A logic puzzle

[[ptx]]

So your solution only works if you ignore that they actually have to figure out their own eye color to leave and that everyone can see everyone else's eye color at all times.


I would agree though that they will all leave during the first night since: "They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly."

Except if the solution needs some kind of system that takes multiple nights to complete

Let me help you to a bit of colour

In any case, me and Vibe already agreed that i made a sort of a "trick answer", since it renders most of the given information as irrelevant and that it should be made clear that it is their eye colour during day time that they need to figure out or some such.

[Overdriven]

That makes no sense to me 

[Tomas]

That makes no sense to me 

I said it was convoluted

[Meow]

Let me help you to a bit of colour

In any case, me and Vibe already agreed that i made a sort of a "trick answer", since it renders most of the given information as irrelevant and that it should be made clear that it is their eye colour during day time that they need to figure out or some such.

Sounds like you just keep refusing to accept that your answer is plain wrong according to the rules stated in the first paragraph but yeah, now I know a lot more about colors

[SixThumbs]

Let's get to the real problem at hand and that's that after the 101st night the guru is going to be very lonely.

[Vibe]

Let's get to the real problem at hand and that's that after the 101st night the guru is going to be very lonely.

But there's also brown eyed people on the island.

[Kafein]

This seems interesting to share :

Let's say N is the number of blue and brown eyed people (there are N blue, N brown and 1 guru).

Case N=1 is trivial. The blue leaves after the guru speech.

Case N=2 is simple. Let's say there's A blue and B blue. A knows that B knows A's color. Therefore if B does not leave the first day, that means that B could not conclude anything from what the guru said, which means A is blue. B does the same and both leave the 2nd day.

Case N=3 is equivalent to N=100, but easier to understand.

Let's say we have A,B and C blue. A knows that both B and C are blue, and that they know A's color and the other one's color, which is blue.

That means that A knows that from the point of view of B and C, there is always at least one other blue which is not A. In other words, B will see C is blue, and C will see B is blue. Everybody sees at least one blue, and that is for certain with only the information that one blue person has. It's the same from the perspective of a brown.

Therefore what the guru says will never have an impact on what they know, since it doesn't change their own direct knowledge, nor what they know of what the others see.

[Vibe]

Err - this could get a little convoluted but i'll try and explian my logic - hopefully it is correct and i didn't just get lucky


- Imagine I am a blue eyed person on the Island
- I can see 99 people with blue eyes, 101 with brown eyes and 1 with green eyes
- There are therefore either 99 Blue eyed people or there are 100, depending on my own eye colour.
- Since the Blue eyed people didn't leave on the 99th night, then those 99 people must still not have been sure that they were blue eyed themselves which means I must have been confusuing them
- Therefore on the 100th night, I will know I have blue eyes for certain since there can't be 101 of us, so I will leave along with all the other blue eyed people who realise the same thing at the same time

You're thinking in the right direction Tomas.

BUT there can be 101 blue eyes.

There can be 101 blue, 99 brown.
There can be 99 blue, 101 brown.
Or there can be 100 blue and 100 brown.

----------------

Kafein pretty much told you the solution. Did you come this by yourself Kafein?

[Teeth]

It makes so much sense for two people with blue eyes. Person 1 and person 2 know that there is 1 person with blue eyes. If after the first night 1 sees that 2 hasn't left, 1 knows that 2 wasn't sure because 1 himself must have blue eyes. Same thing the other way around.

I get how that works. But I don't understand how the 100 people find out their eye color after 100 days, when there is no new information at all since day 1.

Let's say there are 3 persons. You know the two others have blue eyes and you are unsure if you have blue eyes. Its the same for them. Then after one night, nobody will have left. That confirms that there are not one but two people with blue eyes. You already knew that.

So when comparing day 2 and day 3, there is no new information. Why do you know your eye color in night 3 and not at night 2, when nothing changed and there was no new information?

Oh yeah, and what Kafein said, the Guru seems quite the useless bitch. Kafein didnt tell the solution, he only described what every person in a 3 person situation sees. There is my problem, three people have the same thing where they are not sure about. Seems to be that from n > 2 there is no one leaving.

[Vibe]

Here's the full solution for better understanding:

The answer is that on the 100th day, all 100 blue-eyed people will leave.

If you consider the case of just one blue-eyed person on the island, you can show that he obviously leaves the first night, because he knows he's the only one the Guru could be talking about. He looks around and sees no one else, and knows he should leave. So: [THEOREM 1] If there is one blue-eyed person, he leaves the first night.

If there are two blue-eyed people, they will each look at the other. They will each realize that "if I don't have blue eyes [HYPOTHESIS 1], then that guy is the only blue-eyed person. And if he's the only person, by THEOREM 1 he will leave tonight." They each wait and see, and when neither of them leave the first night, each realizes "My HYPOTHESIS 1 was incorrect. I must have blue eyes." And each leaves the second night.

