Open
Description
Hi,
The solution provided will asset that subset([a,b,c,a], [a, b, c]). is true.
It is missing checking that the first parameter is a set itself.
Here I propose a fix using the set definition implemented in a previous chapter.
set([], []).
set([H|T], X):- member(H, T), !, set(T, X).
set([H|T], [H|X]):- set(T, X).
subset_check(SS, S):-
set(SS, SS),
all_members(SS, S).
all_members([], _).
all_members([H|T], S):-
member(H, S),
!,
all_members(T, S)
Thanks a lot for making this repository.
Metadata
Metadata
Assignees
Labels
No labels