algorithm for generating number combinations without repetition

Question

I checked almost every similar post in here but I couldn't figure out how I can do what I want. What I am trying is to give an input in a C program, let say number 4, and the program return the following numbers in an array:

1
2
3
4
12
13
14
23
24
34
123
134
124
1234

To be more clear: If the input number is 4 then i want to use digits 1-4 and generate all the possible combinations of digits(from 1-digit combinations to 4-digit combinations) without digit repetitions.

I tried the following code:

#include <stdio.h>

/* Prints out a combination like {1, 2} */
void printc(int comb[], int k) {
    printf("{");
    int i;
    for (i = 0; i < k; ++i)
        printf("%d, ", comb[i] + 1);
    printf("\\b\\b}\\n");
}


    int next_comb(int comb[], int k, int n) {
    int i = k - 1;
    ++comb[i];
    while ((i >= 0) && (comb[i] >= n - k + 1 + i)) {
        --i;
        ++comb[i];
    }

    if (comb[0] > n - k) /* Combination (n-k, n-k+1, ..., n) reached */
        return 0; /* No more combinations can be generated */

    /* comb now looks like (..., x, n, n, n, ..., n).
    Turn it into (..., x, x + 1, x + 2, ...) */
    for (i = i + 1; i < k; ++i)
        comb[i] = comb[i - 1] + 1;

    return 1;
}

int main(int argc, char *argv[]) {
    int n = 5; /* The size of the set; for {1, 2, 3, 4} it's 4 */
    int k = 3; /* The size of the subsets; for {1, 2}, {1, 3}, ... it's 2 */
    int comb[16]; /* comb[i] is the index of the i-th element in the
            combination */

    /* Setup comb for the initial combination */
    int i;
    for (i = 0; i < k; ++i)
        comb[i] = i;

    /* Print the first combination */
    printc(comb, k);

    /* Generate and print all the other combinations */
    while (next_comb(comb, k, n))
        printc(comb, k);

    return 0;
}

The above program prints the result. I want to get the result somehow.. but I can't because the above code prints the result in a strange manner.

Solution 1

We use a int to represent a set. For the i-th bit, if it is 1, then i is in the set and vice versa.

Take a example:1010(2)={4,2} 1111(2)={4,3,2,1}

For every element that will be considered, there is two choices: in or not in the set.

So, there are 2^n different set in total. And in my code, I just enumerate every possible int which is corresponding a set, and output the corresponding set.

So we get this code:

for(int i=1;i<(1<<n);i++)
{
    for(int j=0;j<n;j++)
        if ((1<<j)&i) printf("%d",j+1);
    puts("");
}

when n=4, output:

1
2
12
3
13
23
123
4
14
24
124
34
134
234
1234

If you want to output the answer as the order of giving, just make them string and put these string in vector and sort.

If n is large, you can use bitset. But when n>30,it may not be terminates in hours. So int is efficient.

Solution 2

Here is a program that generates combinations of numbers. It is written in C. But it could be rewritten in any other language. For now, just compile it and try it !

#include <stdio.h>
#include <stdlib.h>
int v[100], stack[100];
int sp,i,n,g;
int main()
{
printf("Dimension of V:");
scanf( "%d",&n);
//Input numbers
for (i=0 ; i<n ; i++) {
printf("V[%d]=",i);
scanf( "%d",&g);
v[i]=g;
}
printf("running...\n");
sp=-1;
while(1) {
// stack ones until stack overflow
    while(sp<n-1)  {
      sp=sp+1;
      stack[sp]=1;
      for (i=0 ; i<=sp ; i++) if (stack[i]==1) printf("v[%d]=%d ",i,v[i]);
      printf("\n");
   }
// unstack all the zeros until top of stack is equal one
while (stack[sp]==0 && sp>=0) sp=sp-1;
// if Bottom of stack is reached then stop
if (sp<0) break;
// set top of stack from one to zero
      stack[sp]=0;
  }
return 0;
}

run for n=4 :

[oldache@localhost fin]$ ./comb
Dimension of V:4
V[0]=10
V[1]=20
V[2]=30
V[3]=40
running...
v[0]=10
v[0]=10 v[1]=20
v[0]=10 v[1]=20 v[2]=30
v[0]=10 v[1]=20 v[2]=30 v[3]=40
v[0]=10 v[1]=20 v[3]=40
v[0]=10 v[2]=30
v[0]=10 v[2]=30 v[3]=40
v[0]=10 v[3]=40
v[1]=20
v[1]=20 v[2]=30
v[1]=20 v[2]=30 v[3]=40
v[1]=20 v[3]=40
v[2]=30
v[2]=30 v[3]=40
v[3]=40

Comments

[WhozCraig]

For solving the specific problem the OP is having, this is wonderful. I honestly not sure it is standard-portable (I think it is, but would need to read up a bit). if so, it will work for numbers up to (sizeof(int)*8)-2, and you could squeeze that extra bit out if you used unsigned int. If larger sizes were needed a byte-array with an extra loop would easily be incorporated. All in all, a nice answer.

[rliu]

I think adding a description at the top for why the algorithm works would be nice. This algorithm generates the (unordered) sets and represents them with a set of bits. So {1, 2, 3, 4} = 1111 and {} = 0000 and {1, 3} = 1010, etc. It goes through all of the bit sets and generates the corresponding set of numbers by checking if that bit is "on" (i.e. == 1) and prints it if that's the case. Ah I just noticed it doesn't print the empty set, but that may or may not be desirable.

[rliu]

Ah, I screwed up the ordering. It's more like {4, 3, 2, 1} = 1111 (instead of {1, 2, 3, 4} = 1111)

[DeepS1X]

I find your lack of indentation... Disturbing.

[GioGio]

Great short solution, is there a way to change it such that it generates the combinations in order? Such as 1,2,3,4,12,13,23,14,24,34,123,124,134,234,1234