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Russell Impagliazzo ( IAS & UCSD ) Ragesh Jaiswal ( Columbia U. ) Valentine Kabanets ( IAS & SFU ) Avi Wigderson ( IAS ) ( based on [IJKW08, IKW09] )

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f g hard on fraction of inputs hard on 1- fraction of inputs

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f s f is δ -hard (for size s), if every circuit (of size s) fails to compute f on δ inputs. 2 n {0,1} n

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Intuition If on a random x one can compute f(x) on < ( 1- ) fraction of inputs, then on k independent random ( x 1,…, x k ), one can compute all ( f(x 1 ),…, f(x k ) ) on < ( 1- ) k exp(- k) fraction of inputs.

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For f: U R, its k-wise DP function is f k : U k R k where: f k ( x 1, …, x k ) = ( f(x 1 ), …, f(x k ) )

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If: f is - hard ( for size s ) Then: f k is exp(- k)- hard ( for size s * poly(, ) ) 2 n {0,1} n 2 n {0,1} nk

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U UkUk f f kf k 0 1 1 0 1 0 011 010 110 010

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U UkUk f fkfk Local encoding Local approximate decoding Poor distance… Distance amplification: far-away messages are mapped to farther-away codewords Superpoly rate… Derandomized DP Code with poly rate.

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Given: circuit C ( of size s * poly(, ) ) exp(- k)-computing f k Construct: circuit C ( of size s ) ( 1- )-computing f. 2 n {0,1} nk 2 n {0,1} n

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Proof: Pick L = 1/ functions f 1, …, f L. Partition inputs into L blocks of size each. Define C to agree with f i k on block i. Theorem: To decode from agreement, need the list size (1/ ).

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r1r1 X rkrk C X b1b1 bbkbk if enough b i = f(r i ), then output b b [GNW, IW97,…]: List-size > exp(1/ ). [IJK06]: poly(1/ ) list size for large, but still sub-optimal & complicated.

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Features: list-size O(1/ ) ( tight ! ) simple algorithm (and analysis) generalizes to Derandomized DP Code

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B1B1 B2B2 A Pick random k-set x Randomly partition: |A|=|B 1 |=k/2 Freeze these random choices On input x, Pick random k-set (A,B 2 ) containing x. If C is consistent ( C(B 1,A)| A = C(A,B 2 )| A ), output C(A,B 2 )| x. Else re-sample B 2 ( O((1/ ) log 1/ ) times ).

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Pick random k-set Randomly partition: |A|=|B 1 |=k/2 On input x, Pick random k-set (A,B 2 ) containing x. If C is consistent ( C(B 1,A)| A = C(A,B 2 )| A ), output C(A,B 2 )| x. Else re-sample B 2 ( O((1/ ) log 1/ ) times ). Preprocessing Algo A,B 1 (x):

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Pick random k-set Randomly partition: |A|=|B 1 |=k/2 Preprocessing On input x, Pick random k-set (A,B 2 ) containing x. If C is consistent ( C(B 1,A)| A = C(A,B 2 )| A ), output C(A,B 2 )| x. Else re-sample B 2 ( O((1/ ) log 1/ ) times ). Algo A,B 1 (x): Theorem: With probability ( ² ) over (B 1,A), the resulting circuit Algo (1- )-computes f.

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Flower: determined by S=(A,B) Core: A Core values: α =C(A,B) A Consistent petals: { (A,B) | C(A,B) A = α } [IJKW08]: Flower analysis B B4B4 AA B2B2 B3B3 B1B1 B5B5

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Then: There are many ( ² /2) flowers determined by S=(A,B) that are nice. A flower is nice if it has correct core ( C(S) = f k (S) ), many ( /2) consistent petals. Also: In a random nice flower, almost all consistent petals are mostly correct (C ¼ f k ) B B4B4 AA B2B2 B3B3 B1B1 B5B5 Assume: C ² -agrees with f k Consistency Correctness

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There are many ( ² /4) flowers determined by S=(A,B) that have: correct core ( C(S) = f k (S) ), many ( /4) consistent petals, almost all consistent petals are mostly correct (C ¼ f k ) Preprocessing likely picks a nice flower Algo A,B likely does not time-out Algo A,B (x) likely equals f(x)

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Then: There are many ( ² /2) flowers determined by S=(A,B) that are nice. A flower is nice if it has correct core ( C(S) = f k (S) ), many ( /2) consistent petals. Also: In a random nice flower, almost all consistent petals are mostly correct (C ¼ f k ) B B4B4 AA B2B2 B3B3 B1B1 B5B5 Assume: C ² -agrees with f k Averaging Symmetry

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Pr A,B [ ( (A,B) correct ) & ( A has /2 correct extensions (A,B) ) ] = Pr A,B [(A,B) correct ] - Pr A,B [( (A,B) correct ) & ( A has /2 correct extensions B ) ]

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Pr A,B [ ( (A,B) correct ) & ( A has /2 correct extensions (A,B) ) ] Pr A,B [(A,B) correct ] - Pr A,B [( (A,B) correct ) | ( A has /2 correct extensions B ) ] /2 = /2.

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Idea: A highly incorrect set S cant be a consistent petal in a random flower with correct core B S A

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B S A

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Idea: A highly incorrect set S cant be a consistent petal in a random flower with correct core. B S A f(A) = C(B,A) A & C(B,A) A = C(S) A & C(S) A f(A). Contradiction !

