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?
>>> ----------------------------------------------------------------------
>>> For information about J forums see http://www.jsoftware.com/forums.htm
>>>
>> ----------------------------------------------------------------------
>> For information about J forums see http://www.jsoftware.com/forums.htm
>
> ----------------------------------------------------------------------
> For information about J forums see http://www.jsoftware.com/forums.htm
>
----------------------------------------------------------------------
For information about J forums see http://www.jsoftware.com/forums.htm
