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Parallel Repetition of Two Prover Games

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

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Parallel Repetition of Two Prover Games

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

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

  3. Example: • Player A getsx 2R {1,2} • Player B getsy2R {3,4} • A answersa=A(x) 2 {1,2,3,4} • B answersb=B(y) 2 {1,2,3,4} • They win ifa=b=x or a=b=y • Val(G) = ½ • (protocol: a=x, b 2R {1,2}) • (alternatively : b=y, a 2R {3,4})

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

  5. 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

  6. Parallel Repetition: • A getsx = (x1,..,xn) • B getsy = (y1,..,yn) • (xi,yi) 2R the original distribution • A answersa=(a1,..,an)=A(x) • B answersb=(b1,..,bn)=B(y) • V(x,y,a,b) =1 iff8i V(xi,yi,ai,bi)=1 • Val(Gn) = MaxA,B Prx,y [V(x,y,a,b)=1]

  7. Parallel Repetition: • A getsx = (x1,..,xn) • B getsy = (y1,..,yn) • (xi,yi) 2R the original distribution • A answersa=(a1,..,an)=A(x) • B answersb=(b1,..,bn)=B(y) • V(x,y,a,b) =1 iff8i V(xi,yi,ai,bi)=1 • Val(Gn) = MaxA,B Prx,y [V(x,y,a,b)=1] • Val(G) ¸ Val(Gn) ¸ Val(G)n • Is Val(Gn) = Val(G)n?

  8. Example [Feige]: • A getsx1,x22R {1,2} • B getsy1,y22R {3,4} • A answersa1,a22 {1,2,3,4} • B answersb1,b22 {1,2,3,4} • They win if8iai=bi=xior ai=bi=yi • Val(G2) = ½ = Val(G) • By: a1=x1, b1=y2-2, a2=x1+2, b2=y2 • (they win iff x1=y2-2)

  9. Parallel Repetition Theorem [R-95]: • (Conjectured by Feige and Lovasz) • 8G Val(G) < 1 ) 9 w < 1 • (s= length of answers inG) • Assume thatVal(G) = 1- • What can we say about w ?

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

  11. 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-,

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

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

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

  15. 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(Vaccepts) = 1 statement has noproof ) P(V¼accepts) ≤ 1-ε

  16. 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(Vaccepts) = 1 statement has noproof ) P(V¼accepts) ≤ 1-ε

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

  18. 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: • 9¼ solution that satisfies >1-ε fraction and • every solution satisfies <½+ε fraction

  19. proof is correct ) P(Vaccepts) = 1 statement has noproof ) P(V¼accepts) ≤ 1-ε (for some constant ε > 0) • PCP Theorem (2 Queries): • Original: • UsingParallelRepetition: proof is correct ) P(Vaccepts) = 1 statement has noproof ) P(V¼accepts) ≤ ε (for any constant ε > 0)

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

  21. Cubical Foams: • The unit cube C = tiles Rn by • the lattice Zn. Rn = C £ Zn • What is the minimal • surface area of a • (volume 1) object D • that tiles Rnby Zn ? • (Rn = D £ Zn )

  22. Best Cubical Foams: • Minimal surface area of a (volume 1) • object that tiles Rnby Zn : • Upper bound:2n : unit cube • Lower bound: : volume 1 ball

  23. Best Cubical Foams: • Minimal surface area of a (volume 1) • object that tiles Rnby Zn : • 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 Rn as a cube

  24. Odd Cycle Game [CHTW,FKO]: • A getsx 2R {1,..,m} (m is odd) • B getsy2R {x,x-1,x+1} (mod m) • A answersa=A(x) 2 {0,1} • B answersb=B(y) 2 {0,1} • They win ifx=y , a=b

  25. Odd Cycle Game [CHTW,FKO]: • A getsx 2R {1,..,m} (m is odd) • B getsy2R {x,x-1,x+1} (mod m) • A answersa=A(x) 2 {0,1} • B answersb=B(y) 2 {0,1} • They win ifx=y , a=b • To win with probability 1, the players need • to 2-color the odd cycle consistently 1 0 0 1 1

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

  27. n 2 3 1 • Parallel Repetition of OCG: • A getsx1,..,xn2R {1,..,m} • B getsy1,..,yn2R {1,..,m} • 8 i yi2R {xi,xi-1,xi+1} (mod m) • A answersa1,..,an2 {0,1} • B answersb1,..,bn2 {0,1} • They win if8 i xi=yi, ai=bi

  28. n 2 3 1 • Parallel Repetition of OCG: • A getsx1,..,xn2R {1,..,m} • B getsy1,..,yn2R {1,..,m} • 8 i yi2R {xi,xi-1,xi+1} (mod m) • A answersa1,..,an2 {0,1} • B answersb1,..,bn2 {0,1} • They win if8 i xi=yi, ai=bi • We think of the game as played on the • n dimensional torus

  29. Parallel Repetition of OCG: • A getsx1,..,xn2R {1,..,m} • B getsy1,..,yn2R {1,..,m} • 8 i yi2R {xi,xi-1,xi+1} (mod m) • A answersa1,..,an2 {0,1} • B answersb1,..,bn2 {0,1} • They win if8 i xi=yi, ai=bi • A,B answer by a color • for each coordinate. • They win if the colors • are consistent on all • coordinates

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

  31. 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)

  32. Cubical Foams: • Minimal surface area of a (volume 1) • cell that tiles Rnby Zn : • 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 Rn as a cube

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

  34. Entangled Two Prover Games: • A,B share entangled quantum state |siA,B • A gets x, B gets y • A measures A,B measures B • A answers a, B answers b • They win ifV(x,y,a,b)=1 • ValQ(G) = MaxA,B Prx,y [V(x,y,a,b)=1]

  35. Bell Inequalities (EPR Paradox): • 9 Gs.t.ValQ(G) > Val(G) • [CHTW 04]: 9 Gs.t. • ValQ(G) = 1and Val(G) ·1- • (for some constant  > 0) • Using Parallel Repetition: 9 Gs.t. • ValQ(G) = 1and Val(G) · • (for any constant  > 0)

  36. 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)

  37. 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, ValQ(G) = Val(G)

  38. Bell Inequalities (EPR Paradox): • 9 Gs.t.ValQ(G) > Val(G) • [CHTW 04]: 9 Gs.t. • ValQ(G) = 1and Val(G) ·1- • (for some constant  > 0) • Using Parallel Repetition: 9 Gs.t. • ValQ(G) = 1and Val(G) · • (for any constant  > 0)

  39. Thank You!

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