https://www.explainxkcd.com/wiki/api.php?action=feedcontributions&user=141.101.99.216&feedformat=atomexplain xkcd - User contributions [en]2020-04-09T00:55:16ZUser contributionsMediaWiki 1.30.0https://www.explainxkcd.com/wiki/index.php?title=Blue_Eyes&diff=75987Blue Eyes2014-09-17T09:50:12Z<p>141.101.99.216: correcting wikilink</p>
<hr />
<div>XKCD's [http://xkcd.com/blue_eyes.html Blue eyes] puzzle is a logic puzzle posted around the same time as comic [[169]]. [[Randall]] calls it "the hardest logic puzzle in the world" on its page; whether or not it really is the hardest is up to speculation.<br />
<br />
The page contains two comics. On the top is [[82: Frame]], and on bottom is [[37: Hyphen]].<br />
<br />
==The Puzzle==<br />
<br />
A group of people with assorted eye colors live on an island. They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. Any islanders who have figured out the color of their own eyes then leave the island, and the rest stay. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph.<br />
<br />
On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru (she happens to have green eyes). So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 101 brown and 99 blue. Or 100 brown, 99 blue, and he could have red eyes.<br />
<br />
The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following:<br />
<br />
"I can see someone who has blue eyes."<br />
<br />
Who leaves the island, and on what night?<br />
<br />
==Solution==<br />
<br />
Randal's solution is at [http://xkcd.com/solution.html xkcd.com/solution.html].<br />
<br />
Here are some observations that help simplify the problem.<br />
<br />
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.<br />
<br />
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.<br />
<br />
The problem is symmetrical for all blue-eyed people, so this means they will either all leave at once or all stay forever.<br />
<br />
'''Theorem:''' If there are N blue-eyed people, they will all leave on the Nth night.<br />
<br />
'''Intuitive Proof.'''<br />
<br />
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. <br />
<br />
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.<br />
<br />
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.<br />
<br />
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.<br />
<br />
'''Formal Proof.'''<br />
<br />
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.<br />
<br />
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.<br />
<br />
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."<br />
<br />
This suffices to prove our theorem. The base case tells us the theorem holds for N=1. That together with the inductive step tells us that it therefore holds for N=2, and that together with the inductive step again tells us that it holds for N=3, and so on... In particular, it holds for the case the original puzzle asked about, N=100, so we get the conclusion that the 100 blue-eyed people will leave on the 100th night.<br />
<br />
==Randall's thought-provoking questions==<br />
<br />
After giving his solution, Randall posed three questions for further thought about the puzzle. (I'll answer them in a different order than he asked.)<br />
<br />
'' '''Question 2.''' Each person knows, from the beginning, that there are no less than 99 blue-eyed people on the island. How, then, is considering the 1 and 2-person cases relevant, if they can all rule them out immediately as possibilities?''<br />
<br />
Blue-eyed people can't see their own faces, so blue-eyed people can see one less blue-eyed face than non-blue-eyed people can. Even though I can see that there are at least 99 blue-eyed people, I don't know that they can see that, so I need to imagine people who see only 98, who would base their actions in part by imagining people who can see only 97 who would base their actions in part by imagining people who can see only 96, and so on... All the levels are relevant. (It's like [https://www.youtube.com/watch?v=U_eZmEiyTo0 the Princess Bride scene] where Vizzini is trying to think about what Wesley would choose in part based upon Wesley thinking about what Vizzini would choose in part based upon... "So clearly I cannot choose the one in front of me!") Each layer of thinking about what someone else might be thinking about can decrement by 1 the number of blue-eyed people visible to the lattermost imagined person, so it turns out that even the base case with N=1 blue-eyed person is relevant. As the days go by, some of the more far-fetched "he might be thinking that I might be thinking that he might be thinking that I might be thinking that..." hypotheses get ruled out. But it's only after night N-1 that the blue-eyed people rule out all the possibilities in which they have brown eyes, whereas the brown-eyed people only learn on night number N that they don't have blue eyes.<br />
<br />
It might help to think of all the different situations people might be in. (Remember brown-eyed people always are situated where they can see one more blue-eyed face than blue-eyed people can.)<br />
<br />
'''Situation 0.''' If I see 0 blue-eyed people, I can leave right after the announcement on night 1.<br />
'''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.<br />
'''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.<br />
:<br />
:<br />
'''Situation N.''' If I see N blue-eyed people, they might be in situation N-1 watching to see whether any people in situation N-2 leave on night N-1 (I know nobody will leave that night, but they wouldn't know this), in which case they would leave together on night N; or else they might be in situation N just like me, in which case we'll all leave together on night N+1.<br />
:<br />
:<br />
<br />
Even though I start out in situation 99, I need to worry that the blue-eyed people might be in situation 98, so I need to wait long enough for people in situation 98 to figure out what's going on, and then see whether they act like they are indeed in situation 98. But if they're in situation 98, then they're worrying about whether all the blue-eyed people might be in situation 97, so they're going to need to wait long enough for people in situation 97 to figure out what's going on. Of course, that requires waiting long enough for people in situation 96 to figure out what's going on, and so on, down all the way to situation 0. All the levels are relevant, and it takes a separate day to eliminate each level, which is why the whole process takes N days.<br />
<br />
'' '''Question 3.''' Why do they have to wait 99 nights if, on the first 98 or so of these nights, they're simply verifying something that they already know?<br />
''<br />
<br />
Consider an analogy. I've heard that miners used to take canaries down into mines because canaries pass out more quickly in poor air than miners do. Suppose you know the canary will do fine for 98 or so seconds, and then pass out if the air is bad. As you watch the canary for those 98 seconds, there's a sense in which you're just verifying something you already know (it'll do fine), but it seems more accurate to say that your best detector for the quality of the air takes 98 seconds to give you a reading, and you're waiting 98 seconds to see what that reading is.<br />
<br />
When the blue-eyed people wait 98 or so days to leave, that's because their best available detector of their own eye-color takes 98 or so days to give a reading. (This detector involves watching what the other blue-eyed people do, and of course they themselves are waiting on a detector that takes 97 or so days to yield its result...) There's a sense in which they're "simply verifying something that they already know", but it seems more accurate to say that they're waiting for their best available detector of their own eye-color to deliver its reading. <br />
<br />
'' '''Question 1.''' What is the quantified piece of information that the Guru provides that each person did not already have?''<br />
<br />
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 or not 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.<br />
