s-n+1,s-n+2,\ldots n$, which obviously constitute a 1-factor of $K_{(p+1)\times q}$. Due to cyclic rotation of rows and columns we obtain two orthogonal one factorizations of $K_{(p+1)\times q}$. $\square$

An elementary verification shows that the following sets $SA_{s,n}$ are Howell starters and adders for Howell designs of type (s, 2n) whose underlying graphs are $K_{4\times q}$, where $q=n/2$. Notice that none of the pairs in starters has the difference of elements divisible by 3.

Construction 1

$s\equiv 3\pmod {24}$, $s\ge 27$

Let $s=24m+3$ and $n=16m+2$, $m\ge 1$.

$SA_{s,n}=\{\{8m-2i,8m+2+4i\}^{16m+2-i},\, \{8m-1-2i,8m+4+4i\}^{4m-i},\, \{16m+1-4j,16m+2+2j\}^{20m+3+j},\, \{16m-1-4i,16m+3+2i\}^{8m+2+i},\, \{20m+3+4j\}^{13m+3-j},\, \{20m+4+4i\}^{7m+2-i},\, \{20m+5+4i\}^{m+1-i},\, \{20m+6+4i\}^{19m+3-i}:\; i=0,1,\ldots ,2m-1,\; j=0,1,\ldots ,2m\}$.

Construction 2

$s\equiv 9\pmod {24}$, $s\ge 33$

Let $s=24m+9$ and $n=16m+6$, $m\ge 1$.

$SA_{s,n}=\{\{8m+2-2j,8m+4+4j\}^{16m+6-j},\, \{8m+1-2i,8m+6+4i\}^{4m+1-i},\, \{16m+5-4j,16m+6+2j\}^{20m+8+j},\, \{16m+3-4j,16m+7+2j\}^{8m+4+j},\, \{20m+8+4j\}^{7m+4-j},\, \{20m+9+4j\}^{13m+6-j},\, \{20m+10+4j\}^{19m+8-j},\, \{20m+11+4i\}^{m+1-i}:\; i=0,1,\ldots ,2m-1,\; j=0,1,\ldots ,2m\}$.

Construction 3

$s\equiv 15\pmod {24}$, $s\ge 15$

Let $s=24m+15$ and $n=16m+10$, $m\ge 0$.

$SA_{s,n}=\{\{8m+4-2j,8m+6+4j\}^{16m+10-j},\, \{8m+3-2j,8m+8+4j\}^{4m+2-j},\, \{16m+9-4k,16m+10+2k\}^{20m+13+k},\, \{16m+7-4j,16m+11+2j\}^{8m+6+j},\, \{20m+13+4k\}^{7m+5-k},\, \{20m+14+4j\}^{m+1-j},\, \{20m+15+4j\}^{13m+9-j},\, \{20m+16+4j\}^{19m+11-j}:\; j=0,1,\ldots ,2m,\; k=0,1,\ldots ,2m+1\}$.

Construction 4

$s\equiv 21\pmod {24}$, $s\ge 21$

Let $s=24m+21$ and $n=16m+14$, $m\ge 0$.

$SA_{s,n}=\{\{8m+6-2k,8m+8+4k\}^{16m+14-k},\, \{8m+5-2j,8m+10+4j\}^{4m+3-j},\, \{16m+13-4k,16m+14+2k\}^{20m+18+k},\, \{16m+11-4k,16m+15+2k\}^{8m+8+k},\, \{20m+18+4k\}^{7m+6-k},\, \{20m+19+4k\}^{13m+11-k},\, \{20m+20+4k\}^{19m+16-k},\, \{20m+21+4j\}^{m-j}:\; j=0,1,\ldots ,2m,\; k=0,1,\ldots ,2m+1\}$.

Some examples of small order have to be constructed separately.

Example 1

Two orthogonal one-factorizations of $K_{3\times 4}$.

The starter-adder for a Howell design H(8, 12) is $SA_{8,6}=\{\{0,1\}^1,\{2,5\}^2,\{3\}^5,\{4\}^7,$$\{6\}^0,\{7\}^6\}$.

Example 2

Two orthogonal one-factorizations of $K_{4\times 3}$.

