Does anyone know if the solution is posted anywhere? It seems that the only thing the stranger does is to give them a reference point to figure out how many blue-eyed people there are, and not much else.
The solution is on that page although the way the puzzle is worded on xkcd is better I think.
The key piece that xkcd adds is that each islander keeps a count of how many people of what eye color he/she sees.
It's easier if we assume there are only 2 islanders with blue eyes. That means that each blue-eyed person can see one blue-eyed person.
The day after the announcement each blue-eyed islander expects that if they themselves have brown eyes that the single blue-eyed person that they see would kill themselves, since if that person is the only blue eyed person they would only see other islanders with brown eyes. Since that doesn't happen, they know the day after the expected kill day that they themselves must also have blue eyes.
As a logic puzzle it is very tenuous. The proposed solution (which is lower down the blog post) is that because any blue eyed Islander is able to count the number of blue eyed people (99), if on day 99 those people do not kill themselves it means that there is one further blue eyed Islander, and it is them.
Brown eyed Islanders see 100 blue eyes, and so know that if mass suicide occurs on day 100 then they are not blue eyed.
It's only a short reprieve though; because either the browns know that only two colours exist (and so on day 100 immediately know they are brown) or they face the same conundrum as blue, and commit suicide on day 900.
The problem with the puzzle is not really the logical reasoning (which is sound, as described in the puzzle). The problem is that the Islanders, who are described as very logical, should know this information already. That they are still alive when the foreigner arrives is illogical.
EDIT: to downvoters; consider the problem & solution very very carefully (or read my subsequent comment). Initially it seems a little illogical that the browns would die too, but it is ultimately the same basic problem that the blues face :)
> It's only a short reprieve though; because either the browns know that only two colours exist (and so on day 100 immediately know they are brown) or they face the same conundrum as blue, and commit suicide on day 900.
In the case where they aren't told that only two colours exist, why would they kill themselves on day 900? Suppose instead that one of the brown eyed people had green eyes, I don't see how the brown eyed people or green eyed person could deduce the colour of their own eyes.
Imagine the problem like this. The island is populated by just 100 Blue eyed people.
An individual can therefore see 99 blue eyes. Their logical reasoning is that if they are not blue eyed then on day 99 all of the blue eyed people will commit suicide (because they can see 98 blue and 1 of another colour). Because the others don't commit suicide on day 99 the only logical inference is that everyone else can see 99 blue eyes too. And so they too must be blue.
The reason the solution is so complex to comprehend is because the logical inference has little to do with what eye colours exist in the tribe. It is, ultimately, simple mathematics. If a tribe has n blue eyes, y brown eyes and z green eyes any of the individuals in the tribe will eventually be able to logically infer which colour they have, regardless of having any starter information.
Another way to think of it is this; they know that there are either n people with X eyes. Or n+1. Where n is the number of people they can see with X eyes and the unknown factor is their own eye colour. For everyone without X eyes n is greater than those with X eyes.
You originally said that the people with brown eyes would commit suicide after 900 days. Can you explain to me how any of them can know for sure that they don't have green eyes?
I have done; I was using the "blue eyes on their own" as an example to demonstrate it.
I'll try again (this can take a long while to get your head round :D).
Once the Blue eyed people are dead each islander now knows that there are definitely at least 899 brown eyed people, with the unknown factor their own eye colour.
Their hypothesis must be as follows (imagine this is the logical reasoning of ONE islander laying out the scenario):
- If I do not have Brown eyes that means each of the brown eyed people can see 898 browns. If that is the case, then on day 899 all of those people will kill themselves.
- If I did have Brown eyes then each of the brown eyed people can also see 899 brown eyes. If that is the case then no one will kill themselves on day 899, therefore as I can only see 899 browns, I must have brown
Here is where most people struggle to understand: The number of brown eyes each individual can see is fixed at 899. The only unknown factor is their own eye colour. The solution is that everyone knows each other unknown factor. You know that either n or n+1 brown eyes exist (where n is the number of brown eyes they can observe). And so does everyone else (and, crucially, you know they know).
If everyone's n matches yours, then on day n+1 you can infer what the unknown factor is.
