- 89 名前：デフォルトの名無しさん mailto:sage [2008/05/24(土) 15:33:59 ]
- prolog で解いてみた。いろいろ微妙。
xor( A, B ) :-
A -> ¥+B ; B.
truth(Is) :-
Is = [ [rabbit, g], [hatman, n], [yamane, n] ] ;
Is = [ [rabbit, n], [hatman, g], [yamane, n] ] ;
Is = [ [rabbit, n], [hatman, n], [yamane, g] ] .
rabbit_says(Is) :-
truth(Is),
member( [rabbit, n], Is ).
hatman_says(Is) :-
truth(Is),
member( [hatman, n], Is ).
yamane_says(Is) :-
xor( rabbit_says(Is), hatman_says(Is) ).
more_research(Is) :-
xor( rabbit_says(Is), yamane_says(Is) ).
solve(It) :-
truth(It),
yamane_says(It),
more_research(It).
?- findall(It, solve(It), All).
All = [[[rabbit, g], [hatman, n], [yamane, n]]].
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