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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应该衡量亲属关系。但:

  1. 对于查询,我不断得到trues connected(x,y)
  2. N而且我的distance(x,y,N)查询越来越多。有什么建议么?

以下是我的事实:

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,我想我现在已经设法解决了一半的问题。接受@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),!.

但是现在出现了一系列新问题

  1. 它不能接受任意查询,例如distance(X,y,n)
  2. 类似connected(X,y)返回重复结果的查询

我认为可以通过使用该findall/3谓词来删除重复的结果,但我对如何实际实现它一无所知。

4

1 回答 1

5

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联系的所有方式,这将是一个非常大的数字。由于您可能只对他们是否完全连接感兴趣,因此您应该在子句末尾添加剪切,以connected/3防止在您成功确定两个人连接后回溯,例如:

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 N 的无限增加值,但distance/3谓词仍然会确定两个人如何连接的不同路径。根据人的不同,最小距离不会是第一个被计算的。我会将 的定义更改为distance/3如下所示:

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

这是重用您的minimum/2谓词。请注意,您应该在此谓词的前两个子句中添加削减,以避免虚假的选择点。

关于您的其他问题:

3)你要区分寻找最小的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distance/4. 这与添加到的削减不同connected/3

可能有更好的方法可以避免区分这两种模式,但目前我能想到的只是使用某种广度优先搜索(以保证最小程度)......

4)我没有得到重复的答案,比如?- connected(X,victoria).你有例子吗?

于 2012-05-03T08:42:29.067 回答