Linear Systems With Composite Moduli Arkadev Chattopadhyay (University of Toronto) Joint with: Avi...

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Transcript of Linear Systems With Composite Moduli Arkadev Chattopadhyay (University of Toronto) Joint with: Avi...

Linear Systems With Composite Moduli

Arkadev Chattopadhyay (University of Toronto)

Joint with:Avi Wigderson

The Problem.

Question: What can we say about the boolean solution set of such systems?

Outline of Talk.

Motivation. Natural problem. Circuits with MOD Gates . Surprising power of composite moduli.

Our Result. Some Circuit Consequences. High Level Argument.

Circuits With MOD Gates.

Theorem (Razborov’87, Smolensky’87). Addition of MODp gates to bounded-depth circuits, does not help to compute function MODq , if (p,q)=1 and p is a prime power.

Nagging Question: Is ‘and p is a prime power’ essential?

Smolensky’s Conjecture.

Conjecture: MODq needs exponential size circuits of constant depth having AND/OR/MODm gates if (m,q)=1.

Not known even for m=6.

Barrier: Prove any non-trivial lower bounds for AND/OR/MOD6.

The Weakness of Primes.

MODp Gates

Conclusion: AND cannot be computed by constant-depthcircuits having only MODp gates (in any size).

Fermat’s Gift for prime p:

The Power of Composites.

MODm MODm MODm

MODm

C

Fact: Every function can be computed by depth-two circuits having only MODm gates in exponential size, when m is a product of two distinct primes.

Power of Polynomials Modulo Composites.

Defn: Let P(x) reperesent f over Zm, w.r.t A:

Def: The MODm -degree of f is the degree of minimal degree P representing f, w.r.t. A.

Fact: The MODm -degree of OR is (n).

Power of Composite Moduli.

Theorem(Barrington-Beigel-Rudich’92): MODm-degree of OR is O(n1/t) if m has t distinct prime factors, i.e. for m=6 it is .

Theorem(Green’95, BBR’92): MODm -degree of MODq is (n).

Theorem(Hansen’06): Let m,q be co-prime. MODm-degree of MODq is O(n1/t) if m has t distinct prime factors, as long as m satisfies certain condition, i.e. MOD35 – degree of PARITY is .

Can Many Polynomials Help?

Defn: P represents f if:

Question: What is the relationship of t and deg(P)?Observation: n linear polynomials can represent AND and NOR functions.

Linear Systems: Our Result.

Aiµ Zm

Theorem: The boolean solution set, , lookspseudorandom to the MODq function.

(independent of t)

Circuit Consequence.

Corollary: Exponential size needed by MAJ ± AND ± MODm to compute MODq, if m=p1p2 and m,q co-prime.

(Solves Beigel-Maciel’97 for such m).

Remark: Obtaining exponential lower bounds on size ofMAJ ± MODm ± AND is wide open.

Proof Strategy.

Gradual generalization leading to result. Singleton Accepting Sets. Low rank systems. Low rigid rank

Deal with high rigid rank separately.

Exponential sums

(Extend Grigoriev-Razborov).

of Bourgain.

Singleton Accepting Set.

Assume Ai={0} Set ofBoolean solns

A linear form

Fourier Expansion

Finishing Off For Singleton Accepting Set.

Exponential sum reduction

(Goldman, Green)

Non-Singleton Accepting Sets.

+

j · (m-1)t singleton systems

+

Union Bound:

Low Rank Systems.

Shouldn’t High Rank be Easy?

Tempting Intuition from linear algebra: If L has high rank, then the size of the solution set BL should be a small fraction of the universe, and hence correlation w.r.t MODq is small.

Caveat: Our universe is only the boolean cube!

Example:

rank is n.

BL ´ {0,1}n

Sparse Linear Systems.

Observation: For each i, there exists a polynomial Pi over Zm of degree at most k, such that

Polynomial Systems With Singleton Accepting Set.

Degree · k

Relevant Sum for Correlation:

Bourgain’s breakthrough:

Low Rigid Systems.

We can combine low rank and sparsity into rigidity:

rank=r k-sparse(k,r)-sparse

Strategy:

Rank With Respect To Individual Prime Factors.

Chinese Remaindering

Low Rigidity Over Prime Fields is Enough.

Otherwise: High Rigid Rank.

Theorem: If L does not admit a partition into L1 [ L2 such that L1 (and L2) has k-rigid rank over Z (resp. Z ) at most r. Then,

Extends ideas of Grigoriev-Razborov for arithmetic circuits.

Combining the Two, We Are Done.

Question: What about m=30?

Answer: Recently, in joint work with Lovett, we deal with arbitrary m.

THANK YOU!