1. A Counterexample to Strong Parallel Repetition Ran Raz Weizmann Institute

2. Two Prover Games: Player A gets x Player B gets y (x,y) 2 R publicly known distribution Player A answers a=A(x) Player B answers b=B(y) They win if V(x,y,a,b)=1 (V is a publicly known function) Val(G) = Max A,B Pr x,y [V(x,y,a,b)=1]

3. Example: Player A gets x 2 R {1,2} Player B gets y 2 R {3,4} A answers a=A(x) 2 {1,2,3,4} B answers b=B(y) 2 {1,2,3,4} They win if a=b=x or a=b=y Val(G) = ½ (protocol: a=x, b 2 R {1,2}) (alternatively : b=y, a 2 R {3,4})

4. Parallel Repetition: A gets x = (x 1,..,x n ) B gets y = (y 1,..,y n ) (x i,y i ) 2 R the original distribution A answers a=(a 1,..,a n ) =A(x) B answers b=(b 1,..,b n ) =B(y) V(x,y,a,b) =1 iff 8 i V(x i,y i,a i,b i )=1 Val(G n ) = Max A,B Pr x,y [V(x,y,a,b)=1]

5. Parallel Repetition: A gets x = (x 1,..,x n ) B gets y = (y 1,..,y n ) (x i,y i ) 2 R the original distribution A answers a=(a 1,..,a n ) =A(x) B answers b=(b 1,..,b n ) =B(y) V(x,y,a,b) =1 iff 8 i V(x i,y i,a i,b i )=1 Val(G n ) = Max A,B Pr x,y [V(x,y,a,b)=1] Val(G) ¸ Val(G n ) ¸ Val(G) n Is Val(G n ) = Val(G) n ?

6. Example: A gets x 1,x 2 2 R {1,2} B gets y 1,y 2 2 R {3,4} A answers a 1,a 2 2 {1,2,3,4} B answers b 1,b 2 2 {1,2,3,4} They win if 8 i a i =b i =x i or a i =b i =y i Val(G 2 ) = ½ = Val(G) By: a 1 =x 1, b 1 =y 2 -2, a 2 =x 1 +2, b 2 =y 2 (they win iff x 1 =y 2 -2)

7. Parallel Repetition Theorem [R95]: 8 G Val(G) < 1 ) 9 w < 1 (s = length of answers in G) Assume that Val(G) = 1-  What can we say about w ?

8. Parallel Repetition Theorem: Val(G) = 1- ,  (  < ½) ) [R-95]: [Hol-06]: For unique and projection games: [Rao-07]: (s = length of answers in G)

9. Strong Parallel Repetition Problem: Is the following true ? Val(G) = 1- ,  (  < ½) ) (for any game or for interesting special cases) Our Result: G s.t.: Val(G) = 1- ,

10. Applications of Parallel Repetition: 1) Communication Complexity: direct product results [PRW] 2) Geometry: understanding foams, tiling the space R n [FKO] 3) Quantum Computation: strong EPR paradoxes [CHTW] 4) Hardness of Approximation: [BGS],[Has],[Fei],[Kho],...

11. EPR Paradox: 9 G s.t. Val Q (G) > Val(G) Val Q (G) = value when the provers share entangled quantum states [CHTW 04]: 9 G s.t. Val Q (G) = 1 and Val(G) · 1-  (for some constant  > 0) Using Parallel Repetition: 9 G s.t. Val Q (G) = 1 and Val(G) ·  (for any constant  > 0)

12. PCP Theorem [BFL,FGLSS,AS,ALMSS]: Given G (with constant answer size) It is NP hard to distinguish between : Val(G) = 1 and Val(G) · 1-  (for some constant  > 0) Using Parallel Repetition: It is NP hard to distinguish between : Val(G) = 1 and Val(G) ·  (for any constant  > 0)

13. Unique Games (UG): G is a UG if V(x,y,a,b) satisfies : 8 x,y,a 9 unique b, V(x,y,a,b) = 1 8 x,y,b 9 unique a, V(x,y,a,b) = 1 Unique Games Conjecture [Khot]: 8 constant  > 0, 9 constant s, s.t. Given a UG G (with answer size s) It is NP hard to distinguish between : Val(G) ¸ 1-  and Val(G) · 

14. UGC and Max-Cut [KKMO]: UGC ) 8  > 0, given a graph G, It is NP hard to distinguish between : Max-Cut(G) ¸ 1-  2 and Max-Cut(G) · 1-2  Using Strong Parallel Repetition: UGC, 8  > 0, given a graph G, It is NP hard to distinguish between : Max-Cut(G) ¸ 1-  2 and Max-Cut(G) · 1-2 

