Impossible for me to solve: Simple recursion does not take integers for indexing a list....source_sink does not exist? - prolog

Question

This recursion should slice IL to IR out of the list Lin and hand result LOut...
slice(_,IL,IR,LOut) :-
IR<IL,
[LOut].
slice(Lin,IL,IR,LOut) :-
nth0(IL,Lin,X),
append(LOut,[X],LOut2),
IK is IL + 1,
slice(Lin,IK,IR,LOut2).
Input / Output:
?- slice([1,2,3,4],2,3,X).
ERROR: source_sink `'3'' does not exist
ERROR: source_sink `'4'' does not exist
X = [] .
I m also new to Prolog, but I think this recursion must somehow work. Also I'm not really known to the error codes in Prolog, but after checking multiple times I just have to give up... I hope you guys can help me.

Answer 1

slice(_,IL,IR,LOut) :-
IR<IL,
[LOut]. % <-- this line causes source_sink error.
That syntax [name] tries to load the file name.pl as Prolog source code. By the time your code gets there, LOut is [3,4] so it tries to load the files 3.pl and 4.pl, and they don't exist (thankfully, or else who knows what they could do).
I think this recursion must somehow work
It won't; you are appending to a list as you go down into the recursion, which means you will never see the result.
The following might be a close version which works, at least one way:
slice(_,IL,IR,[]) :-
IR < IL.
slice(Lin,IL,IR,[X|LOut]) :-
IR >= IL,
nth0(IL,Lin,X),
IK is IL + 1,
slice(Lin,IK,IR,LOut).
?- slice([0,1,2,3,4,5,6,7,8,9], 2, 5, X).
X = [2, 3, 4, 5]
See how [X|LOut] in the second rule's header puts X in the result that you get, and append/3 is not needed, and LOut finishes down in the recursion eventually as [] the empty list from the first rule, and all the X's are prepended on the front of it to make the result on the way down into the recursion, which is tail recursion, so it doesn't need to go back up, only forward, since there's nothing left to be done after the recursive call.
Since the "cons" is done before the recursion, this is known as "tail recursion modulo cons" in other languages, but in Prolog it is just tail, and the list is being built top-down on the way forward, as opposed to being built bottom up on the way back:
Lin=[0,1,2,3,4,5,6,7,8,9], slice( Lin, 2, 5, R)
:-
nth0(2,Lin,X2), R=[X2|R2], slice( Lin, 3, 5, R2)
:-
nth0(3,Lin,X3), R2=[X3|R3], slice( Lin, 4, 5, R3)
:-
nth0(4,Lin,X4), R3=[X4|R4], slice( Lin, 5, 5, R4)
:-
nth0(5,Lin,X5), R4=[X5|R5], slice( Lin, 6, 5, R5)
:-
R5 = [].

Answer 2

I think findall/3 provides a readable readable solution for your problem:
slice(Lin,IL,IR,LOut) :-
findall(E,(nth0(P,Lin,E),between(IL,IR,P)),LOut).
yields
?- slice([1,2,3,4],2,3,X).
X = [3, 4].
If you expect a different outcome, use standard arithmetic comparison operators (=<,>=) instead of between/3.

Answer 3

I think you want:
list_elems_slice(Start, End, Lst, Slice) :-
list_elems_slice_(Lst, 1, Start, End, Slice).
list_elems_slice_([H|T], N, N, End, [H|Slice]) :-
list_elems_slice_capture_(T, N, End, Slice).
list_elems_slice_([_|T], N, Start, End, Slice) :-
N1 is N + 1,
list_elems_slice_(T, N1, Start, End, Slice).
list_elems_slice_capture_(_, N, N, []).
list_elems_slice_capture_([H|T], N, End, [H|Slice]) :-
N1 is N + 1,
list_elems_slice_capture_(T, N1, End, Slice).
Result in swi-prolog:
?- list_elems_slice(S, E, [a,b,c], Slice).
S = E, E = 1,
Slice = [a] ;
S = 1,
E = 2,
Slice = [a, b] ;
S = 1,
E = 3,
Slice = [a, b, c] ;
S = E, E = 2,
Slice = [b] ;
S = 2,
E = 3,
Slice = [b, c] ;
S = E, E = 3,
Slice = [c] ;
false.

Answer 4

Assuming that the point of this exercise is to teach you to think recursively, I would approach the problem as follows.
To get what you want is essentially two separate operations:
You first must discard some number of items from the beginning of the list, and then
Take some number of items from what's left over
That gives us discard/3:
discard( Xs , 0 , Xs ) .
discard( [_|Xs] , N , Ys ) :- N > 0 , N1 is N-1, discard(Xs,N1,Ys) .
and take/3, very nearly the same operation:
take( _ , 0 , [] ) .
take( [X|Xs] , N , [Y|Ys] ) :- N > 0 , N1 is N-1, take(Xs,N1,Ys) .
Once you have those two simple predicates, slice/4 itself is pretty trivial:
%
% slice( List , Left, Right, Sublist )
%
slice( Xs, L, R, Ys ) :- % to slice a list,
L =< R, % - the left offset must first be less than or equal to the right offset
N is R-L, % - compute the number of items required, and then
discard(Xs,L,X1), % - discard the first L items, and
take(X1,N,Ys). % - take the next N items
. % Easy!
Another approach would be to use append/3:
slice( Xs , L, R, Ys ) :-
length(Pfx,L), % - construct of list of the length to be discarded
append(Pfx,Sfx,Xs), % - use append to split Xs
N is R-L, % - compute the number of items required
length(Ys,N), % - ensure Ys is the required length
append(Ys,_,Sfx) % - use append to split off Ys
. % Easy!