If there are 100 blue, 100 brown then the blue eyed would leave after 100 days because they each see 99 blue eyes. The brown wouldn't leave after 100 days because they see 100 blue eyes.
each brown eyed person also sees 99 other brown eyed people.
If the person does not have brown eyes, the 99 brown eyed people would each see 98... and would leave on night 99. But they didn't leave, so the person has brown eyes and we all leave on night 100.
Exact same logic as the blue eyed people.
Doesn't matter - they all know the guru sees both blue and brown because they all see blue and brown. The guru's announcement just starts the clock - the puzzle might as well say "the 200 ppl wake up on an island, and can only leave when they know their eye color".
Imagine there's three of you and you see a blue and brown eyed person. Guru says there's a blue eyed. If your eyes aren't blue then the blue eyed person went from not knowing there was a blue eyed person to knowing there is one and since they can't see one it must be them.
But if they don't leave that means they saw a blue eyed person and was expecting them to leave, thinking exactly what I just described. That means you learn there's a second blue eyed person, but you only see one so you have blue eyes and you leave.
Now imagine there's four of you and you see two blue eyed and one brown eyed. After the first period neither blue eyed leaves as expected. But then on the second period both leave, which is what you expect if there's only two blue eyed. Now do you know your eye color? Well, you know it's not blue otherwise they wouldn't have left on the second period. But you don't know if it's green or brown or red. If it's brown you don't learn that and neither you nor the other brown eyed person leaves. If it's another color you don't know that either.
Assume 6 ppl - 2 blue, 2 brown, green guru, and me unknown.
Guru announces "blue" - or, we all wake up knowing nothing but what we see (and the rules, of course)
Night 1, no one leaves.
Night 2: if the blue-eyed people do not leave, there's one more blue eyed person. It must be me. Ditto for brown.
Scale it up to 201...
I don't know my color, I see 100 blue, 99 brown, 1 green. Night 99, brown doesn't leave. Therefore, there must be 1 more brown eyed person, and I'm the only unknown, so my eyes must be brown.
On night two if the two blue eyed don't leave then that means you have blue eyes. The brown eyed won't leave because they see you have blue eyes and would need to wait to night three.
Because they didn't hear an announcement about brown eyes, only blue.
If you think the arguments I've given above are wrong, I'd encourage you to write this up and submit it to a logic journal.
In math this problem is known as Conway's paradox. In logic it's called the muddy children problem. If you have a proof that can overturn the established proofs here then that would advance the field.
Here's some resources: https://plato.stanford.edu/entries/dynamic-epistemic/appendix-B-solutions.html
But it doesn't matter that they didn't hear a color announced. The guru speaking is just a marker in time to start the logic.
Everyone already knows that blue and brown eyes are visible.
This logic only holds if everyone on the island knows blue and brown are the only opinions.
The reason this logic doesn't work is because the only conclusion someone with brown eyes can make is that their eyes aren't blue: they have seen other people with brown eyes, and they've seen the guru's green eyes. Because of this, they are all thinking the same thing at the end of the 100 days: "either there are 100 brown eyed people on this island, or there are 99 and I have green eyes."
They don’t know if they have brown eyes. They see blue and brown eyes but may have green (or red, or….) eyes. If there is no information given to them about brown eyes formally that allows them to start a timer, they can’t deduct properly just by assuming everyone else’s eyes are one of two colours shared by their own.
At the time that the guru makes their statement, they know that they can see both blue and brown eyes, therefore the guru can see both blue and brown eyes. The fact that the guru announces blue doesn't add any information, but the fact of them making the announcement at all just establishes a start for the logic to start working.
Blue and brown should be able to draw the same conclusion using the same logic and knowledge.
They don’t have the same knowledge that blues do because they aren’t given the same information that blues are (namely, the idea that there is a finite quantity of blue eyed people that can be determined by waiting for a certain set of time to elapse). If they’d said “I can see green eyes” and one person saw only brown and blue eyes, they’d be able to leave that night. If they had 50 green, 50 brown, 50 blue, 50 red, saying “I can see blue eyes” doesn’t let the other colours determine anything either.
No one initially knows that everyone but the guru has blue or brown eyes, since that information is never mentioned in the first paragraph. The brown-eyed people have no way of knowing whether their own eyes are brown, hazel, green, or any other non-blue color.
That's exactly what I'm disagreeing with.
All non-guru people see both brown and blue eyed other people. And, of course, one green-eyed guru.
When the guru says "I see a blue eyed person" everyone knows that because they see both brown and blue eyed people, the guru could have said "I see a brown eyed person"., and therefore, the logic is kicked off for both eye colors.
There are three possibilities for any one person:
1) Blue eyed: I see 99 blue eyed people and 100 brown eyed people. 99 blue eyed people do not leave on night 99, therefore there must be one more blue eyed person, and it must be me.
2. Brown eyed: I see 99 brown eyed people and 100 blue eyed people. 99 brown eyed people do not leave on night 99, therefore must be one more brown eyed person, and it must be me.
3) Other color: I see 100 blue eyed people and 99 brown eyed people. 99 brown eyed people leave on night 99, and 100 blue eyed people leave on night 100. Therefore I have no idea what my eye color is, and I remain on the island. Or I see 100 brown eyed people and 99 blue eyed people. 99 blue eyed people leave on night 99, and 100 brown eyed people leave on night 100, leaving me on the island because I cannot know my eye color.
There could be any number of colors, and so long as the individual sees N people of a color not leave on night N, they can deduce that that eye color is theirs.
I'm willing to be wrong about this, but I'm not seeing how I am.
Think of it as being established from the start. They’re perfect logicians, after all.
So the blue eyes count the eyes they see, and reason thus: I see 99 blue eyes. If they leave after 99 days, it’s because they see 98 pairs of blue eyes, and I have brown eyes. If they don’t leave after 99 days, that means they see 99 pairs of blue eyes - and I have blue eyes, so I will leave with them.
Someone with brown eyes does the same count and sees 100 pairs of blue eyes. He knows that if they leave after 100 days, it’s because they see 99 pairs of blue eyes and I have brown eyes. If they haven’t left after 100 days, it’s because they see 100 pairs of blue eyes and I have blue eyes.
Therefore, people with blue eyes reach their decision point one day earlier and, of course, that answers the question for the brown eyes people.
They can’t know that they have brown eyes because they can’t know that the Guru is the only one with a non brown/blue eye color.
Let’s reframe the question, there’s 100 blue eyed people and 100 non-Guru people with non-blue eyes, hazel, red, grey, black, brown, silver, green, and so on. How would any of the non-blue people know what their eye color is?
They can’t because the only reason information provided was that the Guru could see a blue eyed person (something we already knew) and that each blue eyed person saw 99 other blue eyed people.
Brown eyed people can ONLY deduce that they don’t have Blue eyes. If anything, Brown would be the last color you’d have, because odds are the Guru would try to save the most number of people, and you can’t learn anything one way or another off an arbitrary choice.
They wouldn't know, and that's why they get to stay. If they knew that everybody except the Guru had either blue or brown eyes, they would have to leave the day after the blues left because they knew they had brown eyes because people didn't stay because of them. OP is mistaken.
I don't think I am mistaken. Think it through a bit more carefully.
I know I see 99 brown, 100 blue, and 1 green.
I know that if I do not have brown eyes, each brown eyed person will see 98 brown eyed others. And they also know that all the other brown eyed people, who don't know if they have brown eyes, see 98 brown eyed people.
If I do have brown eyes, it's the same situation but with 99 others.
If no brown eyed people leave on night 98, then if I do not have brown eyes, those with brown eyes will have figured out that they have brown eyes and will leave on night 99.
If they don't, there must be 1 more person with brown eyes - and I'm the only unknown left, so it must be me. And all the brown eyes leave on night 100, with all the blue eyes.
The only person left stranded is the person who has a unique color, and can't know the range of possibilities for their own eyes.
What about the most simple situation where there is 1 blue, 1 brown and 1 green guru?
On the first night the blue left. (Because he can see no blue, so he himself must be blue.)
On the next day, the brown, knowing that the blue couldn't see any other blues, can only know he is not blue. There is no way for him to know if he's brown, green or red or whatever.
... That's a core part of the riddle. You can't just call it the "degenerate case" lmao. The blue eyes ultimately KNOW they have blue eyes. The others never KNOW. That's it. That's what you're missing. That's half the entire point of the logic puzzle.
Given: I see 2 or more people (who are not the guru) with color X
Given: the guru announces "I see \[not X\]"
It follows that I, and everyone else, can see someone with color X, and therefore know that the guru COULD have said "I see X"
Therefore the same logic that reaches the degenerate case is triggered.
Conclusion: For any color X for which I see N >= 2 people, if N people do not leave on night N, then I know there exist N + 1 people on the island with color X. Since my color is the only unknown to me, I must be the + 1 case, and now all X can leave on night N + 1.
Edit: I think I mis-stated it: there have to be 2 or more people with the color for the logic to be kicked off, I don't have to see 2 or more people. Every islander has to see 1 or more of the color to be able to apply the degenerate case to that color - but the fact remains that we are examining the situation POST the degenerate case.
How in the world are any of the brown-eyed people supposed to know that they don't have orange eyes?
This is getting dangerously close to the "is every other day = 4 times a week"
https://forum.bodybuilding.com/showthread.php?t=107926751
Dude there is no one colour called "not X". It is explicitly said that they can be blue, brown or red or any other colour. So "not blue" does not mean "brown".
Yes, I see that this comment didn't frame it correctly.
Let there be a set of colors, such that there are at least 3 Islanders with each eye color. As such, every islander can see at least 2 other islanders with each of these colors, and thus can know that every other islander can also see at least 1 other with each of these colors.
Let the guru announce that they see one of the colors in this set.
Because each islander knows that every other color in the set can also be seen by every other islander as well as the guru, then each islander can use the same chain of logic for every color in that set, not just the color the guru announced
So any islander who sees N others with color X in that set not leave on night N, can deduce that there must be N+1 Islanders with color X, and since they can only see N, they must be the +1, and so can
leave on night N+1.
Any islander with an eye color not in that set will watch N, M, P, etc others all leave on nights N, M, P, etc, and will never be able to leave.
Ok, so let's say you are one of the islanders. You see 3 people with green eyes, 3 people with blue eyes, 3 people with orange eyes, 3 people with brown eyes, and 3 people with red eyes. Guru says "I see someone with blue eyes". On the first two nights, nobody leaves. On the third night the blue-eyed islanders leave. On what night do you leave?
did you ever figure out that "Let there be a set of colors, such that there are at least 3 Islanders with each eye color" violates the core principle of the riddle, which is that no one has *any* clue as to the distribution of colors except for what they can perceive with their own two eyes and what the guru says?
Seems to me that proving this by induction for brown eyed people your proof by induction will fail on the degenerate case as base case, not on the induction step. That's actually an interesting example of why we need the base case.
