Blue Eyes - Revision history

Giraffequeries: Took this page out of My Hobby. if someone can explain what the circumstances were for putting this in said category in the first place please contact me

2026-02-07T20:39:09Z

Took this page out of My Hobby. if someone can explain what the circumstances were for putting this in said category in the first place please contact me

92.23.2.208: Undo revision 402563 by 2001:8003:DC65:7E01:4460:4138:3D1A:2B74 (talk) This is a Talk comment (though not sure how relevent, even then)

2025-12-27T18:20:00Z

Undo revision 402563 by 2001:8003:DC65:7E01:4460:4138:3D1A:2B74 (talk) This is a Talk comment (though not sure how relevent, even then)

2001:8003:DC65:7E01:4460:4138:3D1A:2B74: /* Explanation */

2025-12-27T09:58:34Z

‎Explanation

84.43.20.118: /* Question 1 / Slightly less "reduce[d...] to" than before, but more accurate. Known (seen) blue-eyes must believe otherwise. Unknown (self) blue-eyeses can believe that* conclusion, without it (yet) being reason to reconsider their own applicability.

2025-08-22T01:12:50Z

‎Question 1: Slightly less "reduce[d...] to" than before, but more accurate. Known (seen) blue-eyes must believe otherwise. Unknown (self) blue-eyeses can believe *that* conclusion, without it (yet) being reason to reconsider their own applicability.

FaviFake: Using gender-neutral pronouns

2025-07-24T13:51:08Z

Using gender-neutral pronouns

FaviFake: /* Question 2 */

2025-07-24T13:35:25Z

‎Question 2

FaviFake: /* Explanation */

2025-06-17T15:54:36Z

‎Explanation

FaviFake: removed Category:Comics edited after their publication using HotCat why'd i add that

2025-04-23T16:44:35Z

removed Category:Comics edited after their publication using HotCat why'd i add that

FaviFake: vandalism makes people actually contribute!

2025-03-10T15:42:53Z

vandalism makes people actually contribute!

172.68.205.178: /* Transcript */ ...dunno how that happened.

2025-03-09T23:57:43Z

‎Transcript: ...dunno how that happened.

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 {{comic discussion}}
 [[Category:Multiple Cueballs]]
 [[Category:No title text]]
 [[Category:No date]]
 [[Category:Comics with lowercase text]]

← Older revision		Revision as of 18:20, 27 December 2025
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	Randall cites "some dude on the streets in Boston named Joel" as his source for the comic idea, although he's rewritten it.		Randall cites "some dude on the streets in Boston named Joel" as his source for the comic idea, although he's rewritten it.
−
−
−	~~I think that the use of the word 'Guru' here confuses Zen and Indian mysticism. The puzzle depends on intentionality, a necessary condition which is not made clear by the word 'Guru'~~

	==Puzzle==		==Puzzle==

← Older revision		Revision as of 09:58, 27 December 2025
Line 47:		Line 47:

	Randall cites "some dude on the streets in Boston named Joel" as his source for the comic idea, although he's rewritten it.		Randall cites "some dude on the streets in Boston named Joel" as his source for the comic idea, although he's rewritten it.
		+
		+
		+	I think that the use of the word 'Guru' here confuses Zen and Indian mysticism. The puzzle depends on intentionality, a necessary condition which is not made clear by the word 'Guru'

