序言列表差異例程

我試圖在序言中實現列表差異例程。出於某種原因，下面的失敗：序言列表差異例程

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).

當嘗試：

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

我得到這個錯誤：

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

來源

2009-10-06 TheOne

您的使用A1是[X | A]不正確。謂詞是僅用於算術運算。順便說一句，SWI-Prolog有內置減謂：

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.

這是你需要什麼？

來源

2009-10-06 12:52:05 Xonix

使用發現所有的解決方案變得很明顯：

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

來源

2009-10-06 02:52:31 TheOne

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]

來源

2011-07-08 18:41:04 DarkAnemone

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) 
).

來源

2012-12-13 17:54:12 ilia

序言列表差異例程

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