Once upon a time there was a monastery in which lived a number of happy monks. B
ID: 3205547 • Letter: O
Question
Once upon a time there was a monastery in which lived a number of happy monks. Being monks, there was, naturally, no communication of any kind between them. Starting one day (let's call it day 1) all the monks gathered in one place for morning prayers. Each monk looked at every other monk, but no one said a word, exchanged a note or made a gesture. The rest of the day the monks were isolated; no monk saw any other. Although the monks were generally nice, they were alas susceptible to sin. Any time a monk sinned badly, a mark instantaneously appeared magically on his forehead. One of the house rules was that when a monk realized he had a mark on his forehead, he left the monastery that same day. Now, happy monks cannot tell when they have sinned badly and our monks could not see whether or not they had a mark on their forehead. Since the monks had renounced vanity, there were no mirrors or reflecting surfaces available inside the monastery. No monk would leave until he had proof that he was marked. Our monks were master logicians and could figure things out instantaneously. And, finally, you may assume that none of our monks was blind and that no monks sinned after day 1 morning prayers. If exactly one monk sinned, what will happen? If exactly two monks sinned before morning prayers on day 1, what will happen? Will both monks eventually leave, and, if so, on which days will they leave? Now suppose that a stranger arrives for morning prayers on day 1 and truthfully announces "I see that at least one of you has a mark on his forehead." How does this stranger's announcement change your answers to parts (a) and (b)? Describe the events on the days after the stranger's announcement when k > 2 monks sinned before day 1 morning prayers. Prove your claim by induction.Explanation / Answer
a) The monk with the mark on his forehead will see that no one has any mark , but everyone can see the mark on the monk but as they cannot communicate between them there is no way the monk can conclude that he has the mark on his head. HENCE HE WILL NEVER KNOW ABOUT THE MARK.
b) In case of two monks having mark both will see that one more has the mark. But they won't be able to conclude that they themselves have the mark even on further days hence no onw will leave.
c) In the case when they are told that atleat one of them have the mark
in case of one monk having the mark will see that no one has the mark so he can conclude that he has the mark on his head and will leace on day 1.
in case of two monks having mark they wont leave on day 1 as they will expect the other one to leave. But when again seeing each other in the second day they can conclude that there is one more monk having the mark other than the monk they can see having the mark hence they both can conclude that they both have the mark and both will leave on the second day.
d) Following the same routine for 3 monks having marks they would expect the other two leaving on the second day but on seeing themselves on the third day they can conclude that three of them have mark and will leave on the third day.
Hence bu induction we can conclude that if k monks are having marks thay will leave on the kth day.
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