Yakk wrote: On island 0 everyone leaves.

Only if they assume that there are only two possible eye-colours - blue and not-blue - otherwise they don't know which of the not-blue colours their own eyes are.

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Yakk wrote: On island 0 everyone leaves.

Only if they assume that there are only two possible eye-colours - blue and not-blue - otherwise they don't know which of the not-blue colours their own eyes are.

- Yakk
- Poster with most posts but no title.
**Posts:**11128**Joined:**Sat Jan 27, 2007 7:27 pm UTC**Location:**E pur si muove

rmsgrey wrote:Yakk wrote: On island 0 everyone leaves.

Only if they assume that there are only two possible eye-colours - blue and not-blue - otherwise they don't know which of the not-blue colours their own eyes are.

Yakk wrote: (Assume the islanders know everyone has brown or blue eyes).

I think that covers that? Really, the digression into "what about green eyes" is not all that interesting.

One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

If blue eyes leave, and Island 0 has 0 blue eyes, and the guru (who is known to be truthful) says that is the case, why would the person on island 0 leave? The person on island 1 would, because guru says 1 blue eye, and there's only 1, so it must be him. Add in island 2, they both leave on day 2, because the other didn't leave on day 1, etc up to island X. Whether green, brown, purple or red other eye colors exist is irrelevant.

I understand this portion of the solution. However, this only works if the participants are not 'perfect' logicians and/or cannot see each other at all times. We are told they can.

This being the case, they will be able to deduce that the minimum number of blue eyes each other person on their island sees is X-1. The rebuttal says that each would see progressively X-1, all the way down to 0, diffusing the solution, however, I can see EVERYONE, and I know EVERYONE sees X-1 (at least), so this is not the case. The other rebuttal is that they will not agree on when day 0 is. However, any day a person sees all the other people seeing all the other people is at LEAST day 0 (possibly a higher day). So, the same situation arises as if the guru spoke.

However, additionally, assuming each person does not WANT to leave, they can simply obscure the timeline or information of the situation (if possible, maybe they can see through walls? How else do they see each other all the time?) to reset their own counter.

I understand this portion of the solution. However, this only works if the participants are not 'perfect' logicians and/or cannot see each other at all times. We are told they can.

This being the case, they will be able to deduce that the minimum number of blue eyes each other person on their island sees is X-1. The rebuttal says that each would see progressively X-1, all the way down to 0, diffusing the solution, however, I can see EVERYONE, and I know EVERYONE sees X-1 (at least), so this is not the case. The other rebuttal is that they will not agree on when day 0 is. However, any day a person sees all the other people seeing all the other people is at LEAST day 0 (possibly a higher day). So, the same situation arises as if the guru spoke.

However, additionally, assuming each person does not WANT to leave, they can simply obscure the timeline or information of the situation (if possible, maybe they can see through walls? How else do they see each other all the time?) to reset their own counter.

- Kingreaper
**Posts:**170**Joined:**Sun Jan 27, 2008 4:23 pm UTC

Not at all, it works fine with them all seeing each other, as long as they can't read each other's minds.Reithan wrote:I understand this portion of the solution. However, this only works if the participants are not 'perfect' logicians and/or cannot see each other at all times. We are told they can.

This being the case, they will be able to deduce that the minimum number of blue eyes each other person on their island sees is X-1. The rebuttal says that each would see progressively X-1, all the way down to 0, diffusing the solution, however, I can see EVERYONE, and I know EVERYONE sees X-1 (at least), so this is not the case.

So you know that everyone sees at least X-1. Which means everyone knows that everyone sees at least X-2. Which means everyone knows that everyone knows that everyone sees at least X-3. That's all good, but what if you can only see three blue eyed people? Clearly it's not true to say that "Everyone knows that everyone knows that everyone sees at least one blue eyed person".

They may not be able to read each other's minds, but they can see if they can see each other.

The recursion would only occur if we only saw that 1 other saw all others. However, we can see that ALL others see ALL others. Thereby if I am person one, who sees X-1 blue eyeds, and I see person 2 see X-2 minimum blue-eyeds, I don't need to infer that he sees person 3 see X-3 blue eyeds, because I also can see that he sees all seeing at minimum X-2 blue eyeds, and so on for each of the X-1 others. So, all X-1 others see at minimum X-2 blue-eyeds, no less.

The recursion would only occur if we only saw that 1 other saw all others. However, we can see that ALL others see ALL others. Thereby if I am person one, who sees X-1 blue eyeds, and I see person 2 see X-2 minimum blue-eyeds, I don't need to infer that he sees person 3 see X-3 blue eyeds, because I also can see that he sees all seeing at minimum X-2 blue eyeds, and so on for each of the X-1 others. So, all X-1 others see at minimum X-2 blue-eyeds, no less.

Reithan wrote:This being the case, they will be able to deduce that the minimum number of blue eyes each other person on their island sees is X-1. The rebuttal says that each would see progressively X-1, all the way down to 0, diffusing the solution, however, I can see EVERYONE, and I know EVERYONE sees X-1 (at least), so this is not the case. The other rebuttal is that they will not agree on when day 0 is. However, any day a person sees all the other people seeing all the other people is at LEAST day 0 (possibly a higher day). So, the same situation arises as if the guru spoke.

