Link to the Problem
Manikandan's solution was almost correct and Yash correctly modified it!
Here's my solution:
In such type of problems always try decreasing the number of parameters (i.e. Use induction).
If there was only 1 blue eyed person, he would have left at the end of the first day itself as he doesn't see any other blue eyed person there (and he knows that there is at least one blue eyed person).
If there were 2 blue eyed people, then the first would see one other blue eyed person on the first day and vice versa and hence both won't leave on the first day (as they can't determine their own eye colour). However, on the second day they know that there are more than one blue eyed persons (if there was only one blue eyed person then he would have left the first day itself). And since the blue eyed persons see only one other blue eyed person, each of them themselves is the other blue eyed person creating confusion on the first day. So the two blue eyed persons know that their eye colour is blue on the second day and hence would leave at the end of that day.
Similarly, if there were 3 blue eyed people, these three blue eyed people would be confused for the first two days as they see 2 other blue eyed people not leaving the island, however on the third day they would know there is one more blue eyed person and it could be only be he himself (as they know that if there were only 2 blue eyed people, they would have left at the end of the 2nd day by above argument). So these three blue eyed people would leave at the end of the 3rd day.
By similar argument we can prove that if there are N blue eyed people, they would be confused till N-1 days but on the Nth day they would all come to know that their eye colour is blue! And hence all the N blue eyed people leave at the end of the Nth day.
Hence in this case all the 100 Blue eyed people leave on the 100th day and no one else leaves!
3 comments:
By the logic you mentioned all the logicians (regardless of eye color) will know that all blue eyed people will leave on the same day (X days where X = no. of blue eyed people). So on the X + 1'th day when the logicians see only brown eyed people, they will know they also have brown eyes and will leave on that night!
the no of colors possible is not fixed. Even if they see all brown eyed people and maybe 1 green eyed person, they cannot be sure of their eye color since they dont know all people have only 3 possible colors ...
All your answers leave me with a feeling of deja vu in the end :)
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