Question

我是Prolog的新手，我不知道该如何解决。我必须写一个兄弟姐妹规则。一直到代码底部，我都有兄弟姐妹规则。规则的问题在于它会返回兄弟姐妹两次。例如，如果我想得到茱莉亚的兄弟姐妹，我会两次获得“罗莎”。我认为正在发生的事情是该程序对每个父级（alex和lina）进行评估。那么，我该如何修正自己的规则，使我只得到朱莉娅（或任何人）的兄弟姐妹一次？

parent(alex,julia).
parent(alex,rosa).
parent(lina,julia).
parent(lina,rosa).
parent(romeo,peter).
parent(julia,peter).
parent(rosa,silvia).
parent(oscar,ida).
parent(eva,ida).
parent(eva,bruno).
parent(peter,bruno).
parent(peter,georg).
parent(peter,irma).
parent(ruth,georg).
parent(ruth,irma).
parent(silvia,otto).
parent(silvia,pascal).
parent(irma,olga).
parent(irma,jean).
parent(otto,olga).
parent(otto,jean).
parent(jean,tina).
parent(marie,tina).

male(alex).
male(romeo).
male(oscar).
male(peter).
male(bruno).
male(georg).
male(otto).
male(pascal).
male(jean).


husband(alex,lina).
husband(romeo,julia).
husband(oscar,eva).
husband(peter,ruth).
husband(otto,irma).
husband(jean,marie).

% father(X,Y) :- X is the father of Y.

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

% grandfather(X,Y) :- X is the grandfather of Y.

grandfather(X,Y) :- father(X,P), parent(P,Y).

% brother(X,Y) :- X is the brother of Y.

brother(X,Y) :- parent(P,X), parent(P,Y), male(X), X \= Y.

% uncle(X,Y) :- X is the uncle of Y.

uncle(X,Y) :- brother(X,P), parent(P,Y).

% female(X) :- X is a female person.

female(X) :- \+ male(X).

% sister(X,Y) :- X is the sister of Y.

sister(X,Y) :- parent(P,X), parent(P,Y), female(X), X \= Y.

% has_son(X) :- the person X has a son.

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

% married(X,Y) :- X and Y are married to each other.

married(X,Y) :- husband(X,Y).
married(X,Y) :- husband(Y,X).

% brother_in_law(X,Y) :- X is the brother-in-law of Y.

brother_in_law(X,Y) :- brother(X,P), married(P,Y).
brother_in_law(X,Y) :- husband(X,W), sister(W,Y).

% ancestor(X,Y) :- X is an ancestor of Y.

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

% relatives(X,Y) :- X and Y are relatives (related by blood to each other).

relatives(X,X).
relatives(X,Y) :- ancestor(X,Y).
relatives(X,Y) :- ancestor(Y,X).
relatives(X,Y) :- ancestor(A,X), ancestor(A,Y).

% ancestors(As,X) :- As is the set of all (known) ancestors of X.

ancestors(As,X) :- setof(A,ancestor(A,X),As).

% descendants(Ds,X) :- Ds is the set of all (known) descendants of X.

descendants(Ds,X) :-setof(D,ancestor(X,D),Ds).

% ancestor(X,Y,L) :- X is an ancestor of Y, and L is the list of names
% leading from X to Y.

ancestor(X,Y,[X,Y]) :- parent(X,Y).
ancestor(X,Y,[X|L]) :- parent(X,P), ancestor(P,Y,L).


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Exercise 1
% Write a rule that determines if mother (M) is mother of (C) child.
% mother(M, C) :-

mother(M,C) :- female(M), parent(M,C).

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Exercise 2
% Write a rule that infers the siblings of certain (X) person
% siblings. Ss is the siblings list.
% siblings(Ss, X) :-

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

Answer 1

您的怀疑是正确。您的谓词存在的问题是，一个人（通常）有两个父母：mother/2和father/2。

这意味着，如果rosa和julia是（ real ）兄弟姐妹（则它们共享相同的mother/2和father/2）。 / p>

如果我们trace.看到重复出现了这个原因：

?- trace.
true.

[trace]  ?- sibling(rosa, julia).
   Call: (8) sibling(rosa, julia) ? creep
   Call: (9) dif:dif(rosa, julia) ? creep
   Exit: (9) dif:dif(rosa, julia) ? creep
   Call: (9) parent(_5278, rosa) ? creep
   Exit: (9) parent(alex, rosa) ? creep
   Call: (9) parent(alex, julia) ? creep
   Exit: (9) parent(alex, julia) ? creep
   Exit: (8) sibling(rosa, julia) ? creep
true ;
   Redo: (9) parent(alex, julia) ? creep
   Fail: (9) parent(alex, julia) ? creep
   Redo: (9) parent(_5278, rosa) ? creep
   Exit: (9) parent(lina, rosa) ? creep
   Call: (9) parent(lina, julia) ? creep
   Exit: (9) parent(lina, julia) ? creep
   Exit: (8) sibling(rosa, julia) ? creep
true.

因此，首先我们通过sibling/2（两者中的alex）找到father/2关系，然后我们通过sibling/2找到lina关系（两者中的mother/2。

如果这里没有半兄弟姐妹（两个人共享相同的father/2或mother/2，但不共享相同的mother/2或{{1} }），那么我们可以通过仅选择两个父级之一来“消除重复”，例如father/2：

mother/2

根据注释，您的sibling(A, B) :- dif(A, B), mother(M, A), mother(M, B).谓词似乎无法正常工作。因此，我建议您尝试解决此问题。

或通过female/1：

father/2

如果我们考虑一个有半兄弟姐妹的家庭，我们可以通过同时检查sibling(A, B) :- dif(A, B), father(F, A), father(F, B).和sibling/2来实现一个考虑mother/2的谓词。

家庭序言练习

1 个答案: