Hamiltonicity of $1$-tough $(P_2\cup kP_1)$-free graphs (2303.09741v2)

Abstract: Given a graph $H$, a graph $G$ is $H$-free if $G$ does not contain $H$ as an induced subgraph. For a positive real number $t$, a non-complete graph $G$ is said to be $t$-tough if for every vertex cut $S$ of $G$, the ratio of $|S|$ to the number of components of $G-S$ is at least $t$. A complete graph is said to be $t$-tough for any $t>0$. Chv\'{a}tal's toughness conjecture, stating that there exists a constant $t_0$ such that every $t_0$-tough graph with at least three vertices is Hamiltonian, is still open in general. Chv\'{a}tal and Erd\"{o}s \cite{CE} proved that, for any integer $k\ge 1$, every $\max{2,k}$-connected $(k+1)P_1$-free graph on at least three vertices is Hamiltonian. Along the Chv\'{a}tal-Erd\"{o}s theorem, Shi and Shan \cite{SS} proved that, for any integer $k\ge 4$, every $4$-tough $2k$-connected $(P_2\cup kP_1)$-free graph with at least three vertices is Hamiltonian, and furthermore, they proposed a conjecture that for any integer $k\ge 1$, any $1$-tough $2k$-connected $(P_2\cup kP_1)$-free graph is Hamiltonian. In this paper, we confirm the conjecture, and furthermore, we show that if $k\ge 3$, then the condition $2k$-connected' may be weakened to be$2(k-1)$-connected'. As an immediate consequence, for any integer $k\ge 3$, every $(k-1)$-tough $(P_2\cup kP_1)$-free graph is Hamiltonian. This improves the result of Hatfield and Grimm \cite{HG}, stating that every $3$-tough $(P_2\cup 3P_1)$-free graph is Hamiltonian.