@Nitin: Answer to question 3 is 50. On Mon, Sep 19, 2011 at 11:44 AM, praveen raj <praveen0...@gmail.com> wrote:
> @nitin Plz explain how u have reached answer of question no. 4 and 6 > > On 19-Sep-2011 12:26 AM, "Nitin Garg" <nitin.garg.i...@gmail.com> wrote: > > > > Answer 3 - 100 > > Answer 6 - 103 > > Answer 4 - 194 total processes including the parent > > Answer 7 - 12 km south, 12 km east > > > > > > On Sun, Sep 18, 2011 at 11:53 PM, Ashima . <ashima.b...@gmail.com> > wrote: > >> > >> @malay: how cm n+logn-2? > >> cn u explain the logic ? > >> > >> Ashima > >> M.Sc.(Tech)Information Systems > >> 4th year > >> BITS Pilani > >> Rajasthan > >> > >> > >> > >> > >> On Sun, Sep 18, 2011 at 11:07 AM, Ashima . <ashima.b...@gmail.com> > wrote: > >>> > >>> rite! 62.5% > >>> > >>> Ashima > >>> M.Sc.(Tech)Information Systems > >>> 4th year > >>> BITS Pilani > >>> Rajasthan > >>> > >>> > >>> > >>> > >>> On Sat, Sep 17, 2011 at 9:04 PM, malay chakrabarti <m1234...@gmail.com> > wrote: > >>>> > >>>> create a tournament tree.in each round one value is eliminated to > obtain in the process the winner or the highest value in n-1 comparisons. > Then check the queue of the winner which contains log(n) entries of the > values beaten by the winner which implicitly will contain the runners > up.Then log(n)-1 comparisons to find the highest among all the losers whom > the winner had beaten. So all over complexity will be n-1 +log(n) -1 = > n+log(n)-2. Hp that answers ur query. nice question btw :) > >>>> > >>>> > >>>> On Sun, Sep 18, 2011 at 8:02 AM, VIHARRI <viharri....@gmail.com> > wrote: > >>>>> > >>>>> hey i'm also thinking n + logn -2.. but couldnt able to figure out > >>>>> how??? can you please explain the logic > >>>>> > >>>>> -- > >>>>> 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?hl=en. > >>>>> > >>>> > >>>> -- > >>>> 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?hl=en. > >>> > >>> > >> > >> -- > >> 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?hl=en. > > > > > > > > > > -- > > Nitin Garg > > > > "Personality can open doors... but only Character can keep them open" > > > > -- > > 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?hl=en. > > -- > 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?hl=en. > -- Bhanu Chowdary -- 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?hl=en.
