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MAJORITY-3SAT (and Related Problems) in Polynomial Time (2107.02748v2)

Published 6 Jul 2021 in cs.CC, cs.AI, and cs.DS

Abstract: Majority-SAT is the problem of determining whether an input $n$-variable formula in conjunctive normal form (CNF) has at least $2{n-1}$ satisfying assignments. Majority-SAT and related problems have been studied extensively in various AI communities interested in the complexity of probabilistic planning and inference. Although Majority-SAT has been known to be PP-complete for over 40 years, the complexity of a natural variant has remained open: Majority-$k$SAT, where the input CNF formula is restricted to have clause width at most $k$. We prove that for every $k$, Majority-$k$SAT is in P. In fact, for any positive integer $k$ and rational $\rho \in (0,1)$ with bounded denominator, we give an algorithm that can determine whether a given $k$-CNF has at least $\rho \cdot 2n$ satisfying assignments, in deterministic linear time (whereas the previous best-known algorithm ran in exponential time). Our algorithms have interesting positive implications for counting complexity and the complexity of inference, significantly reducing the known complexities of related problems such as E-MAJ-$k$SAT and MAJ-MAJ-$k$SAT. At the heart of our approach is an efficient method for solving threshold counting problems by extracting sunflowers found in the corresponding set system of a $k$-CNF. We also show that the tractability of Majority-$k$SAT is somewhat fragile. For the closely related GtMajority-SAT problem (where we ask whether a given formula has greater than $2{n-1}$ satisfying assignments) which is known to be PP-complete, we show that GtMajority-$k$SAT is in P for $k\le 3$, but becomes NP-complete for $k\geq 4$. These results are counterintuitive, because the ``natural'' classifications of these problems would have been PP-completeness, and because there is a stark difference in the complexity of GtMajority-$k$SAT and Majority-$k$SAT for all $k\ge 4$.

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