Exact algorithms for Kayles

In the game of KAYLES, two players select alternatingly a vertex from a given graph G, but may never choose a vertex that is adjacent or equal to an already chosen vertex. The last player that can select a vertex wins the game. In this paper, we give an exact algorithm to determine which player has a winning strategy in this game. To analyze the running time of the algorithm, we introduce the notion of a K-set: a nonempty set of vertices W subset of V is a K-set in a graph G = (V, E), if G[W] is connected and there exists an independent set X such that W = V - N[X]. The running time of the algorithm is bounded by a polynomial factor times the number of K-sets in G. We prove that the number of K-sets in a graph with n vertices is bounded by O(1.6052(n)). A computer-generated case analysis improves this bound to O(1.6031(n)) K-sets, and thus we have an upper bound of O(1.6031(n)) on the running time of the algorithm for KAYLES. We also show that the number of K-sets in a tree is bounded by n center dot 3(n/3) and thus KAYLEs can be solved on trees in O(1.4423(n)) time. We show that apart from a polynomial factor, the number of K-sets in a tree is sharp. As corollaries, we obtain that determining which player has a winning strategy in the games G(avoid)(POSDNF2) and G(seek)(POSDNF3) can also be determined in O(1.6031(n)) time. In G(avoid)(POSDNF2), we have a positive formula F on n Boolean variables in Disjunctive Normal Form with two variables per clause. Initially, all variables are false, and players alternately set a variable from false to true; the first player that makes F true loses the game. The game G(seek)(POSDNF3) is similar, but now there are three variables per clause, and the first player that makes F true wins the game. (C) 2014 Elsevier B.V. All rights reserved.