213 lines
		
	
	
		
			4.9 KiB
		
	
	
	
		
			C
		
	
	
	
	
	
		
		
			
		
	
	
			213 lines
		
	
	
		
			4.9 KiB
		
	
	
	
		
			C
		
	
	
	
	
	
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								/*
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								 * Copyright (c) 2003, 2007-14 Matteo Frigo
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								 * Copyright (c) 2003, 2007-14 Massachusetts Institute of Technology
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								 *
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								 * This program 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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								 *
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								 * This program 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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								 *
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								 * You should have received a copy of the GNU General Public License
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								 * along with this program; if not, write to the Free Software
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								 * Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA  02110-1301  USA
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								 *
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								 */
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								#include "kernel/ifftw.h"
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								/***************************************************************************/
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								/* Rader's algorithm requires lots of modular arithmetic, and if we
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								   aren't careful we can have errors due to integer overflows. */
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								/* Compute (x * y) mod p, but watch out for integer overflows; we must
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								   have 0 <= {x, y} < p.
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								   If overflow is common, this routine is somewhat slower than
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								   e.g. using 'long long' arithmetic.  However, it has the advantage
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								   of working when INT is 64 bits, and is also faster when overflow is
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								   rare.  FFTW calls this via the MULMOD macro, which further
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								   optimizes for the case of small integers. 
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								*/
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								#define ADD_MOD(x, y, p) ((x) >= (p) - (y)) ? ((x) + ((y) - (p))) : ((x) + (y))
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								INT X(safe_mulmod)(INT x, INT y, INT p)
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								{
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								     INT r;
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								     if (y > x) 
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									  return X(safe_mulmod)(y, x, p);
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								     A(0 <= y && x < p);
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								     r = 0;
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								     while (y) {
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									  r = ADD_MOD(r, x*(y&1), p); y >>= 1;
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									  x = ADD_MOD(x, x, p);
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								     }
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								     return r;
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								}
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								/***************************************************************************/
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								/* Compute n^m mod p, where m >= 0 and p > 0.  If we really cared, we
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								   could make this tail-recursive. */
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								INT X(power_mod)(INT n, INT m, INT p)
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								{
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								     A(p > 0);
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								     if (m == 0)
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									  return 1;
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								     else if (m % 2 == 0) {
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									  INT x = X(power_mod)(n, m / 2, p);
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									  return MULMOD(x, x, p);
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								     }
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								     else
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									  return MULMOD(n, X(power_mod)(n, m - 1, p), p);
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								}
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								/* the following two routines were contributed by Greg Dionne. */
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								static INT get_prime_factors(INT n, INT *primef)
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								{
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								     INT i;
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								     INT size = 0;
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								     A(n % 2 == 0); /* this routine is designed only for even n */
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								     primef[size++] = (INT)2;
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								     do {
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									  n >>= 1;
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								     } while ((n & 1) == 0);
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								     if (n == 1)
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									  return size;
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								     for (i = 3; i * i <= n; i += 2)
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									  if (!(n % i)) {
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									       primef[size++] = i;
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									       do {
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										    n /= i;
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									       } while (!(n % i));
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									  }
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								     if (n == 1)
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									  return size;
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								     primef[size++] = n;
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								     return size;
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								}
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								INT X(find_generator)(INT p)
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								{
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								    INT n, i, size;
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								    INT primef[16];     /* smallest number = 32589158477190044730 > 2^64 */
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								    INT pm1 = p - 1;
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								    if (p == 2)
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									 return 1;
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								    size = get_prime_factors(pm1, primef);
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								    n = 2;
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								    for (i = 0; i < size; i++)
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								        if (X(power_mod)(n, pm1 / primef[i], p) == 1) {
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								            i = -1;
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								            n++;
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								        }
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								    return n;
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								}
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								/* Return first prime divisor of n  (It would be at best slightly faster to
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								   search a static table of primes; there are 6542 primes < 2^16.)  */
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								INT X(first_divisor)(INT n)
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								{
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								     INT i;
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								     if (n <= 1)
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									  return n;
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								     if (n % 2 == 0)
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									  return 2;
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								     for (i = 3; i*i <= n; i += 2)
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									  if (n % i == 0)
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									       return i;
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								     return n;
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								}
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								int X(is_prime)(INT n)
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								{
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								     return(n > 1 && X(first_divisor)(n) == n);
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								}
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								INT X(next_prime)(INT n)
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								{
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								     while (!X(is_prime)(n)) ++n;
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								     return n;
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								}
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								int X(factors_into)(INT n, const INT *primes)
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								{
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								     for (; *primes != 0; ++primes) 
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									  while ((n % *primes) == 0) 
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									       n /= *primes;
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								     return (n == 1);
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								}
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								/* integer square root.  Return floor(sqrt(N)) */
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								INT X(isqrt)(INT n)
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								{
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								     INT guess, iguess;
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								     A(n >= 0);
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								     if (n == 0) return 0;
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								     guess = n; iguess = 1;
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								     do {
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								          guess = (guess + iguess) / 2;
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									  iguess = n / guess;
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								     } while (guess > iguess);
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								     return guess;
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								}
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								static INT isqrt_maybe(INT n)
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								{
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								     INT guess = X(isqrt)(n);
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								     return guess * guess == n ? guess : 0;
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								}
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								#define divides(a, b) (((b) % (a)) == 0)
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								INT X(choose_radix)(INT r, INT n)
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								{
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								     if (r > 0) {
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									  if (divides(r, n)) return r;
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									  return 0;
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								     } else if (r == 0) {
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									  return X(first_divisor)(n);
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								     } else {
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									  /* r is negative.  If n = (-r) * q^2, take q as the radix */
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									  r = 0 - r;
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									  return (n > r && divides(r, n)) ? isqrt_maybe(n / r) : 0;
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								     }
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								}
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								/* return A mod N, works for all A including A < 0 */
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								INT X(modulo)(INT a, INT n)
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								{
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								     A(n > 0);
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								     if (a >= 0)
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									  return a % n;
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								     else
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									  return (n - 1) - ((-(a + (INT)1)) % n);
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								}
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								/* TRUE if N factors into small primes */
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								int X(factors_into_small_primes)(INT n)
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								{
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								     static const INT primes[] = { 2, 3, 5, 0 };
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								     return X(factors_into)(n, primes);
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								}
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