They now know that at least one person has blue eyes.
Working logically, each 'blue-eyed' person will look and count the amount of blue eyed people, which will be 99 for blue eyed and 100 for brown. But they don't know that they are blue straight away.
If there was one blue eyed person, they would look at the others, see no blues and leave on first night.
If there were two, they would see the other blue and expect them to leave on night 1. When they didn't, they must know that they too have blue eyes and will both leave on night two.
I'm assuming that this will hold true as the pattern increases, meaning that when the 99 blue eyed people they have seen don't leave on the 99th night, there will be a realisation among all the blue eyed people that they too have blue eyes and will thus leave on night 100.
That's the only 'logical' explanation I can think of, though you'd think there must be a way to work it out sooner. The Night 100 looks too conveniently symmetrical for me.
You are correct.
to my knowledge that is the most efficient answer. If you can find a faster answer than you know it better than I do
Proof by induction. If there were only one blue-eyed person, he would leave on day 1. Since nobody leaves on day 1, there must be at least 2 with blue eyes, etc. Once everybody knows there are at least 100 with blue eyes, they can all leave. But not the brown-eyed people; their eyes could be pink for all they know.
Here's a pretty simple one that I like (pretty likely some of you have heard it before):
You're in a dark room with 100 coins on a table, 12 are heads and the rest are tails, and the two faces are completely indistinguishable in the dark. How do you separate the coins into two piles so that the number of face up heads in each pile are equal?
(again, no dumb solutions like "turn the lights on")
(January 24th, 2013, 15:47)zakalwe Wrote: I claim the most concise proof! Using an iPad helps.
Well, you haven't properly stated the proof so I'm not sure you can claim that. I found a properly styled proof for it though:
==BASE CASE==
TO PROVE: F(1) = G(1). If exactly one person had blue eyes, he would leave on day one
1 GIVEN at least one person on the island has blue eyes (from the oracle, who speaks only truth)
2 BASE HYPOTHESIS exactly one person has blue eyes
3 BY EXCLUSION AND 2 The person with blue eyes sees no one else with blue eyes
4 BY ELIMINATION AND 3 AND 1 The person with blue eyes concludes that he must have blue eyes
5 QED: If exactly one person had blue eyes, he would leave on day one. F(1) = G(1)
==INDUCTIVE CASE==
TO PROVE: If F(n) = G(n) then F(n+1) = G(n+1). (If n people have blue eyes, they will leave on day n) implies (If n+1 people have blue eyes, they will leave on day n+1)
6 GIVEN No one leaves until he knows he has blue eyes
7 ASSUMING n+1 people have blue eyes
8 INDUCTIVE HYPOTHESIS n people with blue eyes will leave on day n
9 BY EXCLUSION AND 7 Each person with blue eyes will see n people with blue eyes
10 BY 6 AND 7 AND 8 No one leaves on day n
11 _BY 8 AND 10 _The number of people with blue eyes cannot be n
12 BY ELIMINATION AND 9 AND 11 Each person with blue eyes reasons that he must have blue eyes (to make the total not equal to n)
13 QED: (If n people have blue eyes, they will leave on day n) implies (If n+1 people have blue eyes, they will leave on day n+1). If F(n) = G(n) then F(n+1) = G(n+1)
==INDUCTIVE CONCLUSION==
14 BY INDUCTION AND 5 AND 13 x people with blue eyes will leave on day x
(January 24th, 2013, 15:48)scooter Wrote: You're in a dark room with 100 coins on a table, 12 are heads and the rest are tails, and the two faces are completely indistinguishable in the dark. How do you separate the coins into two piles so that the number of face up heads in each pile are equal?
(January 24th, 2013, 15:48)scooter Wrote: You're in a dark room with 100 coins on a table, 12 are heads and the rest are tails, and the two faces are completely indistinguishable in the dark. How do you separate the coins into two piles so that the number of face up heads in each pile are equal?
Okay, I was not correct about the Guru speaking not giving any information. Without the Guru speaking, they can indeed tell that there are people with blue eyes, which is exactly what she says. But without the Guru saying it, the chain never starts. If there was just one with blue eyes, and there was no Guru saying that she sees blue eyes, then that one could not figure out that he has blue eyes, but if the Guru speaks, then he can. Basically, when there are 100 blue-eyed people, they are all the "100th" and have to wait until the 99th night before they know. Why? They see 99 others with blue eyes. But they could have brown eyes themselves, it's exactly what a brown eyed would see if there were only 99 blue-eyed. So if there were 99 blue-eyed, they would leave at the 99th night. But when they don't leave, it must mean that there are 100 blue-eyed people, and since you can only see 99, you are the 100th. Well, everyone with blue eyes is the 100th.
Everyone who has blue eyes knows from the beginning that there are either 99 or 100 blue-eyed people, and everyone who has brown eyes knows from the beginning that there are either 100 or 101 blue-eyed people. They know this, but to be SURE what color their OWN eyes are, they have to wait until the 99th or 100th night and see if people leave, then leave the next if they didn't.
