Complexity of k-SAT

The problem of k-SAT is to determine if the given k-CNF has a satisfying solution. It is a celebrated open question as to whether it requires exponential time to solve k-SAT for k greater than or equal to 3. Define s(k) (for k greater than or equal to 3) to be the infimum of {delta : there exists an O(2(delta n)) algorithm for solving k-SAT}. Define ETH (Exponential-Time Hypothesis) for k-SAT as follows: for k greater than or equal to 3, s(k) > 0. In other words, for k greater than or equal to 3, k-SAT does not have a subexponential-time algorithm. In this paper; we show that sk is an increasing sequence assuming ETH for k-SAT. Let s(infinity) be the limit of s(k). We will in fact show that s(k) less than or equal to (I - d/k)s(infinity) for some constant d > 0.