Blue-eyes logic puzzle

Factual Questions
41

Let’s reduce the number of people to make it simpler. Let’s say there are only six people besides the guru on the island: Al, Bob, Chuck, Dan, Ed, and Frank.

The guru says he sees at least one blue-eyed person.

Case #1 - Let’s assume Al looks around and sees that Bob, Chuck, Dan, Ed, and Frank all have brown eyes. Al realizes he must have blue eyes. So Al gets on the boat on the first day.

Case #2 - Let’s assume that Al looks around and sees that Bob has blue eyes and Chuck, Dan, Ed, and Frank have brown eyes. Al figures that if he has brown eyes that Bob is seeing all brown-eyed people. So Bob is in the position Al was in in Case #1. Bob is as smart as Al so he will figure out that he must have blue eyes. So Bob will get on the boat on the first day.

But suppose Bob doesn’t get on the boat on the first day. Al will then realize that Bob must not be seeing only brown-eyed people. Bob must be seeing a blue-eyed person. And Al can see that Chuck, Dan, Ed, and Frank have brown eyes. So Al will figure out he must be the blue-eyed person Bob is seeing. So Al now knows he has blue eyes and he can get on the boat on the second day. And Bob, having the same information as Al, will also get on the boat on the second day.

Case #3 - Let’s assume Al looks around and sees that Bob and Chuck have blue eyes and Dan, Ed, and Frank have brown eyes. Al figures if he has brown eyes then Chuck is seeing one other blue-eyed person, Bob. So Chuck’s in the same position Bob was in in Case #2. Chuck, being smart, will figure this out and will get on the boat on the second day and so will Bob.

But if Chuck and Bob don’t get on the boat on the second day, then Al will know that they must be seeing more than one other blue-eyed person. And Al can see Dan, Ed, and Frank have brown eyes. So he’ll figure out he must be the other blue-eyed person. So Al will get on the boat on the third day, as will Bob and Chuck.

Case #4 - Let’s assume Al looks around and sees that Bob, Chuck, and Dan have blue eyes and Ed and Frank have brown eyes. Al figures if he had brown eyes then Dan is seeing two blue-eyed people, Bob and Chuck. So Dan is in the position that Chuck was in in Case #3 and will figure it out and get in the boat on the third day, along with Bob and Chuck.

Once again, if Bob, Chuck, and Dan don’t get in the boat on the third day, then Al will know there is another blue-eyed person. And once again he’ll know he’s the other blue-eyed person and will get in the boat on the fourth day, along with Bob, Chuck, and Dan.

Case #N - It scales up. No matter how many blue-eyed people there are, you will reach a point where all of the blue-eyed people will be able to follow a chain of logic and establish that they have blue eyes. The only person who’s unaccounted for is you.

So you count up how many blue-eyed people you see and call this N. If on the Nth day, all the blue-eyed people get on the boat, you’ll know that N is the total number of blue-eyed people and you have brown eyes. If the blue-eyed people you see don’t get on the boat on the Nth day, you’ll know there must be more than N blue-eyed people. You’ll know there are N+1 blue-eyed people and you must be the +1 blue-eyed person.

42

If you are an islander, the Guru saying “There is at least one person with blue eyes” has two effects on you
A) It tells you that there is at least one person with blue eyes. Well, actually, this is no effect, because you already knew that…
B) It brings about a situation where you know that everyone else knows A). And that everyone else knows B) as well! [So everyone knows that everyone knows A). And everyone knows that everyone knows that everyone knows A). And so on, ad infinitum]. In technical jargon, it turns A) into “common knowledge”, which it wasn’t before. Frylock’s example with two people is particularly illustrative.

Everyone who thinks the Guru’s statement is uninformative is only looking at A). It’s true that A) is nothing you don’t already know. The value of the Guru’s pronouncement is B).

It is vital for this, mind you, that the Guru makes this pronouncement publicly. If the Guru just went around whispering privately to each islander “I see at least one blue-eyed person”, this would only inform each islander of A), and thus have no effect. It’s the public nature of the pronouncement that allows it to bring about effect B).

