On the Generalized Honeymoon Oberwolfach Problem

TL;DR

This paper investigates the generalized Honeymoon Oberwolfach Problem with multiple round tables, providing solutions for specific cases based on modular arithmetic conditions and small table sizes, advancing combinatorial seating arrangements.

Contribution

It develops new solutions for the generalized HOP with multiple round tables, especially for cases with two tables and small table sizes, under specific modular conditions.

Findings

Solutions exist when n ≡ 1 or m₁ + m₂ mod (2m₁ + 2m₂).

HOP solutions are found for small tables with m ≤ 10, n odd, and n(n-1) divisible by 2m.

The paper extends the understanding of seating arrangements in complex combinatorial problems.

Abstract

The generalized Honeymoon Oberwolfach Problem (HOP) asks whether it is possible to seat $2n$ participants consisting of $n$ newlywed couples at a conference with $s$ tables of size $2$ and $t$ ''round'' tables of sizes $2m_1, 2m_2, \ldots, 2m_t$, where $n = s + \sum_{i=1}^{t} m_i $ with all $m_i \geq 2$, over several nights so that each participant sits next to their spouse every time and next to each other participant exactly once. We denote this problem by $\mathrm{HOP}(2^{\langle s \rangle}, 2m_1, \ldots, 2m_t)$. This paper is the first of two papers investigating the generalized HOP. While the second paper will deal with the generalized HOP with a single round table (i.e. table of size at least $4$), the present work develops solutions for the generalized HOP with multiple round tables. In particular, we present solutions to certain cases with two round tables, showing that a solution to $\mathrm{HOP}(2^{\langle s \rangle}, 2m_1, 2m_2)$ exists when $n \equiv 1 \pmod{(2m_1 + 2m_2)}$ or $n \equiv m_1 + m_2 \pmod{(2m_1 + 2m_2)}$. We also develop solutions for cases with small round tables, showing that $\mathrm{HOP}(2^{\langle s \rangle}, 2m_1, \dots, 2m_t)$ has a solution whenever $m = m_1 + \dots + m_t \leq 10$, $n = s + m$ is odd, and $n(n - 1) \equiv 0 \pmod{2m}$.