<br />
[[Category:Meta]]</div>141.101.99.216https://www.explainxkcd.com/wiki/index.php?title=Blue_Eyes&diff=75986Blue Eyes2014-09-17T09:45:42Z<p>141.101.99.216: /* Randall's thought-provoking questions */ Correcting answer to Q 1</p>
<hr />
<div>XKCD's [http://xkcd.com/blue_eyes.html Blue eyes] puzzle is a logic puzzle posted around the same time as comic [[169]]. [[Randall]] calls it "the hardest logic puzzle in the world" on its page; whether or not it really is the hardest is up to speculation.<br />
<br />
The page contains two comics. On the top is [[82: Frame]], and on bottom is [[37: Hyphen]].<br />
<br />
==The Puzzle==<br />
<br />
A group of people with assorted eye colors live on an island. They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. Any islanders who have figured out the color of their own eyes then leave the island, and the rest stay. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph.<br />
<br />
On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru (she happens to have green eyes). So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 101 brown and 99 blue. Or 100 brown, 99 blue, and he could have red eyes.<br />
<br />
The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following:<br />
<br />
"I can see someone who has blue eyes."<br />
<br />
Who leaves the island, and on what night?<br />
<br />
==Solution==<br />
<br />
Randal's solution is at [http://xkcd.com/solution.html xkcd.com/solution.html].<br />
<br />
Here are some observations that help simplify the problem.<br />
<br />
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.<br />
<br />
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.<br />
<br />
The problem is symmetrical for all blue-eyed people, so this means they will either all leave at once or all stay forever.<br />
<br />
'''Theorem:''' If there are N blue-eyed people, they will all leave on the Nth night.<br />
<br />
'''Intuitive Proof.'''<br />
<br />
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. <br />
<br />
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.<br />
<br />
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.<br />
<br />
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.<br />
<br />
'''Formal Proof.'''<br />
<br />
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.<br />
<br />
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.<br />
<br />
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."<br />
<br />
This suffices to prove our theorem. The base case tells us the theorem holds for N=1. That together with the inductive step tells us that it therefore holds for N=2, and that together with the inductive step again tells us that it holds for N=3, and so on... In particular, it holds for the case the original puzzle asked about, N=100, so we get the conclusion that the 100 blue-eyed people will leave on the 100th night.<br />
<br />
==Randall's thought-provoking questions==<br />
<br />
After giving his solution, Randall posed three questions for further thought about the puzzle. (I'll answer them in a different order than he asked.)<br />
<br />
'' '''Question 2.''' Each person knows, from the beginning, that there are no less than 99 blue-eyed people on the island. How, then, is considering the 1 and 2-person cases relevant, if they can all rule them out immediately as possibilities?''<br />
<br />
Blue-eyed people can't see their own faces, so blue-eyed people can see one less blue-eyed face than non-blue-eyed people can. Even though I can see that there are at least 99 blue-eyed people, I don't know that they can see that, so I need to imagine people who see only 98, who would base their actions in part by imagining people who can see only 97 who would base their actions in part by imagining people who can see only 96, and so on... All the levels are relevant. (It's like [https://www.youtube.com/watch?v=U_eZmEiyTo0 the Princess Bride scene] where Vizzini is trying to think about what Wesley would choose in part based upon Wesley thinking about what Vizzini would choose in part based upon... "So clearly I cannot choose the one in front of me!") Each layer of thinking about what someone else might be thinking about can decrement by 1 the number of blue-eyed people visible to the lattermost imagined person, so it turns out that even the base case with N=1 blue-eyed person is relevant. As the days go by, some of the more far-fetched "he might be thinking that I might be thinking that he might be thinking that I might be thinking that..." hypotheses get ruled out. But it's only after night N-1 that the blue-eyed people rule out all the possibilities in which they have brown eyes, whereas the brown-eyed people only learn on night number N that they don't have blue eyes.<br />
<br />
It might help to think of all the different situations people might be in. (Remember brown-eyed people always are situated where they can see one more blue-eyed face than blue-eyed people can.)<br />
<br />
'''Situation 0.''' If I see 0 blue-eyed people, I can leave right after the announcement on night 1.<br />
'''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.<br />
'''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.<br />
:<br />
:<br />
'''Situation N.''' If I see N blue-eyed people, they might be in situation N-1 watching to see whether any people in situation N-2 leave on night N-1 (I know nobody will leave that night, but they wouldn't know this), in which case they would leave together on night N; or else they might be in situation N just like me, in which case we'll all leave together on night N+1.<br />
:<br />
:<br />
<br />
Even though I start out in situation 99, I need to worry that the blue-eyed people might be in situation 98, so I need to wait long enough for people in situation 98 to figure out what's going on, and then see whether they act like they are indeed in situation 98. But if they're in situation 98, then they're worrying about whether all the blue-eyed people might be in situation 97, so they're going to need to wait long enough for people in situation 97 to figure out what's going on. Of course, that requires waiting long enough for people in situation 96 to figure out what's going on, and so on, down all the way to situation 0. All the levels are relevant, and it takes a separate day to eliminate each level, which is why the whole process takes N days.<br />
<br />
'' '''Question 3.''' Why do they have to wait 99 nights if, on the first 98 or so of these nights, they're simply verifying something that they already know?<br />
''<br />
<br />
Consider an analogy. I've heard that miners used to take canaries down into mines because canaries pass out more quickly in poor air than miners do. Suppose you know the canary will do fine for 98 or so seconds, and then pass out if the air is bad. As you watch the canary for those 98 seconds, there's a sense in which you're just verifying something you already know (it'll do fine), but it seems more accurate to say that your best detector for the quality of the air takes 98 seconds to give you a reading, and you're waiting 98 seconds to see what that reading is.<br />
<br />
When the blue-eyed people wait 98 or so days to leave, that's because their best available detector of their own eye-color takes 98 or so days to give a reading. (This detector involves watching what the other blue-eyed people do, and of course they themselves are waiting on a detector that takes 97 or so days to yield its result...) There's a sense in which they're "simply verifying something that they already know", but it seems more accurate to say that they're waiting for their best available detector of their own eye-color to deliver its reading. <br />
<br />
'' '''Question 1.''' What is the quantified piece of information that the Guru provides that each person did not already have?''<br />
<br />
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 or not they are blue eyed. This means it is not {{w|Common knowledge|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.<br />
<br />
