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```					Hardness of Reconstructing
Multivariate Polynomials.

Parikshit Gopalan   U. Washington
Subhash Khot        NYU/Gatech
Rishi Saket         Gatech/NYU
Curve Fitting

Problem: Given data points, find a low degree
polynomial that fits best.
Easy if there is a perfect fit.
Well studied problem …
Curve Fitting through the ages
Curve Fitting through the ages
Curve Fitting through the ages
!
Statistics: Least Squares
Computational                    Coding
Learning                       Theory

Polynomial
Reconstruction

PCPs
Cryptography

Pseudorandom
-ness
The Reconstruction Problem

Input: Degree d.           Points   Values
x1      f(x1)
xi     f(xi)
xm      f(xm)

Output: A degree d polynomial that best fits
the data.
In this talk: Finite fields, Hamming distance.
The Reconstruction Problem

Input: Degree d, set S, values f(x) for x 2 S.
Output: A degree d polynomial that best fits
the data.

Parameters that matter:
1. Degree d, Field F.

2. Set S.

3. How good is the best fit? (error-rate )
Algorithms for Reconstruction

Univariate Case [Sudan, Guruswami-Sudan]:

Multivariate Case [Goldreich-Levin, Goldreich-

Can tolerate very high error rate .
Are these algorithms optimal?
Hardness Results: Univariate Case
Degree d polynomials, n points in F.

[Guruswami-Vardy]: NP-hard to tell if some
degree d poly. has d +2 agreements.

[Guruswami-Sudan]: Can tell if some degree d
poly. has (nd)0.5 agreement.
Hardness Results: Multivariate Case
Linear polynomials over F2
 Some linear poly. satisfies 1-  fraction of
points.
 Every linear poly. satisfies less than 0.5 + 
fraction of points.

Extends to any F and d =1.
Implies something for d < F.

d ¸ 2 over F2: Nothing known.
Our Results
Over F2 for any d, NP-hard to tell whether
 Some linear polynomial satisfies 1-  fraction
of points.
 Every degree d polynomial satisfies at most
1 -2-d +  fraction of points.
d=1   ½+
SZ Lemma: For a degree d poly P  0 over F2,
d=2 ¾ + 
Prx[ P(x)  0] ¸ 2-d.
Our Results
Over Fq for any d, NP-hard to tell whether
 Some linear polynomial satisfies 1-  fraction
of points.
 Every degree d polynomial satisfies at most
c(d,q) +  fraction of points.

c(d,q): Schwartz-Zippel for polynomials of total
degree d over Fq.
Overview of Reduction
 Reducing from Label-Cover.
 Dictatorship Testing.
 Consistency Testing.
 Putting it all together.
Label Cover
1
Graph: G(V,E), |V| =n.         n                 2
Labels: [k]
3
Edges: pe ½ [k] £ [k]
Goal: Find a labeling satisfying all edges.

Thm [PCP + Raz]: It is NP-hard to tell if
• Some labeling satisfies all edges.
• Every labeling satisfies ·  frac. of edges.
The Reduction
Henceforth d =2, field = F2.
X11 X12 … X1k

Xn 1 Xn 2 … Xn k                 X21 X22 … X2k

X31 X32 … X3k

Constraints: Points in {0,1}nk + values.
Yes Case: Some L satisfies most constraints.
No Case: No Q satisfies many constraints.
The Reduction
X11 X12 … X1k

Xn1 Xn2 … Xnk                X21 X22 … X2k

X31 X32 … X3k

• If l(v) is a good labelling, then L = v Xvl(v)
will satisfy most points.
The Reduction
X11 X12 … X1k

Xn1 Xn2 … Xnk                X21 X22 … X2k

X31 X32 … X3k

• If l(v) is a good labelling, then L = v Xvl(v)
will satisfy most points.
• Any Q that does ¾ +  gives a labelling
satisfying ’ fraction of edges.
Overview of Reduction
3, 71, 99
Dictatorship:                            p
Q1 = Q(X11,…,X1k,0,..,0).       17, 45
Q1 looks like a Dictator X1j.
Will settle for small list.
Constant
independent
of k.
Consistency:
Some pair of labels in the list satisfy p.
Overview of Reduction
3, 71, 99
Dictatorship:                             p
Q1 = Q(X11,…,X1k,0,..,0).       17, 45
Q1 looks like a Dictator X1j.
Will settle for small list.
Can enforce this for  frac. of vertices.

