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Parallel Repetition From Fortification Dana Moshkovitz MIT

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Coloring Two Prover Projection Game: Given graph G=(V,E), 1.Pick uniformly v V and two edges {v,u},{v,u’} E. 2.Each prover gets an edge. Answers: colors for endpoints. 3.Check that the two provers agree on the color of v. {v,u}{v,u’} v, uv, u’ val( G ) = max P(verifier accepts) NP-hardness: It is NP-hard, given a game, to distinguish between val( G )=1 and val( G )<1. PCP Theorem: It is NP-hard, given a game, to distinguish between val( G )=1 and val( G )<0.99. 0.01?

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Parallel Repetition: Sequential repetition with two provers?? Product game G k e 1,…,e k e 1 ’,…,e k ’ a 1,…,a k a 1 ’,…,a k ’ val( G k ) val( G ) k The Parallel Repetition Problem: val( G k ) ≤ val( G ) k ??

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Raz: 2 is tight! Special cases analyzed Twenty Five Years of Parallel Repetition Research 1990 1994 Problem posed by Fortnow, Rompel, Sipser Feige,Kilian: Engineer G so val( G k ) poly(1/k). Raz: If val( G )=1- & players’ answers in , val( G k ) (1- ( 32 )) k/2log| |. 2007 Holenstein simplifies! If val( G )=1- & players’ answers in , val( G k ) (1- ( 3 )) k/2log| |. Rao: For projection games, if val( G )=1- & players’ answers in , val( G k ) (1- ( 2 )) Ω(k). Impagliazzo,Kabanets,Wigderson: Feige-Kilian engineering of G yields val( G k ) exp(-Ω( k)). 2014 Partial results Dinur,Steurer: For projection games, val( G k ) (2val( G )) k/2. Current result: Can engineer projection G so val( G k ) val( G ) k + Raz-Rosen: If G projection game on expander & val( G )=1- , val( G k ) (1- ) Ω(k).

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Basics of Two Prover games [BGKW’88] Hardness of approximation is based on projection games (aka Label Cover). Val( G ) – determines the hardness factor. Size( G )=|R|- determines reduction blow-up. Alphabet size = | | - also effects the blow-up. R = randomness strings. PickTest : R X X. : R {accept,reject}. A game G is defined by: X = set of questions. = set of answers. there is a set Y, labels L for Y, a bipartite graph G=(X,Y,E), and functions {f e : L}. PickTest picks a random y Y and two random neighbors x,x’ X ; (r,a,a’) accepts if f (x,y) (a)=f (x’,y) (a’).

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Parallel Repetition Might Not Decrease Value Feige’s Non-Interactive Agreement Verifier picks random bits as x, x’. Each player should respond (player,bit). Verifier accepts if both answered same player and his input bit. 01 Wang,0 Wang,0 WangMang val(NIA) = ½. val(NIA 2 ) = ?

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val(NIA 2 ) = val(NIA) = 1/2 x 1,x 2 Wang,x 1 Mang,x 1 Wang,x 2 ’ Mang,x 2 ’ WangMang x 1 ’,x 2 ’ Verifier accepts in first round with prob 1/2. Conditioned on acceptance in first round, x 1 =x 2 ’, i.e., probability 1 of acceptance in second round.

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Parallel Repetition is Subt le val( G 2 )=P(x 1,x 1 ’ agree) P(x 2,x 2 ’ agree|x 1,x 1 ’ agree) We focus on a sub-game of G w here x 1,x 1 ’ agree. Its value might be much higher than val( G ).

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Compare to val(NIA k ) val(NIA) k/2 This Work: Engineer The Game so Parallel Repetition Sequential Repetition -Fortification: Simple, natural, transformation on projection games; maintains the value of the game, somewhat increases size and alphabet. Parallel repetition theorem: for -fortified G, ≤ poly( )(#labels) -k val( G k ) ≤ (val( G )+O(k )) k No Round Left Behind® Combinatorial fortified G G repeat

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Implication to PCP – Combinatorial PCP with Low Error Starting from [Dinur 05]: combinatorial projection PCP with arbitrarily arbitrarily small constant error. – Implies that for any >0, it is NP-hard to approximate Max-SAT to within 7/8 + [Håstad 97]. – In general: sufficient to determine approximation threshold for many optimization problems.

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Fortification The verifier picks questions to provers x, x’ as before. Picks extra questions {x 1,…,x w }, x {x 1,…,x w } and {x 1 ’,…,x w ’}, x’ {x 1 ’,…,x w ’}, where w=poly(1/ ). – Sets of questions picked using extractor on X. – E.g., random walk on a constant-degree expander on X. The provers answer all questions; the verifier only checks the answers to x, x’. x 1,…,x w x 1 ’,…,x w ’ a 1,…,a w a 1 ’,…,a w ’ - In Feige-Kilian’s “Confuse and Compare” w=2. - In parallel repetition independence between the k tests seems essential.

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Dfn: Fraction rectangular sub-games: questions of each prover restricted to fraction subset. Dfn: -fortified: value of all fraction rectangular sub-games of G at most val( G )+ . Lem: The transformed game is -fortified (from extractor property). Extractors: Not every projection game on extractor is fortified. Size: Fortification increases size before repeating, but (size poly(1/ )) k less than size 2k. Alphabet: Provers give w times more answers where w=poly(1/ ). In repetition only k log(1/ ) times more answers. Projection: Fortification preserves projection, but not necessarily uniqueness.

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Squaring: For -fortified G, ≤ /(2 #colors ) val( G 2 ) ≤ val( G ) 2 +2 Proof: Assume by way of contradiction a strategy for G 2 that does better. 1.Consider questions x 1,x 1 ’,y 1 & label to y 1. Define: S:= { x 2 | (x 1,x 2 ) assigns y 1 } T:= { x 2 ‘| (x 1 ’, x 2 ‘) assigns y 1 } 2.Small Prob Events: Probability of (|S|< |X| & x 2 S) or (|T|< |X| & x 2 ’ T) is 2 #colors ≤ . 3.Conditioning: P(agree on y 2 & |S| |X|& |T| |X| |agree on y 1 ) > val( G )+ . 4.Fortification: If |S| |X|& |T| |X|, value of G restricted to S,T is ≤ val( G )+ . 5.Overall: P(agree on y 2 & agree on y 1 )≤val( G ) 2 +2 .

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The Influence of Parallel Repetition PCP and Hardness of Approximation: Soundness amplification [Raz]. Cryptography: zero-knowledge two prover protocols [BenOr-Goldwasser-Kilian-Wigderson], arguments [Haitner]. Quantum Computing: Amplifying Bell’s inequality. Communication complexity: Direct sum theorems [Krachmer-Raz-Wigderson], compression [Barak-Braverman-Chen-Rao]. Geometry: Tiling of R n by volume 1 tiles with surface area sphere [Feige-Kindler-O’Donnell, Kindler-O’Donnell-Rao-Wigderson].

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