# Direct-Product testing Parallel Repetitions And Foams Avi Wigderson IAS.

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Direct-Product testing Parallel Repetitions And Foams Avi Wigderson IAS

Parallel Repetition of Games and Periodic Foams

Isoperimetric problem: Minimize surface area given volume. One bubble. Best solution: Sphere

Many bubbles Isoperimetric problem: Minimize surface area given volume. Why? Physics, Chemistry, Engineering, Math… Best solution?: Consider R 3 Lord Kelvin 1873 Optimal… Wearie-Phelan 1994 Even better

Our Problem Minimum surface area of body tiling R d with period Z d ? Volume=1 d=2 area: 4>4Choe89: Optimal!

Bounds in d dimensions OPT [Kindler,ODonnell, Rao,Wigderson] Rao,Wigderson] OPT Spherical Cubes exist! Probabilistic construction! (simpler analysis [Alon-Klartag]) OPEN: Explicit?

Randomized Rounding Round points in R d to points in Z d such that for every x,y 1. 2. x y 1 Bound does not depend on d

Spine Torus Surface blocking all cycles that wrap around

Probabilistic construction of spine Step 1 Randomly construct B in [0,1) d, which in expectation satisfies B Step 2 Sample independent translations of B until [0,1) d is covered, adding new boundaries to spine.

PCPs & Linear equations over GF(2) m linear equations: Az = b in n variables: z 1,z 2,…,z n Given (A,b) 1) Does there exist z satisfying all m equations? Easy – Gaussian elimination 2) Does there exist z satisfying.9m equations? NP-hard 3) Does there exist z satisfying.5m equations? Easy – YES! [Hastad] δ >0, it is NP-hard to distinguish (A,b) which are not (½+ δ )-satisfiable, from those (1- δ )-satisfiable!

Linear equations as Games 2n variables: X 1,X 2,…,X n, Y 1,Y 2,…,Y n m linear equations: Xi 1 + Yi 1 = b 1 Xi 2 + Yi 2 = b 2 ….. Xi m + Yi m = b m Promise: no setting of the X i,Y i satisfy more than (1- δ )m of all equations Game G Draw j [m] at random Xi j Yi j Alice Bob α j β j Check if α j + β j = b j Pr [YES] 1- δ

Hardness amplification by parallel repetition 2n variables: X 1,X 2,…,X n, Y 1,Y 2,…,Y n m linear equations: Xi 1 + Yi 1 = b 1 Xi 2 + Yi 2 = b 2 ….. Xi m + Yi m = b m Promise: no setting of the X i,Y i satisfy more than (1- δ )m of all equations Game G k Draw j 1,j 2,…j k [m] at random Xi j1 Xi j2 Xi jk Yi j1 Yi j2 Yi jk Alice Bob α j1 α j2 α jk β j1 β j2 β jk Check if α jt + β jt = b jt t [k] Pr[YES] (1- δ 2 ) k [Raz,Holenstein,Rao] Pr[YES] (1- δ 2 ) k [Feige-Kindler-ODonnell] Spherical Cubes [Raz] X [KORW]Spherical Cubes

Hardness amplification by other means? Xi 1 + Yi 1 = b 1 Xi 2 + Yi 2 = b 2 ….. Xi m + Yi m = b m Promise: no setting of the X i,Y i satisfy more than (1- δ )m of all equations Amplification Xi j1 … Xi jk Yi j1 … Yi jk Alice Bob α j1 … α jk β j1 … β jk Test: α jt + β jt = b jt t ? Pr[YES] (1- δ 2 ) k [Raz,Holenstein,Rao] Pr[YES] (1- δ 2 ) k [Raz] Major open question: Is there Test s.t. Pr[YES] (1- δ ) k ? [Khot] Unique games conjecture Idea: force each player to answer consistently - e.g. make Alice commit to one assignment of Xis [ Impagliazzo-Kabanets-W ] New Test with Pr[YES] (1- δ )k

Direct-product testing Part of - local testing of codes - property testing - discrete rigidity / stability Related to - local decoding of codes - Yaos XOR lemma

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 ) ). [Impagliazzo02, Trevisan03]: 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 …

Direct-Product Testing Given an oracle C : U k R k Test makes few queries to C, and (1) Accept if C = f k. (2) Reject if C is far away from any f k (2) [Inverse Thm] Pr [ Test accepts C ] > C f k on > ( ) of inputs. - Minimize # queries e.g. 2, 3,.. ? - Analyze small e.g. < 1/k, < exp(-k) ? - Reduce rate/Derandomize e.g.|C| = poly (|U|) ?

DP Testing History Given an oracle C : U k R k, is C ¼ g k ? #queries acc prob Goldreich-Safra00* 20.99 Dinur-Reingold06 2.99 Dinur-Goldenberg08 2 1/k α Dinur-Goldenberg08 2 1/k Impagliazzo-Kabanets-W08 3 exp(-k α ) Impagliazzo-Kabanets-W08* 2 1/k α / *Derandomization

Consistency tests

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 [DG08]: If V-Test accepts with probability ² > 1/k α, then there is g : U R s.t. C ¼ g k on at least ² / 4 fraction of k-sets. [IKW09]: Derandomize [DG08]: V-Test fails for ² <1/k S1S1 S2S2

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 [IKW09]: If Z-Test accepts with probability ² > exp(-k α ), then there is g : U R s.t. C ¼ g k on at least ² / 4 fraction of k-sets. B1B1 B2B2 A1A1 A2A2 S1S1 S2S2 S3S3

Proof Ideas

Proof steps 1. Pr [ Test accepts C ] > structure 2. Structure local agreement 3. local agreement global agreement Agreement: there is g : U R s.t. C ¼ g k on at least ² / 4 fraction of k-sets.

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 for DP-decoding. Symmetry arguments! B B4B4 AA B2B2 B3B3 B1B1 B5B5

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

V-Test: Harmony Almost all B 1 = (E,D 1 ) and B 2 = (E,D 2 ) in Cons (with |E|=|A|) satisfy C(A, B 1 ) E C(A, B 2 ) E 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!

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. Challenge: Our analysis gets stuck in ² ¼ exp(-k) Can one get ² ¼ exp(-k) ??

Local DP structure across U k 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 AA2A2 BiBi AAiAi B AA B3B3 AA3A3 B1B1 AA1A1

From local DP to global DP Q: How to glue local solutions ? A: If a typical S has two disjoint large, harmonious As ² > 1/k α high probability (2 queries) [DG] ² > exp(-k α ) Z-test (3 queries) [IKW]

Derandomization DP code whose length is poly (|U|), instead of |U| k

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

Derandomized DP Test Derandomized DP: U=(F q ) d Encode f k (S), S subspace of const dimension (as [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!

Summary Spherical cubes exist Power of consistency

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

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