Tuesday, December 16, 2008

Objections to the blue-eyed islander problem

I've always found the classical answer to the blue-eyed islander problem vaguely unsatisfying. For background, here's one formulation of the puzzle:



"I can see someone who has blue eyes."

Who leaves the island, and on what night?
The classical answer is that all blue-eyed islanders leave on the 100th night. The proof proceeds inductively - here's a nice explanation by a commenter on Terry Tao's blog. Note that this explanation uses slightly different formulation of the puzzle in which islanders must commit suicide (!!) if they learn their eye color (rather than simply leave the island).

Let’s use the letter k to represent the number of blue-eyed islanders, of which you will be one of. In the argument “I” is used so that you may read this aloud to yourself as if it were your own dilema.

If k is 1, then I cannot see any blue-eyed people on the island. This was fine until the visitor told us that they exist. But now, knowing that they do exist and not being able to see any, I can only conclude that it must be me! Thus, I must die one day from now.

If k is 2, then I can see only one other blue-eyed person, Joe. I know that Joe is as logical as I am so by the reasoning when k is 1, he’s going to have to kill himself. I check on him tomorrow and see that he hasn’t, but how can that be? The only logical conclusion is that the above reasoning did not apply for Joe because when he looked around he must have seen someone else with blue eyes. But I don’t see such a person so it means that it must be me! Thus the following day (two days after the information)I must die. But since Joe is as logical as I am and has been faced with the same situation, he must also have discovered, exactly as I did, that he has blue eyes. Thus he will kill himself at the same time as I do.

If k is 3, then I can see see two blue-eyed people, Joe and Ted. Now, here is where you see the argument continue to depend on the earlier cases (this is what is meant by induction). If Joe and Mike are truly the only two people on the island, then based on the exact same reasoning when k is 2, they will both be dead after two days exactly. However, after two days I see that both are still living. Again, the only logical conlusion is that when Joe and Ted looked around they must have seen someone other than each other with blue eyes. Since I can’t see such a person it means that it must be me! Thus I must kill myself the next day (now three days after the information). But, since Joe and Ted are equally logical, also have blue eyes, and have been part of the same situation, they must have learned exactly as I did that they too have blue eyes. They will join me at the suicide ritual on the same day.

At this point you can see the repetitive nature of the argument. For each higher value of k the argument is quickly formed in terms of our previous argument for k-1. The induction proof is merely a concise formation of this which (correctly) reasons in terms of a general value of k. The visitor’s information is essential because it sets up the “base case” where k is 1.

The induction certainly "works" in the sense that if start with your base case (k=1) and reason inductively, you are led to the theorem that after k days, all k blue-eyed people leave the island. But note what you must do to start off this inductive process - you must consider what would happen if k were 1. But in reality, no matter your eye color, you are certain it is impossible for k to be 1. I claim that considering the k=1 case is an arbitrary thing to do. Maybe it is natural that you happen to consider the k=1 case in light of the guru's announcement, but you are not logically compelled to do so. It is only if we assume the announcement will cause everyone to consider k=1, and further, that everyone knows the announcement will have this effect, that the proof goes through. As given, the problem is unclear.

Note that we could do the exact same proof to argue that all j brown-eyed islanders would leave after j days, provided we assume that everyone will consider the j=1 case, and that everyone knows everyone else will consider this case. The only effect of the announcement (since it reveals no new information) is simply to act as a coordination device to determine which chain of reasoning is to be considered. In the absense of some way of coordinating which chain of reasoning to use, no one can be convinced that others will follow through with the reasoning, and the induction cannot proceed.


Anonymous said...

Interesting, yet morbid post. You frame the initial problem in terms of being able to leave the island, but then, all of a sudden, the islanders turn suicidal, and instead of leaving on the ferry, they commit seppuku. I understand that you are commingling two different formulations of the problem here, one benign, and one decidedly less so, but it is a bit jarring to read.

Stefan W. said...

If we assume that every blue eyed person sees at least two other blue eyed persons, every blue eyed person can assume that there is at least one blue eyed person, so the guru isn't telling anything new.

Lenoxus said...

From the OP:


Here's your error, kinda. None of the islanders are thinking "K might be 1!" The blue-eyed ones (henceforth "blues") are instead thinking "K might be 99. I wonder what would happen if it were."

Everyone agrees that it works when K = 2. Well, suppose K is 3. In that case the islanders can simply ignore the fact that it happens to work when K = 1. Each of the 3 blue-eyed islanders just has to remember the rule "When K = 2, all 2 blues will leave on Day 2." It's an absolute ironclad rule. From it, it follows that when K = 3, the 3 blues will leave on Day 3. And they did it without ever thinking "K might be 1."

Because of the previous paragraph's rules, a group of 4 blues will each know that IF there are only 3 blues, they have to leave on Day 3. So that's another ironclad rule. The 4 could even forget all about the K=1 and K=2 cases (which at no point do they consider possible. The point is that the situation necessarily works when K = 3, and thus the 4 can deduce the situation once Day 3 arrives and no one's left.