Suppose you and I and KingReaper are on the island. You look around and see that KingReaper and I both have blue eyes, and so do 7 other people. So you know that there are at least 9 blue-eyed people, possibly 10 if you're one yourself. So you can also deduce that I see either 8 or 9 blue-eyed people, so I know there are at least 8 blue-eyed people on the island, and might know there are at least 9, depending on your eye-colour. Now if you try to figure out what I know about what KingReaper knows, you know that I know that KingReaper knows about the other 7 blue-eyed people. You also know that I don't know that KingReaper actually sees an 8th blue-eyed person - me - and you don't know what KingReaper and I both know about your eye colour, so you don't know whether I know KingReaper sees at least the 7 blue-eyed people we all know about, or at least 8 including you, so you know for sure that I know that KingReaper sees at least 7 blue-eyed people. Add in one of the 7 and you know that I know that KingReaper knows that that person sees at least 6 blue-eyed people - the other 6 of the 7. Extend the chain to cover all ten of us, and you don't know whether I know that KingReaper knows that the first of the 7 knows that the ... knows that the seventh of the 7 sees anyone with blue eyes - at least not until the Guru gives it away...

Reithan wrote:They may not be able to read each other's minds, but they can see if they can see each other.

The recursion would only occur if we only saw that 1 other saw all others. However, we can see that ALL others see ALL others. Thereby if I am person one, who sees X-1 blue eyeds, and I see person 2 see X-2 minimum blue-eyeds, I don't need to infer that he sees person 3 see X-3 blue eyeds, because I also can see that he sees all seeing at minimum X-2 blue eyeds, and so on for each of the X-1 others. So, all X-1 others see at minimum X-2 blue-eyeds, no less.

If person 2 only sees X-2 (that is, if person 1 doesn't have blue eyes after all) then he sees person 3 only seeing X-3 minimum. You can only be sure that person 2 knows that person 3 sees at least X-2 if you either know that you have blue eyes, or know that person 2 knows he has blue eyes.

There are three people you don't know that person 2 knows that person 3 knows them to have blue eyes - yourself (because you don't know whether you do or not), person 2 (because they don't know that they do) and person 3 (because you know that they don't know their own eye colour)

Reithan wrote:This being the case, they will be able to deduce that the minimum number of blue eyes each other person on their island sees is X-1. The rebuttal says that each would see progressively X-1, all the way down to 0, diffusing the solution, however, I can see EVERYONE, and I know EVERYONE sees X-1 (at least), so this is not the case.

Consider the following statements:

1) Everyone knows at least X people have blue eyes.

2) Everyone knows statement 1.

3) Everyone knows statement 2.

4) Everyone knows statement 3.

5) Everyone knows statement 4.

6) Everyone knows statement 5.

7) Everyone knows statement 6.

8) Everyone knows statement 7.

etc.

Each of these statements is meaningfully different. If N is the number of blue-eyed people on the island and X = N-1, then statement 1 is clearly true. Statement 2 is not, however, and no amount of logic or observation of other people can change that without some form of communication. If you change X to be N-2, then statement 2 becomes true - but statement 3 is still false. Each time you decrement X, one and only one additional statement in this sequence becomes true - and at every step, you need precisely one additional statement to be true without decrementing X in order for the puzzle's solution to work.

The Guru's statement and, most importantly, the fact that it is made at a communal gathering where everyone can see (that everyone can see that everyone can see... ad infinitum) that everyone is present, makes the full infinite series of numbered statement true for X = 1, where previously only the first 99 of them were true.

Your point of confusion appears to be about misunderstanding the recursion to be all about statement 1 with decrementing X values. The decrementing X value happens, not because we're arbitrarily excluding people, but because we're advancing to increasingly higher-numbered statements in this series. To refute that X's value must be decremented to maintain truth of the statement, you must show that this is despite shifting from statement 1 to statement 2, but what you're actually doing is just cycling back to statement 1.

- Yakk
- Poster with most posts but no title.
**Posts:**11128**Joined:**Sat Jan 27, 2007 7:27 pm UTC**Location:**E pur si muove

Reithan wrote:If blue eyes leave, and Island 0 has 0 blue eyes, and the guru (who is known to be truthful) says that is the case, why would the person on island 0 leave?

Everyone knows their eyes are blue or brown.

Thus after "I see no blue eyed people" they leave.

I understand this portion of the solution. However, this only works if the participants are not 'perfect' logicians and/or cannot see each other at all times. We are told they can.

So you think they can leave earlier without the guru. Got it.

This being the case, they will be able to deduce that the minimum number of blue eyes each other person on their island sees is X-1. The rebuttal says that each would see progressively X-1, all the way down to 0, diffusing the solution, however, I can see EVERYONE, and I know EVERYONE sees X-1 (at least), so this is not the case. The other rebuttal is that they will not agree on when day 0 is. However, any day a person sees all the other people seeing all the other people is at LEAST day 0 (possibly a higher day). So, the same situation arises as if the guru spoke.

On island 1, Doug (with blue eyes) can see no blue eyed people. Doug does not know that everyone else can see blue eyes.

On island 2, Bob (with blue eyes) can see one blue eyed person. You do not know that everyone can see blue eyes.

On island 3, Alice (with blue eyes) can see two blue eyed people. Alice knows that everyone else can see blue eyes, but Alice does NOT know that the blue eyed people she can see knows that everyone can see blue eyes.

On island 4, Tracy (with blue eyes) can see three blue eyed people. She knows that everyone she can see can see blue eyes. She knows that everyone she can see know that everyone they can see can see blue eyes (ie, the blue eyed people are in state Alice or better). But she does not know that everyone she can see knows that everyone they can see knows that everyone they can see can see blue eyes. Ie, Tracy might be on island 3 (with brown eyes) and looking at Alice, so she cannot know that Alice can see more than 2 sets of blue eyes, which means she cannot know that the people she is looking at know more than Alice.

On island 5, anyone with blue eyes cannot know that anyone with blue eyes they are looking at knows more than Tracy does.

