And I suppose it's fitting to close it on the 300th post! I must say I was terribly impressed by the level of analysis exhibited by the readers of this thread, despite the extreme subtleties, I strongly doubt that any major aspect of this puzzle was overlooked. Well done all! (Even the doubters, who helped extend the post count and made sure there were enough explanations posted such that everyone, or almost everyone, could get that 'aha' moment.)

Delighted you gave us this trip to the island. It was a colorful eye-opener for all of us.

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Lighten up! This is a stellar board!

If someone could explain this version to me, I'd appreciate it. Thanks.

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Everything I need to know I learned through Googling.

I'd be happy to. OK, so we have a tribe of 3 blue-eyed people, and nothing else. The three meet each morning to conduct business (which isn't much for a 3-person tribe), but they also must commit ritual suicide if they know their eye color (and they are well known by all to be brilliant logicians). Now, #1 sees that #2 can see that #3 has blue eyes, so #1 knows that #2 knows there are blue eyed people. However, #1 does not know if #2 knows if #3 knows that there are blue eyed people. To know this would require that #1 has blue eyes, which would force his/her suicide. So the three carry on. But one day, a visitor says to the three, all together, "Hey, I notice there is at least one blue eyed person among you". Now, this is perfectly obvious as a statement, but nevertheless it delivers killing information. For now #1 knows that #2 knows that #3 knows there are blue eyed people. Thus if #1 did not have blue eyes, then he/she would know that #2 would be expecting #3 to commit suicide the next day, unless #2 has blue eyes. #2 would then notice that #3 does not kill himself/herself on the first day, and conclude that #2 must have blue eyes (if #1 is not blue, that is). Then #2 and #3 would die on day 2. As it is, nobody kills themself on day 2, so #1 concludes they must all be blue. By symmetry, they all go through the same reasoning, and are toast on day #3 unless something is done.

What a <not sure how strict the rules are> religion these island folks have...

It's easiest, I suppose, with two to illustrate. [I've had fun with this at the Christmas parties, and end up with using two to trigger the logic.] If both have blue eyes and learn at least one has blue eyes, each person would watch to see if the other person took the knife. Neither would because the other person has blue, but when that doesn't happen, each will realize the other person is seeing blue eyes. Then, after a period (one day in this game), they will conclude they both have blue. Had one of them had brown, day one would have seen the blue eye go and the other would know that his own eyes had to be brown, so he goes next.

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Lighten up! This is a stellar board!

Would you believe I just found this thread?!

Question:

I'm a member of the tribe. After hearing the anthropologist's statement, I look around and see 15 brown-eyed people and 4 blue-eyed people. I know that no-one is going to commit suicide on the first day. More importantly, I know that everyone knows that no-one is going to commit suicide on the first day.

So, when there's no suicide on the first day, why would I then begin to wonder if that fact provided any new information to anybody else? After all, I can see that everybody, regardless of their own eye color, was expecting no suicide...

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SeanF

"Ask to understand, but don't challenge unless you have the knowledge."--NEOWatcher

The contents of this post are ©2009 by SeanF and may not be copied or retransmitted in any form without the express written consent of SeanF

The problem is stopping at what you know everyone knows. You have to keep going, to what you know that everyone knows that everyone knows, etc. With 4 blues, the killing information, if you are blue #4, let's say, is: you know that #3 knows that #2 knows that #1 knows that there are blue eyed people. You would not have known that prior to the visitor's comment. This is more complicated then necessary, start with 2 blues (as in George's post) and then to get the full flavor, extend to 3 (as in my last post). Once there, 4 and more just extend by levels of induction. Don't be surprised if your brain hurts a few times along the way, eventually you will be convinced.

But all the anthropologist told them was there were both brown and blue eyes, that everybody knew it, and that everybody knew everybody knew it. He didn't (explicitly) tell them everybody knew everybody knew everybody knew, or any deeper.

Now, if I can see four (other) people with blue eyes, don't I already know that there are both brown and blue, that everybody knows it, and that everybody knows everybody knows it?

If I'm supposed to conclude I'm a blue-eyed person when the other four I see haven't killed themselves four days after the anthropologist made his statement, why shouldn't I have concluded I was a blue-eyed person four days after I saw there were only four (other) blue-eyed people in the first place?

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SeanF

"Ask to understand, but don't challenge unless you have the knowledge."--NEOWatcher

The contents of this post are ©2009 by SeanF and may not be copied or retransmitted in any form without the express written consent of SeanF

Hold it.

If I see four blue-eyed people, then I know:

A) Nobody thinks there might be no blue eyes.
B) Nobody thinks that anybody thinks there might be no blue eyes
C) Nobody thinks that anybody thinks that anybody thinks there might be no blue eyes

It is, however, possible that:

D) Somebody might think that somebody might think that somebody might think that somebody might think that there might be no blue eyes.

