I’m leaving after the 100th comment. And yes, I do have green eyes … at least that’s what my passport claims.
There is no new information in either statement, whether it is “at least one” or “everybody you can see”.
Every one of the logicians would be able to give the correct answer to those questions beforehand, and if you would tell them one by one instead of everybody together, it would not help them.
The information is not in the statement themselves, it’s in the fact that they now know that everybody knows this. And in the “at least one” version it also matters that they all know that everybody knows it now, at this moment.
There’s a few assumptions made about the prisoners;
- none of them are colour blind (also the guards with the flashlight) or just regular blind
- none of them are deaf
- they all understand your language (but hey, maybe you recognise them all and know this, you are on a mission to free an entire convention of logicians after all)
- they all were individually able to work this out (I know that they’re logicians, but a statician would probably say that at least one of them didn’t get it or would make a mistake or lose count of the days).
Initially I thought that this was a trick by the dictator to ensure that even the greeneyed prisoners would in the dead of night have differently coloured eyes.
Would generalissimo have been enraged if you’d said “there is an even number of greeneyed prisoners”? (or odd for when the puzzle would perhaps involve an odd number) cos that would have had everyone out on night one.
Everyone has always known that at least one prisoner is green-eyed.
The facts:
- No prisoners are not green-eyed.
- Everyone has always known that at most one prisoner is not green-eyed.
- Everyone has always known that everyone has always known that at most two prisoners are not green-eyed.
- Everyone has always known that everyone has always known that everyone has always known that at most three prisoners are not green-eyed.
- For all i, (Everyone has always known)^i that at most i prisoners are not green eyed.
What is the information contained in hearing the statement “at least one person is green-eyed” addressed to everyone?
- At most n-1 prisoners are not green-eyed
- Everyone knows that at most n-1 prisoners are not green-eyed
- For all i, (Everyone knows)^i that at most n-1 prisoners are not green-eyed
- Specifically, a new piece of information: (Everyone knows)^n that at most n-1 prisoners are not green-eyed
I wasn’t really disputing it, I was just pointing out that 4 was special since in any case where there are at least 4 green-eyed people, everyone knows there is at least one green eyed person and knows that everyone else knows that too. So they all would have left 3 days after the fourth logician arrived so the whole story behind the puzzle is flawed. The story as told does work if there are only three green eyed people. In that case everyone knows there is one green eyed person but doesn’t know everyone else knows that.
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So it sounds like you are disputing it. My argument above (and the argument in the video, and in other solutions to the puzzle) says that four people cannot leave until someone tells them publically that there is at least one green-eyed person.
Again, work it out by induction:
- You agree that three green-eyed prisoners can’t leave the island by themselves, until they’re told the information.
- If there are three green-eyed prisoners and one brown-eyed prisoner arrives, they are in the same situation. They can’t leave until they are told the information.
- If you are a new prisoner and you arrive and see three green-eyed people, as far as you know you might be in the situation above: three green-eyed people and you, a brown-eyed person. Therefore, you can’t leave after any number of days.
- All four green-eyed people arriving together will have the same thought-process. There is nothing to tell them that they have green eyes
- Therefore four green-eyed people can’t leave the island by themselves, until they’re told the information.
Again, you’ve got to put yourself in the position of one of the four prisoners. What would it take to convince you that you have green eyes? If you’re in a situation where you think you might have brown eyes, you can’t leave.
Like what Daneel said, I would add that changing the phrase “everyone you see has green eyes” to “everyone EVERYONE sees has green eyes” is basically the same statement in this case, except the second statement seems more obviously to be giving new information. Maybe I’m looking at this wrong, but I just thought I’d throw that idea out there.
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The puzzle is famously difficult because the answer is incorrect.
Each logician sees 99 green eyed logicians. They know there are either 100 or 99 green eyed logicians. Telling them there is at least 1 green eyed logician will not set them free through the chain of events as it was laid out.