On island N, anyone with blue eyes cannot know that anyone with blue eyes they are looking at knows more than the person on island N-1 with blue eyes knows.

The recursion gets endlessly complicated and difficult to grasp, but it never falls down.

Do you understand that you have too much confidence in your chain of reasoning? Are you willing to put up a minimum N at which your chain of reasoning shows that the above recursion fails to show the guru is needed? And if someone shows that the N must be lower, you would conclude your chain of reasoning is wrong? And if someone shows that at that N, the guru is needed, you'd conclude your chain of reasoning is wrong?

It is really easy to say "yes, but what about something else irrelevant". I'm asking you to make a concrete prediction, strong enough that if your prediction is wrong you are willing to conclude your approach is fundamentally flawed.

This is a non interesting objection.However, additionally, assuming each person does not WANT to leave, they can simply obscure the timeline or information of the situation (if possible, maybe they can see through walls? How else do they see each other all the time?) to reset their own counter.

One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

Working from N=3, If I see 2 other blue-eyeds see me and see each other both, then IF I have blue they both see 2. If I do not, they both see one.

I know there are only 2 possibilities.

As we are told, we are all perfect logicians.

Possibility 1: I have blue eyes

They both see 2, and know the other sees 2 or 1, their knowledge is both identical to my own.

Possibility 2: I have on-blue eyes.

They both see 1. They know that the other sees either 0 or 1 blue-eyeds. Depending on their own condition.

Possibility 2.1: "I" (the other's internal monologue) have blue eyes.

They both see 1, same all me, all knowledge is identical.

Possibility 2.2: "I" (the other's internal monologue) have non-blue eyes.

No one sees blue eyes.

Now, this, to me, seems to be the recursion you're speaking of, this chain of internal knowledge that leads to a possibility of no one seeing blue eyes. However, myself, I see that everyone can see at least N-2 blue eyes, so Possibility 2.2 and the preceding recursions are impossible. Or so I think?

Also, is Yakk always so tactless?

I know there are only 2 possibilities.

As we are told, we are all perfect logicians.

Possibility 1: I have blue eyes

They both see 2, and know the other sees 2 or 1, their knowledge is both identical to my own.

Possibility 2: I have on-blue eyes.

They both see 1. They know that the other sees either 0 or 1 blue-eyeds. Depending on their own condition.

Possibility 2.1: "I" (the other's internal monologue) have blue eyes.

They both see 1, same all me, all knowledge is identical.

Possibility 2.2: "I" (the other's internal monologue) have non-blue eyes.

No one sees blue eyes.

Now, this, to me, seems to be the recursion you're speaking of, this chain of internal knowledge that leads to a possibility of no one seeing blue eyes. However, myself, I see that everyone can see at least N-2 blue eyes, so Possibility 2.2 and the preceding recursions are impossible. Or so I think?

Also, is Yakk always so tactless?

Reithan,

Can you explain why "someone has blue eyes" is common knowledge for the N=2 case?

If not, then how does a third blue-eyed person learn from the fact that the other two blue-eyed people didn't leave on day 2?

Can you explain why "someone has blue eyes" is common knowledge for the N=2 case?

If not, then how does a third blue-eyed person learn from the fact that the other two blue-eyed people didn't leave on day 2?

That's where my sticking point is. I understand working from N=0 up to N=whatever, however, working backwards, I'm having a harder time with.

As I noted in the N=3 example, since I know that each other sees at least 1 blue eyed person, and based on my line of reasoning in the previous post, I can rule out anyone else seeing less that 1 blue eyed person, and they can as well, then unlike in the N=2 case, there is no chance of anyone seeing 0 blue-eyeds. So in the case of N=2 each person knows the other can deduce the existence of at least 0 blue-eyeds, which gives you nothing to work with. However, in the case of N=3, you know each other can deduce the existence of at least 1 blue-eyed.

As I noted in the N=3 example, since I know that each other sees at least 1 blue eyed person, and based on my line of reasoning in the previous post, I can rule out anyone else seeing less that 1 blue eyed person, and they can as well, then unlike in the N=2 case, there is no chance of anyone seeing 0 blue-eyeds. So in the case of N=2 each person knows the other can deduce the existence of at least 0 blue-eyeds, which gives you nothing to work with. However, in the case of N=3, you know each other can deduce the existence of at least 1 blue-eyed.

Reithan wrote:That's where my sticking point is. I understand working from N=0 up to N=whatever, however, working backwards, I'm having a harder time with.

As I noted in the N=3 example, since I know that each other sees at least 1 blue eyed person, and based on my line of reasoning in the previous post, I can rule out anyone else seeing less that 1 blue eyed person, and they can as well, then unlike in the N=2 case, there is no chance of anyone seeing 0 blue-eyeds. So in the case of N=2 each person knows the other can deduce the existence of at least 0 blue-eyeds, which gives you nothing to work with. However, in the case of N=3, you know each other can deduce the existence of at least 1 blue-eyed.

How does a blue-eyed islander in the N=3 case know that it's N=3 (and that he can therefore deduce the existence of at least 1 blue eyed person) and not N=2 (and that he has non-blue eyes)?

Basically, your line of reasoning is giving the islanders knowledge that they don't have. In the N=3 case, the reasoning for person A is:

"Either I have blue eyes, or N=2. If N=2, then B is thinking X."

And if X is anything other than exactly what A would think if N=2, then it's wrong. So what you have is:

"Either I have blue eyes, or N=2. If N=2, then B is thinking "Either I have blue eyes, or N=1. If N=1, then C is thinking "Either I have blue eyes, or N=0."""

Since what is in bold is exactly what both islanders think in the N=2 case, and you admit that in the N=2 case they don't have common knowledge, then how does islander A come to a common knowledge conclusion?

