An even simpler hard variant of Not-All-Equal 3-SAT

TL;DR

This paper investigates the computational complexity of a restricted version of Not-All-Equal 3-SAT, showing NP-completeness for certain parameters and polynomial-time solvability for others, with implications for hypergraph coloring.

Contribution

It establishes NP-completeness for k-partitioned instances with k≥4 and polynomial-time solvability for k≤3, refining previous results and connecting to hypergraph bicolorability.

Findings

NP-completeness for k≥4 partitions

Polynomial-time solvability for k≤3

Hypergraph bicolorability complexity results

Abstract

We show that Not-All-Equal 3-Sat remains NP-complete when restricted to instances that simultaneously satisfy the following properties: (i) The clauses are given as the disjoint union of k partitions, for any fixed $k \geq 4$, of the variable set into subsets of size 3, and (ii) each pair of distinct clauses shares at most one variable. Property (i) implies that each variable appears in exactly $k$ clauses and each clause consists of exactly 3 unnegated variables. Therewith, we improve upon our earlier result (Darmann and D\"ocker, 2020). Complementing the hardness result for at least $4$ partitions, we show that for $k\leq 3$ the corresponding decision problem is in P. In particular, for $k\in \{1,2\}$, all instances that satisfy Property (i) are nae-satisfiable. By the well-known correspondence between Not-All-Equal 3-Sat and hypergraph coloring, we obtain the following corollary of our results: For $k\geq 4$, Bicolorability is NP-complete for linear 3-uniform $k$-regular hypergraphs even if the edges are given as a decomposition into $k$ perfect matchings; with the same restrictions, for $k \leq 3$ Bicolorability is in P, and for $k \in \{1,2\}$ all such hypergraphs are bicolorable. Finally, we deduce from a construction in the work by Pilz (Pilz, 2019) that every instance of Positive Planar Not-All-Equal Sat with at least three distinct variables per clause is nae-satisfiable. Hence, when restricted to instances with a planar incidence graph, each of the above variants of Not-All-Equal 3-Sat turns into a trivial decision problem.