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 2 $n for some $>0. In this paper, assuming that, for k 3, k-SAT requires exponential time complexity, we show that the complexity of k-SAT increases as k increases. More precisely, for k 3, define s k=inf[$: there exists 2 $n algorithm for solving k-SAT]. Define ETH (Exponential-Time Hypothesis) for k-SAT as follows: for k 3, s k>0. In this paper, we show that s k is increasing infinitely often assuming ETH for k-SAT. Let s be the limit of s k. We will in fact show that s k (1&d k) s for some constant d>0. We prove this result by bringing together the ideas of critical clauses and the Sparsification Lemma to reduce the satisfiability of a k-CNF to the satisfiability of a disjunction of 2 =n k$-CNFs in fewer variables for some k $ k and arbitrarily small =>0. We also show that such a disjunction can be computed in time 2 =n for arbitrarily small =>0.

Citations