Which algorithm?

-- 
Raul

On Tue, Nov 22, 2011 at 6:16 AM, David Vaughan <[email protected]
> wrote:

> I see that the algorithm you suggested only works on linked lists. Is
> there a way to model linked lists in J?
>
> On 20 Nov 2011, at 15:05, Henry Rich wrote:
>
> > You have to use ~. so that the list doesn't keep growing when a match is
> > found.  ^:_ would run forever then, since it stops when two successive
> > results are identical.
> >
> > This method takes O(*:n) time because of the ~.@,  .  If that's a
> > problem, you might want to look into the very elegant Floyd's Cycle
> > Algorithm.
> >
> > Henry Rich
> >
> > On 11/20/2011 8:51 AM, David Vaughan wrote:
> >> This seems to have the behaviour I'm looking for, thanks.
> >>
> >> I don't understand why though. I can see that each result of f is being
> stored in a list as the verb is being applied each time, but I don't
> understand why ~.@, enables this.
> >>
> >> On 20 Nov 2011, at 13:31, Raul Miller wrote:
> >>
> >>> You can do
> >>> (~.@, f@{:)^:_ y
> >>>
> >>> But I am not sure how to work<x into this.
> >>>
> >>> --
> >>> Raul
> >>>
> >>> On Sun, Nov 20, 2011 at 5:36 AM, David Vaughan<
> [email protected]
> >>>> wrote:
> >>>
> >>>> If I'm doing f^:(<x) y, can I write an expression so that f is
> repeatedly
> >>>> applied to y until the result has come up before?
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