Dara Moazzami
Abstract
The tenacity of a graph $G$, $T(G)$, is defined by$T(G) = min\{\frac{\mid S\mid +\tau(G-S)}{\omega(G-S)}\}$, where theminimum is taken over all vertex cutsets $S$ of $G$. We define$\tau(G - S)$ to be the number of the vertices in the largestcomponent of the graph $G-S$, and $\omega(G-S)$ be the number ...
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The tenacity of a graph $G$, $T(G)$, is defined by$T(G) = min\{\frac{\mid S\mid +\tau(G-S)}{\omega(G-S)}\}$, where theminimum is taken over all vertex cutsets $S$ of $G$. We define$\tau(G - S)$ to be the number of the vertices in the largestcomponent of the graph $G-S$, and $\omega(G-S)$ be the number ofcomponents of $G-S$. In this paperwe consider the relationship between the minimum degree $\delta (G)$ of a graph and the complexityof recognizing if a graph is $T$-tenacious. Let $T\geq 1$ be a rational number. We first show that if$\delta(G)\geq \frac{Tn}{T+1}$, then $G$ is $T$-tenacious. On the other hand, for any fixed $\epsilon>0$, weshow that it is $NP$-hard to determine if $G$ is $T$-tenacious, even for the class of graphs with $\delta(G)\geq(\frac{T}{T+1}-\epsilon )n$.