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	<title>Comments on: Monks with Dots</title>
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		<title>By: January Aggregated I &#187; House of Naked</title>
		<link>http://www.houseofnaked.com/2008/01/07/monks-with-dots/comment-page-1/#comment-717</link>
		<dc:creator>January Aggregated I &#187; House of Naked</dc:creator>
		<pubDate>Thu, 17 Jan 2008 22:39:18 +0000</pubDate>
		<guid isPermaLink="false">http://www.houseofnaked.com/2008/01/07/monks-with-dots/#comment-717</guid>
		<description>[...] drives us all mad with his monk riddle (he already did it once at MT&#8217;s [...]</description>
		<content:encoded><![CDATA[<p>[...] drives us all mad with his monk riddle (he already did it once at MT&#8217;s [...]</p>
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		<title>By: noah</title>
		<link>http://www.houseofnaked.com/2008/01/07/monks-with-dots/comment-page-1/#comment-675</link>
		<dc:creator>noah</dc:creator>
		<pubDate>Mon, 07 Jan 2008 22:56:22 +0000</pubDate>
		<guid isPermaLink="false">http://www.houseofnaked.com/2008/01/07/monks-with-dots/#comment-675</guid>
		<description>well done greg . . . you win.</description>
		<content:encoded><![CDATA[<p>well done greg . . . you win.</p>
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		<title>By: Greg</title>
		<link>http://www.houseofnaked.com/2008/01/07/monks-with-dots/comment-page-1/#comment-673</link>
		<dc:creator>Greg</dc:creator>
		<pubDate>Mon, 07 Jan 2008 20:21:17 +0000</pubDate>
		<guid isPermaLink="false">http://www.houseofnaked.com/2008/01/07/monks-with-dots/#comment-673</guid>
		<description>Well, if there&#039;s only Monk A, he&#039;s got to have a dot, so that equals one night.

If there&#039;s two monks, there&#039;s two cases:

- one monk with a dot, one without
- two monks with dots

If a monk sees the other monk&#039;s got no dots, he therefore knows he has to have a dot and kills himself.  One day. 

If both monks have dots, they learn nothing about which case it is - until they wake up the next day, both still alive.  Then they realize that the other monk must see a dot, and they both kill themselves that night.  Two days.

If there&#039;s three monks, there&#039;s three cases:

- one monk with dot, two without
- two monks with dot, one without
- three monks with dots

If there&#039;s only one monk with the dot, he&#039;ll see the undotted heads of the other two and kill himself that night.

If there&#039;s two monks with the dot, then the undotted monk sees two dotted monks, and the two dotted monks see one dotted monk, and one undotted.  The two dotted monks both sleep on it.  When they wake up the next day, and see the other dotted monk still alive, they realize there&#039;s more than one dot (since otherwise the dotted monk would&#039;ve seen two undotted heads and killed himself).  Since they still see the undotted monk, they realize there must be two dots, and they&#039;ve got them.  They kill themselves, two days.

Say all three have dots.  Each monk sees two dotted monks and doesn&#039;t know anything about his own head.  However, he knows that if he&#039;s undotted, that&#039;s the above two-dots-one-undotted case, and the other two monks will clue in and kill themselves after two days.  So he waits for two days.  If he wakes up on the third day, and the other two monks are still alive, he knows he must have a dot too.  They all kill themselves after three days.

This is a recursive problem.  Don&#039;t ask me to do the exact math, but it looks like the number of dots = the number of days.  (And no matter how many dots, all the monks are going to do the deed on the same day - there&#039;s no way any monk can find out earlier than any other.)

Interesting side effect: if God lies, and places dots on *none* of the monks, every single monk will kill himself that night. Doh!</description>
		<content:encoded><![CDATA[<p>Well, if there&#8217;s only Monk A, he&#8217;s got to have a dot, so that equals one night.</p>
<p>If there&#8217;s two monks, there&#8217;s two cases:</p>
<p>- one monk with a dot, one without<br />
- two monks with dots</p>
<p>If a monk sees the other monk&#8217;s got no dots, he therefore knows he has to have a dot and kills himself.  One day. </p>
<p>If both monks have dots, they learn nothing about which case it is &#8211; until they wake up the next day, both still alive.  Then they realize that the other monk must see a dot, and they both kill themselves that night.  Two days.</p>
<p>If there&#8217;s three monks, there&#8217;s three cases:</p>
<p>- one monk with dot, two without<br />
- two monks with dot, one without<br />
- three monks with dots</p>
<p>If there&#8217;s only one monk with the dot, he&#8217;ll see the undotted heads of the other two and kill himself that night.</p>
<p>If there&#8217;s two monks with the dot, then the undotted monk sees two dotted monks, and the two dotted monks see one dotted monk, and one undotted.  The two dotted monks both sleep on it.  When they wake up the next day, and see the other dotted monk still alive, they realize there&#8217;s more than one dot (since otherwise the dotted monk would&#8217;ve seen two undotted heads and killed himself).  Since they still see the undotted monk, they realize there must be two dots, and they&#8217;ve got them.  They kill themselves, two days.</p>
<p>Say all three have dots.  Each monk sees two dotted monks and doesn&#8217;t know anything about his own head.  However, he knows that if he&#8217;s undotted, that&#8217;s the above two-dots-one-undotted case, and the other two monks will clue in and kill themselves after two days.  So he waits for two days.  If he wakes up on the third day, and the other two monks are still alive, he knows he must have a dot too.  They all kill themselves after three days.</p>
<p>This is a recursive problem.  Don&#8217;t ask me to do the exact math, but it looks like the number of dots = the number of days.  (And no matter how many dots, all the monks are going to do the deed on the same day &#8211; there&#8217;s no way any monk can find out earlier than any other.)</p>
<p>Interesting side effect: if God lies, and places dots on *none* of the monks, every single monk will kill himself that night. Doh!</p>
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		<title>By: Monks with Dots &#187; House of Naked</title>
		<link>http://www.houseofnaked.com/2008/01/07/monks-with-dots/comment-page-1/#comment-672</link>
		<dc:creator>Monks with Dots &#187; House of Naked</dc:creator>
		<pubDate>Mon, 07 Jan 2008 19:31:05 +0000</pubDate>
		<guid isPermaLink="false">http://www.houseofnaked.com/2008/01/07/monks-with-dots/#comment-672</guid>
		<description>[...] posted this over at the Naked Communications blog and thought it would be fun to share here as well. It&#8217;s a bit [...]</description>
		<content:encoded><![CDATA[<p>[...] posted this over at the Naked Communications blog and thought it would be fun to share here as well. It&#8217;s a bit [...]</p>
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