/* 

  Crystal Maze problem in Picat.

  From
  http://phys.org/news/2015-11-problems-ai-future-british-gameshow.html
  """
  On the wall was written a clue: "No consecutive letters in adjacent circles". 
  The letters A to H were printed on circular plates which could be fitted onto each circle.
  
        ?  ?
    ?   A  H ?
        ?  ? 
  
   [
  
     Represented with numbers, the cell are:
      
         2 3
      1  4 5 6 
         7 8
  
     And the connections:
    
      1,2,
      1,4,
      1,7,
      2,3,
      2,4,
      2,5,
      3,5,
      3,6,
      4,5,
      4,7,
      4,8,
      5,6,
      5,7,
      5,8,
      6,8,
      7,8  
   ]   
  """

  There are  four solutions:

           D    F
      G    A    H    B
           C    E

  ----------

           E    B
      G    A    H    D
           C    F

  ----------

           C    E
      G    A    H    B
           D    F

  ----------

           D    B
      G    A    H    F
           E    C

  ----------


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

*/

import cp.


main => go.

go ?=>
  L = [A,_B,_C,_D,_E,_F,_G,H],
  N = L.len,
  L = 1..N,

  S = ["A","B","C","D","E","F","G","H"],

Connections = [
              [1,2],
              [1,4],
              [1,7],
              [2,3],
              [2,4],
              [2,5],
              [3,5],
              [3,6],
              [4,5],
              [4,7],
              [4,8],
              [5,6],
              [5,7],
              [5,8],
              [6,8],
              [7,8]],
  NumConnections = Connections.len,

  Hints = [
   0, % 1
   0, % 2
   0, % 3
   A, % 4
   H, % 5
   0, % 6
   0, % 7
   0  % 8
  ],
  

  X = new_list(N),
  X :: 1..N,

  all_different(X),

  foreach(I in 1..NumConnections)
     abs(X[Connections[I,1]] - X[Connections[I,2]]) #> 1
  end,

  foreach(I in 1..N) 
    Hints[I] #> 0 #=> X[I] #= Hints[I]
  end,

  solve(X),

  println(x=X),
  printf("\n"),
  printf("         %s    %s\n",S[X[2]],S[X[3]]),
  printf("    %s    %s    %s    %s\n", S[X[1]],S[X[4]],S[X[5]],S[X[6]]),
  printf("         %s    %s\n", S[X[7]],S[X[8]]),
  printf("\n"),


  fail,
  nl.

go => true.