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 -~----------~----~----~----~------~----~------~--~---