Another way of thinking about it is this; an individual can logically reason that only two types of people exist on the island. Those who see 899 browns (him, at the least), those who see 898 browns (potentially everyone else; on the hypothesis that his eye colour differs). If on day 899 no one commits suicide then each individual knows that only the former exists. And therefore they have brown eyes.
(this is horribly hard to put across :S which is why it is such a delicious problem! The main thing to remember is that to the knowledge of one individual there is a known number of brown eyes, and a figure for the minimum and maximum number of brown eyes any other individual can see. If the number of days exceeds the number of browns you can see, you must be brown)
Suppose instead of being 900 brown eyed people there are X brown eyed people.
Suppose X is one, then the brown eyed person appears to be stuck, all the blue eyed people are dead so he knows he doesn't have blue eyes, but he can't determine what colour his eyes are (are they brown?, are they green?).
If X is two then we're left with two people after all the blues are dead. Each one can see a brown eyed person, but again, this doesn't help them determine what colour their own eyes are. Again, why can they not be green?
The case for X>2 seems similar to the case for X=2, there's no magical point at which they can suddenly all say 'I don't have green eyes, I must have brown eyes'
With all respect, I think you are underestimating the problem. :)
Suppose X is one, then the brown eyed person appears to be stuck
Yes, the problem specifically doesn't work if the number of people with an eye colour is below a certain amount.
The case for X>2 seems similar to the case for X=2, there's no magical point at which they can suddenly all say
In a scenario where there is, say, just brown and green (regardless of whether they know this or not), then X>3 is enough.
Where X=3 or less there is not enough information to go on (you successfully point out the flaws) and they could conceivably be any other eye colour. But above that you know it is impossible for there to be an individual who can see only one or two browns, therefore it can be inferred properly.
One of the main problems comes from the explanation of the issue which involves the suggestion that you recurse the problem back to X=1, this is inaccurate because there is always a known absolute minimum people that could have brown eyes (n, the brown eyed people you can see) and a known minimum number that any one person could see (n - 1). So long as for a brown eyed person (n - 1) > 2 the the inference is possible.
Remember; it does not matter what alternative eye colour they might be (and whether they can see it or not). Only that they are either brown or not brown. The logical inference is based on what the others must be seeing.
Try jotting it down and then reasoning it out as an individual in the tribe. Once it clicks it will seem simple :)
OK, so we agree that if there are X <= 3 people then they're stuck and cannot kill themselves.
You claim that if X = 4 and everyone has brown eyes then on day 4 they will all kill themselves. Suppose I am one of these islanders. I can see three people with brown eyes. On days one, two and three no one kills themselves. On day four I get up and kill myself because I know that I have brown eyes.
Fine, now let's consider the case where there are three people with brown eyes, and one person with green eyes:
As I think we've agreed above, the three people with brown eyes cannot infer that they have brown eyes (i.e. the X < 3 case). So they are not able to kill themselves.
But consider the green eyed person: he can see three people with brown eyes and no one kills themselves on days one, two or three.
He is in exactly the same situation at this point as the brown eyed person we considered in the four-brown-eyed-people case. So at this point, by your logic, he must know that he has brown eyes. Which is a contradiction.
I think you are wrong that induction is not involved in establishing eye colour. (unless you can convince me in the X = 4 case that is :) ).
This isn't a problem I'm unfamiliar with it, a colleague asked me it when I was being interviewed for my current job, and it gets rediscussed periodically. I really do think you're wrong I'm afraid.
Umm, we defined X as the number of people with brown eyes. So on an island with 4 people, three of whom have brown eyes, then X=3 and, yep, there is a problem.
Where X=4 (i.e. there are 4 people with brown eyes) it works.
In your case, where the number of people is 4, but X=3, then the fourth possibly incorrectly infers that he has brown eyes and so, on his own, kills himself. On the other hand they are "highly logical" so I argue they would realise that there were too few people to know.
But that's precisely the point, if they're all "highly logical" then in the island-wtih-3-brown-and-1-green case the person with green eyes can't correctly know their eye colour so can't kill themself. But they are in exactly the same situation as a person with brown eyes in the island-with-four-brown-people case. So if the green eyed person in the first case can't deduce their eye colour, then neither can the brown eyed person in the second case.
