049: Number Partitioning

/*

  Set partition problem in Comet.
  
  Problem formulation from
    http://www.koalog.com/resources/samples/PartitionProblem.java.html
  """
   This is a partition problem.
   Given the set S = {1, 2, ..., n},
   it consists in finding two sets A and B such that:
 
     A U B = S,
     |A| = |B|,
     sum(A) = sum(B),
     sum_squares(A) = sum_squares(B)
 
  """

  This model uses a binary matrix to represent the sets.

  
  Also, compare with other models which uses var sets:
  * MiniZinc: http://www.hakank.org/minizinc/set_partition.mzn
  * Gecode/R: http://www.hakank.org/gecode_r/set_partition.rb 

  This Comet model was created by Hakan Kjellerstrand (hakank@gmail.com)
  Also, see my Comet page: http://www.hakank.org/comet 

 */

// Licenced under CC-BY-4.0 : http://creativecommons.org/licenses/by/4.0/

import cotfd;

int t0 = System.getCPUTime();

int n = 16;
int num_sets = 2;

assert(n % num_sets == 0); // sanity: is the partition possible?

Solver<CP> m();

var<CP>{bool} a[1..num_sets, 1..n](m); // the set


Integer num_solutions(0);

exploreall<m> {

  forall(i in 1..num_sets, j in i+1..num_sets) {

    // same cardinality
    m.post(sum(k in 1..n) a[i,k] == sum(k in 1..n) a[j,k]);
    
    // same sum
    m.post(sum(k in 1..n) k*a[i,k] == sum(k in 1..n) k*a[j,k]);
    
    // same sum squared
    m.post((sum(k in 1..n) (k*a[i,k])^2) == (sum(k in 1..n) (k*a[j,k])^2));
    
  }

  partition_sets(a);
  
  // symmetry breaking (for num_sets == 2)
  if (num_sets == 2)
    m.post(a[1,1] == true);


} using {
      
  labelFF(m);

  num_solutions := num_solutions + 1;

  cout << "sums: " << sum(j in 1..n) j*a[1,j] << endl;
  cout << "sums squared: " << (sum(j in 1..n) (j*a[1,j])^2) << endl;

  // show the partitions
  forall(i in 1..num_sets) {
    if ( sum(j in 1..n) a[i,j] > 0) {
      cout << i << ": ";
      forall(j in 1..n) {
        if (a[i,j])
          cout << j << " ";
      }
      cout << endl;
    }
  }
  cout << endl;


}

cout << "\nnum_solutions: " << num_solutions << endl;
    
int t1 = System.getCPUTime();
cout << "time:      " << (t1-t0) << endl;
cout << "#choices = " << m.getNChoice() << endl;
cout << "#fail    = " << m.getNFail() << endl;
cout << "#propag  = " << m.getNPropag() << endl;


//
// Partition the sets (binary matrix representation).
//
function void partition_sets(var<CP>{bool} [,] x) {
  Solver<CP> m = x[1,1].getSolver();
  range R1 = x.getRange(0);
  range R2 = x.getRange(1);

  forall(i in R1, j in R1 : i != j)
      m.post(sum(k in R2) (x[i,k] && x[j,k]) == 0);

  // ensure that all integers is in (exactly) one partition
  m.post((sum(i in R1, j in R2) x[i,j]) == R2.getSize());
}