How'd they figure that? (logic puzzle)

[belliott4488]

That statement is an example of logical induction and isn't necessarily true. This is very different from mathematical induction, which I'm, unfortunately, not familiar with. It's been ages since my last math class, but I did realize that some mathematical principle was being used here. I wish others would confirm or deny the validity of mathematical induction in this logic problem. It might not really help me understand the answer, but I would feel more comfortable accepting it. In the meantime, I'll consider how this functions in the problem.

I hope we're not getting too far afield, here, but ... I don't know how "logical induction" and "mathematical induction" are two "very different" things. I would have thought the second is just a specific example of the first.

In any case, induction is a way of proving the truth of a statement for all values of some parameter in the statement. To do it you have to prove two things: 1. If the statement is true for any given value n of the parameter, then it is also true for n+1, and 2. It is true for a specific value of the parameter, typically one or zero.

The people with green eyes and the people with blue eyes would each see a different number of people with green eyes. When N = the number of greenies that they see and nobody kills themselves then they can be sure that they are a greenie also. All the greenies would realize this at the same time. I get that. I think at this point I'm trying to reason how the induction can be true if there is never a point when someone can look around and see only one other greenie. If they can't verify the premise then how can they use induction? Does it matter?

I'm afraid I'm not following you. What do you mean, "there is never a point when someone can look around and see only one other greenie"? If there are two greenies, then each one looks around and sees only one greenie. If there are three, then they each one wonders if the other two see only one, but they learn otherwise one day two.
Beyond that, you don't need anyone to look around and see only one greenie. All you need to know is that if there is only one greenie, then he dies on day one - that gives you point 2. above. Point 1. above comes from realizing that if N green guys kill themselves on day N, then N+1 green guys must kill themselves on day N+1 (because they observed that the N greenies that they see didn't kill themselves on day N, and thus conclude that there must be N+1 greenies, i.e. they are green-eyed themselves).

Sorry again. I worded my statement incorrectly. I should have said that when there are 4 greenies then the minimum number of greenies that can be assumed to be seen by anyone is 2.

Not really. Since everyone could look around and see at least 100 greenies they could arbitrarily say N = 100. I think there is a minimum number of greenies that can be assumed to exist that can be arrived at mathematically. N could be set to that number.

If there are 200 greenies and 800 blueys

If I have blue eyes I would see 200 greenies. I am looking for a minimum number of greenies that everyone can see so I'll assume I have blue eyes. (Without knowing it, I am fortunate that I am correct.) Then I would consider the perspective of a greenie.

From the perspective of any greenie he would see 199 greenies. He is looking for a minimum number of greenies that everyone can see so he'll assume he has blue eyes. (Without knowing it, he is unfortunate that he is incorrect.) So he would consider the perspective of one of the 199 greenies that he sees.

From his perspective he would consider himself a bluey and see 198 greenies. The first greenie knows this greenie is incorrect about the color of his own eyes, but this greenie doesn't know that. He would need to consider someone elses perspective. Any other greenie will also see this guy has green eyes and will calculate the same total number of greenies. Thus the minimum number of greenies that anyone can be assumed to see is 198. N = 198.

This is incorrect, although I admit to having a difficult time following your argument, especially in terms of which guy your pronouns are referring to ("he" and "his").
In any case, whatever reasoning you apply to the blue eyed guy will apply equally well to the green-eyed guy so that whatever number the blue-eyed guy suggests will be one more than the number the green-eyed guy suggests, simply because the blue-eyed guy counts one more green-eyed guy. That will allow them to figure out which of them sees more greenies, which is all they need to know to figure out their own eye color.
You even state that one of them (I didn't follow which) knows that the other is incorrect. That is true, but it's the key to the fact that they will be able to determine exactly how many greenies that other guys see.
I have a feeling this won't convince you, so let's try something: Suppose you're in this position, only let's say you see 153 green-eyed guys. What number would you suggest as the minimum that we all see? If you tell me what number you'd suggest, then I'll tell you how many greenies you see and the color of my own eyes.
In case you think it matters, you can see that I'm blue-eyed (although I don't know that, of course). Of course, I know your eye color, but I'm not telling you. (I've already decided what color eyes I see that you have, as well as how many greenies I see, but I'm not telling you that until after you respond and I tell you my reasoning - I promise I won't change my mind!)

