but you need some kind of proof, for that.i alsow see from first sight that it is O(n^2), but i wane just fo verify that.
2009/4/1 Ajinkya Kale <kaleajin...@gmail.com> > 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 -~----------~----~----~----~------~----~------~--~---
