Multiple known algorithmic paradigms (backtracking, local search and the polynomial method) only yield a 2n(1−1/O(k)) time algorithm for k-SAT in the worst case. For this reason, it has been hypothesized that the worst-case k-SAT problem cannot be solved in 2n(1−f(k)/k) time for any unbounded function f. This hypothesis has been called the “Super-Strong ETH”, modeled after the ETH and the Strong ETH. We give two results on the Super-Strong ETH:
1. It has also been hypothesized that k-SAT is hard to solve for randomly chosen instances near the “critical threshold”, where the clause-to-variable ratio is 2^kln2−Θ(1). We give a randomized algorithm which refutes the Super-Strong ETH for the case of random k-SAT and planted k-SAT for any clause-to-variable ratio. For example, given any random k-SAT instance F with n variables and m clauses, our algorithm decides satisfiability for F in 2^n(1−Ω(logk)/k) time, with high probability (over the choice of the formula and the randomness of the algorithm). It turns out that a well-known algorithm from the literature on SAT algorithms does the job: the PPZ algorithm of Paturi, Pudlák and Zane [17].
2. The Unique k-SAT problem is the special case where there is at most one satisfying assignment. Improving prior reductions, we show that the Super-Strong ETHs for Unique k-SAT and k-SAT are equivalent. More precisely, we show the time complexities of Unique k-SAT and k-SAT are very tightly correlated: if Unique k-SAT is in 2^n(1−f(k)/k) time for an unbounded f, then k-SAT is in 2^n(1−f(k)(1−ε)/k) time for every ε>0.
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Faster Random k-CNF Satisfiability
We describe an algorithm to solve the problem of Boolean CNF-Satisfiability when the input formula is chosen randomly. We build upon the algorithms of Schöning 1999 and Dantsin et al. in 2002. The Schöning algorithm works by trying many possible random assignments, and for each one searching systematically in the neighborhood of that assignment for a satisfying solution. Previous algorithms for this problem run in time O(2^(n (1- Ω(1)/k))). Our improvement is simple: we count how many clauses are satisfied by each randomly sampled assignment, and only search in the neighborhoods of assignments with abnormally many satisfied clauses. We show that assignments like these are significantly more likely to be near a satisfying assignment. This improvement saves a factor of 2^(n Ω(lg² k)/k), resulting in an overall runtime of O(2^(n (1- Ω(lg² k)/k))) for random k-SAT.
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- Award ID(s):
- 1909429
- NSF-PAR ID:
- 10178923
- Date Published:
- Journal Name:
- Leibniz international proceedings in informatics
- Volume:
- 168
- ISSN:
- 1868-8969
- Page Range / eLocation ID:
- 78:1--78:12
- Format(s):
- Medium: X
- Sponsoring Org:
- National Science Foundation
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Multiple known algorithmic paradigms (backtracking, local search and the polynomial method) only yield a 2n(1-1/O(k)) time algorithm for k-SAT in the worst case. For this reason, it has been hypothesized that the worst-case k-SAT problem cannot be solved in 2n(1-f(k)/k) time for any unbounded function f. This hypothesis has been called the "Super-Strong ETH", modelled after the ETH and the Strong ETH. It has also been hypothesized that k-SAT is hard to solve for randomly chosen instances near the "critical threshold", where the clause-to-variable ratio is such that randomly chosen instances are satisfiable with probability 1/2. We give a randomized algorithm which refutes the Super-Strong ETH for the case of random k-SAT and planted k-SAT for any clause-to-variable ratio. For example, given any random k-SAT instance F with n variables and m clauses, our algorithm decides satisfiability for F in 2n(1-c*log(k)/k) time with high probability (over the choice of the formula and the randomness of the algorithm). It turns out that a well-known algorithm from the literature on SAT algorithms does the job: the PPZ algorithm of Paturi, Pudlak, and Zane (1999). The Unique k-SAT problem is the special case where there is at most one satisfying assignment. Improving prior reductions, we show that the Super-Strong ETHs for Unique k-SAT and k-SAT are equivalent. More precisely, we show the time complexities of Unique k-SAT and k-SAT are very tightly correlated: if Unique k-SAT is in 2n(1-f(k)/k) time for an unbounded f, then k-SAT is in 2n(1-f(k)/(2k)) time.more » « less
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