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Tight Complexity Bounds for Counting Generalized Dominating Sets in Bounded-Treewidth Graphs Part I: Algorithmic Results (2211.04278v3)

Published 8 Nov 2022 in cs.CC and cs.DS

Abstract: We investigate how efficiently a well-studied family of domination-type problems can be solved on bounded-treewidth graphs. For sets $\sigma,\rho$ of non-negative integers, a $(\sigma,\rho)$-set of a graph $G$ is a set $S$ of vertices such that $|N(u)\cap S|\in \sigma$ for every $u\in S$, and $|N(v)\cap S|\in \rho$ for every $v\not\in S$. The problem of finding a $(\sigma,\rho)$-set (of a certain size) unifies standard problems such as Independent Set, Dominating Set, Independent Dominating Set, and many others. For all pairs of finite or cofinite sets $(\sigma,\rho)$, we determine (under standard complexity assumptions) the best possible value $c_{\sigma,\rho}$ such that there is an algorithm that counts $(\sigma,\rho)$-sets in time $c_{\sigma,\rho}{\sf tw}\cdot n{O(1)}$ (if a tree decomposition of width ${\sf tw}$ is given in the input). For example, for the Exact Independent Dominating Set problem (also known as Perfect Code) corresponding to $\sigma={0}$ and $\rho={1}$, we improve the $3{\sf tw}\cdot n{O(1)}$ algorithm of [van Rooij, 2020] to $2{\sf tw}\cdot n{O(1)}$. Despite the unusually delicate definition of $c_{\sigma,\rho}$, an accompanying paper shows that our algorithms are most likely optimal, that is, for any pair $(\sigma, \rho)$ of finite or cofinite sets where the problem is non-trivial, and any $\varepsilon>0$, a $(c_{\sigma,\rho}-\varepsilon){\sf tw}\cdot n{O(1)}$-algorithm counting the number of $(\sigma,\rho)$-sets would violate the Counting Strong Exponential-Time Hypothesis (#SETH). For finite sets $\sigma$ and $\rho$, these lower bounds also extend to the decision version, and hence, our algorithms are optimal in this setting as well. In contrast, for many cofinite sets, we show that further significant improvements for the decision and optimization versions are possible using the technique of representative sets.

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