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1001

In the canonical solution, the counting is just shorthand. The logic works because very day they are ruling out the base case of “if there are N blue islanders, they would leave today.” When that is ruled out, the islanders know a new syllogism, which says “if there were N+1 islanders, they would leave today.” If’s not important that they’re counting, it’s important what they’re counting.

If they aren’t counting the right thing, it doesn’t matter if they start counting on the same number. If the guru just shouts out “start counting from 98!” and they start counting, it doesn’t get anywhere. The following is not a true conditional:
everyone starts counting at 98 - > if there were only 98 blues on the island they would leave the first day
The guru has to provide the information that “there are at least 98 blues” for 98 blues to leave on the first day.

1002

In other words, you still don’t know of any number between 0-99 that wouldn’t work.

1003

Sorry, did you get “I don’t know” when I said

Because this doesn’t mean “I don’t know”. It means it depends, and you need to be clearer. Since I know you to be able to present your case as clearly as you like, if you can’t be clearer I must assume it’s because clarifying your position will make it more clearly untenable.

1004

Well, since others know what I said and I showed an explicit example, which you seemingly ignored, I have to conclude that you choose to merely make every effort to avoid the clarity. But nevermind. It has become more than obvious that you just don’t care. No problem. There is no requirement that anyone care.

For those who do care, I have presented the argument that there are other more efficient means to ensure that everyone on the island can deduce their eye color (with examples). And since this puzzle is a derivative of the old prisoner puzzle offering either escape from prison for deducing properly or death for guessing improperly, we can take it that the islanders actually want to “escape” the island and have incentive to do so, else they just close their ears and ignore even the guru.

So since there are more efficient means to deduce their color, they cannot depend upon the others to follow any one particular scheme which yields the canonized scheme defunct. All knowing perfect logicians would know which scheme was the most efficient and use that one instead of the canonized version.

1005

This is not what we’re dealing with. We’re talking about perfect logicians, defined as those who will deduce any conclusion logically deducible (whether they want to or not).

Moreover, as if it’s relevant, another early version of this problem deals with a tribe of perfect logicians whose religion requires that they commit suicide if they learn the color of their own eyes. The logic still works there, even though they certainly don’t want to kill themselves.

Interestingly, and significantly for the problem at hand, part of the consequence of the religious tribe version is that after the guru (there, a foreigner who inadvertently mentions that he sees blue eyes) gives the tribe the necessary information, he can defuse the situation by pointing to a specific individual with blue eyes and saying “that guy has blue eyes”. After that, ‘that guy’ kills himself and everyone else is saved.

Let me try this tack: I will describe what I believe your argument to be, so that you can correct what I get wrong, and as a show of good faith. I’d appreciate the same from you (and I don’t mean the argument for the canonical solution, which I take it you understand and accept; rather, present my counter-argument to yours as you read it).

Mike’s version of James’ argument:
For any number X, if X islanders with blue eyes know that there are X islanders with blue eyes, they will leave the island on the next ferry.
I can see that there are 99 blues, so I know that that there are at least 97 blues who know that there are at least 97 blues. Therefore, they will leave on the next ferry. When they don’t, they learn that there must be 98. When those 98 don’t leave, they learn that there must be 99. When those 99 don’t leave, they know that there must be 100, and so everyone leaves on the 4th day.

Is that right?

1006

No.

1007

It seems an awful lot like you’re no longer participating in this discussion in good faith. Is that right?

1008

I kinda wish I had never posted this, although at the same time I am glad I did, there’s a word for that.

Maths is fun people, sometimes at least I think so. :wink: :slight_smile:

:popcorn:

1009

I don’t think anyone changed their mind, but I learned a lot about the logic behind the problem from debating it.

A question for anyone who accepts the canonical answer (or anyone who doesn’t and wants to pretend they do for the purposes of this question): Do the rules in the first paragraph need to be presented publicly in order for it to work? In particular, is it enough to know that everyone else is a perfect logician, or do I need to know that everyone knows that everyone knows that everyone knows… It’s easily solved by posting the rules or having the guru read the text of the first paragraph before she says she sees someone with blue eyes, but I’m not entirely sure that it’s even a problem.

