Ouch.... I got the question completely wrong assuming the inner disc is continuous.Sorry for the confusion.
On 3/25/07, Prunthaban Kanthakumar <[EMAIL PROTECTED]> wrote: > > On 3/25/07, Rajiv Mathews <[EMAIL PROTECTED]> wrote: > > > > > > On 3/25/07, Prunthaban Kanthakumar <[EMAIL PROTECTED]> wrote: > > > If you see carefully his proof does not assume anything about > > "sections > > > colored continuously". His proof assumes only one thing "Half of them > > are > > > red and half of them are white" > > > Half does not mean it should be continuous. So the proof still works > > > correct unmodified even if the "halves" are not continuous. > > > > > > > Could you elaborate please. > > His proof contains, Quote: > > "If r >= R-r, match half1 with Red half of outer disk. > > Total matching = r + 100 - R + r = 100 - R + 2*r" > > How do you justify this if the sections aren't contiguous? > > I think the proof elaborated by _stone_ is correct and apt. > > > There is an "equivalence" > > It is simple.Just consider, > Half1 = All the sections in the outer disc painted red (This is not > continuous. But nothing prevents you from assuming a non-continuous 100 red > sections as a logical half) > Half2 = All the sections in the outer disc painted white > > Now with this interpretation, read his proof. Just remember that when you > say 'half' of inner disc it means the sections corresponding to the half in > the outer disc as defined above. This is the key to establish equivalence). > > Regards, > Prunthaban > > > -- > > > > > > Regards, > > Rajiv Mathews > > > > > > > > > --~--~---------~--~----~------------~-------~--~----~ 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 -~----------~----~----~----~------~----~------~--~---
