/* 

  Latin square card puzzle in Picat.

  Problem from Mario Livio's book about group theory
  "The Equation that couldn't be solved",
  page 22
  """
  "... Incidentally, you may get a kick out of solving this
  eighteenth century card puzzle: Arrange all the jacks,
  queens, kings, and aces from a deck of cards in a square so that 
  no suit or value would appear twice in any row, column, or the
  two main diagonals.
  """
  
  Also see
  - http://en.wikipedia.org/wiki/Graeco-Latin_square
  - http://en.wikipedia.org/wiki/Thirty-six_officers_problem


  This Picat model was created by Hakan Kjellerstrand, hakank@gmail.com
  See also my Picat page: http://www.hakank.org/picat/

*/


import sat.

main => go.


go =>
  N = 4,
  All = findall(X,card_puzzle(N, X)),
  foreach(X in All)
    print_matrix(N,X)
  end,
  println(sols=All.length),
  nl.

go2 =>
  foreach(N in 2..20)
    println(n=N),
    ( card_puzzle(N,X), print_matrix(N,X) ; true),
    nl
  end,
  nl.


print_matrix(N,X) =>
  % println(x=X),
  M = (N*(N+1)) div 2,
  foreach(I in 1..N) 
    foreach(J in 1..N)
      printf("%2d/%2d ", X[I,J] div M, X[I,J] mod M)
    end,
    nl
  end,
  nl.

% for n=4 (and m = 10)
% cards = 
% values: i mod 10 
%      0, 1, 2, 3,  % suite 1: 0 div 10
%     10,11,12,13,  % suite 2: 1 div 10
%     20,21,22,23,  % suite 3: 2 div 10
%     30,31,32,33   % suite 4: 3 div 10
%
card_puzzle(N, X) =>
  println($card_puzzle(N, X)),
  M = (N*(N+1)) div 2,
  Cards = [I + M*J : I in 0..N-1, J in 0..N-1],
  println(cards=Cards),

  % decision variables
  X = new_array(N,N),
  X :: Cards,

  % all values must be different 
  % all_different(X.vars()),
  all_different($[X[I,J] : I in 1..N, J in 1..N]),

  % diagonals1
  all_different($[X[I,I] div M : I in 1..N ]),
  all_different($[X[I,I] mod M : I in 1..N ]),

  % diagonal2
  println(here1),
  all_different($[X[I,N-I+1] div M : I in 1..N ]),
  println(here2),
  all_different($[X[I,N-I+1] mod M : I in 1..N ]),
  println(here3),

  % rows, columns, 
  foreach(I in 1..N)
    all_different($[X[I,J] div M : J in 1..N ]),
    all_different($[X[J,I] div M : J in 1..N ]),

    all_different($[X[I,J] mod M : J in 1..N ]),
    all_different($[X[J,I] mod M : J in 1..N ])

  end,
  println(xxx),
  % symmetry breaking
  X[1,1] #= 0,

  solve([], X).