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Towards Fast Puiseux Series Computation

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Towards Fast Puiseux Series Computation Towards fast Puiseux series computation Adrien Poteaux?, Marc Rybowiczy ?: LIFL - Université Lille 1 y: XLIM - Université de Limoges Computer algebra and polynomials Workshop, Linz November 25th, 2013 adrien.poteaux@lifl.fr Puiseux series 1 / 16 Algebraic plane curves: a projective point of view y3 = (α) a number field K Q y2′ F (X ; Y ) K[X ; Y ] y2′′ 2 2 = (x; y) C F (x; y) = 0 C f 2 j g Let x0 C : 2 y2 y1′′ Fiber at x0: (x0) = roots of F (x0; Y ) = 0 . F f g y1′ y1 Regular point : # (x0) = dY . F Critical point : # (x0) < dY . F = roots of RF = ResY (F ; FY ). ) x0 α1 α2 adrien.poteaux@lifl.fr Puiseux series 1 / 16 x0 critical Theorem (Puiseux) 1 k X jk e There are dY seriesY ij (X ) = αik ζ (X x0) i s.t. ei − k=ni F (X ; Yij (X )) = 0 for all 1 j ei , 1 i s, with ≤ ≤ ≤ ≤ ζei primitive ei -th root of unity, ei e1;:::; es partition of dY (ramification indices). Puiseux series: a generalization of formal power series x0 regular ; (x0) = y1; ; yd . F f ··· Y g Theorem (Implicit function theorem) 1 X k There are dY series Yi (X ) = αik (X x0) s.t − k=0 F (X ; Yi (X )) = 0 around x0 and Yi (x0) = yi . adrien.poteaux@lifl.fr Definition 2 / 16 Puiseux series: a generalization of formal power series x0 regular ; (x0) = y1; ; yd . F f ··· Y g Theorem (Implicit function theorem) 1 X k There are dY series Yi (X ) = αik (X x0) s.t − k=0 F (X ; Yi (X )) = 0 around x0 and Yi (x0) = yi . x0 critical Theorem (Puiseux) 1 k X jk e There are dY seriesY ij (X ) = αik ζ (X x0) i s.t. ei − k=ni F (X ; Yij (X )) = 0 for all 1 j ei , 1 i s, with ≤ ≤ ≤ ≤ ζei primitive ei -th root of unity, ei e1;:::; es partition of dY (ramification indices). adrien.poteaux@lifl.fr Definition 2 / 16 Long term goal Puiseux series w w Monodromy group w w Effective Abel-Jacobi theorem .& Computer Algebra Physics Integral basis computation.s KP & KdV equations Algebraic solutions of ODEs Differential Galois theory adrien.poteaux@lifl.fr Motivations 3 / 16 Difficult part is the singular part 1 k X jk e Sij (X x0) = αik ζ (X x0) i − ei − k=ni rij k X jk e = αik ζ (X x0) i + next terms ei − k=ni rij is the regularity index; ri = rij for 1 j ei ≤ ≤ Next terms can be computed using quadratic Newton iterations Kung & Traub 1978, All Algebraic Functions Can Be Computed Fast adrien.poteaux@lifl.fr Singular part 4 / 16 Singulart part computation: the Newton-Puiseux algorithm Newton, 1676 introduction of the concept. ! Puiseux, 1850 rediscovers ; first procedure. ! Chystov, 1986 “Newton-Puiseux bit complexity is polynomial”. ! Duval, 1989 rational algorithm ; arithmetic complexity O(D8). ! Walsh, 2000 bit complexity O~(D36) (classical algorithm). ! Walsh, 1999 polynomial size for rational coefficients, no algorithm. ! Rybowicz & P., 2008 improved arithmetic complexity: O~(D5). ! adrien.poteaux@lifl.fr Newton-Puiseux algorithm 5 / 16 2 Supp(F)= (i; j) N aij = 0 × f 2 j 6 g — (F ): lower part of the convex hullN of Supp(F). Characteristic polynomial: X i−i0 φ∆(T ) = aij T q (i; j)2∆ The Newton-Puiseux algorithm: main tools F (X ; Y ) = Y 7 + Y 5X 2 Y 4X + 5 Y 4X 3 + 4 Y 2X 2 + X 6 − adrien.poteaux@lifl.fr Newton-Puiseux algorithm 6 / 16 — (F ): lower part of the convex hullN of Supp(F). Characteristic polynomial: X i−i0 φ∆(T ) = aij T q (i; j)2∆ Support of a polynomial F (X ; Y ) = Y 7 X 0+Y 5 X 1 2 Y 4X 1+5 Y 3X 4 Y 3X 2+4 Y 2X 2+Y 0X 6 − − 2 Supp(F)= (i; j) N aij = 0 6 × f 2 j 6 g 4 2 1 0 2 3 4 5 7 adrien.poteaux@lifl.fr Newton-Puiseux algorithm 6 / 16 Characteristic polynomial: X i−i0 φ∆(T ) = aij T q (i; j)2∆ Newton polygon X i j F (X ; Y ) = aij Y X i; j 2 Supp(F)= (i; j) N aij = 0 6 × f 2 j 6 g (F ) N — (F ): lower part of the convex hullN of Supp(F). 