1. Polynomial Interpolation over Composites. Parikshit Gopalan Georgia Tech.

2. 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 Z m = {0,1,.., m-1}: A degree d polynomial could have many roots. Eg:32 ·X = 0 mod 64(0,2,4 …, 62) X 6 = 0 mod 64 (0,2,4 …, 62)

3. Polynomials modulo Composites Eg: Find P(X 1 ··· X n ) which is 0 at the origin but non-zero elsewhere on {0,1} n. Over Z 2, requires degree Ω (n). Over Z 6, can get degree O( n). [Barrington-Beigel-Rudich94] Complexity Theory: –Boolean function representations. [BBR94, …, Bhatnagar-Gopalan-Lipton03] Combinatorics: –Ramsey graphs. [Grolmusz97, Gopalan06] –Extremal set theory. [Grolmusz97] Algorithms: –Attacks on RSA [Coppersmith96, …] –Primality testing [Agrawal-Biswas, AKS02] –Factoring [Shamir]

4. Interpolation over Z m Interpolation over composites can be easier than primes w.r.t degree, query complexity. Low degree polys. can be 0 everywhere. (X-1) 6 X 6 = 0 mod 64. P(X) = Q(X) (X-1) 6 X 6 + 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 Z 64 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).

5. Our Results Problem: Given a set I of inputs, compute P(X) from its evaluations at points in I. Main Result: Interpolation algorithm over Z m : 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 = h 1 h 2 … h t (h i, h j ) = 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 …]

6. 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 I S

7. Finding Interpolating Sets The Prime Power Case Problem: Given a set I Z m where m = p e, compute a minimum interpolating set S. Viewing I as a tree: Let m = 27. 01 2 mod 3 603 mod 9 09121114 012 012 0 1 If x and y are close, so are P(x) and P(y). |P(x) – P(y)| |x – y|

8. Finding Interpolating Sets The Prime Power Case Problem: Given a set I Z m where m = p e, compute a minimum interpolating set S. Main Idea: Greedily add the point that maximizes sum of distances from already chosen points. v p [n] = Highest power of p dividing n. Algorithm IntSet Pick a 0 I arbitrarily. Given S = {a 0, …, a i-1 } Let N(X) = (X – a 0 )… (X – a i-1 ). If N(X) is 0 over I, output S, halt. Else Find a i I s.t. v p [N(a i )] is minimum; Add it to S.

9. 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.

10. Finding Interpolating Sets Problem: Given a set I Z m, 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 = {a 0, …, a i-1 }. Let N(X) = (X – a 0 )… (X – a i-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(a i ) | g(a) for all a I, Add a i to S. Else Factor m = h 1 h 2 s.t. (h 1,h 2 ) =1. Recurse, combine solutions by CRT. p 2 q, pq 2 m =p e q f

11. Conclusions Understanding polynomials on {0,1} n over Z p. 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 Z p are the same as univariate polynomials mod p e. Can get tight lower bounds for symmetric polynomials. Used to rule out some algebraic approaches to better Ramsey graphs [G06]. Asymmetric case is wide open.