[[Category:Meta]]</div>141.101.99.216https://www.explainxkcd.com/wiki/index.php?title=Talk:1402:_Harpoons&diff=72688Talk:1402: Harpoons2014-08-01T13:07:42Z<p>141.101.99.216: upload issues</p>
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<div>Harpoon is a brand of rum. Did a bottle make it into space? [[Special:Contributions/108.162.219.196|108.162.219.196]] 12:55, 1 August 2014 (UTC)<br />
<br />
Did this comic upload quite late in the day for anyone else? Is anyone else experiencing or did anyone else experience that "Latest Comic" is still going to 1401 as ix XKCD.com and XKCD.com/#</div>141.101.99.216https://www.explainxkcd.com/wiki/index.php?title=Talk:1386:_People_are_Stupid&diff=71715Talk:1386: People are Stupid2014-07-15T13:15:24Z<p>141.101.99.216: </p>
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<div>On average yes, an individual is of average intelligence. But taken as a population of a whole, well, that's a different story entirely. Randall needs a vacation, ever since he jumped the shark with the dead baby it just feels like the downward trend is getting steeper. --[[Special:Contributions/108.162.210.135|108.162.210.135]] 13:20, 25 June 2014 (UTC)<br />
:I don't really think that he jumped the shark. I don't quite get what you are trying to say, and individual can't be of average intelligence. You must first define the average, if we take the mean intelligence of the whole population, then take a person from the sample, then we say that the individual is of average intelligence. You can't say people is stupid while referring to the whole population, because of the definition of stupid, if we take a sample of low IQ people then those people are going to be of average intelligence within the sample, the same goes to the whole population. So this comic is perfectly valid. --[[Special:Contributions/108.162.212.192|108.162.212.192]] 04:50, 27 June 2014 (UTC)<br />
<br />
Isn't that a reference to the Montgomery Burns Award for Outstanding Achievement in the Field of Excellence? [[Special:Contributions/103.22.200.119|103.22.200.119]] 04:49, 25 June 2014 (UTC)krayZpaving<br />
<br />
White Hat being burned? This certainly will not end here.--[[Special:Contributions/141.101.102.208|141.101.102.208]] 04:52, 25 June 2014 (UTC)<br />
<br />
'''''Explain xkcd: It's 'cause you're dumb.''''' This wiki is founded on the very principle that people are stupid. [[Special:Contributions/108.162.223.29|108.162.223.29]] 05:35, 25 June 2014 (UTC)<br />
: You make an intelligent point, which I both appreciate and like. [[Special:Contributions/108.162.222.50|108.162.222.50]] 13:41, 25 June 2014 (UTC)<br />
::Awww, it's just a joke, it's not personal or anything! '''[[User:Davidy22|<u>{{Color|#707|David}}<font color=#070 size=3>y</font></u><font color=#508 size=4>²²</font>]]'''[[User talk:Davidy22|<tt>[talk]</tt>]] 13:43, 25 June 2014 (UTC)<br />
<br />
This comment is one that makes me scratch my head and wonder... surely Randall is able to see that intelligence is not a relative but rather an absolute thing (if one were to kill the 10% most intelligent people the rest wouldn't get dumber, nor smarter). Surely intelligence is not to be measured in units of the common denominator. Surely it is obvious that 2nd panel is a pure strawman. Sigh...<br />
Oh and btw an IQ of 100 is the median, not the average. [[Special:Contributions/141.101.104.17|141.101.104.17]] 09:18, 25 June 2014 (UTC)<br />
: I am wondering if the explanation should not include a mention of the Median/Mean problem because it is entirely possible for a majority of a population to be above or below some mean (average) statistic depending on the distribution. Also stupidity is a standard that is not dependent on either median or mean.[[User:Sturmovik|Sturmovik]] ([[User talk:Sturmovik|talk]]) 11:46, 25 June 2014 (UTC)<br />
: The IQ of 100 is actually defined to be the median AND the average (and also the mode). It is also defined that the distibution around the IQ of 100 is a perfect bell curve. The IQ just tells you how many people in the world have your IQ (It is also defined that two values that have same distance from hundred, e.g. 80 and 120 have the same amount of people, 'cause it's a perfect bell curve (this means that there are as many people with IQ 120 as people with IQ 80). If the overall population gets more intelligent they have to make the IQ tests harder, so that 100 is again the average and median (This really happened). This and some other things are reasons why I think that IQ tests are BS. --[[Special:Contributions/141.101.93.219|141.101.93.219]] 14:01, 25 June 2014 (UTC)<br />
:: "A test device with numerous correlates measures an amount of environmental influences beside innate determinants, therefore bullshit"... What are your other objections to I.Q. testing? [[Special:Contributions/141.101.89.221|141.101.89.221]] 14:17, 25 June 2014 (UTC)<br />
<br />
The mocking "award", which is an analogy of saying "intelligence isn't everything" (an EXTREMELY common cliche), reflects the fact that Randall, like just about anyone, is oblivious to the magnitude of the totality of positive correlates of intelligence, and even (TRIGGER WARNING, TABOO CONCEPT AHEAD) I.Q. Intelligence, I.Q., not only makes you happier, it also makes you more helpful to other people, more creative, more socially stable, better-to-do, less susceptible to mental illnesses, more likely to remember events in your life, etc. etc. etc... Basically, there isn't a positive trait or quality of life with which intelligence doesn't correlate. But people positively LOATHE awareness of how highly intelligence, in fact, matters. Hence the vehement denial whenever someone indicates its importance, all the "I know an intelligent person who is miserable/mean/...", all stressing of exceptions, all ridicule of the notion of intelligence in general, all the "don't think about it"-mentality, all writing off of I.Q. as "antiquated, grossly limited, racist, metric" rather than the extremely potent predictor that it is. tl;dr Randall at all, take time to actually STUDY intelligence or the g factor before you mock it like that. [[Special:Contributions/141.101.89.211|141.101.89.211]] 09:25, 25 June 2014 (UTC)<br />
: In other words (and this is going to be my last addendum to this note, because it is a vast subject), whenever people say (or imply, as in the comic's case) that "intelligence isn't everything", the question to ask in return is, "okay, now what is the degree to which intelligence enables, facilitates, contributes to, 'the rest' to which you're opposing intelligence here?". People minimise the depth and breadth of the intellectual substrate of achievement. [[Special:Contributions/141.101.89.211|141.101.89.211]] 09:33, 25 June 2014 (UTC)<br />
: Also, Randall (and everyone saying that) is being highly unjust in equating "people aren't smart" with "people aren't as smart as me". A perfectly valid alternative sense is, "people aren't as smart as to be rationally expected to contribute to rather than damage the discussion/situation/position at hand"--having the objective good, the objective recognition that certain situations (for instance, a certain online conversation which is expected to be competent) require certain minimal intellectual thresholds (for instance, an I.Q. of 120), in mind rather than egotic comparison. Lower intelligence, deny it all you please, comes with temperamental problems for instance. Selection for intelligence will largely filter them out. [[Special:Contributions/141.101.89.211|141.101.89.211]] 09:46, 25 June 2014 (UTC)<br />
:: tl;dr of my entire production here: people must learn that BOTH situations of the Dunning-Kruger are equally harmful, the one that's less often considered perhaps actually even more so. Mistaken self-perception as intelligent is bad for the individual, but refusal to acknowledge the importance of one's own cognitive capacity (which is as good as universal in intelligent people--"I am not that smart" (who hasn't heard that one innumerable times?), "I just like doing thing x, my proficiency in it has nothing to do with my intelligence or I.Q.", "I have areas in which I'm 'stupid' too", "effort counts too") has societal consequences, of contributing to erroneous dismissal of the notions of intelligence & I.Q. & g etc. Shutting up for good now. Night. [[Special:Contributions/141.101.89.211|141.101.89.211]] 10:11, 25 June 2014 (UTC)<br />
