https://www.explainxkcd.com/wiki/api.php?action=feedcontributions&user=Bonob&feedformat=atomexplain xkcd - User contributions [en]2024-03-28T13:38:12ZUser contributionsMediaWiki 1.30.0https://www.explainxkcd.com/wiki/index.php?title=Talk:334:_Wasteland&diff=53500Talk:334: Wasteland2013-11-23T11:59:32Z<p>Bonob: Created page with "Isn't the meaning of the title text that it's so difficult to forget her that forgetting anything else is easy in comparison?"</p>
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<div>Isn't the meaning of the title text that it's so difficult to forget her that forgetting anything else is easy in comparison?</div>Bonobhttps://www.explainxkcd.com/wiki/index.php?title=445:_I_Am_Not_Good_with_Boomerangs&diff=52563445: I Am Not Good with Boomerangs2013-11-12T16:03:44Z<p>Bonob: /* Explanation */</p>
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<div>{{comic<br />
| number = 445<br />
| date = July 4, 2008<br />
| title = I Am Not Good with Boomerangs<br />
| image = i_am_not_good_with_boomerangs.png<br />
| titletext = Bonus strip: just read the rightmost panels straight down.<br />
}}<br />
<br />
==Explanation==<br />
The strip shows [[Cueball]] throwing a boomerang four times, each time finding a difficulty in catching it.<br />
<br />
The first time, it merely hits him in the head.<br />
<br />
The second time, about six boomerangs come after him.<br />
<br />
The third time, a shark somehow returns to him.<br />
<br />
The fourth and final time, his girlfriend [[Megan]] appears, stating "I'm leaving you."<br />
<br />
The title text refers to a bonus strip - if one reads the rightmost panels straight down, you get a strip that suggests that Megan threw multiple things at him out of anger before breaking up with him. This seems to mirror the plot of the 2006 comedy film {{w|My Super Ex-Girlfriend}} where the titular character throws multiple things at the protagonist (including a shark) prior to breaking up with him.<br />
<br />
Boomerang returns in [[475: Further Boomerang Difficulties]]<br />
<br />
==Transcript==<br />
:[Cueball throws a boomerang, but it hits him in the head when it returns.]<br />
<br />
:[Cueball throws the boomerang again, but this time several boomerangs chase after him.]<br />
<br />
:[Cueball throws the boomerang once more, and this time a shark inexplicably appears.]<br />
<br />
:[Cueball throws the boomerang a final time, and Megan appears, hovering.]<br />
:Megan: I'm leaving you.<br />
<br />
{{comic discussion}}<br />
[[Category:Comics featuring Cueball]]<br />
[[Category:Comics featuring Megan]]<br />
[[Category:Boomerangs]]</div>Bonobhttps://www.explainxkcd.com/wiki/index.php?title=469:_Improvised&diff=52441469: Improvised2013-11-10T16:04:18Z<p>Bonob: /* Explanation */</p>
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<div>{{comic<br />
| number = 469<br />
| date = August 29, 2008<br />
| title = Improvised<br />
| image = improvised.png<br />
| titletext = Oh, your brother is Luke. Sorry, should've mentioned that first.<br />
}}<br />
<br />
==Explanation==<br />
{{incomplete}}<br />
In the film "{{w|Star Wars}} Episode V: {{w|The Empire Strikes Back}}", just before {{w|Han Solo}}, portrayed by {{w|Harrison Ford}}, is frozen in carbonite, the following conversation occurs:<br />
:Leia: I love you.<br />
:Han: I know.<br />
<br />
The original script had Han Solo respond with "I love you, too", but Harrison Ford felt that the character would not give such a cliched response, even in the face of likely death and ad-libbed the "I know" line that was actually used in the finished film. The ad-libbed line is generally thought to be better than the original would have been.<br />
<br />
The comic presents several alternative ad-libs that Ford could have made in that conversation as well as at various points throughout the trilogy.<br />