In terms of the archipelago metaphor, prior to The Guru’s public pronouncement, it was possible to get from the island of all blue-eyes to the island of no blue-eyes by taking N many bridges. After the Guru’s public pronouncement that there is at least one blue-eyed person, the island of no blue-eyes has been bombed out of the world [at least, it’s been cut off from that portion of the world compatible with this pronouncement], and the only islands left accessible all have at least one person with blue eyes. That is the pronouncement’s value; to bring about that action.

43

Very clearly explained, Little Nemo.

The reason that you can’t ever skip a day is that you don’t know what anyone else is really seeing (due to your lack of information about your own eye color). Because that is true for everyone, the logical bootstrap can’t skip any days.

44

I wish they’d just start speaking to each other again. C’mon guys, forgive and forget!

45

[QUOTE=xkcd]
Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph.
[/QUOTE]
Given that there are at least 3 blue eyes on the island, everyone knows everyone can see at least 2 blue eyes. Anyone that can see 2 blue eyes can deduce that everyone can see one blue eyes. There are at least 3 blue eyes, so everyone can deduce that everyone sees one blue eyes.

Example: Bob, Fred, Joe have blue eyes. Bob sees Joe and Fred. Bob knows that Joe sees Fred, so Bob knows that Joe knows one person has blue eyes. Mutatis mutandis for the other cases.

I don’t understand what the Guru has added.

46

The Guru has added that everyone knows that everyone knows that everyone knows that there is at least one blue-eyed person. And so on.

In your case, Bob thinks he might have brown eyes. Because Bob thinks he might have brown eyes, Bob thinks Fred might see Bob with brown eyes. Bob also knows that Fred doesn’t know his own eye color. So Bob thinks it might be the case that Fred thinks it might be the case that both Bob and Fred have brown eyes. Which is to say, Bob thinks it might be the case that Fred thinks it might be the case that Joe sees only brown eyes. Which is to say, Bob thinks it might be the case that Fred thinks it might be the case that Joe thinks it might be the case that everyone has brown eyes.

After the Guru makes his announcement, that no longer holds. After the announcement, Bob knows that Fred knows that Joe knows that at least one person has blue eyes.

Originally, there is a path of length 3 from island (1, 1, 1) to an island with no blue eyes. The Guru’s announcement has the effect of destroying the island with no blue eyes, so that every path of length 3 out of (1, 1, 1) still lands at an island with blue eyes.

47

This isn’t right; given that everyone knows that there are at least 3 blue eyes, everyone knows everyone can see at least 2 blue eyes.

But this doesn’t resolve my confusion.

48

Let me try that again. The important assumption is that everone knows that everyone can see everyone but themselves.

Case: 4 blues on the island.

  1. Everyone can see 3 blue eyes, because there are 4 blue eyes.
  2. So everyone knows that there are at least 3 blue eyes. (by 1)
  3. Everyone knows everyone can see 2 blue eyes. (by 2)
  4. So everyone knows everyone knows everyone can see 1 blue eyes. (by 3)

I can’t see which step is wrong.

49

Right, if everyone knows that there are at least 3 blue eyes, everyone knows everyone can see at least 2 blue eyes, and thus everyone knows that everyone knows that everyone can see at least one 1 blue eye, and thus everyone knows that everyone knows that everyone knows that there is at least 1 blue eye.

And if everyone knows there are at least 4 blue eyes, then everyone knows that everyone knows that everyone knows that everyone knows that there is at least 1 blue eye.

And in general, if everyone sees at least N blue eyes, then everyone knows (iterated N times) that there is at least 1 blue eye. But it won’t be the case that everyone knows (iterated N + 1 times) that there is at least 1 blue eye.

If we start with 100 islanders, each islander sees 99 blue eyes. Everyone knows (iterated 99 times) that there is at least 1 blue eye. But, to start with, before the Guru steps in, it isn’t the case that everyone knows (iterated 100 times) that there is at least 1 blue eye. The Guru’s PUBLIC announcement makes that happen.

50

What you’ve written is correct. With 4 people, you can get to “everyone knows everyone knows everyone can see 1 blue eye”. But you can’t get to “Everyone knows everyone knows everyone knows everyone can see 1 blue eye”.

