puzzle.prolog

% A keypad lock on a door has a code of five different digits. Can you use the six clues below to determine the correct code to get in?
%
% 1. There are no zeroes in the five-digit code.
% 2. The first digit is smaller than the second digit.
% 3. Exactly two of the five digits are odd.
% 4. The middile digit is a perfect square.
% 5. The two-digit number formed by the fourth and fifth digits is a perfect square.
% 6. The two-digit number formed by the third and fourth digits is equal to the first digit multiplied by ythe second digit.
is_digit(1).
is_digit(2).
is_digit(3).
is_digit(4).
is_digit(5).
is_digit(6).
is_digit(7).
is_digit(8).
is_digit(9).

is_perfect_square(X) :- is_digit(T), X is T * T.

odd(1).
odd(N) :- N > 0, K is N - 2, odd(K).

two_odd([A,B,_,_,_]) :- odd(A), odd(B).
two_odd([A,_,C,_,_]) :- odd(A), odd(C).
two_odd([A,_,_,D,_]) :- odd(A), odd(D).
two_odd([A,_,_,_,E]) :- odd(A), odd(E).
two_odd([_,B,C,_,_]) :- odd(B), odd(C).
two_odd([_,B,_,D,_]) :- odd(B), odd(D).
two_odd([_,B,_,_,E]) :- odd(B), odd(E).
two_odd([_,_,C,D,_]) :- odd(C), odd(D).
two_odd([_,_,C,_,E]) :- odd(C), odd(E).
two_odd([_,_,_,D,E]) :- odd(D), odd(E).

all_diff([A,B,C,D,E]) :-
    A \= B, A \= C, A \= D, A \= E,
    B \= C, B \= D, B \= E,
    C \= D, C \= E,
    D \= E.

solution(Code) :-
    Code = [A,B,C,D,E],

    is_digit(A),
    is_digit(B),
    is_digit(C),
    is_digit(D),
    is_digit(E),

    all_diff(Code),

    A < B,

    two_odd(Code),

    C \= 1,
    is_perfect_square(C),

    P0 is D * 10 + E,
    is_perfect_square(P0),

    P1 is A * B,
    P1 is C * 10 + D.