<br />
; Well, duh<br />
:Absurd answer to serious a remark<br />
; Han Solo in the cockpit of the Millennium Falcon.<br />
:(explanation needed)<br />
; Oh! Hey, that explains the kissing earlier.<br />
:Another absurd answer to a serious remark<br />
; I'm nailing your brother.<br />
: So far Leia doesn't know she has a brother. Also, since ''to nail'' means to penetrate, it should be a shock for her to know that Han is either gay or bisexual.<br />
; Hokey religions and ancient weapons are no match for scissors, though they do beat paper and rock.<br />
: He is referring to {{w|Rock-paper-scissors}}<br />
<br />
;Cool. Listen, this thing is really, REALLY cold.<br />
:Han is in a freezing chamber.<br />
; Wowzers<br />
:''Wowzers'' is an expression used by {{w|Inspector Gadget}}<br />
; General Solo, is your strike team assembled?<br />
:(explanation needed)<br />
;I'd just as soon kiss a wookie.<br />
:''Chewwie'' is a nickname for {{w|Chewbacca}}<br />
<br />
;Title text <br />
It refers to the plot twist that {{w|Luke Skywalker}} is princess {{w|Princess Leia|Leia's}} brother, which would not be revealed until the next film in the series. How Han Solo knows this twist at this point in the story is unknown, but he must at least know that Leia has a brother in the center left panel.<br />
<br />
==Transcript==<br />
:Harrison Ford Famously Improvised His "I know" Line in E.S.B. (The Empire Strikes Back). Here are a few of his less-successful ad-libs:<br />
:[Han Solo stands in front of Princess Leia on the Cloud City Carbon Freezing Chamber.]<br />
:Leia: I Love You.<br />
:Han: Well, Duh<br />
:[Han Solo in the cockpit of the Millennium Falcon.]<br />
:C-3P0: Sir, the possibility of successfully navigating an asteroid field is approximately 3720 to 1!<br />
:Han: Seriously? ...Christ<br />
:[Han Solo stands in front of Princess Leia on the Cloud City Carbon Freezing Chamber.]<br />
:Leia: I Love You.<br />
:Han: Oh! Hey, that explains the kissing earlier.<br />
:[Han Solo stands in front of Princess Leia on the Cloud City Carbon Freezing Chamber.]<br />
:Leia: I Love You.<br />
:Han: I'm Nailing Your Brother.<br />
:[Han Solo standing in front of Luke Skywalker, who is holding the blast shield helmet. The training droid hovers between them.]<br />
:Han: Hokey Religions and ancient weapons are no match for scissors, though they do beat paper and rock.<br />
:[Han Solo stands in front of Princess Leia on the Cloud City Carbon Freezing Chamber.]<br />
:Leia: I Love You.<br />
:Han: Cool. Listen, this thing is really, REALLY cold.<br />
:[Han Solo stands in front of Princess Leia on the Cloud City Carbon Freezing Chamber.]<br />
:Leia: I Love You.<br />
:Han: Wowzers.<br />
:[Han Solo sits with two others. General Madine approaches.]<br />
:Madine: General Solo, is your strike team assembled?<br />
:Han: Barely.<br />
:Han: They're pretty drunk.<br />
:[Han Solo and Princess Leia stand in an Ice Tunnel of Hoth.]<br />
:Leia: I'd just as soon kiss a wookie.<br />
:Han: Man, me too, but chewie never seems interested.<br />
:Han: Maybe I should Grow My hair out.<br />
<br />
{{comic discussion}}<br />
[[Category:Star Wars]]</div>Bonobhttps://www.explainxkcd.com/wiki/index.php?title=525:_I_Know_You%27re_Listening&diff=52408525: I Know You're Listening2013-11-09T07:26:52Z<p>Bonob: /* Explanation */</p>
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<div>{{comic<br />
| number = 525<br />
| date = January 2, 2009<br />
| title = I Know You're Listening<br />
| image = i know youre listening.png<br />
| titletext = Basically it's Pascal's Wager for the paranoid prankster.<br />
}}<br />
<br />
==Explanation==<br />
[[Cueball]] periodically says "I know you're listening" aloud in empty rooms. The idea is, that if nobody is listening he doesn't lose anything, but if somebody ''is'' listening he gains by freaking them out.<br />