So on the 2nd morning after the laws were passed they would all kill themselves? If induction doesn't work backwards in this problem, might it not work forwards also?

No, that's one of the weird things. On the day the laws are passed, no one states that there are green-eyed people, so the process doesn't begin.
Again, if there were only one green-eyed guy, he wouldn't have any reason to kill himself on day 1, since no one told him there is a green-eyed guy.
That means if there were two, they wouldn't expect the other guy to kill himself on day 1, so they wouldn't have any reason to kill themselves on day 2.
That means if there were three, they wouldn't expect the other two to kill themselves on day 2, so they wouldn't have any reason to kill themselves on day 3.
That means if there were four, they wouldn't expect the other three to kill themselves on day 3, so they wouldn't have any reason to kill themselves on day 4.
Etc.

 

[belliott4488]

I still don't understand what significance the stranger holds in this puzzle. I would think they would start counting the days as soon as the laws were passed, since they already know that some villagers have green eyes.

Sorry - I posted my long response before seeing the edits you added.
The thing about the stranger is that it's the announcement that there are green-eyed people that is the key, not the simple knowledge. The announcement makes the knowledge general, whereas people's observations give them different pieces of information, i.e. any green-eyed people are aware of one fewer green-eyed person than any non-green-eyed people. The lack of shared understanding seems to be key, here.
Again, the simple case of one green-eyed guy makes this clear. In that case, he's the only one who is explicitly unaware that there is a green-eyed guy. For the two or more, you have to do the inductive thing.

 

[Huckleberry]

A common example of logical induction would be a statement like "Every crow I've ever seen has been black, therefore, all crows are black." They are often reasonable statements, but they are not inherently factual. They seem different to me from mathematical principals.

No, that's one of the weird things. On the day the laws are passed, no one states that there are green-eyed people

If there were one person he would never know his eye color.

If there were two greenies then when they look at each other they would wonder why the person doesn't kill themselves at the next ritual. They would realize that he doesn't know his eyes are green. He may be the only greenie.

If there are 3 greenies then they would all see 2 other greenies. They might assume that those greenies are the only 2 on the island, who are looking at each other thinking the last greenie doesn't know his eye color.

Ok, this may be an example of induction also. Hmm, but once the stranger makes his statement then 1 no longer is true. Then 1 greenie would know he is the only greenie on the island. And if there were 2 then the second would know he had green eyes after the first doesn't kill himself, etc. It seems that after the stranger makes his statement it can't be assumed that people don't know their eye color any more. So the only option when they don't kill themselves is that one must have green eyes. I'm still not sure that's correct, but it is very strange. I have a hard time believing that people would look around and see many green-eyed people and not come to the conclusion themselves that there are green-eyed people on the island.

What do you mean, "there is never a point when someone can look around and see only one other greenie

I think this is the last sticking point for me on this riddle and the reason I question the validity of induction here.

If there are 4 greenies then they all can assume they have blue eyes and look at the perspective of another greenie. They see the 2nd greenie and know that he is wrong to assume he has blue eyes. The most they can assume is that the 2nd greenie sees 2 other greenies. So all the greenies would realize that every greenie knows that there are more than 1 greenie on the island.

Also, if there are 4 greenies and none of them ever see only 1 other greenie then there is no original greenie to draw the first conclusion that they are the only greenie on the island. If all the greenies know that there never was the original greenie who could deduce they had green eyes, then how can they use induction to be certain if they have green eyes.

I don't know. Very possible I'm wrong here. I'm not sure what you could say that would be new to somehow explain it to me. You have been very patient so far, but I'm sure you must be tired of me by now and I think I need a break from this problem for a while. Thanks for sharing the puzzle. I enjoyed it very much.

 

[belliott4488]

I think this is the last sticking point for me on this riddle and the reason I question the validity of induction here.

If there are 4 greenies then they all can assume they have blue eyes and look at the perspective of another greenie. They see the 2nd greenie and know that he is wrong to assume he has blue eyes. The most they can assume is that the 2nd greenie sees 2 other greenies. So all the greenies would realize that every greenie knows that there are more than 1 greenie on the island.

