Sub-Constant Error Low Degree Test of Almost-Linear Size

Sub-Constant Error Low Degree Testof Almost-Linear Size Dana Moshkovitz Weizmann Institute Ran Raz Weizmann Institute

n s(n) Motivation: Probabilistically Checkable Proofs (PCP) [AS92,ALMSS92] “Claim: formula  is satisfiable.” size NP proof PCP  alphabet • Probabilistic Checking of Proofs: • Pick at randomq=O(1) places in proof. • Read only them and decide accept/reject. error • Completeness:sat. )9A, Pr[accept] = 1. • Soundness:notsat. )8A, Pr[accept] ·.

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.

Error • Note:¸1/||q. • Remark: Not tight! easy! “The Sliding Scale Conjecture” [BGLR93] sub-const?? [AS92] [ALMSS92] [D06] [ArSu97] [RaSa97] [DFKRS99] error 2-log1-n 8>0 0.99

Size size • [AS92,ALMSS92]: s(n)=nc for large constant c. • [GS02,BSVW03,BGHSV04]: almost-linear sizen1+o(1) PCPs • [D06] (based on [BS05]): s(n)=n¢polylog n • Only constant error! nc n1+o(1) almost linear?? n

Our Motivation size Want: PCP with both • sub-constant error and • almost-linear size nc n1+o(1) almost linear?? n sub-const?? error o(1)

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)

Fm Low Degree Testing Finite field F.f: Fm!F (m¿|F|). Def: the agreement of f with degree d(d¿|F|): agrmd(f) = maxQ,deg·dPx( f (x)=Q(x) ) f Q(x1,…,xm) deg Q ·d

y z Restriction of Polynomials to Affine Subspaces Definitions: • Affine subspace of dimension k, for translation z2Fmand (linearly independent) directions y1,…,yk2Fm, s={z+t1¢y1 + tk¢yk | t1,…,tk2F} • Restriction off :Fm!F to s is f|s(t1,…,tk)=f (z+t1¢y1 + tk¢yk) Observation: ForQ:Fm!F of degree ·d, for any s of any dimension k, have agrkd(Q|s)=1.

agreement with degree d: agrmd(f) = maxQ,deg·dPx( f (x)=Q(x) ) Low Degree Testing Low Degree Testing Theorems: For some family Smk of affine subspaces in Fm of dimension k=O(1), agrmd(f)¼ Es2Smkagrkd(f|s) • [RuSu90],[AS92],…,[FS93]: For k=1 and Smk = all lines, • Gives large additive error ¸7/8. • [RaSa97]: For k=2 and Smk = all planes, • Gives additive error mO(1)(d/|F|)(1). • [ArSu97]: For k=1 and Smk = all lines, • Gives additive error mO(1)¢dO(1)(1/|F|)(1).

k-variate poly of deg ·d LDT Thm ) Low Degree Tester Task: Given input f :Fm!F, d, probabilisticallytest whether f is close to degree d by performing O(1) queries to f and to proofA. • pick uniformly at random s2Smk and x2s. • accept iff A(s)(x)=f(x). Fm Smk f A Subspace vs. Point Tester f,A: • Completeness:agrmd(f)=1 )9A,Pr[accept]=1. • Soundness:agrmd(f)·)8A, Pr[accept] / .

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

Our Results Thm (LDT, [MR06]):8m,d,0>0, for infinitely many finite fields F, for k=3, 9 explicit Smk of size |Smk|=|F|m¢(1/0)O(m), such that agrmd(f) = Es2Smk agrkd(f|s) where e = mO(1)¢(d/|F|)1/4 +mO(1)¢0. ) for m(1) · 1/0· |F|o(1), get sub-constant error and almost-linear size. Thm (PCP, [MR07]):90<<1, 9PCP: on input size n, queries O(1) places in proof of size n¢2O((logn)1-) over symbols with O((logn)1-) bits and achieves error 2-((logn)).

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:Fm! F, there are few polynomials that agree with f on many points.

Fm Proving LDT Theorem Need to show: • agrmd(f) / Es2Smkagrkd(f|s). • agrmd(f)' Es2Smkagrkd(f|s). Note:(2) is the main part of the analysis. (1) is easy provided that Smk samples well, i.e., for any AµFm, it holds that Es2Smk[|sÅA|/|s|]¼|A|/|Fm|. f Q(x1,…,xm)

Previous Work [on size reduction] size • [GS02]: For k=1, pick small Smk at random. Show with high probability, 8f:Fm!F, Es2Smkagrkd(f|s)¼Eline sagrkd(f|s) • [BSVW03]: • Fix YµFm, ′-biased for 1/′=poly(m,log|F|). • Take k=1 and Smk={x+ty | x2Fm, y2Y}. • Show that Smk samples well. • Analysis gives additive error >½. n2 n1+o(1) n almost linear??

y1 y2 Our Work Main Observation:The set of directions should not be pseudo-random! y1,y22Y

Our Idea Fix subfieldHµF of size (1/0). Set Y=HmµFm. Take k=3. Smk={t0¢z+t1¢y1+t2¢y | z2Fm,y1,y22Y} • Useful:Can take F=GF(2g1¢g2) for g1=log(1/0). • Short: Indeed |Smk|=|Fm|¢(1/0)O(m). • Natural:H=F ! standard testers. • Different:Y=HmµFm has large bias when HF. Note:8y1,y22Y, 8t1,t22HµF, t1¢y1+t2¢y22Y

Fm Sampling Lemma (Sampling): Let AµFm. Let  = |A|/|Fm| and =1/|H|. Pick random z2Fm, y2Y. Let l= { z+t¢y | t2F } and X=|lÅA|/|l| (hitting). Then, for any >0(hitting ¼ true fraction): P[ | X -  | ¸ ] · 1/2¢¢ Proof: Via Fourier analysis.

Sub-Constant Error Low Degree Test of Almost-Linear Size