pqc/external/flint-2.4.3/padic/log.c

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2014-05-18 22:03:37 +00:00
/*=============================================================================
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 <gmp.h>
#include "flint.h"
#include "fmpz.h"
#include "padic.h"
#include "ulong_extras.h"
/*
Assumes that $1 \leq v$ or $2 \leq v$ as $p$ is even
or odd, respectively, and that $v < N < 2^{f-2}$ where
$f$ is \code{FLINT_BITS}.
Under the assumption that $1 \leq v < N$, or $2 \leq v < N$,
one can easily prove that with $c = N - \floor{\log_p v}$
the number $b = \ceil{(c + \ceil{\log_p c} + 1) / b}$ is such
that for all $i \geq b$, $i v - \ord_p(i) \geq N$.
Under the additional condition that $N < 2^{f-2}$ one can
show that the code branch for primes that fit into a
\code{slong} does not cause overflow. Moreover,
independently of this, it follows that the above value $b$
is less than $2^{f-1}$.
In the first branch, we have that $b v - log_p(b) \geq N$.
We need to show that we can replace $\log_p$ by $\ord_p$ here.
That is, we need that $iv - \ord_p(i) \geq iv - \log_p(i) \geq N$,
i.e., $\log_p(i) \geq \ord_p(i)$, which is true. We then work
backwards to find the first $i$ such that this fails, then
using that the function is strictly increasing for $i \geq 2$.
In the second branch we use that using signed indices in the
summation is still sufficient and hence that all terms $1/i$
are units.
Then $ord_p(x^i/i) \geq N$ provided that $i v \geq N$.
*/
slong _padic_log_bound(slong v, slong N, const fmpz_t prime)
{
if (N >= (WORD(1) << (FLINT_BITS - 2)))
{
flint_printf("Exception (_padic_log_bound). N = %wd is too large.\n", N);
abort();
}
if (fmpz_fits_si(prime))
{
slong b, c, p = fmpz_get_si(prime);
c = N - n_flog(v, p);
b = ((c + n_clog(c, p) + 1) + (v - 1)) / v;
while (--b >= 2)
{
slong t = b * v - n_clog(b, p);
if (t < N)
return b + 1;
}
return 2;
}
else
{
return (N + v - 1) / v;
}
}
void _padic_log(fmpz_t z, const fmpz_t y, slong v, const fmpz_t p, slong N)
{
if (N < (WORD(1) << 9) / (slong) fmpz_bits(p))
{
_padic_log_rectangular(z, y, v, p, N);
}
else
{
_padic_log_balanced(z, y, v, p, N);
}
}
int padic_log(padic_t rop, const padic_t op, const padic_ctx_t ctx)
{
const fmpz *p = ctx->p;
const slong N = padic_prec(rop);
if (padic_val(op) < 0)
{
return 0;
}
else
{
fmpz_t x;
int ans;
fmpz_init(x);
padic_get_fmpz(x, op, ctx);
fmpz_sub_ui(x, x, 1);
fmpz_neg(x, x);
if (fmpz_is_zero(x))
{
padic_zero(rop);
ans = 1;
}
else
{
fmpz_t t;
slong v;
fmpz_init(t);
v = fmpz_remove(t, x, p);
fmpz_clear(t);
if (v >= 2 || (!fmpz_equal_ui(p, 2) && v >= 1))
{
if (v >= N)
{
padic_zero(rop);
}
else
{
_padic_log(padic_unit(rop), x, v, p, N);
padic_val(rop) = 0;
_padic_canonicalise(rop, ctx);
}
ans = 1;
}
else
{
ans = 0;
}
}
fmpz_clear(x);
return ans;
}
}