Why almost all satisfiable k - CNF formulas are easy? Danny Vilenchik Joint work with A. Coja-Oghlan and M. Krivelevich

SAT – Basic Notions 3CNF form: F = ( x 1 Ç x 2 Ç ¬x 5 ) Æ ( x 3 Ç ¬x 4 Ç ¬x 1 ) Æ ( x 1 Ç x 2 Ç x 6 ) Æ … Ã F = ( F Ç F Ç T ) Æ ( T Ç T Ç T ) Æ ( T Ç F Ç T ) Æ … x6x6 x5x5 x4x4 x3x3 x2x2 x1x1 TFFTFF x 5 supports this clause w.r.t. Ã Goal: algorithm that produces optimal result, efficient, and works for all inputs

SAT – Some Background  Finding a satisfying assignment is NP Hard [Cook’71]  No approximation for MAX-SAT with factor better than 7/8 [Hastad’01]  How to proceed?  Hardness results only show that there exist hard instances  The heuristical approach - relaxes the universality requirement  Typical instance?  One possibility: random models Heuristic is a polynomial time algorithm that produces optimal results on typical instances Heuristic is a polynomial time algorithm that produces optimal results on typical instances

Random 3SAT  Random 3SAT:  Fix m, n  Pick m clauses uniformly at random (over the n variables)  Threshold: there exists a constant d such that [Fri99]  m/n ¸ d : most 3CNF s are not satisfiable (4.506)  m/n<d : most 3CNF s are satisfiable (3.52)  Near-threshold 3CNF s are apparently “hard” for many SAT heuristics  Possible reason: complicated structure of solution space (clustering)

Near Threshold Clustering Phenomenon Conjectured solution space of Random k-SAT just below the threshold: (part of this picture was rigorously proved for k ¸ 8, [AR06,MMZ05])  All assignments within a cluster are “close”  A linear number of variables are “frozen”  Every two clusters are “far” from each other  Exponentially many clusters

Our Result Rigorously characterize the structure of the solution space of Random 3SAT, m/n some constant above the threshold:  Single cluster of satisfying assignments  Size of the cluster is exponential in n  (1-e -  (m/n) )n variables are frozen

Our Results  Rigorously complement results for the very sparse case:  When clustering is simple – the problem is easy  When clustering is “complicated” – the problem is harder (?)  Improving the exponential time algorithm for uniform satisfiable 3CNF s in this regime (only one known so far, [Chen03]) Almost all k-CNF formulas are easy ! Theorem: There exists a deterministic polynomial time algorithm that finds a satisfying assignment for almost all satisfiable 3CNF formulas with m/n > C, C a sufficiently large constant Theorem: There exists a deterministic polynomial time algorithm that finds a satisfying assignment for almost all satisfiable 3CNF formulas with m/n > C, C a sufficiently large constant

The Planted Distribution  Planted 3SAT distribution with parameters m, n :  Fix an assignment   Pick u.a.r. m clauses out of all clauses that are satisfied by   Planted 3SAT was analyzed in several papers:  [Fla03] shows a spectral algorithm for solving sparse instances  Ben-Sasson et. al. for m/n=  (logn) (planted and uniform coincide)  Planted models also “fashionable” for graph coloring, max clique, max independent set, min bisection …  Planted models are more approachable – clauses are practically independent  Open question: how does the planted model compare with the uniform?

Our Result  We show that the planted and uniform distributions share many structural properties (“close”)  In particular, same structure of the solution space  Justifying the somewhat unnatural usage of planted-solution models  Flaxman’s algorithm [Fla03] works for the uniform distribution as well

SAT and Message Passing  [FMV06] Warning Propagation was shown to solve planted 3SAT instances with m/n>C, C some sufficiently large constant  Our work implies – WP works in the uniform setting as well  Reinforces the following thesis:  When clustering is complicated ) formulas are hard ) sophisticated algorithms needed: Survey Propagation  When clustering is simple ) formulas are easy ) naïve algorithms work: Warning Propagation

Clustering: Proof Technique  Recall: uniform distribution over satisfiable 3CNF s with m clauses  Why more difficult than the planted distribution?  Edges are not independent  For starters, consider the planted 3SAT distribution  m/n sufficiently large constant  Every variable is expected to support 3m/(7n) clauses w.r.t. planted  Pr[x supports C]=Pr[x supports C | x appears in C]Pr[x appears in C] Fact 1: whp there is no subformula H on h variables s.t. h<n/100 and there are at least hm/(10n) clauses containing two variables from H Fact 1: whp there is no subformula H on h variables s.t. h<n/100 and there are at least hm/(10n) clauses containing two variables from H Fact 2: whp there are no two satisfying assignments at distance greater than n/100

Clustering: Proof Technique Claim: suppose that every variable has the expected support, and Facts 1 and 2 hold, then F is uniquely satisfiable Claim: suppose that every variable has the expected support, and Facts 1 and 2 hold, then F is uniquely satisfiable Proof: suppose not,  Let  be the planted assignment and à some other satisfying assignment  Take x s.t. Ã(x)  (x), x supports 3m/(7n) clauses w.r.t.   Consdier such clause (T Ç F Ç F)  Define H= { x : Ã(x)  (x) }, h= | H | <n/100 (Fact 1)  There exists 3hm/(7n) clauses containing two variables from H  This contradicts Fact 2. Proof: suppose not,  Let  be the planted assignment and à some other satisfying assignment  Take x s.t. Ã(x)  (x), x supports 3m/(7n) clauses w.r.t.   Consdier such clause (T Ç F Ç F)  Define H= { x : Ã(x)  (x) }, h= | H | <n/100 (Fact 1)  There exists 3hm/(7n) clauses containing two variables from H  This contradicts Fact 2. FT Ã:Ã:

Clustering: Proof Technique  This picture is whp the case when m/n>Clog n  When m/n=O(1) - whp not the case (some variables have 0 support) Definition: Given a 3CNF F and a satisfying assignment Ã, a set C is called a core of F if 8 x 2 C, x supports at least m/(4n) clauses in F[C] Definition: Given a 3CNF F and a satisfying assignment Ã, a set C is called a core of F if 8 x 2 C, x supports at least m/(4n) clauses in F[C] Claim: For F in the planted distribution, m/n sufficiently large constant there exists a core C s.t.  | V(C) | >(1-e -  m/n  )n  C is frozen in F Claim: For F in the planted distribution, m/n sufficiently large constant there exists a core C s.t.  | V(C) | >(1-e -  m/n  )n  C is frozen in F Corollary: one-cluster structure

Moving to the Uniform Case  A – a “bad” structural property (in our case: no big core)   –expected number of satisfying assignments of planted 3CNF Claim: Pr uniform [A] <  ¢ Pr planted [A] Claim: Pr uniform [no big core] <  ¢ Pr planted [no big core]< ¹ ¢ e -nc Claim: ¹<e nc ’, c ’ <c Corollary: Pr uniform [no big core] = o(1)

Further Research m/nm/n solution space 4.26 cclogn