Derandomized parallel repetition theorems for free

games

Ronen Shaltiel, University of Haifa

Parallel repetition/direct product

To what extent is it harder to solve many independent instances of the same problem compared to solving a single random instance?

Asked in many computational models.

This talk: 2-prover 1-round games.

Example: the setting of polynomial size circuits [GILRZ,Imp,IW,IJKW]

For function f and integer n define: “parallel repetition of f” by f(n)(x1,,xn) = (f(x1),,f(xn)).

Parallel repetition/direct product theorem: 8fIf 8poly-size circuit C, on random X, Pr[C(X)=f(X)]≤1-². Then 8poly-size circuit D, on random X1,,Xn Pr[D(X1,,Xn)=f(n)(X1,,Xn)]≤

Application: Hardness amplification: “f mildly hard” ⇒ “f(n) very hard”.

Weakness: input length blows up by a factor of n. Derandomized parallel repetition: Generate

(correlated) X1,,Xn from few random bits by G(X’)=(X1,,Xn).

Prove theorem for fG(x’)=f(n)(G(x’)).

(1-²)n +little bit

Outline for this talk Starting point: There is a parallel

repetition theorem for 2P1R games [Raz].

Goal: derandomized version. Our results: a derandomized version for

the subfamily of “free games”.

2P1R Games A game G between two cooperating players. Referee samples x,y2{0,1}m according to a known

distribution ¹ on pairs. First player receives “input” x and responds with

“answer” a=a(x)2{0,1}L. Second player receives “input” y and responds with

“answer” b=b(y)2{0,1}L. No communication between players. A strategy is a pair of functions (a(¢),b(¢)). Players win if they satisfy a known predicate

V(x,y,a,b).

Val(G) = success probability in best strategy. Rand(G) = number of random bits tossed by referee. Free game: ¹ is the uniform distribution.

(Rand(G)=2m).

Background and disclaimer

2P1R games capture the interaction between an honest verifier and cheating provers in a 2-prover 1-round multi-prover system.

Important for PCP, Hardness of approximation.

Important Disclaimer: Results in this talk are only for free games. The games that come up in PCP are not free.

Parallel repetition of 2P1R Games

For a game G we define the parallel repetition game Gn.

8i 2 [n], referee independently samples (xi,yi) according to the distribution ¹ of the initial game G.

First player receives x1,,xn and responds with “answers” a1=a1(x1,,xn),, an=an(x1,,xn)2{0,1}L.

Second player receives y1,,yn and responds with “answers” b1=b1 (y1,,yn),, bn=bn(y1,,yn)2{0,1}L.

Players win if they win all n games.

Observations: Rand(Gn) = n ¢ Rand(G). If G is free then Gn is free.

Parallel repetition theorem [Raz,Hol]

Let G be a game with Val(G)≤1-², for ²≤½. How large should n be so that Val(Gn)≤(1-²)t ? Naïve guess: n=t suffice. Wrong even for free games [For,Fei]. No function n(t,²) will do (even for free games) [FV]. n=O(t¢L/²C) repetitions suffice for every game [Raz]. Dependence on L is optimal up to log factors [FV]. Dependence on ²: C=2 suffices [Hol] , C=1 suffices

for free games [BRRRS], C=1 necessary for general games [Raz].

Amplifcation from 1-² to (1-²)t currently requires multiplying randomness complexity by n=O(t¢L/²C).

This work: derandomized parallel repetition for free games. Multiplies the randomness complexity by O(t) (in case L=O(m)).

Marketing: In terms of randomness complexity, amplification for free games

can be done at the correct rate!*certain restrictions apply.

Our results Let G be a free game with Val(G)≤1-², for ²≤½. Let E:{0,1}r £ [n] ! {0,1}m be a function.

Define the (derandomized) game GE as follows: Referee chooses x’,y’ uniformly from {0,1}r. First player receives x’ and 8i2[n], sets xi=E(x’,i). Second player receives y’ and 8i2[n], sets yi=E(y’,i). The players play Gn on x1,, xn and y1,,yn.

If E is a strong extractor (with suitable parameters) then

Val(GE) ≤ (1-²)t . Rand(GE) = O(t(m+L)) . For L=O(m), Rand(GE)=O(t) ¢ Rand(G). n=O(t(m+L)/²2), (no cheats with # of repetitions).

