Approximate and Randomized Algorithms for Computing a Second Hamiltonian Cycle

Abstract

AbstractIn 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 $$C^{\prime } \ne C$$ C ′ ≠ 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 $$\delta $$ δ and maximum degree $$\Delta $$ Δ . We provide a linear-time algorithm for computing a cycle $$C^{\prime } \ne C$$ C ′ ≠ C of length at least $$n-4\alpha (\sqrt{n}+2\alpha )+8$$ n - 4 α ( n + 2 α ) + 8 , where $$\alpha = \frac{\Delta -2}{\delta -2}$$ α = Δ - 2 δ - 2 . This results provides a constructive proof of a recent result by Girão, Kittipassorn, and Narayanan in the regime of $$\frac{\Delta }{\delta } = o(\sqrt{n})$$ Δ δ = o ( 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 $$0<p\le 0.02$$ 0 < p ≤ 0.02 , if the minimum degree of G is at least $$\frac{8}{p} \log \sqrt{8}n + 4$$ 8 p log 8 n + 4 , then a second Hamiltonian cycle can be computed with probability at least $$1 - \frac{1}{n}\left( \frac{50}{p^4} + 1 \right) $$ 1 - 1 n 50 p 4 + 1 in $$poly(n) \cdot 2^{4pn}$$ p o l y ( n ) · 2 4 p n time. This result implies that, when the minimum degree $$\delta $$ δ is sufficiently large, we can compute with high probability a second Hamiltonian cycle faster than any known deterministic algorithm. In particular, when $$\delta = \omega (\log n)$$ δ = ω ( log n ) , our probabilistic algorithm works in $$2^{o(n)}$$ 2 o ( n ) time.