Prolog陷入无限循环

Question

我有两个问题困扰了我好几个小时.connected/2应该判断两个人是否有联系; distance/3应该衡量血缘关系.但:

1. 我一直在true寻求无限的查询connected(x,y);
2. 而且我N的distance(x,y,N)查询无限增加.有什么建议？

这是我的事实:

male(ted).
male(barney).
male(ranjit).
male(marshall).
male(tony).
male(swarley).
male(steve).
male(chuck).
male(john).
male(devon).
male(morgan).

female(robin).
female(lily).
female(wendy).
female(stellar).
female(abby).
female(victoria).
female(carina).
female(sarah).
female(ellie).

married(ted,      robin).
married(marshall, lily).
married(ranjit,   wendy).
married(stellar,  tony).
married(steve,    carina).
married(sarah,    chuck).
married(ellie,    devon).

father(ted,      barney).
father(ted,      ranjit).
father(marshall, wendy).
father(ranjit,   stellar).
father(tony,     abby).
father(tony,     swarley).
father(tony,     victoria).
father(steve,    chuck).
father(steve,    ellie).
father(chuck,    john).
father(devon,    morgan).

mother(robin,    barney).
mother(robin,    ranjit).
mother(lily,     wendy).
mother(wendy,    stellar).
mother(stellar,  abby).
mother(stellar,  swarley).
mother(stellar,  victoria).
mother(carina,   chuck).
mother(carina,   ellie).
mother(sarah,    john).
mother(ellie,    morgan).

现在,我的谓词:

parent(X,Y) :- father(X,Y).
parent(X,Y) :- mother(X,Y).

son(X,Y) :-
    male(X),
    parent(Y,X).

daughter(X,Y) :-
    female(X),
    parent(Y,X).

sibling(X,Y) :-
    parent(Z,X),
    parent(Z,Y).

cousin(X,Y) :-
    parent(Z,X),
    parent(W,Y),
    parent(G,Z),
    parent(G,W).

ancestor(X,Y) :-
    parent(X,Z),
    ancestor(Z,Y).
ancestor(X,Y) :- parent(X,Y).

notmember(X,[]).
notmember(X,[H|T]) :- 
    X \= H,
    notmember(X,T).

connected(X,Y,_) :- X == Y.
connected(X,Y,Visited) :- 
    ancestor(X,Z),
    notmember(Z,Visited),
    connected(Z,Y,[Z|Visited]).
connected(X,Y,Visited) :- 
    ancestor(Z,X),
    notmember(Z,Visited),
    connected(Z,Y,[Z|Visited]).
connected(X,Y,Visited) :- 
    sibling(X,Z),
    notmember(Z,Visited),
    connected(Z,Y,[Z|Visited]).
connected(X,Y,Visited) :- 
    married(X,Z),
    notmember(Z,Visited),
    connected(Z,Y,[Z|Visited]).
connected(X,Y) :- connected(X,Y,[X]).

minimum(X,[X]).
minimum(X,[M,H|T]) :- 
    M =< H,
    minimum(X,[M|T]).
minimum(X,[M,H|T]) :-
    M > H,
    minimum(X,[H|T]).

distance(X,X,_,0).
distance(X,Y,Visited,N) :- 
    parent(X,Z),
    notmember(Z,Visited),
    distance(Z,Y,[Z|Visited],N1),
    N is N1+1.
distance(X,Y,Visited,N) :- 
    parent(Z,X),
    notmember(Z,Visited),
    distance(Z,Y,[Z|Visited],N1),
    N is N1+1.
distance(X,Y,N) :- distance(X,Y,[],N).

编辑:谢谢,我想我已经设法解决了问题的一半.按照@ twinterer的建议,我已经修复了这样的谓词

connected(X,Y,_) :- X == Y.
connected(X,Y,V) :-
    married(X,Z),
    notmember(Z,V),
    connected(Z,Y,[Z|V]),!.
connected(X,Y,V) :-
    sibling(X,Z),
    notmember(Z,V),
    connected(Z,Y,[Z|V]),!.
connected(X,Y,V) :-
    parent(X,Z),
    notmember(Z,V),
    connected(Z,Y,[Z|V]),!.
connected(X,Y,V) :-
    parent(Z,X),
    notmember(Z,V),
    connected(Z,Y,[Z|V]),!.
connected(X,Y) :- connected(X,Y,[X]).

minimum(X,[X]).
minimum(X,[M,H|T]) :- 
    M =< H,
    minimum(X,[M|T]).
minimum(X,[M,H|T]) :-
    M > H,
    minimum(X,[H|T]).

count(X,[],0).
count(X,[X|T],N) :-
    count(X,T,N1),
    N is N1+1.
count(X,[H|T],N) :-
    X \== H,
    count(X,T,N1),
    N is N1.

distance(X,X,Visited,0) :-
    count(X,Visited,N),
    N =< 1, !.
distance(X,Y,Visited,N) :- 
    parent(X,Z),
    (notmember(Z,Visited)->
        distance(Z,Y,[Z|Visited],N1),
        N is N1+1
    ;
        fail
    ),!.
distance(X,Y,Visited,N) :- 
    parent(Z,X),
    (notmember(Z,Visited)->
        distance(Z,Y,[Z|Visited],N1),
        N is N1+1
    ;
        fail
    ),!.
distance(X,Y,N) :- 
    findall(N1,distance(X,Y,[X],N1),L),!,
    minimum(N,L),!.

但现在出现了一系列新问题

1. 它不能像任意查询一样 distance(X,y,n)
2. 查询connected(X,y)返回重复结果

我认为通过使用该findall/3谓词可以实现删除重复结果,但我对如何实际实现它毫无头绪.

Answer 1

1)我不认为你以无限循环结束,但是你让你的程序探索两个人连接的所有方式,这将是一个非常大的数字.因为你很可能只对它们是否在所有连接的兴趣,你应该在末尾加上削减connected/3,一旦你已经成功地确定了两个人连接起来,例如,以防止回溯条款:

connected(X,Y,_) :- X == Y,!.
connected(X,Y,Visited) :- 
    ancestor(X,Z),
    notmember(Z,Visited),
    connected(Z,Y,[Z|Visited]),!.
...

2)当我测试你的代码时,我没有得到无限增加的N值,但是distance/3谓词仍然会确定两个人如何连接的不同路径.取决于人,最小距离不是第一个被计算的.我会将定义distance/3改为这样的:

distance(X,Y,N) :- 
    findall(N0, distance(X,Y,[],N0), Ns), !,
    minimum(N, Ns).

这是重用你的minimum/2谓词.请注意,您应该为此谓词的前两个子句添加剪切,以避免虚假的选择点.

关于您的其他问题:

3)你必须区分寻找最小N和寻找匹配度N的人:

distance(X,Y,N) :-
  nonground(N),!,
  findall(N0, distance(X,Y,[],N0), Ns), !,
  minimum(N, Ns).
distance(X,Y,N) :-
  distance(X,Y,[X],N).

此外,您需要删除添加到的剪切distance/4.这与削减加入不同connected/3!

可能有一种更好的方法可以避免区分这两种模式,但我现在能想到的只是使用某种广度优先搜索(为了保证最小程度)......

4)我没有得到重复的答案,例如?- connected(X,victoria).你有一个例子吗？