Compatible Hamilton cycles in graphs with large minimum degree

TL;DR

This paper investigates the conditions under which large graphs with high minimum degree contain compatible Hamilton cycles, focusing on the bounds of forbidden edge pairs that still guarantee such cycles.

Contribution

The authors improve the bounds on the incompatibility parameter  for guaranteeing compatible Hamilton cycles in graphs with large minimum degree.

Findings

=1/8 suffices for graphs with minimum degree /2 +  n

 /6 is necessary for such graphs

Provided bounds are functions of the ratio (G)/n

Abstract

The renowned theorem of Dirac states that if $G$ is a graph with minimum degree at least $n/2$ then $G$ has a Hamilton cycle. A natural generalisation asks what properties of an edge-colouring of $G$ guarantee the existence of a properly edge-coloured Hamilton cycle in $G$. This concept can be further generalised as follows: an \emph{incompatibility system} for $G$ is a set~$\mathcal{F}$ of `forbidden' pairs of adjacent edges, that is, $\mathcal{F}\subseteq \{\{uv,vw\}\in \binom{E(G)}2\}$. A cycle in $G$ is then \emph{compatible} if no two of its edges form a pair in $\mathcal{F}$. The system $\mathcal{F}$ is called \emph{$\mu n$-bounded} if for all $v\in V(G)$ and $uv\in E(G)$, there are at most $\mu n$ pairs $\{uv,vw\}\in \mathcal{F}$. How small must $\mu$ be to guarantee the existence of a compatible Hamilton cycle in $G$? Krivelevich, Lee and Sudakov showed that $\mu=10^{-16}$ suffices (for $n$ large), while an example of Bollob\'as and Erd\H{o}s shows that $\mu\leq 1/4$ is necessary. We significantly reduce this gap for large graphs of minimum degree at least $(1/2+\varepsilon)n$, by showing that $\mu=1/8$ suffices but $\mu\leq 1/6$ is necessary for such graphs. In fact, we give more precise bounds which are functions of $\delta(G)/n$.