Prolog列表差异例程

Question

我正在尝试在Prolog中实现列表差异处理程序。 不知何故，以下内容会失败：

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).

当尝试执行以下操作时：

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

我遇到了这个错误：

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

- TheOne

Answer 1

你的使用方式 A1 是不正确的，[X|A] 中的谓词is只用于算术运算。 顺便说一下，SWI-Prolog 内置了减法谓词：

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.

这是你需要的吗？

- Volodymyr Gubarkov

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]

- DarkAnemone

Answer 3

使用find，所有的解决方案都变得显而易见：

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

- TheOne

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)
).

- ilia