/* 

  Self Referencing Quiz puzzle in Picat.

  http://www.lcc.uma.es/~afdez/srq/

  Model and description from GProlog ExamplesFD/srq.pl
  """
  Name           : srq.pl                                                 
  Title          : Self-Referential Quiz puzzle                           
  Original Source: M. Henz                                                
  Adapted by     : Daniel Diaz for GNU Prolog                             
  Date           : February 1997                                          
                                                                          
                                                                          
  Q1 : the first question whose answer is A is                            
       (A) 4   (B) 3   (C) 2   (D) 1   (E) none of the above          
     
  Q2 : the only two consecutive questions with identical answers are      
       (A) 3 and 4   (B) 4 and 5   (C) 5 and 6   (D) 6 and 7   (E) 7 and 8

  Q3 : the next question with answer A is                                 
       (A) 4   (B) 5   (C) 6   (D) 7   (E) 8                              

  Q4 : the first even numbered question with answer B is                  
       (A) 2   (B) 4   (C) 6   (D) 8   (E) 10                             

  Q5 : the only odd numbered question with answer C is                    
       (A) 1   (B) 3   (C) 5   (D) 7   (E) 9                              

  Q6 : a question with answer D                                           
       (A) comes before this one but not after this one                   
       (B) comes after this one but not before this one                   
       (C) comes before and after this one                                
       (D) does not occur at all                                          
       (E) none of the above                                              

  Q7 : the last question whose answer is E is                             
       (A) 5   (B) 6   (C) 7   (D) 8   (E) 9                              

  Q8 : the number of questions whose answers are conconants is            
       (A) 7   (B) 6   (C) 5   (D) 4   (E) 3                              

  Q9 : the number of questions whose answers are vowels is                
       (A) 0   (B) 1   (C) 2   (D) 3   (E) 4                              

  Q10: the answer of this question is                                     
       (A) A   (B) B   (C) C   (D) D   (E) E                              
  """
                                                                          
  Solution:                                                               
  [3,1,2,2,1,2,5,2,5,4]                                                   
   C,A,B,B,A,B,E,B,E,D                                                    


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

*/


% import util.
import cp.


main => go.


go ?=>
   N = 5,

   Qs = [Q1,Q1,Q2,Q3,Q4,Q5,Q6,Q7,Q8,Q9,Q10],
   Qs :: 1..5,

   Bools = [BeforeQ4,AfterQ4,BCD1,BCD2,BCD3,BCD4,BCD5,BCD6,BCD7,BCD8,BCD9,BCD10],
   Bools :: 0..1,

   BCD :: 0..10,
   AE :: 0..10,

   Bools2 = [AE1,AE2,AE3,AE4,AE5,AE6,AE7,AE8,AE9,AE10],
   Bools2 :: 0..1,

   LD = [Q1,Q2,Q3,Q4,Q5,Q6,Q7,Q8,Q9,Q10],
   LD :: 1..N,

   %
   % The two commented lines was commented in the GProlog source.
   %
   (Q1#=1 #<=> (Q4#=1 #/\ Q1 #!= 1 #/\ Q2 #!= 1 #/\ Q3 #!= 1)),
   (Q1#=2 #<=> (Q3#=1 #/\ Q1 #!= 1 #/\ Q2 #!= 1)),
   (Q1#=3 #<=> (Q2#=1 #/\ Q1 #!= 1)),
   (Q1#=4 #<=> Q1#=1),
   %(Q1#=5 #<=> (Q1 #!= 1 #/\ Q2 #!= 1 #/\ Q3 #!= 1 #/\ Q4 #!= 1)),


   (Q2#=1 #<=> Q3#=Q4),
   (Q2#=2 #<=> Q4#=Q5),
   (Q2#=3 #<=> Q5#=Q6),
   (Q2#=4 #<=> Q6#=Q7),
   (Q2#=5 #<=> Q7#=Q8),


   (Q3#=1 #<=> Q4#=1),
   (Q3#=2 #<=> (Q5#=1 #/\ Q4 #!= 1)),
   (Q3#=3 #<=> (Q6#=1 #/\ Q4 #!= 1 #/\ Q5 #!= 1)),
   (Q3#=4 #<=> (Q7#=1 #/\ Q4 #!= 1 #/\ Q5 #!= 1 #/\ Q6 #!= 1)),
   (Q3#=5 #<=> (Q8#=1 #/\ Q4 #!= 1 #/\ Q5 #!= 1 #/\ Q6 #!= 1 #/\ Q7 #!= 1)),


   (Q4#=1 #<=> Q2#=2),
   (Q4#=2 #<=> (Q4#=2 #/\ Q2 #!= 2)),
   (Q4#=3 #<=> (Q6#=2 #/\ Q2 #!= 2 #/\ Q4 #!= 2)),
   (Q4#=4 #<=> (Q8#=2 #/\ Q2 #!= 2 #/\ Q4 #!= 2 #/\ Q6 #!= 2)),
   (Q4#=5 #<=> (Q10#=2 #/\ Q2 #!= 2 #/\ Q4 #!= 2 #/\ Q6 #!= 2 #/\ Q8 #!= 2)),


