srq10.pl

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/***********************************************************************
 * A solution to SRQ (Self-referential quizzes) implemented in B-Prolog
 * Solution similar to N-queens problem solutions. 
 * This involves only 10 FD variables in the domain 1..5.
 * Antonio J. Fernandez                      
 * *********************************************************************/

q:- statistics(runtime,_),srq(L), 
    statistics(runtime,[_,Y]),
    write(L), nl,
    write('time : '), write(Y).

srq(Q):-
    Q=[Q1,Q2,Q3,Q4,Q5,Q6,Q7,Q8,Q9,Q10],
    domain(Q,1,5),

%%  Question 1  
    (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),

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

%% Question 3 
    (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),

%% Question 4 
    (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),

%% Question 5 
    (Q5#=1)#<=>(Q1#=3),
    (Q5#=2)#<=>(Q3#=3),
    (Q5#=3)#<=>(Q5#=3),
    (Q5#=4)#<=>(Q7#=3),
    (Q5#=5)#<=>(Q9#=3),

%% Question 6  
    Bef in 0..1, Aft in 0..1,

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

    SmD in 0..10,  
    exactly(SmD,Q,4), 

    (Q6#=1)#<=> Bef #/\ (#\ Aft),
    (Q6#=2)#<=> Aft #/\ (#\ Bef),
    (Q6#=3)#<=> Bef #/\ Aft,
    (Q6#=4)#<=>(SmD#=0),
    (Q6#=5)#<=>(Q6#=4),

%%  Question 7 
    (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),

%%  Question 8 
    SmBCD in 0..10,
    exactly(SmB,Q,2),
    exactly(SmC,Q,3),

    SmBCD #= SmB + SmC + SmD,

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

%%  Question 9      
    SmA in 0..10, SmE in 0..10, SmAE in 0..10,

    10 #= SmAE + SmBCD,   

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

%%    10 #= SmAE + SmBCD,

%% labeling.
%%
%%  FIRST FAIL
%%    labeling([ff],Q).

%%  MIN Size, More constrained.
%%  labeling([ffc],Q).

%%  NORMAL
    labeling([],Q).

/*****************************************************************************
IMPORTANT.
exactly(X,L,N) is true if X occurs exactly N times in the list L. 
******************************************************************************/

exactly(_, [], 0).
exactly(X, [Y|L], N) :-
    X #= Y #<=> B,
    N #= M+B,
    exactly(X, L, M).