How can I delete every occurrence of a sublist from a list in prolog?

Question

This is the code for deleting or removing an element from a given list:

remove_elem(X,[],[]). 
remove_elem(X,L1,L2) :-
   L1 = [H|T],
   X == H,
   remove_el         


        

Answer 1

Inspired by @CapelliC's implementation I wrote the following code based on and_t/3:

append_t([]    ,Ys,Ys, true).
append_t([X|Xs],Ys,Zs,Truth) :-
   append_aux_t(Zs,Ys,Xs,X,Truth).

append_aux_t([]    ,_ ,_ ,_,false).     % aux pred for using 1st argument indexing
append_aux_t([Z|Zs],Ys,Xs,X,Truth) :-
   and_t(X=Z, append_t(Xs,Ys,Zs), Truth).

One append_t/4 goal can replace two prefix_of_t/3 and append/3 goals.

Because of that, the implementation of list_sublist_removed/3 gets a bit simpler than before:

list_sublist_removed([]    ,[_|_] ,[]).
list_sublist_removed([X|Xs],[L|Ls],Zs) :-
    if_(append_t([L|Ls],Xs0,[X|Xs]),
        (Zs =    Zs0 , Xs1 = Xs0),
        (Zs = [X|Zs0], Xs1 = Xs)),
    list_sublist_removed(Xs1,[L|Ls],Zs0).

Still deterministic?

?- list_sublist_removed([1,2,3,4,1,2,5,6,1,2,1],[1,2],L).
L = [3,4,5,6,1].

Yes! What about the following?

?-  list_sublist_removed([1,2,3,4,1,2,5,6,1,2,1],X,[3,4,5,6,1]).
X = [1,2] ;                             % succeeds with useless choice-point
false.

Nope. So there is still room for potential improvement...

Answer 2

This logically pure implementation is based on the predicates if_/3 and (=)/3.

First, we build a reified version of prefix_of/2:

prefix_of_t([],_,true).
prefix_of_t([X|Xs],Zs,T) :-
   prefix_of_t__aux(Zs,X,Xs,T).

prefix_of_t__aux([],_,_,false).
prefix_of_t__aux([Z|Zs],X,Xs,T) :-
   if_(X=Z, prefix_of_t(Xs,Zs,T), T=false).

Then, on to the main predicate list_sublist_removed/3:

list_sublist_removed([],[_|_],[]).
list_sublist_removed([X|Xs],[L|Ls],Zs) :-
    if_(prefix_of_t([L|Ls],[X|Xs]),                 % test
        (Zs = Zs0,     append([L|Ls],Xs0,[X|Xs])),  % case 1
        (Zs = [X|Zs0], Xs0 = Xs)),                  % case 2
    list_sublist_removed(Xs0,[L|Ls],Zs0).

A few operational notes on the recursive clause of list_sublist_removed/3:

1. First (test), we check if [L|Ls] is a prefix of [X|Xs].

2. If it is present (case 1), we strip it off [X|Xs] yielding Xs0 and add nothing to Zs.

3. If it is absent (case 2), we strip X off [X|Xs] and add X to Zs.

4. We recurse on the rest of [X|Xs] until no more items are left to process.

---

Onwards to some queries!

1. The use case you gave in your question:

   ?- list_sublist_removed([1,2,3,4,1,2,5,6,1,2,1],[1,2],L).
   L = [3,4,5,6,1].                           % succeeds deterministically
   

2. Two queries that try to find the sublist that was removed:

   ?- list_sublist_removed([1,2,3,4,1,2,5,6,1,2,1],Sub,[  3,4,5,6,1]).
   Sub = [1,2] ? ;
   no
   
   ?- list_sublist_removed([1,2,3,4,1,2,5,6,1,2,1],Sub,[1,3,4,5,6,1]).
   no
   

3. Next, let's find a suitable Ls in this query:

      
   ?- list_sublist_removed(Ls,[1,2],[3,4,5,6,1]).
   % a lot of time passes ... and nothing happens!
   

   Non-termination! This is unfortunate, but within expectations, as the solution set is infinite. However, by a-priori constraining the length of Ls, we can get all expected results:

   ?- length(Ls,_), list_sublist_removed(Ls,[1,2],[3,4,5,6,1]).
     Ls = [      3,4,5,6,1]     ?
   ; Ls = [1,2,  3,4,5,6,1]     ?
   ; Ls = [3, 1,2, 4,5,6,1]     ?
   ; Ls = [3,4, 1,2, 5,6,1]     ?
   ; Ls = [3,4,5, 1,2, 6,1]     ?
   ; Ls = [3,4,5,6, 1,2, 1]     ?
   ; Ls = [3,4,5,6,1, 1,2 ]     ?
   ; Ls = [1,2, 1,2, 3,4,5,6,1] ? ...
   

Answer 3

<rant>

So many years I study Prolog, still it deserves some surprises... your problem it's quite simple to solve, when you know the list library, and you have a specific mode (like the one you posted as example). But can also be also quite complex to generalize, and it's unclear to me if the approach proposed by @repeat, based on @false suggestion (if_/3 and friends) can be 'ported' to plain, old Prolog (a-la Clocksin-Mellish, just to say).

</rant>

A solution, that has been not so easy to find, based on old-school Prolog

list_sublist_removed(L, S, R) :-
    append([A, S, B], L),
    S \= [],
    list_sublist_removed(B, S, T),
    append(A, T, R),
    !
    ; L = R.

some test:

?- list_sublist_removed([1,2,3,4,1,2,5,6,1,2,1],[1,2],L).
L = [3, 4, 5, 6, 1].

?- list_sublist_removed([1,2,3,4,1,2,5,6,1,2,1],X,[3, 4, 5, 6, 1]).
X = [1, 2].

?- length(X,_), list_sublist_removed(X,[1,2],[3, 4, 5, 6, 1]).
X = [3, 4, 5, 6, 1] ;
X = [3, 4, 5, 6, 1, 2, 1] ...