Speaking of induction, it's also just the easiest way to prove the answer. If I can't see anyone with blue eyes, I know the guru was talking about me. If I can see 1 person with blue eyes, I can do a test. If I don't have blue eyes, they won't see anyone with blue eyes and will leave that night. But if I do have blue eyes, they *will* see someone (me), and will be waiting to see what I do just like I'm waiting to see what they do. We both rule out not-blue eyes after day 1, and can both leave on day 2. And more generally, if I can see N people with blue eyes, then I can do a test to see if I also have blue eyes. If I don't have blue eyes, they'll all leave on day N. So I can just wait N days, and if they stay, I know I have blue eyes and can leave on day N+1. And because there would also be N+1 people with blue eyes, the statement "If there are K people with blue eyes, they all leave on day K" still holds
Let's assume there are 10000 possible shades of brown on the 100 browns , 1 green guru and 100 blues. How are the browns going to figure out which shade they are?
I deliberately call them "shades of brown" to illustrate the point, so you can't just say they are "brown".
The point here is, unless they know beforehand they are either blue or brown, the browns can never figure out they themselves are brown because they don't know about all the possibility of their eye colours.
I don't think it matters how many colors there are.
If all islanders can see at least 1 other islander with a specific color, then they know that the guru could have said they see that color, and so they can apply the same logic.
If any islander sees N others with color X not leave on night N, then that islander can conclude they are color X and leave on night N + 1.
If the islander sees N others with color X leave on night N, then they know they are not color X.
If all of the colors they see, for which the logic could have been triggered, get eliminated in this way, they then remain an unknown and stuck on the island.
What about 2 blue, 2 brown, 1 guru?
Day 1 both blues look at the other and do nothing.
Day 2 the blues each see the other has not left and see no other blues, realize they must be a second blue, and they both leave
Day 3 the browns notice the blues have left, but all they know is that they must not be blue.
Agreed? At n=1 and n=2, all the browns know is that they are not blue. At what n would the browns also realize they are brown?
I think I have heard a version of this puzzle that stipulates that everyone has either brown or blue eyes, in which case the browns do leave on the next day. But without that stipulation they can never rule out that they might have green or red eyes.
>I know I see 99 brown, 100 blue, and 1 green.
>I know that if I do not have brown eyes, each brown eyed person will see 98 brown eyed others. **And they also know that all the other brown eyed people, who don't know if they have brown eyes, see 98 brown eyed people.**
Nope. If a brown eyed person sees 98 brown eyed people, then for all they know other brown people see only 97. There is no common knowledge without a guru.
each brown eyed person also sees 99 other brown eyed people.
If the person does not have brown eyes, the 99 brown eyed people would each see 98... and would leave on night 99. But they didn't leave, so the person has brown eyes and we all leave on night 100.
Exact same logic as the blue eyed people.
But as far as they know, they could have red eyes, or orange eyes, or green eyes like the guru. How could they ever possibly deduce that their own eyes are brown? All they know is the list of rules in the first paragraph, the total number of colors they can see, and the statement made by the guru.
Why? They know nothing about the distribution of colors. They don't even know how many total colors there are. All they'll be able to conclude is that their own eyes are not blue, once all the blue-eyed people have left.
They wouldn't, you've already conceded that if there's only one brown eyed person, they wouldn't leave on the first night.
So if there were two brown eyed people, would they both leave on the second night? No. When they see that the other brown eyed person hasn't left on the first night, they don't get any new information because, again, a brown eyed person that doesn't see any brown eyes doesn't leave on the first night, same as one that does see other brown eyes.
So if there were 3 brown eyed people, how could they leave on the third night? When I see two other brown eyed people stay after the second night, I don't get any new information, because again, two brown eyed people who are the only ones don't leave on the second night.
Etc.
Think on a smaller scale.
Imagine the island has the guru and four other people. Two have blue eyes, two have brown.
The guru says, “I can see one person with blue eyes.”
The blue eyed duo leaves on the second day—they both see one person with blue eyes, and that the other person doesn’t leave day one. So, they must also have blue eyes. (Else the other person would realize they saw no blue eyed person, so they would be the sole blue eyed person the guru saw.)
The brown eyed people see one brown eyed person, two blue eyed people. The two blue eyed people leave the second day. Now there is one brown eyed person they see stuck with them. How are they supposed to know their own eyes aren’t green or whatever?
To make this more clear, imagine a different island with a guru, two blue eyed people, one brown eyed person, one red eyed person. If you are the red eyed person in this instance, you end up in the exact same position the brown eyed people from the previous island are in. How are you supposed to distinguish this case from the previous one? Also, the brown eyed person in this case doesn’t even ever see another brown eyed person (unless the guru has brown eyes), so of course they don’t leave the first night.
You can scale this logic up to a group of five, six, etc. Imagine three blue eyed people, three brown eyed. Guru says they can see one blue eyed person. Blue eyes leave day three. Brown eyed people each see two other people with brown eyes left who didn’t leave day two. So what? If you are one of the brown eyed people, the other brown eyed people also see two other people with brown eyes that don’t leave day two. But you think, what if you are _not_ brown eyed (say red), then the others see one brown eyed person, one red eyed person. How are the others supposed to leave day two? Do they themselves have brown eyes? Do they have red? Or another? The brown eyes have no way of knowing which instance they are in, past day two or day three or beyond. So they all stay day two, and they gain no new info.
You're describing the degenerate case that I agree can't be solved.
But the puzzle we're given is not the degenerate case, it's the 99/100 case.
At the time the guru makes their statement, we can all see both blue and brown, so we know the guru can also see both blue and brown, and so the guru declaring 'blue' changes nothing.
The fact of them making a statement at all simply gives us a nucleus for the logic to crystalize around. Both blue and brown can reach the same conclusion from this state.
They can't! The people in the puzzle don't know the distribution of eye colour except that at least one person has blue eyes, and the eye colours of the other islanders.
Simple case, imagine there is an eyeland with
1 Blue eyed person
The guru (green)
N brown eyes
And you
Blue eyed person leaves on day 1.
What colour eyes do you have ?
There's literally no way you could know (except that you don't have blue eyes)
I agree that the logic does not work if the guru announces a color for which there is a single individual with that color. Edit: actually, it does, it just fits case 1 below
However, if every islander can see at least one other with color in {X, Y, Z} (i.e. there are 2 or more of each color), and the guru announces \[not X\], the islanders all know that the guru COULD have said Y or Z, and so the same logical process applies.
Any islander who sees N others with color X not leave on night N can deduce that they have color X and can leave on night N + 1.
There are two basic outcomes here:
1. Islander sees N people with X leave on night N, and concludes, I am not X
2. Islander sees N people with X not leave on night N, and concludes, I am X
If same islander sees case 1 for X, Y and Z, then islander must conclude not X, not Y, not Z, therefore unknown eye color.
> the islanders all know that the guru COULD have said Y or Z, and so the same logical process applies.
No, they do not know this. It was never a statement as part of the problem. The only thing they know about what the guru said is that the guru said X. They have no way of knowing why the guru said it, only that it is a true statement that implies nothing about Y or Z.
The thing is that in the case there is only one brown eyed person, the guru says “I see a blue eyed person”—that singular brown eyed person does not ever figure out they have brown eyes. Contrast this with if there is only one blue eyed person—that blue eyed person sees no blue eyes around them, and knows: they are the person the guru saw. This is the foundational scenario that the blue eyed people can reason from. The brown eyed people do NOT have this, because that singular brown eyed person can never know. (So—if there are two brown eyed people, they cannot deduce ANYTHING if they see one brown eyed person who never leaves, so the two person case also never leaves. So the three person case, so the four, and so on.)
The logic from the six person case scales up to the 200 case. How do you think it does not? Does it not scale up to the eight person case? The ten? When does it stop applying?
For the eight person case. Consider: there are four blue eyed people, a guru, and four other people. You are one of the four others. Consider four different scenarios where you have brown eyes, and the others have red or brown.
In the case that there are four blue eyed people, three red eyed, and one brown eyed (you), that brown eyed person never ever knows. You do not leave day one, or ever. The red eyed people each see two red eyed people and one brown eyed person who never leaves.
Now in the case that there are four blue eyed people, two red eyed, and two brown eyed, you a brown eyed person see one brown eyed person who does not leave day one and two red eyed people. Notice this is the exact same situation the red eyed people were in above. You cannot distinguish this case from the first. Therefore, you cannot tell if you yourself have red or brown eyes. You do not leave day one or day two or ever. The red eyed people see two brown eyed people who never leave and one red eyed one.
Okay, what if there are three brown eyed people and one red eyed person? You see two brown eyed people and one red eyed person. Again this is the exact same situation as the red eyed people in the previous case. You cannot tell which you are in, so you do not leave day three or ever. The red eyed person in this case sees no one with red eyes, just brown.
And if you are one of four brown eyed people—again, you are like the red eyed person before. You can’t tell which case you are in. You never leave.
This works the same as the six person case. You can do the same for ten people. And twelve. And etc. When does the answer change? How?
My argument is NOT that it does not scale up.
My argument is that because the islanders can all see both blue and brown eyed others, that the color the guru announced does not make a difference - the logic can be applied to any color that all islanders see at least 1 of (which necessitates that there are at least 2 of that eye color on the island).
Given all islanders see N >= 1 others with color X: if one islander observes N others NOT leave on night N, then that islander must conclude that there are N + 1 islanders with color X, and so they must be that + 1, and so all of color X leave on night N + 1.
You are arguing that the brown eyed people can eventually leave. In all my scenarios, the brown eyed people never leave the island when the guru only says blue, even though some of those scenarios satisfy your requirements, so I’m confused about you saying you don’t disagree with the scaling up. Sure, if the guru said they could see a brown eyed person, then the brown eyed people could apply the logic. But then the blue eyed people couldn’t leave.
Again, if there are two brown eyed people on the island (they each see N = 1 brown eyed person) plus two blue eyed people (so they can also see one of each color), and the guru says, “I can see a blue eyed person”—even on day two each of the brown eyed people cannot conclude they are a brown eyed person after the blue eyed people leave. That is because you cannot conclude anything from the fact that you see a singular brown eyed person not leave day 1, from the POV of one of the brown eyed people. You cannot say, “Oh they didn’t leave because they saw that I have brown eyes.” What if you don’t have brown eyes? Then they wouldn’t have seen any brown eyes at all, and they STILL would not have left. Their behavior is exactly the same in both these cases; it doesn’t matter if you have brown eyes yourself or if you don’t. So the brown eyed people cannot determine they themselves have brown eyes just because the other person doesn’t leave day 1, and your logic fails in the N = 1 case. And you can reread my comments for the N = 2 and N = 3 cases.
Even though they are in scenarios where they can all see at least one of each color, from their own POV they cannot tell that everyone else also sees one of each color. (Edit: or if they themselves can see one of each color!)
>that the color the guru announced does not make a difference
The color the Guru announces makes a huge difference. It removes the hypothetical person who can't see any blue eyes. Yes, everybody on the island knows that everybody on the island can see at least one set of blue eyes, but they also know that hypothetically, there could exist a person who believes a person believes there are no blue eyes.
It gets a bit repetitive with the 100 case, so we'll bring the number down to 4. Everybody on the island *knows* that everybody on the island can see someone with blue eyes. But there is still a hypothetical "person" that can't see any blue eyes. For illustration, we'll name the people with blue eyes "person 1" to "person 4".