::: GAHHHHH just one more thing. Consider this: the fact that people dismiss I.Q. is the best indicator of how important a trait it really is. Thing is, people would not feel compelled by modesty to deny its importance had it not been vitally integral to many, many things. We deny what we value, so to give hope to those who lack that thing (to comfort those who lack intelligence). [[Special:Contributions/141.101.89.211|141.101.89.211]] 10:15, 25 June 2014 (UTC)<br />
:::: Hey 141.101.89.211... I wonder if you have something to say, but despite my best efforts, I'm having trouble following everything you're saying - I have a feeling you were a bit emotional (perhaps tired?) when writing that, or you might have had fewer "more things" immediately following "I'm done" statements. If you're up for it, I'd appreciate you taking the time to make sure you're saying what you want to say, and ''then'' say it, because you seem to at least have good grammar (though there ''were'' a few British spellings... :-D), so I suspect you probably have a good point. It's also conceivable that I'm just not smart enough to get what you're saying (?) or perhaps it's just too ''early'' for me. BTW the best way of making sure I see what you're saying would probably be to let me know on my [[User talk:Brettpeirce|talk page]]... might even have the conversation there if you'd prefer. Thanks for your time. [[User:Brettpeirce|Brettpeirce]] ([[User talk:Brettpeirce|talk]]) 11:25, 25 June 2014 (UTC)<br />
: I don't know why you think that 141.101.89.211... No where does the comic say that. The mocking award is simply mocking people who '''may or may not''' have higher intelligence than the people they're addressing taking a Better Than Thou attitude because they think they do. In other words: "Higher intelligence doesn't give you an excuse to act like a jerk." I'm sure you can agree with that too [[Special:Contributions/108.162.245.218|108.162.245.218]] 04:42, 26 June 2014 (UTC)<br />
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I would add one "people are stupid" angle not yet mentioned: judging by behavior, most groups of people are less intelligent that any member of that group individually. This is valid even for the "all people" group - just look at the planet. Surprisingly, judging by content of most wikis, the "editors of wiki" groups seems to immune. -- [[User:Hkmaly|Hkmaly]] ([[User talk:Hkmaly|talk]]) 10:05, 25 June 2014 (UTC)<br />
: Good point--conforming to pressures of one's group or one's position to the detriment of one's judgment is a separate personality trait. The phenomenon is remedied by intelligence, but independent from it. [[Special:Contributions/141.101.89.211|141.101.89.211]] 10:11, 25 June 2014 (UTC)<br />
: Beat me to it. I'd like to add that even individual people have their occasional stupid and intelligent moments, with the stupid ones typically being of greater magnitude. Thus, it's not unreasonable to say that the average actions of people are at least slightly less intelligent than the average intelligence of most people on most days. [[Special:Contributions/173.245.55.83|173.245.55.83]] 12:13, 25 June 2014 (UTC)<br />
: Similar to the statement in the film "Men In Black". Agent J says, "Why the big secret [about the aliens among us]? People are smart. They can handle it." Agent K responds, "A person is smart. People are dumb, panicky dangerous animals and you know it." [[Special:Contributions/108.162.221.45|108.162.221.45]] 01:15, 26 June 2014 (UTC)<br />
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I can't believe people say things like that, man, people are stupid [[User:Halfhat|Halfhat]] ([[User talk:Halfhat|talk]]) 10:52, 25 June 2014 (UTC)<br />
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Thanks for the Lake Wobegon references. Not only is it on-target, but I take personal joy seeing mentions of uniquely Minnesotan culture anywhere I can find them. --BigMal27, Minnesota-born, Minnesotan-raised // [[Special:Contributions/173.245.55.88|173.245.55.88]] 11:53, 25 June 2014 (UTC)<br />
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Instead of saying, "People are stupid," we would do better to say "People make poor decisions / statements / judgments." And this, for multiple reasons, few of them I suspect tied to basal intelligence. Stage of life, level of health and stress, experience relative to the topic, level of education and the quality of that education, cultural idiotic beliefs that interfere with optimal choices, and a zillion others. Plus, as a large percentage of humans are either just coming online in experience and education, or are winding down in health and mental function, we are guaranteed to see a large percentage of stupid decisions right across the IQ landscape. No help for it. [[Special:Contributions/108.162.246.217|108.162.246.217]] 13:04, 25 June 2014 (UTC)<br />
: I.Q. affects level of health and stress, rate of acquisition of experience, level of education, quality of education obtained, preference of cultural beliefs. It doesn't seem to defy reason that it affects the zillion other factors, too. [[Special:Contributions/141.101.89.221|141.101.89.221]] 13:17, 25 June 2014 (UTC)<br />
: Remember, in interaction between psychological and social factors, the question is never of *existence* of a connection, but of its magnitude. It is fine to posit a multitude of environmental factors that determine (ir)rationality, but as long as such position keeps people from connecting I.Q. with those factors' actual occurrence (how much I.Q. does it take to finish a good school? to develop a habit of reading a book every month? this is not at all trivial question, and it needs to be resolved with more than anecdotal evidence of "I know an intelligent illiterate person"), there might be an elephant buried underneath the room which no one knows about. [[Special:Contributions/141.101.89.221|141.101.89.221]] 13:25, 25 June 2014 (UTC)<br />
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I know Cueball's explanation can be construed to illustrate otherwise; but I doubt the comic was meant to be a comment on the relative intelligence of humanity. It seems more likely, to me, that the purpose of the comic was to comment on the stonewalling that the mindset, "I'm better than you," induces. [[Special:Contributions/108.162.216.35|108.162.216.35]] 15:12, 25 June 2014 (UTC)<br />
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The cartoon never mentions I.Q. at all, Just "Average Intelligence", so the Mean/Median discussion is moot. As for the other discussion on this page, I'm just going to quote Blaise Pascal: "I would have written a shorter letter, but I did not have the time" [[User:Jim E|Jim E]] ([[User talk:Jim E|talk]]) 16:00, 25 June 2014 (UTC)<br />
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As mentioned above, in other comments that it's hard to find a way to indent from, there's a difference between different 'average's. (To compare "the median" with "the average" is not a good way of doing it, because one needn't know whether you're talking mean or mode in the second sense. I could even say that I have more than the average number of arms, for a human.) The assumption that the median [i]and[/i] mean (and, perhaps, also mode) are a single location at which 100IQ can be placed is dependant upon the bell curve being symmetrical. Just one hyper-intelligent could skew the mean well above the median. (Ok, so we're talking about comic-book "hyper"ness, to make it significant, in a world's worth of population, but the principle still stands for any more manageable population.) And about IQ tests being recalibrated... there is already a common convention that there's a score-adjuster (or a look-up table, based on this) that gives you different IQs for the same number of correct answers but for people of different ages (and sometimes male/female). Which seems to me like "we give up trying to be demographically neutral, let's just find how well different people answer in our test and then work out where their own arbitrary sub-group's bell-curve stradles". That said, I like IQ tests. I do well in them, and have fun doing them, even if I don't actually believe in them any more than I believe in Sudoku puzzles! And, sorry, I ended up typing far more than I had intended... [[Special:Contributions/141.101.99.193|141.101.99.193]] 16:31, 25 June 2014 (UTC)<br />