Et cetera.

(To be more accurate, these islanders do in fact consider the K=1 case upon the Guru's statement. That's because "if a conclusion can be logically deduced, they will do it instantly". They can't not think about how it would work when K=1. However, they likely already thought of it ages ago. Once the Guru speaks, they have cached thoughts about how it works when K=2, K=3, K=4, and so on.)

Lenoxus said...

Part 2… Now, what if the Guru says nothing other than "Today is Day Zero for blue eyes"? Then there's no base case. A lone blue wouldn't leave under those circumstances. (Would you?) Hence, two blues wouldn't leave. Hence, three blues wouldn't leave. And so forth.

In general, here's how it goes down:

1. I see exactly N people with blue eyes (aka "blues").
2. If my eyes are blue, then exactly N+1 blues exist; otherwise, exactly N do.
3. In order to deduce that I personally am a blue, I have to exclude the possibility that exactly N blues exist.
4. Because my only relevant source of information is people leaving the island, I can ONLY exclude the possibility of exactly N blues based on people NOT leaving by a certain day.
5. Suppose I have deduced that if there were N blues, they would leave on Day X, no sooner or later.
6. So if Day X arrives and no blues leave, then there must not be exactly N blues.
7. If there are not exactly N blues, then there are N+1, and I am a blue.
8. Thus, if Day X arrives and no blues leave, I will deduce my own eye color to be blue and leave on Day X+1.
9. In general, any blue-eyed person on the same island must reason exactly the same way as the above, since they have identical information.
10. Thus, in the case where there are exactly N blues, they would conclude themselves to be blue if and only if they observed Day X-1 come and go with no blues leaving. (Day X-1 is "their" Day X.)

Extrapolating this argument backwards, we find that a group of any number of blues will eventually leave if and only if some smaller number of blues would have left ONE DAY after having been given some piece of information. That smaller group may be of a large size. For example, if the Guru says "There are at least 95 people with blue eyes" on an island with exactly 95 blues, then all the blues will leave the next day. Following from this, a group of 96 blues would leave on Day 2, 97 on Day 3, and so on.

However, if the Guru says something like "95 is a number, and blue is a color. Start counting!", then no one will ever leave, because there's no reason 95 blues would leave (would you?), and hence their not leaving says nothing.

Lenoxus said...

Part 3…

So, what is the information the Guru provides? It's not in the content of the statement. If the Guru went to each individual islander and said "At least one person here has blue eyes", then no one would ever leave (assuming there is more than 1 blue). The key is the fact that the statement was made publicly. This causes everyone to know that everyone knows that everyone knows (times INFINITY) that at least one islander has blue eyes. Before this, they only knew it times 99. Now, for ordinary humans, 99th-order knowledge seems no different from like infinitieth-order knowledge. Then again, for most of us, there's little conceivable difference between what it might be like to be a trillionaire and what it might be like to have an infinite supply of money. These islanders, with their super-prefect logic, are like accountants considering the difference in that money.

In any case, even if it feels like the Guru gave no information, the basic logic is irrefutable. To dispute it, you have to name a number of blue-eyed islanders at which it "breaks down", and it can always be shown with simple logic that the number you name would not in fact break down.

For example, Stefan suggested that the breakdown number is three. So exactly two blues would leave on Day 2, but three would not leave on Day 3. All right, say you're on the island yourself, not knowing the color of your eyes, and you see exactly two blues. If your eyes are not blue, should the blues that you see leave on Day 2? (Saying "no" is inconsistent with agreeing that it works with two.) If your eyes are blue, then should the blues you see leave on Day 2? (An answer of "yes" suggests that those two blues are somehow better informed then you, even though in this instance you are three blues and your knowledge should be symmetric.)

Paul Chiusano said...

@Lenoxus - I'm still not really convinced. :) You have offered more or less the same explanation as I quoted in the original post. And that reasoning is of the form "If X were true, then Y follows". Since everyone knows X is not true, no one is logically compelled to even consider this implication.

The puzzle is ambiguous, which is why people keep talking past each other when talking about it. There is no logical implication between the Guru's statement and following the given chain of reasoning. Any implication drawn from it is subjective, based on an assumption of how the words will be interpreted by everyone.

If the Guru had said something like: "Everyone, reason inductively on the number of blue-eyed islanders to determine a protocol for when blue-eyed islanders should leave the island" then I don't think there would be much controversy. :)

Simon Youl said...


Respectfully, you have not understood how the solution is supposed to work.

The islanders are perfect logicians. That means that if here is an islander, Bob, and Bob knows a bunch of stuff X, and there is a valid proof whose assumptions are X and whose conclusions are Y, then Bob will instantly know Y.

If bob is a blue-eyed islander, then on the fateful night, he will have the information X, he will run the proof, he will know Y, and he will leave the island.

So it's isn't "arbitrary" that Bob decides to do the proof by induction. Bob does every possible proof whenever he can, because he is a perfect logician. He stumbles upon the proof by induction while scanning over all possible proofs, runs the proof in his head, and the rest is history.