Polynomial equivalence problem for sums of affine powers ISSAC 2018 - PowerPoint PPT Presentation

Polynomial equivalence problem for sums of affine powers ISSAC 2018 Ignacio Garcia-Marco 1 , Pascal Koiran 2 , Timoth´ ee Pecatte 2 1 Universidad de la Laguna, 2 LIP, ´ Ecole Normale Sup´ erieure de Lyon 1 / 21

Models of interest 1 / 21

Classical models: the Waring model Model (Waring decomposition) k � α i ( x − a i ) d with α i , a i ∈ F , e i ∈ N , and d = deg( f ) i =1 2 / 21

Classical models: the Waring model Model (Waring decomposition) k � α i ( x − a i ) d with α i , a i ∈ F , e i ∈ N , and d = deg( f ) i =1 Reconstruction problem: Given f , one wants an optimal expression of f in this model (with the minimum value of k possible). 2 / 21

Classical models: the Waring model Model (Waring decomposition) k � α i ( x − a i ) d with α i , a i ∈ F , e i ∈ N , and d = deg( f ) i =1 Reconstruction problem: Given f , one wants an optimal expression of f in this model (with the minimum value of k possible). Solution: Sylvester (1851) 2 / 21

Classical models: the Waring model Model (Waring decomposition) k � α i ( x − a i ) d with α i , a i ∈ F , e i ∈ N , and d = deg( f ) i =1 Reconstruction problem: Given f , one wants an optimal expression of f in this model (with the minimum value of k possible). Solution: Sylvester (1851) Lots of partial solutions for the multivariate version: Auslander, Hirschowitz, Boij, Carlini, Geramita, Oeding, Landsberg, Sturmfels, . . . 2 / 21

Classical models: the Sparsest Shift model Model (Sparsest shift) k � α i ( x − a ) e i with α i , a ∈ F , e i ∈ N . i =1 3 / 21

Classical models: the Sparsest Shift model Model (Sparsest shift) k � α i ( x − a ) e i with α i , a ∈ F , e i ∈ N . i =1 Reconstruction problem: Given f , one wants an optimal expression of f in this model. 3 / 21

Classical models: the Sparsest Shift model Model (Sparsest shift) k � α i ( x − a ) e i with α i , a ∈ F , e i ∈ N . i =1 Reconstruction problem: Given f , one wants an optimal expression of f in this model. Solution: Borodin-Tiwari (1991), Giesbrecht-Roche (2010) 3 / 21

Classical models: the Sparsest Shift model Model (Sparsest shift) k � α i ( x − a ) e i with α i , a ∈ F , e i ∈ N . i =1 Reconstruction problem: Given f , one wants an optimal expression of f in this model. Solution: Borodin-Tiwari (1991), Giesbrecht-Roche (2010) For the multivariate version: Grigoriev-Karpinski (1993), Giesbrecht-Kaltofen-Lee (2003). 3 / 21

Sums of affine powers Model (Univariate Σ ∧ Σ) k � α i ( x − a i ) e i with α i , a i ∈ F i =1 4 / 21

Sums of affine powers Model (Univariate Σ ∧ Σ) k � α i ( x − a i ) e i with α i , a i ∈ F i =1 We now consider f an multivariate polynomial with coefficients in F , this is, f ∈ F [ X ]. 4 / 21

Sums of affine powers Model (Univariate Σ ∧ Σ) k � α i ( x − a i ) e i with α i , a i ∈ F i =1 We now consider f an multivariate polynomial with coefficients in F , this is, f ∈ F [ X ]. Model (Multivariate Σ ∧ Σ) k � α i ( a i , 1 x 1 + . . . + a i , n x n + a i , 0 ) e i i =1 4 / 21

Sums of affine powers Model (Univariate Σ ∧ Σ) k � α i ( x − a i ) e i with α i , a i ∈ F i =1 We now consider f an multivariate polynomial with coefficients in F , this is, f ∈ F [ X ]. Model (Multivariate Σ ∧ Σ) k � ℓ e i with ℓ i ∈ F [ X ] , deg( ℓ i ) ≤ 1 i i =1 4 / 21

