Re: Sorting a list of lists in the order of ascending length

Not sure about the relative efficiency, maybe non backtrackable assignment
could help:

:- use_module(library(rbtrees)).
:- use_module(library(nb_rbtrees)).

 ordkey :- rb_empty(R),
   forall(member(W, [ls,mkdir,cd,ftp]),
     ( atom_length(W, K),
       (  nb_rb_get_node(R, K, N)
       -> nb_rb_node_value(N, Ws),
          nb_rb_set_node_value(N, [W|Ws])
       ;  nb_rb_insert(R, K, [W])
       )
    )), rb_visit(R, L), writeln(L).

yields

ordkey.
[2-[cd,ls],3-[ftp],5-[mkdir]]

bye Carlo


2013/9/24 Norbert E. Fuchs <[email protected]>

>
> Hi
>
> I need all subsets of a list in the order of ascending length of the
> subsets.
>
> Thus I wrote a predicate
>
>   generate_subset(+List, -SubList) that generates a SubList of List
>
> and that is also used for other purposes, and a variant of insertion sort
>
>   sort_list_of_lists_in_ascending_length(+ListOfLists, -SortedListOfList)
> that sorts ListOfLists into SortedListOfList
>
> and then called
>
>   findall(SubList, generate_subset(List, SubList), SubLists),
> sort_list_of_lists_in_ascending_length(SubLists, SortedSubLists)
>
> This works quite nicely and efficiently, but nevertheless I wonder whether
> there isn't a simpler way to perform this operation.
>
> Regards.
>
>    --- nef
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