When both disks are not painted continuous 'equivalence' does not work. Because in your proof when one half does not give the answer, you just take take the other half and align it. But for arbitrary configuration, when one configuration does not work, you cannot align the other half. It will not fit unless the sections are painted symmetrically.
On 3/25/07, Vishal <[EMAIL PROTECTED]> wrote: > > I did assume that the outer disk is painted half (contiguous) red and half > white. > However the 'equivalence' should do the trick and the same proof applies. > As far as Stone's proof goes, I did not understand - > For each inner section,no matter white or black ,there is 100 > color-matching events. > Can somebody explain? > > ~Vishal > > On 3/25/07, Prunthaban Kanthakumar <[EMAIL PROTECTED]> wrote: > > > > 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 -~----------~----~----~----~------~----~------~--~---