Consistency:
Some pair of labels in the list satisfy p.
Can enforce this for all edges.
Overview of Reduction
3, 71, 99
p
17, 45

If Q does ¾ + 
•Small list for  frac. of vertices.
•Consistency for all edges.
Assign random labels from list.
Satisfies constant fraction of edges.
Overview of Reduction
 Dictatorship Testing.
 Consistency Testing.
 Putting it all together.
Overview of Reduction
 Dictatorship Testing.
 Consistency Testing.
 Putting it all together.
Dictatorship Testing for low-
degree Polynomials.
Input: Q(X1,…,Xk) of degree 2.
Goal: Design a test s.t
 Every dictatorship Xi passes w.p close to 1.

 If Q does better than ¾, it is close to a
dictatorship.
Small List

Test: Pick a random point x 2 {0,1}k.
Check if Q(x) = y.
Mini reconstruction problem!
Dictatorship Testing for low-
degree Polynomials.

All polys.

Dictatorships
Dictatorship Testing
Hard to do with
just 2 queries.

All polys.

Dictatorships
Dictatorship Testing for low-
degree Polynomials.
 Poly. is of low degree.
 Allowed one query (!)

Dictatorships
Dictatorship Test
Dictatorship Test:
 Pick  2 {0,1}k from the -biased distribution.
 Check if Q() = 0.               Each i =1
independently.
w.p 
polys. are 3:1 balanced.
• -biased: Dictatorships are
highly skewed.

(0,…,0)               • Is there a converse?
Dictatorship Test
Dictatorship Test:
 Pick  2 {0,1}k from the -biased distribution.
 Check if Q() = 0.

 Xi passes w.p 1- .
 XiXj passes w.p 1- 2.
 X1(X1 + … + Xk) + X2(X1 + …) passes w.p 1 - 2
Dictatorship Test
Dictatorship Test:
 Pick  2 {0,1}k from the -biased distribution.
 Check if Q() = 0.

Define G(Q) to be the graph of Q.         2

Q = X1X2 + X2X3,       G(Q) =         1       3

Thm: If Q passes w.p ¾ + , then G(Q) has no
large matchings.
Dictatorship Test
Dictatorship Test:
 Pick  2 {0,1}k from the -biased distribution.
 Check if Q() = 0.
Thm: If Q passes w.p ¾ + , then G(Q) has no large matchings.

1. Large matching:

Independent monomials.

2. Only small matchings:

Small vertex cover. L + X L
X1 1   2 2
Dictatorship Test
Dictatorship Test:
 Pick  2 {0,1}k from the -biased distribution.
 Check if Q() = 0.
Thm: If Q does better than ¾, then G(Q) has no
large matchings.
Xi = 0 w.p 1- 2        Xi 2R {0,1}
Q                      Q’                 c =? 0

• If G(Q) has a large matching, then Q’  0 w.h.p.
• If Q’  0, then c =1 w.p ¸ ¼ (SZ lemma).
• If Q does well, G(Q) has no large matchings.
Dictatorship Test
Dictatorship Test:
 Pick  2 {0,1}k from the -biased distribution.
 Check if Q() = 0.
Thm: If Q does better than ¾, then G(Q) has no
large matchings.
If G(Q) has a large matching, then Q’  0 w.h.p.

• Each edge survives w.p 42.

