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Unformatted text preview: CS151 Complexity Theory Lecture 14 May 12, 2011 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 a c 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 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 dont know how to show GNI in NP GNI in coNP Theorem (LFKN): coNP IP May 12, 2011 6 The power of IP Proof idea: input: (x 1 , x 2 , , x n ) prover: I claim has k satisfying assignments true iff (0, x 2 , , x n ) has k satisfying assignments (1, x 2 , , x n ) has k 1 satisfying assignments k = k + k 1 prover sends k , k 1 verifier sends random c {0,1} prover recursively proves = (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 : (x 1 , x 2 , , x n ) has k satisfying assignments accept with prob. 1 Soundness : (x 1 , x 2 , , x n ) does not have k satisfying assigns. accept prob. 1 2n 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 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 = { (, k): CNF has exactly k satisfying assignments} is in IP May 12, 2011 9 The power of IP First step: arithmetization transform (x 1 , x n ) into polynomial p (x 1 , x 2 , x n ) of degree d over a field F q ; q prime > 2 n recursively: x i x i (1  p ) (p )(p ) 1  (1  p )(1  p ) for all x {0,1} n we have p (x) = (x) degree d  can compute p (x) in poly time from and x May 12, 2011 10 The power of IP Prover wishes to prove: k = x 1 = 0, 1 x 2 = 0,1 x n = 0, 1 p (x 1 , x 2 , , x n ) Define: k z = x 2 = 0,1 x n = 0, 1 p ( z , x 2 , , x n ) prover sends: k z for all z...
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This document was uploaded on 01/05/2012.
 Fall '09

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