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/*============================================================================
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);
}
}
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