As a logician on the island I know that every other logician can see at least 98 green eyed people (100 minus myself and themselves). Every single logician on that island has that information and I know that.
No one ever leaves.
The mathematical proof has an error somewhere, I think because it’s cumulative.
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I think you’re missing something SamSam.
A. If you arrive and see three green eyed people then you know each of them can see at least two green eyed people (it doesn’t matter what colour eyes you have for this bit).
B. But anyone who can see at least two green eyed people knows that everyone can see at least one green-eyed person (all except the first green-eyed person can see their green eyes and they can see the second green-eyed person).
So from A and B you know that everyone knows that everyone can see at least one green-eyed person. You never needed anyone to come along and tell you. That’s my point.
Then you just need to wait three days to find out if your eyes or green or not depending on whether everyone else left on the third day or not. If they didn’t you must have green eyes too and you all leave on the fourth day and the island has nobody with green eyes.
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A can see three green eyed people, so A knows that B can see at least two, so A knows that B knows that C can see at least one, but A doesn’t know that B knows that C knows that D can see any - until X tells everyone that at least one person has green eyes.
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Oops, Fail.
Sorry, the logic here with “at least one logician has green eyes” will only work for two people.
(but then it will give them new information - that the other logician also knows at least one has green eyes)
for 3 logicians or more, it will not provide new information, because a logician already knows that the other two logicians know that there is at least one with green eyes.
The induction for the analog puzzle with the cheating wives works because you provide them with new information!
Edit.
my bad. it works fine.
although the visitor information is only needed in the case of one or two logicians.
It seems to me that the new information is the act of starting the clock. Everyone is born on the island, so there’s no set date at which they can start the inductive reasoning unless if it’s somehow coordinated - the fact that all of the other prisoners are still on the island has no meaning until you hear that statement as the signal. If everyone leaves after 99 days, then you know that you have differently coloured eyes and are doomed to stay there. One possible flaw: prisoners are there from birth, so presumably there would be people who may genetically have perfect logic, but can’t understand spoken language yet. Do you ignore them in your count? If they are only brought into the main group at a particular moment, this would have the same effect as the visitor’s statement - the group now has a date to start the clock. If the rule is only introduced after everyone has come of age, this date is when the clock starts. I suppose it could work if everyone were told the rule individually or if it was announced every day and there was no salient starting date for everyone.
What if one of the prisoners has red eyes like the guards? Can they just identify as green-eyed? (since they grew up surrounded by 99 prisoners with green eyes)
Daneel, if you want everyone to know that everyone knows that everyone can see someone with green eyes, then you just need to be able to see four people with green eyes, not three. If you can see four people with green eyes, then you know each of them can see at least three people with green eyes. So each of them knows that everyone can see at least two people with green eyes. So each of them knows that everyone knows that everyone can see someone with green eyes which is surely the most information the announcement gave us.
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buddy , i think the puzzle works beautifully . (sorry if iam wrong) , lets take 4 people into scene . name them 1,2,3,4 . first night , no one leaves .now put yourself in 4’s shoes . your thinking " if i have lets say red eyes , 1,2,3 should leave on the third night " , why? here’s how - 3 is thinking - " 4 has red eyes , so no need to put him in the equation and if i have red eyes , since the speaker said “atleast one of you has green eyes” and this should tell both 1 and 2 if they didnt leave the first night , that both of them have green eyes. otherwise , if 1 was the only green eyed person , 1 would have left thinking that the speaker was referring to him .
second night passes , no one leaves . On the third day’s headcount , 3 is thinking , " 1 and 2 arent sure that 1 stayed back because of 2 or viceversa , because i too have green eyes " .
so 1,2,3 should leave on the third night . again , thats what 4 was thinking if he had red eyes .
BUT, on the fourth day , 4 sees no one went . Which means that 3 is not sure that 1 and 2 stayed back because of him because 4 too has green eyes .
thats how all of them leaves the fourth night .
see , that sentence “atleast one of has green eyes” is the root of all this .
