I dont think you even need to solve the recursion .. by looking at it it seems to be O(n^2) right ?
On Wed, Apr 1, 2009 at 2:18 PM, Miroslav Balaz <gpsla...@googlemail.com>wrote: > no that is just asymptotic recursion. > the answer is between n^2 and n^2log n > > of coure the answer is n^2; > > T(n)=n^2/2-n^2/4+n^2=n^2/4+n^2=T(n/2)+n^2=by master theorem n^2 > > T(n)< B(n)=2B(n/2)+n^2 what is by master theorem n^2 log n > > n > n/2 n/2 > n/4 n/4 n/4 n/4 > > T(n/2)=2T(n/4)-4T(n/8)+n^2/4 > > T(n)=4T(n/4)-8T(n/8)+n^2/2-4T(n/4)+n^2=-8T(n/8)+3n^2/2 > > you may pick up what is solution > > 2009/3/31 Arunachalam <arunachala...@gmail.com> > > What is the base value of this recursion? Without a base value the >> recursion is not solvable? >> >> There should be some base value like T(x) = 1 where x <= 1. >> >> regards, >> Arun. >> >> On Mon, Mar 30, 2009 at 12:35 AM, nikoo <shaker.far...@gmail.com> wrote: >> >>> >>> Hello everybody >>> >>> I need the solution to the following recursion equation >>> >>> T(n) = 2 T (n/2) - 4 T (n/4) + n^2 >>> >>> does anybody know how to solve this equation? >>> I appreciate any help >>> >>> thanks >>> nikoo >>> >>> >> >> >> -- >> =================================== >> want to know more about me >> http"//ww.livejournal.com/users/arunachalam >> >> >> >> > > > > -- 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 algogeeks+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/algogeeks -~----------~----~----~----~------~----~------~--~---