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C : U k R k Is C = f k, for some f : U R ? Fact: C = f k iff for all pairs of intersecting k-sets (S,S), with A=S S, C(S)| A = C(S)| A

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C : U k R k Test for one random pair of intersecting k-sets (S,S), with A=S S, if C(S)| A = C(S)| A A S S

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On C : U k R k Test makes few queries, and (1) Accepts if C = f k. (2) Rejects if C is far away from any f k (2) Pr [ Test accepts C ] > C f k on > ( ) of inputs. - Minimize # queries ( 2 ? 3 ? ) - Analyze small ( < 1/k ? < exp(-k) ? )

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Given C : U k R k, is C 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 need > 2 1/k IKW 09 3 exp(-k α ) IKW 09* 2 1/k α * Derandomization *

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Randomly pick two k-sets S 1 =(B 1,A) and S 2 =(A,B 2 ) (with |A| = k 1/2 ). B1B1 B2B2 A S1S1 S2S2 Accept if C( S 1 ) A = C( S 2 ) A

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Flower: determined by S=(A,B) Core: A Core values: α =C(A,B) A Consistent petals: { (A,B) | C(A,B) A = α } B B4B4 AA B2B2 B3B3 B1B1 B5B5

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There are many ( ² /2) flowers determined by S=(A,B) such that: There is g : U R so that on almost all consistent petals B i, C (B i ) g k (B i ). B B4B4 AA B2B2 B3B3 B1B1 B5B5 Assume: V-Test accepts with prob ² Locally C is DP

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There are many ( ² /2) harmonious flowers determined by S=(A,B). Harmonious flower: many ( /2) consistent petals, on consistent petals, V- test accepts almost certainly ( 1-poly( ² ) ). B B4B4 AA B2B2 B3B3 B1B1 B5B5 E C(A, B 1 ) E ¼ C(A, B 2 ) E, with |E| = |A| Assume: V-Test accepts with prob ² Proof by symmetry arguments (as in Decoding)

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Harmonious flower: many ( /2) consistent petals, on consistent petals, V-test accepts almost certainly ( 1-poly( ² ) ). B B4B4 AA B2B2 B3B3 B1B1 B5B5 Main Lemma: Define g(x) = PLURALITY { C( S ) x | consistent petals S, x S }. Then C(S) ¼ g k (S) for almost all (1-poly( ² )) consistent petals S.

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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 B1B1 B2B2

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There are many harmonious flowers with: many consistent petals, restricted to consistent petals, V-Test accepts almost surely. V-Test ² -accepts C Conclude: C(S) ¼ g k (S) for almost all consistent petals S of the flower. There are many nice flowers with: correct core, many consistent petals. C ² -computes f k Define: g(x) = PLURALITY { C( S ) x } consistent petals S : x 2 S Conclude: g(x) = f(x) for almost all inputs x. Consistency Correctness Harmony DP

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Field of flowers (A i,B i ) Each with its own Local DP function g i Global g ? B AA B2B2 AA BiBi AA B3B3 AA B1B1 AA

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Yes, if ² > 1/k a [DG08] [we re-prove it] ( can glue together many flowers ) No, if ² < 1/k [DG08]No But, with one extra query, get ² = exp( - k a ) !

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Randomly pick k-sets S 1 =(B 1,A 1 ), S 2 =(A 1,B 2 ), S 3 =(B 2,A 2 ) ( |A 1 | = |A 2 | = m = k 1/2 ). B1B1 B2B2 A1A1 A2A2 S1S1 S2S2 S3S3 Accept if C( S 1 ) A 1 = C( S 2 ) A 1 and C( S 2 ) B 2 = C( S 3 ) B 2

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U = ( F q ) m T U k T = { 8-dim affine subspaces of U } ( k = q 8 ) T = { 8-dim affine subspaces of U } ( k = q 8 ) Same list-decoding algo (from = 1/poly( k) agreement) Same DP Test (V-Test) ( for = 1/poly(k) acc prob ) Same list-decoding algo (from = 1/poly( k) agreement) Same DP Test (V-Test) ( for = 1/poly(k) acc prob ) Corollary: Polynomial-rate locally (approximately) list-decodable and locally testable code.

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All k-setsAll d-dim subspaces -approx list-decodable from agreement: exp ( - k ) 1/poly( k ) 2-query testable, acc prob > 1/poly( k ). 3-query testable, acc prob > exp (-k 1/2 ) Analysis: sampling properties of DP graphs Chernoff Chebyshev (full independence) (2-wise independence)

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( Derandomized ) DP Code Decoding and Testing Analysis of V-Test : Either reject C, or verify that locally C = g k ( for some g ), and get g(x 1 ), …, g(x k ) for independent random x i s. Application to 2-Query PCP: Parallel k- repetition for restricted games.

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Yaos XOR binary ECC: f k (x 1, …, x k ) = f(x 1 ) … f(x k ) - approximately locally list-decodable from agreement ½ + > exp(- k), with list size O(1/ 2 ) (tight)

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Other examples of derandomized DP Codes ? Better parameters ??? Derandomized 2-PCP : Re-proving / improving [Moshkovitz-Raz08, Dinur- Harsha09] using similar techniques. ( recent progress [Dinur-Meir09] )

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