Merging two ordered lists ProLog - prolog

Question

Hey so this is my code so far. I am only a begginer in prolog but i need it for school
firstElement([_|_], [Elem1|List1], [Elem2|List2]):-
Elem1 =< Elem2, merge([Elem1] , List1, [Elem2|List2]);
merge([], [Elem2], List2).
merge([Head|Tail], [Elem1|List1], [Elem2|List2]):-
Elem1 =< Elem2,!, add(Elem1,[Head|Tail],[Head|Tail1]),
merge([Head|Tail1], List1, [Elem2|List2]);
add(Elem2,[Head|Tail],[Head|Tail1]),
merge([Head|Tail1], [Elem1|List1], List2).
merge([Head|Tail], [], [Elem2|List2]):-
add(Elem2,[Head|Tail],[Head|Tail1]).
merge([Head|Tail], [Elem1|List1], []):-
add(Elem1,[Head|Tail],[Head|Tail1]).
merge([Head|Tail], [], []).
add(X,[],[X]).
add(X,[Y|Tail],[Y|Tail1]):-
add(X,Tail,Tail1).
I found out that everytime it gets out of a merge it keeps forgetting the last number so it gets back to nothing in the end.

Answer 1

I think you’ve gotten very mixed up here with your code. A complete solution can be had without helpers and with only a few clauses.
First let us discuss the two base cases involving empty lists:
merge(X, [], X).
merge([], X, X).
You don’t quite have these, but I see some sort of recognition that you need to handle empty lists specially in your second and third clauses, but I think you got confused and overcomplicated them. There’s really three scenarios covered by these two clauses. The case where both lists are empty is a freebie covered by both of them, but since that case would work out to merge([], [], []), it’s covered. The big idea here is that if you exhaust either list, because they were sorted, what you have left in the other list is your result. Think about it.
This leaves the interesting case, which is one where we have some items in both lists. Essentially what you want to do is select the smaller of the two, and then recur on the entire other list and the remainder of the one you selected the smaller value from. This is one clause for that:
merge([L|Ls], [R|Rs], [L|Merged]) :-
L #< R,
merge(Ls, [R|Rs], Merged).
Here’s what you should note:
The “result” has L prepended to the recursively constructed remainder.
The recursive call to merge rebuilds the entire second list, using [R|Rs].
It should be possible to build the other clause by looking at this.
As an intermediate Prolog user, I would be naturally a bit suspicious of using two clauses to do this work, because it’s going to create unnecessary choice points. As a beginner, you will be tempted to erase those choice points using cuts, which will go badly for you. A more intermediate approach would be to subsume both of the necessary clauses into one using a conditional operator:
merge([L|Ls], [R|Rs], [N|Ns]) :-
( L #< R ->
N = L, merge(Ls, [R|Rs], Ns)
; —- other case goes here
).
An expert would probably build it using if_/3 instead:
#<(X,Y,true) :- X #< Y.
#<(X,Y,false) :- X #>= Y.
merge([L|Ls], [R|Rs], [N|Ns]) :-
if_(#<(L,R),
(N = L, merge(Ls, [R|Rs], Ns)),
( -- other case here )).
Anyway, I hope this helps illustrate the situation.