lec15 - CS151 Complexity Theory Lecture 15 Arthur-Merlin...

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CS151 Complexity Theory Lecture 15 May 17, 2011
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May 17, 2011 2 Arthur-Merlin Games IP permits verifier to keep coin-flips private necessary feature? GNI protocol breaks without it Arthur-Merlin game : interactive protocol in which coin-flips are public Arthur (verifier) may as well just send results of coin-flips and ask Merlin (prover) to perform any computation Arthur would have done
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May 17, 2011 3 Arthur-Merlin Games Clearly Arthur-Merlin IP “private coins are at least as powerful as public coins” Proof that IP = PSPACE actually shows PSPACE Arthur-Merlin IP PSPACE “public coins are at least as powerful as private coins” !
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May 17, 2011 4 Arthur-Merlin Games Delimiting # of rounds: AM[k] = Arthur-Merlin game with k rounds, Arthur (verifier) goes first MA[k] = Arthur-Merlin game with k rounds, Merlin (prover) goes first Theorem : AM[k] ( MA[k] ) equals AM[k] ( MA[k] ) with perfect completeness. Theorem : for all constant k 2 AM[k] = AM[2] .
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May 17, 2011 5 MA and AM Two important classes: MA = MA[2] AM = AM[2] definitions without reference to interaction: L MA iff 9 poly-time language R x L 9 m Pr r [(x, m, r) R] = 1 x L 8 m Pr r [(x, m, r) R] ½ L AM iff 9 poly-time language R x L Pr r [ 9 m (x, m, r) R] = 1 x L Pr r [ 9 m (x, m, r) R] ½
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May 17, 2011 6 MA and AM P NP coNP Σ 2 Π 2 AM coAM MA coMA
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May 17, 2011 7 MA and AM Theorem : coNP AM PH = AM . Proof: – suffices to show Σ 2 AM (and use AM Π 2 ) – L Σ 2 iff 9 poly-time language R x L 9 y 8 z (x, y, z) R x L 8 y 9 z (x, y, z) R Merlin sends y 1 AM exchange decides coNP query: 8 z (x, y, z) R ? 3 rounds; in AM
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May 17, 2011 8 MA and AM We know Arthur-Merlin = IP . “public coins = private coins” Theorem (GS): IP[k] AM[O(k)] stronger result implies for all constant k 2, IP[k] = AM[O(k)] = AM[2] So, GNI IP[2] = AM
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May 17, 2011 9 Back to Graph Isomorphism The payoff: not known if GI is NP -complete. previous Theorems: if GI is NP -complete then PH = AM unlikely! Proof: GI NP -complete GNI coNP -complete coNP AM PH = AM
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Derandomization revisited L MA iff 9 poly-time language R x L 9 m Pr r [(x, m, r) R] = 1 x L 8 m Pr r [(x, m, r) R] ½ Recall PRGs: for all circuits C of size at most s : |Pr y [C(y) = 1] – Pr z [C(G(z)) = 1]| ≤ ε May 17, 2011 10 NP AM MA s e e d o utput s tring G t b its u  b its
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Using PRGs for MA L MA iff 9 poly-time language R x L 9 m Pr r [(x, m, r) R] = 1 x L 8 m Pr r [(x, m, r) R] ½ produce poly-size circuit C such that C(x, m, r) = 1 , (x,m,r) 2 R • for each x, m can hardwire to get C x,m 9 m Pr y [C x,m (y) = 1] = 1 (“yes”) 8 m Pr y [C x,m (y) = 1] ≤ 1/2 (“no”) May 17, 2011 11
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lec15 - CS151 Complexity Theory Lecture 15 Arthur-Merlin...

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