1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196
|
/*
* Copyright (c) 2003, 2006 Matteo Frigo
* Copyright (c) 2003, 2006 Massachusetts Institute of Technology
*
* This program 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.
*
* This program 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 this program; if not, write to the Free Software
* Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA
*
*/
/* $Id: primes.c,v 1.25 2006-02-05 23:19:55 athena Exp $ */
#include "ifftw.h"
/***************************************************************************/
/* Rader's algorithm requires lots of modular arithmetic, and if we
aren't careful we can have errors due to integer overflows. */
/* Compute (x * y) mod p, but watch out for integer overflows; we must
have 0 <= {x, y} < p.
If overflow is common, this routine is somewhat slower than
e.g. using 'long long' arithmetic. However, it has the advantage
of working when INT is 64 bits, and is also faster when overflow is
rare. FFTW calls this via the MULMOD macro, which further
optimizes for the case of small integers.
*/
#define ADD_MOD(x, y, p) ((x) >= (p) - (y)) ? ((x) + ((y) - (p))) : ((x) + (y))
INT X(safe_mulmod)(INT x, INT y, INT p)
{
INT r;
if (y > x)
return X(safe_mulmod)(y, x, p);
A(0 <= y && x < p);
r = 0;
while (y) {
r = ADD_MOD(r, x*(y&1), p); y >>= 1;
x = ADD_MOD(x, x, p);
}
return r;
}
/***************************************************************************/
/* Compute n^m mod p, where m >= 0 and p > 0. If we really cared, we
could make this tail-recursive. */
INT X(power_mod)(INT n, INT m, INT p)
{
A(p > 0);
if (m == 0)
return 1;
else if (m % 2 == 0) {
INT x = X(power_mod)(n, m / 2, p);
return MULMOD(x, x, p);
}
else
return MULMOD(n, X(power_mod)(n, m - 1, p), p);
}
/* the following two routines were contributed by Greg Dionne. */
static INT get_prime_factors(INT n, INT *primef)
{
INT i;
INT size = 0;
A(n % 2 == 0); /* this routine is designed only for even n */
primef[size++] = (INT)2;
do
n >>= 1;
while ((n & 1) == 0);
if (n == 1)
return size;
for (i = 3; i * i <= n; i += 2)
if (!(n % i)) {
primef[size++] = i;
do
n /= i;
while (!(n % i));
}
if (n == 1)
return size;
primef[size++] = n;
return size;
}
INT X(find_generator)(INT p)
{
INT n, i, size;
INT primef[16]; /* smallest number = 32589158477190044730 > 2^64 */
INT pm1 = p - 1;
if (p == 2)
return 1;
size = get_prime_factors(pm1, primef);
n = 2;
for (i = 0; i < size; i++)
if (X(power_mod)(n, pm1 / primef[i], p) == 1) {
i = -1;
n++;
}
return n;
}
/* Return first prime divisor of n (It would be at best slightly faster to
search a static table of primes; there are 6542 primes < 2^16.) */
INT X(first_divisor)(INT n)
{
INT i;
if (n <= 1)
return n;
if (n % 2 == 0)
return 2;
for (i = 3; i*i <= n; i += 2)
if (n % i == 0)
return i;
return n;
}
int X(is_prime)(INT n)
{
return(n > 1 && X(first_divisor)(n) == n);
}
INT X(next_prime)(INT n)
{
while (!X(is_prime)(n)) ++n;
return n;
}
int X(factors_into)(INT n, const INT *primes)
{
for (; *primes != 0; ++primes)
while ((n % *primes) == 0)
n /= *primes;
return (n == 1);
}
/* integer square root. Return floor(sqrt(N)) */
INT X(isqrt)(INT n)
{
INT guess, iguess;
A(n >= 0);
if (n == 0) return 0;
guess = n; iguess = 1;
do {
guess = (guess + iguess) / 2;
iguess = n / guess;
} while (guess > iguess);
return guess;
}
static INT isqrt_maybe(INT n)
{
INT guess = X(isqrt)(n);
return guess * guess == n ? guess : 0;
}
#define divides(a, b) (((b) % (a)) == 0)
INT X(choose_radix)(INT r, INT n)
{
if (r > 0) {
if (divides(r, n)) return r;
return 0;
} else if (r == 0) {
return X(first_divisor)(n);
} else {
/* r is negative. If n = (-r) * q^2, take q as the radix */
r = 0 - r;
return (n > r && divides(r, n)) ? isqrt_maybe(n / r) : 0;
}
}
|