Simple Listing Procedure

Question

I am trying to implement a list markup procedure in a prolog. For some reason, the following fails:

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). 

Upon attempt:

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

I get this error:

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

Answer 1

Your use of A1 is [X | A] wrong. Predicate used only for arithmetic. Btw, SWI-Prolog has a built-in subtraction predicate:

 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. 

That's what you need?

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([],_,[]). 

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

Answer 3

 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) ). 

Answer 4

Using find all solution becomes obvious:

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