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Parallel Repetition of Two Prover Games Ran Raz Weizmann Institute and IAS

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Two Prover Games [BGKW]: 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]

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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})

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Projection Games: 8 x,y, 8 a, 9 unique b, such that V(x,y,a,b)=1

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Projection Games: 8 x,y, 8 a, 9 unique b, such that V(x,y,a,b)=1 Unique Games: In addition: 8 x,y, 8 b, 9 unique a, such that V(x,y,a,b)=1

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Probabilistic Protocols: Player A answers a=A(x) Player B answers b=B(y) Where A(x),B(y) may depend on a common random string However: Probabilistic Value = Deterministic Value

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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]

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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 ?

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Example [Feige]: 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)

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Parallel Repetition Theorem [R-95]: (Conjectured by Feige and Lovasz) 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 ?

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

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Strong Parallel Repetition Problem: Is the following true ? Val(G) = 1-, ( < ½) ) (for any game or for interesting special cases) [R-08]: G s.t.: Val(G) = 1-,

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Over Expander Graphs [RR-10]: Val(G) = 1-, ( < ½) ) For general games: For projection games:

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Applications of Parallel Repetition: 1) PCP & Hardness of Approximation: [BGS],[Has],[Fei],[Kho],... 2) Geometry: understanding foams, tiling the space R n [FKO],[KORW],[AK] 3) Quantum Information: strong EPR paradoxes [CHTW] 4) Communication Complexity: direct sum/product results [PRW],[BBCR] 5) Cryptography

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Parallel Repetition, PCP, and Hardness of Approximation [BGS],[Has],[Fei],[Kho],...

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PCP Theorem [BFL,FGLSS,AS,ALMSS]: Any (length n) proof can be rewritten as a length poly(n) proof that can be (probabilistically) verified by reading a constant number of bits. proof is correct ) P(V accepts) = 1 statement has no proof ) P(V ¼ accepts) 1-ε

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Two Query PCP: Any (length n) proof can be rewritten as a length poly(n) proof that can be verified by reading only 2 symbols (O(1) bits) proof is correct ) P(V accepts) = 1 statement has no proof ) P(V ¼ accepts) 1-ε

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PCP as a Two-Prover Game : A (length poly(n)) PCP can be translated into a 2-prover game with questions length O(log n) and answers length O(1) xy A(x)B(y) proof is correct ) P(V accepts) = 1 statement has no proof ) P(V ¼ accepts) 1-ε

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PCP by Parallel Repetition: Any (length n) proof can be rewritten as a length poly(n) proof that can be verified by reading only 2 symbols (O(1) bits) proof is correct ) P(V accepts) = 1 statement has no proof ) P(V ¼ accepts) ε

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Two Query PCP: Φ 2 SAT can be proved by a poly-size proof that can be verified by reading only 2 of its symbols (O(1) bits) Φ sat ) 9 ¼ proof, P(V ¼ accepts) = 1 Φ not sat ) 8 ¼ proof, P(V ¼ accepts) 1-ε

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PCP as a Two-Prover Proof : Φ 2 SAT can be proved by a 2-prover proof with questions length O(log n) and answers length O(1) Φ sat ) 9 ¼ proof, P(V ¼ accepts) = 1 Φ not sat ) 8 ¼ proof, P(V ¼ accepts) 1-ε xy A(x)B(y)

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PCP Theorem (2 Queries): Original: Using Parallel Repetition: Φ sat ) 9 ¼ proof, P(V ¼ accepts) = 1 Φ not sat ) 8 ¼ proof, P(V ¼ accepts) ε (for any constant ε > 0) Φ sat ) 9 ¼ proof, P(V ¼ accepts) = 1 Φ not sat ) 8 ¼ proof, P(V ¼ accepts) 1-ε (for some constant ε > 0)

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PCP as Hardness of Approximation: Given a two-prover game G (with constant answer size) It is NP hard to distinguish between: Val(G) = 1 and Val(G) · 1- (for some constant > 0) [FGLSS] Using Parallel Repetition: It is NP hard to distinguish between: Val(G) = 1 and Val(G) · (for any constant > 0)

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Hardness of Approximation Results: [BGS],[Has],[Fei],[Kho],... Optimal hardness results for: 3SAT, 3LIN, Set-Cover,… Example: 3LIN [Has-98]: Given a set of linear equations over GF[2] It is NP hard to distinguish between: 9 ¼ solution that satisfies > 1-ε fraction and every solution satisfies < ½+ε fraction

