Question

I am implementing the algorithm for subset sum :

SUBSET SUM(X[1 .. n], T ):
if T = 0
return T RUE
else if T < 0 or n = 0
return FALSE
else
return SUBSET SUM(X[2 .. n], T ) ∨ SUBSET SUM(X[2 .. n], T − X[1])

Please help me out in how can I pass the reduced array X[2…n] when recursing ?

This is the code I wrote and it causes a segmentation fault:

#include <stdio.h>

    int subsetsum(int a[], int sum, int size)
    {
            if(sum==0)
            return 1;
            else if (sum<0 || size <0)
            return  0;
            else
            return (subsetsum(a+1 , sum, size-1) || subsetsum(a+1, sum - *a, size-1));
    }
    `

    int main(int argc, char **argv)
    {
            int a[]={2,4,1,3,5},x;
            x=subsetsum(a,6,5);
            printf("%d",x);
            return 0;
    }

Answer 1

Array’s in C/C++ are implicitly decayed into pointers to the original memory buffer representing the array when used as arguments to functions. So in order to pass X[2...n], you simply can increment the array pointer argument by 1. For instance, you could do something like the following:

bool subset_sum(const int* array, const int array_size, int* sum)
{
    //...your code for the rest of the algorithm

    subset_sum(array+1, array_size, sum) //one of the recursive calls
}

Passing the argument array+1 on the next recursive call will increment the array pointer by one, and take a pointer pointing to the array X[1...n] and now make it point to the array X[2...n]. You would use the array_size argument to detect the end of the array.

Finally, you would call subset_sum like so:

int array_numbers = {1, 2, 3, 4, 5, 6, 7};
int sum = 0;

subset_sum(array_numbers, sizeof(array_numbers)/sizeof(int), &sum);