On Super Strong ETH
Author(s) -
Nikhil Vyas,
Ryan Williams
Publication year - 2021
Publication title -
journal of artificial intelligence research
Language(s) - English
Resource type - Journals
SCImago Journal Rank - 0.79
H-Index - 123
eISSN - 1943-5037
pISSN - 1076-9757
DOI - 10.1613/jair.1.11859
Subject(s) - randomness , satisfiability , backtracking , mathematics , combinatorics , randomized algorithm , time complexity , discrete mathematics , deterministic algorithm , function (biology) , boolean satisfiability problem , maximum satisfiability problem , variable (mathematics) , true quantified boolean formula , algorithm , boolean function , statistics , mathematical analysis , evolutionary biology , biology
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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