Let's consider the two-person case further, as you state that two people with blue eyes can not deduce their own eye color. Given that this is true, imagine you (a brown-eyed person) standing with those two people. Can they deduce their eye color now?

Since you don't know your own eye color, you know everyone can see one pair of blue eyes, but you can imagine that neighbor A can imagine having brown eyes, only seeing the one pair of blue eyes, and thinking neighbor B doesn't see any. Now, even if you have blue eyes, you don't know that, so your thinking is the same. (If you know your own eye color, it sorta ruins the bit).

Since you don't know your own eye color, you know everyone can see one pair of blue eyes, but you can imagine that neighbor A can imagine having brown eyes, only seeing the one pair of blue eyes, and thinking neighbor B doesn't see any. Now, even if you have blue eyes, you don't know that, so your thinking is the same. (If you know your own eye color, it sorta ruins the bit).

Ah, the imagination phrasing worked, I see it now. Thanks.

- Potatoberg
**Posts:**19**Joined:**Wed Nov 05, 2014 10:30 pm UTC

mward wrote:Potatoberg wrote:it's the only valid solution that does not require perfect logicians to assume stuff (aka a contradiction).

I would love to see your perfectly logical proof of your assertion that it is a contradiction for a perfect logician to make assumptions. Do not make any assumptions!

It doesn't take a perfect logician to understand how a perfect logician thinks.

Gwydion wrote:Potatoberg -

Coming to a pages-long discussion and insisting not only that everyone here is wrong, but that the entire subforum is invalid, is trolling. Please don't take personal offense - this claim comes from your behavior and not your character. You may not have called any of us names, but saying things like "I'm ok with you guys doing puzzles for fun and games, just don't call them logic puzzles" is very insulting to me as someone who has spent more time in this forum than any other, and I suspect to others as well.

Me having the audacity to question the logic in your logic puzzles is offensive to you? I'm sorry, but your offense is totally your fault. However, calling me a troll is an indisputable offense and it only serves to discredit you and this forum (if you spend so much time here thinking about logic, how is it that in a debate you resort to name calling?).

Is it my avatar? I would expect people who work with logic to not fall for such a bias (halo effect).

Last edited by Potatoberg on Wed Dec 10, 2014 2:37 am UTC, edited 1 time in total.

- TheGrammarBolshevik
**Posts:**4878**Joined:**Mon Jun 30, 2008 2:12 am UTC**Location:**Going to and fro in the earth, and walking up and down in it.

And what is your basis for thinking that perfect logicians cannot make assumptions?

Nothing rhymes with orange,

Not even sporange.

Not even sporange.

- Potatoberg
**Posts:**19**Joined:**Wed Nov 05, 2014 10:30 pm UTC

TheGrammarBolshevik wrote:And what is your basis for thinking that perfect logicians cannot make assumptions?

By definition, an assumption is not based on evidence or logic.

- TheGrammarBolshevik
**Posts:**4878**Joined:**Mon Jun 30, 2008 2:12 am UTC**Location:**Going to and fro in the earth, and walking up and down in it.

That doesn't dispose of the question, because you haven't said why each of a perfect logician's beliefs must be based on evidence and logic.

But let me put it this way. Suppose we define "shmogician" as follows:

Note that if I say "Jessica is a perfect logician," that might imply that Jessica doesn't make unfounded assumptions. But if I say "Jessica is a shmogician," I imply no such thing. Shmogicians can make as many unfounded assumptions as they want, so long as they also believe all the logical consequences of those assumptions.

Now suppose I amend the original problem statement in the following two ways:

Would your objection to the puzzle still stand? Certainly the problem couldn't be that the islanders' assumptions are not based on evidence and logic, because my definition of "shmogician" allows that the islanders can have beliefs that are not based on evidence and logic.

And if there is no objection to this version of the problem statement, it seems to me that this is merely a case of verbal disagreement. The term "perfect logicians" means one thing to you.* It means a different thing to most others here.** And so the puzzle that you get out of the problem statement is one puzzle, while the puzzle that other people get out of the problem statement is a different puzzle. But that wouldn't undermine the solution that's been given to the puzzle, because the solution is only meant as a solution to a particular puzzle that can be extracted by reading the problem statement in a certain way. It is not meant to be a solution to every puzzle that could be extracted from every reading of the problem statement.

*Namely, someone who spends too much time reading crappy lists of fallacies on the Internet and too little time talking to professional logicians.

**Namely, the same thing that "shmogician" means, as the problem statement itself seems to indicate when it says that "if a conclusion can be logically deduced, [the islanders] will do it instantly."

But let me put it this way. Suppose we define "shmogician" as follows:

- A person, X, is a shmogician if and only if: Whenever X believes each proposition in a set Σ, and Σ logically implies a proposition p, then X also believes p.

Note that if I say "Jessica is a perfect logician," that might imply that Jessica doesn't make unfounded assumptions. But if I say "Jessica is a shmogician," I imply no such thing. Shmogicians can make as many unfounded assumptions as they want, so long as they also believe all the logical consequences of those assumptions.

Now suppose I amend the original problem statement in the following two ways:

- "Every islander knows that the Guru's statements are all true" is added to the first paragraph.
- The term "perfect logicians" in the first paragraph is replaced with "shmogicians."

Would your objection to the puzzle still stand? Certainly the problem couldn't be that the islanders' assumptions are not based on evidence and logic, because my definition of "shmogician" allows that the islanders can have beliefs that are not based on evidence and logic.

