All known SAT-solving paradigms (backtracking, local search, and the polynomial method) only yield a 2n(1−1/O(k)) time algorithm for solving k-SAT in the worst case, where the big-O constant is independent of k. For this reason, it has been hypothesized that k-SAT cannot be solved in worst-case 2n(1−f(k)/k) time, for any unbounded ƒ : ℕ → ℕ. This hypothesis has been called the “Super-Strong Exponential Time Hypothesis” (Super Strong ETH), modeled after the ETH and the Strong ETH. We prove two results concerning 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 2k ln 2 −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. In particular, given any random k-SAT instance F with n variables and m clauses, our algorithm decides satisfiability for F in 2n(1−Ω( 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 (1998).2. The Unique k-SAT problem is the special case where there is at most one satisfying assignment. It is natural to hypothesize that the worst-case (exponential-time) complexity of Unique k-SAT is substantially less than that of k-SAT. Improving prior reductions, we show the time complexities of Unique k-SAT and k-SAT are very tightly related: 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)(1−ɛ)/k) time for every ɛ > 0. Thus, refuting Super Strong ETH in the unique solution case would refute Super Strong ETH in general.
Improving Efficiency of 3-SAT-Solving Tile Systems
