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The relation between Hamiltonian and $1$-tough properties of the Cartesian product graphs

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 Added by Louis Kao
 Publication date 2020
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and research's language is English




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The relation between Hamiltonicity and toughness of a graph is a long standing research problem. The paper studies the Hamiltonicity of the Cartesian product graph $G_1square G_2$ of graphs $G_1$ and $G_2$ satisfying that $G_1$ is traceable and $G_2$ is connected with a path factor. Let Pn be the path of order $n$ and $H$ be a connected bipartite graph. With certain requirements of $n$, we show that the following three statements are equivalent: (i) $P_nsquare H$ is Hamiltonian; (ii) $P_nsquare H$ is $1$-tough; and (iii) $H$ has a path factor.



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101 - Yuping Gao , Songling Shan 2021
The toughness of a noncomplete graph $G$ is the maximum real number $t$ such that the ratio of $|S|$ to the number of components of $G-S$ is at least $t$ for every cutset $S$ of $G$, and the toughness of a complete graph is defined to be $infty$. Determining the toughness for a given graph is NP-hard. Chv{a}tals toughness conjecture, stating that there exists a constant $t_0$ such that every graph with toughness at least $t_0$ is hamiltonian, is still open for general graphs. A graph is called $(P_3cup 2P_1)$-free if it does not contain any induced subgraph isomorphic to $P_3cup 2P_1$, the disjoint union of $P_3$ and two isolated vertices. In this paper, we confirm Chv{a}tals toughness conjecture for $(P_3cup 2P_1)$-free graphs by showing that every 7-tough $(P_3cup 2P_1)$-free graph on at least three vertices is hamiltonian.
111 - Songling Shan 2021
Let $G$ be a $t$-tough graph on $nge 3$ vertices for some $t>0$. It was shown by Bauer et al. in 1995 that if the minimum degree of $G$ is greater than $frac{n}{t+1}-1$, then $G$ is hamiltonian. In terms of Ores conditions in this direction, the problem was only studied when $t$ is between 1 and 2. In this paper, we show that if the degree sum of any two nonadjacent vertices of $G$ is greater than $frac{2n}{t+1}+t-2$, then $G$ is hamiltonian.
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