Pack of consecutive duplicates of list items in sublists

Question

I need help. I looked through the database and I found one question already asked about this example, but the answers really didn't help me, so I decided to post my own question.

The challenge is to pack consecutive duplicates of list items into sublists:

% ?- pack([a,a,a,a,b,c,c,a,a,d,e,e,e,e],X).
% X = [[a,a,a,a],[b],[c,c],[a,a],[d],[e,e,e,e]].

Here is what I got:

pack([], []).
pack([X], [[X]]).
pack(Liste, Ergebnis):-
   Liste = [H, T|TS],
   H \= T,
   pack([T|TS], Ergebnis1),
   append([[H]], Ergebnis1, Ergebnis).
pack([H, H|HS], Ergebnis):-
   pack([H|HS], Ergebnis1),
   append([H|HS], Ergebnis1, Ergebnis).

The first case works very well (the case when H \ = T). The second is not, and I really don't know why. Can someone help me and explain the problem according to my solution?

thank

Answer 1

[H|HS] , , [H|HS] - . -

pack([H, H|HS], [[H|TFR]|TR]):-
    pack([H|HS], [TFR|TR]).

, , , , (.. H) , pack.

, , Liste, "" "" "inlined" , Ergebnis1. :

pack([H, T|TS], [[H]|TR]):-
    H \= T,
    pack([T|TS], TR).

.

Answer 2

.

. , [H,T|TS] H \= T, [H], [T|TS].

- :

pack([H, H|HS], Ergebnis):-
    pack([H|HS], Ergebnis1),
    append([H|HS], Ergebnis1, Ergebnis).

, [H,H|HS] [H|HS], [H|HS]. . , [H|HS], , : [[H,...,H]|X], H . , :

pack([H, H|HS], Ergebnis):-
    pack([H|HS], [HH|T]),
    Ergebnis = [[H|HH]|T].

:

pack([], []).
pack([X], [[X]]).
pack([H,T|TS], [[H]|PTS]):-
   H \= T,
   pack([T|TS], PTS).
pack([H, H|HS], [[H|HH]|T]):-
   pack([H|HS], [HH|T]).

...

| ?- pack([a,a,a,a,b,c,c,a,a,d,e,e,e,e],X).

X = [[a,a,a,a],[b],[c,c],[a,a],[d],[e,e,e,e]] ? a

no
| ?-

Answer 3

, , . .

. - splitlistIfAdj/3, if_/3, @false .

, splitlistIfAdj/3 dif/3, reified dif/2:

?- splitlistIfAdj(dif,[a,a,a,a,b,c,c,a,a,d,e,e,e,e],Lists).
Lists = [[a,a,a,a],[b],[c,c],[a,a],[d],[e,e,e,e]]. % succeeds deterministically

, , :

?- splitlistIfAdj(dif,[A,B,C],Lists).
A = B,    B = C,    Lists = [[C, C, C]]    ;
A = B,    dif(B,C), Lists = [[B, B], [C]]  ;
dif(A,C), B = C,    Lists = [[A], [C, C]]  ;
dif(A,B), dif(B,C), Lists = [[A], [B], [C]].

Answer 4

I would change the problem a bit and decompose the problem into two simpler tasks.

First, we need to split the list into a prefix consisting of a list of consecutive elements at the head of the list and its suffix (everything else):

partition( []      , []    , []     ) .
partition( [X]     , [X]   , []     ) .
partition( [X,Y|Z] , [X]   , [Y|Z]  ) :- X \= Y .
partition( [X,Y|Z] , [X|P] , S      ) :- X  = Y , partition([Y|Z],P,S) .

Once we succeed, the rest is easy:

pack([],[]) .               % packing an empty list yields the empty list
pack([X|Xs],[P,Ps]) :-      % packing a non-empty list consists of
  partition([X|Xs],P,T) ,   % - partitioning it into its prefix and suffix, and
  pack(T,Ps)                % - packing the suffix
  .                         %

Answer 5

To avoid adding, you can use dcg :

pack(L, LP) :-                  % pack/2
    phrase(pack(L), LP).

pack([]) --> [].                % pack//1
pack([X]) --> [[X]].
pack([H, H1 | T]) -->
     {H \= H1},
     [[H]],
     pack([H1 | T]).
pack([H, H | T]) -->
    pack([H | T], [H]).

pack([H, H | T], P) -->         % pack//2
    pack([H | T], [H|P]).
pack([H, H1 | T], P) -->
    {H \= H1},
    [[H | P]],
    pack([H1 | T]).
pack([H, H], P) -->
    [[H,H|P]].