If you don't agree, please explain to me what extra piece of information the brown eyed person in the second case has which means he can kill himself on day four. As far as I can see they are both in the situation that they can see three people with brown eyes, and no one has yet killed themselves. What is the extra piece of information that allows the brown eyed person in the second case to deduce that he has brown eyes?
Try jotting it down and then reasoning it out as an individual in the tribe.
(1) Suppose there are 4 browns, b1 to b4, and b1 is trying to figure out if his eyes might not be brown.
(2) b1 wants to know if it's logically possible for his eyes not to be brown and for the others not to know their eye colour.
(3) b1 therefore supposes b1 has green eyes. If this were the case, b2 would see b1's green eyes and both b3 and b4 having brown eyes.
(4) If then b2 imagined b2 had green eyes, b3 would see b1 and b2's green eyes, and b4's brown eyes.
(5) If then b3 imagined b3 had green eyes, b4 would see three sets of green eyes.
---
(6) So b1 can imagine a situation, where b2 imagines a situation, where b3 imagines a situation, where b4 sees three sets of green eyes.
(7) If everybody on the island knew at least one person had brown eyes then in b1's imaginary scenario, b4 would know they had brown eyes and would commit suicide on day 1.
(8) If this didn't happen then we'd know that at least 2 people had brown eyes by day 2, and the induction takes off.
Since they don't know that at least one person has brown eyes, then b1's imaginary scenario is possible, and it's therefore possible that b1 doesn't have brown eyes.
The brown eyes must certainly entertain suicidal thoughts. Their religion compels them to commit ritual suicide if they know their own eye color and they know that the odds their eyes are brown are 999 to 1 ;)
If I do not have Brown eyes that means each of the brown eyed people can see 898 browns. If that is the case, then on day 899 all of those people will kill themselves.
No - if I do not have brown eyes then the knowledge that everyone shares is that there are at least 898 brown eyes, not that there are at least 899. This is because the others would then see only 898, since by hypothesis my eyes are not brown.
The solution (to the paradox which the OP is about, not the xkcd version, which the OP is not about) is not about what each individual knows after the foreigner has imparted the fact that there are people with blue eyes, but what each individual knows about what the others know after that.
For n=1, after the foreigner has spoken, any other islander would know that the islander who can see no blue eyes (i.e. the one with blue eyes) will kill himself.
For n=2, after one day, by the logic in the OP, an islander will know if the one blue-eyed islander has not yet committed suicide, then he himself must have blue eyes and commit suicide.
For n+1 it follows the OP.
The solution to the paradox therefore is before the foreigner speaks, each islander knows there are some islanders with blue eyes, after the foreigner speaks, each islander knows that each other islander knows that there are islanders with blue eyes. (The existence of blue eyes has become common knowledge http://en.wikipedia.org/wiki/Common_knowledge_%28logic%29 )
Aside: I am not sure an English language description really covers the contortions of this logical conundrum.
Solution 2 is correct, Solution 1 is wrong because there is always transfer of information in the stranger's speech.
Suppose there was only 1 blue-eyed person:
[before]: Not everyone knows there exist blue eyed people
(the blue eyed person doesn't know it)
[after]: Everyone knows there exist blue eyed people.
Suppose there are 2 blue-eyed persons:
[before]: Everyone knows there are blue eyed persons.
Not everyone knows that everyone knows that there are blue eyed persons
(the 2 blue eyed don't know).
If Alice and Bob are blue eyed, Alice knows Bob is blue eyed,
but thinks that Bob thinks nobody has blue eyes.
[after]: Everyone knows and knows that everyone knows etc.
Suppose there are 3 blue-eyed persons:
[before]: Everyone knows there are blue eyed people.
Everyone knows everyone knows there are blue eyed people.
Not everyone knows everyone knows everyone knows there are blue eyed people.
This last statement changes after the speech.
Similarly for any n:
[before]: (everyone knows that)*(n-1) there are blue eyed people.
[after]: everyone knows everything.
Incidentally, this is why dictatorships don't allow gatherings of people (or free broadcast media). Everyone knows the regime is bad but not everyone knows that everyone knows etc. When the whole hierarchy collapses, revolutions break out.