<br />
As mentioned in the title text, this is similar to {{w|Pascal's Wager}}. Blaise Pascal was a French philosopher and mathematician who discussed the issue of the possibility that God actually does exist or not. A rational person should believe in God because he wouldn't lose anything if this is wrong, but if this belief is correct he would gain immensely by going to heaven at his afterlife due to being a Christian.<br />
<br />
==Transcript==<br />
:Now and then, I announce "I know you're listening" to empty rooms.<br />
:[Cueball is sitting in an armchair, reading. He murmurs something.]<br />
:[Second man in front of a large computer terminal jumps out of chair after hearing the first man mumble. His chair has fallen over.]<br />
:If I'm wrong, no one knows. And if I'm right, maybe I just freaked the hell out of some secret organization.<br />
<br />
{{comic discussion}}<br />
[[Category:Comics featuring Cueball]]</div>Bonobhttps://www.explainxkcd.com/wiki/index.php?title=539:_Boyfriend&diff=52349539: Boyfriend2013-11-08T17:26:04Z<p>Bonob: /* Explanation */</p>
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<div>{{comic<br />
| number = 539<br />
| date = February 4, 2009<br />
| title = Boyfriend<br />
| image = boyfriend.png<br />
| titletext = ...okay, but because you said that, we're breaking up.<br />
}}<br />
<br />
==Explanation==<br />
In statistics, {{w|statistical significance}} is used to measure how well a set of data demonstrates a particular hypothesis or statement. The term {{w|significant other}} is used to refer to a person's intimate relation, typically a spouse or a boyfriend/girlfriend.<br />
<br />
In this comic, [[Megan]] asserts her claim that [[Cueball]] is her boyfriend by presenting the time that he spends with people in the form of a {{w|box plot}}, with her data point lying far ahead of the rest of the chart. Cueball accepts her claim, to which she responds with a pun on the word "significant," combining the phrases "statistically significant" and "significant other" in a so bad pun that in the title text, Cueball breaks up with her over.<br />
<br />
==Transcript==<br />
:[Megan is on the phone.]<br />
:Megan: Can my boyfriend come along?<br />
:Cueball: I'm not your boyfriend!<br />
:Megan: You totally are.<br />
:Cueball: I'm casually dating a number of people.<br />
:[Megan points to a chart.]<br />
:Megan: But you spend twice as much time with me as with anyone else. I'm a clear outlier.<br />
:Cueball: Your math is irrefutable.<br />
:Megan: Face it - I'm your statistically significant other.<br />
<br />
{{comic discussion}}<br />
[[Category:Comics featuring Cueball]]<br />
[[Category:Comics featuring Megan]]<br />
[[Category:Math]]<br />
[[Category:Romance]]<br />
[[Category:Charts]]</div>Bonobhttps://www.explainxkcd.com/wiki/index.php?title=569:_Borders&diff=52217569: Borders2013-11-07T17:12:37Z<p>Bonob: Close quote</p>
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<div>{{comic<br />
| number = 569<br />
| date = April 15, 2009<br />
| title = Borders<br />
| image = borders.png<br />
| titletext = Eventually a UN is set up. And then a lone rebel runs down the line of flags in front of it, runs back to his base, and gets a kajillion points.<br />
}}<br />
<br />
==Explanation==<br />
{{incomplete}}<br />
Capture the flag (CTF) is a common way of playing games where the objective is to capture the opponent's flag, while protecting your own team's flag. This comic describes a CTF server where peace has been established, and no one is trying to capture each other's flags, therefore making the game pointless.<br />
<br />
The title text refers to the line of flags common in front of UN buildings, where the flags of all the teams has been gathered. One could then quickly capture all the flags and reach a very high score. A {{Wiktionary|kajillion}} is slang for "an unspecified large number" of something.<br />