	==Puzzle==		==Puzzle==

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 Before the Guru speaks, the hypothetical chain of A imagining B imaging C imagining D... imagining Z seeing N blue-eyed people cannot terminate uniquely. Z seeing no blue-eyed people can consider whether they are blue-eyed. This means it is not {{w|Common knowledge (logic)|common knowledge}} that there are blue eyes. Once the Guru makes their pronouncement it is common knowledge, and every chain of reasoning must terminate at 1 blue-eyed person and Z above would have to conclude that they had blue eyes. From then on, every midnight the common knowledge that there are N blue-eyed people increments by 1 as everyone sees nobody leaving on the ferry.
-Stated another way, there's only one stable set of beliefs for the blue-eyed people that would allow them to have so many exist on the island indefinitely. That is if each blue-eyed person believed not only that they have brown eyes, but also that every other blue-eyed person believed, incorrectly, that they had brown eyes. Logic reduces this to "all blue-eyes believe that all blues-eyes have brown eyes". The Guru eliminates that particular possibility.
+Stated another way, there's only one stable set of beliefs for the blue-eyed people that would allow them to have so many exist on the island indefinitely. That is if each blue-eyed person believed not only that they have brown eyes, but also that every other blue-eyed person believed, incorrectly, that they had brown eyes. Logic reduces this to "all unknowing blue-eyes believes that all known blues-eyes believe that they have brown eyes". The Guru eliminates that particular possibility.
 Another straightforward way to understand why the Guru's information is important is thus. Each blue-eyed person knows two sets of information: what the actual situation is on the island (both now and in the past), and what would happen in a hypothetical situation. Each blue-eyed person then needs only to compare the actual situation to a known hypothetical one, and if it matches up, then they take the corresponding action. Consider this: If there were only one blue-eyed person, and the Guru never made the announcement, they would not leave on day 1 because they would not know that N is greater than or equal to 1. Now let's add a second blue-eyed person. Blue-eyes 2 would not be able to inductively determine whether or not to leave on night 2, because blue-eyes 2's knowledge of whether or not to leave on night 2 is dependent on what blue-eyes 1 does on night 1 if and only if blue-eyes 1 knows what to do on night 1.  If blue-eyes 1 doesn't know that N is greater than or equal to 1, then blue-eyes 1 doesn't know what to do on night 1. Thus, their lack of leaving gives blue-eyes 2 no new information, since it was an uninformed action and blue-eyes 2's inductive reasoning was dependent on blue-eyes 1 knowing what to do, and so the inductive process never takes off for the hypothetical situation. This means a hypothetical situation for N people cannot be induced. As such, blue-eyes 100 does not have certain knowledge of the hypothetical situation that would occur on nights 99 and 100, and so even though they know N = either 99 or 100, they can't take action on either of those nights, because they have no certain hypothetical situation to compare reality to, and as such cannot have certainty about the actions they should take.

