On Saturday, 30 of July 2005 03:35, Alan Stern wrote:
> On Fri, 29 Jul 2005, Michal Schmidt wrote:
>
> > Rafael J. Wysocki wrote:
> > > On Friday, 29 of July 2005 21:46, Michal Schmidt wrote:
> > >
> > >>The function calc_nr uses an iterative algorithm to calculate the number
> > >>of pages needed for the image and the pagedir. Exactly the same result
> > >>can be obtained with a one-line expression.
> > >
> > >
> > > Could you please post the proof?
> > >
> > > Rafael
> >
> > OK, attached is a proof-by-brute-force program. It compares the results
> > of the original function and the simplified one.
>
> Here's a more general proof.
>
> As I understand it, calc_nr is given nr_copy, the number of data pages
> that need to be written out, and it has to return the number of pages
> needed to hold the image data plus a bunch of PBE pagedir indexes, where
> each page gets one index (and pages containing PBEs need their own indexes
> as well).
>
> For brevity, let n = nr_copy, let p = PBES_PER_PAGE, and let x be the
> number of pagdir pages needed. Since each page can hold p PBEs, there
> will be room to store px PBEs. The total number of pages is n + x, so
> the routine needs to find the smallest value of x for which
>
> 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?
Rafael
--
- Would you tell me, please, which way I ought to go from here?
- That depends a good deal on where you want to get to.
-- Lewis Carroll "Alice's Adventures in Wonderland"
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