Thanks. But on the face of it the verb does
not do the same as your description:
dice=: 9999&$: : (4 : 0)"0
y=. x #.^:_1 y-1
label_again.
l=.y =&{. r=.?1+{.y
while. l *. r<&#y do.
if. d>f=.y{~_1+#r=.r,d=.?x do. goto_again. end.
l=.d=f
end.
x #. r,?x#~y-&#r
)
That is, "dice" does not generate n random digits.
What it does is it generates random digits one
by one, and depending on the digit d:
d<i{y - keep it, and generate n-1-i more digits
without further checking
d=i{y - increment i and generate the next digit
d>i{y - discard all digits and start from scratch
Perhaps this is equivalent to your description,
but the equivalence is not obvious to me.
As well, if we "tweak" the verb and omit the
"discard all digits" part but instead just
discard the current digit d and generate
another digit d1 and check again, that seems to
satisfy your description of "uniform finite
average number of discards". But you must have
a reason for not doing that, otherwise you would
have (I guess) gotten rid of the distasteful goto.
----- Original Message -----
From: Andrew Nikitin <[EMAIL PROTECTED]>
Date: Thursday, July 13, 2006 1:24 pm
Subject: [Jbeta] monadic '?' (and #. slowdown
> R&S HUI:
> >So why are the numbers uniform?
>
> First, y is converted to a number base x (n digit vector) 0&{ is
> highest"digit", _1&{ is the "lowest". Then we genereate an n-
> "digit" random number
> in
> a range 0.._1+x^n and if it is less than y we keep it and if it
> >:y we drop
> it
> and start over. Assuming uniform and independent distribution of
> random"digits" we get uniformly distributed result. The way we
> generate and check
> digits promises uniformly finite average number of discarded
> digits per
> result (I
> estimate this limit to be either 1 or 2).
----------------------------------------------------------------------
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