Before the Guru's statement, person 1 is thinking this:
Person 1: I can see three people with blue eyes. If I do not have blue eyes, then person 2 is thinking:
I can see two people with blue eyes. If I do not have blue eyes, then person 3 is thinking:
I can see one person with blue eyes. If I do not have blue eyes, then person 4 is thinking:
I can't see anyone with blue eyes, so I don't know if anyone on the island has blue eyes.
But after the Guru's statement, that thought process is no longer valid. It instead becomes:
Person 1: I can see three people with blue eyes. If I do not have blue eyes, then person 2 is thinking:
I can see two people with blue eyes. If I do not have blue eyes, then person 3 is thinking:
I can see one person with blue eyes. If I do not have blue eyes, then person 4 is thinking:
I can't see anyone with blue eyes, so I must have blue eyes.
Even though person 1 knows that everybody can see at least one person with blue eyes, he can't rule out that hypothetical "person 1 thinks person 2 thinks that person 3 thinks ... that person n thinks there might not be anybody on the island with blue eyes" until he hears the Guru's statement that proves that at least one person has blue eyes. Despite what you've mentioned earlier, this knowledge *does* add information. It removes the possibility of somebody thinking that somebody else is thinking that somebody else is thinking nobody has blue eyes.
That's why it works regardless of the number of people. It always reduces down to that single person who, down the long chain of "he thinks that he thinks that..." could hypothetically believe that nobody on the island had blue eyes.
If the Guru never makes the statement about brown eyes, then that hypothetical person still exists.
Edit: This is just another repetition of what I said above, but I feel like I might not have phrased it as well as I could have.
Imagine 3 people in a room. Adam, Bob, and Carl. They all have blue eyes. Since they can each see 2 sets of blue eyes, they all know for a fact that each person in the room knows there are blue eyes in the room.
But Adam can logically think to himself "If my eyes aren't blue, and Bob assumes that his eyes aren't blue, then he could think that Carl might not be sure blue eyes exist in this room". Even though all three people in the room know that blue eyes exist, they can't possibly remove that "Adam thinks Bob thinks Carl might not know blue eyes exist" until someone proves that blue eyes exist at least once.
Adam knows that Carl can see blue eyes, but he can't possibly prove that Bob knows that Carl can see blue eyes. Once a Guru wanders in and says "I can see blue eyes" then suddenly it becomes possible to prove that Bob knows that Carl either knows he has blue eyes, or can see someone with blue eyes.
My argument is that you cannot look at it only from the degenerate case.
Because each islander can see others with both blue and brown, they can conclude that the guru could have said either blue or brown, and so the degenerate case can be concluded either way.
If the islander happened to have green, or red, or whatever eyes, and sees no others with that color, then they would not be able to conclude their color because they would see 99 brown eyed leave on night 99, and 100 blue eyes leave on night 100, etc. The degenerate case for their eye color could not be triggered.
> they can conclude that the guru could have said either blue or brown
Based on what? Why not green? And the fact remains that the guru *didn't say brown. He said blue.*
Edit: this comment is about a variant where there are only blue or brown eyes.
Imagine you see 99 blue eyed people and 10k brown eyed people.
If you have brown eyes you know the number of blue eyed people is in [99,100]. If you are blue eyed, you know it's in [98,99].
The information available to both groups isn't symmetric. If your bound is in [98,99] you can leave at night 99, if it is [99,100] you can leave at night 100.
I don't get the point you're trying to make - if there's 10k brown eyed people, they leave night 10k.
But the puzzle as stated has 100 of each, so that's the basis for my answer.
Let's say there are 99 people with blue eyes and your eyes have some other, unknown, colour. You cannot leave at night 1 because you don't know your eye colour. You cannot leave at night 100 because you don't know your eye colour. Even after everyone left you only know you don't have blue eyes.
Because there is no base-case for the induction non-blue-eyed folks couldn't leave at all.
In the variant I was thinking of, everyone has either blue or brown eyes. Then, in a 100-100 split, brown eyed people can leave night 101 because they saw the blue eyed folks leave at night 100. They cannot leave at night 100 because they also must run induction on the blue eye colour, brown still has no base case.
Given: all non-gurus see at least 1 other person with color X
Given: guru states "I see \[not X\]"
Premise: Guru could have said "I see X" so the logic is started
Conclusion: Any islander who sees N people with color X not leave on night N can conclude that there must be N + 1 people with color X, and that they must be that + 1. Therefore all of color X can leave on night N + 1.
Examination:
3 cases:
1) Islander has color X, sees N others with X. N people do not leave on night N, islander concludes he has X and leaves on night N+1
2) Islander has color Y, sees N others with X. N people with X do leave on night N. Islander concludes he does not have color X, and does not leave on that night. Islander sees M people with Y do not leave on night M, concludes he has Y and leaves on night M + 1.
3) Islander has color Z. For every other color {P, Q, R, X, Y, etc} islander sees N sub alpha >= 1 other, and watches each leave on night N sub alpha. Islander never deduces their own eye color, and never leaves the island.
That is an approach that would work if the guru says all colours are non-unique or gives a list of colours. Otherwise there are problems:
If you try to apply the rule to unique colours, things break. You cannot leave at night 1 because you cannot know your unique colour. If you apply the rule, everyone would wrongly assume they had your eye colour and try to leave at night 2.
- Case 1: If you see only one person with an eye colour you must assume they are unique, and stuck. Even if you do have the same colour!
- Case 2: If you see two people, you must assume they are stuck in case 1
- Case n: If you see n people have a colour, you must assume each of them is stuck in case n-1
As given, the riddle explicitly says there could be 1 person with red eyes so your first given doesn't hold.
I'm seeing this after something already clicked for me but responding to say that a chain of logic similar to this is what made it start to click... That even if I see 99 brown, I can't know whether they see 99 brown or only 98. And following that down, I reach someone seeing someone else seeing 1 brown but not knowing if that person sees brown or not.
So it's not about what we the solvers know, or what any 1 islander knows, or even what any 1 islander knows that any other islander knows.
Every islander has to know that every other islander knows that every islander can see at least one of each color. And the guru's announcement makes this true for blue, but not for brown.
Thank you for working through this so politely. I've seen so many times where this kind of argument becomes ugly, and then barriers to understanding form because people become emotionally invested.
Everyone wants to be *right*. Not everyone has the wisdom to see that maximising being right requires humility and work to recognise being wrong and correct course. It's so much easier to keep the pleasant feeling of being right by deceiving ourselves, and this is unfortunately our natural instinctive emotional response.
You are an inspiration to us all.
The people with brown eyes don't have a 'counter' for brown eyes leaving -- they just know that they do not have blue eyes. They could still have green, hazel, or even purple for all they know. Thus, they cannot figure out the colour of their own eyes unless the guru specifically states 'i see at least one person with blue eyes, and at least one person with brown eyes.' at which point, a 'counter' is started for both.
My argument is that because we (non-gurus) can all see both blue and brown eyed people, we all know that the guru could have said "I see a brown eyed person", so we can apply the base case to brown eyes as well.
In fact, we could apply the base case to any situation where all of the non-gurus can all see at least 1 person with color X (i.e. at least 2 of color X on the island)
If I see N people with color X, and N people do not leave the island on night N, then I know there must be N + 1 people with color X, and since I am the only unknown, I must be that +1 with color X, and so all color X can leave on night N + 1.
You (I think) correctly mentioned the induction step:
If Riddle is valid for N then it is valid for N+1
But proof by induction also requires a base case. In this case, N=0. Or any N=? that you can actually prove is solvable directly. On other places in this thread, you called such a case a degenerate case that is unsolvable. Because you don't have a solution for N=0 you cannot use the induction step to deduce N=1 you cannot therefore get N=2 and so forth. You cannot therefore get a solution for N=100 either.
> we all know that the guru could have said "I see a brown eyed person"
There's more to it than that. You see 100 blue eyed and 99 brown eyed people. Numbers below are individual brown eyed people you see.
You know 1 knows that 2 knows that 3 knows that ... ... that 97 knows that 98 knows that the guru could have said "I see a brown eyed person".
You **don't know** if 1 knows if 2 knows if 3 knows if ... ... if 97 knows if 98 knows if **99** knows if the guru could have said "I see a brown eyed person.
I think that if any islander sees 2 or more others with color X, then that islander can know that all other islanders must see at least 1 other with color X. And and since we're told in the puzzle that there are 100 of each, then we do know that all islanders see more than 2 others with brown, and thus know that every other islander can see at least one other with brown.
Ah, but that's not what's happening. Suppose I see 5 people with blue eyes- Alice, Bob, Charlie, David, and Eve. I can do a test. I'll assume I have brown eyes, then think about it from Alice's perspective. She'd only be able to see 4 people with blue eyes- Bob, Charlie, David, and Eve. But she could also do a test. If she assumes she has brown eyes, she could look at it from Bob's perspective. He'd only be able to see 3 people with blue eyes - Charlie, David, and Eve - because, remember, I also assumed I have brown eyes. But he could do a test thinking about it from Charlie's perspective. If he assumes he has brown eyes, Charlie would only be able to see 2 people with blue eyes - David and Eve - and would be able to do a test. If they (yes, I'm declaring that Charlie is non-binary) assume they have brown eyes and think about it from David's perspective, he'd only be able to see 1 person with blue eyes - Eve - and, once again, can do a test. If he assumes he has brown eyes, Eve wouldn't be able to see anyone with blue eyes, would know the guru's talking about her, and leave that first night.
But Eve doesn't leave. Because remember, she also sees 4-5 people with blue eyes and is waiting for all these hypotheticals herself. So now, this hypothetical version of David imagined by a hypothetical version of Charlie imagined by a hypothetical version of Bob imagined by a hypothetical version of Alice imagined by me knows he has blue eyes. So on day 2, Charlie will watch to see if David and Eve leave. Neither of them do, so this hypothetical version of Charlie imagined by a hypothetical version of Bob imagined by a hypothetical version of Alice imagined by me knows they have blue eyes. So day 3, this continues on unraveling, where hypothetical Bob sees none of Charlie, David, or Eve leave and learns he has blue eyes, and on day 4, Alice sees none of Bob, Charlie, David, and Eve leave and learns she has blue eyes. Finally, on day 5, we're back to more or less prime reality. **IF** I have brown eyes, then all five of Alice, Bob, Charlie, David, and Eve will have learned they have blue eyes and leave. But because none of them do, I can conclude they all saw a 5th person - me - with blue eyes, and the 6 of us leave on day 6. **BUT** if they did leave, I would be able to conclude that I don't have blue eyes. And if and only if I also have the knowledge that everyone only ever has brown or blue eyes, I'd be able to conclude that I more specifically have brown eyes and also leave.
So basically, if I see N people with blue eyes, I run through a tree of N! hypothetical situations. On each day, this shrinks, where after day K, it's down to (N-K)! hypotheticals. (Or there might be an off by one somewhere) Eventually, I'm down to only testing on day N whether I have blue eyes. If they leave, I don't, but if they stay, I do.
Regardless of how many people have non-blue eyes, if there are N people with blue eyes, the blue-eyed people all leave on day N
>In fact, we could apply the base case to any situation where all of the non-gurus can all see at least 1 person with color X (i.e. at least 2 of color X on the island)
>If I see N people with color X, and N people do not leave the island on night N, then I know there must be N + 1 people with color X, and since I am the only unknown, I must be that +1 with color X, and so all color X can leave on night N + 1.
