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On obtaining a polynomial as a projection of another polynomial. Neeraj Kayal Microsoft Research. A dream. Conjecture #1: The determinantal complexity of the permanent is superpolynomial Conjecture #2: The arithmetic complexity of matrix multiplication is

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On obtaining a polynomial as a projection of another polynomial

On obtaining a polynomial as a projection of another polynomial

NeerajKayalMicrosoft Research


A dream
A dream polynomial

  • Conjecture #1: The determinantal complexity of the permanent is superpolynomial

  • Conjecture #2: The arithmetic complexity of matrix multiplication is

  • Conjecture #3: The depth three complexity of the permanent is


Projections of polynomials
Projections of polynomials polynomial

  • Definition (Valiant): An n-variate polynomial is said to be a projection of an m-variate polynomial iffor some matrix A and vector . That is

    ,

    where the ’s are affine functions.

    (Affine function : )

    We will say that such a projection is invertible if is invertible.


Some families of polynomials
Some families of polynomials polynomial

  • Determinant:

  • Permanent:

  • Sum of Products:


Some families of polynomials1
Some families of polynomials polynomial

  • Trace of Matrix Multiplicationwhere X, Y and Z are matrices

  • Sum of powers:

  • Elementary Symmetric Polynomial:


The dream
The dream polynomial

  • Conjecture #1: The determinantal complexity of the permanent is superpolynomial

  • Conjecture #2: The arithmetic complexity of matrix multiplication is

  • Conjecture #3: The depth three complexity of the permanent is


The dream1
The dream polynomial

  • Conjecture #1:If is a projection of then m is superpolynomial in n.

  • Conjecture #2: The arithmetic complexity of matrix multiplication is

  • Conjecture #3: The depth three complexity of the permanent is


The dream2
The dream polynomial

  • Conjecture #1:If is a projection of then m is superpolynomial in n.

  • Conjecture #2: can be expressed as a projection of for m =

  • Conjecture #3: The depth three complexity of the permanent is


The dream3
The dream polynomial

  • Conjecture #1: If is a projection of then m is superpolynomial in n.

  • Conjecture #2: can be expressed as a projection of for m =

  • Conjecture #3:If is a projection of then .


A computational problem
A computational problem polynomial

  • POLY_PROJECTION: Given polynomials and , determine if is a projection of and if so find A and b such that


Some conventions
Some conventions polynomial

  • We consider polynomials over , the field of complex numbers.

  • Polynomials are encoded as arithmetic circuits (unless mentioned otherwise).


Unfortunately
Unfortunately … polynomial

Theorem: POLY_PROJECTION is NP-complete.


A more modest ambition
A more modest ambition polynomial

  • Given a polynomial and an integer n determine if is a projection of say .

  • Given a polynomial and integers n,d determine if is a projection of .


A conjecture
A conjecture polynomial

Conjecture (Scott Aaronson): A random low rank projection of is indistinguishable from a truly random polynomial.

More precisely, if are random m-variate affine functions then is indistuinguishable from a random m-variate polynomial of degree n.


Polynomial equivalence
Polynomial Equivalence polynomial

POLY_EQUIVALENCE: Given polynomials and , determine if for some invertible matrix .

Theorem (Agrawal-Saxena): POLY_EQUIVALENCE is at least as hard as graph isomorphism.

(Probably much harder than graph isomorphism.)


Lowering our sights further polynomial

  • Given a polynomial and an integer n determine if is equivalent to .

  • Given a polynomial and integers n,d determine if is equivalent to .


Polynomial equivalence results
Polynomial Equivalence Results polynomial

Theorem #1: There is an efficient randomized algorithm that determines if a given polynomial is an invertible projection of .

Theorem #2: There is an efficient randomized algorithm that determines if a given polynomial is an invertible projection of .


Invertible projections
Invertible Projections polynomial

Theorem #3: There is an efficient randomized algorithm that determines if a given polynomial is an invertible projection of .

Theorem #4: There is an efficient randomized algorithm that determines if a given polynomial is an invertible projection of .


Invertible projections1
Invertible Projections polynomial

Theorem #3: There is an efficient randomized algorithm that determines if a given polynomial is an invertible projection of .

Theorem #4: There is an efficient randomized algorithm that determines if a given polynomial is an invertible projection of .

… and so on for the other families of polynomials


Preliminaries
Preliminaries polynomial

Fact:Given polynomials in randomized polynomial time we can determine a basis of the vector space

Fact:Given an n-variate polynomial , in randomized polynomial time we can find an invertible matrix A such that has fewer than n variables, if such an exists.


