Lecture 25

Given a DNF formula φ with n variables, DNF counting is the problem finding the number of satisfying assignments for φ. Note that in general satisfiability for DNF is easy as we need only satisfy a single clause, but the counting problem is hard. Indeed, if we could do this, then given any 3-CNF formula f with n variables, we could take its negation, count how many satisfying assignments its negation has, and if that was equal to 2 then ¬f is a tautology so f is unsatisfiable, otherwise, it is satisfiable, so we have solved 3-SAT. The counting problem is hard for a class called #P , which is a very strong form of intractability. We seek to find approximate solutions to this problem. The obvious approach is the following: sample random assignments uniformly at random, and take the estimate to be p2 where p was the fraction of assignments that satisfied φ. Let Xi be the random variable which is 1 if the ith assignment we chose satisfied F , zero otherwise, and suppose we sample m times. Then p = 1 m ∑m i=1Xi. Let μ be true fraction of satisfying assignments. Then for fixed , δ > 0 if we want Pr(|X − μ| ≥ μ) ≤ δ