With 5 people, you could get to that but you couldn’t get to “Everyone knows that everyone knows everyone knows everyone knows everyone can see 1 blue eye”.

With N people, you can get to “Everyone knows (iterated less than N times) that there is at least 1 blue eye”, but you can’t get to “Everyone knows (iterated N times) that there is at least 1 blue eye”. Until the Guru steps in, and makes his public announcement, which has the effect of letting you get to “Everyone knows (iterated however many times you like) that there is at least 1 blue eye”.

In terms of the archipelago: The distance from (1, 1, 1, 1, …) to (0, 0, 0, 0, …) is the number of islanders (per island). If there are N islanders, it takes N bridge-hops to get from “All blue eyes” to “No blue eyes” by bridge-hopping. But it can be done! The Guru’s public announcement has the effect of making it impossible to get from “All blue eyes” to “No blue eyes” by bridge-hopping.

51

Very clever.

Anyone know if anyone put this one (or similar puzzles) into logical form, where that means fits one of the common knowledge logics.

52

Take it to step 5:

  1. So everyone knows everyone knows everyone knows everyone can see at least 0 blue eyes. (by 4)

But that tells you nothing, because of course everyone can see “at least zero” blue eyes. Step 5 is useless, which I guess is why you stopped at step 4. The guru’s statement, however, allows you to say

  1. So everyone knows everyone knows everyone knows everyone can see at least 1 blue eyes.

and

  1. So everyone knows everyone knows everyone knows everyone knows everyone can see at least 1 blue eyes.

and so on ad infinitum. That is what **Indistinguishable **means by it “turning into common knowledge”.

[edit] OK, well somehow I missed that he had already dealt with your question

53

It was already common knowledge. If you can see more than 1 person with Blue eyes then there is no possible eye color that my own eye color can be that means that there is no one on the island that can see no blue eyes.

If everyone is a perfect logician then they should realise at the beginning that everyone on the island already knows that the only possible combinations are 99 Blue, 101 Brown, 1 Green OR 100 Blue, 100 Brown, 1 Green or 100 Blue, 100 Brown and 1 Unknown.

the webpage makes this clear “Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph.”

everyone knows the rules means everyone knows that everyone else has made a count and knows the potential possibilities. So the Guru’s pronouncement adds nothing. No one ever leaves.

54

No, it wasn’t. “Common knowledge”, in this more formal sense, means the you can extend the “everyone knows that everyone knows […] …that everyone can see at least one blue-eyes” statement for as long as you like. Before the Guru speaks, you cannot do that.

55

Nonsense, the rules as written make clear that everyone on the islands has to be aware of the only possible combinations I’ve spelt out. There is no chain of logic that would lead anyone of the people to believe that there is no blue eyes on the island in the scenario as written.

56

You need the Guru, even in a case where there are multiple blue-eyed people and everyone can see blue-eyed people are there. Because you need to build up a chain of logically supported hypotheticals to be able to prove what color your eyes are and the first hypothetical has to be that there’s only one blue-eyed person.

57

You said “if you can see more than 1 person with Blue eyes then there is no possible eye color that my own eye color can be that means that there is no one on the island that can see no blue eyes.” That is correct, but it is only your knowledge of the number of blue-eyed people. Do you know that the two people you can see who have blue eyes know the same as you? No. As far as you know, they don’t know that that the other blue-eyed person can see any blue-eyes.

“Common knowledge” doesn’t merely mean “something that everybody knows”. It’s more than that.

58

I’m not even buying that. Everyone has to have worked out that there are only three possible combinations of eye numbers at the start. The minimum number of blue eyes is 99 and everyone already knows that. So waiting 100 days and then deducing that I must have blue eyes and then everyone leaves makes no sense.

The possibilities of there being less than 99 blue eyes never existed and everyone knew that all along.

59

Yes they do, because if I can see 2 or more Blue eyes then I know that everyone else can see a minimum of 1 pair. And everyone else has made the same conclusion because at it’s stated, they are perfect logicians.

60

But in order to reach that conclusion, you had to be able see two blue-eyes. You don’t know that they can see two blue-eyes, so you don’t know that they can reach the same conclusion.