Equivalence to
Equivalence to polynomial

Problem restatement: Given an n-variate polynomial of degree d, determine if there exist n linearly independent affine functions such that


Introducing the hessian
Introducing the Hessian polynomial

Definition: The hessian of an n-variate polynomial g is the following matrix:

(here g is a shorthand for the second order derivative )

Property:If then


Using the hessian
Using the Hessian polynomial

Property: If then

In particular and

Fact:


Using the hessian1
Using the Hessian polynomial

In particular, if

In particular, then

This gives the algorithm for equivalence to


Equivalence to1
Equivalence to polynomial

Problem restatement: Given an n-variate polynomial of degree d, determine if there exist n linearly independent affine functions such that

Fact: is a multilinear polynomial, so

Fact: All the other second-order partial derivatives of are linearly independent.


Outline of algorithm
Outline of Algorithm polynomial

  • Input: A polynomial

  • Compute all the second order partial derivatives of .

  • Compute all the linear dependencies between these second order partials of f.

  • This gives us a linear space of second order differential operators which vanish at f.

  • A second order differential operator naturally corresponds to a matrix. Find a basis of the corresponding linear space of matrices consisting of rank one matrices.


Equivalence to2
Equivalence to polynomial

Problem restatement: Given an -variate polynomial of degree n, determine if there exist n linearly independent affine functions such that

The following approach was suggested by Mulmuley and Sohoni.

Fact: The group of symmetries of is continuous.


Symmetries of
Symmetries of polynomial

Fact:If we take to be aribtarily close to 1 then we get a matrix A arbitarily close to identity such that


Symmetries of1
Symmetries of polynomial

Let be a formal variable with . Then there exist nontrivial matrices A such that

Fact: For any polynomial , the set of matrices A such that

forms a vector space.

Fact: For a given , a basis for this space can be computed in random polynomial time.


Outline of algorithm1
Outline of Algorithm polynomial

Input: A polynomial

Compute a basis for the space of matrices A satisfying .

Let the basis be . These matrices act on an dimensional vector space V.

Find all subspaces such that for all .

Infer the appropriate equivalence of and from the 1-dimensional invariant subspaces U.


That’s all fine, but what about POLY_PROJECTION? polynomial

Theorem (Kaltofen): Projections of can be reconstructed efficiently. (This is just polynomial factoring)

Theorem (follows quickly from the work of Kleppe): Given a univariatepolynomial of degree d and an integer s, we can efficiently compute ’s and ’s (if they exist) such that


Scott s conjecture
Scott’s conjecture polynomial

Conjecture (Scott Aaronson): If are random m-variate affine functions then

is indistuinguishable from a random m-variate polynomial of degree n.

In other words, solving random instances of projections of is conjecturally hard.


Digression background of scott s conjecture
Digression: Background of Scott’s conjecture polynomial

Observation: All the known bound proofs (for projections of a family of polynomials ) follow the following strategy:

Step 1. Find an “efficiently computable” property P that is satisfied by projections of small polynomials from .

Step 2. (Relatively easy) Find an explicit polynomial not having the property P.


Example mignon and ressayre
Example: Mignon and polynomialRessayre

Theorem (Mignon and Ressayre):

If is a projection of then m is at least .

Lemma: Let be a projection of . Then for any zero of ,


Example projections of
Example: Projections of polynomial

Lemma (implicit in Grigoriev-Karpinski): Let be a projection of . Then the set of all possible partial derivatives of contain at most linearly independent polynomials.


Consequences of scott s conjecture
Consequences of Scott’s conjecture polynomial

  • If the conjecture is true then no current proof technique will yield superpolynomial formula size lower bound.

  • If the conjecture is false then we would have obtained a good handle on the determinant versus permanent conjecture.


Probing a little deeper
Probing a little deeper polynomial

  • For families of polynomials such as we have lower bounds.

  • Given , where the ’s are m-variate affine functions randomly chosen, can we efficiently recover the ’s?


Poly projection on the average
POLY_PROJECTION on the average polynomial

Theorem: Projections of can be reconstructed efficiently on the average.

Theorem: For bounded n, projections of can be reconstructed efficiently on the average.


Summary
Summary polynomial

  • POLY_PROJECTION and POLY_EQUIVALENCE are difficult computational problems in general.

  • Empirically, for most families of polynomials that we actually care about and/or encounter in practice, we can solve POLY_EQUIVALENCE efficiently.

  • Empirically, efficient average-case algorithms for POLY_PROJECTION of some family seem to be closely related to lower bounds for projections of .


Conclusion an easier open problem
Conclusion: An easier(?) open problem polynomial

Conjecture (Amir Shpilka): If is a projection of then .

THE END


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