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: 8f
If 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)]≤(1-²)n + little bit
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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’)).
Outline for this talk
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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
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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
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2P1R games capture the interaction between
an honest verifier and cheating provers in a 2prover 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
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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:
n
 Rand(G ) = n ¢ Rand(G).
n
 If G is free then G is free.
Marketing: In terms of randomness complexity,
Parallel
repetition
theorem
amplification for free games can be done at the
[Raz,Hol] correct rate!
*certain restrictions apply.
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].
C
 n=O(t¢L/² ) 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
C
randomness complexity by n=O(t¢L/² ).
 This work: derandomized parallel repetition for free games.
Multiplies the randomness complexity by O(t) (in case
L=O(m)).
Our results
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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:
r
 Referee chooses x’,y’ uniformly from {0,1} .
 First player receives x’ and 8i2[n], sets xi=E(x’,i).
 Second player receives y’ and 8i2[n], sets yi=E(y’,i).
n
 The players play G on x1,, xn and y1,,yn.
If E is a strong extractor (with suitable parameters) then
E
t
 Val(G ) ≤ (1-²) .
E
 Rand(G ) = O(t(m+L)) .
E
 For L=O(m), Rand(G )=O(t) ¢ Rand(G).
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 n=O(t(m+L)/² ), (no cheats with # of repetitions).
Perspective (and disclaimers)
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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.
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 Let W be an event such that Pr[Z 2 W] ≥ 2 .
2
 Assume that n ≥ a/² .
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
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“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)=Z
Using better extractors we can generate Z=(Z1,,Zn)
with similar properties from
r = O(m + a + log(1/²)) random bits
i
Rather than r = O(m ¢ a /²2 ).
 Let Z=(Z1,,Zn) be i.i.d. random variables where each Zi is
uniform over {0,1}m.
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 Let W be an event such that Pr[Z 2 W] ≥ 2 .
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 Assume that n ≥ a/² . ⇒ entropy: n¢m-a >> m, output: m
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!
Generating Z=(Z1,,Zn) using
few random bits
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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:
-a
 Let W be an event such that Pr[Z 2 W] ≥ 2 .
 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).
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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).
Conclusion
82P1R 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).
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[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