On the regular 2-connected 2-path Hamiltonian graphs

TL;DR

This paper establishes conditions under which 2-connected, k-regular graphs are 2-path Hamiltonian, demonstrating that if the graph minus a specific 2-path remains connected, a Hamiltonian cycle containing that path exists.

Contribution

It introduces a new sufficient condition for 2-connected, k-regular graphs to be 2-path Hamiltonian based on connectivity after removing a 2-path.

Findings

Hamiltonian cycle containing a 2-path exists if G minus the 2-path is connected

The condition is nearly sharp with the bound at 2k+1 vertices

Provides a new criterion for 2-path Hamiltonicity in regular graphs

Abstract

A graph $G$ is $l$-path Hamiltonian if every path of length not exceeding $l$ is contained in a Hamiltonian cycle. It is well known that a 2-connected, $k$-regular graph $G$ on at most $3k-1$ vertices is edge-Hamiltonian if for every edge $uv$ of $G$, $\{u,v\}$ is not a cut-set. Thus $G$ is 1-path Hamiltonian if $G\setminus \{u,v\}$ is connected for every edge $uv$ of $G$. Let $P=uvz$ be a 2-path of a 2-connected, $k$-regular graph $G$ on at most $2k$ vertices. In this paper, we show that there is a Hamiltonian cycle containing the 2-path $P$ if $G\setminus V(P)$ is connected. Therefore, the work implies a condition for a 2-connected, $k$-regular graph to be 2-path Hamiltonian. An example shows that the $2k$ is almost sharp, i.e., the number is at most $2k+1$.