Yes, it's beautiful. Thank you very much, you're great! On Jul 21, 7:06 pm, Ashesh <[EMAIL PROTECTED]> wrote: > On Jul 21, 8:38 pm, zpedja <[EMAIL PROTECTED]> wrote:
> > I'm now dealing with your solution. > > I've managed to prove that the argument in k-th recursion acts like > > ceil(n/log(n)^k) , but I don't know how you obtain k=Θ( log(n)/ > > log(log(n)) ). > > Asymptotically speaking, the recurrence reaches an end when n/log(n)^k > = ϴ(1) = c, where c is a constant. So, c.log(n)^k = n > or, k.log(log(n)) + logc = log(n) > getting rid of the constant, the asymptotic expression for k would > then be: k=ϴ( log(n) / log(log(n)) ). > --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/algogeeks -~----------~----~----~----~------~----~------~--~---
