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Direct-product testing, and a new 2-query PCP Russell Impagliazzo (IAS & UCSD) Valentine Kabanets (SFU) Avi Wigderson (IAS)

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Presentation on theme: "Direct-product testing, and a new 2-query PCP Russell Impagliazzo (IAS & UCSD) Valentine Kabanets (SFU) Avi Wigderson (IAS)"— Presentation transcript:

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2 Direct-product testing, and a new 2-query PCP Russell Impagliazzo (IAS & UCSD) Valentine Kabanets (SFU) Avi Wigderson (IAS)

3 Direct Product: Definition  For f : U  R, the k -wise direct product f k : U k  R k is f k (x 1,…, x k ) = ( f(x 1 ), …, f(x k ) ). [Impagliazzo’02, Trevisan’03]: DP Code TT ( f k ) is DP Encoding of TT ( f ) Rate and distance of DP Code are “bad”, but the code is still very useful in Complexity …

4 DP Code: Two Basic Questions - Decoding: Given C ¼ f k, “find” f. (useful for Hardness Amplification) - Testing: Given C, test if C ¼ f k. (useful for PCP constructions) C is given as oracle Decoding vs. Testing Promise on C no promise Search problem decision problem Small # queries Minimal # queries

5 Decoding: Hardness Amplification f k is harder to compute on average than f Motivation: Cryptography Pseudorandomness, Computational Complexity, PCPs DP Theorem/ XOR Lemma: [Yao82, Levin87, GL89, I94, GNW95, IW97, T03, IJK06, IJKW08] If C computes f k on ² of all (x 1,…, x k )  U k Then C’ computes f on 1-δ of all x  U ² = exp(-δk)

6 Direct-Product Testing  Given an oracle C : U k  R k Test makes some queries to C, and (1) Accept if C = f k. (2) Reject if C is “far away” from any f k (2’) If Test accepts C with “high” probability ², then C must be “close” to some f k. - Want to minimize number of queries to C. - Want to minimize acceptance probability ²

7 DP Testing History  Given an oracle C : U k  R k, is C ¼ g k ? #queries acc. prob. Goldreich-Safra 00* 20.99 Dinur-Reingold 06 2.99 Dinur-Goldenberg 08 2 1/k α Dinur-Goldenberg 08 2 1/k New 3 exp(-k α ) New* 2 1/k α * Derandomization /

8 Consistency tests Test: Query C(S 1 ), C(S 2 ), … check consistency on common values. Thm: If Test accepts oracle C with prob ² then there is a function g: U  R such that for ≈ ² of k-tuples S, C (S) ¼ g k (S) [C(S) = g k (S) in all but 1/k  elements in S] Proof: g(x) = Plurality { C (S) x | x 2 S} g(x) = Plurality { C (S) x | x 2 S & C(S) A =a } Unique DecodingList Decoding

9 Consistency tests

10 V-Test [GS00,FK00,DR06,DG08] Pick two random k-sets S 1 = (B 1,A), S 2 = (A,B 2 ) with m = k 1/2 common elements A. Check if C(S 1 ) A = C(S 2 ) A B1B1 B2B2 A Theorem [DG08]: If V-Test accepts with probability ² > 1/k , then there is g : U  R s.t. C ¼ g k on at least ² fraction of k-sets. When ² < 1/k, the V-Test does not work. S1S1 S2S2

11 Z-Test Pick three random k-sets S 1 =(B 1, A 1 ), S 2 =(A 1,B 2 ), S 3 =(B 2, A 2 ) with |A 1 | = |A 2 | = m = k 1/2. Check if C(S 1 ) A 1 = C(S 2 ) A 1 and C(S 2 ) B 2 = C(S 3 ) B 2 Theorem (main result): If Z-Test accepts with probability ² > exp(-k  ), then there is g : U  R s.t. C ¼ g k on at least ² fraction of k-sets. B1B1 B2B2 A1A1 A2A2 S1S1 S2S2 S3S3

12 Proof Ideas

13 Flowers, cores, petals Flower: determined by S=(A,B) Core: A Core values: α =C(A,B) A Petals: Cons A, B = { (A,B’) | C(A,B’) A = α } In a flower, all petals agree on core values! [IJKW08]:Flower analysis B B4B4 AA B2B2 B3B3 B1B1 B5B5

