Reputation: 37
Ex: If I have been given two list [1,4,3,2,5,6] and [1,2,3] the final list should be [4,5,6].
i.e., Del([1,4,3,2,5,6], [1,2,3], Result).
----should output Result=[4,5,6].
I have tried something like this:
delete1(A, [A|B], B).
delete1(A, [B, C|D], [B|E]) :- delete1(A, [C|D], E).
But the output I'm getting is by deleting the element being passed as an parameter and not a list.
Output:
delete1(a,[a,b,c,d],Res).
(0) Call: delete1(a,[a,b,c,d],_h210) ?
(0) Exit: delete1(a,[a,b,c,d],[b,c,d]) ?
Res = [b,c,d]
Can anyone please help me how to go about this ?
Upvotes: 1
Views: 123
Reputation: 18726
Pure and simple: Use meta-predicate tfilter/3
in tandem with list_nonmember_t/3
!
Like we did with memberd_t/3
, we define list_nonmember_t/3
based on if_/3
and (=)/3
:
list_nonmember_t([],_,true).
list_nonmember_t([E|Es],X,T) :-
if_(E=X, T=false, list_nonmember_t(Es,X,T)).
Let's put it together!
?- tfilter(list_nonmember_t([1,2,3]), [1,4,3,2,5,6], Xs).
Xs = [4,5,6]. % succeeds deterministically
Upvotes: 2
Reputation: 18331
So your delete1
is a good start, it allows to delete single element from the list, but it is not handling all of the cases. For example the lists that are not containing the element to be deleted.
So the right one would be :
delete1(_, [], []).
delete1(A, [A|B], B).
delete1(A, [C|B], [C|E]) :- delete1(A, B, E).
and then using this, you can define your del1
by applying delete1
recursively on the whole list:
del1([], _, []).
del1(L, [], L).
del1(L, [H|T], R) :-
delete1(H, L, R1),
del1(R1, T, R).
And of course you can use builtin list predicates as stated in the other answer.
Upvotes: 0
Reputation: 5565
del1([], _, []).
del1([A|L], B, R) :- member(A, B), del1(L, B, R).
del1([A|L], B, [A|R]) :- not(member(A,B)), del1(L, B, R).
Upvotes: 0