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

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camgirls@hush.com

Posts: 12
Registered: 4/8/11
Re: Australian Mathematical Society - Hat Puzzle
Posted: Mar 3, 2013 3:05 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

On Mar 4, 5:35 am, Pratik Poddar <pratik.ph...@gmail.com> wrote:
> Posted on:http://pratikpoddarcse.blogspot.com/2013/02/mad-hat-party.html
>
> 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?
>
> About the blog: CSE Blog (http://pratikpoddarcse.blogspot.com) is a collection of 225+ Quant, Math & Computer Science Puzzles for Interview Preparation & Brain Teasing with Solution. It has had more than 300,000 page views since last 3 years. I hope you will enjoy it. Cheers!
>
>



On Mar 4, 5:35 am, Pratik Poddar <pratik.ph...@gmail.com> wrote:
> Posted on:http://pratikpoddarcse.blogspot.com/2013/02/mad-hat-party.html
>
> 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?
>
> About the blog: CSE Blog (http://pratikpoddarcse.blogspot.com) is a collection of 225+ Quant, Math & Computer Science Puzzles for Interview Preparation & Brain Teasing with Solution. It has had more than 300,000 page views since last 3 years. I hope you will enjoy it. Cheers!



2 guests = 1 hatswap

B
-
A

A
-
B

---- Add a 3rd guest C - swaps with A

C
-
A

B
- C now has B's hat
C

----

C
-
A

B
- C and B swap
B

A
-
C

----- Along comes D , leaving B alone, problem repeats...


============ ATTEMPT 2 =============

2 guests = 1 hatswap

B
-
A

A
-
B

---- Along come C & D

B C
- -
A C

A D
- -
B D

---- 2 Swap with 2


C B
- -
A C

D A
- -
B D

-----Diagonal Swap AGHAST!

A D
- -
A C

B C
- -
B D


----- C swap with D

A C
- -
A C

B D
- -
B D



I'm going to hazard a guess and say the Mad Hatter doubled the number
of guests!

G. Cooper (BInfTech)
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