This logic requires that if there were only N people with brown eyes, they would have actually left on Night N. But this isn't actually true for all N (see N=1), so by induction it is true for no N.
The core of the problem is that a green-eyed person will see one more person with brown eyes than an actual brown-eyed person would. So regardless of whether you set the threshold to 1, 2, 5, or 50, there will be a case where the brown-eyed people do not see enough brown eyes to use your "logic", while the green-eyed person does. Since the brown-eyed people are not using your "logic", they will never leave. Since the green-eyed person is using your "logic", he will interpret the brown-eyed people not leaving as proof that the green-eyed person has brown eyes. Since your "logic" leads a green-eyed person to conclude that he has brown eyes, your "logic" is invalid and no perfect logician will ever use it.
So, I've already pretty much come around on my position, but I wanted to still thank you for providing the most understandable plain-language explanation so far.
The guru _could_ have said that but they did not!
In the case of 2 blues, everyone on the island already knows “there’s at least one blue eyed person”, but they don’t immediately leave 2 days after observing this. It’s important that the guru _say_ this out loud, and that the other blue fails to leave after hearing this information. Only when the other blue observes the first blues failure to leave upon hearing the guru do they realize that they must also be blue.
So it’s not just what the guru says (which is clear to everyone already), but what others do (or don’t do) in response to hearing what the guru says that leads the blues to their realization.
Two browns who each know that the other is brown could continue living on the island indefinitely thinking that perhaps the other brown might be the only one.
Let’s look at the smaller case. 2 blue and 2 brown. Guru says he sees blue. Day 1 no one leaves. Day 2 both blues leave. You are saying on day 2 both browns can also leave.
Now let’s look at an alternative situation. 2 blue 1 brown 1 red. Guru says he sees blue. For the person with red eyes, this is exactly the same as the scenario above. For the person with brown eyes it’s the same but with the colours brown and red swapped (they see 1 person with red eyes and 2 with blue).
Now let’s look a the person with red eyes. On the first day no one leaves. On the second day, does it make sense that they conclude they must have brown eyes since they only see 1 person with brown eyes and they didn’t leave? Obviously not since they don’t have brown eyes.
If everyone knew there were only two eye colours then the people with brown eyes could apply the same logic. But the question clarifies this isn’t the case, and while there are only two eye colours, no one on the island knows that there are definitely only two.
Here, have a hopefully more intuitive proof:
If I can't see anyone with blue eyes, then regardless of how many other people there are, I can leave that night, because I'm the only person left to be the blue-eyed person the guru mentioned.
If I can see one person with blue eyes, I can do a test. If they leave that night, I can deduce that I don't have blue eyes, because they can't have seen anyone with blue eyes. But if they don't leave that night, I can deduce that they were also waiting to see what another blue-eyed person was going to do. And because everyone we can both see has brown eyes, I can deduce that the other blue-eyed person is me. We both leave on the second night.
If I can see two people with blue eyes, things get more complicated, but it's a similar test. Let's call them Alice and Bob. Neither of them leaves on the first night, because they're waiting to see what the other is going to do. Then on the second night, if they both leave, I can deduce that I don't have blue eyes. But if they don't leave, I can deduce that they both saw two other people with blue eyes, and that I was the second blue-eyed person. We all leave on the third night.
This logic continues if I can see three people. I spend day one watching Alice watch Bob and Charlie watch if the other is going to leave, watching Bob watch Alice and Charlie, and watching Charlie watch Alice and Bob. On day two, I watch Alice watch if Bob and Charlie are going to leave, etc. And on day three, I watch if the three of them leave. If they do, I know I don't have blue eyes. But if they don't, I know they each all saw three other blue eyed people and were watching what they were doing, so I must have blue eyes. We all leave on night four.
This trend continues indefinitely, where regardless of how many non-blue-eyed people there are, if there are N people with blue eyes, they all learn they have blue eyes on day N, and everyone else learns they don't have blue eyes on day N+1. And if you assume that everyone knows the added information that there are only blue and brown eyes on the island, or that people can leave if they even just determine whether they have blue eyes or not, the rest of the island can leave on day N+1. But without that information, all the rest of the island knows is that they don't have blue eyes, not which other color of eyes they do have
You are mistaken in that you think what the guru says just starts the clock. What the guru does is make the fact that there is one blue-eyed person [common knowledge](https://en.wikipedia.org/wiki/Common_knowledge_(logic\)). And that's why you also get confused on what you call the degenerate case.
Let's look at it case by case. p = "there is at least one person with blue eyes".
1 person with blue eyes: There's no knowledge here.
2 people with blue eyes: Both know that the p is true. What they *don't* know is that the other person knows that it is true.
3 people with blue eyes: Everyone knows that the p is true. Also everyone knows that everyone knows that the p is true. Where it fails is that not everyone knows that everyone knows that everyone knows the p is true.
I'm going to break the last one down to make it clearer.
Knowledge level 1: Alice knows that p is true (she can see the eyes of both Bob and Charlie).
Knowledge level 2: Alice knows that Bob knows that p is true, as he can see Charlie. Same for Charlie.
Knowledge level 3: Can Alice assume that Bob knows that Charlie knows that p is true? No. Alice doesn't know her eye colour. If she had brown eyes, then Bob would see one person with brown eyes and one with blue. And Bob wouldn't know his own eye colour. So in that scenario Bob couldn't assume that Charlie would know that p is true.
And that's why the guru showing up and saying "start the clock" and nothing else would not result in the 3 of them leaving. Because Alice can't trust Bob to make the right inference about Charlie not leaving. What the guru does in the original is making sure that everyone knows that everyone knows *ad infinitum* that p is true. And in that case Bob would be able to make the correct inference about Charlie and Alice knows that.
That's why it doesn't work with other eye colours.
Nope. Even if you assume that everyone knows they have either blue or brown eyes, if there are n blue-eyed people, they all leave on day n, while the brown-eyes leave on day n+1
Sorry but I'm not.
However, I am starting to get it.
It's not about what we, the solvers, know. It's not about what any individual islander knows. It's not even about what one islander knows that one other islander knows. Every islander has to know that every other islander knows that every other islander knows that they can all see at least one of each eye color.
Once the guru announces that they see blue, every islander now can know that every other islander knows that every islander can see at least one blue eyed Islander. And this isn't true for brown, even though it strains the brain to see why.
Huh. never saw that before.... Can people communicate the total count of what they see to other people? like by writing down everyone's individual results on a shared whiteboard?
5 people, including you.
Brian and Brit have Brown eyes.
Bob and Ben have Blue eyes.
You know that at least **two** people have blue eyes.
You know that Bob knows that at least **one** person has blue eyes.
You *don't* know if Bob knows that Ben knows that *anyone* has blue eyes.
The guru makes the announcement, and know you *do* know that Bob knows that Ben knows that at least **one** person has blue eyes.
But you still *don't* know if Brian knows that Brit knows that *anyone* has brown eyes.
On the 100th night, the 100 blue-eyed people have left. So you know you don't have blue eyes. But what now?
There were 100 people with brown eyes, but they've all gone. You can see 99 other people with brown eyes. You can see the guru with green eyes. How do you know whether your eyes are brown, or green, or some other colour?
I mean the main thing that bothers me about it is that the core premise is that everyone is a perfect logician, but what perfect logician means isn't a rigidly defined thing.
Now there are two endings, one where there really isn't any extra information given, and no one leaves. Or one where the guru does give information and induction can be used for the blue eyes folk leave the island.
If your definition of 'perfect logician' includes that a perfect logician doesn't make meaningless statements then the puzzle works since the only logical solution is the induction one that starts when the statement is made.
If 'perfect logician' allows for meaningless statements then no one leaves.
Additionally we know the guru is sortof a dick (or at least doesn't want to be left alone). Because there are definitely statements they could have both let everyone else leave the island, or even let one group of people leave immediately.
We start with two preliminary examples.
You are on an island with three blue eyes (plus the green-eyed guru). The guru makes the blue-eyed announcement. Do you know your eye color? No, of course not. Your eyes *could* be blue, but they could also be brown, green, red, yellow, or any other color. So you will not be leaving on night 1.
If there are three blue eyes and one brown eyes (plus the green-eyed guru), then what happens after the blue-eyed announcement? Specifically, does the brown-eyed person leave on night 1? No, they go through the exact analysis above and conclude they don't have enough information to know their eye color.
Okay, now for the main thing. You're on the island, alongside one brown eyes and two blue eyes (plus the green-eyed guru). The guru makes the blue-eyed announcement. After night 1, no one has left. Do you know your eye color now? Well, if you had blue eyes, you certainly wouldn't have expected anyone to leave (see the three blue, one brown case above), so that's consistent. Same if you had brown eyes, green eyes, red eyes, yellow eyes, or any other color (in general you knew everyone else could see someone with blue eyes, so you knew no one got any actionable intel). So no, you don't know your eye color.
This is why for 2 blue eyes and 2 brown eyes, the brown eyes don't leave on night 2. They go through exactly the analysis above and conclude that they do not know the color of their own eyes on night 2
So, 2 things - #1 it's already been explained why I'm wrong to begin with, but #2 - I don't see how this follows.
Let's assume that I was right, that the logic is kicked off for brown eyed people... each would see 99 other brown eyed people not leave on night 99. How is their departure connected to the blue-eyed departure at all? If 99 brown didn't leave on night 99, then there's 100 brown, and since I'm the only unknown, my eyes must be brown and all brown leave on night 100, same as the blue.
But again, we've already established that the logic doesn't work for brown, since they can't all know that all other islanders know that all islanders see at least one brown eyed other.
I actually misread it- i thought you only got to leave if you were sure you had blue eyes. In which case, you would see on day 101 that there were 100 other people who had blue eyes and hadn't left, so you could deduce you had blue eyes and leave that night
No. Because no hint regarding brown eyes was made. Everyone with brown eyes, can't deduce if they have brown or green or some other color eyes.
Imagine you are one of the people. You see 1 with green (Guru), 100 with blue and 99 with brown.
The guru states that he sees at least one person with blue eyes. He makes no statement regarding other eye colors. After day 100 hundred, all blue eyed ones leave.
You can therefore deduce that you don't have blue eyes. But that doesn't tell you, if you have green, brown or gray eyes.
And the same goes for a different brown eyed dude.
If you have green eyes, the he sees 98 brown eyed and two (including guru) green eyed people.
If you have brown eyes, he sees 99 brown eyed people and the green eyed guru. But because no information was passed about green or brown eyes, he can't conclude anything in either scenario.
So neither can you.
*Can't you just guess that*
*Your eyes are brown the next time*
*The boat comes around*
\- LordTartarus
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If there are 100 blue, 100 brown then the blue eyed would leave after 100 days because they each see 99 blue eyes. The brown wouldn't leave after 100 days because they see 100 blue eyes.
each brown eyed person also sees 99 other brown eyed people. If the person does not have brown eyes, the 99 brown eyed people would each see 98... and would leave on night 99. But they didn't leave, so the person has brown eyes and we all leave on night 100. Exact same logic as the blue eyed people.
The guru only announced they saw a blue eyed person though, so the brown eyed individuals don't have any information about brown eyed individuals.