Once the anthropologist makes his statement, though, nobody can think that, or at any level of repetition, because everybody knows there are blue eyes, and everybody knows everybody knows, and everybody knows everybody knows everybody knows, ad infinitum.

I gets it now.

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SeanF

"Ask to understand, but don't challenge unless you have the knowledge."--NEOWatcher

The contents of this post are ©2009 by SeanF and may not be copied or retransmitted in any form without the express written consent of SeanF

Right on Sean. I think this is a good puzzle to start from a place of skepticism, rather than be convinced without really thinking it through. Now you'll be in a better position to hande the "what ifs" if you try this with someone else!

They should kill that stupid anthropologist.
(And make David Bowie their tribal god.)

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Something occurred to me though, in regards to the discussion about preventing suicides by publicly "exiling" one blue-eyed person so nobody knows why or if that person did or didn't commit suicide.

And maybe this has already been mentioned somewhere in the thread.

Wouldn't you also have to exile one brown-eyed person, in order to prevent the browns from committing suicide after the 15th day (or however many browns there are)?

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SeanF

"Ask to understand, but don't challenge unless you have the knowledge."--NEOWatcher

The contents of this post are ©2009 by SeanF and may not be copied or retransmitted in any form without the express written consent of SeanF

Yes you would, and yes it was explored in the thread. What, you didn't read all 11 pages?

I saw Sam5's name in there, and I was afraid to.

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SeanF

"Ask to understand, but don't challenge unless you have the knowledge."--NEOWatcher

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11 pages - bah! That's nothing!

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- Learn a lot teaching others.

How long did the 1=0.99999 thread get?

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Everything I need to know I learned through Googling.

Well, the way you wrote them, the two numbers are not equal.

1. in 15 days the tribe will be all dead. The blue eyed ones will be all dead in five days. The Brown eyed ones in 15 days.

2. The anthropologist told them that they had two eye colors among their tribe. Perhaps you ought make that a lot more clear like, "Wow, I notice that your tribe has two eye colors, blue and brown - my people also have those colors, but, my people also have OTHER colors, etc." Otherwise, it opens the door to ifs ands and butts - or things like "if they are supremely logical, why would they believe a stranger......"

I don't think this problem is hard. This following was also touted as the hardest problem - it's very similar and was posted into an egroup I'm in. I did not find it hard at all since I reduced it to 2 blue and 2 brown plus the guru.

Here it is:

Blue Eyes:
The Hardest Logic Puzzle in the World

A group of people live on an island. They are all perfect logicians --
if a conclusion can be logically deduced, they will do it instantly.
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?


There are no mirrors or reflecting surfaces, nothing dumb, It is not a
trick question, and the answer is logical. It doesn't depend on tricky
wording, and it doesn't involve people doing something silly like
creating a sign language or doing genetics. The Guru is not making eye
contact with anyone in particular; she's simply saying "I count at
least one blue-eyed person on this island who isn't me."

And lastly, the answer is not "no one leaves."

I've done my best to make the wording as precise and unambiguious as
possible (after working through the explanation with many people), but
if you're confused about anything, please let me know. A word of
warning: The answer is not simple. This is an exercise in serious
inductive logic, not a lateral thinking riddle. There is not a
quick-and-easy answer, and really understanding it takes some effort.

===================

I tried to email Ken G - but had no luck. I joined this egroup only to post this answer. And I copied his puzzle to the same egroup.

A harder problem, imo, is the 12 coin problem - you have 12 coins, all the same weight except for one. That one coin is EITHER lighter or heavier. You have 3 chances to weigh them to determine which coin is wrong and tell is it's lighter or heavier. Since that got handed out (the answer did - to cheaters) - I made it harder. How many can you weigh the same way if you have 5 chances, how about 7 chances? One physics professor got it - it took him 3 days, he said. Did I think it was hard? Uh - yeah. It took me about one hour, maybe a bit more since I got up to make coffee and etc. But, it generates a mathematical pattern.

These problems, yours and the so-called "hardest puzzle ever..." above are more like problems where you have to consider what people know and know others know - and what they also know others do not know based on their actions. Is this math? IS there a math way to spell this out? Like X versus X plus 1 or something like that?

Well, C U L8R - probably not. My favorite hang out, when I hang out, is Stargate Atlantis board.

Greetings from clop.

I have a new solution for you.

The entire tribe can save themselves by one of the women giving birth to a blue-eyed child before the meeting on day 4.

Furthermore, the anthropologist did not have to say that he could see both brown and blue eyes. He could have just said that he could see blue eyes (or brown eyes).

clop

Phew!

I read until page 7, where I thought I had it, but came to post and went - hang on..

Thankyou to Fortunate, posts 190 + 214
and Wierd Dave, post 194.
These were the clinchers for me, but I should really give credit to everyone that took the time to explain it..
I did agree with Sam5 for quite a while, until I got my head around things I have never been taught.. I still have some doubts that the premise would work if it were 6 or 7, rather than 5, but I'm glad I made it this far - so I'll leave it there!

