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

Published 6 Jul 2021 in cs.CC, cs.AI, and cs.DS | (2107.02748v2)

Abstract: Majority-SAT is the problem of determining whether an input nn-variable formula in conjunctive normal form (CNF) has at least 2<sup>n−12<sup>{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-kkSAT, where the input CNF formula is restricted to have clause width at most kk. We prove that for every kk, Majority-kkSAT is in P. In fact, for any positive integer kk and rational ρ∈(0,1)\rho \in (0,1) with bounded denominator, we give an algorithm that can determine whether a given kk-CNF has at least ρ⋅2<sup>n\rho \cdot 2<sup>n 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-kkSAT and MAJ-MAJ-kkSAT. 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 kk-CNF. We also show that the tractability of Majority-kkSAT is somewhat fragile. For the closely related GtMajority-SAT problem (where we ask whether a given formula has greater than 2<sup>n−12<sup>{n-1} satisfying assignments) which is known to be PP-complete, we show that GtMajority-kkSAT is in P for k≤3k\le 3, but becomes NP-complete for k≥4k\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-kkSAT and Majority-kkSAT for all k≥4k\ge 4.

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