/* __gmpfr_isqrt && __gmpfr_cuberoot -- Integer square root and cube root Copyright 2004, 2005, 2006, 2007, 2008 Free Software Foundation, Inc. Contributed by the Arenaire and Cacao projects, INRIA. This file is part of the MPFR Library. The 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 2.1 of the License, or (at your option) any later version. The 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 MPFR Library; see the file COPYING.LIB. If not, 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; }