On the minisymposium problem

TL;DR

This paper introduces the minisymposium problem, a variant of the Oberwolfach problem involving a subset of vertices, and provides solutions under specific modular and cycle-length conditions, expanding the known cases.

Contribution

It extends the classical Oberwolfach problem by incorporating a subsystem on a subset of vertices and offers new solutions using the Hilton-Johnson method for specific parameters.

Findings

Solutions when v ≡ m ≡ 2 mod 4 with even cycle lengths.

Extensive results for equal cycle lengths k, especially when m divides v.

Most cases solved for equal cycle lengths, except possibly when k is odd and v is even.

Abstract

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.