English

The Honeymoon Oberwolfach Problem: small cases

Combinatorics 2024-07-02 v1

Abstract

The Honeymoon Oberwolfach Problem HOP(2m1,2m2,,2mt)(2m_1,2m_2,\ldots,2m_t) asks the following question. Given n=m1+m2++mtn=m_1+m_2+\ldots +m_t newlywed couples at a conference and tt round tables of sizes 2m1,2m2,,2mt2m_1,2m_2,\ldots,2m_t, is it possible to arrange the 2n2n participants at these tables for 2n22n-2 meals so that each participant sits next to their spouse at every meal, and sits next to every other participant exactly once? A solution to HOP(2m1,2m2,,2mt)(2m_1,2m_2,\ldots,2m_t) is a decomposition of K2n+(2n3)IK_{2n}+(2n-3)I, the complete graph K2nK_{2n} with 2n32n-3 additional copies of a fixed 1-factor II, into 2-factors, each consisting of disjoint II-alternating cycles of lengths 2m1,2m2,,2mt2m_1,2m_2,\ldots,2m_t. The Honeymoon Oberwolfach Problem was introduced in a 2019 paper by Lepine and \v{S}ajna. The authors conjectured that HOP(2m1,2m2,,(2m_1,2m_2,\ldots, 2mt)2m_t) has a solution whenever the obvious necessary conditions are satisfied, and proved the conjecture for several large cases, including the uniform cycle length case m1==mtm_1=\ldots=m_t, and the small cases with n9n \le 9. In the present paper, we extend the latter result to all cases with n20n \le 20 using a computer search.

Cite

@article{arxiv.2407.00204,
  title  = {The Honeymoon Oberwolfach Problem: small cases},
  author = {Marie Rose Jerade and Mateja Šajna},
  journal= {arXiv preprint arXiv:2407.00204},
  year   = {2024}
}
R2 v1 2026-06-28T17:23:15.748Z