1 / 17

On Parallel Repetition

Thomas Holenstein Microsoft Research, SVC Dagstuhl 15.9.2008. On Parallel Repetition. Parallel Repetition. Problem Statement and Connection to PCP Consistent Sampling Lemma Raz’s Counterexample [ Raz 08] Parallel Repetition Theorem . Problem Statement and Theorem.

fynn
Download Presentation

On Parallel Repetition

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Thomas Holenstein Microsoft Research, SVC Dagstuhl 15.9.2008 On Parallel Repetition

  2. Parallel Repetition • Problem Statement and Connection to PCP • Consistent Sampling Lemma • Raz’s Counterexample [Raz 08] • Parallel Repetition Theorem

  3. Problem Statement and Theorem i=1,…,n: (xi,yi) ← PXY (x,y) ← PXY x1,…,xn x y1,…,yn y a1,…,an a b1,…,bn b For all i: Q(xi,yi,ai,bi)? Q(x,y,a,b)? Does this decrease the winning probability? Theorem[Raz 95, H 07]:Yes, from 1 - γ to (1 - O(γ3))n/log(|A||B|) Why would I care about this? Why is this not trivial? Theorem[Raz 08]:G with v(G)= 1 – γ, v(Gn)=(1-O(γ2))n Players may not answer independently. (1 – γ)n not true!

  4. The PCP Theorem (strong version) 7 8 It is NP-hard to satisfy –+ of the clauses of a satisfiable 3SAT formula [Håstad 01] Proof sketch: 3-SAT is NP-hard [Cook71, Levin73] Gap AmplificationIt is NP-hard to satisfy 99% of the clauses of a satisfiable 3-SAT formula [BFL91, BFLS91, FGLSS 91, AS92, ALMSS92, Dinur07] Parallel Repetition It is NP-hard to win in a game as before with probability  [Raz98, H] Fourier Techniques It is NP-hard to satisfy 7/8+  of the clauses of a 3-SAT formula

  5. SAT-formulas 1 2 207 921 (x1x2x3)  (x1x5x9)    (x2x6x9)  …  (x16x18x41) Pick a random clause, send clause to Alice, Variable to Bob 9 207 2 0 (1,0,0) Fraction of unsatisfied clauses = O(rejection probability)

  6. Overview • Problem Statement and connection to PCP • Consistent Sampling Lemma • Raz’s Counterexample [Raz 08] • Parallel Repetition Theorem

  7. Consistent Sampling Subtask: Alice gets PS, Bob gets PS’, minimize Pr[SS’] x1,…,xn y1,…,yn 1 4 a1,…,an b1,…,bn 1 ½ 2 3 A s4 5 B s3 6 0 s2 s4 s1 s3

  8. Overview • Problem Statement and connection to PCP • Consistent Sampling Lemma • Raz’s Counterexample [Raz 08] • Parallel Repetition Theorem

  9. A Counterexample to Strong Parallel Repetition [Raz, FOCS 08]  G: v(G) = 1- and v(Gn) = (1-O(2))n v(G) = 1 – O(1/k) v(Gn)  ½ for n = O(k2) Odd cycle game (2k+1): x  {1, ..., 2k+1}, y  {x-1,x,x+1} Q(x,y,a,b)  (x = y) = (a = b)

  10. A Counterexample to Strong Parallel Repetition [Raz, FOCS 08] Start with a single instance (2k+1 = 11): 1 2 3 4 5 6 7 8 9 10 11 Pick an edge with the consistent sampling lemma. x=7 y=8 0 1 0 0 1 0 1 0 1 0 1 Parameters are such that: ||PS - PT|| = O(1/k), Pr[Bad edge] = O(1/k2) ||PSn- PTn|| = O(1) for n = O(k2) QED!

  11. Overview • Problem Statement and connection to PCP • Consistent Sampling Lemma • Raz’s Counterexample [Raz 08] • Parallel Repetition Theorem

  12. Parallel Repetition Theorem i=1,…,n: (xi,yi) ← PXY x1,…,xn y1,…,yn a1,…,an b1,…,bn v(G) = 1 - γ implies v(Gn)  (1-O(γ3))n/log(|A||B|)

  13. Decomposition Goal: show Pr[W1  W2  ... Wn] << v x1,…,xn y1,…,yn Pr[W1  ... Wn] = Pr[W1] Pr[W2|W1] ... Pr[Wn|W1  ... Wn-1] a1,…,an b1,…,bn It is sufficient to show that Pr[Wm+1|W1  ... Wm] ≤ v + ε Also enough: for all i1, …,im there exists a j such that Pr[Wj|Wi1  ... Wim] ≤ v + ε ε =O( log(|A||B|) – log(Pr[Wi1  ... Wim]) mn Increasing n decreases ε!

  14. Proof Given: strategy of Alice and Bob, indices i1,…,imGoal: show that for some j: Pr[Wj|Wi1  ... Wim] ≤ v + ε x1,…,xn y1,…,yn Assume otherwise and get a strategy for the initial game. a1,…,an b1,…,bn x y Can only work if: • xj = x and yj = y x1,…,xn y1,…,yn • (x1,…,xn,y1,…,yn) have correct distribution aj bj a1,…,an b1,…,bn

  15. Embedding x1,…,xn y1,…,yn PXY X Y a1,…,an b1,…,bn Goal: embed into (X,S) PS,T|X=x,Y=y Goal: embed into (Y,T) I know of two sufficient conditions: • S ↔ X ↔ Y ↔ T • S = T and PS|X=x,Y=y ≈ PS|X=x ≈ PS|Y=y

  16. Main Lemma Assume: Pr[W5 | W1 W2] is big x1,…,xn y1,…,yn x y Won games Won games a1,…,an b1,…,bn x1 x2 x3 x4 x x6 x7 x1 x2 x3 x4 x6 y1 y2 y7 y1 y2 y3 y4 y y6 y7 a5 a1 a2 a1 a2 b5 b1 b2 b1 b2 Send the corresponding answer to the referee Apply the given strategy for multiple games Forget part of the generated information Using PS|X=xY=y ≈ PS|X=x ≈ PS|Y=y Using S ↔ X ↔ Y ↔ T a1 a2 a3 a4 a6 a7 b1 b2 b3 b4 b6 b7

  17. Overview v(G) = 1 – γ v(Gn)  (1 – O(γ3))n/log(|A||B|) [Raz 98,H07] v(P) = 1 – γ v(Pn)  (1 – O(γ2))n [Rao 08] G: v(G) = 1 – γ and v(Gn)  (1 – O(γ2))n [Raz 08] ~ G: v(G) = ¾ and v(Gn)  (½)n/O(log(|A| |B|)) [FV 02] • Connection with Foams: [FKO 07, KORW 08] Parallel Repetition of Unique Games: [BHHRRS ’08] Thank you!

More Related