Polynomial Interpolation over Composites.

Parikshit GopalanGeorgia Tech.

Polynomial Interpolation

Problem: Given a set I of inputs, compute P(X) from its evaluations at points in I.

Black-Box model: Only have access to the values of the polynomial.

• Minimize the degree of P(X).

• Minimize the number of black-box queries.

Polynomials over fields: [Newton, Lagrange]

A degree d polynomial has at most d roots.

Crucial for applications.

Polynomials over Zm = {0,1,.., m-1}:

A degree d polynomial could have many roots.

Eg: 32 ·X = 0 mod 64(0,2,4 …, 62)

X6 = 0 mod 64 (0,2,4 …, 62)

Polynomials modulo Composites

Eg: Find P(X1 ··· Xn) which is 0 at the origin but non-zero elsewhere on {0,1}n.

Over Z2, requires degree Ω(n).

Over Z6, can get degree O(n). [Barrington-Beigel-Rudich’94]

Complexity Theory:– Boolean function representations.

[BBR’94, … , Bhatnagar-Gopalan-Lipton’03]

Combinatorics: – Ramsey graphs. [Grolmusz’97, Gopalan’06]– Extremal set theory. [Grolmusz’97]

Algorithms:– Attacks on RSA [Coppersmith’96, …]– Primality testing [Agrawal-Biswas, AKS’02]– Factoring [Shamir]

Interpolation over Zm

Interpolation over composites can be easier than primes w.r.t degree, query complexity.

• Low degree polys. can be 0 everywhere.

(X-1)6X6 = 0 mod 64.

P(X) = Q(X) (X-1)6X6 + R(X)

R(X) agrees with P(X) everywhere.

Every polynomial is equivalent to a low degree polynomial.

• Values at various points are dependent.

Let x, y Z64

If x y mod 2 then P(x) P(y) mod 2

If x y mod 4 then P(x) P(y) mod 4

P(x) might tell us something about P(y).

Our ResultsProblem: Given a set I of inputs, compute

P(X) from its evaluations at points in I.

Main Result: Interpolation algorithm over Zm:

Minimizes degree. Number of queries can vary between log|

I| and |I|. Minimizing queries is NP-complete. Algorithm gives a log m approximation. On termination, we get a factorization

m = h1 h2 … ht (hi, hj) = 1

Approximation factor is bounded by t.

Corollaries: Algorithms for

Exact learning under uniform distribution.

PAC-learning.

Query efficient zero-testing.

[Karpinski-Shparlinski, Bshouty et.al …]

Let q(I) be the size of the smallest interpolating set for I.

• Values of P(X) over I are fixed by the values at S.

• Set of queries of any algorithm must be an interpolating set.

• Computing q(I) is NP-complete.

d(I) ≤ q(I) ≤ d(I)·(#factors of m)

Let d(I) be the min. degree of a monic polynomial which is 0 over I.

log|I| ≤ d(I) ≤ |I|

Can be computed efficiently.

Def: S I is an interpolating set for I if every polynomial that is 0 over S is 0 over I.

Our Results

IS

Finding Interpolating Sets The Prime Power Case

Problem: Given a set I Zm where m = pe, compute a minimum interpolating set S.

Viewing I as a tree:

Let m = 27.

0 1 2 mod 3

60 3

mod 9

0 9 1 2 11 14

0 1 2

0 1 2

0 1

If x and y are close, so are P(x) and P(y).

|P(x) – P(y)| ≤ |x – y|

Finding Interpolating Sets The Prime Power Case

Problem: Given a set I Zm where m = pe, compute a minimum interpolating set S.

Main Idea: Greedily add the point that maximizes sum of distances from already chosen points.

vp[n] = Highest power of p dividing n.

Algorithm IntSet

Pick a0 I arbitrarily.

Given S = {a0, …, ai-1}

Let N(X) = (X – a0)… (X – ai-1).

If N(X) is 0 over I, output S, halt.

Else

Find ai I s.t. vp [N(ai)] is minimum;

Add it to S.

Characterizing Interpolating Sets:

What objective function is being optimized?

Interpolating sets are all sets S I of size d(I) s.t. the sum of pair-wise distances is maximized.

Interpolating sets are all sets S I of size d(I) s.t. the power of p dividing the Vandermonde determinant is minimized.

Many other characterizations …

Ultrametric:

d(x,z) ≤ max(d(x,y), d(y,z))

Some properties hold for any ultrametric.

Proof follows from proof for polynomials.

Finding Interpolating Sets

Problem: Given a set I Zm, compute a minimum interpolating set S.

Main Idea: Try to simulate the prime-power algorithm for factors of m.Problem: What are the factors of m?

Algorithm IntSet

Given S = {a0, …, ai-1}.

Let N(X) = (X – a0)… (X – ai-1).

If N(X) is 0 over I, output S, halt.

Else

For each a I, let g(a) = (m, N(a)).

If some g(ai) | g(a) for all a I,

Add ai to S.

Else

Factor m = h1h2 s.t. (h1,h2) =1.

Recurse, combine solutions by CRT.

p2q, pq2

m =peqf

Conclusions

Understanding polynomials on {0,1}n over Zp.

Problem: Given I {0,1}n, let d(I) be the smallest degree polynomial which is 0 on I. Prove lower bounds on d(I).

Symmetric polynomials over Zp are the same as univariate polynomials mod pe.

Can get tight lower bounds for symmetric polynomials.

Used to rule out some algebraic approaches to better Ramsey graphs [G’06].

Asymmetric case is wide open.