Limitations of the Shpilka-Volkovich generator Arpita Korwar joint work with Herv´ e Fournier Universit´ e Denis Diderot - Paris 7 March 16, 2018 Arpita Korwar (Universit´ e Denis Diderot - Paris 7) Limitations of the Shpilka-Volkovich generator March 16, 2018 1 / 23
1 Polynomial Identity Testing 2 Shpilka-Volkovich (SV) Generator 3 Finding the Annihilating polynomial Arpita Korwar (Universit´ e Denis Diderot - Paris 7) Limitations of the Shpilka-Volkovich generator March 16, 2018 2 / 23
Polynomial Identity Testing Section 1 Polynomial Identity Testing Arpita Korwar (Universit´ e Denis Diderot - Paris 7) Limitations of the Shpilka-Volkovich generator March 16, 2018 3 / 23
Polynomial Identity Testing Polynomial Identity Testing ( PIT ) PIT: Is a given input polynomial identically zero? Arpita Korwar (Universit´ e Denis Diderot - Paris 7) Limitations of the Shpilka-Volkovich generator March 16, 2018 4 / 23
Polynomial Identity Testing Input model: Arithmetic circuits + × × y x 5 A natural and succinct representation of a polynomial. Arpita Korwar (Universit´ e Denis Diderot - Paris 7) Limitations of the Shpilka-Volkovich generator March 16, 2018 5 / 23
Polynomial Identity Testing Blackbox test (a.k.a. Hitting set) PIT can be classified according to how the polynomial is given to the algorithm. Arpita Korwar (Universit´ e Denis Diderot - Paris 7) Limitations of the Shpilka-Volkovich generator March 16, 2018 6 / 23
Polynomial Identity Testing Blackbox test (a.k.a. Hitting set) PIT can be classified according to how the polynomial is given to the algorithm. a 1 , a 2 , . . . , a h P ( a 1 ) , P ( a 2 ) , . . . , P ( a h ) P ∈ P a i = ( a i 1 , a i 2 , . . . , a in ). Arpita Korwar (Universit´ e Denis Diderot - Paris 7) Limitations of the Shpilka-Volkovich generator March 16, 2018 6 / 23
Polynomial Identity Testing Blackbox test (a.k.a. Hitting set) PIT can be classified according to how the polynomial is given to the algorithm. a 1 , a 2 , . . . , a h P ( a 1 ) , P ( a 2 ) , . . . , P ( a h ) P ∈ P a i = ( a i 1 , a i 2 , . . . , a in ). Example: PIT for univariate polynomials of degree bounded by d . Arpita Korwar (Universit´ e Denis Diderot - Paris 7) Limitations of the Shpilka-Volkovich generator March 16, 2018 6 / 23
Polynomial Identity Testing Blackbox test (a.k.a. Hitting set) PIT can be classified according to how the polynomial is given to the algorithm. a 1 , a 2 , . . . , a h P ( a 1 ) , P ( a 2 ) , . . . , P ( a h ) P ∈ P a i = ( a i 1 , a i 2 , . . . , a in ). Example: PIT for univariate polynomials of degree bounded by d . For n -variate P , a small-degree univariate substitution is enough. Arpita Korwar (Universit´ e Denis Diderot - Paris 7) Limitations of the Shpilka-Volkovich generator March 16, 2018 6 / 23
Polynomial Identity Testing Hitting set generator For a family P of n -variate, a polynomial map to k -variate polynomials ( f 1 ( y ) , f 2 ( y ) , . . . , f n ( y )) is a hitting set generator if for all polynomials P ( x 1 , x 2 , . . . , x n ) � = 0 ∈ P , P ( f 1 ( y ) , f 2 ( y ) , . . . , f n ( y )) � = 0. Final time complexity = ( δ d + 1) k , where d is the degree of f i s and the polys in P are of degree δ . Poly when k is constant. Quasipoly when k is log n . Arpita Korwar (Universit´ e Denis Diderot - Paris 7) Limitations of the Shpilka-Volkovich generator March 16, 2018 7 / 23
Shpilka-Volkovich (SV) Generator Section 2 Shpilka-Volkovich (SV) Generator Arpita Korwar (Universit´ e Denis Diderot - Paris 7) Limitations of the Shpilka-Volkovich generator March 16, 2018 8 / 23
Shpilka-Volkovich (SV) Generator Applications of the SV generator s O (1) -size hitting set for Read-once formulas [Shpilka and Volkovich, 2009, Minahan and Volkovich, 2016]. s O (1) -size hitting set for Constant-read multilinear formulas [Anderson et al., 2015]. s O (log log s ) -size hitting set for Commutative Read-once ABPs [Forbes et al., 2014]. Arpita Korwar (Universit´ e Denis Diderot - Paris 7) Limitations of the Shpilka-Volkovich generator March 16, 2018 9 / 23
Shpilka-Volkovich (SV) Generator Polynomials for Lagrange interpolation Building blocks of the SV generator. Choose ( a 1 , a 2 , . . . , a n ) such that all a i s are unique. ( y − a j ) L r ( y ) := � ( a r − a j ) . j � = r � 1 if b = a r , L r ( b ) = 0 if b ∈ { a 1 , a 2 , . . . , a n } , b � = a r . Arpita Korwar (Universit´ e Denis Diderot - Paris 7) Limitations of the Shpilka-Volkovich generator March 16, 2018 10 / 23
