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"
