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Sub-Constant Error Low Degree Test of Almost-Linear Size Dana Moshkovitz Weizmann Institute Ran Raz Weizmann Institute.

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Presentation on theme: "Sub-Constant Error Low Degree Test of Almost-Linear Size Dana Moshkovitz Weizmann Institute Ran Raz Weizmann Institute."— Presentation transcript:

1 Sub-Constant Error Low Degree Test of Almost-Linear Size Dana Moshkovitz Weizmann Institute Ran Raz Weizmann Institute

2 2 Probabilistic Checking of Proofs: Pick at random q=O(1) places in proof. Read only them and decide accept/reject. Motivation: Probabilistically Checkable Proofs (PCP) [AS92,ALMSS92] “ Claim: formula  is satisfiable. ” NP proofPCP ns(n) Completeness:  sat. ) 9A, Pr[accept] = 1. Soundness:  not sat. ) 8A, Pr[accept] · . siz e error  alphabet

3 3 Importance of PCP Theorem Surprising insight to the power of verification and NP. But it ’ s even more important than that! [FGLSS91, … ]: Enables hardness of approximation results. [FS93,GS02, … ]: Yields codes with local testing/decoding properties.

4 4 Error Note:  ¸ 1/|  | q. Remark: Not tight! q |||| easy! “The Sliding Scale Conjecture” [BGLR93] error 0.992 -log 1-  n 8  >0 [AS92] [ALMSS92] [D06] [ArSu97] [RaSa97] [DFKRS99] O(1) |  |=(1/  ) O(1) O(log(1/  )) O(1) sub-const?? 2O(1)

5 5 Size size ncnc n 1+o(1) almost linear?? [AS92,ALMSS92]: s(n)=n c for large constant c. [GS02,BSVW03,BGHSV04]: almost-linear size n 1+o(1) PCPs [D06] (based on [BS05]) : s(n)=n ¢ polylog n Only constant error! n

6 6 Our Motivation Want: PCP with both sub-constant error and almost-linear size error o(1) sub-const?? size ncnc n 1+o(1) n almost linear??

7 7 Our Work We show: [STOC ’ 06] Low Degree Testing Theorem (LDT) with sub- constant error and almost-linear size. Mathematical Thm of independent interest Core of PCP Subsequent work: [ECCC ’ 07] (our) LDT ) PCP (with sub-const error, almost linear size)

8 8 Low Degree Testing Finite field F. f : F m ! F (m ¿ |F|). Def: the agreement of f with degree d (d ¿ |F|) : F f Q(x 1, …,x m ) deg Q · d agr m d ( f ) = max Q,deg · d P x ( f (x)=Q(x) ) FmFm

9 9 Restriction of Polynomials to Affine Subspaces Definitions: Affine subspace of dimension k, for translation z 2 F m and (linearly independent) directions y 1, …,y k 2 F m, s={z+t 1 ¢ y 1 + t k ¢ y k | t 1, …,t k 2 F} Restriction of f :F m ! F to s is f| s (t 1, …,t k )= f (z+t 1 ¢ y 1 + t k ¢ y k ) Observation: For Q:F m ! F of degree · d, for any s of any dimension k, have agr k d (Q| s )=1. y z

10 10 Low Degree Testing Low Degree Testing Theorems: For some family S m k of affine subspaces in F m of dimension k=O(1), agr m d ( f ) ¼ E s 2 S m k agr k d ( f | s ) agreement with degree d: agr m d ( f ) = max Q,deg · d P x ( f (x)=Q(x) ) [RuSu90],[AS92],…,[FS93]: For k=1 and S m k = all lines, Gives large additive error ¸ 7/8. [RaSa97]: For k=2 and S m k = all planes, Gives additive error m O(1) (d/|F|)  (1). [ArSu97]: For k=1 and S m k = all lines, Gives additive error m O(1) ¢ d O(1) (1/|F|)  (1).

