Post on 30-Dec-2015
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.
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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
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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
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Our Motivation
Want: PCP with both sub-constant error
and almost-linear size
erroro(1)
sub-const??
sizenc
n1+o(1)
n
almost linear??
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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 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