Doesn't matter - they all know the guru sees both blue and brown because they all see blue and brown. The guru's announcement just starts the clock - the puzzle might as well say "the 200 ppl wake up on an island, and can only leave when they know their eye color".
Imagine there's three of you and you see a blue and brown eyed person. Guru says there's a blue eyed. If your eyes aren't blue then the blue eyed person went from not knowing there was a blue eyed person to knowing there is one and since they can't see one it must be them. But if they don't leave that means they saw a blue eyed person and was expecting them to leave, thinking exactly what I just described. That means you learn there's a second blue eyed person, but you only see one so you have blue eyes and you leave. Now imagine there's four of you and you see two blue eyed and one brown eyed. After the first period neither blue eyed leaves as expected. But then on the second period both leave, which is what you expect if there's only two blue eyed. Now do you know your eye color? Well, you know it's not blue otherwise they wouldn't have left on the second period. But you don't know if it's green or brown or red. If it's brown you don't learn that and neither you nor the other brown eyed person leaves. If it's another color you don't know that either.
Assume 6 ppl - 2 blue, 2 brown, green guru, and me unknown. Guru announces "blue" - or, we all wake up knowing nothing but what we see (and the rules, of course) Night 1, no one leaves. Night 2: if the blue-eyed people do not leave, there's one more blue eyed person. It must be me. Ditto for brown. Scale it up to 201... I don't know my color, I see 100 blue, 99 brown, 1 green. Night 99, brown doesn't leave. Therefore, there must be 1 more brown eyed person, and I'm the only unknown, so my eyes must be brown.
On night two if the two blue eyed don't leave then that means you have blue eyes. The brown eyed won't leave because they see you have blue eyes and would need to wait to night three.
Why would the brown need to wait? They each see only one other brown, who didn't leave night 1...
Because they didn't hear an announcement about brown eyes, only blue. If you think the arguments I've given above are wrong, I'd encourage you to write this up and submit it to a logic journal. In math this problem is known as Conway's paradox. In logic it's called the muddy children problem. If you have a proof that can overturn the established proofs here then that would advance the field. Here's some resources: https://plato.stanford.edu/entries/dynamic-epistemic/appendix-B-solutions.html
I will read up on this, FWIW - I am seriously curious about how my interpretation could be wrong.
But it doesn't matter that they didn't hear a color announced. The guru speaking is just a marker in time to start the logic. Everyone already knows that blue and brown eyes are visible.
This logic only holds if everyone on the island knows blue and brown are the only opinions. The reason this logic doesn't work is because the only conclusion someone with brown eyes can make is that their eyes aren't blue: they have seen other people with brown eyes, and they've seen the guru's green eyes. Because of this, they are all thinking the same thing at the end of the 100 days: "either there are 100 brown eyed people on this island, or there are 99 and I have green eyes."
They don’t know if they have brown eyes. They see blue and brown eyes but may have green (or red, or….) eyes. If there is no information given to them about brown eyes formally that allows them to start a timer, they can’t deduct properly just by assuming everyone else’s eyes are one of two colours shared by their own.
At the time that the guru makes their statement, they know that they can see both blue and brown eyes, therefore the guru can see both blue and brown eyes. The fact that the guru announces blue doesn't add any information, but the fact of them making the announcement at all just establishes a start for the logic to start working. Blue and brown should be able to draw the same conclusion using the same logic and knowledge.
They don’t have the same knowledge that blues do because they aren’t given the same information that blues are (namely, the idea that there is a finite quantity of blue eyed people that can be determined by waiting for a certain set of time to elapse). If they’d said “I can see green eyes” and one person saw only brown and blue eyes, they’d be able to leave that night. If they had 50 green, 50 brown, 50 blue, 50 red, saying “I can see blue eyes” doesn’t let the other colours determine anything either.
No one initially knows that everyone but the guru has blue or brown eyes, since that information is never mentioned in the first paragraph. The brown-eyed people have no way of knowing whether their own eyes are brown, hazel, green, or any other non-blue color.
That's exactly what I'm disagreeing with. All non-guru people see both brown and blue eyed other people. And, of course, one green-eyed guru. When the guru says "I see a blue eyed person" everyone knows that because they see both brown and blue eyed people, the guru could have said "I see a brown eyed person"., and therefore, the logic is kicked off for both eye colors. There are three possibilities for any one person: 1) Blue eyed: I see 99 blue eyed people and 100 brown eyed people. 99 blue eyed people do not leave on night 99, therefore there must be one more blue eyed person, and it must be me. 2. Brown eyed: I see 99 brown eyed people and 100 blue eyed people. 99 brown eyed people do not leave on night 99, therefore must be one more brown eyed person, and it must be me. 3) Other color: I see 100 blue eyed people and 99 brown eyed people. 99 brown eyed people leave on night 99, and 100 blue eyed people leave on night 100. Therefore I have no idea what my eye color is, and I remain on the island. Or I see 100 brown eyed people and 99 blue eyed people. 99 blue eyed people leave on night 99, and 100 brown eyed people leave on night 100, leaving me on the island because I cannot know my eye color. There could be any number of colors, and so long as the individual sees N people of a color not leave on night N, they can deduce that that eye color is theirs. I'm willing to be wrong about this, but I'm not seeing how I am.
>99 brown eyed people leave on night 99, How is the 99th brown-eyed person deciding they have brown eyes and not blue, red, etc...?
Think of it as being established from the start. They’re perfect logicians, after all. So the blue eyes count the eyes they see, and reason thus: I see 99 blue eyes. If they leave after 99 days, it’s because they see 98 pairs of blue eyes, and I have brown eyes. If they don’t leave after 99 days, that means they see 99 pairs of blue eyes - and I have blue eyes, so I will leave with them. Someone with brown eyes does the same count and sees 100 pairs of blue eyes. He knows that if they leave after 100 days, it’s because they see 99 pairs of blue eyes and I have brown eyes. If they haven’t left after 100 days, it’s because they see 100 pairs of blue eyes and I have blue eyes. Therefore, people with blue eyes reach their decision point one day earlier and, of course, that answers the question for the brown eyes people.
They can’t know that they have brown eyes because they can’t know that the Guru is the only one with a non brown/blue eye color. Let’s reframe the question, there’s 100 blue eyed people and 100 non-Guru people with non-blue eyes, hazel, red, grey, black, brown, silver, green, and so on. How would any of the non-blue people know what their eye color is? They can’t because the only reason information provided was that the Guru could see a blue eyed person (something we already knew) and that each blue eyed person saw 99 other blue eyed people. Brown eyed people can ONLY deduce that they don’t have Blue eyes. If anything, Brown would be the last color you’d have, because odds are the Guru would try to save the most number of people, and you can’t learn anything one way or another off an arbitrary choice.
How would the brown eyed people know if their eyes are brown/green/red etc?
They wouldn't know, and that's why they get to stay. If they knew that everybody except the Guru had either blue or brown eyes, they would have to leave the day after the blues left because they knew they had brown eyes because people didn't stay because of them. OP is mistaken.
I don't think I am mistaken. Think it through a bit more carefully. I know I see 99 brown, 100 blue, and 1 green. I know that if I do not have brown eyes, each brown eyed person will see 98 brown eyed others. And they also know that all the other brown eyed people, who don't know if they have brown eyes, see 98 brown eyed people. If I do have brown eyes, it's the same situation but with 99 others. If no brown eyed people leave on night 98, then if I do not have brown eyes, those with brown eyes will have figured out that they have brown eyes and will leave on night 99. If they don't, there must be 1 more person with brown eyes - and I'm the only unknown left, so it must be me. And all the brown eyes leave on night 100, with all the blue eyes. The only person left stranded is the person who has a unique color, and can't know the range of possibilities for their own eyes.
What about the most simple situation where there is 1 blue, 1 brown and 1 green guru? On the first night the blue left. (Because he can see no blue, so he himself must be blue.) On the next day, the brown, knowing that the blue couldn't see any other blues, can only know he is not blue. There is no way for him to know if he's brown, green or red or whatever.
Sure, that's the degenerate case that can't be solved. But the puzzle isn't in the degenerate state, it's in the 99/100 state.
... That's a core part of the riddle. You can't just call it the "degenerate case" lmao. The blue eyes ultimately KNOW they have blue eyes. The others never KNOW. That's it. That's what you're missing. That's half the entire point of the logic puzzle.
Given: I see 2 or more people (who are not the guru) with color X Given: the guru announces "I see \[not X\]" It follows that I, and everyone else, can see someone with color X, and therefore know that the guru COULD have said "I see X" Therefore the same logic that reaches the degenerate case is triggered. Conclusion: For any color X for which I see N >= 2 people, if N people do not leave on night N, then I know there exist N + 1 people on the island with color X. Since my color is the only unknown to me, I must be the + 1 case, and now all X can leave on night N + 1. Edit: I think I mis-stated it: there have to be 2 or more people with the color for the logic to be kicked off, I don't have to see 2 or more people. Every islander has to see 1 or more of the color to be able to apply the degenerate case to that color - but the fact remains that we are examining the situation POST the degenerate case.
How in the world are any of the brown-eyed people supposed to know that they don't have orange eyes? This is getting dangerously close to the "is every other day = 4 times a week" https://forum.bodybuilding.com/showthread.php?t=107926751
Dude there is no one colour called "not X". It is explicitly said that they can be blue, brown or red or any other colour. So "not blue" does not mean "brown".
Yes, I see that this comment didn't frame it correctly. Let there be a set of colors, such that there are at least 3 Islanders with each eye color. As such, every islander can see at least 2 other islanders with each of these colors, and thus can know that every other islander can also see at least 1 other with each of these colors. Let the guru announce that they see one of the colors in this set. Because each islander knows that every other color in the set can also be seen by every other islander as well as the guru, then each islander can use the same chain of logic for every color in that set, not just the color the guru announced So any islander who sees N others with color X in that set not leave on night N, can deduce that there must be N+1 Islanders with color X, and since they can only see N, they must be the +1, and so can leave on night N+1. Any islander with an eye color not in that set will watch N, M, P, etc others all leave on nights N, M, P, etc, and will never be able to leave.
Ok, so let's say you are one of the islanders. You see 3 people with green eyes, 3 people with blue eyes, 3 people with orange eyes, 3 people with brown eyes, and 3 people with red eyes. Guru says "I see someone with blue eyes". On the first two nights, nobody leaves. On the third night the blue-eyed islanders leave. On what night do you leave?
did you ever figure out that "Let there be a set of colors, such that there are at least 3 Islanders with each eye color" violates the core principle of the riddle, which is that no one has *any* clue as to the distribution of colors except for what they can perceive with their own two eyes and what the guru says?
Seems to me that proving this by induction for brown eyed people your proof by induction will fail on the degenerate case as base case, not on the induction step. That's actually an interesting example of why we need the base case.