And if there is no objection to this version of the problem statement, it seems to me that this is merely a case of verbal disagreement. The term "perfect logicians" means one thing to you.* It means a different thing to most others here.** And so the puzzle that you get out of the problem statement is one puzzle, while the puzzle that other people get out of the problem statement is a different puzzle. But that wouldn't undermine the solution that's been given to the puzzle, because the solution is only meant as a solution to a particular puzzle that can be extracted by reading the problem statement in a certain way. It is not meant to be a solution to every puzzle that could be extracted from every reading of the problem statement.

*Namely, someone who spends too much time reading crappy lists of fallacies on the Internet and too little time talking to professional logicians.

**Namely, the same thing that "shmogician" means, as the problem statement itself seems to indicate when it says that "if a conclusion can be logically deduced, [the islanders] will do it instantly."

Nothing rhymes with orange,

Not even sporange.

Not even sporange.

- Potatoberg
**Posts:**19**Joined:**Wed Nov 05, 2014 10:30 pm UTC

TheGrammarBolshevik wrote:shmogician

A schmogician may or may not induce the solution that says they all leave on the 100th night. If you change the term "perfect logicians" for "schmogicians", to reach said solution you would have to assume (along with all the other assumptions I previously pointed out) that the islanders can figure all of this out immediately and thus leave on the 100th night instead of the 105th night because some are slow to figure it all out.

PS: why are you attacking my knowledge of argumentative fallacies? How do you know how much time I spend talking to professional logicians? How do you know that I am not one myself? But more importantly, why are you focusing on my person instead of on my arguments?

- TheGrammarBolshevik
**Posts:**4878**Joined:**Mon Jun 30, 2008 2:12 am UTC**Location:**Going to and fro in the earth, and walking up and down in it.

Potatoberg wrote:A schmogician may or may not induce the solution that says they all leave on the 100th night. If you change the term "perfect logicians" for "schmogicians", to reach said solution you would have to assume (along with all the other assumptions I previously pointed out) that the islanders can figure all of this out immediately and thus leave on the 100th night instead of the 105th night because some are slow to figure it all out.

This is specifically stipulated in my definition of "shmogician." A shmogician believes the logical consequences of Σ whenever she believes Σ. I didn't say that she believes the logical consequences five days later.

Potatoberg wrote:PS: why are you attacking my knowledge of argumentative fallacies? How do you know how much time I spend talking to professional logicians? How do you know that I am not one myself? But more importantly, why are you focusing on my person instead of on my arguments?

I don't know in what sense I could be said to be "focusing" on your person rather than your arguments. I wrote several paragraphs addressing your argument, with a footnote suggesting that you're confused about what logic is. As for why I included that footnote, here is much of the reason: You have repeatedly insisted in this discussion that perfect logicians must live up to certain criteria. You have been challenged to defend those criteria, but you have spent all of one sentence actually defending them. Instead, you primarily seem to be asking us to take it on your authority that perfect logicians must not make assumptions, and so on.

In that context, it seems entirely relevant whether you are actually an authority on logic, or just someone with an opinion on the Internet.

The rest of the reason, I suppose, is that you're mostly ignoring the pertinent things that people are saying to you here, and when you ignore people who take the time to respond to you, they tend to get annoyed and say impolite things. I don't know why you're focusing on that, rather than on the responses to your arguments. After all, we have moderators here, and you can report posts if you feel that they've violated the forum rules.

Nothing rhymes with orange,

Not even sporange.

Not even sporange.

- Potatoberg
**Posts:**19**Joined:**Wed Nov 05, 2014 10:30 pm UTC

TheGrammarBolshevik wrote:You have been challenged to defend those criteria, but you have spent all of one sentence actually defending them.

A perfect logician comes to conclusions based solely on valid logic. Assumptions are by definition things you take for granted without questioning their logical validity. Thus a perfect logician cannot assume.

You don't need more than a sentence to explain it.

- TheGrammarBolshevik
**Posts:**4878**Joined:**Mon Jun 30, 2008 2:12 am UTC**Location:**Going to and fro in the earth, and walking up and down in it.

Do you even, without looking it up, know what "valid logic" means? Why should we take your bare assertions seriously?

Last edited by TheGrammarBolshevik on Wed Dec 10, 2014 6:58 pm UTC, edited 1 time in total.

Nothing rhymes with orange,

Not even sporange.

Not even sporange.

Potatoberg wrote:A perfect logician comes to conclusions based solely on valid logic. Assumptions are by definition things you take for granted without questioning their logical validity. Thus a perfect logician cannot assume.

In your first sentence your perfect logician takes for granted that logic is valid (i.e. he/she assumes the axioms and rules of inference of logic). Therefore, according to your second sentence, a perfect logician has to assume some things. Thus your assertion that "a perfect logician cannot assume" is incorrect.

Potatoberg, could you, for everyone's sake (or maybe just me, I don't know if anyone else wants this), express again in a clear way exactly what it is you

a) take issue with, and

b) propose as an alternative solution?

If you believe there is no solution, that counts as an answer for b, but please then instead explain why you believe no solution exists (note that this is different from claiming the solution is wrong, so please don't bother explaining why this solution (link: https://xkcd.com/solution.html) is wrong in your answer to why you believe no solution exists.)

Please keep it to the fundamental logic problem itself (link: https://xkcd.com/blue_eyes.html) and the fundamental solution provided by xkcd (link: https://xkcd.com/solution.html).

If you could be as precise as possible in both your answers to a and b above, I, at least, would really appreciate it.

Please don't bother referencing any specific posts by anyone in the xkcd logic thread (link: viewtopic.php?f=3&t=3) in your answers to a and b to keep things as unclouded as possible.

Thanks.

a) take issue with, and

b) propose as an alternative solution?

