Prolog的列表差异常规

Question

我想实现在序言列表差异程序。 出于某种原因，下面的失败：

difference(Xs,Ys,D) :- difference(Xs,Ys,[],D).
difference([],_,A,D) :- D is A, !.
difference([X|Xs],Ys,A,D) :-
  not(member(X,Ys)),
  A1 is [X|A],
  difference(Xs,Ys,A1,D).

当尝试：

?- difference([1,2],[],D).

我得到这个错误：

ERROR: '.'/2: Type error: `[]' expected, found `1' ("x" must hold one character)
^  Exception: (10) _L161 is [2|1] ? 

Answer 1

您使用A1是[X | A]是不正确。谓词的是仅用于运算。 顺便说一句，SWI-Prolog的具有内置的减法谓词：

1 ?- subtract([1,2,3,a,b],[2,a],R).
R = [1, 3, b].

2 ?- listing(subtract).
subtract([], _, []) :- !.
subtract([A|C], B, D) :-
        memberchk(A, B), !,
        subtract(C, B, D).
subtract([A|B], C, [A|D]) :-
        subtract(B, C, D).

true.

这是你需要什么？

Answer 2

minus([H|T1],L2,[H|L3]):-
    not(member(H,L2)),
    minus(T1,L2,L3).
minus([H|T1],L2,L3):-
    member(H,L2),
    minus(T1,L2,L3).
minus([],_,[]). 

---

minus([1,2,3,4,3], [1,3], L).

output: L=[2,4]

Answer 3

使用发现所有的溶液变得明显：

difference(Xs,Ys,D) :- 
  findall(X,(member(X,Xs),not(member(X,Ys))),D).

Answer 4

always (subtructLists(List, [Head|Rest], Result): -
       ( 
          delete_element(Head, List, Subtructed)
        , !
        , subtructLists(Subtructed, Rest, Result)
       ) ; (
          subtructLists(List, Rest, Result)
       )
).

always (subtructLists(List, [], List)).

always( delete_element(X, [X|Tail], Tail)).

always( delete_element(X, [Y|Tail1], [Y|Tail2]): -
        delete_element(X, Tail1, Tail2)
).