15. Odd Cycle Game [CHTW,FKO]: A gets x 2 R {1,..,m} (m is odd) B gets y 2 R {x,x-1,x+1} (mod m) A answers a=A(x) 2 {0,1} B answers b=B(y) 2 {0,1} They win if x=y, a  b

16. Odd Cycle Game [CHTW,FKO]: A gets x 2 R {1,..,m} (m is odd) B gets y 2 R {x,x-1,x+1} (mod m) A answers a=A(x) 2 {0,1} B answers b=B(y) 2 {0,1} They win if x=y, a  b 1 0 1 1 0

17. Parallel Repetition of OCG: A gets x 1,..,x n 2 R {1,..,m} B gets y 1,..,y n 2 R {1,..,m} 8 i y i 2 R {x i,x i -1,x i +1} (mod m) A answers a 1,..,a n 2 {0,1} B answers b 1,..,b n 2 {0,1} They win if 8 i x i =y i, a i  b i 1 23 n

18. Parallel Repetition of OCG: A gets x 1,..,x n 2 R {1,..,m} B gets y 1,..,y n 2 R {1,..,m} 8 i y i 2 R {x i,x i -1,x i +1} (mod m) A answers a 1,..,a n 2 {0,1} B answers b 1,..,b n 2 {0,1} They win if 8 i x i =y i, a i  b i Motivation [FKO]: Max-Cut vs. UGC, Understanding foams, Tiling the space 1 23 n

19. Our Results: (match an upper bound of [FKO]) For n ¼ m 2, For n ¸  (m 2 ), 1 23 n

20. Odd Cycle Game: A gets x, B gets y. If they can agree on an edge e that doesn’t touch x,y, they win ! x y e 0 0 0 0 1 1 1 1 1

21. Parallel Repetition of OCG: A gets x 1,..,x n, B gets y 1,..,y n. If they can agree on edges e 1,..,e n that don’t touch x 1,..,x n, y 1,..,y n, they win !

22. Holenstein’s Lemma [B,KT]: A has f: W ! R, B has g: W ! R, s.t., |f-g| 1 · O(  ) Using shared randomness, A can choose: u 2 f W, and B: v 2 g W, s.t. Prob[u=v] ¸ 1-O(  )

23. Distribution P: m=2k+1, P:[-k,k] ! R (symmetric) : 1) P(i) ¼ (k+1-|i|) 2 / m 3 2) P(0) = £ (1/m) 3) P(k) = P(-k) = £ (1/m 3 ) (negligible) 0 -kk 1/m 1/m 3

24. Distribution P: m=2k+1, P:[-k,k] ! R (symmetric) : 1) P(0) = £ (1/m) 2) P(k) = P(-k) = £ (1/m 3 ) (negligible) 3) 0 -kk 1/m 1/m 3

25. Distributions on the Odd Cycle: E = Edges of the odd cycle. Given x, A defines f x : E ! R : f x =P, concentrated on the edge opposite to x Given y, B defines g y : E ! R : g y =P, concentrated on the edge opposite to y f x ¼ g y (since x,y are adjacent) |f x -g y | 1 · O(1/m) x g y (e)=P(0) f x (e)=P(0) y

26. Holenstein’s Lemma: A has f: W ! R, B has g: W ! R, s.t., |f-g| 1 · O(  ) Using shared randomness, A can choose: u 2 f W, and B: v 2 g W, s.t. Prob[u=v] ¸ 1-O(  ) OCG: Using f x,g y, A,B can agree on an edge e that doesn’t touch x,y, with probability ¸ 1-O(1/m)

27. Our Protocol: Given x=(x 1,..,x n ), A defines f x : E n ! R, Given y=(y 1,..,y n ), B defines g y : E n ! R, Lemma: Using Holenstein’s lemma, A,B agree on edges e 1,..,e n that don’t touch x 1,..,x n, y 1,..,y n, with probability ¸

28. Proof Idea: Typically: in coordinates and in coordinates n/3 coordinates cancel each other. We are left with distance

29. Proof Idea: Hence, typically:

30. Follow Up Works: 1) Generalizations to unique games [BHHRRS] : Protocols for parallel repetition of any unique game 2) Tiling the space R n [KORW,AK] : R n can be tiled (with translations in Z n ), by objects with surface area similar to the one of the sphere (with volume 1)

31. The End