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I see a lot of discussion on intelligence, but nothing on "losing faith in humanity". The way I see it everywhere is not in response to stupid people, but to acts of inhumanity. Random acts of violence and hate, for example. Or not random, but large scale. "Restored my faith in humanity" comments often refer to the opposite (in my experience) which involve random acts of kindness, or large-scale altruism. [[Special:Contributions/108.162.237.161|108.162.237.161]] 08:48, 26 June 2014 (UTC)<br />
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What about people using Facebook, Twitter, Whatsapp and any other "social network web 2.0" thing? They certainly aren't an individual or small group, they are stupid and I've lost my faith in them. :) {{unsigned ip|173.245.56.166}}<br />
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There are distributions where majority of the population would indeed be below average. Luckily for humanity, intelligence is on a bell curve! I am happy beyond words that this is the case. {{unsigned ip|108.162.216.31}}<br />
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This has to be one of the most entertaining boring conversations I've ever come across! Brilliant! (Or not.) [[User:Taibhse|Taibhse]] ([[User talk:Taibhse|talk]]) 14:12, 26 June 2014 (UTC)<br />
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I think when someone says "people are stupid", they actually usually mean something like "people systematically make mistakes that I feel are readily avoidable", rather than making an actual judgement regarding general intelligence. So this comic feels rather off to me. [[Special:Contributions/173.245.48.113|173.245.48.113]] 08:01, 27 June 2014 (UTC)<br />
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:If you read xkcd long enough, you'll find a lot of Randall's comics feel "off." {{unsigned ip|108.162.212.215}}<br />
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Interestingly, the people making comments about average people being stupid tend to be, eh, below-average-smart themselves. [[Special:Contributions/108.162.246.217|108.162.246.217]] 00:47, 28 June 2014 (UTC)<br />
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:"Interestingly," huh? You sound smart. [[Special:Contributions/108.162.212.215|108.162.212.215]] 14:39, 30 June 2014 (UTC)<br />
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When I say "People are stupid" I mean that a group of people making a decision is much stupider than a person. [[Special:Contributions/108.162.246.215|108.162.246.215]] 04:33, 28 June 2014 (UTC)<br />
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'''"No, people aren't stupid. On average, people are of average intelligence."'''<br />
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Hey, guys. Consider that average intelligence ''is'' stupid. [[Special:Contributions/108.162.212.215|108.162.212.215]] 14:39, 30 June 2014 (UTC)<br />
: Yeah, this is how I've always interpreted "People are stupid" it means, considering we all think we're a smart species, our average intelligence is really low. It's not "I'm better than everybody/average/most people" but "Everybody/the average person/most people is/are worse than most people believe" [[Special:Contributions/141.101.99.216|141.101.99.216]] 13:15, 15 July 2014 (UTC)<br />
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What if the distribution of intelligence is bimodal? If no one is of "average" intelligence, might the more extreme stupidity of a large portion of the population give the impression that the actual average is lower than it appears? [[User:Bppubjr|Bppubjr]] ([[User talk:Bppubjr|talk]]) 14:48, 1 July 2014 (UTC)</div>141.101.99.216https://www.explainxkcd.com/wiki/index.php?title=1313:_Regex_Golf&diff=567341313: Regex Golf2014-01-06T12:31:32Z<p>141.101.99.216: /* Explanation */</p>
<hr />
<div>{{comic<br />
| number = 1313<br />
| date = January 6, 2014<br />
| title = Regex Golf<br />
| image = regex_golf.png<br />
| titletext = <nowiki>/bu|[rn]t|[coy]e|[mtg]a|j|iso|n[hl]|[ae]d|lev|sh|[lnd]i|[po]o|ls/ matches the last names of elected US presidents but not their opponents.</nowiki><br />
}}<br />
<br />
==Explanation==<br />
This comic revolves around a set of increasingly complicated {{w|regular expressions}}, which are patterns used to search through text for blocks of text matching the pattern. There is a saying in professional programming that goes like this (see [[1171]]):<br />
:Some people, when confronted with a problem, think “I know, I'll use regular expressions.” Now they have two problems.<br />
The comic exemplifies this as Megan's problems grow increasingly more convoluted - originally she was writing regex as a game, then moved on to automatically building regex on arbitrary lists of text, to searching through her files for code that appears to be a regex golf generator. At the end, Cueball quips that she now has "Infinite Problems" as a result of her efforts, tying back to the saying above.<br />
Code golf is a game where by programmings attempt to solve a problem using as few characters as possible, analogous to the number of golf shots took to reach the goal.<br />
<br />
===Regular Expressions===<br />
Regular expressions are a way for programmers to specify a textual pattern. You can later search for the pattern inside a text string - if the pattern is found it's said that the pattern "matches" the string, if it's not found it's said they do not match. The "Regex Golf" challenge is to make a regular expression that matches all of the strings in one group and none of the strings in another. As in "Code golf" the challenge is to find the shortest possible Regex that does this.<br />
The first regex Megan uses is /m | [tn]|b/, said to match Star Wars subtitles but not Star Trek. Subtitles are the secondary titles of the movies, after the "Star Trek: " or "Star Wars Episode N: ". For example, in "Star Wars Episode I: The Phantom Menace" the subtitle is "The Phantom Menace".<br />
The forward slashed just mark the start and end of the regex. The | character means "or", so the regex matches any string that contains the patterns "m ", " [tn]" or "b" (including the spaces). The square brackets match one of the enclosed characters, meaning that " [tn]" matches either " t" or " n". The regex is appearently non case sensitive because it wouldn't work otherwise.<br />
The star wars subtitles match the parts of the regex in the following way:<br />
* "The Phanto<u>m </u>Menace" is matched by "m ".<br />
* "Attack of<u> t</u>he Clones" is matched by " [tn]".<br />
* "Revenge of<u> t</u>he Sith" is matched by " [tn]".<br />
* "A<u> N</u>ew Hope" is matched by " [tn]".<br />
* "The Empire Strikes <u>B</u>ack" is matched by "b".<br />
* "Return of<u> t</u>he Jedi" is matched by " [tn]".<br />
Note that the animated film "Star Wars: The Clone Wars" is not included.<br />
<br />
On the other hand, none of the Star Trek subtitles contains an M followed by a space, a T or an N preceded by a space, or any B, so the regex does not match any of them. Note that in the original series all subtitles start with a "T" but it's the first character so it's not preceded by a space. Here is the list that Megan probably used:<br />
* Original series:<br />
