Download presentation

Presentation is loading. Please wait.

Published byEzekiel Parris Modified over 3 years ago

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

5
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

6
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

7
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]

8
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 ?

9
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)

10
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 ?

11
Parallel Repetition Theorem: Val(G) = 1-, ( < ½) ) [R-95]: [Hol-06]: For projection games: [Rao-07]: (s = length of answers in G)

12
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-,

13
Over Expander Graphs [RR-10]: Val(G) = 1-, ( < ½) ) For general games: For projection games:

14
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

15
Parallel Repetition, PCP, and Hardness of Approximation [BGS],[Has],[Fei],[Kho],...

16
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

17
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

18
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

19
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

20
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

21
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)

22
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)

23
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)

24
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

25
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)

26
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)

27
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,…

28
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) ·

29
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]

30
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...

31
Tiling R n, Cubical Foams, and Parallel Repetition of the Odd Cycle Game [FKO],[KORW],[AK],[A]...

32
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 )

33
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

36
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

37
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

38
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

39
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

40
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

41
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

42
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

43
Odd Cycle Game and Cubical Foams: The players can color consistently every cell. Err on edges that cross the surface

44
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)

45
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

47
Thank You!

51
Parallel Repetition, Bell Inequalities, and the EPR Paradox [CHTW]

52
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]

53
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)

54
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)

55
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)

56
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)

57
Thank You!

Similar presentations

OK

Quantum Information and the PCP Theorem Ran Raz Weizmann Institute.

Quantum Information and the PCP Theorem Ran Raz Weizmann Institute.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on earthquake and tsunami Ppt on social etiquettes Chapter 16 dilutive securities and earnings per share ppt online Download ppt on levels of management Ppt on supply chain management of mcdonald Ppt on principles of peace building commission Ppt on biological pest control Ppt on cloud based mobile social tv Ppt on low level language examples Ppt on power grid failure