189 lines
5.3 KiB
C
189 lines
5.3 KiB
C
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/*============================================================================
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This file is part of FLINT.
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FLINT is free software; you can redistribute it and/or modify
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it under the terms of the GNU General Public License as published by
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the Free Software Foundation; either version 2 of the License, or
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(at your option) any later version.
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FLINT is distributed in the hope that it will be useful,
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but WITHOUT ANY WARRANTY; without even the implied warranty of
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MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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GNU General Public License for more details.
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You should have received a copy of the GNU General Public License
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along with FLINT; if not, write to the Free Software
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Foundation, Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA
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=============================================================================*/
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/******************************************************************************
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Copyright (C) 2012 Sebastian Pancratz
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******************************************************************************/
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#include "fmpz_mod_poly.h"
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#include "padic_poly.h"
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/*
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TODO: Move this bit of code into "padic".
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*/
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static void __padic_reduce(fmpz_t u, slong *v, slong N, const padic_ctx_t ctx)
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{
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if (!fmpz_is_zero(u))
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{
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if (*v < N)
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{
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int alloc;
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fmpz_t pow;
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alloc = _padic_ctx_pow_ui(pow, N - *v, ctx);
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fmpz_mod(u, u, pow);
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if (alloc)
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fmpz_clear(pow);
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}
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else
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{
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fmpz_zero(u);
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*v = 0;
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}
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}
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}
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/*
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Evaluates the polynomial $F(x) = p^w f(x)$ at $x = p^b a$, setting
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$y = p^v u$ to the result reduced modulo $p^N$.
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Suppose first that $b \geq 0$, in which case we can quickly relay
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the call to the \code{fmpz_mod_poly} module. Namely, we need to
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compute $f(x) \bmod {p^{N-w}}$ where we know $f(x)$ to be integral.
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Otherwise, suppose now that $b < 0$ and we still wish to evaluate
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$f(x) \bmod {p^{N-w}}$.
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\begin{align*}
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f(x) & = \sum_{i = 0}^{n} a_i x^i \\
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& = \sum_{i = 0}^{n} a_i p^{i b} a^i \\
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\intertext{Multiplying through by $p^{- n b} \in \mathbf{Z}$, }
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p^{-nb} f(x) & = \sum_{i = 0}^{n} a_i p^{-(n-i)b} a
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\end{align*}
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which leaves the right hand side integral. As we want to
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compute $f(x)$ to precision $N-w$, we have to compute
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$p^{-nb} f(x)$ to precision $N-w-nb$.
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*/
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void _padic_poly_evaluate_padic(fmpz_t u, slong *v, slong N,
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const fmpz *poly, slong val, slong len,
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const fmpz_t a, slong b, const padic_ctx_t ctx)
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{
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if (len == 0)
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{
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fmpz_zero(u);
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*v = 0;
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}
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else if (len == 1)
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{
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fmpz_set(u, poly);
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*v = val;
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__padic_reduce(u, v, N, ctx);
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}
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else if (b >= 0)
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{
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if (val >= N)
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{
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fmpz_zero(u);
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*v = 0;
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}
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else
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{
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fmpz_t x;
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fmpz_t pow;
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int alloc;
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fmpz_init(x);
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alloc = _padic_ctx_pow_ui(pow, N - val, ctx);
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fmpz_pow_ui(x, ctx->p, b);
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fmpz_mul(x, x, a);
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_fmpz_mod_poly_evaluate_fmpz(u, poly, len, x, pow);
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if (!fmpz_is_zero(u))
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*v = val + _fmpz_remove(u, ctx->p, ctx->pinv);
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else
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*v = 0;
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fmpz_clear(x);
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if (alloc)
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fmpz_clear(pow);
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}
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}
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else /* b < 0 */
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{
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const slong n = len - 1;
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if (val + n*b >= N)
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{
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fmpz_zero(u);
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*v = 0;
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}
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else
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{
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fmpz_t pow;
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int alloc;
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slong i;
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fmpz_t s, t;
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fmpz *vec = _fmpz_vec_init(len);
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fmpz_init(s);
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fmpz_init(t);
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alloc = _padic_ctx_pow_ui(pow, N - val - n*b, ctx);
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fmpz_pow_ui(s, ctx->p, -b);
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fmpz_one(t);
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fmpz_set(vec + (len - 1), poly + (len - 1));
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for (i = len - 2; i >= 0; i--)
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{
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fmpz_mul(t, t, s);
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fmpz_mul(vec + i, poly + i, t);
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}
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_fmpz_mod_poly_evaluate_fmpz(u, vec, len, a, pow);
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if (!fmpz_is_zero(u))
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*v = val + n*b + _fmpz_remove(u, ctx->p, ctx->pinv);
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else
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*v = 0;
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if (alloc)
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fmpz_clear(pow);
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fmpz_clear(s);
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fmpz_clear(t);
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_fmpz_vec_clear(vec, len);
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}
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}
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}
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void padic_poly_evaluate_padic(padic_t y, const padic_poly_t poly,
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const padic_t x, const padic_ctx_t ctx)
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{
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if (y == x)
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{
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padic_t t;
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padic_init2(t, padic_prec(y));
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_padic_poly_evaluate_padic(padic_unit(t), &padic_val(t), padic_prec(t),
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poly->coeffs, poly->val, poly->length,
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padic_unit(x), padic_val(x), ctx);
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padic_swap(y, t);
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padic_clear(t);
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}
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else
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{
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_padic_poly_evaluate_padic(padic_unit(y), &padic_val(y), padic_prec(y),
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poly->coeffs, poly->val, poly->length,
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padic_unit(x), padic_val(x), ctx);
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}
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}
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