** The Motion Picture<br />
** The Wrath of Khan<br />
** The Search For Spock<br />
** The Voyage Home<br />
** The Final Frontier<br />
** The Undiscovered Country<br />
* The Next Generation:<br />
** Generations<br />
** First Contact<br />
** Insurrection<br />
** Nemesis<br />
* Reboot series:<br />
** - the one without a subtitle -<br />
** Into Darkness<br />
<br />
In the last panel "and beyond" Megan uses the regular expression "/(meta-)*regex golf/" to describe her problem. * means "zero or more" or the preceding character/group (parentheses group characters). So this regex matches "regex golf", "meta-regex golf", "meta-meta-regex golf" etc.. In a way this is regex golf in itself, matching all levels of meta-regex golf while not matching anything else.<br />
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In the title text there is a long regex that is the solution of another regex golf challenge - matching the last names of all elected US presidents but not their opponents. Note that the list of opponents include some people who were previously or later became presidents, so taken literally this is impossible. To make this work the list of opponents must exclude anyone who was also president.<br />
<br />
==Transcript==<br />
:Regex golf:<br />
:[Megan is sitting at a laptop. Cueball is standing behind her.]<br />
:Megan: You try to match one group but not the other.<br />
:Megan: /m | [tn]|b/ matches ''Star Wars'' subtitles but not ''Star Trek''.<br />
:Cueball: Cool.<br />
<br />
:Meta-regex golf:<br />
:Megan: So I wrote a program that plays regex golf with arbitrary lists...<br />
:Cueball: Uh oh...<br />
<br />
:Meta-meta-regex golf:<br />
:Megan: ...But I lost my code, so I'm grepping for files that look like regex golf solvers.<br />
<br />
:...And beyond:<br />
:Megan: Really, this is all /(meta-)*regex golf/.<br />
:Cueball: Now you have ''infinite'' problems.<br />
:Megan: No, I had those already.<br />
<br />
{{comic discussion}}<br />
[[Category:Comics featuring Cueball]]<br />
[[Category:Comics featuring Megan]]<br />
[[Category:Computers]]</div>141.101.99.216https://www.explainxkcd.com/wiki/index.php?title=1310:_Goldbach_Conjectures&diff=563131310: Goldbach Conjectures2014-01-01T05:40:29Z<p>141.101.99.216: /* Explanation */</p>
<hr />
<div>{{comic<br />
| number = 1310<br />
| date = December 30, 2013<br />
| title = Goldbach Conjectures<br />
| image = goldbach_conjectures.png<br />
| titletext = The weak twin primes conjecture states that there are infinitely many pairs of primes. The strong twin primes conjecture states that every prime p has a twin prime (p+2), although (p+2) may not look prime at first. The tautological prime conjecture states that the tautological prime conjecture is true.<br />
}}<br />
<br />
==Explanation==<br />
{{w|Goldbach's conjecture}} and the {{w|Twin prime|twin prime conjecture}} are unsolved problems in mathematics relating to {{w|prime numbers}} (numbers whose only {{w|divisors}} are 1 and itself). A claimed proof of {{w|Goldbach's weak conjecture}} is currently under review.<br />
<br />
Randall is riffing on the relationship between "strong" and "weak" logical statements, which are an interplay between the boldness or usefulness of a statement and the ease with which it might be proven to be true. For example, if Goldbach's conjecture (given in the comic under the label "strong") could be proven to be true, it would automatically imply that Goldbach's weak conjecture (given in the comic under the label "weak") is also true, because any odd number greater than 5 can be expressed as 3 (a prime number) plus an even number greater than 2 (which, per the strong conjecture, would itself be the sum of two prime numbers), resulting in a way to express the original odd number as the sum of three prime numbers. The weak conjecture does not, however, imply the strong conjecture. <br />
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Mathematicians have been solving related problems that are "weaker" than the weak conjecture, and working towards "stronger" ones. For example, in 1937 the weak conjecture was proven for odd numbers greater than 3<sup>14348907</sup>. In 1995 a version was proven based on the sum of no more than seven prime numbers, and in 2012 the ceiling was lowered to five primes. In 2013 the weak conjecture was claimed proven for numbers greater than 10<sup>30</sup>, while all numbers below 10<sup>30</sup> have been verified by supercomputer to satisfy the conjecture; these together imply that the weak conjecture is true (although there is no ''general'' proof of it for all numbers). Goldbach's strong conjecture remains unsolved.<br />
<br />
This comic plays on the "strong" and "weak" naming of Goldbach's conjectures by extending it to further degrees of strength or weakness. The "very weak" and "extremely weak" conjectures are indeed implied by Goldbach's weak conjecture, just as the weak conjecture is implied by the strong one. The "very strong" and "extremely strong" conjectures are extensions of Goldbach's strong conjecture, even as it is an extension of the weak conjecture. However, the "very weak" and "extremely weak" conjectures are so obviously true that they are hardly worth stating, while the "very strong" and "extremely strong" conjectures make such bold claims that they are obviously false.<br />
<br />
Moreover, the "extremely weak" and "extremely strong" conjectures contradict each other, even though they're both derived (albeit in opposite directions) from the same initial conjectures. <br />
<br />
The title text refers to the twin prime conjecture, which states that there are an infinite number of pairs of primes that differ by 2, and then applies the same spectrum of "weak" and "strong" statements to it.<br />
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[[Randall]]'s weak twin prime conjecture states that there are an infinite number of pairs of primes. This is clearly true. Per {{w|Euclid's theorem}}, there are an infinite number of primes. Unlike the actual twin prime conjecture (which specifies a distance of two), this conjecture does not specify a required distance. Thus, any pair from the infinite set of primes suffices. An example is 5 and 13.<br />
<br />
His strong twin prime conjecture states that every prime is 2 less than another prime. This statement is obviously false, as there are many possible counter-examples to this statement (thus Randall's humorous {{w|hedge (linguistics)|hedge}} that some prime numbers "may not look prime at first"). <br />
<br />
The tautological prime conjecture states that it itself is true, while making no statement about primes. It is not technically a {{w|tautology}} but more of a plain assertion. Randall has mentioned tautologies before in [[703: Honor Societies]].<br />
<br />
==Transcript==<br />
:'''Goldbach Conjectures'''<br />
:'''Weak'''<br />
:Every odd number greater than 5 is the sum of three primes<br />
:'''Strong'''<br />
:Every even number greater than 2 is the sum of two primes<br />
:'''Very weak'''<br />
:Every number greater than 7 is the sum of two other numbers<br />
:'''Very strong'''<br />
:Every odd number is prime<br />
:'''Extremely weak'''<br />
:Numbers just ''keep going''<br />
:'''Extremely strong'''<br />
:There are no numbers above 7<br />
<br />
{{comic discussion}}<br />
<br />
[[Category:Math]]</div>141.101.99.216https://www.explainxkcd.com/wiki/index.php?title=761:_DFS&diff=56079761: DFS2013-12-28T16:59:01Z<p>141.101.99.216: /* Transcript */</p>
<hr />
<div>{{comic<br />
| number = 761<br />
| date = July 2, 2010<br />
| title = DFS<br />
| image = dfs.png<br />
| titletext = A breadth-first search makes a lot of sense for dating in general, actually; it suggests dating a bunch of people casually before getting serious, rather than having a series of five-year relationships one after the other.<br />
}}<br />
<br />
==Explanation==<br />
