Partition Problems Brute Force Algorithm

Question

I am trying to do the pseudocode for the partition problem below in bruteforce.

a set of integers X and an integer k (k >1). Find k subsets of X such that the numbers in each subset sum to the same amount and no two subsets have an element in common, or conclude that no such k subsets exist. The problem is NP-Complete

For example, with X = {2, 5, 4, 9, 1, 7, 6, 8} and k = 3, a possible solution would be: {2, 5, 7}, {4, 9, 1}, {6, 8} because all of them sum up to 14.

for exhaustive search I know typically we would have to search every possible solution and see if the target is similar. But as this is partition problem this could be tricky.

The algorithm brute force :

Subset= X.sum/K //I had a guess this would make the parition
For int i==1; I <subset; i++ // this would change partition if not found in the first one
If (j=0; I<n; i++)
    Sum == s[i]
    If sum == target 
        Display “found”
    Else 
    “not found”

Answer 1

Here is an example in JavaScript that asssumes positive array elements. The algorithm pops the stack and outputs the result if it is valid by checking the count of completed parts; otherwise, it takes each array element in turn and adds another set of parameters to the stack, one where the array element is the first addition to an empty part, and one where it's added in turn to each of the parts yet filled. (For convenience, result accrues as a string where the part index precedes each array element.)

var arr = [2,5,4,9,1,7,6,8]
var k = 3;

var n = arr.length;
var target = arr.reduce( (prev, curr) => prev + curr ) / k;
var sums = [];
for (var i=0; i<k; i++){
  sums[i] = 0;
}

var stack = [[0,sums,0,""]];

while (stack[0] !== undefined){
  var params = stack.pop();

  var i = params[0];
  var sums = params[1];
  var done = params[2];
  var result = params[3];

  if (done == k){
    console.log(result);
    continue;
  } else if (i == n){
    continue;
  }

  var was_first_element = false;

  for (var j=0; j<k; j++){
    if (!was_first_element && sums[j] == 0){
      was_first_element = true;
      var _sums = sums.slice();
      _sums[j] += arr[i];
      stack.push([i + 1,_sums,done + (_sums[j] == target ? 1 : 0),result + j + ": " + arr[i] +", "]);
    } else if (sums[j] != 0 && arr[i] + sums[j] < target && i < n - 1){
      var _sums = sums.slice();
      _sums[j] += arr[i];
      stack.push([i + 1,_sums,done,result + j + ": " + arr[i] +", "]);
    } else if (sums[j] != 0 && arr[i] + sums[j] == target){
      var _sums = sums.slice();
      _sums[j] += arr[i];
      stack.push([i + 1,_sums,done + 1,result + j + ": " + arr[i] +", "]);
    }
  }
}

Output:

/*
0: 2, 1: 5, 0: 4, 1: 9, 2: 1, 2: 7, 2: 6, 0: 8
{2,4,8} {5,9} {1,7,6}

0: 2, 1: 5, 0: 4, 1: 9, 0: 1, 0: 7, 2: 6, 2: 8
{2,4,1,7} {5,9} {6,8}

0: 2, 0: 5, 1: 4, 1: 9, 1: 1, 0: 7, 2: 6, 2: 8
{2,5,7} {4,9,1} {6,8}
*/