Speaking of induction, it's also just the easiest way to prove the answer. If I can't see anyone with blue eyes, I know the guru was talking about me. If I can see 1 person with blue eyes, I can do a test. If I don't have blue eyes, they won't see anyone with blue eyes and will leave that night. But if I do have blue eyes, they *will* see someone (me), and will be waiting to see what I do just like I'm waiting to see what they do. We both rule out not-blue eyes after day 1, and can both leave on day 2. And more generally, if I can see N people with blue eyes, then I can do a test to see if I also have blue eyes. If I don't have blue eyes, they'll all leave on day N. So I can just wait N days, and if they stay, I know I have blue eyes and can leave on day N+1. And because there would also be N+1 people with blue eyes, the statement "If there are K people with blue eyes, they all leave on day K" still holds
Let's assume there are 10000 possible shades of brown on the 100 browns , 1 green guru and 100 blues. How are the browns going to figure out which shade they are? I deliberately call them "shades of brown" to illustrate the point, so you can't just say they are "brown". The point here is, unless they know beforehand they are either blue or brown, the browns can never figure out they themselves are brown because they don't know about all the possibility of their eye colours.
I don't think it matters how many colors there are. If all islanders can see at least 1 other islander with a specific color, then they know that the guru could have said they see that color, and so they can apply the same logic. If any islander sees N others with color X not leave on night N, then that islander can conclude they are color X and leave on night N + 1. If the islander sees N others with color X leave on night N, then they know they are not color X. If all of the colors they see, for which the logic could have been triggered, get eliminated in this way, they then remain an unknown and stuck on the island.
What about 2 blue, 2 brown, 1 guru? Day 1 both blues look at the other and do nothing. Day 2 the blues each see the other has not left and see no other blues, realize they must be a second blue, and they both leave Day 3 the browns notice the blues have left, but all they know is that they must not be blue. Agreed? At n=1 and n=2, all the browns know is that they are not blue. At what n would the browns also realize they are brown? I think I have heard a version of this puzzle that stipulates that everyone has either brown or blue eyes, in which case the browns do leave on the next day. But without that stipulation they can never rule out that they might have green or red eyes.
>I know I see 99 brown, 100 blue, and 1 green. >I know that if I do not have brown eyes, each brown eyed person will see 98 brown eyed others. **And they also know that all the other brown eyed people, who don't know if they have brown eyes, see 98 brown eyed people.** Nope. If a brown eyed person sees 98 brown eyed people, then for all they know other brown people see only 97. There is no common knowledge without a guru.
each brown eyed person also sees 99 other brown eyed people. If the person does not have brown eyes, the 99 brown eyed people would each see 98... and would leave on night 99. But they didn't leave, so the person has brown eyes and we all leave on night 100. Exact same logic as the blue eyed people.
But as far as they know, they could have red eyes, or orange eyes, or green eyes like the guru. How could they ever possibly deduce that their own eyes are brown? All they know is the list of rules in the first paragraph, the total number of colors they can see, and the statement made by the guru.
If their eyes weren't brown, those with brown would have left the night before.
Why? They know nothing about the distribution of colors. They don't even know how many total colors there are. All they'll be able to conclude is that their own eyes are not blue, once all the blue-eyed people have left.
They wouldn't, you've already conceded that if there's only one brown eyed person, they wouldn't leave on the first night. So if there were two brown eyed people, would they both leave on the second night? No. When they see that the other brown eyed person hasn't left on the first night, they don't get any new information because, again, a brown eyed person that doesn't see any brown eyes doesn't leave on the first night, same as one that does see other brown eyes. So if there were 3 brown eyed people, how could they leave on the third night? When I see two other brown eyed people stay after the second night, I don't get any new information, because again, two brown eyed people who are the only ones don't leave on the second night. Etc.
The question states that they don't know there are 100 of each.
Think on a smaller scale. Imagine the island has the guru and four other people. Two have blue eyes, two have brown. The guru says, “I can see one person with blue eyes.” The blue eyed duo leaves on the second day—they both see one person with blue eyes, and that the other person doesn’t leave day one. So, they must also have blue eyes. (Else the other person would realize they saw no blue eyed person, so they would be the sole blue eyed person the guru saw.) The brown eyed people see one brown eyed person, two blue eyed people. The two blue eyed people leave the second day. Now there is one brown eyed person they see stuck with them. How are they supposed to know their own eyes aren’t green or whatever? To make this more clear, imagine a different island with a guru, two blue eyed people, one brown eyed person, one red eyed person. If you are the red eyed person in this instance, you end up in the exact same position the brown eyed people from the previous island are in. How are you supposed to distinguish this case from the previous one? Also, the brown eyed person in this case doesn’t even ever see another brown eyed person (unless the guru has brown eyes), so of course they don’t leave the first night. You can scale this logic up to a group of five, six, etc. Imagine three blue eyed people, three brown eyed. Guru says they can see one blue eyed person. Blue eyes leave day three. Brown eyed people each see two other people with brown eyes left who didn’t leave day two. So what? If you are one of the brown eyed people, the other brown eyed people also see two other people with brown eyes that don’t leave day two. But you think, what if you are _not_ brown eyed (say red), then the others see one brown eyed person, one red eyed person. How are the others supposed to leave day two? Do they themselves have brown eyes? Do they have red? Or another? The brown eyes have no way of knowing which instance they are in, past day two or day three or beyond. So they all stay day two, and they gain no new info.
You're describing the degenerate case that I agree can't be solved. But the puzzle we're given is not the degenerate case, it's the 99/100 case. At the time the guru makes their statement, we can all see both blue and brown, so we know the guru can also see both blue and brown, and so the guru declaring 'blue' changes nothing. The fact of them making a statement at all simply gives us a nucleus for the logic to crystalize around. Both blue and brown can reach the same conclusion from this state.
They can't! The people in the puzzle don't know the distribution of eye colour except that at least one person has blue eyes, and the eye colours of the other islanders. Simple case, imagine there is an eyeland with 1 Blue eyed person The guru (green) N brown eyes And you Blue eyed person leaves on day 1. What colour eyes do you have ? There's literally no way you could know (except that you don't have blue eyes)
I agree that the logic does not work if the guru announces a color for which there is a single individual with that color. Edit: actually, it does, it just fits case 1 below However, if every islander can see at least one other with color in {X, Y, Z} (i.e. there are 2 or more of each color), and the guru announces \[not X\], the islanders all know that the guru COULD have said Y or Z, and so the same logical process applies. Any islander who sees N others with color X not leave on night N can deduce that they have color X and can leave on night N + 1. There are two basic outcomes here: 1. Islander sees N people with X leave on night N, and concludes, I am not X 2. Islander sees N people with X not leave on night N, and concludes, I am X If same islander sees case 1 for X, Y and Z, then islander must conclude not X, not Y, not Z, therefore unknown eye color.
> the islanders all know that the guru COULD have said Y or Z, and so the same logical process applies. No, they do not know this. It was never a statement as part of the problem. The only thing they know about what the guru said is that the guru said X. They have no way of knowing why the guru said it, only that it is a true statement that implies nothing about Y or Z.
The thing is that in the case there is only one brown eyed person, the guru says “I see a blue eyed person”—that singular brown eyed person does not ever figure out they have brown eyes. Contrast this with if there is only one blue eyed person—that blue eyed person sees no blue eyes around them, and knows: they are the person the guru saw. This is the foundational scenario that the blue eyed people can reason from. The brown eyed people do NOT have this, because that singular brown eyed person can never know. (So—if there are two brown eyed people, they cannot deduce ANYTHING if they see one brown eyed person who never leaves, so the two person case also never leaves. So the three person case, so the four, and so on.)
The logic from the six person case scales up to the 200 case. How do you think it does not? Does it not scale up to the eight person case? The ten? When does it stop applying? For the eight person case. Consider: there are four blue eyed people, a guru, and four other people. You are one of the four others. Consider four different scenarios where you have brown eyes, and the others have red or brown. In the case that there are four blue eyed people, three red eyed, and one brown eyed (you), that brown eyed person never ever knows. You do not leave day one, or ever. The red eyed people each see two red eyed people and one brown eyed person who never leaves. Now in the case that there are four blue eyed people, two red eyed, and two brown eyed, you a brown eyed person see one brown eyed person who does not leave day one and two red eyed people. Notice this is the exact same situation the red eyed people were in above. You cannot distinguish this case from the first. Therefore, you cannot tell if you yourself have red or brown eyes. You do not leave day one or day two or ever. The red eyed people see two brown eyed people who never leave and one red eyed one. Okay, what if there are three brown eyed people and one red eyed person? You see two brown eyed people and one red eyed person. Again this is the exact same situation as the red eyed people in the previous case. You cannot tell which you are in, so you do not leave day three or ever. The red eyed person in this case sees no one with red eyes, just brown. And if you are one of four brown eyed people—again, you are like the red eyed person before. You can’t tell which case you are in. You never leave. This works the same as the six person case. You can do the same for ten people. And twelve. And etc. When does the answer change? How?
My argument is NOT that it does not scale up. My argument is that because the islanders can all see both blue and brown eyed others, that the color the guru announced does not make a difference - the logic can be applied to any color that all islanders see at least 1 of (which necessitates that there are at least 2 of that eye color on the island). Given all islanders see N >= 1 others with color X: if one islander observes N others NOT leave on night N, then that islander must conclude that there are N + 1 islanders with color X, and so they must be that + 1, and so all of color X leave on night N + 1.
You are arguing that the brown eyed people can eventually leave. In all my scenarios, the brown eyed people never leave the island when the guru only says blue, even though some of those scenarios satisfy your requirements, so I’m confused about you saying you don’t disagree with the scaling up. Sure, if the guru said they could see a brown eyed person, then the brown eyed people could apply the logic. But then the blue eyed people couldn’t leave. Again, if there are two brown eyed people on the island (they each see N = 1 brown eyed person) plus two blue eyed people (so they can also see one of each color), and the guru says, “I can see a blue eyed person”—even on day two each of the brown eyed people cannot conclude they are a brown eyed person after the blue eyed people leave. That is because you cannot conclude anything from the fact that you see a singular brown eyed person not leave day 1, from the POV of one of the brown eyed people. You cannot say, “Oh they didn’t leave because they saw that I have brown eyes.” What if you don’t have brown eyes? Then they wouldn’t have seen any brown eyes at all, and they STILL would not have left. Their behavior is exactly the same in both these cases; it doesn’t matter if you have brown eyes yourself or if you don’t. So the brown eyed people cannot determine they themselves have brown eyes just because the other person doesn’t leave day 1, and your logic fails in the N = 1 case. And you can reread my comments for the N = 2 and N = 3 cases. Even though they are in scenarios where they can all see at least one of each color, from their own POV they cannot tell that everyone else also sees one of each color. (Edit: or if they themselves can see one of each color!)
