Randomzied Unique k-SATR. Paturi, P. Pudlak, M.E. Saks and F. Zane

An Improved Exponential-time Algorithm for k-SATJournal of the ACM, Vol 52, No. 3, pp. 337-364, May 2005

Fenghui Zhang2006, 11, 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(1.307n), much better than the best known algorithm for general 3-SAT, which is O(1.324n)

The algorithm

Resolve all s bounded pairsRepeat 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”

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(C1,C2)=D1 D2, where D1 and D2 are obtained by remove variable from C1 and C2

If none of C1,C2 , R(C1,C2 ) contains more than s literals, we add R(C1,C2) to F

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

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(1.307n)

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

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)|]=vP(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)

Analysis – cont.

Why do s-bounded resolvent? Increase the number of critical clauses for a

variableHow does uniqueness help?

Suppose z=1n is the unique truth assignment Consider C1=x1x2x3x4,001n-2 is not a truth

assignment, there is a clause not satisfied by it that

Either resolvable with C1

Or it is a second critical clause for x1

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.

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(1.307n), i.e., repeat Modify for O(1.307n) many times, we will have a nontrivial chance of finding the truth assignment

The End

Thank you very much

Questions?