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