$(2/2/3)$-SAT problem and its applications in dominating set problems

TL;DR

This paper introduces the $(2/2/3)$-SAT problem, proves its NP-completeness, and applies it to demonstrate NP-completeness in dominating set problems and graph coloring issues.

Contribution

The paper defines a new restricted SAT variant, $(2/2/3)$-SAT, and shows its NP-completeness, then applies this to establish NP-completeness in related graph problems.

Findings

$(2/2/3)$-SAT is NP-complete.

NP-completeness of dominating set problems in regular graphs.

NP-completeness of 4-incidence coloring in cubic graphs.

Abstract

The satisfiability problem is known to be $\mathbf{NP}$-complete in general and for many restricted cases. One way to restrict instances of $k$-SAT is to limit the number of times a variable can be occurred. It was shown that for an instance of 4-SAT with the property that every variable appears in exactly 4 clauses (2 times negated and 2 times not negated), determining whether there is an assignment for variables such that every clause contains exactly two true variables and two false variables is $\mathbf{NP}$-complete. In this work, we show that deciding the satisfiability of 3-SAT with the property that every variable appears in exactly four clauses (two times negated and two times not negated), and each clause contains at least two distinct variables is $ \mathbf{NP} $-complete. We call this problem $(2/2/3)$-SAT. For an $r$-regular graph $G = (V,E)$ with $r\geq 3$, it was asked in [Discrete Appl. Math., 160(15):2142--2146, 2012] to determine whether for a given independent set $T $ there is an independent dominating set $D$ that dominates $T$ such that $ T \cap D =\varnothing $? As an application of $(2/2/3)$-SAT problem we show that for every $r\geq 3$, this problem is $ \mathbf{NP} $-complete. Among other results, we study the relationship between 1-perfect codes and the incidence coloring of graphs and as another application of our complexity results, we prove that for a given cubic graph $G$ deciding whether $G$ is 4-incidence colorable is $ \mathbf{NP} $-complete.