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