So: [THEOREM 2]: If there are two blue-eyed people on the island, they will each leave the 2nd night.

If there are three blue-eyed people, each one will look at the other two and go through a process similar to the one above. Each considers the two possibilities -- "I have blue eyes" or "I don't have blue eyes." He will know that if he doesn't have blue eyes, there are only two blue-eyed people on the island -- the two he sees. So he can wait two nights, and if no one leaves, he knows he must have blue eyes -- THEOREM 2 says that if he didn't, the other guys would have left. When he sees that they didn't, he knows his eyes are blue. All three of them are doing this same process, so they all figure it out on day 3 and leave.

This induction can continue all the way up to THEOREM 99, which each person on the island in the problem will of course know immediately. Then they'll each wait 99 days, see that the rest of the group hasn't gone anywhere, and on the 100th night, they all leave.


Stolen from xkcd.

[Kafein]

(warning : solution)

(click to show/hide)

Let's say there are 3 persons. You know the two others have blue eyes and you are unsure if you have blue eyes. Its the same for them. Then after one night, nobody will have left. That confirms that there are not one but two people with blue eyes. You already knew that.

Yessss ! Thank you btw ! I think I found the explanation.

Let's do it ad absurdum. If A is red (not blue), and sees B and C do not leave after one day, that means each one of them sees only one blue.

Still supposing that A is not blue, that means that B and C are exactly in the same situation as in the N=2 case :

Case N=2 is simple. Let's say there's A blue and B blue. A knows that B knows A's color. Therefore if B does not leave the first day, that means that B could not conclude anything from what the guru said, which means A is blue. B does the same and both leave the 2nd day.

Therefore, from A's point of view, if they do not leave on the second day, that means A is blue !!! A, B and C make the same reasoning and all leave on the third day.

The same principle can be applied recursively all the way down from 100 to 3, therefore this a proof of the N=100 case.

Hell, that was one hell of a tricky problem. The recurrence on 100 people makes attacking it up front nearly impossible.

[Kafein]

Damn you vibe, I swear I got it from my own reasoning. Yet I think I wouldn't have found without teeth :

Then after one night, nobody will have left. That confirms that there are not one but two people with blue eyes.

[Teeth]

Okay I get it, fucking hurts my brain though.

Now I wonder, does the puzzle work without the Guru? If there is no one revealing that there is atleast 1 person with blue eyes, Theorem 1 is false. Theorem 2 is still correct though. Theorem 99 is still correct too. Do you need Theorem 1 to formulate Theorem 2 and up? I don't think you really do, please correct me if I'm wrong. Would make this puzzle even more puzzling.

Gash, its so fucking retarded, 100 blue eyed people need to wait 99 nights to get confirmed that there are not 99 people with blue eyes, but a 100. They all know no one is going to leave up until the 98th night, still they have to wait 98 nights to see what happens in the 99th night. It's really weird, but its still correct.

Damn you vibe, I swear I got it from my own reasoning. Yet I think I wouldn't have found without teeth :

Lol, I stated the obvious information that in the case of N amount of blue eyed people, every blue eyed person always knows for sure that there are N-1 amount of blue people on the island. Because everyone is only unsure about their own color. I'm sure you already thought of that without me as its kinda required to get anywhere.

[Kafein]

Okay I get it, fucking hurts my brain though.

Now I wonder, does the puzzle work without the Guru? If there is no one revealing that there is atleast 1 person with blue eyes, Theorem 1 is false. Theorem 2 is still correct though. Theorem 99 is still correct too. Do you need Theorem 1 to formulate Theorem 2 and up? I don't think you really do, please correct me if I'm wrong. Would make this puzzle even more puzzling.

Gash, its so fucking retarded, 100 blue eyed people need to wait 99 nights to get confirmed that there are not 99 people with blue eyes, but a 100. They all know no one is going to leave up until the 98th night, still they have to wait 98 nights to see what happens in the 99th night. It's really weird, but its still correct.
Lol, I stated the obvious information that in the case of N amount of blue eyed people, every blue eyed person always knows for sure that there are N-1 amount of blue people on the island. Because everyone is only unsure about their own color. I'm sure you already thought of that without me as its kinda required to get anywhere.

Well it sort of pointed out what I knew but what I was forgetting. Made me think that B and C in N=3 are exactly like A and B in N=2, if the A in N=3 is not blue.

And you definetly need the guru. Otherwise, in the N=1 and N=2 case, nobody would leave, because supposing A is not blue, from A's point of view, B has no reason to leave on the first day. B is in the N=1 case, which does not work without the guru.

[Teeth]

True, Theorem 2 isn't correct at all if the Guru is not there, cause Theorem 2 relies on Theorem 1 being correct. Silly me.