How hard is it to find a local optimum? If we are given a graph and want to find a locally maximal cut--meaning that the number of edges in the cut can't be improved by moving a single vertex from one side to the other--then just iterating improving steps finds a local maximum in $ |E|$ steps. If, on the other hand, the edges are weighted, this problem becomes hard for the class PLS (Polynomial Local Search). We are interested in optimization problems with lexicographic costs. For Max-Cut this would mean that the edges $e_1,\dots, e_m$ have costs $c(e_i) = 2^i$. For such a cost function finding a global Max-Cut is easy. In contrast, we show that it is PLS-complete to find an assignment for a 4-CNF formula that is locally maximal (when the clauses have lexicographic weights); and also for a 3-CNF when we allow switching two variables at a time. We use these results to answer a question in Scheder and Tantow, who showed that finding a lexicographic local minimum of a string $s \in \{0,1\}^n$ under the action of a list of given permutations $π_1, \dots, π_k \in S_{n}$ is PLS-complete. They ask whether the problem stays PLS-complete when the $π_1,\dots,π_k$ commute, i.e., generate an Abelian subgroup $G$ of $S_n$. We show that it does, and in fact stays PLS-complete even (1) when every element in $G$ has order two or (2) when $G$ is cyclic. Additionally, we use it to further investigate the complexity of computing pure $α$-Nash equilibria in congestion games. Using lexicographic 4-SAT, we obtain a simple proof of the PLS-completeness originally shown by Skopalik and Vöcking that can be extended to exponential and polynomial delay functions with positive coefficients. The number of strategies per player and players per resource is bounded. However, the degree of the polynomials is not bounded by a constant.

翻译：暂无翻译