Also, if there are 4 greenies and none of them ever see only 1 other greenie then there is no original greenie to draw the first conclusion that they are the only greenie on the island. If all the greenies know that there never was the original greenie who could deduce they had green eyes, then how can they use induction to be certain if they have green eyes.

I don't know. Very possible I'm wrong here. I'm not sure what you could say that would be new to somehow explain it to me. You have been very patient so far, but I'm sure you must be tired of me by now and I think I need a break from this problem for a while. Thanks for sharing the puzzle. I enjoyed it very much.

Well, I agree, we've probably spun this out about as long as we should. ;-) Before I let it go, however, let me address your remaining question, in case you do come back to this thread at some point.
If I understand you, you're concerned that if there are more than 2 greenies, then there is no one to make that initial declaration, "whaddaya mean there are green eyed people - I don't see any! Uh-oh, it must be ME!" That is true, but it's also unnecessary for 2 or more, because all anyone needs to know is that the N green-eyed people he sees didn't kill themselves on Day N. Let me walk through this for the case of 4 greenies that you asked about.
First, the quick answer is that each of the four greenies sees three greenies, and when they don't kill themselves on Day 3, that is all that they need to see, because if there were only three then they'd have to kill themselves on Day 3, for the reasons we've already gone over. They don't have to reason all the way back to the case of one greenie. But ... they could. Here's how:
This is what the blue-eyed people think:
"I hope that I'm not green-eyed, and since I see only four greenies,
..that means I hope they each see three greenies and think to themselves,
...'I hope I'm not green-eyed, and since I see only three greenies,
... that means I hope they each see only two greenies and think to themselves,
..."I hope I'm not green-eyed, and since I see only two greenies,
... that means I hope they each see only one greenie and think to themselves,
..."I hope I'm not green-eyed, and since I see only one greenie,
... that means I hope he sees no greenies and thinks to himself,
..."Oh, crap, I'm screwed."
..."
..."
..."
"
Follow that? ;-) It's not actually that any of them are thinking those things beyond the first two, but rather that they are thinking about what the others are thinking, and specifically, what the others are thinking that the others are thinking (and so forth).

 

[Huckleberry]

They don't have to reason all the way back to the case of one greenie. But ... they could. Here's how:
This is what the blue-eyed people think:
"I hope that I'm not green-eyed, and since I see only four greenies,
...'I hope I'm not green-eyed, and since I see only three greenies,
... that means I hope they each see only two greenies and think to themselves,
..."I hope I'm not green-eyed, and since I see only two greenies,
... that means I hope they each see only one greenie and think to themselves,
..."I hope I'm not green-eyed, and since I see only one greenie,
... that means I hope he sees no greenies and thinks to himself,
..."Oh, crap, I'm screwed."

I think you are wrong here. If there are 4 greenies then all the villagers will know that the other villagers know that there are greenies on the island. That is all that is needed to start counting days to N=3 when all the greenies realize they have green eyes. So the strangers words aren't necessary, and the counting of days would begin at the passing of the law.

All the people with blue eyes will see 4 greenies and all the people with green eyes will see 3 greenies. Since the greenies will always see one less person with green eyes than a bluey does, I'll show how every greenie knows that the other greenies know there are other greenies on the island.

There are 4 greenies.
Greenie4 hopes he has blue eyes and sees 3 other greenies.
Greenie4 hopes that Greenie3 only sees 2 other greenies.
Greenie4 realizes that Greenie2 will see that Greenie3 also has green eyes,
so Greenie2 will include Greenie3 in his count and will hopefully arrive at a total of 2 greenies.

The reason of Greenie4 will be the same reasoning of all the greenies. They can all assume that the other greenies see other greenies. I think it's kind of silly to assume that when there are hundreds of greenies on this island that they can't agree that greenies exist. Induction isn't appropriate for this part of the problem.

 

[belliott4488]

I think you are wrong here. If there are 4 greenies then all the villagers will know that the other villagers know that there are greenies on the island. That is all that is needed to start counting days to N=3 when all the greenies realize they have green eyes. So the strangers words aren't necessary, and the counting of days would begin at the passing of the law.

