Factors of Low Individual Degree Polynomials this Work 3 Background - PowerPoint PPT Presentation

Rafael Oliveira Princeton University Factors of Low Individual Degree Polynomials

this Work 3 Background 2 1 Conclusion Conclusion Introduction Introduction Main Ideas of Main Ideas of & Open & Open and and this Work Questions Questions Background Outline

1 Introduction & Background Arithmetic Circuits and Factoring

Factoring in Real Life Basic routine in many tasks: Fast decoding of Reed Solomon Codes Used to compute: • Primary Decompositions of Ideals • Gröbner Bases, etc. Can be done efficiently in (randomized) poly time! In theory, interested in: • Derandomization • Parallel complexity • Structure of factors

Arithmetic Circuits Definition by picture Main measures: f = y 2 – x 2 × Size = # edges + + -1 x y x Depth = length of longest path from root to leaf Model captures our notion of algebraic computation Many interesting polynomials have succinct rep. in this model, such as . Det n ( X ) , σ k ( x 1 , . . . , x n ) It is a major open question whether has a Perm n succinct rep. in this model.

Polynomial Factorization Problem: Given a circuit for , where P ( x ) P ( x ) = g 1 ( x ) g 2 ( x ) . . . g k ( x ) output circuits for g 1 ( x ) , g 2 ( x ) , . . . , g k ( x ) • [LLL ’82, Kal ’89]: if is computed by a small circuit, then P ( x ) so are the factors . Moreover g 1 ( x ) , g 2 ( x ) , . . . , g k ( x ) Kaltofen gives a randomized algorithm to compute factors • Fundamental consequences to: • Circuit Complexity & Pseudorandomness: [KI ’04, DSY ’09] • Coding Theory: [Sud ’97, GS’06] • Geometric Complexity Theory: [Mul’13]

What About Depth? [Kaltofen ’89]: factorization behaves nicely w.r.t. size. What about depth? More generally: Structure: given polynomial in circuit class , which P ( x ) C classes efficiently compute the factors of ? C ∗ P ( x ) • If has a small depth circuit, do its factors have small P ( x ) depth circuits? • If has a small formula, do its factors have small formula? P ( x )

Gap of Understanding If is a polynomial with monomials and degree P ( x ) d s Kaltofen & depth reduction Factors of computed by formulas of P ( x ) depth and 4 size . exp( ˜ √ O ( d )) General depth reductions [AV’08, Koi’12, GKKS’13, Tav’13] give subexponential gap. Can this be improved?

Why Bound Individual Degrees? Polynomials with bounded ind. deg. form a very rich class, which generalizes multilinear polynomials. Well studied, works of [Raz ’06, RSY ’08, Raz ’09, SV ’10, SV ’11, KS ’15 2 , KCS’15, KCS’16]. Bounded Little is Individual known Degree Step towards understanding Multilinear general case Trivial

This Work Theorem: If is a polynomial which: P ( x ) • has individual degrees bounded by , r • is computed by a circuit (formula) of size & depth d s Then any factor of is computed by a circuit f ( x ) P ( x ) (formula) of size poly ( n r , s ) & depth d + 5 Furthermore, result provides a randomized algorithm for computing all factors of in time poly ( n r , s ) P ( x )

Prior Work [DSY ’09]: if is computed by a circuit of size , depth d P ( x , y ) s is bounded by • deg y ( P ) r Then its factors of the form have circuits of y − g ( x ) depth and size d + 3 poly ( n r , s ) Extend Hardness vs Randomness approach of [KI ’04] to bounded depth circuits. [DSY ’09] noticed that only factors of the form are y − g ( x ) important to extend [KI ’04] to bounded depth.

2 Main Ideas of this Work Lifting Root Approximation Reversal Outline

Lifting Suppose input is: P ( x , y ) = ( y − g 1 ( x ))( y − g 2 ( x )) Where µ 1 = g 1 ( 0 ) , µ 2 = g 2 ( 0 ) and µ 1 6 = µ 2 How do we factor in this case? Can try to build the homogeneous parts of one g i ( x ) at a time.

Lifting Note that: P ( 0 , y ) = ( y − µ 1 )( y − µ 2 ) Which we know how to factor. Hence, found the constant terms of the roots. How to find the linear terms of the roots?

