Transcript A Counterexample to Strong Parallel Repetition Ran Raz Weizmann Institute Two Prover Games: Player A gets x Player B gets y (x,y) 2R publicly known distribution Player A.

A Counterexample to
Strong Parallel Repetition
Ran Raz
Weizmann Institute
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 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) = 1
8 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) · 1-2/p
Using Strong Parallel Repetition:
UGC , 8  > 0, given a graph G,
It is NP hard to distinguish between :
Max-Cut(G) ¸ 1-2
Max-Cut(G) · 1-2/p
and
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
1
0
0
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
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
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 can
agree on an edge e that doesn’t
touch x,y, they win !
0
1
1
e
0
1
x
y
0
1
0
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 !
0
1
1
1
e1
0
1
x1
y1
0
1
0
1
0
1
en
0
1
xn
yn
0
1
0
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)
-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)
-k
k
1/m3 1/m3
1/m
0
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
fx ¼ gy (since x,y are adjacent)
|fx-gy|1 · O(1/m)
gy(e)=P(0)
x
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 on
edges e1,..,en that don’t touch x1,..,xn,
y1,..,yn, with probability ¸
0
1
e1
0
1
x1
y1
0
1
0
1
0
1
1
1
en
0
1
xn
yn
0
1
0
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