Nope. That will work only if someone says out loud, "I see green people" - that's critical - it's not enough that everybody knows it. Think again about the case of only one greenie - I think you agree that he won't kill himself. Without that, the reasoning changes for all other cases, starting with the case of two greenies, neither of whom will expect the other to kill himself on Day 1, and who thus won't learn anything about their own eye color when he doesn't. It doesn't matter that they both knew that there was (at least) one greenie when the law was passed - they don't expect him to do anything special on Day 1. The point is that there has to be some statement made that they all agree on. Until it's voiced, however, it has no effect. Weird, but necessary.

All the people with blue eyes will see 4 greenies and all the people with green eyes will see 3 greenies. Since the greenies will always see one less person with green eyes than a bluey does, I'll show how every greenie knows that the other greenies know there are other greenies on the island.

There are 4 greenies.
Greenie4 hopes he has blue eyes and sees 3 other greenies.
Greenie4 hopes that Greenie3 only sees 2 other greenies.
Greenie4 realizes that Greenie2 will see that Greenie3 also has green eyes,
so Greenie2 will include Greenie3 in his count and will hopefully arrive at a total of 2 greenies.

The reason of Greenie4 will be the same reasoning of all the greenies. They can all assume that the other greenies see other greenies. I think it's kind of silly to assume that when there are hundreds of greenies on this island that they can't agree that greenies exist. Induction isn't appropriate for this part of the problem.

Not quite!
Yes, with four greenies, they all know that there are at least three greenies, because that's what they see. If they could say this aloud, then they could skip the first three days of waiting, but then again, doing so would violate the law, so they'd have to die the next day, anyway - and with good reason, since they could not say this without revealing how many greenies they, themselves, actually see.
Here's how:
You've said that the four greenies all know that everyone sees at least two greenies, so they should just say this aloud and get it over with. Let's say Greenie 4 takes the plunge and makes this prounouncement. Here's what everyone else reasons in response:

Blue guys: "hm ... I see four greenies, and if I'm green-eyed then Greenie 4 should also see 4, but then he'd suggest that we all see at least 3, not 2 - so he must see only three, which means I'm blue-eyed!"

Green guys: "uh-oh - That guy with green eyes just said we all see at least 2 greenies, but if I'm blue-eyed, then he should see only two greenies himself, and he should have said everyone sees at least one greenie, but since he didn't, that means he must see three greenies, which mean I'm green-eyed!"

They can't suggest a minimum number without giving away the number that they see, since everyone else sees either the same number. or one more or less than they do (depending on whether the guy making the suggestion is green-eyed or blue-eyed).

 

[Hurkyl]

So the strangers words aren't necessary

Yes they are. In order to conclude your own eyes are green, you need to know:
(1) You see N green-eyed people.
(2) N days ago, if there was only 1 green-eyed person person, he would have killed himself.

 

[Sleek]

Lets say there is a green eyed person and 2 blue eyed person. If someone says (like the stranger did) that he sees green eyed people, then the green eyed person, knowing there are only three people in total with other two blue eyed, would know he is the only green eyed.

If there are two green eyed people. Then at the first day, the green eyed duo will look at each other, assuming themselves to be blue. But they both won't die the next day. Thus it would imply that the other one thinks someone else is green eyed, but the rest two are blue eyed, so he should also be green eyed. Thus they both die on the second day.

Irrespective of the no. of blue eyed people (>0), n green eyed people would die on nth day.

 

[Huckleberry]

Yes they are. In order to conclude your own eyes are green, you need to know:
(1) You see N green-eyed people.
(2) N days ago, if there was only 1 green-eyed person person, he would have killed himself.

I would think they have all this information before the stranger tells them there are green-eyed people on the island. They only need look around and see that there are other green-eyed people on the island. That satisfies the requirement for (1). If there are 4 green-eyed people then in no case can anyone assume that another person sees less than 2. Since everyone knows that everyone knows that there are more than one green-eyed people it satisfies the requirement for (2).

It doesn't matter if the green-eyed people and blue-eyed people can't agree on N. As soon as there are 4 green-eyed people it can no longer be assumed that someone is the only green-eyed person.

 

[Hurkyl]

I would think they have all this information before the stranger tells them there are green-eyed people on the island.
...
If there are 4 green-eyed people then in no case can anyone assume that another person sees less than 2. Since everyone knows that everyone knows that there are more than one green-eyed people it satisfies the requirement for (2).