When doing a blind search in computing, there are two main tactics - {{w|depth-first search}}, and {{w|breadth-first search}}. Depth-first search, as indicated in the comic, means going as far as you possibly can before looking at other possibilities. This turns out to be a bad idea: instead of preparing for his date, as he should have been doing, he instead spent the whole time doing research on snake venom, to the exclusion of even getting dressed. By contrast, a breadth-first search will look only minimally into a topic before moving on to another; any new depth exposed by this minimal check will be added to a list of stuff to do later. This would have allowed the man to briefly check many more things within the time allotted, and probably still have been able to get dressed.<br />
<br />
{{w|Median lethal dose|LD<sub>50</sub>}}, or median lethal dose, is the dose of a toxin required to kill 50% of the population studied, usually expressed in milligrams of toxin per kilogram of body mass. The {{w|inland taipan}}'s venom does, indeed, have highest median lethal dose among snake venoms. (Fortunately, it is extremely shy in temperament, and will always escape danger rather than bite if it can, which is why it isn't considered to be a particularly dangerous snake.) Incidentally, corn snakes and garter snakes are not even remotely dangerous to humans (in fact they're the most popular pet snakes), and of the four different species commonly known as "copperheads," the only dangerously venomous one is ''{{w|Deinagkistrodon acutus}}'' or sharp-nosed viper.<br />
<br />
The relationship advice given in the title text should probably not be taken too seriously, even though one might be more sure about what kind of person one is looking for after already having dated a few people.<br />
<br />
==Transcript==<br />
:Preparing for a date:<br />
<br />
:[Hairy with wet hair and a towel around his waist thinks with his hand to his chin.]<br />
:Hairy: What situations might I prepare for?<br />
:::1) medical emergency<br />
:::2) dancing<br />
:::3) food too expensive<br />
:::4) bee eating contest...<br />
<br />
:[Close-up on Hairy's face.]<br />
:Hairy: Okay, what kind of emergencies can happen?<br />
:::1)A) snakebite<br />
:::B) lightning strike<br />
:::C) fall from chair<br />
:::D) tracheal bowing...<br />
<br />
:[Still thinking...]<br />
:Hairy: Hmm. Which snakes are dangerous? Let's see...<br />
:::1)A)a) corn snake: ?<br />
:::b) garter snake: ?<br />
:::c) copperhead: ?<br />
:::d) coral snake...<br />
<br />
:[Sits down in a chair with a laptop, still wearing towel.]<br />
:Hairy: The research comparing snake venoms is scattered and inconsistent. I'll make a spreadsheet to organize it.<br />
<br />
:[Bottom panel is larger than top four, and aligned to right.]<br />
:[Ponytail meets Hairy on his front stoop. She is carrying a purse, and looks down at his towel. Hairy holds his arms in the air triumphantly.]<br />
:Ponytail: I'm here to pick you up. You're not dressed?<br />
:Hairy: By LD<sub>50</sub>, the inland taipan has the deadliest venom of ''any'' snake!<br />
<br />
:I really need to stop using depth-first searches.<br />
<br />
{{comic discussion}}<br />
[[Category:Comics featuring Hairy]]<br />
[[Category:Comics featuring Ponytail]]<br />
[[Category:Romance]]</div>141.101.99.216https://www.explainxkcd.com/wiki/index.php?title=177:_Alice_and_Bob&diff=56050177: Alice and Bob2013-12-28T02:12:18Z<p>141.101.99.216: /* Explanation */</p>
<hr />
<div>{{comic<br />
| number = 177<br />
| date = October 30, 2006<br />
| title = Alice and Bob<br />
| image = alice_and_bob.png<br />
| titletext = Yet one more reason I'm barred from speaking at crypto conferences.<br />
}}<br />
<br />
==Explanation==<br />
Any good cryptography presentation will include at least one story about {{w|Alice and Bob}}. They are the canonical "protagonists" of the crypto world, frequently used in illustrations to demonstrate how a cryptographic system works. (The names were mostly chosen to abbreviate to A and B, as well as being of different genders so that they can be distinguished by pronouns alone.)<br />
<br />
Here, Randall casts the story in a different light. Instead of Alice and Bob being perfectly innocent people who just want to communicate in private, Bob is actually having an affair with Alice, and his former partner, upset, cracked the encryption to see what the message contained. Nevertheless, the "gossipy cryptographic protocol specs" all took Alice's side (since the goal of any good crypto system is, of course, to ''succeed'' in this struggle).<br />
<br />
The rest of the comic makes a few other allusions to cryptography:<br />
<br />
*{{w|Bruce Schneier}} and {{w|Ron Rivest}} are two well-known cryptographers. They have written lots of papers that use Alice and Bob as examples (Alice / Bob fanfic, if you will).<br />
*Public and private keys are two extremely large numbers, chosen such that there's a mathematical relation between them, and yet it's extremely hard (i.e. would take many billions of years) to derive the private key from the public key. They're the basis of {{w|asymmetric cryptography}}. A public-key authenticated signature is a way of cryptographically proving that a certain person created a file, as only that person could have possibly generated that signature. One downside is that anybody who has the public key can verify who a message is from, so it removes plausible deniability; Bob's partner clearly knew that Alice and Bob were communicating, on disks marked by lipstick hearts no less.<br />
*A known-plaintext attack is a type of cryptographic attack where the plaintext (i.e. unencrypted text) of a message is known, and the attacker wants to figure out the keys used to encrypt the message. The woman is saying that she should have known all along that the messages were adulterous in nature.<br />
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Finally, in a twist ending, the girlfriend is revealed to be none other than Eve, the ''eave''sdropper, who is also ubiquitous in Alice and Bob stories. Hell hath no fury, indeed.<br />
<br />
The title text continues the theme of Randall getting barred from speaking at conferences due to his unusual take on certain topics.<br />
<br />
To further spice things up, there are many other characters in the Alice/Bob canon, including Mallory, the ''mal''icious, who wants to actually ''alter'' the message with nefarious intent; Craig, the ''cr''acker, who doesn't particularly care about the message but ''does'' care about the passwords used; plod, a law enforcement officer attempting to access key or data, and Chuck, a third party in the communication who secretly has a villainous intent.<br />
<br />
==Transcript==<br />
:[Eve stands in the frame, talking to the reader.]<br />
:Eve: I'm sure you've heard all about this sordid affair in those gossipy cryptographic protocol specs with those busybodies Schneier and Rivest, always taking Alice's side, always labeling me the attacker.<br />
:Eve: Yes, it's true. I broke Bob's private key and extracted the text of her messages. But does anyone realize how much it hurt?<br />
:Eve: He said it was nothing, but everything from the public-key authenticated signatures on the files to the lipstick heart smeared on the disk screamed "Alice."<br />
:Eve: I didn't want to believe. Of course on some level I realized it was a known-plaintext attack. But I couldn't admit it until I saw it for myself.<br />
:[Eve places her hands on her hips.]<br />
:Eve: So before you so quickly label me a third party to the communication, just remember: I loved him first. We had something and she tore it away. She's the attacker, not me. Not Eve.<br />
<br />
{{comic discussion}} <br />
[[Category:Romance]]<br />
[[Category:Computers]]<br />
[[Category:Banned from conferences]]</div>141.101.99.216https://www.explainxkcd.com/wiki/index.php?title=1308:_Christmas_Lights&diff=559151308: Christmas Lights2013-12-25T11:09:35Z<p>141.101.99.216: /* Explanation */ Star spectra don't look like that; also the red and green ones are too monochromatic to be incandescent bulbs so they're probably LEDs too</p>
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<div>{{comic<br />
