Another hats problem -- this one is original General forum

5 replies. Last post: 2017-06-25

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Another hats problem -- this one is original
  • William Fraser at 2017-06-25

    I’ve been looking for the right time to unveil this one (I hope it doesn’t have too many bugs):


    There are an even number of identical gnomes in a completely symmetric circular room (the only exit is a completely symmetric hole in the exact center).

    They are allowed to discuss their strategy, after which they may not communicate.  Then half of them are given red hats and half of them blue hats.

    One of the blue-hatted gnomes is chosen at random to leave the room.

    The remaining gnomes are arranged at random in an evenly spaced circle.

    Each of the red-hatted gnomes is allowed to raise their hand.  Unless exactly one does so, they lose.  His hat is replaced by a blue one.

    The gnome who left the room returns and must point at a blue-hatted gnome.  If it is the gnome who raised his hand, they win.  Otherwise, they lose.

    Can you find a strategy which will let them win?

  • Martyn Hamer at 2017-06-25

    I’m struggling to think of a mathematical answer to this one as the gnomes are identical and randomly placed. A couple of possible answers...

    The gnome who puts his hand up leaves it up until the other gnome returns.

    The gnomes who are given the blue hats turn them inside out or mark them in some way. When the hat is replaced, it will look different to the other blue hats.

  • alihv at 2017-06-25

    The setup looks unnecessarily complicated. I think what you want to say is that you have a binary word of N + 1 zeros and N ones written on a circle and you want to flip a zero into one so that the location of the flip can be identified just by looking at the result. For example, if you have 001 that could be rotated into 010 or 100, you could flip it into 011 and then identify the bit after the 0 as the flipped one.

  • Richard Moxham at 2017-06-25

    Hm.  Sounds like an even funner problem put that way.  I must remember to try it out on the lads in the pub tonight.

    :)

  • William Fraser at 2017-06-25

    alihv has a correct statement of an equivalent problem.


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