The case with 3 people always bugs me in solutions for this puzzle:
If Alice, Bob, and Carol have blue eyes how can they not know that everyone knows that there are blue eyed people?
They know that everyone knows that there are blue eyed people.
What they don't know is that everyone knows this. In other words, they don't know that everyone knows that everyone knows that there are blue eyed people.
Alice thinks like this:
1. there are blue eyed people (e.g. Bob and Carol)
2. everyone knows there are blue eyed people (Bob can see Carol, Carol can see Bob, everyone else can see them)
3. I know that 2 is true, but Bob and Carol might not know. Bob might think that Carol thinks that there are no blue-eyed people. (Of course, in reality, Bob does not think this because he can see Alice as well)
Thanks. That was very helpful. Its easy to see but somehow hard to believe that this applies to 4 people as well (now everyone can see 3 blue eyed people) etc.
I understand the logic behind the solution and it makes full sense to me. What I still struggle with is: what specific piece of information is being passed by the foreigner. I get that it is "everyone knows everything" but for n > 1 everyone already knew the specific fact that was being communicated. What is special about the announcement that drives it? It seems like it is actually "the timer starts now!"
The specific information which was not known and is known after the foreigner's speech is that
"everyone knows that everyone knows that everyone knows that (repeated n or more times) there are people with blue eyes."
This was known only up to n-1 repetitions. To infer this new statement, the islanders have to be aware that the others are listening, they understand, they know that everyone understands, etc. etc.
To infer this new statement, the islanders have to be aware that the others are listening, they understand, they know that everyone understands, etc. etc.
That's deeply unsatisfying because we are told that everyone is capable of making the observation that is communicated right from the beginning for n > 1. What prevents the group from acting as soon as the situation starts but before the foreigner speaks?
The message adds information to the system but it is not that there is one person with blue eyes, i.e. what was explicitly stated. I think it adds information about the time at which an action will or will not occur. The information passed is "you can expect something to happen to the people with blue eyes tomorrow."
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[ 3.1 ms ] story [ 65.0 ms ] threadThe key piece that xkcd adds is that each islander keeps a count of how many people of what eye color he/she sees.
It's easier if we assume there are only 2 islanders with blue eyes. That means that each blue-eyed person can see one blue-eyed person.
The day after the announcement each blue-eyed islander expects that if they themselves have brown eyes that the single blue-eyed person that they see would kill themselves, since if that person is the only blue eyed person they would only see other islanders with brown eyes. Since that doesn't happen, they know the day after the expected kill day that they themselves must also have blue eyes.
So on day two, they both kill themselves.
HTH
Brown eyed Islanders see 100 blue eyes, and so know that if mass suicide occurs on day 100 then they are not blue eyed.
It's only a short reprieve though; because either the browns know that only two colours exist (and so on day 100 immediately know they are brown) or they face the same conundrum as blue, and commit suicide on day 900.
The problem with the puzzle is not really the logical reasoning (which is sound, as described in the puzzle). The problem is that the Islanders, who are described as very logical, should know this information already. That they are still alive when the foreigner arrives is illogical.
EDIT: to downvoters; consider the problem & solution very very carefully (or read my subsequent comment). Initially it seems a little illogical that the browns would die too, but it is ultimately the same basic problem that the blues face :)
In the case where they aren't told that only two colours exist, why would they kill themselves on day 900? Suppose instead that one of the brown eyed people had green eyes, I don't see how the brown eyed people or green eyed person could deduce the colour of their own eyes.
An individual can therefore see 99 blue eyes. Their logical reasoning is that if they are not blue eyed then on day 99 all of the blue eyed people will commit suicide (because they can see 98 blue and 1 of another colour). Because the others don't commit suicide on day 99 the only logical inference is that everyone else can see 99 blue eyes too. And so they too must be blue.
The reason the solution is so complex to comprehend is because the logical inference has little to do with what eye colours exist in the tribe. It is, ultimately, simple mathematics. If a tribe has n blue eyes, y brown eyes and z green eyes any of the individuals in the tribe will eventually be able to logically infer which colour they have, regardless of having any starter information.
