1. Randomzied Unique k-SAT R. Paturi, P. Pudlak, M.E. Saks and F. Zane An Improved Exponential-time Algorithm for k-SAT Journal of the ACM, Vol 52, No. 3, pp. 337-364, May 2005 Fenghui Zhang 2006, 11, 2

2. Introduction Unique k-SAT  Given a CNF formula F with at most one satisfying assignment, decide if F is satisfiable or not Main result to present:  For unique 3-SAT, the running time is O(1.307 n ), much better than the best known algorithm for general 3-SAT, which is O(1.324 n )

3. The algorithm Resolve all s bounded pairs Repeat T times  Let  be a random permutation of [1..n]  Let y be a random assignment to F  Let u be the result of applying y to F following the order in π – Modify (F, , y);  If u satisfies F, return u Return “no”

4. Resolve s bounded pair Two clause form a resolvable pair if they conflict on exactly one variable v, i.e., one of them contains v and the other contains  v. R(C 1,C 2 )=D 1  D 2, where D 1 and D 2 are obtained by remove variable from C 1 and C 2 If none of C 1,C 2, R(C 1,C 2 ) contains more than s literals, we add R(C 1,C 2 ) to F

5. Modify (F, , y) Take the next variable v from π  If the value of v can be decided, assign that value to v When there is a unit clause only involves v  Else v takes the value assigned in y Return the modified assignment

6. What are s, T? s is a large constant or a slowly increasing function of n T is the number of time we run the algorithm, in our unique 3-SAT case, it is O(1.307 n )

7. Analysis Suppose z is a satisfying assignment to F, for given y and , Modify (F, , y) returns z iff y and z agrees on all variables not forced by π, i.e., those not decidable For fixed π, the number of y that Modify (F, , y) will return z is 2 |Forced(G, , z)| The probability that the algorithm finds z is then

8. Analysis – cont. It is bounded by 2 -n+E[|Forced(F, , z)|] We say that a clause C is critical for a variable v and a truth assignment z if v is the only true literal variable in C v  C is in Forced(F, , z) iff in , v is after all other variables of C Hence E[|Forced(F, , z)|]=  v P(v,F,z) where P(v,F,z) is the probability that a random  actually put v as the last variable among the variables of a critical clause for (v, z)

9. Analysis – cont. Why do s-bounded resolvent?  Increase the number of critical clauses for a variable How does uniqueness help?  Suppose z=1 n is the unique truth assignment  Consider C 1 =x 1  x 2  x 3  x 4,001 n-2 is not a truth assignment, there is a clause not satisfied by it that Either resolvable with C 1 Or it is a second critical clause for x 1

10. Analysis – cont. P (v, F, z)>0.613 This is because we assume that F has at most one truth assignment, thus for any integer d between 1 and n, flip any d variables will give us a false assignment, hence we will have many critical clauses for each variable, hence the probability of  actually put v as the last in one of its critical clause increases. Detail skipped.

11. Analysis – cont. Hence the success probability of Modify (F, , y) is at least 2 -n+0.613n =1.307 -n Hence if we let T=O(1.307 n ), i.e., repeat Modify for O(1.307 n ) many times, we will have a nontrivial chance of finding the truth assignment

12. The End Thank you very much Questions?