024: Langford's number problem

/*
  Langford's number problem in Comet


  Langford's number problem (CSP lib problem 24)
  http://www.csplib.org/prob/prob024/
  """
  Arrange 2 sets of positive integers 1..k to a sequence,
  such that, following the first occurence of an integer i, 
  each subsequent occurrence of i, appears i+1 indices later
  than the last. 
  For example, for k=4, a solution would be 41312432
  """
  
  * John E. Miller: Langford's Problem
    http://www.lclark.edu/~miller/langford.html
  
  * Encyclopedia of Integer Sequences for the number of solutions for each k
    http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=014552
 

  Also, see the following models:
  - MiniZinc: http://www.hakank.org/minizinc/langford2.mzn
  - Gecode/R: http://www.hakank.org/gecode_r/langford.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;
// import cotls;
// import cotln;
int t0 = System.getCPUTime();

int k = 4;
if (System.argc() >= 3) {
  k = System.getArgs()[2].toInt();
}


range positionDomain = 1..2*k;

Solver m();
var{int} position[i in positionDomain](m, positionDomain);
var{int} solution[i in positionDomain](m, 1..k);

Integer num_solutions(0);

// explore {
exploreall {

  forall(i in 1..k) {
    m.post(position[i+k] == position[i] + i+1 );
    m.post(solution[position[i]] == i );
    m.post(solution[position[k+i]] == i);
  }

  m.post(alldifferent(position));
  // symmetry breaking
  m.post(solution[1] < solution[2*k]);


  } using {
      
    // label(position);
    // label(solution);
    /*
    forall(i in positionDomain : !position[i].bound()) {// by (-position[i].getSize()) {
      label(position[i]);
    }

    forall(i in positionDomain : !solution[i].bound()) {// by (-solution[i].getSize()) {
      label(solution[i]);
    }
    */
    labelFF(m);
   
    // cout << position << endl;
    cout << solution << endl;
    
    num_solutions := num_solutions + 1;

}

cout << "num_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;