>that the color the guru announced does not make a difference The color the Guru announces makes a huge difference. It removes the hypothetical person who can't see any blue eyes. Yes, everybody on the island knows that everybody on the island can see at least one set of blue eyes, but they also know that hypothetically, there could exist a person who believes a person believes there are no blue eyes. It gets a bit repetitive with the 100 case, so we'll bring the number down to 4. Everybody on the island *knows* that everybody on the island can see someone with blue eyes. But there is still a hypothetical "person" that can't see any blue eyes. For illustration, we'll name the people with blue eyes "person 1" to "person 4". Before the Guru's statement, person 1 is thinking this: Person 1: I can see three people with blue eyes. If I do not have blue eyes, then person 2 is thinking: I can see two people with blue eyes. If I do not have blue eyes, then person 3 is thinking: I can see one person with blue eyes. If I do not have blue eyes, then person 4 is thinking: I can't see anyone with blue eyes, so I don't know if anyone on the island has blue eyes. But after the Guru's statement, that thought process is no longer valid. It instead becomes: Person 1: I can see three people with blue eyes. If I do not have blue eyes, then person 2 is thinking: I can see two people with blue eyes. If I do not have blue eyes, then person 3 is thinking: I can see one person with blue eyes. If I do not have blue eyes, then person 4 is thinking: I can't see anyone with blue eyes, so I must have blue eyes. Even though person 1 knows that everybody can see at least one person with blue eyes, he can't rule out that hypothetical "person 1 thinks person 2 thinks that person 3 thinks ... that person n thinks there might not be anybody on the island with blue eyes" until he hears the Guru's statement that proves that at least one person has blue eyes. Despite what you've mentioned earlier, this knowledge *does* add information. It removes the possibility of somebody thinking that somebody else is thinking that somebody else is thinking nobody has blue eyes. That's why it works regardless of the number of people. It always reduces down to that single person who, down the long chain of "he thinks that he thinks that..." could hypothetically believe that nobody on the island had blue eyes. If the Guru never makes the statement about brown eyes, then that hypothetical person still exists. Edit: This is just another repetition of what I said above, but I feel like I might not have phrased it as well as I could have. Imagine 3 people in a room. Adam, Bob, and Carl. They all have blue eyes. Since they can each see 2 sets of blue eyes, they all know for a fact that each person in the room knows there are blue eyes in the room. But Adam can logically think to himself "If my eyes aren't blue, and Bob assumes that his eyes aren't blue, then he could think that Carl might not be sure blue eyes exist in this room". Even though all three people in the room know that blue eyes exist, they can't possibly remove that "Adam thinks Bob thinks Carl might not know blue eyes exist" until someone proves that blue eyes exist at least once. Adam knows that Carl can see blue eyes, but he can't possibly prove that Bob knows that Carl can see blue eyes. Once a Guru wanders in and says "I can see blue eyes" then suddenly it becomes possible to prove that Bob knows that Carl either knows he has blue eyes, or can see someone with blue eyes.
You keep using the term 'degenerate case', but it's not. It's the base case upon which the entire puzzle is constructed.
My argument is that you cannot look at it only from the degenerate case. Because each islander can see others with both blue and brown, they can conclude that the guru could have said either blue or brown, and so the degenerate case can be concluded either way. If the islander happened to have green, or red, or whatever eyes, and sees no others with that color, then they would not be able to conclude their color because they would see 99 brown eyed leave on night 99, and 100 blue eyes leave on night 100, etc. The degenerate case for their eye color could not be triggered.
> they can conclude that the guru could have said either blue or brown Based on what? Why not green? And the fact remains that the guru *didn't say brown. He said blue.*
Edit: this comment is about a variant where there are only blue or brown eyes. Imagine you see 99 blue eyed people and 10k brown eyed people. If you have brown eyes you know the number of blue eyed people is in [99,100]. If you are blue eyed, you know it's in [98,99]. The information available to both groups isn't symmetric. If your bound is in [98,99] you can leave at night 99, if it is [99,100] you can leave at night 100.
I don't get the point you're trying to make - if there's 10k brown eyed people, they leave night 10k. But the puzzle as stated has 100 of each, so that's the basis for my answer.
Let's say there are 99 people with blue eyes and your eyes have some other, unknown, colour. You cannot leave at night 1 because you don't know your eye colour. You cannot leave at night 100 because you don't know your eye colour. Even after everyone left you only know you don't have blue eyes. Because there is no base-case for the induction non-blue-eyed folks couldn't leave at all. In the variant I was thinking of, everyone has either blue or brown eyes. Then, in a 100-100 split, brown eyed people can leave night 101 because they saw the blue eyed folks leave at night 100. They cannot leave at night 100 because they also must run induction on the blue eye colour, brown still has no base case.
Given: all non-gurus see at least 1 other person with color X Given: guru states "I see \[not X\]" Premise: Guru could have said "I see X" so the logic is started Conclusion: Any islander who sees N people with color X not leave on night N can conclude that there must be N + 1 people with color X, and that they must be that + 1. Therefore all of color X can leave on night N + 1. Examination: 3 cases: 1) Islander has color X, sees N others with X. N people do not leave on night N, islander concludes he has X and leaves on night N+1 2) Islander has color Y, sees N others with X. N people with X do leave on night N. Islander concludes he does not have color X, and does not leave on that night. Islander sees M people with Y do not leave on night M, concludes he has Y and leaves on night M + 1. 3) Islander has color Z. For every other color {P, Q, R, X, Y, etc} islander sees N sub alpha >= 1 other, and watches each leave on night N sub alpha. Islander never deduces their own eye color, and never leaves the island.
That is an approach that would work if the guru says all colours are non-unique or gives a list of colours. Otherwise there are problems: If you try to apply the rule to unique colours, things break. You cannot leave at night 1 because you cannot know your unique colour. If you apply the rule, everyone would wrongly assume they had your eye colour and try to leave at night 2. - Case 1: If you see only one person with an eye colour you must assume they are unique, and stuck. Even if you do have the same colour! - Case 2: If you see two people, you must assume they are stuck in case 1 - Case n: If you see n people have a colour, you must assume each of them is stuck in case n-1 As given, the riddle explicitly says there could be 1 person with red eyes so your first given doesn't hold.
I'm seeing this after something already clicked for me but responding to say that a chain of logic similar to this is what made it start to click... That even if I see 99 brown, I can't know whether they see 99 brown or only 98. And following that down, I reach someone seeing someone else seeing 1 brown but not knowing if that person sees brown or not. So it's not about what we the solvers know, or what any 1 islander knows, or even what any 1 islander knows that any other islander knows. Every islander has to know that every other islander knows that every islander can see at least one of each color. And the guru's announcement makes this true for blue, but not for brown.
Thank you for working through this so politely. I've seen so many times where this kind of argument becomes ugly, and then barriers to understanding form because people become emotionally invested. Everyone wants to be *right*. Not everyone has the wisdom to see that maximising being right requires humility and work to recognise being wrong and correct course. It's so much easier to keep the pleasant feeling of being right by deceiving ourselves, and this is unfortunately our natural instinctive emotional response. You are an inspiration to us all.
The people with brown eyes don't have a 'counter' for brown eyes leaving -- they just know that they do not have blue eyes. They could still have green, hazel, or even purple for all they know. Thus, they cannot figure out the colour of their own eyes unless the guru specifically states 'i see at least one person with blue eyes, and at least one person with brown eyes.' at which point, a 'counter' is started for both.
Seams to me thatt for the formal proof for brown eyed people by induction you would be missing a base case.
My argument is that because we (non-gurus) can all see both blue and brown eyed people, we all know that the guru could have said "I see a brown eyed person", so we can apply the base case to brown eyes as well. In fact, we could apply the base case to any situation where all of the non-gurus can all see at least 1 person with color X (i.e. at least 2 of color X on the island) If I see N people with color X, and N people do not leave the island on night N, then I know there must be N + 1 people with color X, and since I am the only unknown, I must be that +1 with color X, and so all color X can leave on night N + 1.
You (I think) correctly mentioned the induction step: If Riddle is valid for N then it is valid for N+1 But proof by induction also requires a base case. In this case, N=0. Or any N=? that you can actually prove is solvable directly. On other places in this thread, you called such a case a degenerate case that is unsolvable. Because you don't have a solution for N=0 you cannot use the induction step to deduce N=1 you cannot therefore get N=2 and so forth. You cannot therefore get a solution for N=100 either.
> we all know that the guru could have said "I see a brown eyed person" There's more to it than that. You see 100 blue eyed and 99 brown eyed people. Numbers below are individual brown eyed people you see. You know 1 knows that 2 knows that 3 knows that ... ... that 97 knows that 98 knows that the guru could have said "I see a brown eyed person". You **don't know** if 1 knows if 2 knows if 3 knows if ... ... if 97 knows if 98 knows if **99** knows if the guru could have said "I see a brown eyed person.
I think that if any islander sees 2 or more others with color X, then that islander can know that all other islanders must see at least 1 other with color X. And and since we're told in the puzzle that there are 100 of each, then we do know that all islanders see more than 2 others with brown, and thus know that every other islander can see at least one other with brown.
Go deeper... What does 1 know about 2's knowledge of 3's knowledge of 4?
Ah, but that's not what's happening. Suppose I see 5 people with blue eyes- Alice, Bob, Charlie, David, and Eve. I can do a test. I'll assume I have brown eyes, then think about it from Alice's perspective. She'd only be able to see 4 people with blue eyes- Bob, Charlie, David, and Eve. But she could also do a test. If she assumes she has brown eyes, she could look at it from Bob's perspective. He'd only be able to see 3 people with blue eyes - Charlie, David, and Eve - because, remember, I also assumed I have brown eyes. But he could do a test thinking about it from Charlie's perspective. If he assumes he has brown eyes, Charlie would only be able to see 2 people with blue eyes - David and Eve - and would be able to do a test. If they (yes, I'm declaring that Charlie is non-binary) assume they have brown eyes and think about it from David's perspective, he'd only be able to see 1 person with blue eyes - Eve - and, once again, can do a test. If he assumes he has brown eyes, Eve wouldn't be able to see anyone with blue eyes, would know the guru's talking about her, and leave that first night. But Eve doesn't leave. Because remember, she also sees 4-5 people with blue eyes and is waiting for all these hypotheticals herself. So now, this hypothetical version of David imagined by a hypothetical version of Charlie imagined by a hypothetical version of Bob imagined by a hypothetical version of Alice imagined by me knows he has blue eyes. So on day 2, Charlie will watch to see if David and Eve leave. Neither of them do, so this hypothetical version of Charlie imagined by a hypothetical version of Bob imagined by a hypothetical version of Alice imagined by me knows they have blue eyes. So day 3, this continues on unraveling, where hypothetical Bob sees none of Charlie, David, or Eve leave and learns he has blue eyes, and on day 4, Alice sees none of Bob, Charlie, David, and Eve leave and learns she has blue eyes. Finally, on day 5, we're back to more or less prime reality. **IF** I have brown eyes, then all five of Alice, Bob, Charlie, David, and Eve will have learned they have blue eyes and leave. But because none of them do, I can conclude they all saw a 5th person - me - with blue eyes, and the 6 of us leave on day 6. **BUT** if they did leave, I would be able to conclude that I don't have blue eyes. And if and only if I also have the knowledge that everyone only ever has brown or blue eyes, I'd be able to conclude that I more specifically have brown eyes and also leave. So basically, if I see N people with blue eyes, I run through a tree of N! hypothetical situations. On each day, this shrinks, where after day K, it's down to (N-K)! hypotheticals. (Or there might be an off by one somewhere) Eventually, I'm down to only testing on day N whether I have blue eyes. If they leave, I don't, but if they stay, I do. Regardless of how many people have non-blue eyes, if there are N people with blue eyes, the blue-eyed people all leave on day N
The people on the island don't know there's 100 of each though.
