/* __gmpfr_isqrt && __gmpfr_cuberoot -- Integer square root and cube root Copyright 2004-2019 Free Software Foundation, Inc. Contributed by the AriC and Caramba projects, INRIA. This file is part of the GNU MPFR Library. The GNU MPFR Library is free software; you can redistribute it and/or modify it under the terms of the GNU Lesser General Public License as published by the Free Software Foundation; either version 3 of the License, or (at your option) any later version. The GNU MPFR Library 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 Lesser General Public License for more details. You should have received a copy of the GNU Lesser General Public License along with the GNU MPFR Library; see the file COPYING.LESSER. If not, see https://www.gnu.org/licenses/ or write to the Free Software Foundation, Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301, USA. */ #include "mpfr-impl.h" /* returns floor(sqrt(n)) */ unsigned long __gmpfr_isqrt (unsigned long n) { unsigned long i, s; /* First find an approximation to floor(sqrt(n)) of the form 2^k. */ i = n; s = 1; while (i >= 2) { i >>= 2; s <<= 1; } do { s = (s + n / s) / 2; } while (!(s*s <= n && (s*s > s*(s+2) || n <= s*(s+2)))); /* Short explanation: As mathematically s*(s+2) < 2*ULONG_MAX, the condition s*s > s*(s+2) is evaluated as true when s*(s+2) "overflows" but not s*s. This implies that mathematically, one has s*s <= n <= s*(s+2). If s*s "overflows", this means that n is "large" and the inequality n <= s*(s+2) cannot be satisfied. */ return s; } /* returns floor(n^(1/3)) */ unsigned long __gmpfr_cuberoot (unsigned long n) { unsigned long i, s; /* First find an approximation to floor(cbrt(n)) of the form 2^k. */ i = n; s = 1; while (i >= 4) { i >>= 3; s <<= 1; } /* Improve the approximation (this is necessary if n is large, so that mathematically (s+1)*(s+1)*(s+1) isn't much larger than ULONG_MAX). */ if (n >= 256) { s = (2 * s + n / (s * s)) / 3; s = (2 * s + n / (s * s)) / 3; s = (2 * s + n / (s * s)) / 3; } do { s = (2 * s + n / (s * s)) / 3; } while (!(s*s*s <= n && (s*s*s > (s+1)*(s+1)*(s+1) || n < (s+1)*(s+1)*(s+1)))); return s; }