Finding consecutive sublists of a list - prolog

Question

I want to write a predicate split/2 that generates all consecutive lists found inside another list.
Example: split([1,2,3,4],X) should return
X = [4], X = [2,3],X = [1,2], X = [1,2,3] etc.
So far I only have a predicate that returns all possible sublists of a list:
sublist([],[]).
sublist([H|T], [H|R]) :-
sublist(T,R).
sublist([_|T], R) :-
sublist(T,R).
However, with the query from the example this predicate includes unwanted answers like X = [2,4] and X = [1,3] that aren't found consecutively in [1,2,3,4].

Answer 1

Usually a problem is easier if you split it in subproblems. We can first construct a predicate that will construct all suffixes for a given list.
We can construct such predicate as follows:
suffix(_, []).
suffix([H|T], [H|T2]) :-
suffix(T, T2).
So for each point in the list, we can decide to stop (with the empty list), or emit the next item. For the given sample list, we thus get:
?- suffix([1,2,3,4],X).
X = [] ;
X = [1] ;
X = [1, 2] ;
X = [1, 2, 3] ;
X = [1, 2, 3, 4].
Now we only need to decide when we start the suffix. For each item in the list, we can decide to start at that point, and enumerate over all suffixes that we then append to that item:
split([H|T], [H|S]) :-
suffix(T, S).
split([_|T], S) :-
split(T, S).
For example:
?- split([1,2,3,4],X).
X = [1] ;
X = [1, 2] ;
X = [1, 2, 3] ;
X = [1, 2, 3, 4] ;
X = [2] ;
X = [2, 3] ;
X = [2, 3, 4] ;
X = [3] ;
X = [3, 4] ;
X = [4] ;
false.
The nice thing is that we got a second predicate "for free": we can also obtain all suffixes for a list.
We might want to include the empty list as well. I leave this as an exercise.