If you believe there is no solution, that counts as an answer for b, but please then instead explain why you believe no solution exists (note that this is different from claiming the solution is wrong, so please don't bother explaining why this solution (link: https://xkcd.com/solution.html) is wrong in your answer to why you believe no solution exists.)

Please keep it to the fundamental logic problem itself (link: https://xkcd.com/blue_eyes.html) and the fundamental solution provided by xkcd (link: https://xkcd.com/solution.html).

If you could be as precise as possible in both your answers to a and b above, I, at least, would really appreciate it.

Please don't bother referencing any specific posts by anyone in the xkcd logic thread (link: viewtopic.php?f=3&t=3) in your answers to a and b to keep things as unclouded as possible.

Thanks.

- Potatoberg
**Posts:**19**Joined:**Wed Nov 05, 2014 10:30 pm UTC

marzis wrote:Potatoberg, could you, for everyone's sake (or maybe just me, I don't know if anyone else wants this), express again in a clear way exactly what it is you

a) take issue with, and

b) propose as an alternative solution?

If you believe there is no solution, that counts as an answer for b, but please then instead explain why you believe no solution exists (note that this is different from claiming the solution is wrong, so please don't bother explaining why this solution (link: https://xkcd.com/solution.html) is wrong in your answer to why you believe no solution exists.)

I've explained all this before multiple times, but I'll explain again:

a) I take issue with perfect logicians assuming that the claims of the guru are true. A perfect logician needs a reason to accept something as valid. Assumption is accepting that something without valid reasons; so in effect, a perfect logician would know nothing and could know nothing. Anything less than a perfect logician and we would have to assume that all these less-than-perfect logicians think up the same hypothesis at the same time. We would have no reason to accept that, and I believe that is why the puzzle specifically stipulates that these are perfect logicians.

b) A perfect logician would require evidence. Because of this there is no solution (you cannot solve the puzzle if you cannot take into account the claims of the guru).

Also, any rewording that could solve this problem would be admitting that the currently accepted solution was reached through assumptions and was thus invalid (not necessarily wrong, but looks bad when we're meant to be using logic).

- TheGrammarBolshevik
**Posts:**4878**Joined:**Mon Jun 30, 2008 2:12 am UTC**Location:**Going to and fro in the earth, and walking up and down in it.

TheGrammarBolshevik wrote:And if there is no objection to this version of the problem statement, it seems to me that this is merely a case of verbal disagreement. The term "perfect logicians" means one thing to you. It means a different thing to most others here. And so the puzzle that you get out of the problem statement is one puzzle, while the puzzle that other people get out of the problem statement is a different puzzle.

Nothing rhymes with orange,

Not even sporange.

Not even sporange.

Potatoberg wrote:I've explained all this before multiple times, but I'll explain again:

a) I take issue with perfect logicians assuming that the claims of the guru are true.

I've asked you this multiple times, but I'll ask again:

Would you take issue with the player assuming that the perfect logicians accept the claims of the guru as true?

There is a difference between me saying "If the islanders assume X, then the solution is Y" and me saying "If I assume X, then the solution is Y." And the player assuming that the guru's words are accepted by the islanders is something that doesn't negate their perfect logicianitude, and is a perfectly reasonable assumption to make given the scope of puzzles, and avoids this problem:

A perfect logician would require evidence. Because of this there is no solution (you cannot solve the puzzle if you cannot take into account the claims of the guru).

This is also consistent with it being a logic puzzle, since players make assumptions about the contexts of logic puzzles all the time. What if both guards lie? What if the prison warden included a third color hat?

Potatoberg wrote: A perfect logician needs a reason to accept something as valid.

Then we disagree about the definition of "perfect logician". By your definition a "perfect logician" does not believe anything at all. As several people, including myself, have explained: you can only find truth with logic if you have already found truth without it. The "perfect logician" does not beleve that the other people on the island have blue eyes (since he might be colour blind), he does not even believe that there are other people on the island (they might be figments of a deranged imagination), he does not even believe that there is an island.

Given that the whole concept of a "logic puzzle" makes no sense with your definition of "perfect logician": did it not occur to you that you might have misunderstood the puzzle setter's intent in describing the islanders as "perfect logicians"?

Hi All,

So strange that you're arguing about something that logically would never get into the phase where anyone would ever leave the island. Relevant wordings used in the problem are incorrect: 1) "the guru is allowed to speake once, on one day in all their endless years" means only once ever and not once each day. 2) Second, "I can see someone who has blue eyes" only means that there is someone on the island with blue eyes. So even if ad 1) would mean that the guru can speak once every day and she repeats a 100 times that she can see someone with blue eyes that wouldn't mean that there are a hundred blue-eyed persons on the island, she could have seen and referred to the same blue-eyed person each time.

The case would have been different would she have said : "I can see someone who has blue eyes" on the first day and on next days : "I can see another person with blue eyes" - but in that case the riddle would have been too easy I guess.

To sum it up, this problem has become a problem because wordings aren't precise enough and because someone wanted to desperately create a problem...

So strange that you're arguing about something that logically would never get into the phase where anyone would ever leave the island. Relevant wordings used in the problem are incorrect: 1) "the guru is allowed to speake once, on one day in all their endless years" means only once ever and not once each day. 2) Second, "I can see someone who has blue eyes" only means that there is someone on the island with blue eyes. So even if ad 1) would mean that the guru can speak once every day and she repeats a 100 times that she can see someone with blue eyes that wouldn't mean that there are a hundred blue-eyed persons on the island, she could have seen and referred to the same blue-eyed person each time.

The case would have been different would she have said : "I can see someone who has blue eyes" on the first day and on next days : "I can see another person with blue eyes" - but in that case the riddle would have been too easy I guess.