1010

I think that everyone would have to know that everyone knows that everyone… concerning many details including whether they wanted to leave the island or stay. If they wanted to stay, being perfect logicians, no one would leave regardless of the guru.

1011

Mike’s version of James’ argument, take two:

The guru says there’s someone with blue eyes
The other guy has blue eyes
If he doesn’t leave on the first day, then I would have to leave.
I don’t want to leave.
Therefore the guru didn’t say anything.

QED.

1012

Carleas you seem to love the games of dodge ball and strawman (and who is Mike?).

If they didn’t want to leave, they would merely reason;

I know that I heard the guru, but maybe the 99 that I see didn’t.
And if the 99 did, maybe they are thinking that the 98 didn’t
And if the 98 did, maybe they are thinking that the 97 didn’t
And if the 97 did, maybe they are thinking that the 96 didn’t
And if the 96 did, maybe they are thinking that the 95 didn’t
.
…
And if the 2 did, maybe they are thinking that the 1 didn’t
Since no one can prove that the one did hear the guru, the
2 couldn’t know that the 1 knew and
3 couldn’t know that the 2 knew and
4 couldn’t know that the 3 knew and
5 couldn’t know that the 4 knew and
.
.
Thus I can’t know that any of them are ever going to leave, so I can’t deduce shit.

1013

@Carleas, everyone has to know that everyone else is a perfect logician, and that everyone else knows that everyone else is a perfect logician as well. Everyone has to have heard the guru, and know that everyone else heard the guru, and know that everyone else knows everyone else heard the guru on top of that. Everyone has to be able to see everyone else’s eye color, and know that everyone else can see everyone else’s eye color.

1014

Back to what I said that I liked about RM… RM requires the verification of the lack of alternatives.
No one can be 100% certain of anything until there is a total lack of alternatives.

The mental tendency is to presume the first thought;
“everyone knows that everyone heard the guru”
“everyone knows that everyone is a perfect logician”
“everyone knows that everyone can see everyone”
“everyone knows that everyone is thinking the same scheme”

None can depend on the behavior of the others until all of them know that there is no alternative to all of those.
…and know that all of the others do too.

The guru changed nothing.

1015

My latest analysis of the correct solution is that it relies on the assumption that “deducing based on what is known for sure is most logical, when it is only knowing something (one’s own eye colour) for sure that will make a difference”.

Given only two possibilities (to each blue) that there are either 100 or 99 blues (which is exhaustively equivalent to knowing what their own eye colour is), they would therefore start deducing based on the possibility that contains only that which they can initially know for sure: the 99 blues they definitely can see.

For example:
Each of the 100 blues does not see 100 blues. They DO know for sure that they definitely see 99, so they start their deduction based on this rather than anything that contains a variable that could go either way. Importantly, they begin their deductions from the point of view of any blue of the 99 they see - knowing only what any of the 99 would know.

Any of the 100 blues in turn knows for sure that 99 blue-eyed islanders would know for sure that they would see 98 blue-eyed islanders. Deducing ONLY based on what these 98 would see for sure, knowing the only thing that any of these 98 would only know for sure, the series of deductions continues to the point that 1 blue-eyed islander would know for sure that he sees ZERO blue-eyed islanders, based ONLY on what this 1 blue-eyed islander would know for sure, based ONLY on an examination by any of the 100 blues of what can be known FOR SURE.

At this point, the ONLY other thing that can be known for sure (the Guru’s words) CHANGES the only sure knowledge that can be obtained without the Guru’s words, to “1 blue-eyed islander would know for sure that there is ONE blue-eyed islander (whom they cannot see)”. The only islander whose eyes he would not be able to see is his own, therefore he has blue eyes.

And then the rest:
(He can now know for sure that IF there were no other blue-eyed islanders, he would leave on the ferry the next midnight. He can know for sure that IF there was only 1 blue-eyed islander who he could see that did not leave on the ferry the next midnight, that blue-eyed islander that he could see sees another blue-eyed islander, who could only be himself, so they both leave on the ferry on the 2nd midnight. If there were 2 that he could see, they would act in the previously described way if there were in fact only 2 blue-eyed islanders, and if they do not, that leaves the only possibility that hey each see he himself who therefore has blue-eyes, and they all leave on the 3rd midnight. This continues all the way up to what is ACTUALLY the critical juncture: whether 99 leave on the 99th midnight, and if not then 100 leave on the 100th midnight.)

