Thomas Holenstein Microsoft Research, SVC Dagstuhl

 Problem Statement and Connection to PCP  Consistent Sampling Lemma  Raz’s Counterexample [Raz 08]  Parallel Repetition Theorem

(x,y) ← P XY xy ab Q(x,y,a,b)? i=1,…,n: (x i,y i ) ← P XY x 1,…,x n a 1,…,a n y 1,…,y n b 1,…,b n For all i: Q(x i,y i,a i,b i )? Does this decrease the winning probability? Theorem[Raz 95, H 07]:Yes, from 1 - γ to  (1 - O(γ 3 )) n/log(|A||B|) Why is this not trivial? Players may not answer independently. (1 – γ) n not true! Why would I care about this? Theorem[Raz 08]:  G with v(G)= 1 – γ, v(G n )=(1-O(γ 2 )) n

Proof sketch: 3-SAT is NP-hard [Cook71, Levin73] Gap Amplification It is NP-hard to satisfy 99% of the clauses of a satisfiable 3-SAT formula [BFL91, BFLS91, FGLSS 91, AS92, ALMSS92, Dinur07] Parallel Repetition It is NP-hard to win in a game as before with probability  [Raz98, H] Fourier Techniques It is NP-hard to satisfy 7/8+  of the clauses of a 3-SAT formula [Håstad 01] 7878 It is NP-hard to satisfy –+  of the clauses of a satisfiable 3SAT formula

(x 1  x 2  x 3 )  (x 1  x 5  x 9 )    (x 2  x 6  x 9 )  …  (x 16  x 18  x 41 ) ( 1,0,0 ) 9 0 Fraction of unsatisfied clauses = O(rejection probability) Pick a random clause, send clause to Alice, Variable to Bob

 Problem Statement and connection to PCP  Consistent Sampling Lemma  Raz’s Counterexample [Raz 08]  Parallel Repetition Theorem

s1s1 s4s4 s3s3 s2s2 1 0 ½ A B 6 x 1,…,x n a 1,…,a n y 1,…,y n b 1,…,b n s4s4 s3s3 Subtask: Alice gets P S, Bob gets P S’, minimize Pr[S  S’]

 Problem Statement and connection to PCP  Consistent Sampling Lemma  Raz’s Counterexample [Raz 08]  Parallel Repetition Theorem

 G: v(G) = 1-  and v(G n ) = (1-O(  2 )) n Odd cycle game (2k+1): x  {1,..., 2k+1}, y  {x-1,x,x+1}Q(x,y,a,b)  (x = y) = (a = b) v(G) = 1 – O(1/k) v(G n )  ½ for n = O(k 2 )

Start with a single instance (2k+1 = 11): x=7y=8 Pick an edge with the consistent sampling lemma Parameters are such that: ||P S - P T || = O(1/k), Pr[Bad edge] = O(1/k 2 ) ||P S n - P T n || = O(1) for n = O(k 2 )QED!

 Problem Statement and connection to PCP  Consistent Sampling Lemma  Raz’s Counterexample [Raz 08]  Parallel Repetition Theorem

i=1,…,n: (x i,y i ) ← P XY x 1,…,x n a 1,…,a n y 1,…,y n b 1,…,b n v(G) = 1 - γ implies v(G n )  (1-O(γ 3 )) n/log(|A||B|)

Pr[W 1 ...  W n ] = Pr[W 1 ] Pr[W 2 |W 1 ]... Pr[W n |W 1 ...  W n-1 ] It is sufficient to show that Pr[W m+1 |W 1 ...  W m ] ≤ v + ε Also enough: for all i 1, …,i m there exists a j such that Pr[W j |W i 1 ...  W i m ] ≤ v + ε Goal: show Pr[W 1  W 2 ...  W n ] << v x 1,…,x n a 1,…,a n y 1,…,y n b 1,…,b n mnmn ε = O ( log(|A||B|) – log(Pr[W i 1 ...  W i m ] ) Increasing n decreases ε!

Fix a strategy, and indices i 1,…,i m. There exists j such that: Pr[W j |W i 1 ...  W i m ] ≤ v + ε mnmn Increasing n decreases ε! ε = O ( log(|A||B|) – log(Pr[W i 1 ...  W i m ] ) x 1,…,x n a 1,…,a n y 1,…,y n b 1,…,b n

Given: strategy of Alice and Bob, indices i 1,…,i m Goal: show that for some j: Pr[W j |W i 1 ...  W i m ] ≤ v + ε xy x 1,…,x n y 1,…,y n a 1,…,a n b 1,…,b n bjbj ajaj Assume otherwise and get a strategy for the initial game. x j = x and y j = y (x 1,…,x n,y 1,…,y n ) have correct distribution Can only work if: x 1,…,x n a 1,…,a n y 1,…,y n b 1,…,b n

P S,T|X=x,Y=y S ↔ X ↔ Y ↔ T S = T and P S|X=x,Y=y ≈ P S|X=x ≈ P S|Y=y I know of two sufficient conditions: Goal: embed into (X,S)Goal: embed into (Y,T) X Y P XY x 1,…,x n a 1,…,a n y 1,…,y n b 1,…,b n

Send the corresponding answer to the refereeForget part of the generated informationApply the given strategy for multiple games xy Won games x y x7x7 y4y4 y3y3 y6y6 y1y1 y2y2 y7y7 x1x1 x4x4 x3x3 x2x2 x6x6 a1a1 a2a2 b1b1 b2b2 x1x1 x2x2 x3x3 x4x4 x6x6 y1y1 y2y2 y7y7 a2a2 b2b2 a1a1 b1b1 Using P S|X=xY=y ≈ P S|X=x ≈ P S|Y=y Using S ↔ X ↔ Y ↔ T x 1,…,x n a 1,…,a n y 1,…,y n b 1,…,b n b1b1 b7b7 b6b6 b4b4 b3b3 b2b2 a1a1 a7a7 a6a6 a4a4 a3a3 a2a2 a5a5 b5b5 Assume: Pr[W 5 | W 1  W 2 ] is big

v(G) = 1 – γ  v(G n )  (1 – O(γ 3 )) n/log(|A||B|) [Raz 98,H07]  G: v(G) = 1 – γ and v(G n )  (1 – O(γ 2 )) n [Raz 08] v(P) = 1 – γ  v(P n )  (1 – O(γ 2 )) n [Rao 08]  G: v(G) = ¾ and v(G n )  (½) n/O(log(|A| |B|)) [FV 02] Connection with Foams: [FKO 07, KORW 08] ~ Parallel Repetition of Unique Games: [BHHRRS ’08] Thank you!