Question

I need to write a predicate f(L,R) that succeeds if and only if L is a list containing all terms in R that are not lists.
For example:

f(L,[1,2,3,[4,5,6],[[7,8,9]],[]]).

Should give:

L = [1,2,3,4,5,6,7,8,9]

I wrote a predicate that gives the following result instead:

L = [1,2,3,4,5,6,7,8,9,[]]

Empty lists should not be present in the result. My predicate is the following:

f([],[]).
f(V,[H|T]):-  H = [_|_] -> append(L,R,V),
              f(L,H),  f(R,T),!;
              V = [H1|T1], H1=H, f(T1,T).

I have two doubts. First of all, the empty lists should not be present in the result. Also I don’t know why it does not work if I don’t put the cut (!). In fact, if I don’t put the cut it gives me the result as above, but if I ask for another result it loops forever. I really don’t understand why this should loops.

Answer 1

To remove the empty list, handle that case (discard it).

About the loop: I think the cause could be that you’re calling append(L,R,V) with all arguments not instantiated: move append after the recursive calls.

Finally, maybe you don’t use rightly the ‘if then else’ construct: I’ve indented using the usual SWI-Prolog source style, using indentation to highlight ‘sequential’ calls

f([], []).
f(V, [H|T]) :-
    (   H = []      % if H = []
    ->  f(V, T)     %  then discard
    ;   H = [_|_]   % else if H is list
    ->  f(L,H),     %  flat head
        f(R,T),     %  ...
        append(L,R,V)
    ;   V = [H|T1], % else 
        f(T1,T)     %  ...
    ).