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 And lastly, the answer is not "no one leaves."
-<font color="#BB3333">I've done my best to make the wording as precise and unambiguious as possible (after working through the explanation with many people), but if you're confused about anything, please let me know.  A word of warning:  The answer is not simple.  This is an exercise in serious logic, not a lateral thinking riddle.  There is not a quick-and-easy answer, and really understanding it takes some effort.</font></big>
+<font color="#BB3333">I've done my best to make the wording as precise and unambiguious as possible (after working through the explanation with many people), but if you're confused about anything, please let me know.  A word of warning:  The answer is not simple.  This is an exercise in serious logic, not a lateral thinking riddle. There is not a quick-and-easy answer, and really understanding it takes some effort.</font></big>
 <center>[[File:hyphen.jpg|link=https://www.xkcd.com/]]</center>
@@ Line 37: / Line 37: @@
 ==Explanation==
 {{incomplete|
-*This is incorrect, the comic was available long before October 11, 2006. The earliest date we have is [https://web.archive.org/web/20041024201125/http://68.57.186.221:8080/ October 24th, 2004] (see fourth link on the page), and the earliest version of the comic is from [https://web.archive.org/web/20041109034300/http://68.57.186.221:8080/blue_eyes.html November 4th, 2004].
+*The date is incorrect, the comic was available long before October 11, 2006. The earliest date we have is [https://web.archive.org/web/20041024201125/http://68.57.186.221:8080/ October 24th, 2004] (see fourth link on the page), and the earliest version of the comic is from [https://web.archive.org/web/20041109034300/http://68.57.186.221:8080/blue_eyes.html November 4th, 2004].
 *Additionally, both the puzzle [https://xkcd.com/solution.html and the solution] (here's an [https://web.archive.org/web/20061102070433/https://xkcd.com/solution.html earlier version of the solution])were modified and updated several times since its release. Update the explanation and the [[#Trivia]] with this info.}}
 {{misc page}}
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 No one without blue eyes will ever leave the island, because they are given no information that can allow them to determine which non-blue eye color they have. The presence of the non-blue-eyed people is not relevant at all. We can ignore them. All that matters is when the blue-eyed people learn that they actually are blue-eyed.
-There are two ways in which blue-eyed people might leave the island. A lone blue-eyed person might leave on the first night because she can see that no one else has blue eyes, so the guru must have been talking about her. Or an accompanied blue-eyed person can leave on a later night, after noticing that other blue-eyed people have behaved in a way that indicates that they have noticed that her eyes are blue too.
+There are two ways in which blue-eyed people might leave the island. A lone blue-eyed person might leave on the first night because they can see that no one else has blue eyes, so the Guru must have been talking about them. Or an accompanied blue-eyed person can leave on a later night, after noticing that other blue-eyed people have behaved in a way that indicates they have noticed that their eyes are also blue.
 The problem is symmetrical for all blue-eyed people, so this means they will either all leave at once or all stay forever.
@@ Line 77: / Line 78: @@
 ===Intuitive proof===
-Imagine a simpler version of the puzzle in which, on day #1 the guru announces that she can see at least 1 blue-eyed person, on day #2 she announces that she can see at least 2 blue-eyed people, and so on until the blue-eyed people leave.
+Imagine a simpler version of the puzzle in which, on day #1, the Guru announces that she can see at least 1 blue-eyed person, on day #2, she announces that she can see at least 2 blue-eyed people, and so on until the blue-eyed people leave.
-So long as the guru's count of blue-eyed people doesn't exceed your own, then her announcement won't prompt you to leave. But as soon as the guru announces having seen more blue-eyed people than you've seen yourself, then you'll know your eyes must be blue too, so you'll leave that night, as will all the other blue-eyed people.  Hence our theorem obviously holds in this simpler puzzle.
+So long as the Guru's count of blue-eyed people doesn't exceed your own, then her announcement won't prompt you to leave. But as soon as the Guru announces having seen more blue-eyed people than you've seen yourself, then you'll know your eyes must be blue too, so you'll leave that night, as will all the other blue-eyed people.  Hence our theorem obviously holds in this simpler puzzle.
-But this "simpler" puzzle is actually perfectly equivalent to the original puzzle. If there were just one blue-eyed person, she would leave on the first night, so if nobody leaves on the first night, then everybody will know there are at least two blue-eyed people, so there's no need for the guru to announce this on the second day. Similarly, if there were just two blue-eyed people, they'd then recognize this and leave on the second night, so if nobody leaves on the second night, then there must be a third blue-eyed person inspiring them to stay, so there's no need for the guru to announce this on the third day. And so on... The guru's announcements on the later days just tell people things they already could have figured out on their own.
