Got it. Stone's solution seems to be right. On 3/25/07, Prunthaban Kanthakumar <[EMAIL PROTECTED]> wrote: > > 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 > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
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