/*
Langford's number problem in Comet

Langford's number problem (CSP lib problem 24)
http://www.csplib.org/Problems/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<CP> m();
var<CP>{int} position[i in positionDomain](m, positionDomain);
var<CP>{int} solution[i in positionDomain](m, 1..k);

Integer num_solutions(0);

// explore<m> {
exploreall<m> {

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;