To clarify with an example:
foo/\40 _30 _30 _30 _10
40 10 41 41 41
So, lining the input vector up with successive intermediate results shows us
the following:
40 _30 _30 _30 _10
1 NB. capped
1 NB. capped
1 NB. capped
41 NB. not capped
>From this we see that catenating negative numbers to the right will keep
feeding a one to be the number added to 40 for the final result. In fact if
the sum of the numbers to the right of this adds up to 100 or less (which
has to be true because of the upper limit), it will get capped to 1 because
the succession of negative numbers will reduce it to that.
On 1/22/08, Raul Miller <[EMAIL PROTECTED]> wrote:
>
> I think you should be using
> plot k=:foo/\.d=:(?200$6) { _1 2 _3 4 _5 6
>
> Remember that 1 foo 2 foo 3 foo 4 is equivalent
> to 1 foo (2 foo (3 foo (4))) but that \ works against
> prefixes.
>
> In other words, you are seeing secondary effects
> of previous intermediate results having been
> clamped when you use foo/\
>
> --
> Raul
> ----------------------------------------------------------------------
> For information about J forums see http://www.jsoftware.com/forums.htm
>
--
Devon McCormick, CFA
^me^ at acm.
org is my
preferred e-mail
----------------------------------------------------------------------
For information about J forums see http://www.jsoftware.com/forums.htm
