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# lec14 - CS151 Complexity Theory Lecture 14 Interactive...

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CS151 Complexity Theory Lecture 14 May 12, 2011

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May 12, 2011 2 Interactive Proofs interactive proof system for L is an interactive protocol (P, V) Pro ve r  Ve rifie r  .   c o m m o n input: x ac c e pt/r e je c t # ro unds  =  po ly(|x| )
May 12, 2011 3 Interactive Proofs interactive proof system for L is an interactive protocol (P, V) completeness : x L Pr[V accepts in (P, V)(x)] 2/3 soundness : x L 8 P* Pr[V accepts in (P*, V)(x)] 1/3 efficiency : V is p.p.t. machine repetition: can reduce error to any ε

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May 12, 2011 4 Interactive Proofs IP = {L : L has an interactive proof system} Observations/questions: philosophically interesting: captures more broadly what it means to be convinced a statement is true clearly NP IP. Potentially larger. How much larger? if larger, randomness is essential (why?)
May 12, 2011 5 The power of IP We showed GNI 2 IP GNI IP suggests IP more powerful than NP , since we don’t know how to show GNI in NP GNI in coNP Theorem (LFKN): coNP IP

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May 12, 2011 6 The power of IP Proof idea: input: φ(x 1 , x 2 , …, x n ) prover: “I claim φ has k satisfying assignments” true iff φ(0, x 2 , …, x n ) has k 0 satisfying assignments φ(1, x 2 , …, x n ) has k 1 satisfying assignments k = k 0 + k 1 prover sends k 0 , k 1 verifier sends random c {0,1} prover recursively proves “ φ’ = φ(c, x 2 , …, x n ) has k c satisfying assignments” at end, verifier can check for itself.
May 12, 2011 7 The power of IP Analysis of proof idea: – Completeness : φ(x 1 , x 2 , …, x n ) has k satisfying assignments accept with prob. 1 – Soundness : φ(x 1 , x 2 , …, x n ) does not have k satisfying assigns. accept prob. 1 – 2 -n Why? It is possible that k is only off by one; verifier only catches prover if coin flips c are successive bits of this assignment

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May 12, 2011 8 The power of IP Solution to problem (ideas): – replace {0,1} n with (F q ) n verifier substitutes random field element at each step vast majority of field elements catch cheating prover (rather than just 1) Theorem : L = { (φ, k): CNF φ has exactly k satisfying assignments} is in IP
May 12, 2011 9 The power of IP First step: arithmetization transform φ(x 1 , … x n ) into polynomial p φ (x 1 , x 2 , … x n ) of degree d over a field F q ; q prime > 2 n recursively: • x i x i ¬ φ (1 - p φ ) • φ φ’ (p φ )(p φ’ ) • φ φ’ 1 - (1 - p φ )(1 - p φ’ ) for all x {0,1} n we have p φ (x) = φ(x) degree d |φ| can compute p φ (x) in poly time from φ and x

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May 12, 2011 10 The power of IP Prover wishes to prove: k = Σ x 1 = 0, 1 Σ x 2 = 0,1 Σ x n = 0, 1 p φ (x 1 , x 2 , …, x n ) Define: k z = Σ x 2 = 0,1 Σ x n = 0, 1 p φ ( z , x 2 , …, x n ) prover sends: k z for all z F q verifier: checks that k 0 + k 1 = k sends random z F q continue with proof that k z = Σ x 2 = 0,1 Σ x n = 0, 1 p φ ( z , x 2 , …, x n ) at end: verifier checks for itself
May 12, 2011 11 The power of IP Prover wishes to prove: k = Σ x 1 = 0, 1 Σ x 2 = 0,1 Σ x n = 0, 1 p φ (x 1 , x

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lec14 - CS151 Complexity Theory Lecture 14 Interactive...

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