To sum it up, this problem has become a problem because wordings aren't precise enough and because someone wanted to desperately create a problem...

kexter wrote:Hi All,

So strange that you're arguing about something that logically would never get into the phase where anyone would ever leave the island. Relevant wordings used in the problem are incorrect: 1) "the guru is allowed to speake once, on one day in all their endless years" means only once ever and not once each day. 2) Second, "I can see someone who has blue eyes" only means that there is someone on the island with blue eyes. So even if ad 1) would mean that the guru can speak once every day and she repeats a 100 times that she can see someone with blue eyes that wouldn't mean that there are a hundred blue-eyed persons on the island, she could have seen and referred to the same blue-eyed person each time.

The case would have been different would she have said : "I can see someone who has blue eyes" on the first day and on next days : "I can see another person with blue eyes" - but in that case the riddle would have been too easy I guess.

To sum it up, this problem has become a problem because wordings aren't precise enough and because someone wanted to desperately create a problem...

Nope. The guru speaks once, ever, and that alone (plus the circumstances of it) is sufficient to allow the solution to work. There might be nitpicky problems with other parts of the wording, but that part is worded exactly and precisely as intended, and it works.

- jestingrabbit
- Factoids are just Datas that haven't grown up yet
**Posts:**5967**Joined:**Tue Nov 28, 2006 9:50 pm UTC**Location:**Sydney

@kexter: I invite you to imagine what happens for small numbers of blue eyed people ie 1, 2, 3 or 4.

ameretrifle wrote:Magic space feudalism is therefore a viable idea.

marzis wrote:.... which then leads to all of the Blue eye'd people leaving on the 100th night

I haven't searched through all the pages of postings, but I like this one the most of the ones I have read. The guru would not exclude half the group so your conclusion is correct. But the original setup of the problem doesn't exclude some of the possibilities: We assume that everyone wants to leave the island and wants this more than anything else, and they are not required to state the truth. so, on the first night everyone says to the ferryperson, "I have blue eyes". They do this because they know that if they are right they will leave; if they are wrong there is no consequence other than staying. So the first night 100 people leave. The second night the 100 remaining people can state either "I have brown eyes" and leave, or "I have (some other color)" and not leave. They have seen 100 blue-eyed people leave and 99 brown-eyed people stay (and one green-eyed person). They would not go with the statistically lower number (green), so on the second night the remaining 100 people leave.

If the riddle setup is that stating the wrong eye color means staying forever...

- phlip
- Restorer of Worlds
**Posts:**7572**Joined:**Sat Sep 23, 2006 3:56 am UTC**Location:**Australia-
**Contact:**

Robin- wrote:We assume that everyone wants to leave the island and wants this more than anything else,

No, we don't assume anything about what the islanders want. We only are given information about what they will do, specifically that if any islander is able to prove their own eye colour, they will leave. Not "can leave", or "are allowed to leave and they really want to", but also not "will be forced to leave against their will", just "will leave".

Robin- wrote:so, on the first night everyone says to the ferryperson, "I have blue eyes". They do this because they know that if they are right they will leave;

No, the requirement for leaving is not "able to guess their own eye colour" but "able to prove their own eye colour". If a blue-eyed islander goes up to the ferryperson on night 1 and says "I have blue eyes", and the ferryperson says "why do you say that?" and the islander says "I'unno, just a guess" then they're staying behind, even though their guess was correct.

Code: Select all

`enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};`

void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}

- TheGrammarBolshevik
**Posts:**4878**Joined:**Mon Jun 30, 2008 2:12 am UTC**Location:**Going to and fro in the earth, and walking up and down in it.

Someone on reddit linked to a blog post that Terry Tao did about the underlying logic of this problem, which I thought would be of interest to some.

https://terrytao.wordpress.com/2011/05/ ... wer-bound/

https://terrytao.wordpress.com/2011/05/ ... wer-bound/

Nothing rhymes with orange,

Not even sporange.

Not even sporange.

I don't want to tread through 37 pages to see if this has been addressed, but what I don't get is why the Guru plays any role. If everyone can see there is someone with blue eyes, then the Guru saying it does not provide any information so long as there is more than one person with blue eyes. Thus, wouldn't the blue eyed and brown eyed people both be employing the same strategy? I don't think this necessarily changes the answer for when people with blue eyes leave, but it would mean the answer given is incomplete: everyone leaves on day 100, not just the people with blue eyes. If no new information is provided, it would suggest either everyone can leave or no one can leave.

EDIT:

1) If there is one blue eyed person, you obviously need the guru.

2) With two, then without the guru you can assume that you have brown eyes and person two is unaware of someone with blue eyes unless they are told by a third party

3) With three people, you can still assume there is only two people with blue eyes and the situation will play out as above only if the guru speaks

4) With four, it is not so clear. Everyone is certain that everyone knows there are multiple people with blue eyes.

EDIT:

1) If there is one blue eyed person, you obviously need the guru.

2) With two, then without the guru you can assume that you have brown eyes and person two is unaware of someone with blue eyes unless they are told by a third party

3) With three people, you can still assume there is only two people with blue eyes and the situation will play out as above only if the guru speaks

4) With four, it is not so clear. Everyone is certain that everyone knows there are multiple people with blue eyes.

Summum ius, summa iniuria.

Thesh wrote:I don't want to tread through 37 pages to see if this has been addressed, but what I don't get is why the Guru plays any role. If everyone can see there is someone with blue eyes, then the Guru saying it does not provide any information so long as there is more than one person with blue eyes.

The most common fallacy about this problem is assuming the knowledge of an outsider looking in on the island is equivalent to the knowledge of each islander. The guru doesn't provide any information for the outsider, but she does provide information for the people on the island.