14 V-Test ) Structure (similar to [FK, DG]) Suppose V-Test accepts with probability ². Cons A, B = { (A,B’) | C(A,B’) A = C(A,B) A } (1) Largeness: Many ( ² /2) flowers (A,B) have many ( ² /2) petals Cons A, B (2) Harmony: In every large flower, almost all pairs of overlapping sets in Cons are almost perfectly consistent. B B4B4 AA B2B2 B3B3 B1B1 B5B5

15 V-Test: Harmony For random B 1 = (E,D 1 ) and B 2 = (E,D 2 ) (|E|=|A|) Pr [B 1 2 Cons & B 2 2 Cons & C(A, B 1 ) E  C(A, B 2 ) E ] < ² 4 << ² B D2D2 D1D1 A E Proof: Symmetry between A and E (few errors in AuE ) Chernoff: ² ¼ exp(-k α ) E A Implication: Restricted to Cons, an approx V-Test on E accepts almost surely: Unique Decode!

16 Harmony ) Local DP Main Lemma: Assume (A,B) is harmonious. Define g(x) = Plurality { C(A,B’) x | B’ 2 Cons & x 2 B’ } Then C(A,B’) B’ ¼ g k (B’), for almost all B’ 2 Cons B AA D2D2 D1D1 E Intuition: g = g (A,B) is the unique (approximate) decoding of C on Cons (A,B) B’ x Idea: Symmetry arguments. Largness guarantees that random selections are near-uniform.

17 Proof Sketch Main Lemma: Assume (A,B) is harmonious. Define g(x) = Plurality { C(A,B’) x | B’ 2 Cons & x 2 B’ } Then C(A,B’) B’ ¼ g k (B’), for almost all B’ 2 Cons Proof: Assume otherwise. A random B 1 in Cons has many “minority” elements x where C(B 1 ) x  g(x). A random E ½ B 1 has many “minority” elements [Chernoff] A random B 2 =(E,D 2 ) is likely s.t. C(B 2 ) E ¼ g(E) [def of g] Then C(B 1 ) E  C(B 2 ) E, Hence no harmony ! B A D2D2 D1D1 E

18 Local DP structure Field of flowers (A i,B i ) For each, g i s.t C(S) ¼ g i k (S) if S 2 Cons (Ai,Bi) Global g? B2B2 AA BiBi AA B AA B3B3 AA B1B1 AA

19 Counterexample [DG] For every x 2 U pick a random g x : U  R For every k-subset S pick a random x(S) 2 S Define C(S) = g x(S) (S) C(S 1 ) A =C(S 2 ) A “iff” x(S 1 )=x(S 2 ) V-test passes with high prob: ² = Pr[C(S 1 ) A =C(S 2 ) A ] ~ m/k 2 No global g if ² < 1/k 2 B1B1 B2B2 A S1S1 S2S2

20 From local DP to global DP How to “glue” local solutions? ² > 1/k α “double excellence” (2 queries) [DG] ² > exp(-k α ) Z-test (3 queries)

21 Local to Global DP: small ² Lemma: (A 1,B 1 ) random (Cons large w.p. ² / 2 ). Define g(x) = Plurality { C (A 1,B’) x | B’ 2 Cons & x 2 B’ } (local) Then C(S) ¼ g k (S), for ¼ ² /4 of all S (global) B1B1 B2B2 A1A1 A2A2 B1B1 B2B2 A1A1 A2A2 B1B1 B2B2 A1A1 A2A2

22 Local to Global DP: Z-test Proof: Cons = Cons A1,B1. Define g(x) = Plurality { C(A 1,B’) x | B’ 2 Cons & x 2 B’ } Harmony implies C(A 1,B’) B’ ¼ g k (B’), for almost all B’ 2 Cons B1B1 B2B2 A1A1 A2A2 Can assume Flower (A 1, B 1 ) is large, (otherwise V-Test rejects) So (A 1, B 1 ) harmonious  have g. Pick random S=(B 2, A 2 ). May assume B 2 in Cons (otherwise V-Test rejects) If g(S) very different from C(S), then g(B 2 )  C(S ) B2 But g(B 2 ) ¼ C(A 1,B 2 ) B2 Z-Test rejects ( S

23 Local to Global DP: large ² “double harmony” B1B1 A1A1 A2A2 B2B2 S Three events all happen with probability > poly(m/k) (1) (A 1, B 1 ) is harmonious,  g 1 (2) (A 2, B 2 ) is harmonious,  g 2 (3) S is consistent with both Get that g 1 (x) = g 2 (x) for most x 2 U.