Shpilka-Volkovich (SV) Generator Shpilka-Volkovich map ( SV n , k )[Shpilka and Volkovich, 2009] SV n , 1 ( y , z ) : F [ x ] − → F [ y , z ], given by SV n , 1 ( y , z ) : x r �− → zL r ( y ) . SV n , 1 is a bivariate map. SV n , k ( y 1 , z 1 , y 2 , z 2 , . . . , y k , z k ) : F [ x ] − → F [ y , z ], given by → � k SV n , k ( y , z ) : x r �− i =1 z i L r ( y i ) . SV n , k is a 2 k -variate map. Arpita Korwar (Universit´ e Denis Diderot - Paris 7) Limitations of the Shpilka-Volkovich generator March 16, 2018 11 / 23
Shpilka-Volkovich (SV) Generator Some properties SV n , k of each x i is a linear form 1 in z . SV n , k is a hitting set generator for 2 k -sparse polynomials. SV n , k is a hitting set generator for degree- k polynomials. 1 constant part of the linear polynomial is 0 Arpita Korwar (Universit´ e Denis Diderot - Paris 7) Limitations of the Shpilka-Volkovich generator March 16, 2018 12 / 23
Shpilka-Volkovich (SV) Generator Question We want f such that SV n , k ( f ) = 0. What is the smallest degree polynomial that evaluates to 0 at SV n , k ? Conjecture: There exists a degree k + 1 multilinear polynomial on n = 2 k + 1 variables that maps to 0 on applying SV n , k . I.e. A multilinear, degree k + 1 annihilating polynomial for SV n , k exists. Arpita Korwar (Universit´ e Denis Diderot - Paris 7) Limitations of the Shpilka-Volkovich generator March 16, 2018 13 / 23
Finding the Annihilating polynomial Section 3 Finding the Annihilating polynomial Arpita Korwar (Universit´ e Denis Diderot - Paris 7) Limitations of the Shpilka-Volkovich generator March 16, 2018 14 / 23
Finding the Annihilating polynomial Finding a small annihilating polynomial - homogeneity Let � � f ( x 1 , x 2 , . . . , x n ) = γ S x r . r ∈ S S : | S |≤ k +1 → � k Recall that SV n , k ( y , z ) : x r �− i =1 z i L r ( y i ) . The polynomial SV n , k ( f ) can be seen as a polynomial in F [ y ][ z ]. After the map is applied, the z -degree of a degree- d monomial is d . So, without loss of generality, f is homogeneous. � � f ( x 1 , x 2 , . . . , x n ) = γ S x r . r ∈ S S : | S | = k +1 Arpita Korwar (Universit´ e Denis Diderot - Paris 7) Limitations of the Shpilka-Volkovich generator March 16, 2018 15 / 23
Finding the Annihilating polynomial Coefficients of each monomials The coefficient of any such monomial after the map should be 0. This gives a set of linear constraints on γ S s. Arpita Korwar (Universit´ e Denis Diderot - Paris 7) Limitations of the Shpilka-Volkovich generator March 16, 2018 16 / 23
Finding the Annihilating polynomial ... After some calculations After cleaning the conditions on the coefficients, our problem reduces to finding ( α R ) R : | R | = k such that the following linear constraints are satisfied: � ∀ S ⊆ [ n ] , | S | = k − 1 : α R = 0 R : | R | = k , S ⊆ R and � a R \ S · α R = 0 . R : | R | = k , S ⊆ R E.g. when k = 1, n = 2 k + 1 = 3. Then, we want to find ( α 1 , α 2 , α 3 ) such that � α i = 0 and � a i · α i = 0. Arpita Korwar (Universit´ e Denis Diderot - Paris 7) Limitations of the Shpilka-Volkovich generator March 16, 2018 17 / 23
Finding the Annihilating polynomial The ( α R ) R s that satisfy the first set of constraints are nullvectors of the inclusion matrix M with the rows indexed by { S : | S | = k − 1 } and the columns indexed by { R : | R | = k } with � 1 if S ⊆ R , M S , R = 0 otherwise. The ( α R ) R s that satisfy the second set of constraints are null vectors of N , where � a R \ S if S ⊆ R , N S , R = 0 otherwise. When k = 1, n = 2 k + 1 = 3, M = [1 1 1] and N = [ a 1 a 2 a 3 ] . Arpita Korwar (Universit´ e Denis Diderot - Paris 7) Limitations of the Shpilka-Volkovich generator March 16, 2018 18 / 23
Finding the Annihilating polynomial N = D ′ MD , where � [ n ] D ′ is a � S , S = � diagonal matrix. D ′ i ∈ S 1 / a i = 1 / a S . k − 1 � [ n ] � diagonal matrix. D R , R = � D is a i ∈ R a i = a R . k D ′ does not affect the nullvector of N . Hence, we need to show that N ( M ) ∩ N ( MD ) � = ∅ . � n � n � � N ( M ) has dimension − and has been described by k k − 1 [Graham et al., 1980] and others. Arpita Korwar (Universit´ e Denis Diderot - Paris 7) Limitations of the Shpilka-Volkovich generator March 16, 2018 19 / 23
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