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.
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Conditional lower bounds for sparse parameterized 2-CSP: A streamlined proof
Assuming the Exponential Time Hypothesis (ETH), a result of Marx (ToC’10) implies that there is no f (k) · n^o(k/ log k) time algorithm that can solve 2-CSPs with k constraints (over a domain of arbitrary large size n) for any computable function f . This lower bound is widely used to show that certain parameterized problems cannot be solved in time f (k) · n^o(k/ log k) time (assuming the ETH). The purpose of this note is to give a streamlined proof of this result.
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- Award ID(s):
- 2313372
- PAR ID:
- 10545202
- Publisher / Repository:
- Society for Industrial and Applied Mathematics
- Date Published:
- ISBN:
- 978-1-61197-793-6
- 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”, 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.more » « less
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