Combinations of positive integers uniq (order not important) - c++

Question

So, I was searching for a good solution for my problem.
I need to generate(print) all the combination of a list of integers, for example:
if the array contain integers from 0 to n-1, where n = 5:
int array[] = {0,1,2,3,4};
the order of integers in the combination are NOT important, meaning {1,1,3}, {1,3,1} and {3,1,1} are actually the same combination because they all contain one 3 and two ones.
so for the above array, all combination of length 3:
0,0,0 -> the 1st combination
0,0,1
0,0,2
0,0,3
0,0,4
0,1,1 -> this combination is 0,1,1, not 0,1,0 because we already have 0,0,1.
0,1,2
0,1,3
0,1,4
0,2,2 -> this combination is 0,2,2, not 0,2,0 because we already have 0,0,2.
0,2,3
.
.
0,4,4
1,1,1 -> this combination is 1,1,1, not 1,0,0 because we already have 0,0,1.
1,1,2
1,1,3
1,1,4
1,2,2 -> this combination is 1,2,2, not 1,2,0 because we already have 0,1,2.
.
.
4,4,4 -> Last combination
For Now I Wrote Code for doing this, but my problem is:
if the numbers in the array are not integer from 0 to n-1, lets say if the array was like this
int array[] = {1,3,6,7};
my code doesn't work on this case, any algorithm or code for solving this problem,,
Here is my code :
unsigned int next_combination(unsigned int *ar, int n, unsigned int k){
unsigned int finished = 0;
unsigned int changed = 0;
unsigned int i;
for (i = k - 1; !finished && !changed; i--) {
if (ar[i] < n - 1) {
/* Increment this element */
ar[i]++;
if (i < k - 1) {
/* Make the elements after it the same */
unsigned int j;
for (j = i + 1; j < k; j++) {
ar[j] = ar[j - 1];
}
}
changed = 1;
}
finished = i == 0;
}
if (!changed) {
/* Reset to first combination */
for (i = 0; i < k; i++){
ar[i] = 0;
}
}
return changed;
}
And this is the main:
int main(){
unsigned int numbers[] = {0, 0, 0, 0, 0};
const unsigned int k = 3;
unsigned int n = 5;
do{
for(int i=0 ; i<k ; ++i)
cout << numbers[i] << " ";
cout << endl;
}while (next_combination(numbers, n, k));
return 0;
}

Answer 1

If you have working code to generate all combinations of numbers from 0 to n-1, then this is very simple. You have your array of numbers:
int array[] = {1,3,6,7};
Now, take n = 4, because there are 4 items in the array. Generate all combinations from 0 to 3, and use those as indices into your array. You now have all combinations of your array values by using all combinations of indices into that array.

Answer 2

This code requires that the "element pool" array be sorted from minimum to maximum, with no duplicate entries.
The function first_combination initializes the result array ("dist") to the first combination. After this, next_combination is called in a loop until it returns false (just like in your example). The "n" and "k" arguments have been replaced with template parameters that pick up the arrays' sizes -- so the enumeration functions need the pool array in addition to the result.
#include <iostream>
template<typename T, int N, int K>
void first_combination(const T (&pool)[N], T (&dist)[K]) {
for(int ki=0; ki<K; ++ki) {
dist[ki] = pool[0];
}
}
template<typename T, int N, int K>
bool next_combination(const T (&pool)[N], T (&dist)[K]) {
int ni = 0;;
int ki = 0;
for(;;) {
const int prev_ni = ni;
// search the pool for the value in this slot
for(ni=0; pool[ni] != dist[ki]; ++ni) {
if(ni == N) return false; // slot contains a value not found in the pool
}
if(++ni < N) break;
ni = 0;
dist[ki] = pool[0];
if(++ki == K) return false;
}
int v = pool[ni];
dist[ki] = v;
// code below assumes pool[] is sorted
for(--ki; ki>=0; --ki) {
if(dist[ki] < v) {
dist[ki] = v;
}
else {
v = dist[ki];
}
}
return true;
}
template<typename T, int COUNT>
void dumparray( T (&dist)[COUNT]) {
std::cout << '{';
for(int i=0; i<COUNT; ++i) {
if(i) std::cout << ',';
std::cout << dist[i];
}
std::cout << '}' << std::endl;
}
int main(int argc, char* argv[]) {
const int pool[] = {1,3,6,7};
int dist[3] = {0};
first_combination(pool, dist);
do {
dumparray(dist);
} while(next_combination(pool, dist));
return 0;
}

Answer 3

So you need program for generating combination (wiki link).
Here you have complete description and even ready to use algorithm: http://compprog.wordpress.com/2007/10/17/generating-combinations-1/