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Mon 3 Mar, 2008 12:01 pm
Once upon a time there was a monastery in which lived several monks. One day a monk from another place came to the monastery, gathered all the monks and said to them:
"Dear friends, i'm afraid i have bad news. I've been told that some of you are infected with a mortal disease. Whoever is infected has a red dot behind the neck and the disease is not contagious. I won't tell who is infected and i ask you to do the same. So, from now on, i ask you not to communicate with each other in any kind of way. I also know that this monastery doesn't have any kind of mirror so you won't be able to see if you have the red mark. But being brilliant logicians as you are i propose the following: each day you will meet by the afternoon where you can look at each other and at night you will each go to your private rooms. If you are positive you are infected you leave the monastery during the night. I'm confident about your success. Farewell my friends!"
And so the monks did!
- In the 1st day the monks gathered in the afternoon and at night went to their rooms and none left the monastery
- In the 2nd day the monks gathered in the afternoon and at night went to their rooms and none left the monastery
- In the 3rd day the monks gathered in the afternoon and at night went to their rooms and all who were infected left the monastery
How many monks were infected? What was their reasoning?
Tell the exact way you solved it.
[size=7]Three monks have been infected
The first night uninfected monks see three red dots, and infected monks see two.
Given some number of monks is infected.
Start with one infected monk An infected monk, would think if I'm the only infected monk then I'll see no red dots if I don't see any dots I'll know I'm infected. (one night)
Two infected monks. If I see one dot the first night then I could be infected . If I'm infected the other infected monk will be back tomorrow night because he too saw a single dot. Uninfected monks see two dots so tomorrow they'll be back. After the second night the infected monks will identify themselves and not show up the third night.
Three infected monks--infected monks see three dots--infected ones see two. The sort will take three days to identify there are three infected monks and self identify the monks, and the three infected monks will not show up the fourth night,
K infected monks. Uninfected monks see k red dots. Infected monks see k-1. Using the same type of logic it will take k days to identify the k infected monks. The k infected monks will not show up the (k+1)th night.
Note, this type of logic used could be used to sort batches. It'll work quite well when sorting small concentrations of contaminants from limited populations--it would become quite tedious for large amounts of contaminants, or with large populations.[/size]
Rap
Sorry, BlackBONs, I don't agree with that answer.
The visiting monk said SOME of you ARE infected, so the monks would have known that there was more than one and so 2 would have left the first night, if there were only 2, 3 the second night and 4 the 3rd night. So I think the answer is 4.
It is exactly the same logic, but the wording gives us a different starting point.
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