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A Counterexample to Strong Parallel Repetition

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

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A Counterexample to Strong Parallel Repetition

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  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) 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. 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]

  5. 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?

  6. Example: • 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)

  7. Parallel Repetition Theorem [R95]: • 8G Val(G) < 1 ) 9 w < 1 • (s= length of answers inG) • Assume thatVal(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 inG)

  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 Rn[FKO] • 3) Quantum Computation: strong EPR paradoxes [CHTW] • 4) Hardness of Approximation: [BGS],[Has],[Fei],[Kho],...

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

  12. PCP Theorem [BFL,FGLSS,AS,ALMSS]: • GivenG(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,a9uniqueb, V(x,y,a,b) = 1 • 8 x,y,b9uniquea, V(x,y,a,b) = 1 • Unique Games Conjecture [Khot]: • 8 constant > 0, 9 constant s, s.t. • Given a UGG(with answer sizes) • It is NP hard to distinguish between : • Val(G) ¸ 1-and Val(G) ·

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

  15. 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

  16. 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 1 0 0 1 1

  17. 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

  18. 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 • Motivation [FKO]:Max-Cut vs. UGC, • Understanding foams, Tiling the space

  19. n 2 3 1 • Our Results: • (match an upper bound of [FKO]) • For n ¼ m2, • For n ¸(m2),

  20. e x y • 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 ! 0 1 1 0 1 0 1 0 1

  21. Parallel Repetition of OCG: • A gets x1,..,xn, B gets y1,..,yn. • If they can agree on edges e1,..,en • that don’t touch x1,..,xn, y1,..,yn, • 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 2f W, and B: v 2g W, • s.t. Prob[u=v] ¸ 1-O()

  23. -k k 1/m3 1/m3 1/m 0 • Distribution P: • m=2k+1, P:[-k,k] ! R (symmetric) : • 1) P(i) ¼ (k+1-|i|)2 / m3 • 2) P(0) = £(1/m) • 3) P(k) = P(-k) = £(1/m3) (negligible)

  24. -k k 1/m3 1/m3 1/m 0 • Distribution P: • m=2k+1, P:[-k,k] ! R (symmetric) : • 1) P(0) = £(1/m) • 2) P(k) = P(-k) = £(1/m3) (negligible) • 3)

  25. gy(e)=P(0) x fx(e)=P(0) y • Distributions on the Odd Cycle: • E = Edges of the odd cycle. • Given x, A defines fx: E ! R : fx=P, • concentrated on the edge opposite to x • Given y, B defines gy: E ! R : gy=P, • concentrated on the edge opposite to y • fx¼ gy (since x,y are adjacent) • |fx-gy|1· O(1/m)

  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 2f W, and B: v 2g W, • s.t. Prob[u=v] ¸ 1-O() • OCG: Using fx,gy, 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=(x1,..,xn), A defines fx: En! R, • Given y=(y1,..,yn), B defines gy: En! R, • Lemma: • Using Holenstein’s lemma, A,B agree on • edges e1,..,en that don’t touch x1,..,xn, • y1,..,yn, 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 Rn[KORW,AK] : • Rn can be tiled (with translations in Zn), • by objects with surface area similar to • the one of the sphere (with volume 1)

  31. The End

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