Abstract

For any integer k⩾3, we consider the following decision problem. Given a simple graph, does there exist a partition of its vertices into two disjoint sets such that every simple k-cycle of G contains vertices in both of these sets? This problem is NP-hard because it admits a polynomial reduction from NAE 3-SAT. We construct a reduction that is polynomial both in the length of the instance and in k, which answers a recent question of Karpiński.

Full Text
Published version (Free)

Talk to us

Join us for a 30 min session where you can share your feedback and ask us any queries you have

Schedule a call