Lifting Setting in the input polynomial: y = µ 1 P ( x , µ 1 ) = ( µ 1 − g 1 ( x ))( µ 1 − g 2 ( x )) µ 1 6 = µ 2 Since , the constant term of µ 1 − g 2 ( x ) is nonzero, whereas the constant term of µ 1 − g 1 ( x ) is zero! Hence, linear term of equals the P ( x , µ 1 ) linear term of , up to a constant factor. g 1 ( x )

Lifting Continuing this way, we can recover the roots and factor the input polynomial. Hensel Lifting/Newton Iteration. Pervasive in factoring algorithms, such as [Zas ’69, Kal ’89, DSY ’09] , and many others. [DSY ’09]: if is computed by a circuit of size , depth d P ( x , y ) s is bounded by • deg y ( P ) r Then its factors of the form have circuits of y − g ( x ) depth and size d + 3 poly ( n r , s )

Lifting Two main issues What if does not factor into linear • P ( x , y ) factors in ? y Approximate roots in algebraic closure of F ( x ) by low degree polynomials in . F [ x ] What if is not monic in ? • P ( x , y ) y Use reversal to reduce the number of variables

2 Main Ideas of this Work Lifting Root Approximation Reversal Outline

Approximation Polynomials Suppose input is: r − 1 P ( x , y ) = y r + X P i ( x ) y i i =0 Which does not factor into linear factors. Let P ( x , y ) = f ( x , y ) Q ( x , y ) where k − 1 f ( x , y ) = y k + X f i ( x ) y i i =0 Is irreducible and does not divide the other factor.

Approximation Polynomials Any polynomial factors completely in the algebraic closure of ! F ( x ) r Y P ( x , y ) = ( y − ϕ i ( x )) i =1 k Y f ( x , y ) = ( y − ϕ i ( x )) i =1 ϕ i ( x ) Where each is a “function” on the variables x

Approximation Polynomials Since and share roots , can ϕ i ( x ) P ( x , y ) f ( x , y ) try to approximate these roots by polynomials g i,t ( x ) of degree such that t f ( x , g i,t ( x )) only has terms of degree higher than . t Definition: we say that f ( x ) = t g ( x ) if the polynomial only has terms of degree f ( x ) − g ( x ) higher than . t

Approximation Polynomials Definition: we say that f ( x ) = t g ( x ) if the polynomial only has terms of degree f ( x ) − g ( x ) higher than . t This definition gives us a topology: - Two polynomials are close if they agree on low degree parts - Can use this topology to derive analogs of Taylor series for elements of !(#). Can “approximate” elements of !(#) by polynomials!

Approximation Polynomials If we can find for each root of ϕ i ( x ) g i,t ( x ) f ( x , y ) such that f ( x , g i,t ( x )) = t 0 Then we can prove the following: Lemma: the polynomials are such that g i,t ( x ) k Y f ( x , y ) = t ( y − g i,t ( x )) i =1 Can convert approximations to the roots into approximations to the factors!

Approximation Polynomials How do we obtain these polynomials ? g i,t ( x ) Since each is also a root of , can ϕ i ( x ) P ( x , y ) obtain from via lifting! g i,t ( x ) P ( x , y ) Looking at our parameters: k Y f ( x , y ) = t ( y − g i,t ( x )) i =1 Depth size d + 4 poly ( n r , s ) With standard techniques, can recover f ( x , y ) Observation: for the general case, need to keep the from k Y product top fan in! ( y − g i,t ( x )) i =1

2 Main Ideas of this Work Lifting Root Approximation Reversal Outline

Set Up Suppose input now is: r X P i ( x ) y i , P 0 ( x ) P r ( x ) 6 = 0 P ( x , y ) = i =0 Let P ( x , y ) = f ( x , y ) Q ( x , y ) where k X f i ( x ) y i f ( x , y ) = i =0 is irreducible and does not divide the other factor.

The Game Plan Reduce to the monic case: ! r − 1 P i ( x ) y r + X P r ( x ) y i P ( x , y ) = P r ( x ) · i =0 k − 1 ! f i ( x ) y k + X f k ( x ) y i f ( x , y ) = f k ( x ) · i =0 1. Recover from by some kind of induction f k ( x ) P r ( x ) 2. Recover the part of that depends on f ( x , y ) y

Naïve Recursion Let have individual degrees , variables P ( x , y ) s r n and computed by circuit of size and depth d Let be such that: T ( s, n ) f ( x , y ) | P ( x , y ) There exists with Φ ( x , y ) Φ ( x , y ) = t f ( x , y ) • depth d + 4 • size ≤ T ( s, n ) • top fan in product gate

Naïve Recursion Our recurrence becomes: T ( s, n ) ≤ T (3 rs, n − 1) + poly ( n r , s ) Recover from f k ( x ) P r ( x ) Size of part depending on y After steps, our recursion would become t T ( s, n ) ≤ T ((3 r ) t s, n − t ) + Ω ( n tr s ) Exponential when ! t ∼ n