How does this solution of the subset sum problem work?

Question

this is a solution for the subset sum problem. It uses backtracking. I have been thinking about it for more than 2 hours now and i just cannot understand it.

edit:I have added some comments to the code based on what i have understood. Please correct me if i am wrong.

#include <iostream>

int n, d, w[10], x[10], count=0;

void subset(int cs, int k, int r)//i dont understand the purpose of cs or of the array x[] 
{
    int i;
    x[k] = 1;

    if(cs + w[k] == d)      //if the first element is equivalent to the sum that is required 
    {
        std::cout<< "\n\nSolution " << ++count << " is:"; 
        for(i=0; i<=k; i++)  // so if the subset criteria are met then the array is printed.
        if(x[i] == 1) 
            std::cout << w[i] << " ";
    }
    else if(cs + w[k] + w[k+1] <= d) //if the node is promising then go to the next node and              
        subset(cs + w[k], k+1, r - w[k]); //check if subset criteria are met

    if(cs + w[k+1] <= d && cs + r - w[k] >= d) //else backtrack
    {
        x[k] = 0;
        subset(cs, k+1, r-w[k]);
    }
}

int main()
{
    int sum=0, i;

    std::cout << "enter n\n";
    std::cin >> n;

    std::cout < <"enter " << n << " elements\n";
    for(i=0; i<n; i++)
        std::cin >> w[i];

    for(i=0; i<n; i++)
        sum += w[i];

    std::cout << "enter value of sum\n";
    std::cin >> d;

    if(sum < d)
        std::cout<<"INVALID SOLUTION\n";
    else
        subset(0,0,sum);

}

Note: This is a working solution. It works when compiled with g++. I am sorry if this seems too obnoxious but i just did not understand much from the code and hence i cannot give much of an explanation.

Why don't you add some comments to the code you posted with your thoughts on the parts you do understand and some highlights of those you don't? — John Zwinck, Nov 20 '11 at 17:51
If you are completely stumped, go through some simple examples step-by-step with a debugger. Inspect the variables to see what's going on. — reima, Nov 20 '11 at 17:57
@JohnZwinck I have added the comments. please correct them if i am wrong. — Chaitanya Nettem, Nov 20 '11 at 18:06
@reima i have been doing that but for some reason i keep getting confused. — Chaitanya Nettem, Nov 20 '11 at 18:07
The indentation and crappy variables names are confusing. Fix that and maybe you'll understand it better. — David Grayson, Nov 20 '11 at 18:33

score 2 · Answer 1 · answered Nov 20 '11 at 18:23

Try this one.

#include<iostream>

int n,d,w[10],used[10],count=0;
int cs = 0; // cs=Current Sum

void subset(int k)
{
    if (k >= n) return;  // boundry check
    int i;
    used[k] = 1; // use element k
    cs += w[k];

    if(cs == d) {
        cout<<"\n\nSolution " << ++count << " is:"; 
        for(i=0;i <= k;i++)  
            if(used[i]==1) 
                cout<<w[i]<<" ";
    }
    if (cs < d)  // only when current sum is not enough
        subset(k + 1); 

    used[k] = 0; // not use element k
    cs -= w[k];
    subset(k+1);
}

void main()
{
    int sum=0,i;

    cout<<"enter n\n";
    cin>>n;

    cout<<"enter "<<n<<" elements\n";
    for(i=0;i<n;i++)
    cin>>w[i];

    for(i=0;i<n;i++)
    sum+=w[i];

    cout<<"enter value of sum\n";
    cin>>d;
    cs = 0;
    if(sum<d)
        cout<<"INVALID SOLUTION\n";
    else
        subset(0);
}

prusswan · Accepted Answer · 2011-11-20T18:43:52.053

cs is the sum of values of the weights chosen so far, while r is the remainder of the summed values of the weights yet to be chosen. w[i] is weight i, x[i] is 1 when weight [i] is chosen. In the subset method there are two main decision branches:

Weight k is chosen:

// adding value of weight k to computed sum (cs) gives required sum, solution found
if(cs+w[k]==d)
{
cout<<"\n\nSolution "<<++count<<" is:";
for(i=0;i<=k;i++)
    if(x[i]==1)
    cout<<w[i]<<" ";
}

// both weight k and weight k+1 can be chosen without exceeding d, 
// so we choose k, and see if there's a solution for weight k+1 onwards
// note that available weight values decreased from r to r-w[k]
else if(cs+w[k]+w[k+1]<=d)
    subset(cs+w[k],k+1,r-w[k]);

Weight k is not chosen (notice that this is explored even when a solution was found right after weight k is chosen):

// weight k+1 is choosable (does not exceed d), and despite not choosing weight k
// there would be sufficient weights in r less w[k], and together with the chosen 
// pool cs to meet the requirement of d.
if(cs+w[k+1]<=d && cs+r-w[k]>=d)
{
x[k]=0;
subset(cs,k+1,r-w[k]);
}

Actually I don't understand it as backtracking, but rather, brute force with intelligent checks in order to fail-fast — prusswan, Nov 20 '11 at 18:40

score 0 · Answer 3 · answered Nov 20 '11 at 21:51

0

cs is the current sum and x[] is an array that has 1 set for those elements that belong to the solution branch that is currently being explored.

answered Nov 20 '11 at 21:51

Antti Huima

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How does this solution of the subset sum problem work?

3 Answers3