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