Monks with Dots
Was given this little riddle during Christmas vacation and after stumping a few Nakedites thought it worth sharing . . .
One or more monks live in a monestary. One day God comes down and says to the monks that he is going to place dots on one or more of their heads. If they are sure they have a dot on their head, when they retire to their room that evening they must kill themselves. The catch: They may not speak to each other and they may not look in a mirror. How many days does it take for all the monks with dots to kill themselves?
There are no tricks to this riddle (like they’re some special religion or the monestary is actually on the moon and they don’t look in reflecting pools). It’s just a regular old logic problem.
Maybe there’s a prize for the first person to answer correctly (but I really don’t know what that will be).
[…] posted this over at the Naked Communications blog and thought it would be fun to share here as well. It’s a bit […]
Well, if there’s only Monk A, he’s got to have a dot, so that equals one night.
If there’s two monks, there’s two cases:
- one monk with a dot, one without
- two monks with dots
If a monk sees the other monk’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’s three monks, there’s three cases:
- one monk with dot, two without
- two monks with dot, one without
- three monks with dots
If there’s only one monk with the dot, he’ll see the undotted heads of the other two and kill himself that night.
If there’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’s more than one dot (since otherwise the dotted monk would’ve seen two undotted heads and killed himself). Since they still see the undotted monk, they realize there must be two dots, and they’ve got them. They kill themselves, two days.
Say all three have dots. Each monk sees two dotted monks and doesn’t know anything about his own head. However, he knows that if he’s undotted, that’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’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’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!
well done greg . . . you win.
[…] drives us all mad with his monk riddle (he already did it once at MT’s […]