But that's not what (2) says. (2) says that if only one person had green eyes, he would have killed himself N days ago.

 

[Huckleberry]

But that's not what (2) says. (2) says that if only one person had green eyes, he would have killed himself N days ago.

As soon as it is known by everyone that 2 greenies see each other then someone should start killing themselves.

 

[Hurkyl]

As soon as it is known by everyone that 2 greenies see each other then someone should start killing themselves.

Why?

 

[Huckleberry]

If it is known that nobody sees less than two greenies then everyone knows the information that the stranger told them, that there are greenies on the island.

 

[Hurkyl]

If it is known that nobody sees less than two greenies then everyone knows the information that the stranger told them, that there are greenies on the island.

But how does that prove people start killing themselves?

 

[Huckleberry]

As long as it is certain that everyone on the island knows that there are greenies then they should start induction. The induction goes on until N days where N is equal to the number of people with green eyes that they see. If those people don't kill themselves the next day then that individual knows he must have green eyes.

If it is not certain that everyone on the island knows that there are greenies then the process of induction can't begin. But since as long as there are 4 greenies everyone is certain that the others know, they can begin induction.

I just don't understand why they need a stranger to tell them anything before they can begin induction.

 

[Hurkyl]

I just don't understand why they need a stranger to tell them anything before they can begin induction.

Because induction requires a base case.

 

[Huckleberry]

There is a difference between knowing that there are green-eyed people on the island and knowing that everyone else knows it also. When the stranger tells everyone that there are green-eyed people on the island then everyone is certain that everyone else knows that information. But as long as there are 4 green-eyed people then everyone can already be certain of that information. He is telling them nothing they don't already know. In fact, since the puzzle states that they had been living this way for years then the stranger would be lying, because there would be nobody on the island to make his statement to.

I don't see how the villagers conclude anything different from the strangers statement than they would from the information they could deduce when there are 4 greenies on the island.

 

[Hurkyl]

I don't see how the villagers conclude anything different from the strangers statement than they would from the information they could deduce when there are 4 greenies on the island.

Because of the base case of the induction.

Or, working from the top down (i.e. descent), with the 4 greenies, the stranger's statement affects the outcome of a hypothetical within a hypothetical within a hypothetical.

 

[Huckleberry]

But it is the same information.

If everyone knows that there are greenies then someone must be a greenie. If a greenie doesn't kill himself then there must be more than 1. etc, until everyone knows they are a greenie or not. The base case is formed when all the greenies know that all the other greenies are aware that there are more than one greenie. This is what they gather from the information that the stranger tells them, which is the same information that they deduce from 4 or more greenies existing on the island.

If there is any possibility that not everyone sees another greenie then there is no case for induction. This will happen if there are 3 or less greenies. In this case the 3rd greenie is looking at the 2nd. The 3rd realizes the 2nd doesn't know his own eye color. The 3rd is hoping that the 2nd greenie is looking at the 1st greenie and thinking that the 1st greenie is the only greenie on the island and just doesn't realize it. There may be at least one person that is unaware that there are greenies on the island.

When there are 4 greenies this is no longer possible. The 4th greenie looks at the 3rd. The 4th greenie realizes the 3rd doesn't know his own eye color. The 4th greenie is hoping that the 3rd greenie is looking at the 1st and 2nd greenie. But then the 1st greenie would be looking at the 2nd and 3rd greenie, and the 2nd greenie would be looking at the 1st and 3rd greenie. Every greenie will see 3 other greenie and realize that the other greenies see at least 2 other greenies.They must now all realize that there are greenies on the island, which is the exact same thing as the stranger told them. If I'm correct, and these villagers have infallible logic, then they wouldn't need to communicate this information to each other.

How does the stranger's statement create a base case for induction?

 

[belliott4488]

<edit> Whoops, I followed an email link that took me to the previous page, and I responded before I realized that there was another whole page of posts ... sorry ... anyway, here's how I responded to that <end edit>

Huck, you're just not getting this, are you? ...

<snip> If there are 4 green-eyed people then in no case can anyone assume that another person sees less than 2.

True, but also irrelevant, since there is no need for anyone to assume that anyone else sees fewer than 2.

Since everyone knows that everyone knows that there are more than one green-eyed people it satisfies the requirement for (2).

