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Old 2005-10-27, 00:55   #1
fetofs
 
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I've seen a nice problem (not so difficult):

A group of people live on an island. They are all perfect logicians. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. If anyone has figured out the color of their own eyes, they [must] leave the island that midnight.
On this island live 100 blue-eyed people, 100 brown-eyed people, and the Guru. The Guru has green eyes, and does not know her own eye color either. Everyone on the island knows the rules (but are not given the total numbers) and is constantly aware of everyone else's eye color. Everyone keeps a constant count of the total number they see of each (excluding themselves). However, they cannot otherwise communicate. So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes, but that does not tell them their own eye color; it could be 101 brown and 99 blue. Or 100 brown, 99 blue, and the one could have red eyes.
The Guru speaks only once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following:

"I can see someone with blue eyes."

Who leaves the island, and on what night?

Solve it yourself!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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Old 2005-10-29, 12:48   #2
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They all leave except the guru.

The alternative claim that they all stay does not hold water.

Last fiddled with by Fusion_power on 2005-10-29 at 12:48
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Old 2005-10-29, 13:40   #3
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Quote:
Originally Posted by Fusion_power
They all leave except the guru.

The alternative claim that they all stay does not hold water.
How did the brown eyed people figure out their own eye color? "Not blue" is possble to figure out, but that includes brown and green and red.
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Old 2005-10-30, 02:35   #4
Orgasmic Troll
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all the blue eyed people leave on the 100th day

if there was one blue eyed person, they would leave the first night
if there were two, then the first night neither of them would leave because they would think it was the other person. (I'm going to name them Alice and Bob, because using unspecific pronouns got a little tricky) Alice sees Bob and thinks that Bob is going to be leaving the first night, when Bob is still around, then Alice realizes that Bob sees another person with blue eyes. Since Alice only sees one other person, she realizes that she's the other person and leaves that night.

if there were three (Alice, Bob and Charlie), then Alice sees Bob and Charlie, thinks that they will be leaving on the second night, and on the third night, realizes she has blue eyes.

etc.

so for n people with blue eyes, they leave on the nth night
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Old 2005-10-30, 02:39   #5
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Alternatively, the blue eyed people accuse the guru of optic profiling and form a rebellious subculture leading to an eventual revolution, whereupon the guru is lynched and all brown eyed people are forced off the island.
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Old 2005-10-30, 06:20   #6
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Travis,

Since they are all perfect logicians, once all the blue eyed people leave, all the brown eyed people will realize that they must not be blue eyed. Since that inherently means they must be brown eyed, they all now know the color of their eyes and therefore they leave too.

Unfortunately, your logic has a flaw. I will wait until someone else points it out.

Fusion
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Old 2005-10-30, 06:47   #7
Orgasmic Troll
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Quote:
Originally Posted by Fusion_power
Travis,

Since they are all perfect logicians, once all the blue eyed people leave, all the brown eyed people will realize that they must not be blue eyed. Since that inherently means they must be brown eyed, they all now know the color of their eyes and therefore they leave too.

Unfortunately, your logic has a flaw. I will wait until someone else points it out.

Fusion

Wrong em, boyo. Just because you and I know that doesn't mean that they do. Your logic has a flaw, see bolded above
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Old 2005-10-31, 19:07   #8
Numbers
 
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I think that TravisT has got his logic right, but his counting is a little bit awry.
Suppose that there are in fact only 6 people on the island, plus the Guru. Let’s
call them A, B, C, D, E, F and A, B, C are blue-eyed while D, E, F are brown-eyed.
I am C.

At noon the Guru says, “I can see a blue-eyed person.” Well that doesn’t come as any
surprise to anyone on the island because A, B, C can all see two blue-eyed folk, while
D, E, F can all see three blue-eyed folk.
At midnight the ferry arrives. Thinking that there are only two blue-eyed people on the
island I expect one of them to get on the ferry because they have figured out the colour
of their eyes. When no one gets on the ferry I immediately realise (because I am a perfect
logician) that both A, and B can see someone else with blue eyes, and that someone is me.
So I get on the ferry. Both A and B (who are perfect logicians) immediately realise why I got
on the ferry, and join me.
Therefore, all the blue-eyed folk get on the ferry the first night.


But I would not have been able to work this out without TravisT’s clue
So Kudos to him for getting the logic right in the first place.
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Old 2005-10-31, 20:28   #9
ewmayer
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All the blue-eyed people leave at midnight on the 100th day.
All the brown-eyed people leave at midnight on the 200th day.

The guru stays, I wash ashore on my life raft, and we make sweet music and many babies of all the various eye colors together. What color are my eyes? Wait, don't tell me - I'm having too much fun.
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Old 2005-10-31, 23:32   #10
ewmayer
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Actually, I was just kidding about the brown-eyed people leaving on the 200th day - I was hoping the 100 remaining non-Guru islanders would read this and vacate the island so I could make time with the sexy she-Guru.
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Old 2005-11-01, 04:24   #11
cheesehead
 
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Numbers,

I think you're right as far as you went, but ...

... A and B, being equally perfect logicians, would simultaneously reach the same conclusion as C and, therefore, all three would step toward and board the ferry simultaneously.

Last fiddled with by cheesehead on 2005-11-01 at 04:38
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