/* 

  The Bomb puzzle in Picat.

  From http://blog.kutterer.at/?p=7
  """
  The bomb has 7 flip switches. To disarm the bomb you must put the 
  switches in their correct position. And you got this inforamtions: 
  the bomb detonates when:

    1. if switch 3 is on plus 2 and 4 are off, boom!
    2. if 1 and 4 off plus 7 is on, boom!
    3. if 1, 3 and 4 are off, boom!
    4. if 6 is off plus 2 and 3 are on, boom!
    5. if 4 and 3 are on, boom!
    6. if 6 is on and if, as long as 7 is on, 1 is on too, boom!
    7. if 6 is on and, provided 7 is on, 1 is on too, boom!
    8. if 1 and 5 are on plus 7 is off, boom!
    9. if 3 is off plus 4 and 5 are on, boom!
   10. if 1 and 7 are on, boom!
   11. if 5 off and if, as long as 2 and 6 are on, 3 is on too, boom!
   12. if 7 off plus 3 or 4 on, boom!
   13. if switch 6 and 7 are in different positions, boom!
   14. if switch 2, 3 and 5 are off, boom!
   15. if switch 1 and 2 are on and switch 5 and 7 are off, boom!
   16. if 6 and 7 are off, boom!
  """
  
  Originally from the German puzzle Die Bombe by Angela and Otto Janko
  http://www.janko.at/Raetsel/Logik/012.a.htm
 

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

*/

import cp.

main => go.

go =>
  N = 7,
  X = new_list(N), % the switches
  X :: 0..1,
  
  #~(

    %   1. if switch 3 is on plus 2 and 4 are off, boom!
        (X[3] #/\ X[2] #= 0 #/\ X[4] #= 0)

    %   2. if 1 and 4 off plus 7 is on, boom!
    #\/ (X[1] #= 0 #/\ X[4] #= 0 #/\ X[7])

    %   3. if 1, 3 and 4 are off, boom!
    #\/ (X[1] #= 0 #/\  X[3] #= 0 #/\ X[4] #= 0)

    %   4. if 6 is off plus 2 and 3 are on, boom!
    #\/ (X[6] #= 0 #/\ X[2] #/\ X[3])

    %   5. if 4 and 3 are on, boom!
    #\/ (X[4] #/\ X[3])

    %   6. if 6 is on and if, as long as 7 is on, 1 is on too, boom!
    #\/ (X[6] #/\ X[7] #/\ X[1])

    %   7. if 6 is on and, provided 7 is on, 1 is on too, boom!
    #\/ (X[6] #/\ X[7] #/\ X[1])

    %   8. if 1 and 5 are on plus 7 is off, boom!
    #\/ (X[1] #/\ X[5] #/\ X[7] #= 0)

    %   9. if 3 is off plus 4 and 5 are on, boom!
    #\/ (X[3] #= 0 #/\ X[4] #/\ X[5]) 

    %  10  if 1 and 7 are on, boom! 
    #\/ (X[1] #/\ X[7]) 

    %  11. if 5 off and if, as long as 2 and 6 are on, 3 is on too, boom!
    #\/ (X[5] #= 0 #/\ X[2] #/\ X[6] #/\ X[3]) 

    %  12. if 7 off plus 3 or 4 on, boom!
    #\/ (X[7] #= 0 #/\ (X[3] #\/ X[4]))

    %  13. if switch 6 and 7 are in different positions, boom!
    #\/ (X[6] #!= X[7])

    %  14. if switch 2, 3 and 5 are off, boom!
    #\/ (X[2] #= 0 #/\ X[3] #= 0 #/\ X[5] #= 0)

    %  15. if switch 1 and 2 are on and switch 5 and 7 are off, boom!
    #\/ (X[1] #/\ X[2] #/\ X[5] #= 0 #/\ X[7] #= 0) 

    %  16. if 6 and 7 are off, boom!
    #\/ (X[6] #= 0 #/\ X[7] #= 0)

  ),
  
  solve(X),
  
  println(X),
  fail,
  
  nl.

% A different approach.
go2 =>
  defuse(L),
  println(L),
  fail.

defuse([A,B,C,D,E,F,G]) :-
  per_bin([_,_,_,_,_,_,_],[A,B,C,D,E,F,G]),
  not(boom([A,B,C,D,E,F,G])).

per_bin([],[]).
per_bin([X|Xs],[X|Ys]) :-
  member(X,[0,1]),
  per_bin(Xs,Ys).


%     1 2 3 4 5 6 7
boom([_,0,1,0,_,_,_]).
boom([0,_,_,0,_,_,1]).
boom([0,_,0,0,_,_,_]).
boom([_,1,1,_,_,0,_]).
boom([_,_,1,1,_,_,_]).
boom([1,_,_,_,_,1,1]).
boom([1,_,_,_,1,_,0]).
boom([_,_,0,1,1,_,_]).
boom([1,_,_,_,_,_,1]).
boom([_,1,1,_,0,1,_]).
boom([_,_,1,1,_,_,0]).
boom([_,_,_,_,_,0,1]).
boom([_,_,_,_,_,1,0]).
boom([_,0,0,_,0,_,_]).
boom([1,1,_,_,0,_,0]).
boom([_,_,_,_,_,0,0]).