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

Parallel Repetition of

Two Prover Games

Ran RazWeizmann Institute

and IAS

Two Prover Games [BGKW]:Player A gets xPlayer B gets y (x,y) 2R publicly known

distributionPlayer 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) = MaxA,B Prx,y

[V(x,y,a,b)=1]

Example:Player A gets x 2R {1,2}

Player B gets y 2R {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 2R {1,2})

(alternatively : b=y, a 2R {3,4})

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

Probabilistic Protocols:Player A answers a=A(x)Player B answers b=B(y)Where A(x),B(y) may depend

on acommon random string

However:Probabilistic Value = Deterministic Value

Parallel Repetition:A gets x = (x1,..,xn)

B gets y = (y1,..,yn)

(xi,yi) 2R the original distribution

A answers a=(a1,..,an) =A(x)

B answers b=(b1,..,bn) =B(y)

V(x,y,a,b) =1 iff 8i V(xi,yi,ai,bi)=1

Val(Gn) = MaxA,B Prx,y

[V(x,y,a,b)=1]

Parallel Repetition:A gets x = (x1,..,xn)

B gets y = (y1,..,yn)

(xi,yi) 2R the original distribution

A answers a=(a1,..,an) =A(x)

B answers b=(b1,..,bn) =B(y)

V(x,y,a,b) =1 iff 8i 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 ?

Example [Feige]:A gets x1,x2 2R {1,2}

B gets y1,y2 2R {3,4}

A answers a1,a2 2 {1,2,3,4}

B answers b1,b2 2 {1,2,3,4}

They win if 8i ai=bi=xi or 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)

Parallel Repetition Theorem [R-95]:

(Conjectured by Feige and Lovasz)

8G Val(G) < 1 ) 9 w < 1

(s = length of answers in G)

Assume that Val(G) = 1-What can we say about w ?

Parallel Repetition Theorem:

Val(G) = 1- , ( < ½) )

[R-95]:

[Hol-06]:

For projection games:

[Rao-07]:

(s = length of answers in G)

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-,

Over Expander Graphs [RR-10]:

Val(G) = 1- , ( < ½) )

For general games:

For projection games:

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

Parallel Repetition, PCP, and

Hardness of Approximation

[BGS],[Has],[Fei],[Kho],...

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) = 1statement has no proof ) P(V¼ accepts) ≤ 1-ε

Two Query PCP:Any (length n) proof can be

rewrittenas a length poly(n) proof that can

beverified by reading only 2 symbols (O(1) bits)

proof is correct ) P(V accepts) = 1statement has no proof ) P(V¼ accepts) ≤ 1-ε

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)

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 andevery solution satisfies < ½+ε

fraction

PCP Theorem (2 Queries):Original:

Using Parallel Repetition:

proof is correct ) P(V accepts) = 1statement has no proof ) P(V¼ accepts) ≤ 1-ε(for some constant ε > 0)

proof is correct ) P(V accepts) = 1statement has no proof ) P(V¼ accepts) ≤ ε(for any constant ε > 0)

Tiling Rn, Cubical Foams, and

Parallel Repetition of the

Odd Cycle Game

[FKO],[KORW],[AK],[A]...

Cubical Foams:

The unit cube C = tiles Rn bythe lattice Zn. Rn = C £ Zn

What is the minimalsurface area of a(volume 1) object Dthat tiles Rn by Zn ?(Rn = D £ Zn )

Best Cubical Foams:

Minimal surface area of a (volume 1)

object that tiles Rn by Zn :

Upper bound: 2n : unit cube

Lower bound: : volume 1 ball

Best Cubical Foams:


object that tiles Rn by Zn :



[KORW-08]: There is an object with surface area

similarto the (volume 1) ball, that tiles Rn as a

cube

Odd Cycle Game [CHTW,FKO]:A gets x 2R {1,..,m} (m is odd)

B gets y 2R {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

Odd Cycle Game [CHTW,FKO]:A gets x 2R {1,..,m} (m is odd)

B gets y 2R {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 consistently1

0

1 1

0

Protocol for the Odd Cycle Game:

A gets x, B gets y. They win if(x,y) e (e breaks the cycle)

xy

e

0

0

0

01

1

1

1

1

Parallel Repetition of OCG:A gets x1,..,xn 2R {1,..,m}

B gets y1,..,yn 2R {1,..,m}

8 i yi 2R {xi,xi-1,xi+1} (mod m)

A answers a1,..,an 2 {0,1}

B answers b1,..,bn 2 {0,1}

They win if 8 i xi=yi , ai=bi

1 2 3 n


B gets y1,..,yn 2R {1,..,m}





We think of the game as played on

then dimensional torus

1 2 3 n


B gets y1,..,yn 2R {1,..,m}





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

Odd Cycle Game and Cubical Foams:

The players can color consistently

everycell. Err on edges that cross the surface

The Shorter Story:

[FKO-07]: Cubical Foams implyprotocols for OCG

[R-08]: (a counterexample to strong par.

rep.)

[KORW-08]: Cubical Foams withsurface area (based on the protocol for

OCG)

Cubical Foams:


cell that tiles Rn by Zn :



[KORW-08]: There is an object with surface area

similarto the (volume 1) ball, that tiles Rn as a

cube

Parallel Repetition, Bell

Inequalities,

and the EPR Paradox

[CHTW]

Entangled Two Prover Games:A,B share entangled quantum state

|siA,B

A gets x, B gets y A measures A, B measures BA answers a, B answers b They win if V(x,y,a,b)=1ValQ(G) = MaxA,B Prx,y

[V(x,y,a,b)=1]

Bell Inequalities (EPR Paradox):

9 G s.t. ValQ(G) > Val(G)

[CHTW 04]: 9 G s.t. ValQ(G) = 1 and Val(G) · 1-(for some constant > 0)Using Parallel Repetition: 9 G

s.t. ValQ(G) = 1 and Val(G) · (for any constant > 0)

Does God Play Dice ?a,b are the outcome of a

quantummeasurement. 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)

a=a(x,y,H), b=b(x,y,H)Assume: x,y are independentA chooses x, B chooses y (at time

t-²)Measurements occur at time tInformation cannot propagate

fasterthan light! Hence,Local Hidden Variables Theory:a=a(x,H), b=b(y,H) Thus, ValQ(G) = Val(G)

Bell Inequalities (EPR Paradox):

9 G s.t. ValQ(G) > Val(G)

[CHTW 04]: 9 G s.t. ValQ(G) = 1 and Val(G) · 1-(for some constant > 0)Using Parallel Repetition: 9 G

s.t. ValQ(G) = 1 and Val(G) · (for any constant > 0)

Thank You!

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

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