Bipartite 2-factorizations of complete multipartite graphs

It is shown that if K is any regular complete multipartite graph of even degree, and F is any bipartite 2-factor of K, then there exists a factorization of K into F; except that there is no factorization of K into F when F is the union of two disjoint 6-cycles

On the Hamilton-Waterloo problem for bipartite 2-factors

Given two 2-regular graphs F1 and F2, both of order n, the Hamilton-Waterloo Problem for F1 and F2 asks for a factorization of the complete graph Kn into a1 copies of F1, a2 copies of F2, and a 1-factor if n is even, for all nonnegative integers a1 and a2 satisfying a1+a2=?n-12?. We settle the Hamilton-Waterloo Problem for all bipartite 2-regular graphs F1 and F2 where F1 can be obtained from F2 by replacing each cycle with a bipartite 2-regular graph of the same order

On the minisymposium problem

The generalized Oberwolfach problem asks for a factorization of the complete graph $K_v$ into prescribed $2$-factors and at most a $1$-factor. When all $2$-factors are pairwise isomorphic and $v$ is odd, we have the classic Oberwolfach problem, which was originally stated as a seating problem: given $v$ attendees at a conference with $t$ circular tables such that the $i$th table seats $a_i$ people and ${\sum_{i=1}^t a_i = v}$, find a seating arrangement over the $\frac{v-1}{2}$ days of the conference, so that every person sits next to each other person exactly once. In this paper we introduce the related {\em minisymposium problem}, which requires a solution to the generalized Oberwolfach problem on $v$ vertices that contains a subsystem on $m$ vertices. That is, the decomposition restricted to the required $m$ vertices is a solution to the generalized Oberwolfach problem on $m$ vertices. In the seating context above, the larger conference contains a minisymposium of $m$ participants, and we also require that pairs of these $m$ participants be seated next to each other for $\left\lfloor\frac{m-1}{2}\right\rfloor$ of the days. When the cycles are as long as possible, i.e.\ $v$, $m$ and $v-m$, a flexible method of Hilton and Johnson provides a solution. We use this result to provide further solutions when $v \equiv m \equiv 2 \pmod 4$ and all cycle lengths are even. In addition, we provide extensive results in the case where all cycle lengths are equal to $k$, solving all cases when $m\mid v$, except possibly when $k$ is odd and $v$ is even.Comment: 25 page

Oberwolfach rectangular table negotiation problem

AbstractWe completely solve certain case of a “two delegation negotiation” version of the Oberwolfach problem, which can be stated as follows. Let H(k,3) be a bipartite graph with bipartition X={x1,x2,…,xk},Y={y1,y2,…,yk} and edges x1y1,x1y2,xkyk−1,xkyk, and xiyi−1,xiyi,xiyi+1 for i=2,3,…,k−1. We completely characterize all complete bipartite graphs Kn,n that can be factorized into factors isomorphic to G=mH(k,3), where k is odd and mH(k,3) is the graph consisting of m disjoint copies of H(k,3)