prolog - 在序言中的真实答案后不停止

Question

我写这个程序是为了给我这个结果:“X=约翰”“Y=简”

likes(john,mary).
likes(mary,jane).
likes(l,k).

likes(X,Y) :- likes(X,Z), likes(Z,Y).

但是如果运行这个程序使得这个结果:我认为程序进入循环!我想在真实答案后停下来!

1 ?- likes(X,Y).
X = john,
Y = mary ;
X = mary,
Y = jane ;
X = l,
Y = k ;
X = john,
Y = jane ;
ERROR: Out of local stack

我有错误!这个问题怎么解决？

通过调试:

[debug] 3 ?- likes(john,Y).
 T Call: (6) likes(john, _G2162)
 T Exit: (6) likes(john, mary)
Y = mary ;
 T Redo: (6) likes(john, _G2162)
 T Call: (7) likes(john, _G2267)
 T Exit: (7) likes(john, mary)
 T Call: (7) likes(mary, _G2162)
 T Exit: (7) likes(mary, jane)
 T Exit: (6) likes(john, jane)
Y = jane ;
 T Redo: (7) likes(mary, _G2162)
 T Call: (8) likes(mary, _G2267)
 T Exit: (8) likes(mary, jane)
 T Call: (8) likes(jane, _G2162)
 T Call: (9) likes(jane, _G2267)
 T Call: (10) likes(jane, _G2267)
 T Call: (11) likes(jane, _G2267)
 T Call: (12) likes(jane, _G2267)
 T Call: (13) likes(jane, _G2267)
 T Call: (14) likes(jane, _G2267)
 T Call: (15) likes(jane, _G2267)

等等..为什么喜欢(简，_G2267)??????

Answer 1

如果你需要 "X=john""Y=jane"那么你必须重命名:

likes(X,Y) :- likes(X,Z), likes(Z,Y).

对于其他名称，例如:

likestransitive(X,Y) :- likes(X,Z), likes(Z,Y).

这样你就可以得到你需要的解决方案:

1 ?- transitivelikes(X,Y).    
X = john,    
Y = jane;    
false.