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Unformatted text preview: 1 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) Prover Verifier . . . common input: x accept/ reject # rounds = poly(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 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 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 satisfying assignments • φ(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 “ φ’ = φ(c, x 2 , …, x n ) has k c satisfying assignments” – at end, verifier can check for itself. 2 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 – 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 = { (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 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 + k 1 = k – sends random z F q • continue with proof that k z = Σ x 2 = 0,1 … Σ x...
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This document was uploaded on 01/05/2012.
 Fall '09

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