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Solving the Hamilton cycle problem fast on average

Michael Anastos

2022Year

Abstract

We present CertifyHAM, a deterministic algorithm that takes a graph G as input and either finds a Hamilton cycle of G or outputs that such a cycle does not exist. If G∼G(n,p)G\sim G(n,p) and p≥100log⁡nnp\displaystyle \geq\frac{100\log n}{n} then the expected running time of CertifyHAM is O(np)O\left(\displaystyle \frac{n}{p}\right) which is best possible. This improves upon previous results due to Gurevich and Shelah, Thomason and Alon, and Krivelevich, who proved analogous results for p being constant, p≥12n−1/3p\geq 12n^{-1/3} and p≥70n−1/2p\geq 70n^{-1/2} respectively.

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