/* 

  Friends around a table in Picat.

  From https://swish.swi-prolog.org/p/seating477.pl
  and discussed in 
  https://swi-prolog.discourse.group/t/find-the-bug-in-this-logic-puzzle/4649
  """
  One of the friends is Bruce
  Alice is sitting directly opposite of David
  Henry is sitting between Greta and Eugene.
  There is one person between Greta and Claire.
  Eugene is sitting immediately to David's left.
  Franny is not next to Alice or David.
  """

  Here is a more complete description of the problem, from 
  https://olympiad.org.za/talent-search/wp-content/uploads/sites/2/2020/11/2020-Talent-Search-Solutions-book.pdf
  Page 61:
  """
  26 Seating Arrangement

  Eight friends are sitting in a circle. They are all facing inwards.
  We know the following facts about where they are sitting:
  1. Alice is sitting directly opposite of David.
  2. Henry is sitting between Greta and Eugene.
  3. Franny is not next to Alice or David.
  4. There is one person between Greta and Claire.
  5. Eugene is sitting immediately to David's left.

  Task:
  Place the friends in the correct places in the circle by dragging the letter next to their
  name to their chair.

  (You can click on the letter to return it to the list. There may be multiple correct
  solutions; you only need to find one. Don't forget to click on Save when done!)
  """

  We represents the positions as 
              1 
           8    2
         7        3
           6    4
              5

  There are 8 solutions:
    friends_position = [alice,greta,henry,eugene,david,bruce,franny,claire]
    friends_position = [claire,alice,greta,henry,eugene,david,bruce,franny]
    friends_position = [franny,claire,alice,greta,henry,eugene,david,bruce]
    friends_position = [bruce,franny,claire,alice,greta,henry,eugene,david]
    friends_position = [eugene,david,bruce,franny,claire,alice,greta,henry]
    friends_position = [henry,eugene,david,bruce,franny,claire,alice,greta]
    friends_position = [greta,henry,eugene,david,bruce,franny,claire,alice]

  With the symmetry breaking that Alice is in position 1, there is one solution:
   friends = [1,6,8,5,4,7,2,3]
   friends_position = [alice,greta,henry,eugene,david,bruce,franny,claire]

                alice
       claire            greta
     franny                 henry
        bruce           eugene
                david


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

*/

import v3_utils.
import cp.

main => go.

%
% CP approach
%
go ?=>
  N = 8,
  Friends = [Alice,_Bruce,Claire,David,Eugene,Franny,Greta,Henry],
  Friends :: 1..N,
  FriendsS = [alice, bruce,claire,david,eugene,franny,greta,henry],

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

  all_different(Friends[1..N]),
  assignment(Friends,FriendsPos),

  % Constraints (the hints)

  % One of the friends is Bruce
  
  % Alice is sitting directly opposite of David
  opposite(Alice,David,N),
  
  % Henry is sitting between Greta and Eugene.
  between2(Greta,Henry,Eugene,N),
  
  
  % There is one person between Greta and Claire.
  between2(Greta,_Someone,Claire,N),

  % Eugene is sitting immediately to David's left.
  David - Eugene #= 1,
  
  % Franny is not next to Alice or David.
  not_next_to(Franny,Alice,N),
  not_next_to(Franny,David,N),

  % Symmetry breaking
  Alice #= 1,

  solve(Friends),
  
  println(friends=Friends),
  Pos = [FriendsS[F] : F in FriendsPos],
  println(friends_position=Pos),

  printf("           %10w\n",Pos[1]),
  printf("   %10w       %10w\n",Pos[8], Pos[2]),
  printf(" %10w            %10w\n",Pos[7], Pos[3]),
  printf("   %10w       %10w\n",Pos[6], Pos[4]),
  printf("           %10w",Pos[5]),      

  nl,
  fail,
  
  nl.
go => true.

%
% CP version
%
opposite(X,Y,N) =>
  abs(X-Y) #= N div 2.

% B is between A and C
between2(A,B,C,N) =>
  Table = [{1+(I-1) mod N , 1+((I) mod N), 1+(I+1) mod N} : I in 1..N] ++
          [{1+(I+1) mod N , 1+((I) mod N), 1+(I-1) mod N} : I in 1..N],
  table_in({A,B,C},Table).

% A is not next to B
not_next_to(A,B,N) =>
  Table=[{1+(I-1) mod N , 1+((I  ) mod N)} : I in 1..N] ++
        [{1+(I  ) mod N , 1+((I-1) mod N)} : I in 1..N],
  table_notin({A,B},Table).