Linear Systems With Composite Moduli

1. Linear Systems With Composite Moduli Arkadev Chattopadhyay (University of Toronto) Joint with: Avi Wigderson TexPoint fonts used in EMF. Read the TexPoint manual before you delete this box.: AAAAAAA

2. The Problem. Question: What can we say about the boolean solution set of such systems?

3. Outline of Talk. • Motivation. • Natural problem. • Circuits with MOD Gates . • Surprising power of composite moduli. • Our Result. • Some Circuit Consequences. • High Level Argument.

4. 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?

5. 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.

6. The Weakness of Primes. MODp Gates Fermat’s Gift for prime p: Conclusion: AND cannot be computed by constant-depth circuits having only MODp gates (in any size).

7. 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.

8. 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).

9. 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 .

10. 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.

11. Linear Systems: Our Result. AiµZm Theorem: The boolean solution set, , looks pseudorandom to the MODq function. (independent of t)

12. 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 of MAJ ± MODm± AND is wide open.

13. Proof Strategy. Gradual generalization leading to result. • Singleton Accepting Sets. • Low rank systems. • Low rigid rank • Deal with high rigid rank separately. Exponential sums of Bourgain. (Extend Grigoriev-Razborov).

14. Singleton Accepting Set. Assume Ai={0} Set of Boolean solns A linear form Fourier Expansion

15. Finishing Off For Singleton Accepting Set. Exponential sum reduction (Goldman, Green)

16. Non-Singleton Accepting Sets. Union Bound: + + j· (m-1)t singleton systems

17. Low Rank Systems.

18. 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

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

20. Polynomial Systems With Singleton Accepting Set. Degree · k Relevant Sum for Correlation: Bourgain’s breakthrough:

21. Low Rigid Systems. We can combine low rank and sparsity into rigidity: rank=r k-sparse (k,r)-sparse Strategy:

22. Rank With Respect To Individual Prime Factors. Chinese Remaindering

23. Low Rigidity Over Prime Fields is Enough.

24. 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.

25. 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!