Notes on Complexity Theory Last updated : November , 2011 Lecture 24

We explore three results related to hardness of counting. Interestingly, at their core each of these results relies on a simple — yet powerful — technique due to Valiant and Vazirani. Does SAT become any easier if we are guaranteed that the formula we are given has at most one solution? Alternately, if we are guaranteed that a given boolean formula has a unique solution does it become any easier to find it? We show here that this is not likely to be the case. Define the following promise problem: USAT def = {φ : φ has exactly one satisfying assignment} USAT def = {φ : φ is unsatisfiable}. Clearly, this problem is in promise-N P. We show that if it is in promise-P, then N P = RP. We begin with a lemma about pairwise-independent hashing. Lemma 1 Let S ⊆ {0, 1} n be an arbitrary set with 2 m ≤ |S| ≤ 2 m+1 , and let H n,m+2 be a family of pairwise-independent hash functions mapping {0, 1} n to {0, 1} m+2. Then Pr h∈H n,m+2 [there is a unique x ∈ S with h(x) = 0 m+2 ] ≥ 1/8. Proof Let 0 def = 0 m+2 , and let p def = 2 −(m+2). Let N be the random variable (over choice of random h ∈ H n,m+2) denoting the number of x ∈ S for which h(x) = 0. Using the inclusion/exclusion principle, we have Pr[N ≥ 1] ≥ x∈S Pr[h(x) = 0] − 1 2 · x =x ∈S Pr[h(x) = h(x) = 0] = |S| · p − |S| 2 p 2 , while Pr[N ≥ 2] ≤ x =x ∈S Pr[h(x) = h(x) = 0] = |S| 2 p 2. So Pr[N = 1] = Pr[N ≥ 1] − Pr[N ≥ 2] ≥ |S| · p − 2 · |S| 2 p 2 ≥ |S|p − |S| 2 p 2 ≥ 1/8, using the fact that |S| · p ∈ [