Comments on: Hats and Rooms
https://blog.tanyakhovanova.com/2010/07/hats-and-rooms/
Mathematics, applications of mathematics to life in general, and my life as a mathematician.Fri, 26 Sep 2014 21:52:10 +0000
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By: Tanya Khovanova’s Math Blog » Blog Archive » Hats and Rooms. Take II
https://blog.tanyakhovanova.com/2010/07/hats-and-rooms/#comment-1500
Fri, 16 Jul 2010 22:19:49 +0000https://blog.tanyakhovanova.com/?p=255#comment-1500[…] recently published a puzzle about wizards, hats of different colors and rooms. Unfortunately, I was too succinct in my description and didn’t explicitly mention several […]
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By: Ilya
https://blog.tanyakhovanova.com/2010/07/hats-and-rooms/#comment-1499
Tue, 13 Jul 2010 16:05:18 +0000https://blog.tanyakhovanova.com/?p=255#comment-1499Tanya, I’m not sure I understand the question. If he doesn’t add the digit in his own place the algorithm can’t work since all people with correct hat (i.e. the number on the hat coincides with the corresponding digit in the chosen representative) will go to the same room…
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By: Tanya Khovanova
https://blog.tanyakhovanova.com/2010/07/hats-and-rooms/#comment-1498
Mon, 12 Jul 2010 19:54:47 +0000https://blog.tanyakhovanova.com/?p=255#comment-1498Ilya,

What will happen if he doesn’t add the digit in his own place?

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By: Ilya
https://blog.tanyakhovanova.com/2010/07/hats-and-rooms/#comment-1497
Mon, 12 Jul 2010 17:36:35 +0000https://blog.tanyakhovanova.com/?p=255#comment-1497In fact what I had in mind allows to find a strategy for a stronger problem: namely, if the rooms are colored in the colors of hats, and all but finitely many persons must get to the correct room. However, the idea seems to be similar. Let me consider for simplicity the case of only 2 colors but infinitely many wise men. let us order the wise men and consider sequences of 0 and 1 ordered by the set of wise men. Say two sequences are equivalent if they coincide everywhere but at finitely many places, and let us choose a sequence in each equivalence class. Now the strategy is the following: the rooms are labeled by 0 and 1. Any wise men sees the whole sequence of 0 and 1 but his own digit, hence he sees the equivalence class. He compares the chosen element with what he sees and counts the number of disagreements. He ads to this number the digit that appears in the chosen representative at his own place, and if the result is even he goes to room 0 otherwise to room 1.
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By: Ilya
https://blog.tanyakhovanova.com/2010/07/hats-and-rooms/#comment-1496
Mon, 12 Jul 2010 12:35:46 +0000https://blog.tanyakhovanova.com/?p=255#comment-1496If the numbers are infinite then I know how to use the axiom of choice to make sure that all but finitely many of them will come to the right rooms (i.e. all but finitely many wise men with the hats of the same color will be in the same room, and in any but finitely many rooms all the hats will be of the same color, and in the remaining rooms all but finitely many hats will be of the same color), but I wonder if the problem as you have stated it has a solution?
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By: JBL
https://blog.tanyakhovanova.com/2010/07/hats-and-rooms/#comment-1495
Sun, 11 Jul 2010 16:06:27 +0000https://blog.tanyakhovanova.com/?p=255#comment-1495Yes, you’re right, my solution requires a finite number of wisemen (but allows any cardinality of the color set).
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By: misha
https://blog.tanyakhovanova.com/2010/07/hats-and-rooms/#comment-1494
Sun, 11 Jul 2010 02:51:04 +0000https://blog.tanyakhovanova.com/?p=255#comment-1494Let us take an example. Assume the hats are black or white, the number of men wearing black hats is infinite, as well as the number of men wearing white hats. What algorithm should they use, JBL?
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By: misha
https://blog.tanyakhovanova.com/2010/07/hats-and-rooms/#comment-1493
Sun, 11 Jul 2010 00:41:47 +0000https://blog.tanyakhovanova.com/?p=255#comment-1493Actually you need infinite sums even if the number of wisemen is infinite, and it’s not clear it works out.
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By: misha
https://blog.tanyakhovanova.com/2010/07/hats-and-rooms/#comment-1492
Sun, 11 Jul 2010 00:35:02 +0000https://blog.tanyakhovanova.com/?p=255#comment-1492@JBL: If the number of the wisemen and the colors are both infinite, you need the infinite sums i.e., some extra structure in your group of colors, i.e. a bit more than an abelian group structure.
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By: JBL
https://blog.tanyakhovanova.com/2010/07/hats-and-rooms/#comment-1491
Thu, 08 Jul 2010 12:37:36 +0000https://blog.tanyakhovanova.com/?p=255#comment-1491Yes, my solution assumes that the wisemen know (an upper bound on) the number of colors. In this case it doesn’t matter if there are more rooms (in my phrasing, buttons) than the number of colors, since they may just agree to consider a subset of the buttons of the same cardinality as the set of colors.
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