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Re: Australian Mathematical Society  Hat Puzzle
Posted:
Mar 4, 2013 7:21 PM


On Mar 4, 6:15 am, Transfer Principle <david.l.wal...@lausd.net> wrote: > On Mar 3, 11:35 am, Pratik Poddar <pratik.ph...@gmail.com> wrote: > > > Problem: The Mad Hatter is holding a hat party, where every > > guest must bring his or her own hat. At the party, whenever two guests greet each other, they have to swap their hats. In order to save time, each pair of guests is only allowed to greet each other at most once. > > After a plethora of greetings, the Mad Hatter notices that it is no longer possible to return all hats to their respective owners through more greetings. To sensibly resolve this maddening conundrum, he decides to bring in even more hat wearing guests, to allow for even more greetings and hat swappings. How many extra guests are needed to return all hats (including the extra ones) to their rightful owners? > > Anyone familiar with the Futurama episode "Prisoner of Benda" knows > that the correct answer is 2. > > http://en.wikipedia.org/wiki/The_Prisoner_of_Benda > > Of course, in that episode the characters were exchanging _bodies_, > not _hats_, but it's the same concept. AccordHing to the Wikipedia > above, some people call this the "Futurama Theorem" or "Keeler's > Theorem," after Ken Keeler, the mathematician who wrote the episode.
I think the Proof is simpler than presented.
Swapper A has Swapper B's Hat. Swapper B has Swapper C's Hat. and so forth.
If the party set is not exhausted then someone in Sequence 1 finishes the Sequence with Swapper A's hat.
CYCLE 1 A > B > C > ... > A CYCLE 2 Z > Y > X > ... > Z .. CYCLE N
Now all that remains in the 2 GlobeTrotters to fix individual cycles.
A G1 B C D G2
Z G1 Y X G2
The method should be straight forward!
Herc  www.BLoCKPROLOG.com