| number = 1308<br />
| date = December 25, 2013<br />
| title = Christmas Lights<br />
| image = christmas_lights.png<br />
| titletext = Merry Christmas from xkcd!<br />
}}<br />
<br />
==Explanation==<br />
{{incomplete|Check grammar and add details about the star spectrum}}<br />
Understanding this comic requires knowing about electromagnetic spectrums.<br />
In a nutshell, an electromagnetic spectrum is the scientific way of representing the amounts and type of radiation than a specific source emits. In this comic apears a subtype of these, a light spectrum, that shows the visible radiation, some infrarred and some ultraviolet. More detailed info on wikipedia({{w|Electromagnetic spectrum}}).<br />
<br />
<br />
There are 4 graphs in this comic:<br />
<br />
In the center of the image, between the couple Cueball and Megan and Beret Guy appears a light spectrum of a fire, notably because it emits a lot of energy in the infrared band (The left zone of the spectrum), and in the red and orange zone (typical colors of a fire)<br />
<br />
In the right of the comic appears some spectrums shaped in the form of a Christmas tree. There are 3 different spectrums in this "Christmas tree":<br />
<br />
In the top appears a complex spectrum, probably that of a white LED (see [http://led-brdf.wikispaces.com/Introduction+to+LEDs here]), representing the tradition of putting a star in the top of the Christmas tree, representing the Star that guided the Three Wise Men in Christianity.<br />
<br />
In the "leaves" there are are two simpler spectrums, one with a peak in the green zone, representing a green LED, and other with a peak in the red zone, representing a red LED.<br />
<br />
==Transcript==<br />
{{incomplete transcript}}<br />
<br />
{{comic discussion}}<br />
[[Category:Comics featuring Megan]]<br />
[[Category:Comics featuring Cueball]]<br />
[[Category:Comics featuring Beret Guy]]<br />
[[Category:Comics with color]]</div>141.101.99.216https://www.explainxkcd.com/wiki/index.php?title=Talk:1299:_I_Don%27t_Own_a_TV&diff=54274Talk:1299: I Don't Own a TV2013-12-04T11:50:32Z<p>141.101.99.216: </p>
<hr />
<div>Annual Data for households between 1958-1970<br />
http://www.tvhistory.tv/Annual_TV_Households_50-78.JPG<br />
<br />
Plotted next to a fitted logarithmic function<br />
http://imgur.com/aVWmQ9z<br />
<br />
The negative second derivative of this function<br />
http://imgur.com/xywpEJZ<br />
<br />
If someone can find more data for television ownership I'd love to see it :) {{unsigned ip|173.245.54.12}}<br />
<br />
Can someone explain why Randall believes smugness at not owning a television is decreasing? [[Special:Contributions/199.27.128.138|199.27.128.138]] 08:31, 4 December 2013 (UTC)<br />
<br />
Because as TVs become less relevant, people don't feel smug for not owning one. [[Special:Contributions/141.101.99.216|141.101.99.216]] 11:44, 4 December 2013 (UTC)<br />
<br />
;Current explanation - logistic curve<br />
<br />
The current explanation is total bullshit. The thing with the negative second derivative is just saying, that the more embarrased people are, the more the change of the TV ownership rate will increase, which just means, more and more people will get themselves TVs.<br />
The other point of view is, the more smug you will look like for not owning a TV, the more the change of the TV ownership rate will decline, which means, that less and less people are buying TVs.<br />
<br />
It has nothing to do with a logistic curve. The function, which second derivative is depicted in this comic is totally irrelevant.<br />
<br />
[[Special:Contributions/108.162.231.19|108.162.231.19]] 08:34, 4 December 2013 (UTC)<br />
<br />
I have the strong feeling he is talking about a sine wave, not a logistic function. It fits the curve in the comic as well as the condition of f"=-f. <br />
Also, it makes way more sense for the smugness to behave like this over time as for the first 30 years TV is culturally extremely significant and you therefore would want to own one in order to participate. But with declining quality of television and the emergence of the internet you might feel as if you were extremely progressive by not owning one anymore.<br />
<br />
[[Special:Contributions/108.162.254.189|108.162.254.189]] 09:25, 4 December 2013 (UTC)<br />
<br />
Yes, it definitely could be a sine curve. (see: [http://www.wolframalpha.com/input/?i=d%5E2%2Fdx%5E2%28sin%28x%29%29 http://www.wolframalpha.com/input/?i=d%5E2%2Fdx%5E2%28sin%28x%29%29]). If one would neglect the beginning of the function for simplicity, this could be a solution.<br />
<br />
[[Special:Contributions/108.162.231.19|108.162.231.19]] 10:07, 4 December 2013 (UTC)<br />
<br />
We bid a tearful farewell to our friend the line break. [[Special:Contributions/141.101.99.216|141.101.99.216]] 11:50, 4 December 2013 (UTC)</div>141.101.99.216https://www.explainxkcd.com/wiki/index.php?title=Talk:1299:_I_Don%27t_Own_a_TV&diff=54273Talk:1299: I Don't Own a TV2013-12-04T11:44:42Z<p>141.101.99.216: </p>
<hr />
<div>Annual Data for households between 1958-1970<br />
http://www.tvhistory.tv/Annual_TV_Households_50-78.JPG<br />
<br />
Plotted next to a fitted logarithmic function<br />
http://imgur.com/aVWmQ9z<br />
<br />
The negative second derivative of this function<br />
http://imgur.com/xywpEJZ<br />
<br />
If someone can find more data for television ownership I'd love to see it :) {{unsigned ip|173.245.54.12}}<br />
<br />
Can someone explain why Randall believes smugness at not owning a television is decreasing? [[Special:Contributions/199.27.128.138|199.27.128.138]] 08:31, 4 December 2013 (UTC)<br />
<br />
Because as TVs become less relevant, people don't feel smug for not owning one. [[Special:Contributions/141.101.99.216|141.101.99.216]] 11:44, 4 December 2013 (UTC)<br />
<br />
;Current explanation - logistic curve<br />
<br />
The current explanation is total bullshit. The thing with the negative second derivative is just saying, that the more embarrased people are, the more the change of the TV ownership rate will increase, which just means, more and more people will get themselves TVs.<br />
The other point of view is, the more smug you will look like for not owning a TV, the more the change of the TV ownership rate will decline, which means, that less and less people are buying TVs.<br />
<br />
It has nothing to do with a logistic curve. The function, which second derivative is depicted in this comic is totally irrelevant.<br />
<br />
[[Special:Contributions/108.162.231.19|108.162.231.19]] 08:34, 4 December 2013 (UTC)<br />
<br />
I have the strong feeling he is talking about a sine wave, not a logistic function. It fits the curve in the comic as well as the condition of f"=-f. <br />
Also, it makes way more sense for the smugness to behave like this over time as for the first 30 years TV is culturally extremely significant and you therefore would want to own one in order to participate. But with declining quality of television and the emergence of the internet you might feel as if you were extremely progressive by not owning one anymore.<br />
<br />
[[Special:Contributions/108.162.254.189|108.162.254.189]] 09:25, 4 December 2013 (UTC)<br />
<br />
Yes, it definitely could be a sine curve. (see: [http://www.wolframalpha.com/input/?i=d%5E2%2Fdx%5E2%28sin%28x%29%29 http://www.wolframalpha.com/input/?i=d%5E2%2Fdx%5E2%28sin%28x%29%29]). If one would neglect the beginning of the function for simplicity, this could be a solution.<br />
<br />
[[Special:Contributions/108.162.231.19|108.162.231.19]] 10:07, 4 December 2013 (UTC)</div>141.101.99.216https://www.explainxkcd.com/wiki/index.php?title=Talk:1288:_Substitutions&diff=52303Talk:1288: Substitutions2013-11-08T11:36:41Z<p>141.101.99.216: Created page with "We need a web plugin that does this automatically, stat! ~~~~"</p>
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<div>We need a web plugin that does this automatically, stat! [[Special:Contributions/141.101.99.216|141.101.99.216]] 11:36, 8 November 2013 (UTC)</div>141.101.99.216