Finding Even Cycles Faster via Capped k-Walks

Abstract

In this paper, we consider the problem of finding a cycle of length 2k (a C2k) in an undirected graph G with n nodes and m edges for constant k2. A classic result by Bondy and Simonovits [J.Comb.Th.'74] implies that if m 100k n1+1/k, then G contains a C2k, further implying that one needs to consider only graphs with m = O(n1+1/k). Previously the best known algorithms were an O(n2) algorithm due to Yuster and Zwick [J.Disc.Math'97] as well as a O(m2-(1+ k/2-1)/(k+1)) algorithm by Alon et al. [Algorithmica'97]. We present an algorithm that uses O(m2k/(k+1)) time and finds a C2k if one exists. This bound is O(n2) exactly when m=(n1+1/k). For 4-cycles our new bound coincides with Alon et al., while for every k>2 our bound yields a polynomial improvement in m. Yuster and Zwick noted that it is "plausible to conjecture that O(n2) is the best possible bound in terms of n". We show "conditional optimality": if this hypothesis holds then our O(m2k/(k+1)) algorithm is tight as well. Furthermore, a folklore reduction implies that no combinatorial algorithm can determine if a graph contains a 6-cycle in time O(m3/2-ε) for any ε>0 under the widely believed combinatorial BMM conjecture. Coupled with our main result, this gives tight bounds for finding 6-cycles combinatorially and also separates the complexity of finding 4- and 6-cycles giving evidence that the exponent of m in the running time should indeed increase with k. The key ingredient in our algorithm is a new notion of capped k-walks, which are walks of length k that visit only nodes according to a fixed ordering. Our main technical contribution is an involved analysis proving several properties of such walks which may be of independent interest.

0

Turn this paper into a lesson

ArcXiv compiles a structured reading guide from this paper's metadata: plain-English importance, contributions, prerequisite concepts, which sections to read first, flashcards, and a quiz. Grounded in the abstract, never invented.

Discussion (0)

Sign in to join the discussion.

Loading comments…