Goal: reconstruction algorithms Problem Given a polynomial f ∈ F [ X ] , compute the exact value s = AffPow F ( f ) and a decomposition with s terms. 5 / 21

Goal: reconstruction algorithms Problem Given a polynomial f ∈ F [ X ] , compute the exact value s = AffPow F ( f ) and a decomposition with s terms. AffPow( f ) = s Algorithm Blackbox for f i =1 ℓ e i � s f = i 5 / 21

Goal: reconstruction algorithms Problem Given a polynomial f ∈ F [ X ] , compute the exact value s = AffPow F ( f ) and a decomposition with s terms. AffPow( f ) = s Algorithm Blackbox for f i =1 ℓ e i � s f = i • Change of basis 5 / 21

Goal: reconstruction algorithms Problem Given a polynomial f ∈ F [ X ] , compute the exact value s = AffPow F ( f ) and a decomposition with s terms. AffPow( f ) = s Algorithm Blackbox for f i =1 ℓ e i � s f = i • Change of basis • Solving linear systems 5 / 21

Goal: reconstruction algorithms Problem Given a polynomial f ∈ F [ X ] , compute the exact value s = AffPow F ( f ) and a decomposition with s terms. AffPow( f ) = s Algorithm Blackbox for f i =1 ℓ e i � s f = i • Change of basis • Solving linear systems • Factorization 5 / 21

Goal: reconstruction algorithms Problem Given a polynomial f ∈ F [ X ] , compute the exact value s = AffPow F ( f ) and a decomposition with s terms. AffPow( f ) = s Algorithm Blackbox for f i =1 ℓ e i � s f = i • Change of basis • PIT • Solving linear systems • Factorization 5 / 21

Goal: reconstruction algorithms Problem Given a polynomial f ∈ F [ X ] , compute the exact value s = AffPow F ( f ) and a decomposition with s terms. AffPow( f ) = s Algorithm Blackbox for f i =1 ℓ e i � s f = i • Change of basis • PIT • Solving linear systems • Derivatives • Factorization 5 / 21

Goal: reconstruction algorithms Problem Given a polynomial f ∈ F [ X ] , compute the exact value s = AffPow F ( f ) and a decomposition with s terms. AffPow( f ) = s Algorithm Blackbox for f i =1 ℓ e i � s f = i • Change of basis • PIT • Solving linear systems • Derivatives • Factorization • Homogeneous components 5 / 21

Essential variables f ( x 1 , x 2 , x 3 ) = x 3 1 + x 2 1 x 2 − 2 x 2 1 x 3 − 2 x 1 x 2 x 3 + x 1 x 2 3 + x 2 x 2 3 6 / 21

Essential variables f ( x 1 , x 2 , x 3 ) = x 3 1 + x 2 1 x 2 − 2 x 2 1 x 3 − 2 x 1 x 2 x 3 + x 1 x 2 3 + x 2 x 2 3 = ( x 2 + x 3 )( x 1 − x 3 ) 2 + ( x 1 − x 3 ) 3 6 / 21

Essential variables f ( x 1 , x 2 , x 3 ) = x 3 1 + x 2 1 x 2 − 2 x 2 1 x 3 − 2 x 1 x 2 x 3 + x 1 x 2 3 + x 2 x 2 3 = ( x 2 + x 3 )( x 1 − x 3 ) 2 + ( x 1 − x 3 ) 3 g ( y 1 , y 2 ) = f ( z 1 , y 1 + y 2 − z 1 , z 1 − y 2 ) = y 1 y 2 2 + y 3 2 6 / 21

Essential variables f ( x 1 , x 2 , x 3 ) = x 3 1 + x 2 1 x 2 − 2 x 2 1 x 3 − 2 x 1 x 2 x 3 + x 1 x 2 3 + x 2 x 2 3 = ( x 2 + x 3 )( x 1 − x 3 ) 2 + ( x 1 − x 3 ) 3 g ( y 1 , y 2 ) = f ( z 1 , y 1 + y 2 − z 1 , z 1 − y 2 ) = y 1 y 2 2 + y 3 2 Proposition (Carlini) For a polynomial f ∈ F [ X ] , we have � ∂ f � EssVar ( f ) = dim F | 1 ≤ i ≤ n ∂ x i 6 / 21