Another way to think of it is this; they know that there are either n people with X eyes. Or n+1. Where n is the number of people they can see with X eyes and the unknown factor is their own eye colour. For everyone without X eyes n is greater than those with X eyes.
I'll try again (this can take a long while to get your head round :D).
Once the Blue eyed people are dead each islander now knows that there are definitely at least 899 brown eyed people, with the unknown factor their own eye colour.
Their hypothesis must be as follows (imagine this is the logical reasoning of ONE islander laying out the scenario):
- If I do not have Brown eyes that means each of the brown eyed people can see 898 browns. If that is the case, then on day 899 all of those people will kill themselves.
- If I did have Brown eyes then each of the brown eyed people can also see 899 brown eyes. If that is the case then no one will kill themselves on day 899, therefore as I can only see 899 browns, I must have brown
Here is where most people struggle to understand: The number of brown eyes each individual can see is fixed at 899. The only unknown factor is their own eye colour. The solution is that everyone knows each other unknown factor. You know that either n or n+1 brown eyes exist (where n is the number of brown eyes they can observe). And so does everyone else (and, crucially, you know they know).
If everyone's n matches yours, then on day n+1 you can infer what the unknown factor is.
Another way of thinking about it is this; an individual can logically reason that only two types of people exist on the island. Those who see 899 browns (him, at the least), those who see 898 browns (potentially everyone else; on the hypothesis that his eye colour differs). If on day 899 no one commits suicide then each individual knows that only the former exists. And therefore they have brown eyes.
(this is horribly hard to put across :S which is why it is such a delicious problem! The main thing to remember is that to the knowledge of one individual there is a known number of brown eyes, and a figure for the minimum and maximum number of brown eyes any other individual can see. If the number of days exceeds the number of browns you can see, you must be brown)
Suppose X is one, then the brown eyed person appears to be stuck, all the blue eyed people are dead so he knows he doesn't have blue eyes, but he can't determine what colour his eyes are (are they brown?, are they green?).
If X is two then we're left with two people after all the blues are dead. Each one can see a brown eyed person, but again, this doesn't help them determine what colour their own eyes are. Again, why can they not be green?
The case for X>2 seems similar to the case for X=2, there's no magical point at which they can suddenly all say 'I don't have green eyes, I must have brown eyes'
With all respect, I think you are underestimating the problem. :)
Yes, the problem specifically doesn't work if the number of people with an eye colour is below a certain amount.
The case for X>2 seems similar to the case for X=2, there's no magical point at which they can suddenly all say
In a scenario where there is, say, just brown and green (regardless of whether they know this or not), then X>3 is enough.
Where X=3 or less there is not enough information to go on (you successfully point out the flaws) and they could conceivably be any other eye colour. But above that you know it is impossible for there to be an individual who can see only one or two browns, therefore it can be inferred properly.
One of the main problems comes from the explanation of the issue which involves the suggestion that you recurse the problem back to X=1, this is inaccurate because there is always a known absolute minimum people that could have brown eyes (n, the brown eyed people you can see) and a known minimum number that any one person could see (n - 1). So long as for a brown eyed person (n - 1) > 2 the the inference is possible.
Remember; it does not matter what alternative eye colour they might be (and whether they can see it or not). Only that they are either brown or not brown. The logical inference is based on what the others must be seeing.
Try jotting it down and then reasoning it out as an individual in the tribe. Once it clicks it will seem simple :)
You claim that if X = 4 and everyone has brown eyes then on day 4 they will all kill themselves. Suppose I am one of these islanders. I can see three people with brown eyes. On days one, two and three no one kills themselves. On day four I get up and kill myself because I know that I have brown eyes.
Fine, now let's consider the case where there are three people with brown eyes, and one person with green eyes:
As I think we've agreed above, the three people with brown eyes cannot infer that they have brown eyes (i.e. the X < 3 case). So they are not able to kill themselves.
But consider the green eyed person: he can see three people with brown eyes and no one kills themselves on days one, two or three.
He is in exactly the same situation at this point as the brown eyed person we considered in the four-brown-eyed-people case. So at this point, by your logic, he must know that he has brown eyes. Which is a contradiction.
