A Counterexample to

Strong Parallel Repetition

Ran RazWeizmann Institute

Two Prover Games: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})

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: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 [R95]:

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 unique and 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)

Our Result: G s.t.: Val(G) = 1-,

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],...

EPR Paradox: 9 G s.t. ValQ(G) > Val(G)

ValQ(G) = value when the provers share entangled quantum states[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)

PCP Theorem [BFL,FGLSS,AS,ALMSS]:

Given 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)

Unique Games (UG):G is a UG if V(x,y,a,b) satisfies :8 x,y,a 9 unique b, V(x,y,a,b) = 18 x,y,b 9 unique a, V(x,y,a,b) = 1

Unique Games Conjecture [Khot]:8 constant > 0, 9 constant s, s.t.Given a UG G (with answer size s)It is NP hard to distinguish between :

Val(G) ¸ 1- and Val(G) ·

UGC and Max-Cut [KKMO]:UGC ) 8 > 0, given a graph G, It is NP hard to distinguish between :Max-Cut(G) ¸ 1-2 and Max-Cut(G) · 1-2Using Strong Parallel

Repetition:UGC , 8 > 0, given a graph G,It is NP hard to distinguish between :Max-Cut(G) ¸ 1-2 and Max-Cut(G) · 1-2

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 , ab

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 , ab

1

0

1 1

0

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 , aibi

1 2 3 n

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 , aibi

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

space

1 2 3 n

Our Results:

(match an upper bound of [FKO])

For n ¼ m2,

For n ¸ (m2),

1 2 3 n

Odd Cycle Game:A gets x, B gets y. If they canagree on an edge e that

doesn’t touch x,y, they win !

xy

e

0

0

0

01

1

1

1

1

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 !

x1y1

e1

0

0

0

01

1

1

1

1

x1y1

e1

0

0

0

01

1

1

1

1

x1y1

e1

x1y1

e1

0

0

0

01

1

1

1

1

xnyn

en

0

0

0

01

1

1

1

1

xnyn

en

0

0

0

01

1

1

1

1

xnyn

en

xnyn

en

0

0

0

01

1

1

1

1

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()

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)

0

-k k

1/m

1/m3 1/m3

Distribution P:m=2k+1, P:[-k,k] ! R (symmetric) :

1) P(0) = £(1/m)2) P(k) = P(-k) = £(1/m3)

(negligible)

3)

0

-k k

1/m

1/m3 1/m3

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 xGiven y, B defines gy: E ! R : gy=P,concentrated on the edge opposite

to yfx ¼ gy (since x,y are adjacent)

|fx-gy|1 · O(1/m)

x

gy(e)=P(0)

fx(e)=P(0)y

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)

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

onedges e1,..,en that don’t touch x1,..,xn,y1,..,yn, with probability ¸

x1y1

e1

0

0

0

01

1

1

1

1

x1y1

e1

0

0

0

01

1

1

1

1

x1y1

e1

x1y1

e1

0

0

0

01

1

1

1

1

xnyn

en

0

0

0

01

1

1

1

1

xnyn

en

0

0

0

01

1

1

1

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xnyn

en

xnyn

en

0

0

0

01

1

1

1

1

Proof Idea:

Typically: in coordinates

and in

coordinates

n/3 coordinates cancel each other. We

are left with distance

Proof Idea:

Hence, typically:

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)

The End