The Curse and Blessing of Not-All-Equal in k-Satisfiability

TL;DR

This paper introduces faster algorithms for NAE-$k$-SAT and MAX-NAE-$k$-SAT, showing that NAE-$k$-SAT can be easier than $k$-SAT exactly, but approximate solutions for MAX-NAE-$k$-SAT may be harder for larger $k$.

Contribution

It presents the first algorithms that outperform $k$-SAT for NAE-$k$-SAT and provides new bounds for approximating MAX-NAE-$k$-SAT, highlighting differences in complexity between NAE-$k$-SAT and $k$-SAT.

Findings

NAE-$k$-SAT can be solved faster than $k$-SAT for all $k \\ge 3$.

New upper bounds for MAX-NAE-$k$-SAT approximation algorithms.

Approximate MAX-NAE-$k$-SAT is harder than MAX-$k$-SAT for $k \\ge 4$.

Abstract

As a natural variant of the $k$-SAT problem, NAE-$k$-SAT additionally requires the literals in each clause to take not-all-equal (NAE) truth values. In this paper, we study the worst-case time complexities of solving NAE-$k$-SAT and MAX-NAE-$k$-SAT approximation, as functions of $k$, the number of variables $n$, and the performance ratio $\delta$. The latter problem asks for a solution of at least $\delta$ times the optimal. Our main results include: (1) A deterministic algorithm for NAE-$k$-SAT that is faster than the best deterministic algorithm for $k$-SAT on all $k \ge 3$. Previously, no NAE-$k$-SAT algorithm is known to be faster than $k$-SAT algorithms. For $k = 3$, we achieve an upper bound of $1.326^n$. The corresponding bound for $3$-SAT is $1.328^n$. (2) A randomized algorithm for MAX-NAE-$k$-SAT approximation, with upper bound $(2 - \epsilon_k(\delta))^n$ where $\epsilon_k(\delta) > 0$ only depends on $k$ and $\delta$. Previously, no upper bound better than the trivial $2^n$ is known for MAX-NAE-$k$-SAT approximation on $k \ge 4$. For $\delta = 0.9$ and $k = 4$, we achieve an upper bound of $1.947^n$. (3) A deterministic algorithm for MAX-NAE-$k$-SAT approximation. For $\delta = 0.9$ and $k = 3$, we achieve an upper bound of $1.698^n$, which is better than the upper bound $1.731^n$ of the exact algorithm for MAX-NAE-$3$-SAT. Our finding sheds new light on the following question: Is NAE-$k$-SAT easier than $k$-SAT? The answer might be affirmative at least on solving the problems exactly and deterministically, while approximately solving MAX-NAE-$k$-SAT might be harder than MAX-$k$-SAT on $k \ge 4$.