Randomness (and Pseudorandomness) Avi Wigderson IAS, Princeton
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!