Two things;
Taking the profs statement as in the OP, if there were actually no blueys, wouldn't they all die the next morning?

If the distribution was half/half, would they all die on the same day? i.e. would their 'fuse' run at the same speed?

In this case, the prof would actually be lying, right? If this were so, then each native would see that everyone else had brown eyes, and would (erroneously) conclude that he/she had blue eyes, and would kill themself the next morning.

Which brings up another dilemma --- can they all kill themself at exactly the same time, without observing (and learning from the observation), that they were all about to simultaneously kill themselves? If they could kill themselves without observing (and learning), then they'd all die erroneously thinking they had blue eyes. If they did observe, then they'd all pause, realize that the prof was lying, conclude that they each had brown eyes, and then wait till the next morning to kill themselves.

Yes

That's where it gets interesting for me; I can understand 5 deciding that they need to off themselves, because all of them are looking at the other 4 and thinking they should only be seeing 3. 3 downwards is the easy part, but can you reduce say 6 to the same argument? Any person out of that 6 will think there are 5, but each of them cannot reduce the problem to 4 or 3, because they all see 5.
I sort-of got the impression that a purely valid logical argument was being used to describe human behaviour, and ignoring (in a way) their actual perception.
I know I said I'd had enough, but I'm curious now..

Ah, ok..
It gets to the point where the person realises that the chain has gone far enough that it includes all the known blueys except themselves. Ok, got it now.

Subtle? That's an understatement!

and kill him instead!

Correct. Note the browns would be dead in 6 days if the anthropologist had said he sees only brown and blue eyes, but the way it is written, the browns can live an extra 9 days.

No, this is not the new information. Each individual could have already known that, it makes no difference. The answer is more subtle, and is contained in the thread, and it shows why this is indeed a very tough puzzle.

This is essentially the identical puzzle, and the same solution applies for the same reasons. All the blues leave on day 100 (depending how you count that first day). The browns never need to leave, as long as people know that green eyes are possible whether or not they see any. The only blues that need to leave are those present at the announcement, which is stipulated to be all 100. The information conveyed by the guru involves not so much the words themselves, which everyone knew already, but rather the setup, and what the setup allows the blues to infer on their own about what the other blues know, recursively. It's all in the above thread.

I've seen that problem, it is very nice. For your extended version, we'd use the fact that a scale (I assume you mean a balance here) can only provide three possible results-- left heavy, right heavy, or balanced. If we think of these as the digit 0,1 and 2, it shows that the total number of possibilities we can distinguish for the singular coin in n weighings would correspond to n-digit base-3 numbers. Thus the number is 3 raised to the n, where n is the number of weighings. So if n=5, you can find the special coin among 243 coins. If n=7, you can find it in 729 coins, though I'd sure hate to be the one who had to do it!

Yes, quite so. I'm not sure if this would be called math, maybe just inductive logic. But there's probably some way to make it into graph theory, based on using line segments to delineate what people know about what other people know. It would just end up being the same solution, I don't know if it would add any insight, but perhaps it would. It seems easier just to get the solution inductively.

No, that won't save the blues, only the blue-eyed baby who is immune to the entire puzzle since they weren't there at the time. (This does raise the issue of, how old must a tribe member be before they become logical enough to be counted on by the others, but that's a detail.) All the blues present when the prof says he sees blue eyes are doomed, unless one of the other solutions is invoked to break the chain. Any blue not present is not at risk.

Yes, any color identified is doomed. I made him mention both just to wipe out the whole tribe, I've a twisted sense of irony.

It is crucial that the prof be believed, on the grounds that he would have no motivation for lying. Logically speaking, that is. If the prof is suspected of being capable of lying, then no one has to die, because any doubt will break the chain. If the prof does lie but the tribe is not prepared for that possibility, then you could get the result you described, both the immediate or contemplated suicide scenarios.

Yup, you got it. We are so unused to using recursive logic that one finds onesself constantly having to be reminded of how this actually goes!

Ken posted:
Hmm, close - but that formula would calculate that you should be able to find the odd coin in the original problem out of 27 coins (3^3).

Actually, one can use a balance scale and three weighings to determine the one odd coin out of a maximum of 13 unkown coins. But it is a bit more tricky than just left or right heavy or balanced...

For 2 weighings ~ 4 coins
for 3 weighing ~ 13 coins

With additional weighings, it gets tricky due to required balance loading so the actual number of coins that can be determined may be less than the number of possible combinations of information "bits" but my first estimate is:

for 4 weighings ~ 41 coins
for 5 weighings ~ 121
for 6 ~ 365? (may only be able to determine 361?)
for 7 ~ 1093?

When I first solved this puzzle for 12 (or 13) coins (~30 years ago) I thought it was so neat I actually caried a copy of the solution in my wallet for a number of years!