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# setcover - PCPs and Inapproxiability CIS 6930 October 5...

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PCPs and Inapproxiability CIS 6930 October 5, 2009 Lecture Hardness of Set Cover Lecturer: Dr. My T. Thai Scribe: Ying Xuan 1 Preliminaries 1.1 Two-Prover-One-Round Proof System A new PCP model 2P1R – Think of the proof system as a game between two provers and one verifier. Prover Model: each tries to cheat – convince the verifier that a ”No” instance is ”Yes”; each cannot communicate with the other on their answers; Verifier Model: tries not to be cheated – ensures the probability to be cheated is upper bounded by some small constant; can cross-check the two provers’ answers; is only allowed to query one position in each of the two proofs. Definition 1 Given three parameters: completeness c , soundness s and the number of random bits provided to the verifier r ( n ) , assume the two proofs are written in two alphabets Σ 1 and Σ 2 respectively, then a language L is in 2P1R c,s ( r ( n )) if there is a polynomial time bounded verifier V that receives O ( r ( n )) truly random bits and satisfies: for every input x L , there is a pair of proofs y 1 Σ * 1 and y 2 Σ * 2 that makes V accept with probability c ; for every input x / L and every pair of proofs y 1 Σ * 1 and y 2 Σ * 2 makes V accept with probability < s ; 2P1R is actually a mechanism to solve Label-Cover: L ( G = ( U, V ; E ) , Σ , Π) u U and v V , we have v : proof P 1 ; u : proof P 2 ; uv E : u , v have answers for the same bit; a labeling L is said to satisfy an edge uv iff Π uv ( L ( u )) = L ( v ) : the two answers are consistent, otherwise they are cheating. Hardness of Set Cover-1

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You can see this from the next proof and the proof of NP-hard for Gap-Max-Label- Cover Σ 1 in previous lecture. Theorem 2 (29.21) There is a constant ϵ p > 0 such that NP = 2P1R 1 , 1 - ϵ p (log( n )) . Proof: 2P1R 1 , 1 - ϵ p (log( n )) NP (exercise); NP 2P1R 1 , 1 - ϵ p (log( n )) (shown here). Given a SAT formula ϕ , we employ a gap-producing reduction to obtain a MAX- E3SAT(E5) instance ψ , that is, given m is the number of clauses in ψ : if ϕ is satisfiable, OPT ( ψ ) = m ; if ϕ is not satisfiable, OPT ( ψ ) < 1 - ϵ p for some constant ϵ p Goal : Prove SAT 2P1R 1 , 1 - ϵ p (log( n )) Protocol: V selects an index of a clause, sends it to the first prover, selects a random variable in the clause, and sends it to the second prover. First prover returns 3 bits (the assignment of the selected clause), second returns 1 bit.
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