Essential variables f ( x 1 , x 2 , x 3 ) = x 3 1 + x 2 1 x 2 − 2 x 2 1 x 3 − 2 x 1 x 2 x 3 + x 1 x 2 3 + x 2 x 2 3 = ( x 2 + x 3 )( x 1 − x 3 ) 2 + ( x 1 − x 3 ) 3 g ( y 1 , y 2 ) = f ( z 1 , y 1 + y 2 − z 1 , z 1 − y 2 ) = y 1 y 2 2 + y 3 2 Proposition (Carlini) For a polynomial f ∈ F [ X ] , we have � ∂ f � EssVar ( f ) = dim F | 1 ≤ i ≤ n ∂ x i Eliminating redundant variables can be done with a randomized polynomial time algorithm [Kayal] ⇒ we will assume that f is regular . 6 / 21

Essential variables f ( x 1 , x 2 , x 3 ) = x 3 1 + x 2 1 x 2 − 2 x 2 1 x 3 − 2 x 1 x 2 x 3 + x 1 x 2 3 + x 2 x 2 3 = ( x 2 + x 3 )( x 1 − x 3 ) 2 + ( x 1 − x 3 ) 3 g ( y 1 , y 2 ) = f ( z 1 , y 1 + y 2 − z 1 , z 1 − y 2 ) = y 1 y 2 2 + y 3 2 Proposition (Carlini) For a polynomial f ∈ F [ X ] , we have � ∂ f � EssVar ( f ) = dim F | 1 ≤ i ≤ n ∂ x i Eliminating redundant variables can be done with a randomized polynomial time algorithm [Kayal] ⇒ we will assume that f is regular . EssVar( f ) ≤ AffPow( f ) 6 / 21

From reconstruction to polynomial equivalence Take f such that EssVar( f ) = AffPow( f ), i.e. f = � n i =1 ℓ e i i . 7 / 21

From reconstruction to polynomial equivalence Take f such that EssVar( f ) = AffPow( f ), i.e. f = � n i =1 ℓ e i i . Set     [ ℓ 1 ] ℓ 1 (0) . . . . A = b =  ,     . .    [ ℓ n ] ℓ n (0) so that 7 / 21

From reconstruction to polynomial equivalence Take f such that EssVar( f ) = AffPow( f ), i.e. f = � n i =1 ℓ e i i . Set     [ ℓ 1 ] ℓ 1 (0) . . . . A = b =  ,     . .    [ ℓ n ] ℓ n (0) so that n � x e i f ( X ) = g ( A · X + b ) with g = i i =1 7 / 21

From reconstruction to polynomial equivalence Take f such that EssVar( f ) = AffPow( f ), i.e. f = � n i =1 ℓ e i i . Set     [ ℓ 1 ] ℓ 1 (0) . . . . A = b =  ,     . .    [ ℓ n ] ℓ n (0) so that n � x e i f ( X ) = g ( A · X + b ) with g = i i =1 Definition (Polynomial equivalence) f ∼ g if f ( X ) = g ( A · X ) with A ∈ GL n ( F ) f ≡ g if f ( X ) = g ( A · X + b ) with A ∈ GL n ( F ) , b ∈ F n 7 / 21

From reconstruction to polynomial equivalence Take f such that EssVar( f ) = AffPow( f ), i.e. f = � n i =1 ℓ e i i . Set     [ ℓ 1 ] ℓ 1 (0) . . . . A = b =  ,     . .    [ ℓ n ] ℓ n (0) so that n � x e i f ( X ) = g ( A · X + b ) with g = i i =1 Definition (Polynomial equivalence) f ∼ g if f ( X ) = g ( A · X ) with A ∈ GL n ( F ) f ≡ g if f ( X ) = g ( A · X + b ) with A ∈ GL n ( F ) , b ∈ F n n � x e i for some ( e i ) ∈ N n AffPow( f ) = EssVar( f ) ⇔ f ≡ g with g = i i =1 7 / 21