如何避免这种情况下

Question

我现在这个工作的重复我所想要的：

male(roelof). 
male(mans). 
male(ronald). 
male(jan). 

female(chantal). 
female(marie). 
female(gerda). 
female(dagmar). 
female(denise). 
female(kimberly). 

parent(mans,gerda). 
parent(mans,roelof). 
parent(marie,gerda). 
parent(marie,roelof). 
parent(dagmar,denise). 
parent(dagmar,kimberly). 
parent(ronald,denise). 
parent(ronald,kimberly). 
parent(chantal,tamara). 
parent(roelof,tamara). 
parent(jan,chantal). 
parent(jan,dagmar). 

% Looks for the father of a child. Gives the name if found. 
% if no father is found then it gives false 
father_child(Child) :- 
parent(Father, Child), 
male(Father). 

% Looks for the mother of a child. Gives the name if found. 
% if no mother is found then it gives false 
mother_child(Child) :- 
parent(Mother, Child), 
female(Mother). 

% Looks if two person has the same father. 
% Gives true if a person is a father of both persons 
% Gives false if no person is a father of both persons. 
same_father(Child, Sibling) :- 
    parent(Father,Child), 
    parent(Father,Sibling), 
    male(Father). 

% Looks if two person has the same mother. 
% Gives true if a person is a mother of both persons 
% Gives false if no person is a mother of both persons. 
same_mother(Child, Sibling) :- 
    parent(Mother,Child), 
    parent(Mother,Sibling), 
    female(Mother). 

% Looks if there are siblings of a person. 
% Persons are siblings if they have the same father or 
% if they have the same mother and not the same father. 
siblings(X,Y) :- 
     ( same_father(X, Y), 
      X \= Y 
     ; same_mother(X, Y), 
     \+ same_father(X, Y) 
    ). 

% Displays the output of siblings(X,Y) and takes care that 
% there are no duplicates. 
display_siblings(Person) :- 
     findall(Person - Y, (siblings(Person,Y), Y @< Person), Sibs), 
     display_the_siblings(Sibs). 

% Display a message if there are no siblings found. 
display_the_siblings([]) :- 
     write('Er zijn geen zussen/broers bekend'). 

display_many([]). 
display_many([H|T]):- 
     writeln('Many elements '-H), display_many(T). 

display_the-siblings([X]):- better_display([X]),!. 
display_the_siblings([H|T]):- better_display([H|T]). 

better_display([X]):- 
    writeln('Single Element '-X). 
better_display([X,Y|T]):- 
    writeln('Many elements '-X), display_many([Y|T]). 

但在这种情况下display_siblings（金佰利）是假的，因为kimberly @< denise失败。 我是否打开规则以便Y @< Person然后display_siblings(kimberly)起作用，但display(gerda)失败。

任何想法摆脱这个混乱？

鲁洛夫

Answer 1

看一看

apropos(setof). 
apropos(findall). 
apropos(member). 

它们的组合可以去除dulicates。

Answer 2

你可以尝试这样的事情

display_siblings(Person) :- 
    setof(Pair, (siblings(Person,Y), sib_pair(Person, Y, Pair)), Sibs), 
    display_the_siblings(Sibs). 

sib_pair(Person, Sib, Person-Sib) :- 
    Person @< Sib. 
sib_pair(Person, Sib, Sib-Person) :- 
    Person @> Sib.