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Why almost all satisfiable k - CNF formulas are easy? Danny Vilenchik Joint work with A. Coja-Oghlan and M. Krivelevich
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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
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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
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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)
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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
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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
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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
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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?
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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
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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
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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
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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:
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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
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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)
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Further Research m/nm/n solution space 4.26 cclogn
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