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Topic: Australian Mathematical Society - Hat Puzzle
Replies: 4   Last Post: Mar 4, 2013 7:21 PM

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Graham Cooper

Posts: 4,495
Registered: 5/20/10
Re: Australian Mathematical Society - Hat Puzzle
Posted: Mar 4, 2013 7:21 PM
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On Mar 4, 6:15 am, Transfer Principle <>
> On Mar 3, 11:35 am, Pratik Poddar <> 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.
> 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

Now all that remains in the 2 GlobeTrotters to fix individual cycles.

A G1
D G2

Z G1
X G2

The method should be straight forward!


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