No, it doesn't. If no one has explicitly stated "There are green-eyed people," then why would anyone assume that 1 green-eyed guy would have killed himself 4 days ago (Requirement 2)?

It doesn't matter if the green-eyed people and blue-eyed people can't agree on N. As soon as there are 4 green-eyed people it can no longer be assumed that someone is the only green-eyed person.

Again, true, but irrelevant. What is necessary is that the four green-eyed people all know that the three green-eyed people they see would have killed themselves on Day 3 if they were the only green-eyed people.

 

[belliott4488]

HUCK! You're missing something crucial here: the "base case" Hurkyl is referring to is the case of one greenie.

Do you agree that if there is only one greenie, he will not kill himself until the stranger arrives to make his pronouncement? If so, then by induction you know that no one else will kill himself unless the stranger makes his pronouncement, because the base case - i.e. hypothetical case of one greenie - fails.

The assertion that N people die N days after the announcement depends entirely on the truth of the statement that 1 hypothetical greenie would kill himself. If he does, then they all do; if he doesn't, none of them do.

 

[belliott4488]

Arrgh! You're so close it's driving me mad!

But it is the same information.

If everyone knows that there are greenies then someone must be a greenie. If a greenie doesn't kill himself then there must be more than 1. etc, until everyone knows they are a greenie or not.

No, that's not how it works. No one expects one greenie to kill himself on day 1, unless that greenie is the only greenie he sees, which is possible only for the cases of 1 or 2 greenies.

The base case is formed when all the greenies know that all the other greenies are aware that there are more than one greenie. This is what they gather from the information that the stranger tells them, which is the same information that they deduce from 4 or more greenies existing on the island.

No, you're misunderstanding what is meant by a "base case". It's the case where there is exactly one greenie. In the case you're describing, the 4 greenies kill themselves on Day 4 because they didn't see the other 3 three greenies kill themselves on Day 3. Period.

If there is any possibility that not everyone sees another greenie then there is no case for induction. This will happen if there are 3 or less greenies. In this case the 3rd greenie is looking at the 2nd. The 3rd realizes the 2nd doesn't know his own eye color. The 3rd is hoping that the 2nd greenie is looking at the 1st greenie and thinking that the 1st greenie is the only greenie on the island and just doesn't realize it. There may be at least one person that is unaware that there are greenies on the island.

ALMOST! That last sentence is false, since all 4 greenies see 3 other greenies, and everyone else sees 4 - NO ONE is unaware that there are greenies. BUT - you're right, the 3rd guy is hoping that the 2nd guy is hoping that the 1st guy sees 0 greenies. Get the difference? - it has to do with the "nested" hypotheticals. Each guy has hopes for what the other guy is thinking that the other guy is thinking that the other guy is thinking ...

When there are 4 greenies this is no longer possible. The 4th greenie looks at the 3rd. The 4th greenie realizes the 3rd doesn't know his own eye color. The 4th greenie is hoping that the 3rd greenie is looking at the 1st and 2nd greenie. But then the 1st greenie would be looking at the 2nd and 3rd greenie, and the 2nd greenie would be looking at the 1st and 3rd greenie.

The 4th guy is hoping that the 3rd guy is going through the exact thought process you described above (i.e. as if there are only 3 greenies), right down to the 2nd guy hoping that 1st sees no greenies.

Every greenie will see 3 other greenie and realize that the other greenies see at least 2 other greenies.

Yes, but they're hoping the other greenies don't know that.

They must now all realize that there are greenies on the island, which is the exact same thing as the stranger told them. If I'm correct, and these villagers have infallible logic, then they wouldn't need to communicate this information to each other.

Again, their knowledge of this is not sufficient. It is crucial that there be some kind of spoken statement, to "start the clock", i.e. to set the condition that would be necessary for one greenie to kill himself, if he were the only greenie.

How does the stranger's statement create a base case for induction?

Think about the necessity of the stranger's statement for the base case, i.e. the case of only one greenie. It ALL stems from that.

 

[Huckleberry]

Arrgh! You're so close it's driving me mad!

Haha, me too! I'm not even sure why I care so much.

ALMOST! That last sentence is false, since all 4 greenies see 3 other greenies

Well, the example this quote was referring to only had 3 greenies in it. That is why the final statement is true.