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PCP Theorem (2 Queries): Original: Using Parallel Repetition: proof is correct ) P(V accepts) = 1 statement has no proof ) P(V ¼ accepts) 1-ε (for some constant ε > 0) proof is correct ) P(V accepts) = 1 statement has no proof ) P(V ¼ accepts) ε (for any constant ε > 0)

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PCP Theorem [BFL,FGLSS,AS,ALMSS]: Given a projection game 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)

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Parallel Repetition PCP: 8 constant > 0, 9 constant s, s.t. Given a projection game G (with answer size s) It is NP hard to distinguish between: Val(G) = 1 and Val(G) · Hardness of Approximation Results: [BGS],[Has],[Fei],[Kho],... Optimal hardness results for: 3SAT, 3LIN, Set-Cover,…

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Parallel Repetition PCP: 8 constant > 0, 9 constant s, s.t. Given a projection game G (with answer size s) It is NP hard to distinguish between: Val(G) = 1 and Val(G) · Unique Games Conjecture [Khot]: 8 constant > 0, 9 constant s, s.t. Given a unique game G (with answer size s) It is NP hard to distinguish between: Val(G) ¸ 1- and Val(G) ·

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2 Step Approach for Proving UGC: 1) Show that: It is NP hard to distinguish between: Val(G) ¸ 1- ® and Val(G) · 1- ¯ (where ® << ¯ ) 2) Error Reduction (possibly by repetition): It is NP hard to distinguish between: Val(G) ¸ 1- and Val(G) · Problem: requires strong parallel repetition Nevertheless: supported by [K],[AIMS],[ABS]

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Predictions: 1) If the UGC is proved any time soon, it will be proved by these 2 steps 2) The UGC is not going to be proved any time soon...

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Tiling R n, Cubical Foams, and Parallel Repetition of the Odd Cycle Game [FKO],[KORW],[AK],[A]...

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Cubical Foams: The unit cube C = tiles R n by the lattice Z n. R n = C £ Z n What is the minimal surface area of a (volume 1) object D that tiles R n by Z n ? (R n = D £ Z n )

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Best Cubical Foams: Minimal surface area of a (volume 1) object that tiles R n by Z n : Upper bound: 2n : unit cube Lower bound: : volume 1 ball

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Best Cubical Foams: Minimal surface area of a (volume 1) object that tiles R n by Z n : Upper bound: 2n : unit cube Lower bound: : volume 1 ball [KORW-08]: There is an object with surface area similar to the (volume 1) ball, that tiles R n as a cube

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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

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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 To win with probability 1, the players need to 2-color the odd cycle consistently 1 0 11 0

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Protocol for the Odd Cycle Game: A gets x, B gets y. They win if (x,y) e (e breaks the cycle) x y e 0 0 0 0 1 1 1 1 1

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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

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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 We think of the game as played on the n dimensional torus 1 23 n

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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 A,B answer by a color for each coordinate. They win if the colors are consistent on all coordinates

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Odd Cycle Game and Cubical Foams: The players can color consistently every cell. Err on edges that cross the surface

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The Shorter Story: [FKO-07]: Cubical Foams imply protocols for OCG [R-08]: (a counterexample to strong par. rep.) [KORW-08]: Cubical Foams with surface area (based on the protocol for OCG)

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Cubical Foams: Minimal surface area of a (volume 1) cell that tiles R n by Z n : Upper bound: 2n : unit cube Lower bound: : volume 1 ball [KORW-08]: There is an object with surface area similar to the (volume 1) ball, that tiles R n as a cube

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Thank You!

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Parallel Repetition, Bell Inequalities, and the EPR Paradox [CHTW]

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Entangled Two Prover Games: A,B share entangled quantum state |s i A,B A gets x, B gets y A measures A, B measures B A answers a, B answers b They win if V(x,y,a,b)=1 Val Q (G) = Max A,B Pr x,y [V(x,y,a,b)=1]

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Bell Inequalities (EPR Paradox): 9 G s.t. Val Q (G) > Val(G) [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)

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Does God Play Dice ? a,b are the outcome of a quantum measurement. Does God play dice ? Hidden Variables Theory: 9 additional variables H, s.t. a=a(x,y,H), b=b(x,y,H) (deterministically) H = outcome of all possible measurements (independent of x,y)

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a=a(x,y,H), b=b(x,y,H) Assume: x,y are independent A chooses x, B chooses y (at time t- ² ) Measurements occur at time t Information cannot propagate faster than light! Hence, Local Hidden Variables Theory: a=a(x,H), b=b(y,H) Thus, Val Q (G) = Val(G)

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Bell Inequalities (EPR Paradox): 9 G s.t. Val Q (G) > Val(G) [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)

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