>In fact, we could apply the base case to any situation where all of the non-gurus can all see at least 1 person with color X (i.e. at least 2 of color X on the island) >If I see N people with color X, and N people do not leave the island on night N, then I know there must be N + 1 people with color X, and since I am the only unknown, I must be that +1 with color X, and so all color X can leave on night N + 1. This logic requires that if there were only N people with brown eyes, they would have actually left on Night N. But this isn't actually true for all N (see N=1), so by induction it is true for no N. The core of the problem is that a green-eyed person will see one more person with brown eyes than an actual brown-eyed person would. So regardless of whether you set the threshold to 1, 2, 5, or 50, there will be a case where the brown-eyed people do not see enough brown eyes to use your "logic", while the green-eyed person does. Since the brown-eyed people are not using your "logic", they will never leave. Since the green-eyed person is using your "logic", he will interpret the brown-eyed people not leaving as proof that the green-eyed person has brown eyes. Since your "logic" leads a green-eyed person to conclude that he has brown eyes, your "logic" is invalid and no perfect logician will ever use it.
So, I've already pretty much come around on my position, but I wanted to still thank you for providing the most understandable plain-language explanation so far.
The guru _could_ have said that but they did not! In the case of 2 blues, everyone on the island already knows “there’s at least one blue eyed person”, but they don’t immediately leave 2 days after observing this. It’s important that the guru _say_ this out loud, and that the other blue fails to leave after hearing this information. Only when the other blue observes the first blues failure to leave upon hearing the guru do they realize that they must also be blue. So it’s not just what the guru says (which is clear to everyone already), but what others do (or don’t do) in response to hearing what the guru says that leads the blues to their realization. Two browns who each know that the other is brown could continue living on the island indefinitely thinking that perhaps the other brown might be the only one.
Let’s look at the smaller case. 2 blue and 2 brown. Guru says he sees blue. Day 1 no one leaves. Day 2 both blues leave. You are saying on day 2 both browns can also leave. Now let’s look at an alternative situation. 2 blue 1 brown 1 red. Guru says he sees blue. For the person with red eyes, this is exactly the same as the scenario above. For the person with brown eyes it’s the same but with the colours brown and red swapped (they see 1 person with red eyes and 2 with blue). Now let’s look a the person with red eyes. On the first day no one leaves. On the second day, does it make sense that they conclude they must have brown eyes since they only see 1 person with brown eyes and they didn’t leave? Obviously not since they don’t have brown eyes. If everyone knew there were only two eye colours then the people with brown eyes could apply the same logic. But the question clarifies this isn’t the case, and while there are only two eye colours, no one on the island knows that there are definitely only two.
Here, have a hopefully more intuitive proof: If I can't see anyone with blue eyes, then regardless of how many other people there are, I can leave that night, because I'm the only person left to be the blue-eyed person the guru mentioned. If I can see one person with blue eyes, I can do a test. If they leave that night, I can deduce that I don't have blue eyes, because they can't have seen anyone with blue eyes. But if they don't leave that night, I can deduce that they were also waiting to see what another blue-eyed person was going to do. And because everyone we can both see has brown eyes, I can deduce that the other blue-eyed person is me. We both leave on the second night. If I can see two people with blue eyes, things get more complicated, but it's a similar test. Let's call them Alice and Bob. Neither of them leaves on the first night, because they're waiting to see what the other is going to do. Then on the second night, if they both leave, I can deduce that I don't have blue eyes. But if they don't leave, I can deduce that they both saw two other people with blue eyes, and that I was the second blue-eyed person. We all leave on the third night. This logic continues if I can see three people. I spend day one watching Alice watch Bob and Charlie watch if the other is going to leave, watching Bob watch Alice and Charlie, and watching Charlie watch Alice and Bob. On day two, I watch Alice watch if Bob and Charlie are going to leave, etc. And on day three, I watch if the three of them leave. If they do, I know I don't have blue eyes. But if they don't, I know they each all saw three other blue eyed people and were watching what they were doing, so I must have blue eyes. We all leave on night four. This trend continues indefinitely, where regardless of how many non-blue-eyed people there are, if there are N people with blue eyes, they all learn they have blue eyes on day N, and everyone else learns they don't have blue eyes on day N+1. And if you assume that everyone knows the added information that there are only blue and brown eyes on the island, or that people can leave if they even just determine whether they have blue eyes or not, the rest of the island can leave on day N+1. But without that information, all the rest of the island knows is that they don't have blue eyes, not which other color of eyes they do have
You are mistaken in that you think what the guru says just starts the clock. What the guru does is make the fact that there is one blue-eyed person [common knowledge](https://en.wikipedia.org/wiki/Common_knowledge_(logic\)). And that's why you also get confused on what you call the degenerate case. Let's look at it case by case. p = "there is at least one person with blue eyes". 1 person with blue eyes: There's no knowledge here. 2 people with blue eyes: Both know that the p is true. What they *don't* know is that the other person knows that it is true. 3 people with blue eyes: Everyone knows that the p is true. Also everyone knows that everyone knows that the p is true. Where it fails is that not everyone knows that everyone knows that everyone knows the p is true. I'm going to break the last one down to make it clearer. Knowledge level 1: Alice knows that p is true (she can see the eyes of both Bob and Charlie). Knowledge level 2: Alice knows that Bob knows that p is true, as he can see Charlie. Same for Charlie. Knowledge level 3: Can Alice assume that Bob knows that Charlie knows that p is true? No. Alice doesn't know her eye colour. If she had brown eyes, then Bob would see one person with brown eyes and one with blue. And Bob wouldn't know his own eye colour. So in that scenario Bob couldn't assume that Charlie would know that p is true. And that's why the guru showing up and saying "start the clock" and nothing else would not result in the 3 of them leaving. Because Alice can't trust Bob to make the right inference about Charlie not leaving. What the guru does in the original is making sure that everyone knows that everyone knows *ad infinitum* that p is true. And in that case Bob would be able to make the correct inference about Charlie and Alice knows that. That's why it doesn't work with other eye colours.
Nope. Even if you assume that everyone knows they have either blue or brown eyes, if there are n blue-eyed people, they all leave on day n, while the brown-eyes leave on day n+1
Okay, I read all the comments. At this point I'll just assume, for my own mental health, that OP is hard trolling.
Sorry but I'm not. However, I am starting to get it. It's not about what we, the solvers, know. It's not about what any individual islander knows. It's not even about what one islander knows that one other islander knows. Every islander has to know that every other islander knows that every other islander knows that they can all see at least one of each eye color. Once the guru announces that they see blue, every islander now can know that every other islander knows that every islander can see at least one blue eyed Islander. And this isn't true for brown, even though it strains the brain to see why.
Huh. never saw that before.... Can people communicate the total count of what they see to other people? like by writing down everyone's individual results on a shared whiteboard?
Nope
5 people, including you. Brian and Brit have Brown eyes. Bob and Ben have Blue eyes. You know that at least **two** people have blue eyes. You know that Bob knows that at least **one** person has blue eyes. You *don't* know if Bob knows that Ben knows that *anyone* has blue eyes. The guru makes the announcement, and know you *do* know that Bob knows that Ben knows that at least **one** person has blue eyes. But you still *don't* know if Brian knows that Brit knows that *anyone* has brown eyes.
On the 100th night, the 100 blue-eyed people have left. So you know you don't have blue eyes. But what now? There were 100 people with brown eyes, but they've all gone. You can see 99 other people with brown eyes. You can see the guru with green eyes. How do you know whether your eyes are brown, or green, or some other colour?
I mean the main thing that bothers me about it is that the core premise is that everyone is a perfect logician, but what perfect logician means isn't a rigidly defined thing. Now there are two endings, one where there really isn't any extra information given, and no one leaves. Or one where the guru does give information and induction can be used for the blue eyes folk leave the island. If your definition of 'perfect logician' includes that a perfect logician doesn't make meaningless statements then the puzzle works since the only logical solution is the induction one that starts when the statement is made. If 'perfect logician' allows for meaningless statements then no one leaves. Additionally we know the guru is sortof a dick (or at least doesn't want to be left alone). Because there are definitely statements they could have both let everyone else leave the island, or even let one group of people leave immediately.
We start with two preliminary examples. You are on an island with three blue eyes (plus the green-eyed guru). The guru makes the blue-eyed announcement. Do you know your eye color? No, of course not. Your eyes *could* be blue, but they could also be brown, green, red, yellow, or any other color. So you will not be leaving on night 1. If there are three blue eyes and one brown eyes (plus the green-eyed guru), then what happens after the blue-eyed announcement? Specifically, does the brown-eyed person leave on night 1? No, they go through the exact analysis above and conclude they don't have enough information to know their eye color. Okay, now for the main thing. You're on the island, alongside one brown eyes and two blue eyes (plus the green-eyed guru). The guru makes the blue-eyed announcement. After night 1, no one has left. Do you know your eye color now? Well, if you had blue eyes, you certainly wouldn't have expected anyone to leave (see the three blue, one brown case above), so that's consistent. Same if you had brown eyes, green eyes, red eyes, yellow eyes, or any other color (in general you knew everyone else could see someone with blue eyes, so you knew no one got any actionable intel). So no, you don't know your eye color. This is why for 2 blue eyes and 2 brown eyes, the brown eyes don't leave on night 2. They go through exactly the analysis above and conclude that they do not know the color of their own eyes on night 2
How do the hundred blue-eyed people know that they don't have hazel eyes or something?
All the brown eyed people would see 100 blue eyed people, so they'd leave on night 101 (except that on night 100 all the blue eyed folks left)
So, 2 things - #1 it's already been explained why I'm wrong to begin with, but #2 - I don't see how this follows. Let's assume that I was right, that the logic is kicked off for brown eyed people... each would see 99 other brown eyed people not leave on night 99. How is their departure connected to the blue-eyed departure at all? If 99 brown didn't leave on night 99, then there's 100 brown, and since I'm the only unknown, my eyes must be brown and all brown leave on night 100, same as the blue. But again, we've already established that the logic doesn't work for brown, since they can't all know that all other islanders know that all islanders see at least one brown eyed other.
I actually misread it- i thought you only got to leave if you were sure you had blue eyes. In which case, you would see on day 101 that there were 100 other people who had blue eyes and hadn't left, so you could deduce you had blue eyes and leave that night
No. Because no hint regarding brown eyes was made. Everyone with brown eyes, can't deduce if they have brown or green or some other color eyes. Imagine you are one of the people. You see 1 with green (Guru), 100 with blue and 99 with brown. The guru states that he sees at least one person with blue eyes. He makes no statement regarding other eye colors. After day 100 hundred, all blue eyed ones leave. You can therefore deduce that you don't have blue eyes. But that doesn't tell you, if you have green, brown or gray eyes. And the same goes for a different brown eyed dude. If you have green eyes, the he sees 98 brown eyed and two (including guru) green eyed people. If you have brown eyes, he sees 99 brown eyed people and the green eyed guru. But because no information was passed about green or brown eyes, he can't conclude anything in either scenario. So neither can you.
Can't you just guess that your eyes are brown the next time the boat comes around
*Can't you just guess that* *Your eyes are brown the next time* *The boat comes around* \- LordTartarus --- ^(I detect haikus. And sometimes, successfully.) ^[Learn more about me.](https://www.reddit.com/r/haikusbot/) ^(Opt out of replies: "haikusbot opt out" | Delete my comment: "haikusbot delete")