The starter-adder for a Howell design H(9, 12) is $SA_{9,6}=\{\{2,4\}^5,\{3,7\}^3,\{5,6\}^7,$$\{0\}^2,\{1\}^4,\{8\}^0\}$.

Example 3

Two orthogonal one-factorizations of $K_{4\times 4}$.

The starter-adder for a Howell design H(12, 16) is $SA_{12,8}=\{\{0,1\}^{11},\{4,11\}^2,\{5,9\}^5,$$\{6,8\}^9,\{2\}^7,\{3\}^4,\{7\}^1,\{10\}^6\}$.

3 Main results

Lemma 10

For every even positive integer q there exist two orthogonal one-factorizations of $K_{3\times q}$.

Proof

We consider separately the following cases. If $q=2$ then $K_{3\times 2}$ is the cocktail-party graph and the assertion immediately holds by Theorem 4. For $q=4$ we use two orthogonal one-factorizations of $K_{3\times 4}$ from Example 1. For $q\ge 6$ and $q\ne 12$ we apply the general recursive construction given in Lemma 5 taking as initial graphs $K_{3\times 2}$ and $K_{\frac{q}{2},\frac{q}{2}}$. If $q=12$ we apply the same construction but we use orthogonal one-factorizations of $K_{3\times 4}$ and $K_{3,3}$. $\square$

Lemma 11

For every integer $q\ge 2$ there exist two orthogonal one-factorizations of $K_{4\times q}$.

Proof

If $q=2$ then two orthogonal one-factorizations of the cocktail party graph $K_{4\times 2}$ exist by Theorem 4. If $q=3$ or $q=4$, two orthogonal one-factorizations of $K_{4\times 3}$ and $K_{4\times 4}$ are given in Examples 2 and 3, respectively. For odd $q\ge 5$ the existence is satisfied by Constructions 1–4 and Lemma 9. For even $q\ge 6$ and $q\ne 12$, the general recursive construction given in Lemma 5 can be used taking as initial graphs $K_{4\times 2}$ and $K_{\frac{q}{2},\frac{q}{2}}$. If $q=12$ we apply the same construction but we use orthogonal one-factorizations of $K_{4\times 4}$ and $K_{3,3}$. $\square$

Lemma 12

Let p, q be integers such that p is odd and $p\ge 5$, q is even and $q\ge 2$. Then there exist two orthogonal one-factorizations of $K_{p\times q}$.

Proof

If $q=2$ then two orthogonal one-factorizations of $K_{p\times 2}$ exist by Theorem 4. For $q\ge 4$ and $q\ne 6$, the existence of two orthogonal one-factorizations of $K_{p\times q}$ follows directly from Lemma 8. If $q=6$, a construction in Lemma 5 can be applied for initial graphs $K_{p\times 2}$ and $K_{3,3}$. $\square$

Lemma 13

Let p, q be integers such that p is even, $p\ge 6$ and $q\ge 2$. Then there exist two orthogonal one-factorizations of $K_{p\times q}$.

Proof

If $q=2$ then $K_{p\times 2}$ is the cocktail-party graph and the assertion holds by Theorem 4. For $q\ge 3$ and $q\ne 6$, the existence of two orthogonal one-factorizations of $K_{p\times q}$ follows from Lemma 8. If $q=6$ then we apply the general recursive construction given in Lemma 5 taking $K_{p\times 2}$ and $K_{3,3}$ as initial graphs. $\square$

Combining Lemmas 10–13 together with Theorems 1 and 2 gives the main result.

Theorem 14

For any integers p and q such that pq is even, $p\ge 2$ and $q\ge 1$, a complete balanced multipartite graph $K_{p\times q}$ admits a pair of orthogonal one-factorizations, except for $(p,q)=(2,2),(2,6),(4,1)$ or (6, 1). $\square$

Corollary 15

Let p and q be integers such that pq is even, $p\ge 2$, $q\ge 1$ and (p, q) is none of the pairs (2, 2), (2, 6), (4, 1) and (6, 1). Then there exists a Howell design of type $(pq-q,pq)$ whose underlying graph is $K_{p\times q}$. $\square$