2021The Electronic Journal of CombinatoricsOpen access

On Factors of Independent Transversals in $k$-Partite Graphs

Raphael Yuster

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Abstract

A $[k,n,1]$-graph is a $k$-partite graph with parts of order $n$ such that the bipartite graph induced by any pair of parts is a matching. An independent transversal in such a graph is an independent set that intersects each part in a single vertex. A factor of independent transversals is a set of $n$ pairwise-disjoint independent transversals. Let $f(k)$ be the smallest integer $n_0$ such that every $[k,n,1]$-graph has a factor of independent transversals assuming $n \geqslant n_0$. Several known conjectures imply that for $k \geqslant 2$, $f(k)=k$ if $k$ is even and $f(k)=k+1$ if $k$ is odd. While a simple greedy algorithm based on iterating Hall's Theorem shows that $f(k) \leqslant 2k-2$, no better bound is known and in fact, there are instances showing that the bound $2k-2$ is tight for the greedy algorithm. Here we significantly improve upon the greedy algorithm bound and prove that $f(k) \leqslant 1.78k$ for all $k$ sufficiently large, answering a question of MacKeigan.

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A $[k,n,1]$-graph is a $k$-partite graph with parts of order $n$ such that the bipartite graph induced by any pair of parts is a matching. An independent transversal in such a graph is an independent set that intersects each part in a single vertex. A factor of independent transversals is a set of $n$ pairwise-disjoint independent transversals. Let $f(k)$ be the smallest integer $n_0$ such that every $[k,n,1]$-graph has a factor of independent transversals assuming $n \geqslant n_0$. Several known conjectures imply that for $k \geqslant 2$, $f(k)=k$ if $k$ is even and $f(k)=k+1$ if $k$ is odd. While a simple greedy algorithm based on iterating Hall's Theorem shows that $f(k) \leqslant 2k-2$, no better bound is known and in fact, there are instances showing that the bound $2k-2$ is tight for the greedy algorithm. Here we significantly improve upon the greedy algorithm bound and prove that $f(k) \leqslant 1.78k$ for all $k$ sufficiently large, answering a question of MacKeigan.

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Available abstract

A $[k,n,1]$-graph is a $k$-partite graph with parts of order $n$ such that the bipartite graph induced by any pair of parts is a matching. An independent transversal in such a graph is an independent set that intersects each part in a single vertex. A factor of independent transversals is a set of $n$ pairwise-disjoint independent transversals. Let $f(k)$ be the smallest integer $n_0$ such that every $[k,n,1]$-graph has a factor of independent transversals assuming $n \geqslant n_0$. Several known conjectures imply that for $k \geqslant 2$, $f(k)=k$ if $k$ is even and $f(k)=k+1$ if $k$ is odd. While a simple greedy algorithm based on iterating Hall's Theorem shows that $f(k) \leqslant 2k-2$, no better bound is known and in fact, there are instances showing that the bound $2k-2$ is tight for the greedy algorithm. Here we significantly improve upon the greedy algorithm bound and prove that $f(k) \leqslant 1.78k$ for all $k$ sufficiently large, answering a question of MacKeigan.

Key concepts: Combinatorics, Mathematics, Bipartite graph, Disjoint sets, Upper and lower bounds, Transversal (combinatorics), Graph, Discrete mathematics

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