11 11 LDT Thm ) Low Degree Tester 1.pick uniformly at random s 2 S m k and x 2 s. 2.accept iff A (s)(x)= f (x). A Subspace vs. Point Tester f, A : Completeness: agr m d ( f )=1 ) 9A, Pr[accept]=1. Soundness: agr m d ( f ) ·  ) 8A, Pr[accept] /  SmkSmk f Task: Given input f :F m ! F, d, probabilistically test whether f is close to degree d by performing O(1) queries to f and to proof A. FmFm k-variate poly of deg · d

12 12 Sub-Constant Error and Almost-Linear Size Sub-const error and almost linear size: Additive approximation m O(1) ¢ (d/|F|)  (1). For k=O(1), small family |S m k |=|F m | 1+o(1). error m O(1) ¢ (d/|F|)  (1) sub-const?? size |F m | 3 |F m | 1+o(1) |F m | almost linear?? |F m | 2 7/8

13 13 Our Results Thm (LDT, [MR06]): 8 m,d,  0 , for infinitely many finite fields F, for k=3, 9 explicit S m k of size |S m k |=|F| m ¢ (1/  0 ) O(m), such that agr m d ( f ) = E s 2 S m k agr k d ( f | s )   where  m O(1) ¢ (d/|F|) 1/4 + m O(1) ¢  0. ) for m  (1) · 1/  0 · |F| o (1), get sub-constant error and almost-linear size. Thm (PCP, [MR07]): 9 0<  <1, 9 PCP: on input size n, queries O(1) places in proof of size n ¢ 2 O((logn) 1-  ) over symbols with O((logn) 1-  ) bits and achieves error 2 -  ((logn)  ).

14 14 The Gap From LDT To PCP Large alphabet:  (d). PCP = testing any polynomial-time verifiable property, rather than closeness to degree d. Main Observations for Polynomials/PCP: Low Degree Extension: Any proof can be described as a polynomial of low degree (i.e., of low ratio d/|F|) over a large enough finite field F. List decoding: For every f :F m ! F, there are few polynomials that agree with f on many points.

15 15 Proving LDT Theorem Need to show: 1)agr m d ( f ) / E s 2 S m k agr k d ( f | s ). 2)agr m d ( f ) ' E s 2 S m k agr k d ( f | s ). Note: (2) is the main part of the analysis. (1) is easy provided that S m k samples well, i.e., for any A µ F m, it holds that E s 2 S m k [|s Å A|/|s|] ¼ |A|/|F m |. F f Q(x 1, …,x m ) FmFm

16 16 Previous Work [on size reduction] [GS02]: For k=1, pick small S m k at random. Show with high probability, 8 f :F m ! F, E s 2 S m k agr k d ( f | s ) ¼ E line s agr k d ( f | s ) [BSVW03]: Fix Y µ F m,  ′ -biased for 1/  ′ =poly(m,log|F|). Take k=1 and S m k ={x+ty | x 2 F m, y 2 Y}. Show that S m k samples well. Analysis gives additive error >½. size n2n2 n 1+o(1) almost linear?? n

17 17 Our Work Main Observation: The set of directions should not be pseudo-random! y 1,y 2 2 Y y1y1 y2y2

18 18 Our Idea Fix subfield H µ F of size  (1/  0 ). Set Y=H m µ F m. Take k=3. S m k ={t 0 ¢ z+t 1 ¢ y 1 +t 2 ¢ y | z 2 F m,y 1,y 2 2 Y} 1.Useful: Can take F=GF(2 g 1 ¢ g 2 ) for g 1 =  log(1/  0 ) . 2.Short:Indeed |S m k |=|F m | ¢ (1/  0 ) O(m). 3.Natural: H=F ! standard testers. 4.Different: Y=H m µ F m has large bias when H  F. Note: 8 y 1,y 2 2 Y, 8 t 1,t 2 2 H µ F, t 1 ¢ y 1 +t 2 ¢ y 2 2 Y

19 19 Sampling Lemma (Sampling): Let A µ F m. Let  = |A|/|F m | and =1/|H|. Pick random z 2 F m, y 2 Y. Let l = { z+t ¢ y | t 2 F } and X=| l Å A|/| l | (hitting). Then, for any  >0 (hitting ¼ true fraction): P[ | X -  | ¸  ] · 1/  2 ¢  ¢ Proof: Via Fourier analysis. FmFm


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