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

Dana MoshkovitzWeizmann Institute

Ran RazWeizmann Institute

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 proof PCPn s(n)

• Completeness: sat. ) 9A, Pr[accept] = 1.

• Soundness: not sat. ) 8A, Pr[accept] · .

size

error

alphabet

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

Error

Note: ¸ 1/||q. Remark: Not

tight!

||q

easy!“The Sliding Scale

Conjecture” [BGLR93]

error0.992-log1-n 8>0

[AS92] [ALMSS92] [D06]

]ArSu97] [RaSa97 []DFKRS99[

||=(1/)O(1)O(1)

O(1)O(log(1/))

sub-const??

O(1)2

5

Sizesize

nc

n1+o(1) almost linear??

[AS92,ALMSS92]: s(n)=nc for large constant c.

[GS02,BSVW03,BGHSV04]: almost-linear size n1+o(1) PCPs

[D06] (based on [BS05]): s(n)=n¢polylog n

Only constant error!n

6

Our Motivation

Want: PCP with both sub-constant error

and almost-linear size

erroro(1)

sub-const??

sizenc

n1+o(1)

n

almost linear??

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)

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Low Degree Testing

Finite field F. f : Fm!F (m¿|F|).

Def: the agreement of f with degree d (d¿|F|):

Ff

Q(x1,…,xm)

deg Q ·d

agrmd(f ) = maxQ,deg·dPx( f (x)=Q(x) )agrmd(f ) = maxQ,deg·dPx( f (x)=Q(x) )

Fm

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Restriction of Polynomials to Affine Subspaces

Definitions: Affine subspace of dimension k, for translation z2Fm and (linearly independent)

directions y1,…,yk2Fm,

s={z+t1¢y1 + tk¢yk | t1,…,tk2F} Restriction of f :Fm!F to s is

f|s(t1,…,tk)=f (z+t1¢y1 + tk¢yk)

Observation:For Q:Fm!F of degree ·d, for any s of any

dimension k, have agrkd(Q|s)=1.

y

z

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Low Degree Testing

Low Degree Testing Theorems: For some family Sm

k of affine subspaces in Fm of dimension k=O(1),

agrmd(f ) ¼ Es2Sm

k agrk

d(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).

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LDT Thm ) Low Degree Tester

1. pick uniformly at random s2Smk and x2s.

2. accept iff A(s)(x)=f(x).

A

Subspace vs. Point Tester f,A :

Completeness: agrmd(f)=1 ) 9A, Pr[accept]=1.

Soundness: agrmd(f)· ) 8A, Pr[accept] /

Smk

f

Task: Given input f :Fm!F, d, probabilistically test whether f is close to degree d by performing O(1) queries to f and to proof A.

Fm

k-variate poly of deg ·d

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Sub-Constant Error and Almost-Linear Size

Sub-const error and almost linear size:

Additive approximation mO(1)¢(d/|F|)(1).

For k=O(1), small family |Sm

k|=|Fm|1+o(1).

errormO(1)¢(d/|F|)(1)

sub-const??

size|Fm|3

|Fm|1+o(1)

|Fm|

almost linear??

|Fm|2

7/8

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Our ResultsThm (LDT, [MR06]): 8 m,d,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 ) = Es2Sm

k agrk

d(f|s)

where 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]): 9 0<<1, 9 PCP: 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)).

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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.

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Proving LDT Theorem

Need to show:

1) agrmd(f ) / Es2Sm

k agrk

d(f|s).

2) agrmd(f ) ' Es2Sm

k agrk

d(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|.

Ff

Q(x1,…,xm)

Fm

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Previous Work [on size reduction]

[GS02]: For k=1, pick small Smk at random.

Show with high probability, 8f:Fm!F,Es2Sm

kagrk

d(f|s)¼Eline sagrkd(f|s)

[BSVW03]: Fix YµFm, ′-biased for 1/′=poly(m,log|F|). Take k=1 and Sm

k={x+ty | x2Fm, y2Y}. Show that Sm

k samples well. Analysis gives additive error >½.

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Our Work

Main Observation: The set of directions should not be pseudo-random!

y1,y22Y y1

y2

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Our Idea

Fix subfield HµF of size (1/0).

Set Y=HmµFm. Take k=3.

Smk={t0¢z+t1¢y1+t2¢y | z2Fm,y1,y22Y}

1. Useful: Can take F=GF(2g1¢g2) for g1=log(1/0).

2. Short: Indeed |Smk|=|Fm|¢(1/0)O(m).

3. Natural: H=F ! standard testers.4. Different: Y=HmµFm has large bias when HF.

Note: 8 y1,y22Y, 8 t1,t22HµF, t1¢y1+t2¢y22Y

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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.

Fm