+But this "simpler" puzzle is actually perfectly equivalent to the original puzzle. If there were just one blue-eyed person, they would leave on the first night, so if nobody leaves on the first night, then everybody will know there are at least two blue-eyed people, so there's no need for the Guru to announce this on the second day. Similarly, if there were just two blue-eyed people, they'd then recognize this and leave on the second night, so if nobody leaves on the second night, then there must be a third blue-eyed person inspiring them to stay, so there's no need for the Guru to announce this on the third day. And so on... The Guru's announcements on the later days just tell people things they already could have figured out on their own.
 It's obvious that our theorem holds for the "simpler" puzzle, and this "simpler" puzzle is perfectly equivalent to the original puzzle, so our theorem must hold for the original puzzle too.
-Another way of looking at it is to use selective attention. Although each blue-eyed person can see each other blue-eyed person on the island, she doesn't need to.  The only thing she needs to know in order to determine whether to leave on night N is whether she can see an Nth person with blue eyes. On night 1, she only needs to see 1 other blue-eyed person to not leave; on night 2, she can see 2 other blue-eyed people, so she doesn't leave; and so on until night 100 when she can't see a 100th blue-eyed person, and then leaves.
+Another way of looking at it is to use selective attention. Although each blue-eyed person can see each other blue-eyed person on the island, they don't need to. The only thing they need to know in order to determine whether to leave on night N is whether they can see an Nth person with blue eyes. On night 1, they only need to see 1 other blue-eyed person to not leave; on night 2, they can see 2 other blue-eyed people, so they don't leave; and so on until night 100, when they can't see a 100th blue-eyed person, and then leave.
 ===Formal proof===
 To prove this more formally, we can use mathematical induction.  To do that, we'll need to show that our theorem holds for the base case of N=1, and we'll need to show that, for any given X, *if* we assume that the theorem holds for any value of N less than X, then it will also hold for N=X. If we can show both these things, then we'll know the theorem is true for N=1 (the base case), for N=2 (using the inductive step once), for N=3 (using the inductive step a second time) and so on, for whatever value of N you want.
-Base case: N=1. If there is just one blue-eyed person, she will see that no one else has blue eyes, know that the guru was talking about her, and leave on the first night.
+Base case: N=1. If there is just one blue-eyed person, they will see that no one else has blue eyes, know that the Guru was talking about them, and leave on the first night.
 Inductive step:  Here we assume that the theorem holds for any value of N less than some arbitrary X (integer greater than 1), and we need to show that it would then hold for N=X too. If there are X blue-eyed people, then each will reason as follows:  "I can see that X-1 other people have blue eyes, so either just those X-1 people have blue eyes, or X people do (them plus me). If there are just X-1 people with blue eyes, then by our assumption, they'll all leave on night number X-1. If they don't all leave on night number X-1, then that means that there is an Xth blue-eyed person in addition to the X-1 that I can see, namely me. So, if they all stay past night number X-1, then I'll know I have blue eyes, so I'll leave on night number X. Of course, they'll also be in exactly the same circumstance as me, so they'll leave on night number X too."
@@ Line 107: / Line 108: @@
   '''Situation 0.''' If I see 0 blue-eyed people, I can leave right after the announcement on night 1.
-  '''Situation 1.''' If I see 1 blue-eyed person, then she might be in situation 0 and about to leave on night 1; or else she might be in situation 1 just like me, in which case we'll both leave together on night 2.
+  '''Situation 1.''' If I see 1 blue-eyed person, then they might be in situation 0 and about to leave on night 1; or else they might be in situation 1 just like me, in which case we'll both leave together on night 2.
   '''Situation 2.''' If I see 2 blue-eyed people, they might each be in situation 1 watching to see whether anyone in situation 0 leaves the first night (I know nobody will leave that night, but they wouldn't know this), in which case they would leave together on night 2; or else they might be in situation 2 just like me, in which case we'll all leave together on night 3.
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 :''What is the quantified piece of information that the Guru provides that each person did not already have?''
-Before the Guru speaks, the hypothetical chain of A imagining B imaging C imagining D... imagining Z seeing N blue-eyed people cannot terminate uniquely. Z seeing no blue-eyed people can consider whether they are blue-eyed. This means it is not {{w|Common knowledge (logic)|common knowledge}} that there are blue eyes. Once the guru makes their pronouncement it is common knowledge, and every chain of reasoning must terminate at 1 blue-eyed person and Z above would have to conclude that they had blue eyes. From then on, every midnight the common knowledge that there are N blue-eyed people increments by 1 as everyone sees nobody leaving on the ferry.