2 1 0 2 4 7 adrien.poteaux@lifl.fr Newton-Puiseux algorithm 6 / 16 Characteristic polynomial X i j F (X ; Y ) = aij Y X i; j 2 Supp(F)= (i; j) N aij = 0 × f 2 j 6 g 6 N (F ) — (F ): lower part of the convex hullN of Supp(F). Characteristic polynomial: 2 ∆, slope − m X i−i0 q φ∆(T ) = aij T q 1 (i; j)2∆ 0 i0 = 2 4 7 adrien.poteaux@lifl.fr Newton-Puiseux algorithm 6 / 16 First turn: initial polygon 0(F ) N Rational Newton-Puiseux algorithm D. Duval 1989, Rational Puiseux Expansions For each edge ∆ of (F ) N s Y Mk – φ∆ = φk k=1 l q – For each φk v q m u F (ξ X ; X (ξ + Y )) ∆, slope − m F (X ; Y ) k k q X l with 0 i0 I(F ) ξk s. t. φk (ξk ) = 0, · (u; v) such that uq vm = 1. · − adrien.poteaux@lifl.fr Newton-Puiseux algorithm 7 / 16 Rational Newton-Puiseux algorithm : first turn For each edge ∆ of 0(F ) N s Y Mk – φ∆ = φk k=1 – For each φk F (ξv X q; X m(ξu + Y )) F (X ; Y ) k k X l with ∆, slope − m ξk s. t. φk (ξk ) = 0, q · i0 dY (u; v) such that uq vm = 1. l · − q First turn: initial polygon 0(F ) N adrien.poteaux@lifl.fr Newton-Puiseux algorithm 7 / 16 Pure symbolic computation is costly H(X ; Y ) = (Y 3 X ) ((Y 1)2 X )(Y 2 X 2) + X 2 Y 5 − − − − − 3 RH (X ) = X P(X ), degX (P) = 23; β s.t. P(β) = 0 Singular parts of Puiseux series of H above β: Si (X ) = αi;0, 1 i 4. ≤ ≤ 1 Si (X ) = αi;0 + αi;1(X β) 2 , 5 i 6. − ≤ ≤ Degree of the extension field i = 5; 6: K(αi;0) = K(αi;1) = K(β) extension of degree 23, ! i = 1;:::; 4: [K(αi;0): K(β)] = 4 extension of degree 92, ! Coefficient growth αi;0 rational number with 98 digits, ! αi;1 rational number with 132 digits. ! adrien.poteaux@lifl.fr Newton-Puiseux algorithm 8 / 16 Numerical computations ? Direct computation: almost useless Guessing the structure ? two difficulties: Finding the correct Newton polygon Factorising “well” φ∆ 4 x2 2:0 x + 0:9999 − 2 =? (x 0:99)(x 1:01) − − 1 =? (x 1:)2 − 0 2 4 7 = Multiplicity structure ? ) = Exact informations needed ! ) adrien.poteaux@lifl.fr Newton-Puiseux algorithm 9 / 16 A symbolic-numeric approach: 1 Compute the singular part of Puiseux series modulo a well chosen prime number p This give us the structure of the Puiseux series, thus: Newton polygons, Multiplicity structures of the φ∆. 2 Use this information to conduct a numerical computation of the Puiseux series coefficients. adrien.poteaux@lifl.fr A modular-numeric approach 10 / 16 Modular part: main results Reduction criteria: 1 One can reduce F mod p, 2 p > dY 3 tc(RF ) 0 mod p. 6≡ If p satisfies that: 1 Puiseux series can be reduced modulo p, 2 The structure computed modulo p is the good one. Bounds for p ; log(p) log(D) with probabilistic algorithms. ' References for details: Poteaux & Rybowicz 2008, On the good reduction of Puiseux series and complexity of the Newton-Puiseux algorithm over finite fields ; Poteaux & Rybowicz 2012, On the good reduction of Puiseux series and Applications ; Poteaux & Rybowicz 2011, Complexity bounds for the rational Newton-Puiseux algorithm over finite fields and related problems adrien.poteaux@lifl.fr A modular-numeric approach 11 / 16 Numerical part: following the structure If P(X ) =( X α)m(X β)m =( X γ~)m (X δ~)m, − − − − α γ~ or α δ~ ? $ $ no answer = group of roots dealt together ) We separate the blocs later on: 1 filter according to the Newton polygon (coefficient of F zero or not) 2 filter according to the multiplicity structure first idea = approximate gcd : singular values One example adrien.poteaux@lifl.fr A modular-numeric approach 12 / 16 Fast computation ? 1 Use only “mandatory” monomials 1 Truncating power of X already used in the D8 of Duval, improved in our D5 version, better ? ! relaxed computations (a-posteriori bounds) 2 Reducing the degree in Y After first turn, we only look roots above (0; 0) Get rid of roots above (0; α 6= 0) ? Factorization 2 Fast computation in between two branch separations 2 3 7 4 9 5 11 35 S(X ) = 2 + X X + X +3X 2 +4X X 2 2X + X 2 +7X 6 + ::: − − − adrien.poteaux@lifl.fr A fast algorithm 13 / 16 A new algorithm 1 Substitutions F (X ; Y ) F (X ; Y + Ad −1(X )) Y (compute common terms at once ; less recursive calls) 2 Factorization of the polynomial during the algorithm (Hensel lemma recursive calls with smaller degrees) ! 3 Using relax algorithms. (no a priori bound required) 4 Improved truncation bounds (thanks to relax algorithms). leads to O~(D4) adrien.poteaux@lifl.fr A fast algorithm 14 / 16 Better again ? The idea: all series do not need the same truncations 1 Compute half of the series in O~(D3) (the one that requires the smallest truncations) 2 Factorize F = GH with G corresponding to the computed series, and H to the other ones.
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