   (Q5#=1 #<=> (Q1#=3 #/\ Q3 #!= 3 #/\ Q5 #!= 3 #/\ Q7 #!= 3 #/\ Q9 #!= 3)),
   (Q5#=2 #<=> (Q3#=3 #/\ Q1 #!= 3 #/\ Q5 #!= 3 #/\ Q7 #!= 3 #/\ Q9 #!= 3)),
   (Q5#=3 #<=> (Q5#=3 #/\ Q1 #!= 3 #/\ Q3 #!= 3 #/\ Q7 #!= 3 #/\ Q9 #!= 3)),
   (Q5#=4 #<=> (Q7#=3 #/\ Q1 #!= 3 #/\ Q2 #!= 3 #/\ Q5 #!= 3 #/\ Q9 #!= 3)),
   (Q5#=5 #<=> (Q9#=3 #/\ Q1 #!= 3 #/\ Q3 #!= 3 #/\ Q5 #!= 3 #/\ Q7 #!= 3)),


   (BeforeQ4 #<=> (Q1#=4 #\/ Q2#=4 #\/ Q3#=4 #\/ Q4#=4 #\/ Q5#=4)),
   (AfterQ4  #<=> (Q7#=4 #\/ Q8#=4 #\/ Q9#=4 #\/ Q10#=4)),

   (Q6#=1 #<=> (BeforeQ4 #/\ #~AfterQ4)),
   (Q6#=2 #<=> (#~ BeforeQ4 #/\ AfterQ4)),
   (Q6#=3 #<=> (BeforeQ4 #/\ AfterQ4)),
   (Q6#=4 #<=> (Q1 #!= 4 #/\ Q2 #!= 4 #/\ Q3 #!= 4 #/\ Q4 #!= 4 #/\ Q5 #!= 4 #/\
              Q6 #!= 4 #/\ Q7 #!= 4 #/\ Q8 #!= 4 #/\ Q9 #!= 4 #/\ Q10 #!= 4)),
   %(Q6#=5 #<=> Q6#=4)


   (Q7#=1 #<=> (Q5#=5 #/\ Q6 #!= 5 #/\ Q7 #!= 5 #/\ Q8 #!= 5 #/\ Q9 #!= 5 #/\ Q10 #!= 5)),
   (Q7#=2 #<=> (Q6#=5 #/\ Q7 #!= 5 #/\ Q8 #!= 5 #/\ Q9 #!= 5 #/\ Q10 #!= 5)),
   (Q7#=3 #<=> (Q7#=5 #/\ Q8 #!= 5 #/\ Q9 #!= 5 #/\ Q10 #!= 5)),
   (Q7#=4 #<=> (Q8#=5 #/\ Q9 #!= 5 #/\ Q10 #!= 5)),
   (Q7#=5 #<=> (Q9#=5 #/\ Q10 #!= 5)),

   % note the extensive grouping...
   ((BCD1 #<=> (Q1 #>= 2 #/\ Q1 #<= 4)) #/\ (AE1 #<=> #~ BCD1)),
   ((BCD2 #<=> (Q2 #>= 2 #/\ Q2 #<= 4)) #/\ (AE2 #<=> #~ BCD2)),
   ((BCD3 #<=> (Q3 #>= 2 #/\ Q3 #<= 4)) #/\ (AE3 #<=> #~ BCD3)),
   ((BCD4 #<=> (Q4 #>= 2 #/\ Q4 #<= 4)) #/\ (AE4 #<=> #~ BCD4)),
   ((BCD5 #<=> (Q5 #>= 2 #/\ Q5 #<= 4)) #/\ (AE5 #<=> #~ BCD5)),
   ((BCD6 #<=> (Q6 #>= 2 #/\ Q6 #<= 4)) #/\ (AE6 #<=> #~ BCD6)),
   ((BCD7 #<=> (Q7 #>= 2 #/\ Q7 #<= 4)) #/\ (AE7 #<=> #~ BCD7)),
   ((BCD8 #<=> (Q8 #>= 2 #/\ Q8 #<= 4)) #/\ (AE8 #<=> #~ BCD8)),
   ((BCD9 #<=> (Q9 #>= 2 #/\ Q9 #<= 4)) #/\ (AE9 #<=> #~ BCD9)),
   ((BCD10#<=> (Q10 #>= 2 #/\ Q10 #<= 4)) #/\ (AE10 #<=> #~ BCD10)),

   (BCD#=BCD1+BCD2+BCD3+BCD4+BCD5+BCD6+BCD7+BCD8+BCD9+BCD10),

   (AE#=AE1+AE2+AE3+AE4+AE5+AE6+AE7+AE8+AE9+AE10),


   (Q8#=1 #<=> BCD#=7),
   (Q8#=2 #<=> BCD#=6),
   (Q8#=3 #<=> BCD#=5),
   (Q8#=4 #<=> BCD#=4),
   (Q8#=5 #<=> BCD#=3),

   (Q9#=1 #<=> AE#=0),
   (Q9#=2 #<=> AE#=1),
   (Q9#=3 #<=> AE#=2),
   (Q9#=4 #<=> AE#=3),
   (Q9#=5 #<=> AE#=4),

  Vars = Qs ++ Bools ++ Bools2 ++ LD,

  solve(Vars),

  Chars = ["A","B","C","D","E"],
  println(LD),
  println([Chars[C] : C in LD]),
  nl.

go => true.