/*============================================================================= 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 #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; } }