On the Hamilton-Waterloo Problem with odd orders

TL;DR

This paper establishes broad conditions under which the Hamilton-Waterloo problem can be solved for odd orders, providing new sufficiency results for various parameter ranges and specific cases.

Contribution

It proves that necessary conditions are sufficient for many cases when the order is a multiple of the product of cycle lengths, and explores factorizations of lexicographic products of cycles.

Findings

Necessary conditions are sufficient for v multiple of mn and v > mn, with some exceptions.

Sufficiency is shown for v=mn when β > (n+5)/2, with specific exceptions.

The lexicographic product of cycles can be factorized into cycle factors under broad conditions.

Abstract

Given non-negative integers $v, m, n, \alpha, \beta$, the Hamilton-Waterloo problem asks for a factorization of the complete graph $K_v$ into $\alpha$ $C_m$-factors and $\beta$ $C_n$-factors. Clearly, $v$ odd, $n,m\geq 3$, $m\mid v$, $n\mid v$ and $\alpha+\beta = (v-1)/2$ are necessary conditions. To date results have only been found for specific values of $m$ and $n$. In this paper we show that for any $m$ and $n$ the necessary conditions are sufficient when $v$ is a multiple of $mn$ and $v>mn$, except possibly when $\beta=1$ or 3, with five additional possible exceptions in $(m,n,\beta)$. For the case where $v=mn$ we show sufficiency when $\beta > (n+5)/2$ except possibly when $(m,\alpha) = (3,2)$, $(3,4)$, with seven further possible exceptions in $(m,n,\alpha,\beta)$. We also show that when $n\geq m\geq 3$ are odd integers, the lexicographic product of $C_m$ with the empty graph of order $n$ has a factorization into $\alpha$ $C_m$-factors and $\beta$ $C_n$-factors for every $0\leq \alpha \leq n$, $\beta = n-\alpha$, except possibly when $\alpha= 2,4$, $\beta = 1, 3$, with three additional possible exceptions in $(m,n,\alpha)$.