That is all the blues. The browns and the 1 green cannot do the same as they can only know for sure the same and only definite knowledge that the blues could use aside from the Guru’s words. The Guru’s words do not change what 1 brown or green would definitely and only know for sure about the amount of browns and greens that they would definitely see. Accordingly, they cannot build the deduction back up like the blues needed to be able to do in order to leave. Additionally, the fact that all the blues left tells the browns and the green nothing new about their own eye colour, as it would remain a possibility to each and everyone of them that the could have, e.g. red eyes.

I have intentionally emphasised:
(a) the only assumption that is being relied upon in order for this series of deductions to unfold,
(b) that there are no other assumptions that can be made from certain knowledge,
(c) that each deduction in the series operates ONLY from the knowledge that the number of blues in question would know (no transfer of knowledge out of context, such as “what 100 would know” being transferred to “what any less than 100 would know (critically to what 1 would know)”.

As such, this is the ONLY solution.
All other proposed solutions are abusing at least one of the above.
As far as I know, this clears up every single grievance in the entire thread, though I can illogically deduce that this will not end the discussion - but that is not my fault. Carleas, FJ and I can now all step out and take a breather. Everyone else, study it well and swallow your pride.

1016

That is a list of your errors.

…until you said that.

1017

Well argued.

1018

The counter argument was already made. You seemed to have ignored it to focus (again) on only the one thought, making the assumptions that the counter argument pointed out (which you then listed as your strong points). It seemed kind of Sil-ly.

1019

My reasoning illustrates why there is only one solution to focus on.
The consequence of this is that other attempts are going to be flawed, and the emphases that I have made make explicit exactly where and why they are.

Is all you have to say about that that your counter attempts have already pointed out why they’re wrong?
And as unoriginal your play on my name is, S’ain’t relevant.

1020

James, are any of the premises of the canonical solution inherently contradictory? Because the canonical solution by its nature excludes all others, so if there is a non-canonical solution, it means there is some internal inconsistency in the problem itself.

Also, it’s worth noting that in your breakdown of the situation where the islanders don’t want to leave, the fact that they didn’t want to leave was not used. If they are uncertain about whether the other islanders heard the guru, they can’t complete the syllogism, whether they want to or not. The canonical solution assumes that there is no doubt about whether the islanders heard and understood the guru.

I have another extension of the suicidal tribe presentation of the puzzle, which I think is interesting (again moving beyond defending the canonical solution):

The suicidal tribe presentation goes like this: there’s a tribe of perfect logicians. Some of them have blue eyes, some have brown eyes, but no one knows their own eye color. In fact their religion requires that if any tribe member learns her eye color, she must kill herself the next day at noon in the middle of the village. As a result, they never talk about eye color. A foreign anthropologist finds the tribe and is welcomed. He spends some times, learns the language, but only too late learns about the religious customs: at a meeting involving the whole village, he makes an off-hand remark that it is strange to see that some members of the tribe have blue eyes.

The solution is the same: all the blue eyed people must kill themselves on the same day, and that day will be X days after the foreigner spoke, where X is the number of blue eyed people.

As I mentioned before, this setup allows the anthropologist to save everyone on the island by pointing to one individual and saying “that’s the guy who has blue eyes”, effectively revoking the information that was used to start the inference. But I think he would have to point to 1 person on the first day; on the second, I think he’d have to name 2; on the third 3 and so on. Does anyone disagree with that?

Also, since in this version, the tribe members can communicate, it’s not their inability communicate, but their actual desire not to know (and their respect for the desires of their fellow tribe mates) that prevents them from telling each other. But does the ability to communicate allow them to short circuit the logic before it completes in such a way that everyone is saved?

My first thought is that two people privately discussing the fact that a third person has blue eyes would save them (i.e., with three people, A and B talk about C, B and C talk about A, and C and A talk about B), but that doesn’t seem to solve it: A and B will still expect C to leave.

The question is essentially asking whether there is a way to limit knowledge by increasing the number of agreed truths (i.e., I’m not interested in answers where they all agree to doubt the foreigner). I will have to think more on this.