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Write and explain sum of subset algorithm

Write and explain sum of subset algorithm for W = {2, 7, 8, 9, 15} M = 17

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  1. Given positive numbers (weight) w_i where (1<=i<=n) and m.
  2. This problem calls for finding all subsets of w_i whose sum is m.
  3. For e.g. if n=4, m=31, (w_1,w_2,w_3,w_4) = (11, 13, 24, 7).
  4. Then the desired subset are (11, 13, 7) and (2, 4, 7).
  5. In this method of giving the solution, the solution vector is given as variable-sized tuple (tuple is collection of element).
  6. The solution may also be expressed as fixed size tuple (x_1,x_(2,),x_3,…….,x_n) such that

    x_i =0 if w_i is not selected

    x_i =1 if w_i is selected

  7. Using this strategy the solution would be (1,1,0,1) and (0,0,1,1)

Algorithm:

  • Let n be the number of elements and let m be the required sum.
  • Let w[1…..n] be an array of weights in ascending order and Let x[1….n] be the solution vector.
  • Method

    SumOfSubset(s, k, r)
    
    {
    x[k] =1;     //Generating left child
    if(s+w[k]==m)
        Display x[1….k]  //and remaining zeros
    else if(s+w[k]+w[k+1]<=m)
        SumOfSubset(s+w[k], k+1, r-w[k])
    if((s+r-w[k]>=m) && (s+w[k+1]<=m))
    {
        x[k]=0;
        SumOfSubset(s, k+1, r-w[k]);
    }
    }
    
  • S → existing sum

  • w[k] → weight of current element
  • w[k+1] → weight of next element
  • s+w[k]+w[k+1]<=m → sum of existing element+weight of current element+weight of next element is less tham m.
  • Thus the next element can be selected for time being.
  • s+r-w[k]>=m → Even if u discard the current element, you may still get the required sum later on.
  • s+w[k+1]<=m → The sum of selected element + weight of next element is less than m . This is even if u take the next element the sum will not exit the required sum(m)
  • Thus next element can be selected.

Problem:

n = 5, W = {2, 7, 8, 9, 15} M = 17

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Hence, there are three subset with given sum m=17

A(111)

B(10001)

C(0010)

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