Again , this is the logic that was shown in the video or this is how i understood it .
I’m not missing anything. I’m afraid you’re still not getting the puzzle.
You say:
Then you just need to wait three days to find out if your eyes or green or not depending on whether everyone else left on the third day or not. If they didn’t you must have green eyes too
This is where you’re mistaken. Without any external information (the speaker announcing that one person has green eyes), there is no way that three green-eyed people can leave on their own.
You believe that, if there are three green-eyed people and one brown-eyed person, that on their own the three green-eyed people will leave after three days.
So let’s see if that’s true. Imagine that you are on the island, with three other people: two green-eyed people and one brown-eyed person. You don’t know the color of your eyes. Can you leave the island? On what day?
The answer is you cannot leave the island, because you have no way of determining the color of your own eyes.
In the case of three green-eyed people and one brown-eyed person, all green-eyed people are in the same situation as you were. So therefore, none of them can leave.
I’m going to try and put it another way.
Let’s say I set up repeating scenarios with the same four people each time, A-D. I’m going to put dots on their foreheads each time, either Brown (B) or Green (G). If someone knows they have a green dot, they can announce it.
A B C D
----------
Round 1: B B B B Everyone looks at each other and sees a brown
dot.
No one can ever leave.
Round 2: G B B B Person A looks around. As far as he is concerned,
he is in the exact same situation as above.
He can't ever leave.
Everyone else sees exactly one green dot.
They can't ever leave
Round 3: G G B B Person A looks around. He sees one green dot.
As far as he is concerned, he is in the exact
same situation as everyone else was above.
Just like they couldn't leave before, A can't
ever leave.
Person B is the same as person A
Everyone else sees two green dots.
They can't ever leave
Round 4: G G G B Person A looks around. He sees twos green dots.
As far as he is concerned, he is in the exact
same situation as person C and D was above.
Just like they couldn't leave, he can't ever leave.
Person B and C is the same as person A
Everyone else sees three green dots.
They can't ever leave
Round 5: G G G G Person A looks around. He sees three green dots.
As far as he is concerned, he is in the exact
same situation as person D was above.
Just like person D couldn't leave, he can't ever
leave.
Person B, C and D is the same as person A
They can't ever leave
The key is that, in round 2, Person A is in exactly the same boat as everyone else was in the previous round. Since they couldn’t leave then, he has no way that he can leave.
In round 3, Person A is in exactly the same boat as person B was before — he sees one green dot, just like everyone else saw one green dot the round before. Since person B couldn’t leave before, person A has no reason to leave.
…continue to on to
In round 5, Person A is in exactly the same boat as person D was before — he sees three green dots, just like person D saw before. Since person D couldn’t leave before, person A has no reason to leave.
If you play the game again, though, with someone saying that one person has green eyes, you get a completely new thing:
A B C D
----------
Round 1: B B B B [This scenario can't happen this time, since the
people are told someone has green eyes]
Round 2: G B B B Person A looks around. No one else has green
eyes.
Therefore he can leave.
Round 3: G G B B Person A looks around. He sees one green dot.
So he can't leave yet.
He knows that Person B could leave immediately
if B were the only green dot.
B doesn't leave immediately.
So A can leave the next turn.
…etc. Now we’re at the regular ol’ puzzle.
If you still think this is wrong, tell me what’s wrong with my description of the rounds above.
Here’s how it works for 4. You see 3 green-eyed people. If you have non-green eyes, then each of these people are seeing 2 green-eyed people and one non-green eyed person, If any one of these 3 people also had non-green eyes, and they don’t know their eye color, then the other two people would be looking at two non-green eyed people. If there are two non-green eyed people, the decision is easy for the remaining two green-eyed people. If the green-eyed person each sees doesn’t leave, then the other person also has green eyes. So, they can both leave. Because no one leaves, you know that everyone has green eyes.
