/*

  Killer sudoku in Comet.

  http://en.wikipedia.org/wiki/Killer_Sudoku
  """
  Killer sudoku (also killer su doku, sumdoku, sum doku, addoku, or 
  samunamupure) is a puzzle that combines elements of sudoku and kakuro. 
  Despite the name, the simpler killer sudokus can be easier to solve 
  than regular sudokus, depending on the solver's skill at mental arithmetic; 
  the hardest ones, however, can take hours to crack.

  ...

  The objective is to fill the grid with numbers from 1 to 9 in a way that 
  the following conditions are met:

    * Each row, column, and nonet contains each number exactly once.
    * The sum of all numbers in a cage must match the small number printed 
      in its corner.
    * No number appears more than once in a cage. (This is the standard rule 
      for killer sudokus, and implies that no cage can include more 
      than 9 cells.)

  In 'Killer X', an additional rule is that each of the long diagonals 
  contains each number once.
  """

  Here we solve the problem from the Wikipedia page, also shown here
  http://en.wikipedia.org/wiki/File:Killersudoku_color.svg

  Note, this model is based on the generalized KenKen model: 
  http://www.hakank.org/comet/kenken2.co
  Killer Sudoku is simpler in that the only mathematical operation is 
  summation.

  The output is:
    2 1 5 6 4 7 3 9 8
    3 6 8 9 5 2 1 7 4
    7 9 4 3 8 1 6 5 2
    5 8 6 2 7 4 9 3 1
    1 4 2 5 9 3 8 6 7
    9 7 3 8 1 6 4 2 5
    8 2 1 7 3 9 5 4 6
    6 5 9 4 2 8 7 1 3
    4 3 7 1 6 5 2 8 9

    num_solutions: 1
    time:      12
    #choices = 56
    #fail    = 136
    #propag  = 3013
    

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

 */

import cotfd;

int t0 = System.getCPUTime();

int n = 9;
range R = 1..n;


tuple cell {
  int r; // row
  int c; // column
} 

tuple P {
  set{cell} cells; // cells
  int res; // result
}

//
// state the problem (without the operation)
//
// http://en.wikipedia.org/wiki/File:Killersudoku_color.svg
int num_p = 28; // number of cages
P problem[1..num_p] = 
  [
   P({cell(1,1), cell(1,2)}, 3),
   P({cell(1,3), cell(1,4), cell(1,5)}, 15),
   P({cell(1,6), cell(2,5), cell(2,6), cell(3,5)}, 22),
   P({cell(1,7), cell(2,7)}, 4),
   P({cell(1,8), cell(2,8)}, 16),
   P({cell(1,9), cell(2,9), cell(3,9), cell(4,9)}, 15),
   P({cell(2,1), cell(2,2), cell(3,1), cell(3,2)}, 25),
   P({cell(2,3), cell(2,4)}, 17),
   P({cell(3,3), cell(3,4), cell(4,4)}, 9),
   P({cell(3,6), cell(4,6),cell(5,6)}, 8),
   P({cell(3,7), cell(3,8),cell(4,7)}, 20),
   P({cell(4,1), cell(5,1)},6),
   P({cell(4,2), cell(4,3)},14),
   P({cell(4,5), cell(5,5),cell(6,5)},17),
   P({cell(5,2), cell(5,3),cell(6,2)},13),
   P({cell(5,4), cell(6,4),cell(7,4)},20),
   P({cell(5,9), cell(6,9)}, 12),
   P({cell(6,1), cell(7,1),cell(8,1),cell(9,1)},27),
   P({cell(6,3), cell(7,2),cell(7,3)},6),
   P({cell(6,6), cell(7,6), cell(7,7)}, 20),
   P({cell(6,7), cell(6,8)},6),
   P({cell(7,5), cell(8,4),cell(8,5),cell(9,4)},10),
   P({cell(7,8), cell(7,9),cell(8,8),cell(8,9)},14),
   P({cell(8,2), cell(9,2)}, 8),
   P({cell(8,3), cell(9,3)},16),
   P({cell(8,6), cell(8,7)},15),
   P({cell(9,5), cell(9,6),cell(9,7)},13),
   P({cell(9,8), cell(9,9)},17)

   ];

Solver<CP> m();

var<CP>{int} x[1..n, 1..n](m, R);

//
// assumption: only the segments with 2 cells can be minus or div.
//
function void calc(Solver<CP> m, set{cell} cc, var<CP>{int}[,] x, int res) {
  m.post(sum(i in cc) x[i.r, i.c] == res);
}


Integer num_solutions(0);

exploreall<m> {

  // all rows, columns, and nonets must be unique
  forall(i in R)
      m.post(alldifferent(all(j in R) x[i,j]));

  forall(j in R)
    m.post(alldifferent(all(i in R) x[i,j]));

  forall(i in 0..2,j in 0..2) {
    m.post(alldifferent(all(r in i*3+1..i*3+3,c in j*3+1..j*3+3)  x[r,c]));
  }
  

  // solve the cages
  forall(i in 1..num_p) {
    calc(m, problem[i].cells, x, problem[i].res);
  }


} using {
      
  // label(m);
  forall(i in 1..n, j in 1..n : !x[i,j].bound())  {
    tryall<m>(v in 1..n  : x[i,j].memberOf(v)) 
      label(x[i,j], v);
  }

  num_solutions := num_solutions + 1;
      
  forall(i in 1..n) {
    forall(j in 1..n) {
      cout << x[i,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;