In such a case, ceil(log(n)) > log(n), and if the base of the logarithm is 2 or less, then 2^(ceil(log(n)) > n, and the question fails to be valid. The base of the logarithm has to be greater than 2.
On Sep 14, 9:26 pm, "Ajinkya Kale" <[EMAIL PROTECTED]> wrote: > Sorry i forgot, it is ceil(log n) so n-2^( ceil(log n) ) is not equal to > zero.. > > On Sun, Sep 14, 2008 at 8:57 AM, Karthik Singaram Lakshmanan < > > [EMAIL PROTECTED]> wrote: > > > Isn't n-2^logn = 0? > > since 2^logn = n if you are talking about log base 2 > > -- > Ciao, > Ajinkya --~--~---------~--~----~------------~-------~--~----~ 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 -~----------~----~----~----~------~----~------~--~---
