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arXiv:1710.00505 [math.CO]AbstractReferencesReviewsResources

Hamiltonicity in random graphs is born resilient

Richard Montgomery

Published 2017-10-02Version 1

Let $\{G_M\}_{M\geq 0}$ be the random graph process, where $G_0$ is the empty graph on $n$ vertices and subsequent graphs in the sequence are obtained by adding a new edge uniformly at random. For each $\varepsilon>0$, we show that, almost surely, any graph $G_M$ with minimum degree at least 2 is not only Hamiltonian (as shown by Bollob\'as), but remains Hamiltonian despite the removal of any set of edges, as long as at most $(1/2-\varepsilon)$ of the edges incident to each vertex are removed. We say that such a graph is $(1/2-\varepsilon)$-resiliently Hamiltonian. Furthermore, for each $\epsilon>0$, we show that, almost surely, each graph $G_M$ is not $(1/2+\varepsilon)$-resiliently Hamiltonian. These results strengthen those by Lee and Sudakov on the likely resilience of Hamiltonicity in the binomial random graph. For each $k$, we denote by $G^{(k)}$ the (possibly empty) maximal subgraph with minimum degree at least $k$ of a graph $G$. That is, the $k$-core of $G$. Krivelevich, Lubetzky and Sudakov have shown that, for each $k\geq 15$, in almost every random graph process $\{G_M\}_{M\geq 0}$, every non-empty $k$-core is Hamiltonian. We show that, for each $\varepsilon>0$ and $k\geq k_0(\varepsilon)$, in almost every random graph process $\{G_M\}_{M\geq 0}$, every non-empty $k$-core is $(1/2-\varepsilon)$-resiliently Hamiltonian, but not $(1/2+\varepsilon)$-resiliently Hamiltonian.

Comments: 18 pages, submitted
Categories: math.CO
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