Again, their knowledge of this is not sufficient. It is crucial that there be some kind of spoken statement, to "start the clock", i.e. to set the condition that would be necessary for one greenie to kill himself, if he were the only greenie.

There is already a written statement to start the clock, from the moment they created the laws.

Think about the necessity of the stranger's statement for the base case, i.e. the case of only one greenie. It ALL stems from that.

I'll think about it. Right now I'm trying to figure out how descending induction works to create the belief that it is possible that there may be only one greenie who doesn't know his eye color, especially when there are 200 greenies on the island.

But I did think that if there were 3 greenies and 1 blue, it would be the same case as if there were 4 greenies. So I'm starting to think the whole 4 greenie theory is wrong. I think all I need to understand this puzzle is to know that the descending induction works and the 4 greenie theory doesn't. I'm out of time right now though, so I'll have to work it out later. I've got an Australian Pink Floyd show to go to.

 

[belliott4488]

Haha, me too! I'm not even sure why I care so much.

Well, the example this quote was referring to only had 3 greenies in it. That is why the final statement is true.

My mistake - I should have referred to one fewer greenie in each case, but your statement is wrong, regardless. Your "final statement" was (for the case of 3 greenies),

There may be at least one person that is unaware that there are greenies on the island.

Not so - in this case everyone sees either 2 greenies (if he is himself a greenie) or 3 greenies (if he's not), but no one is unaware that there are greenies. What there are, are people who think that there might be people who think that there is a person who is unware that there are greenies. That's a very different thing. It does, however, work for any number N of greenies; you just have to stick as many "people who think there are people who think that" phrases in there as it takes to get back to the case of one greenie.

There is already a written statement to start the clock, from the moment they created the laws.

No, not just any statement will start the clock; it has to be one that would cause a single greenie to kill himself on day one, if he were the only greenie. THAT is the base case.

I'll think about it. Right now I'm trying to figure out how descending induction works to create the belief that it is possible that there may be only one greenie who doesn't know his eye color, especially when there are 200 greenies on the island.

No - that's not the result. First, it's not that there is "only one greenie who doesn't know his eye color" - none of the greenies should know their eye color (at first, anyway - they'll figure it out on Day 198, when the greenies they see don't kill themselves). Second, it's not that there really is one greenie who sees no other greenies (the base case guy); it's that the 199 greenies each hope that the 198 they see each hope that the 197 that they see each hope that the 196 that they see ... each hope that the 2 greenies they see each hope that one greenie they see doesn't see any other greenies.

But I did think that if there were 3 greenies and 1 blue, it would be the same case as if there were 4 greenies. So I'm starting to think the whole 4 greenie theory is wrong. I think all I need to understand this puzzle is to know that the descending induction works and the 4 greenie theory doesn't. I'm out of time right now though, so I'll have to work it out later. I've got an Australian Pink Floyd show to go to.

Whoa ... Australian Pink Floyd? It that like, a bunch of Aussies in a Pink Floyd tribute band? Or maybe just a Pink Floyd concert in Australia ... either way, I hope it was fun! I'd say "Wish You Were Here," but really, I wish I were there!

 

[ThienAn]

or i know why they committed suicide is they had eyes. They can see each other' eye color. You should have posted that they are color-blinded, or mute

 

[belliott4488]

or i know why they committed suicide is they had eyes. They can see each other' eye color. You should have posted that they are color-blinded, or mute

Or, better yet, you should have read the original statement of the puzzle more closely!

 

[country boy]

Okay, now how many days until the blue-eyes self-inflict?

 

[Oerg]

Okay, now how many days until the blue-eyes self-inflict?

They do not know the colour of their eyes as blue, all they know is that there are green eyed people on the island, and that their eye is of different colour. In other words, they do not know the word to decribe their eye colour.

 

[Oerg]

I still think its a rather funny puzzle, with a few flaws, with one of the major flaws in that

If say there were 2 green and 1 blue, on the first day of comtemplation at the town centre, if there wasnt any hesitation in all 3 to do a ritual suicide, then they would have known at that moment that the 2 greens were green in eye colour. Why don't they commit suicide then and there?

And if all of the greeny citizens were to think this way, all would commit suicide on the first day.