+Before the Guru speaks, the hypothetical chain of A imagining B imaging C imagining D... imagining Z seeing N blue-eyed people cannot terminate uniquely. Z seeing no blue-eyed people can consider whether they are blue-eyed. This means it is not {{w|Common knowledge (logic)|common knowledge}} that there are blue eyes. Once the Guru makes their pronouncement it is common knowledge, and every chain of reasoning must terminate at 1 blue-eyed person and Z above would have to conclude that they had blue eyes. From then on, every midnight the common knowledge that there are N blue-eyed people increments by 1 as everyone sees nobody leaving on the ferry.
 Stated another way, there's only one stable set of beliefs for the blue-eyed people that would allow them to have so many exist on the island indefinitely. That is if each blue-eyed person believed not only that they have brown eyes, but also that every other blue-eyed person believed, incorrectly, that they had brown eyes. Logic reduces this to "all blue-eyes believe that all blues-eyes have brown eyes". The Guru eliminates that particular possibility.
-Another straightforward way to understand why the Guru's information is important is thus. Each blue-eyed person knows two sets of information: what the actual situation is on the island (both now and in the past), and what would happen in a hypothetical situation. Each blue-eyed person then needs only to compare the actual situation to a known hypothetical one, and if it matches up, then they take the corresponding action. Consider this: If there were only one blue-eyed person, and the guru never made the announcement, she would not leave on day 1 because she would not know that N is greater than or equal to 1. Now let's add a second blue-eyed person. Blue-eyes 2 would not be able to inductively determine whether or not to leave on night 2, because blue-eyes 2's knowledge of whether or not to leave on night 2 is dependent on what blue-eyes 1 does on night 1 if and only if blue-eyes 1 knows what to do on night 1.  If blue-eyes 1 doesn't know that N is greater than or equal to 1, then blue-eyes 1 doesn't know what to do on night 1. So, her lack of leaving gives blue-eyes 2 no new information, since it was an uninformed action and blue-eyes 2's inductive reasoning was dependent on blue-eyes 1 knowing what to do, and so the inductive process never takes off for the hypothetical situation. This means a hypothetical situation for N people cannot be induced. As such, blue-eyes 100 does not have certain knowledge of the hypothetical situation that would occur on nights 99 and 100, and so even though she knows N = either 99 or 100, she can't take action on either of those nights, because she has no certain hypothetical situation to compare reality to, and as such cannot have certainty about the actions she should take.
+Another straightforward way to understand why the Guru's information is important is thus. Each blue-eyed person knows two sets of information: what the actual situation is on the island (both now and in the past), and what would happen in a hypothetical situation. Each blue-eyed person then needs only to compare the actual situation to a known hypothetical one, and if it matches up, then they take the corresponding action. Consider this: If there were only one blue-eyed person, and the Guru never made the announcement, they would not leave on day 1 because they would not know that N is greater than or equal to 1. Now let's add a second blue-eyed person. Blue-eyes 2 would not be able to inductively determine whether or not to leave on night 2, because blue-eyes 2's knowledge of whether or not to leave on night 2 is dependent on what blue-eyes 1 does on night 1 if and only if blue-eyes 1 knows what to do on night 1.  If blue-eyes 1 doesn't know that N is greater than or equal to 1, then blue-eyes 1 doesn't know what to do on night 1. Thus, their lack of leaving gives blue-eyes 2 no new information, since it was an uninformed action and blue-eyes 2's inductive reasoning was dependent on blue-eyes 1 knowing what to do, and so the inductive process never takes off for the hypothetical situation. This means a hypothetical situation for N people cannot be induced. As such, blue-eyes 100 does not have certain knowledge of the hypothetical situation that would occur on nights 99 and 100, and so even though they know N = either 99 or 100, they can't take action on either of those nights, because they have no certain hypothetical situation to compare reality to, and as such cannot have certainty about the actions they should take.
 ==Transcript==

← Older revision		Revision as of 16:44, 23 April 2025
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	[[Category:No date]]		[[Category:No date]]
	[[Category:Comics with lowercase text]]		[[Category:Comics with lowercase text]]
−	~~[[Category:Comics edited after their publication]]~~

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 ==Transcript==
 :[The 'comic page' is mostly text<!-- already set down in the {{comic tab}} -->, with two actual images, neither of which relate to the text.]
-:[The first image is that of the comic [[82: Frame]].]
+:[The first image is that of the comic [[82: Frame]]. It is positioned above the title.]
-:[The second image is that of the comic [[37: Hyphen]].]
+:[The second image is that of the comic [[37: Hyphen]]. It is positioned directly above the last sentence.]
 ==Trivia==