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For d=2 we obtain O^*(1.1193^n) or O^*(1.3248^m), and for general Parity-SAT an O^*(1.1052^L) algorithm."},{"attestation":"unclaimed","claim_id":"C2","kind":"weakest_assumption","source":"verdict.weakest_assumption","status":"machine_extracted","text":"The algorithms assume that parity can be exploited via structural reductions and branching rules that avoid the overhead of exact counting, without hidden costs that would appear only in a full proof or implementation."},{"attestation":"unclaimed","claim_id":"C3","kind":"one_line_summary","source":"verdict.one_line_summary","status":"machine_extracted","text":"New randomized and branching algorithms achieve O*(2^{m(1-1/O(d))}) time for Parity-d-occ-SAT and O*(1.1052^L) time for general Parity-SAT, outperforming exact counting bounds."},{"attestation":"unclaimed","claim_id":"C4","kind":"headline","source":"verdict.pith_extraction.headline","status":"machine_extracted","text":"Randomized algorithms solve Parity-d-occ-SAT in O^*(2^{m(1-1/O(d))}) time for any fixed d."}],"snapshot_sha256":"5b6e6683f75ce51e0a14a6698aca48d6cb89be388ef6ef94fea00260a34a7c85"},"formal_canon":{"evidence_count":0,"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"paper":{"abstract_excerpt":"Parity-SAT is the problem of determining whether a given CNF formula has an odd number of satisfying assignments. As a canonical $\\oplus$P-complete problem, it represents a fundamental variant of the exact model counting problem (#SAT). Under the Strong Exponential Time Hypothesis (SETH), Parity-SAT admits no $O^*((2-\\varepsilon)^n)$-time or $O^*((2-\\varepsilon)^m)$-time algorithm for any constant $\\varepsilon>0$, where $n$ and $m$ denote the numbers of variables and clauses, respectively. 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