Twelve statements: Difference between revisions
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Near miss: {1,2,4,7,8,9}, fail on 8
</pre>
=={{header|Picat}}==
{{trans|Prolog}}
<lang Picat>% {{trans|Prolog}}
go ?=>
puzzle,
fail, % check for more answers
nl.
go => true.
puzzle =>
% 1. This is a numbered list of twelve statements.
L = [A1, A2, A3, A4, A5, A6, A7, A8, A9, A10, A11, A12],
L :: 0..1,
element(1, L, 1),
% 2. Exactly 3 of the last 6 statements are true.
A2 #<=> sum(L[7..12]) #= 3,
% 3. Exactly 2 of the even-numbered statements are true.
A3 #<=> sum([L[I]:I in 1..12,I mod 2 == 0]) #= 2,
% 4. If statement 5 is true, then statements 6 and 7 are both true.
A4 #<=> (A5 #=> (A6 #/\ A7)),
% 5. The 3 preceding statements are all false.
A5 #<=> sum(L[2..4]) #= 0,
% 6. Exactly 4 of the odd-numbered statements are true.
A6 #<=> sum([L[I]:I in 1..12,I mod 2 == 1]) #= 4,
% 7. Either statement 2 or 3 is true, but not both.
A7 #<=> (A2 + A3 #= 1),
% 8. If statement 7 is true, then 5 and 6 are both true.
A8 #<=> (A7 #=> A5 #/\ A6),
% 9. Exactly 3 of the first 6 statements are true.
A9 #<=> sum(L[1..6]) #= 3,
% 10. The next two statements are both true.
A10 #<=> (A11 #/\ A12),
% 11. Exactly 1 of statements 7, 8 and 9 are true.
A11 #<=> (A7 + A8 + A9 #= 1),
% 12. Exactly 4 of the preceding statements are true.
A12 #<=> sum(L[1..11]) #= 4,
solve(L),
println('L'=L),
printf("Statements %w are true.\n", [I.to_string : I in 1..12, L[I] == 1].join(" ")),
nl.</lang>
{{out}}
<pre>L=[1,0,1,1,0,1,1,0,0,0,1,0]
Statements 1 3 4 6 7 11 are true.</pre>
=={{header|Prolog}}==
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