Re: [linux-pm] [PATCH] swsusp: simpler calculation of number of pages in PBE list

["Rafael J. Wysocki" <rjw@sisk.pl>]

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


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