I think you are wrong that induction is not involved in establishing eye colour. (unless you can convince me in the X = 4 case that is :) ).
This isn't a problem I'm unfamiliar with it, a colleague asked me it when I was being interviewed for my current job, and it gets rediscussed periodically. I really do think you're wrong I'm afraid.
Where X=4 (i.e. there are 4 people with brown eyes) it works.
In your case, where the number of people is 4, but X=3, then the fourth possibly incorrectly infers that he has brown eyes and so, on his own, kills himself. On the other hand they are "highly logical" so I argue they would realise that there were too few people to know.
If you don't agree, please explain to me what extra piece of information the brown eyed person in the second case has which means he can kill himself on day four. As far as I can see they are both in the situation that they can see three people with brown eyes, and no one has yet killed themselves. What is the extra piece of information that allows the brown eyed person in the second case to deduce that he has brown eyes?
Try jotting it down and then reasoning it out as an individual in the tribe.
(1) Suppose there are 4 browns, b1 to b4, and b1 is trying to figure out if his eyes might not be brown.
(2) b1 wants to know if it's logically possible for his eyes not to be brown and for the others not to know their eye colour.
(3) b1 therefore supposes b1 has green eyes. If this were the case, b2 would see b1's green eyes and both b3 and b4 having brown eyes.
(4) If then b2 imagined b2 had green eyes, b3 would see b1 and b2's green eyes, and b4's brown eyes.
(5) If then b3 imagined b3 had green eyes, b4 would see three sets of green eyes.
---
(6) So b1 can imagine a situation, where b2 imagines a situation, where b3 imagines a situation, where b4 sees three sets of green eyes.
(7) If everybody on the island knew at least one person had brown eyes then in b1's imaginary scenario, b4 would know they had brown eyes and would commit suicide on day 1.
(8) If this didn't happen then we'd know that at least 2 people had brown eyes by day 2, and the induction takes off.
Since they don't know that at least one person has brown eyes, then b1's imaginary scenario is possible, and it's therefore possible that b1 doesn't have brown eyes.
I didn't manage to make that final leap... and hence see to importance of the foreigner. Subtly brilliant!
Consider me trouted.
No - if I do not have brown eyes then the knowledge that everyone shares is that there are at least 898 brown eyes, not that there are at least 899. This is because the others would then see only 898, since by hypothesis my eyes are not brown.
For n=1, after the foreigner has spoken, any other islander would know that the islander who can see no blue eyes (i.e. the one with blue eyes) will kill himself.
For n=2, after one day, by the logic in the OP, an islander will know if the one blue-eyed islander has not yet committed suicide, then he himself must have blue eyes and commit suicide.
For n+1 it follows the OP.
The solution to the paradox therefore is before the foreigner speaks, each islander knows there are some islanders with blue eyes, after the foreigner speaks, each islander knows that each other islander knows that there are islanders with blue eyes. (The existence of blue eyes has become common knowledge http://en.wikipedia.org/wiki/Common_knowledge_%28logic%29 )
Aside: I am not sure an English language description really covers the contortions of this logical conundrum.
Suppose there was only 1 blue-eyed person:
Suppose there are 2 blue-eyed persons: Suppose there are 3 blue-eyed persons: Similarly for any n: EDIT: reworded for clarity.What they don't know is that everyone knows this. In other words, they don't know that everyone knows that everyone knows that there are blue eyed people.
Alice thinks like this:
1. there are blue eyed people (e.g. Bob and Carol)
2. everyone knows there are blue eyed people (Bob can see Carol, Carol can see Bob, everyone else can see them)
3. I know that 2 is true, but Bob and Carol might not know. Bob might think that Carol thinks that there are no blue-eyed people. (Of course, in reality, Bob does not think this because he can see Alice as well)
That's deeply unsatisfying because we are told that everyone is capable of making the observation that is communicated right from the beginning for n > 1. What prevents the group from acting as soon as the situation starts but before the foreigner speaks?
The message adds information to the system but it is not that there is one person with blue eyes, i.e. what was explicitly stated. I think it adds information about the time at which an action will or will not occur. The information passed is "you can expect something to happen to the people with blue eyes tomorrow."