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/*
* Copyright (C) 2013 Andrea Mazzoleni
*
* 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.
*/
#ifndef __RAID_COMBO_H
#define __RAID_COMBO_H
#include <assert.h>
/**
* Get the first permutation with repetition of r of n elements.
*
* Typical use is with permutation_next() in the form :
*
* int i[R];
* permutation_first(R, N, i);
* do {
* code using i[0], i[1], ..., i[R-1]
* } while (permutation_next(R, N, i));
*
* It's equivalent at the code :
*
* for(i[0]=0;i[0]<N;++i[0])
* for(i[1]=0;i[1]<N;++i[1])
* ...
* for(i[R-2]=0;i[R-2]<N;++i[R-2])
* for(i[R-1]=0;i[R-1]<N;++i[R-1])
* code using i[0], i[1], ..., i[R-1]
*/
static __always_inline void permutation_first(int r, int n, int *c)
{
int i;
(void)n; /* unused, but kept for clarity */
assert(0 < r && r <= n);
for (i = 0; i < r; ++i)
c[i] = 0;
}
/**
* Get the next permutation with repetition of r of n elements.
* Return ==0 when finished.
*/
static __always_inline int permutation_next(int r, int n, int *c)
{
int i = r - 1; /* present position */
recurse:
/* next element at position i */
++c[i];
/* if the position has reached the max */
if (c[i] >= n) {
/* if we are at the first level, we have finished */
if (i == 0)
return 0;
/* increase the previous position */
--i;
goto recurse;
}
++i;
/* initialize all the next positions, if any */
while (i < r) {
c[i] = 0;
++i;
}
return 1;
}
/**
* Get the first combination without repetition of r of n elements.
*
* Typical use is with combination_next() in the form :
*
* int i[R];
* combination_first(R, N, i);
* do {
* code using i[0], i[1], ..., i[R-1]
* } while (combination_next(R, N, i));
*
* It's equivalent at the code :
*
* for(i[0]=0;i[0]<N-(R-1);++i[0])
* for(i[1]=i[0]+1;i[1]<N-(R-2);++i[1])
* ...
* for(i[R-2]=i[R-3]+1;i[R-2]<N-1;++i[R-2])
* for(i[R-1]=i[R-2]+1;i[R-1]<N;++i[R-1])
* code using i[0], i[1], ..., i[R-1]
*/
static __always_inline void combination_first(int r, int n, int *c)
{
int i;
(void)n; /* unused, but kept for clarity */
assert(0 < r && r <= n);
for (i = 0; i < r; ++i)
c[i] = i;
}
/**
* Get the next combination without repetition of r of n elements.
* Return ==0 when finished.
*/
static __always_inline int combination_next(int r, int n, int *c)
{
int i = r - 1; /* present position */
int h = n; /* high limit for this position */
recurse:
/* next element at position i */
++c[i];
/* if the position has reached the max */
if (c[i] >= h) {
/* if we are at the first level, we have finished */
if (i == 0)
return 0;
/* increase the previous position */
--i;
--h;
goto recurse;
}
++i;
/* initialize all the next positions, if any */
while (i < r) {
/* each position start at the next value of the previous one */
c[i] = c[i - 1] + 1;
++i;
}
return 1;
}
#endif
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