Everyone can see there is someone with blue eyes, but everyone does not know that. So your logic about no additional information does not hold for everyone on the island.

A -> B

~A -> ?

A = Everyone can see there is someone with blue eyes. B = guru provides no information

As an outsider, you are given A, but an islander is not given A. So for an islander, the guru providing information does not break logic.

This is a block of text that can be added to posts you make. There is a 300 character limit.

Cradarc wrote:Everyone can see there is someone with blue eyes, but everyone does not know that. So your logic about no additional information does not hold for everyone on the island.

Like I said in my edit, once you get to at least four people, everyone knows that everyone knows that there are people with blue eyes.

Summum ius, summa iniuria.

Thesh wrote:Cradarc wrote:Everyone can see there is someone with blue eyes, but everyone does not know that. So your logic about no additional information does not hold for everyone on the island.

Like I said in my edit, once you get to at least four people, everyone knows that everyone knows that there are people with blue eyes.

Yes everyone knows there are blue eyed people

Yes everyone knows that everyone knows there are blue eyed people.

But not everyone knows that everyone knows that everyone knows there are blue eyed people.

The guru imparts the information that for any n: (everyone knows that)*n there are blue eyed people.

- phlip
- Restorer of Worlds
**Posts:**7572**Joined:**Sat Sep 23, 2006 3:56 am UTC**Location:**Australia-
**Contact:**

Thesh wrote:Like I said in my edit, once you get to at least four people, everyone knows that everyone knows that there are people with blue eyes.

So, you would place the line at: 3 people would leave on day 3, but 4 people will never leave?

In that case, tell me the flaw in the reasoning:

I'm on the island, as one of the islanders, I can see 3 people with blue eyes, and a large number of people with brown eyes. The guru makes an announcement that they can see a blue-eyed person.

I think to myself: if my eyes aren't blue, then the other islanders must clearly leave on day 3, as you've agreed they will.

Day 3 passes, and no-one leaves.

How exactly do I not deduce that my eyes must be blue?

[edit] Since I now see you said "either everyone can leave or no-one can"... I suppose I should also address the other side: and again, where do you draw the line?

Obviously if the island had 100 blue-eyed people, 99 brown-eyed people and one hazel-eyed person, the hazel-eyed person will never leave... they're not even aware that it's an option! So if you want to claim that the n=1 case is they'll never leave but the n=100 case they'll leave on their own without the guru's help, then you'd again need to find a border line, where n people will never leave but n+1 people will leave of their own accord. And then you'll need to provide the reasoning that will allow a brown-eyed person in the n+1-brown-eyed-people case to tell their eyes are brown, to tell that case apart from a hazel-eyed person in the n-brown-eyed-people case (given no brown-eyed people will leave before you are supposedly able to make that deduction, in either case).

Thesh wrote:I don't want to tread through 37 pages to see if this has been addressed

You would be hard pressed to find any one of those 37 pages where this exact point hasn't been addressed. Turns out "the single first thing everyone thinks of when presented with this puzzle" has come up a few times.

Code: Select all

`enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};`

void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}

Here's a variation on the problem:

Every evening, there is a secret poll that is held. Every islander must make two guesses about the number of people on the island that have their eye color. If every person makes the same two guesses, then everyone gets to leave the island. Otherwise, nothing happens.

Assume there's a very large population and guesses are uniformly spread over equiprobable states so the chance that everybody leaving the island by luck is very slim.

Before the Guru speaks everyone follows this algorithm: (Let (X,Y) denote seeing X blue and Y browns)

Guess (X+1,Y+1)

This process will always fail because a blue-eyed person will see a different distribution than a brown-eyed person.

After the Guru speaks, the thought process changes:

If X = 0, guess (2,Y)

Else guess (X+1,Y+1)

This process will always fail unless X = 0 case occurs for one person.

If the above process fails then, X=0 no longer has to be considered, so a X=1 becomes the special case. If that algorithm fails, then X=2 becomes the new special case, etc.

For the nth day:

If X = n-1, guess (n+1,Y)

Else guess (X+1,Y+1)

The original algorithm provides no new information when iterated repeated. When the guru adds the special case, every iteration of the algorithm produces new information (by eliminating a possibility and reducing entropy). The Guru doesn't provide direct information about eye-color but she provides meta-knowledge (ie. altering the algorithm) which allows information to be gained from each passing day.

Every evening, there is a secret poll that is held. Every islander must make two guesses about the number of people on the island that have their eye color. If every person makes the same two guesses, then everyone gets to leave the island. Otherwise, nothing happens.

Assume there's a very large population and guesses are uniformly spread over equiprobable states so the chance that everybody leaving the island by luck is very slim.

Before the Guru speaks everyone follows this algorithm: (Let (X,Y) denote seeing X blue and Y browns)

Guess (X+1,Y+1)

This process will always fail because a blue-eyed person will see a different distribution than a brown-eyed person.

After the Guru speaks, the thought process changes:

If X = 0, guess (2,Y)

Else guess (X+1,Y+1)

This process will always fail unless X = 0 case occurs for one person.

If the above process fails then, X=0 no longer has to be considered, so a X=1 becomes the special case. If that algorithm fails, then X=2 becomes the new special case, etc.

For the nth day:

If X = n-1, guess (n+1,Y)

Else guess (X+1,Y+1)

The original algorithm provides no new information when iterated repeated. When the guru adds the special case, every iteration of the algorithm produces new information (by eliminating a possibility and reducing entropy). The Guru doesn't provide direct information about eye-color but she provides meta-knowledge (ie. altering the algorithm) which allows information to be gained from each passing day.

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