Derandomized parallel repetition theorems for free games

1. Derandomized parallel repetition theorems for free games Ronen Shaltiel, University of Haifa

2. 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.

3. 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: 8f If 8poly-size circuit C, on random X, Pr[C(X)=f(X)]≤1-². Then 8poly-size circuit D, on random X1,,XnPr[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,,Xnfrom few random bits by G(X’)=(X1,,Xn). • Prove theorem for fG(x’)=f(n)(G(x’)). + little bit (1-²)n

4. 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”.

5. 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: ¹is the uniform distribution. (Rand(G)=2m).

6. 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.

7. 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,,xnand responds with “answers” a1=a1(x1,,xn),, an=an(x1,,xn)2{0,1}L. • Second player receives y1,,ynand 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.

8. Marketing: In terms of randomness complexity, amplification for free games can be done at the correct rate! *certain restrictions apply. 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)).

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

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

11. High level idea of the proof We observe that a lemma used in [Raz] can be improved using extractors.

12. 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 i2 [n], • (Zi|W) and Ziare 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.

13. Randomness extractors Daddy, how do computers get random bits?

14. Definition of strong extractors A function E:{0,1}r £ [n] ! {0,1}mis a strong (k,²)-extractor if for every distribution X with min-entropy* ≥k, for a random i2 [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

15. 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 ). 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 i2 [n], • (Zi|W) and Ziare 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)=Ziwe 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

16. Generating Z=(Z1,,Zn) using few random bits • Let E:{0,1}r £ [n] ! {0,1}mbe 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).

17. Conclusion 82P1R free game G with Val(G)≤1-²we 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.

18. Thank You