Perspective (and disclaimers)

We get “correct rate”: Rand(GE)=O(t) ¢ Rand(G). In other setups (e.g. poly-size circuits)

derandomization beats the correct rate [GILRZ,Imp,IW,IJKW].

We could hope for Rand(GE)=O(t) + Rand(G). [FK] rule out such derandomization (or even

beating the correct rate) for general games. It is open whether one can achieve

Rand(GE)=O(t) ¢ Rand(G) for general games. Example of [FK] is for “constant degree” games. It is open whether one can beat the correct rate

for free games.

High level idea of the proof

We observe that a lemma used in [Raz] can be improved using extractors.

Lemma from Raz’s parallel repetition theorem

Let Z=(Z1,,Zn) be i.i.d. random variables where each Zi is uniform over {0,1}m.

Let W be an event such that Pr[Z 2 W] ≥ 2-a. Assume that n ≥ a/²2. Then for a uniformly chosen i 2 [n], (Zi|W) and Zi are ²–close in statistical distance. More formally ExpiÃ[n][DIST( (Zi|W) ; Zi )] ≤ ²

“Let Z be the uniform on r=n¢m bits and assume that a bits of information about Z are revealed. Then for a random i, (Zi|W) is (close to) uniform.”

Useful in other settings.

Randomness extractors

Daddy, how do

computers get random

bits?

Definition of strong extractors

A function E:{0,1}r £ [n] ! {0,1}m is a strong (k,²)-extractor if for every distribution X with min-entropy* ≥k, for a random i 2 [n], (i,E(X,i)) is ²–close to uniform.

Equivalently, ExpiÃ[n][DIST( E(X,i) ; Um )] ≤ ²

* Dfn: X has min-entropy ≥k if for every x 2 {0,1}r, Pr[X=x] ≤ 2-k

Raz’s lemma is an extractor construction by E(Z,i)=Zi

Let Z=(Z1,,Zn) be i.i.d. random variables where each Zi is uniform over {0,1}m.

Let W be an event such that Pr[Z 2 W] ≥ 2-a. Assume that n ≥ a/²2. Then for a uniformly chosen i 2 [n], (Zi|W) and Zi are ²–close in statistical distance. More formally ExpiÃ[n][DIST( (Zi|W) ; Zi )] ≤ ²

Interpretation: Z is the uniform on r=n¢m bits. The distribution X=(Z|W) has min-entropy ≥ r-a = n¢m-a. For E(Z,i)=Zi we have that for random i, E(X,i) is ¼

uniform. Lemma ⇒ function E is a strong (r-a,²)-extractor.

This is not a good extractor in terms of “entropy loss”!Main idea: Replace E with a better extractor!

⇒entropy: n¢m-a >> m, output: m

Using better extractors we can generate Z=(Z1,,Zn) with similar properties from

r = O(m + a + log(1/²)) random bits Rather than r = O(m ¢ a /²2 ).

Generating Z=(Z1,,Zn) using few random bits

Let E:{0,1}r £ [n] ! {0,1}m be a strong (r-a,²)-extractor. Exists for r=m+a+O(log(1/²)) << m¢a/²2. Choose a uniform Z’ 2 {0,1}r. Define Z=(Z1,,Zn) by Zi=E(Z’,i).

This gives the behavior of the lemma, specifically: Let W be an event such that Pr[Z 2 W] ≥ 2-a. Then ExpiÃ[n][DIST( (Zi|W) ; Zi )] ≤ ².

Suffices to adapt Raz’s proof (for free games). In the proof the lemma is applied with a=O((m+L)t).

Sample space Z’ ! (Z1,,Zn) is also an “averaging sampler” [Zuc]. Necessary for derandomization.

The use of this sample space here seems different (and may help in other settings).

Conclusion82P1R free game G with Val(G)≤1-² we define a

derandomized GE with: Val(GE) ≤ (1-²)t . For L=O(m), Rand(GE)=O(t) ¢ Rand(G).

[PRW]: Parallel repetition theorem for communication games“.

In the paper: derandomized version for free games.

Open problem: Show derandomized parallel repetition theorems for general 2P1R games.

The extractor approach makes sense for general games.

Analysis may require additional properties of the extractor.

Thank You