I think what you want to do is something like
plot k=. 1>.100<.+/\"1 d=.(?5 200$6) { _1 2 _3 4 _5 6
(to see 5 lines at a time)
The upward slope is a consequence of this:
+/_1 2 _3 4 _5 6
3
i.e. you have an average bias of positive 3.
And this is a consequence not of sequential thinking but maybe of scalar
thinking.
FWIW, I was initially puzzled by your result as well: your code looked
plausible at first glance.
On 1/22/08, Devon McCormick <[EMAIL PROTECTED]> wrote:
>
> 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
>
--
Devon McCormick, CFA
^me^ at acm.
org is my
preferred e-mail
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