On the complexity of k-SAT

Abstract

The k-SAT problem is to determine if a given k-CNF has a satisfying assignment. It is a celebrated open question as to whether it requires exponential time to solve k-SAT for k ≥ 3. Here exponential time means 2δn for some δ > 0. In this paper, assuming that, for k ≥ 3, k-SAT requires exponential time complexity, we show that the complexity of k-SAT increases as k increases. More precisely, for k ≥ 3, define sk = inf{δ: there exists 2δn algorithm for solving k-SAT}. Define ETH (Exponential-Time Hypothesis) for k-SAT as follows: for k ≥ 3, sk > 0. In this paper, we show that sk is increasing infinitely often assuming ETH for k-SAT. Let s∞ be the limit of sk. We will in fact show that sk ≤ (1 - d/k) s∞ for some constant d > 0. We prove this result by bringing together the ideas of critical clauses and the Sparsification Lemma to reduce the satisfiability of a k-CNF to the satisfiability of a disjunction of 2εn k′-CNFs in fewer variables for some k′ ≥ k and arbitrarily small ε > 0. We also show that such a disjunction can be computed in time 2εn for arbitrarily small ε > 0.