I think this puzzle can only be solved on the basis of many assumptions.

Perhaps it would be better to say it this way,

1. All citizens would gather at the towncentre every day to contemplate the colour of their eye for 5 minutes. After 5 minutes of contemplation, all citizens who know the colour of their eyes are to commit ritual suicide immediately and the rest of the citizens are to return to their daily chores right away and return to the town centre again the next day.

 

[country boy]

Okay, now how many days until the blue-eyes self-inflict?

They do not know the colour of their eyes as blue, all they know is that there are green eyed people on the island, and that their eye is of different colour. In other words, they do not know the word to decribe their eye colour.

I don't think they are language-challenged. They will know about colors and have names for them.

To see the symmetry between green and blue, start with 200 of each (or two of each).

Some of the difficulties talked about here could be solved by stating that the islanders don't know how many colors there are until Robinson Crusoe tells them "Hey, there are only green and blue eyes on this island!"

 

[Oerg]

I don't think they are language-challenged. They will know about colors and have names for them.

Thats the problem with this question, it leaves a lot of information to assumption

 

[belliott4488]

I still think its a rather funny puzzle, with a few flaws, with one of the major flaws in that

If say there were 2 green and 1 blue, on the first day of comtemplation at the town centre, if there wasnt any hesitation in all 3 to do a ritual suicide, then they would have known at that moment that the 2 greens were green in eye colour. Why don't they commit suicide then and there?

And if all of the greeny citizens were to think this way, all would commit suicide on the first day.

I think this puzzle can only be solved on the basis of many assumptions.

Perhaps it would be better to say it this way,

1. All citizens would gather at the towncentre every day to contemplate the colour of their eye for 5 minutes. After 5 minutes of contemplation, all citizens who know the colour of their eyes are to commit ritual suicide immediately and the rest of the citizens are to return to their daily chores right away and return to the town centre again the next day.

I'm not sure I'm following your reasoning. Are you suggesting that if the green guys commit suicide on a given day, then since the blue guys might then learn their own eye color they could also commit suicide, i.e. on the same day? I see how your suggested statement of the rule for meetings solves that, but I don't see how my original statement was not equivalent. The reason I said that they must silently contemplate for ten minutes and only after this period commit suicide was specifically to prevent this problem.

I also don't see why anyone would kill himself on Day 1 in your example; the stranger guy announced only that there were green-eyed people, of which there are two, right? So the green guys would each be hoping that the other green guy was the sole green guy. (Unless you're using the fact that the stranger used the plural, which I think is an unnecessary muddling of the problem - in this case just have him say something stupid like, "there are one or more green-eyed people on this island, although I am not specifying anything about the actual number." )

As for country boy's question about how many days till the blue-eyed guys kill themselves - as the problem was stated in the original post, they never should, since each one can hope that he's the one brown-eyed guy on the island. If the stranger had said, "there are some green, some blue, but no other color of eyes ..." then they'd do themselves in on the next day after the green-eyed guys.

 

[country boy]

As for country boy's question about how many days till the blue-eyed guys kill themselves - as the problem was stated in the original post, they never should, since each one can hope that he's the one brown-eyed guy on the island. If the stranger had said, "there are some green, some blue, but no other color of eyes ..." then they'd do themselves in on the next day after the green-eyed guys.

Actually, law 1 requires that "At noon every day, all 1000 citizens must gather... ." After the 200 die, it is not longer possible to comply. So, is everyone else off the hook?

 

[belliott4488]

Actually, law 1 requires that "At noon every day, all 1000 citizens must gather... ." After the 200 die, it is not longer possible to comply. So, is everyone else off the hook?

Oh, for goodness' sake ... of course the law wouldn't be tied to to a specific population. This is a logic puzzle, not a legal puzzle! You have to exert a little effort to get into the spirit of the problem.

Change that to "all citizens must gather ..."

 

[country boy]

Oh, for goodness' sake ... of course the law wouldn't be tied to to a specific population. This is a logic puzzle, not a legal puzzle! You have to exert a little effort to get into the spirit of the problem.

Change that to "all citizens must gather ..."

Sorry, I thought I was in the spirit. Wording is important in logic puzzles, but we can also have a little fun.

I've enjoyed your puzzle and the discussion. Have a nice holiday!