Hi!
> > px >= n + x
> >
> > or
> >
> > (p-1)x >= n
> >
> > or
> >
> > x >= n / (p-1).
> >
> > The obvious solution is
> >
> > x = ceiling(n / (p-1)),
> >
> > so calc_nr should return n + ceiling(n / (p-1)), which is exactly what
> > Michal's patch computes.
>
> Nice. :-)
>
> Could we perhaps add your proof to the Michal's patch as a comment,
> for reference?
No, thanks. It only proves that it is equivalent to old code, but says
nothing about quality of code, and we really do not want to keep old
code around.
Pavel
--
teflon -- maybe it is a trademark, but it should not be.
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