• Events for each matching
edge are independent.
Dictatorship Test
Dictatorship Test:
 Pick  2 {0,1}k from the -biased distribution.
 Check if Q() = 0.

Define G(Q) to be the graph of Q.         2

Q = X1X2 + X2X3,       G(Q) =         1       3

Thm: If Q passes w.p ¾ + , then G(Q) has no
large matchings.
Small List: Vertex set of a maximal matching.
Overview of Reduction
3, 71, 99
Dictatorship:                            p
Assign a small list to a        17, 45
vertex.

Consistency:
Some pair of labels in the list satisfy p.
Overview of Reduction
 Dictatorship Testing.
 Consistency Testing.
 Putting it all together.
Consistency Testing
l(x) = l(y)
Consistency Testing
l(x) = l(y)
X1 X2 … X k                 Y1 Y2 … Yk

Given Q(X1,…,Xk,Y1,…,Yk) s.t Q(Xi) and Q(Yj)
both pass the dict. Test.
Want Q(X1,..,Xk,0,…,0) = Q(0,…,0,Y1,…,Yk).

Test: Q(r,0) = Q(0,r) for r 2R {0,1}k.

 Two queries!
Consistency via Folding
l(x) = l(y)
X1 X2 … X k                    Y1 Y2 … Yk

•Yes case: Q = Xi + Yi for some i.
• All of them vanish over H = (r,r).
H
• Constant on each coset of H.
• Enforce this on Q even in the No case.
Consistency via Folding
Def: Q is folded over subspace H µ {0,1}k
if Q is constant on every coset of H.
H
Examples: Linear polys., juntas.

Thm: Q is folded over H iff for some nice basis
(1,…,t,1,...,k-t),
Q = R(1,…,t) is a t-junta for t = k – dim(H)

In the nice basis (1,…,t,1,...,k-t)
is: coset of H, js: position in coset.
Template for Folding
Want Q folded over a subspace H.
Compute nice basis (i, j).            H
To test if Q(x) = y
o Let x = (, ); test R() = y.
For analysis: Rewrite R() as Q(x).
Now Q is folded.                      {0,1}n/H
Consistency via Folding
l(x) = l(y)

Fold over H = (r,r) for r 2 {0,1}k.
Polys. folded over H can be written as:
Q(X1,…,Xk,Y1,…,Yk) = R(X1 +Y1, …, Xk + Yk)

Gives Q(X1,…,Xk) = Q(Y1,…,Yk).
List of Xis: Vertex set of maximal matching.
Every two maximal matchings intersect.
Summary of Reduction

Each constraint p gives Hp ½ {0,1}nk.
Fold over the span of all Hp.
Run Dict. test on every vertex.
No explicit consistency tests.

If Q passes w.p ¾ + ,
  fraction of vertices do well on Dict. test.
 Consistency for all edges by folding.
Overview of Reduction
 Dictatorship Testing.
 Consistency Testing.
 Putting it all together.
Projections …
X11 X12 … X1k

Xn1 Xn2 … Xnk               X21 X22 … X2k

X31 X32 … X3k

 Can handle equality, permutations.
 Need perfect completeness: no UGC.
 Have to deal with \$#@%! projections.
Projections …

Decoding is a vertex cover for G(Qi).
Need to show that every two vertex covers
intersect.
Projections …
Do every two vertex covers of G intersect?
No:
Projections …
Do every two vertex covers of G intersect?
No:
… but in any three VCs, some pair intersects.
Main Theorem
Over F2 for any d, NP-hard to tell whether
 Some linear polynomial satisfies 1-  fraction
of points.
 Every degree d polynomial satisfies at most
1 -2-d +  fraction of points.
Better Hardness?
Problem: Can we improve soundness to 0.5 + ?

Bottleneck: Dictatorship test.
Present analysis is optimal in general:
Q = (X1 + ..+ Xk)(Xk+1 + … +X2k) passes w.p ¾.
Can assume that Q is balanced.
Thank You!

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