I haven’t worked it out for bigger numbers yet, I’m a little skeptical that it keeps working, but the what the puzzle is asserting is that if any one person doesn’t have green eyes then someone else will be able to solve it, thus when no one can solve it, everyone has to have green eyes.
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That’s what I think too.
The induction explanation is correct, but the explanation of what the new information is and why the induciton argument only works after this information has been added to the “system” is usually bungled.
The video contains the statement, “Now, besides knowing that everyone else has green eyes, each prisoner also knows…” which seems to imply that they didn’t know that before (but they did).
True. I already mentioned this in my earlier post:
Before the “information” was added the situation is as follows:
For all i, (Everyone knows that)^i at least 100-i prisoners are green eyed.
And for i = 100 (Everybody knows that everybody knows that … everybody knows that at least 0 prisoners are green eyed). So, with the induction that everyone keeps explaining, the base case doesn’t work, because every recursion step needs one step of “knowing that everryone knows”.
When you tell them that “at least one person is green-eyed”, nothing changes except for that base case. After you tell them that there are green-eyed prisoners, everyone knows:
- at least one prisoner is green-eyed (well, they already knew)
- everyone knows that at least one prisoner is green-eyed (nothing new either)
- everyone knows that everyone knows that at least one prisoner is green-eyed (not new)
… - (everyone knows that)^100 that at least one prisoner is green-eyed. THIS IS NEW INFORMATION.
… and suddenly, the induction argument works. And starts working for everyone at the same time.
You’re just not correct.
For starters, let me just say that (every) (other) (published) (solution) agrees with me.
Now, that’s simply argument-from-authority, I realize it doesn’t hold any actual weight, but it should at least make you step back and go through the solutions published and really see if you understand what they’re trying to say.
My first “rounds”-based example above is (to me) the clearest example I could come up with the explain why, absent any external information, no one in a group of four people could ever leave, no matter how many of them have green eyes.
If you disagree with it, quote the line that you disagree and explain what’s wrong with it.
No, see, that’s exactly wrong.
You seem to think that if there were three non-green-eyed people and one green-eyed person, that green-eyed person could leave the first night. But how could he? He has no idea that there are any green-eyed people.
So now imagine that there are two green-eyed people and two non-green-eyed people. Each of the green-eyed people sees one other green-eyed person. You say that they both wait to see if the other one leaves the first night. But how could one person possibly leave the first night? If they were the only green-eyed person, they’d be in the exact same situation as the paragraph above, and couldn’t leave.
If I see one green-eyed person, and no one has said anything, there is no scenario in which that person would leave the first night, because they have no information to allow them to leave.
So, no matter what my eye-color is, waiting for the one green-eyed person to leave is not going to tell me anything new, because they cannot leave.
Therefore, I have no new information when I see that the green-eyed person is still there the next day. So I still can’t leave.
So two green-eyed people can’t ever leave.
The situation is totally different once someone announces that there is a green-eyed person. In that situation, if I see one green-eyed person I know that they can leave the first night if they see all brown eyes. If they’re still there the next day, I know it’s because I must have green eyes too.
I really can’t see any other way to explain this.
I didn’t realize what you were proving,my mistake. I was just explaining how the logic worked with the information. Since I thought what you were saying is that the puzzle didn’t work with four people, and I knew that it did, I didn’t feel the need to read through your proof. Of course, if no one knows their own eye color, there is no way for any individual to conclude that their eyes are green, without the information. If there is one person that doesn’t have green eyes, with the information that at least one person has green eyes, it starts a chain where each person tries the case that they don’t have green eyes. It starts at two, and if there are two, then a smaller amount of people will be testing three, and if there are three, then a smaller amount of people are testing four, until you get to the point where someone can make a decision. If everyone has green eyes, the chain never starts, so everyone knows when nobody can make a decision.
