CNFs and DNFs with Exactly $k$ Solutions

June 08, 2025 ยท The Ethereal ยท ๐Ÿ› International Conference on Theory and Applications of Satisfiability Testing

๐Ÿ”ฎ THE ETHEREAL: The Ethereal
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Authors L. Sunil Chandran, Rishikesh Gajjala, Kuldeep S. Meel arXiv ID 2506.07268 Category cs.DM: Discrete Mathematics Cross-listed cs.DS, cs.LO, math.CO, math.LO Citations 0 Venue International Conference on Theory and Applications of Satisfiability Testing Last Checked 5 months ago
Abstract
Model counting is a fundamental problem that consists of determining the number of satisfying assignments for a given Boolean formula. The weighted variant, which computes the weighted sum of satisfying assignments, has extensive applications in probabilistic reasoning, network reliability, statistical physics, and formal verification. A common approach for solving weighted model counting is to reduce it to unweighted model counting, which raises an important question: {\em What is the minimum number of terms (or clauses) required to construct a DNF (or CNF) formula with exactly $k$ satisfying assignments?} In this paper, we establish both upper and lower bounds on this question. We prove that for any natural number $k$, one can construct a monotone DNF formula with exactly $k$ satisfying assignments using at most $O(\sqrt{\log k}\log\log k)$ terms. This construction represents the first $o(\log k)$ upper bound for this problem. We complement this result by showing that there exist infinitely many values of $k$ for which any DNF or CNF representation requires at least $ฮฉ(\log\log k)$ terms or clauses. These results have significant implications for the efficiency of model counting algorithms based on formula transformations.
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