Approximate and Randomized Algorithms for Computing a Second Hamiltonian Cycle

Argyrios Deligkas, George B. Mertzios, Paul G. Spirakis, Viktor Zamaraev

Abstract: In this paper we consider the following problem: Given a Hamiltonian graph G, and a Hamiltonian cycle C of G, can we compute a second Hamiltonian cycle \(C^{\prime } \ne C\) of G, and if yes, how quickly? If the input graph G satisfies certain conditions (e.g. if every vertex of G is odd, or if the minimum degree is large enough), it is known that such a second Hamiltonian cycle always exists. Despite substantial efforts, no subexponential-time algorithm is known for this problem. In this paper we relax the problem of computing a second Hamiltonian cycle in two ways. First, we consider approximating the length of a second longest cycle on n-vertex graphs with minimum degree \(\delta \) and maximum degree \(\Delta \). We provide a linear-time algorithm for computing a cycle \(C^{\prime } \ne C\) of length at least \(n-4\alpha (\sqrt{n}+2\alpha )+8\), where \(\alpha = \frac{\Delta -2}{\delta -2}\). This results provides a constructive proof of a recent result by Girão, Kittipassorn, and Narayanan in the regime of \(\frac{\Delta }{\delta } = o(\sqrt{n})\). Our second relaxation of the problem is probabilistic. We propose a randomized algorithm which computes a second Hamiltonian cycle with high probability, given that the input graph G has a large enough minimum degree. More specifically, we prove that for every \(0<p\le 0.02\), if the minimum degree of G is at least \(\frac{8}{p} \log \sqrt{8}n + 4\), then a second Hamiltonian cycle can be computed with probability at least \(1 - \frac{1}{n}\left( \frac{50}{p^4} + 1 \right) \) in \(poly(n) \cdot 2^{4pn}\) time. This result implies that, when the minimum degree \(\delta \) is sufficiently large, we can compute with high probability a second Hamiltonian cycle faster than any known deterministic algorithm. In particular, when \(\delta = \omega (\log n)\), our probabilistic algorithm works in \(2^{o(n)}\) time.