Perfect $1$-factorisations of $K_{11,11}$

Fuente: arXiv
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Autores principales: Allsop, Jack, Wanless, Ian M.
Formato: Preprint
Publicado: 2025
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author Allsop, Jack
Wanless, Ian M.
author_facet Allsop, Jack
Wanless, Ian M.
contents A perfect $1$-factorisation of a graph is a decomposition of that graph into $1$-factors such that the union of any two $1$-factors is a Hamiltonian cycle. A Latin square of order $n$ is row-Hamiltonian if for every pair $(r,s)$ of distinct rows, the permutation mapping $r$ to $s$ has a single cycle of length $n$. We report the results of a computer enumeration of the perfect $1$-factorisations of the complete bipartite graph $K_{11,11}$. This also allows us to find all row-Hamiltonian Latin squares of order $11$. Finally, we plug a gap in the literature regarding how many row-Hamiltonian Latin squares are associated with the classical families of perfect $1$-factorisations of complete graphs.
format Preprint
id arxiv_https___arxiv_org_abs_2506_02455
institution arXiv
publishDate 2025
record_format arxiv
spellingShingle Perfect $1$-factorisations of $K_{11,11}$
Allsop, Jack
Wanless, Ian M.
Combinatorics
05C70, 05B15
A perfect $1$-factorisation of a graph is a decomposition of that graph into $1$-factors such that the union of any two $1$-factors is a Hamiltonian cycle. A Latin square of order $n$ is row-Hamiltonian if for every pair $(r,s)$ of distinct rows, the permutation mapping $r$ to $s$ has a single cycle of length $n$. We report the results of a computer enumeration of the perfect $1$-factorisations of the complete bipartite graph $K_{11,11}$. This also allows us to find all row-Hamiltonian Latin squares of order $11$. Finally, we plug a gap in the literature regarding how many row-Hamiltonian Latin squares are associated with the classical families of perfect $1$-factorisations of complete graphs.
title Perfect $1$-factorisations of $K_{11,11}$
topic Combinatorics
05C70, 05B15
url https://arxiv.org/abs/2506.02455