24 Derandomization

25 Inclusion graphs are Samplers Most lemmas analyze sampling properties m-subsets A  Subsets: Chernoff bounds – exponential error Subspaces: Chebychev bounds – polynomial error Cons S k-subsets x elements of U

26 Derandomized DP Test Derandomized DP: f k (S), for linear subspaces S (similar to [IJKW08] ). Theorem (Derandomized V-Test): If derandomized V-Test accepts C with probability ² > poly(1/k), then there is a function g : U  R such that C (S) ¼ g k (S) on poly( ² ) of subspaces S. Corollary: Polynomial rate testable DP-code with [DG] parameters!

27 Application: PCPs

28 Constraint Satisfaction Problem A graph CSP over alphabet § : Given a graph G=(V,E) on n nodes, and edge constraints Á e : § 2  {0,1} ( e 2 E ), is there an assignment f: V  § that satisfies all edge constraints. Example: 3-Colorability ( § = {1,2,3}, Á e (a,b) = 1 iff a  b )

29 PCP Theorem [AS,ALMSS] For some constant 0< ± <1 and constant-size alphabet §, it is NP-hard to distinguish between satisfiable graph CSPs over §, and ± -unsatisfiable ones (where every assignment violates at least ± fraction of edge constraints). 2-query PCP ( with completeness 1, soundness 1- ± ) : PCP proof = assignment f: V  §, Verifier: Accept if f satisfies a random edge Q1Q1 Q2Q2

30 Decreasing soundness by repetition  sequential repetition : proof f: V  § soundness : 1- ±  (1- ± ) k X # queries: 2k  parallel repetition : proof F: V k  § k # queries : 2 X soundness: ? Q1Q1 Q2Q2 Q3Q3 Q4Q4 Q 2k-1 Q 2k Q1Q1 Q2Q2

31 PCP Amplification History  f: V  Σ, F : V k  Σ k |V|=N, t= log |Σ| size #queries soundness Sequential repetition N 2k exp( - ± k ) Verbitsky N k 2 very-slow(k)  0 Raz N k 2 exp( - ± 32 k/ t) Holenstein N k 2 exp( - ± 3 k/ t) Feige-Verbitsky N k 2 t essential Rao N k 2 exp( - ± 2 k ) Raz N k 2 ± 2 essential Feige-Kilian N k 2 1/k α New N k 2 exp ( - ± k 1/2 ) Moshkovitz-Raz N 1+o(1) 2 1/loglog N Parallel repetition Projection games Mix N’ Match

32 Ideas: DP-Test of the PCP proof Given F : V k  § k, test if F = f k for some f: V  § and test random constraints! If F close to f k, we get exponential decay (as sequential-repetition) in soundness ! Combine tests to minimize # of queries. Replace Z-test by V-test (local DP suffices)

33 A New 2-Query PCP (similar to [FK])  For a regular CSP graph G = (V, E), the PCP proof is C E : E k  ( § 2 ) k Accept if (1) C E (Q 1 ) and C E (Q 2 ) agree on common vertices, and (2) all edge constraints are satisfied Q1Q1 Q2Q2

34 The 2-query PCP amplification Theorem: If CSP G=(V,E) is satisfiable, there is a proof C E that is accepted with probability 1. If CSP is ± – unsatisfiable, then no C E is accepted with probability > exp ( - ± k 1/2 ). Corollary: A 2-query PCP over § k, of size n k, perfect completeness, and soundness exp(- k 1/2 ). Q1Q1 Q2Q2

35 Analysis of our PCP construction

36 PCP Analysis From C E : E k  § 2 ) k to the vertex proof C : V k  § k : C(v 1,…, v k ) = C E ( e 1,…, e k ) for random incident edges Consistency of C E, Consistency of C Main Lemma for C yields local DP function g : V  § Back to C E : g is also local DP for C E (symmetry) g (Q 2 ) ¼ C E (Q 2 ) (since Q 2 2 Cons Q1 ) g(Q 2 ) violates > ± edges (by soundness of G & Chernoff) Hence, C E (Q 2 ) violates some edges, and Test rejects Q1Q1 Q2Q2

37 Summary Direct Product Testing: 3 queries & exponentially small acceptance probability Derandomized DP Testing: 2 queries & polynomially small acceptance probability ( derandomized V-Test of [DG08] ) PCP: 2-Prover parallel k-repetition for restricted games, with exponential in k 1/2 decrease in soundness

38 Open Questions Better dependence on k in our Parallel Repetition Theorem : exp ( - ± k) ? Derandomized 2-Query PCP : Obtaining / improving [Moshkovitz-Raz’08, Dinur-Harsha’09] via DP-testing ?


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