<br />
LIATE is an acronym sometimes used when integrating "by parts". The preferred part to integrate is "Logarithmic, Inverse-trig, Algebraic, Trig, Exponential." Yarbis is a Turkish name, but may be an error for Yarbus, a Russian psychologist who investigated how the eyes scan faces and other things (thus the borders were set according by looking around).<br />
<br />
==Transcript==<br />
:[Two people stand on a hill overlooking a great city. Between them and the city stands an embassy flying a red flag.]<br />
:Three years ago, the kingdom of Liate overthrew their old order and established a constitutional monarchy. Our leaders signed a treaty with their queen, and our borders were set by the Yarbis Accords.<br />
:Many said war would be unending, that peace would always be a dream deferred. But today, our flag flies proudly over our embassy in their kingdom, and they walk our lands without fear.<br />
:So come, traveller. Lay down your grudges and join us in brotherhood. It is time not to fight, but to live.<br />
:[Cueball sitting at computer.]<br />
:Cueball: This is the worst capture-the-flag server ever.<br />
<br />
{{comic discussion}}<br />
[[Category:Comics featuring Cueball]]<br />
[[Category:Comics with color]]</div>Bonobhttps://www.explainxkcd.com/wiki/index.php?title=Talk:853:_Consecutive_Vowels&diff=51511Talk:853: Consecutive Vowels2013-10-31T14:31:00Z<p>Bonob: </p>
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<div>Present tense, or {{w|gerund}}? -- [[User:IronyChef|IronyChef]] ([[User talk:IronyChef|talk]]) 14:52, 16 November 2012 (UTC)<br />
<br />
I always thought the voyeur reference was to the statistical voyeurism is http://xkcd.com/563/<br />
<br />
I don't think '''y''' is a vowel in that word. [[Special:Contributions/184.66.160.91|184.66.160.91]] 05:17, 8 July 2013 (UTC)<br />
<br />
:Y is ''always'' a vowel.[[Special:Contributions/76.29.225.28|76.29.225.28]] 15:21, 17 July 2013 (UTC)<br />
::No --[[User:JSekula71|JSekula71]] ([[User talk:JSekula71|talk]]) 05:33, 18 July 2013 (UTC)<br />
<br />
Y has to be a vowel here or it's not funny ~JFreund<br />
<br />
I don't know if it's related, but 'queue' is the french word for 'tail', and it's slang for dick. Queueing sounds like 'queuter', which is slang for 'to fuck'. [[User:Bonob|Bonob]] ([[User talk:Bonob|talk]]) 14:30, 31 October 2013 (UTC)</div>Bonobhttps://www.explainxkcd.com/wiki/index.php?title=Talk:853:_Consecutive_Vowels&diff=51510Talk:853: Consecutive Vowels2013-10-31T14:30:27Z<p>Bonob: </p>
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<div>Present tense, or {{w|gerund}}? -- [[User:IronyChef|IronyChef]] ([[User talk:IronyChef|talk]]) 14:52, 16 November 2012 (UTC)<br />
<br />
I always thought the voyeur reference was to the statistical voyeurism is http://xkcd.com/563/<br />
<br />
I don't think '''y''' is a vowel in that word. [[Special:Contributions/184.66.160.91|184.66.160.91]] 05:17, 8 July 2013 (UTC)<br />
<br />
:Y is ''always'' a vowel.[[Special:Contributions/76.29.225.28|76.29.225.28]] 15:21, 17 July 2013 (UTC)<br />
::No --[[User:JSekula71|JSekula71]] ([[User talk:JSekula71|talk]]) 05:33, 18 July 2013 (UTC)<br />
<br />
Y has to be a vowel here or it's not funny ~JFreund<br />
<br />
I don't know if it's related, but 'queue' is the french word for 'tail', and it's slang for dick. Queueing sounds like 'queuter', which is slang for 'to fuck'.[[User:Bonob|Bonob]] ([[User talk:Bonob|talk]]) 14:30, 31 October 2013 (UTC)</div>Bonobhttps://www.explainxkcd.com/wiki/index.php?title=Talk:1266:_Halting_Problem&diff=50880Talk:1266: Halting Problem2013-10-18T19:05:23Z<p>Bonob: </p>
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<div>I wrote an explanation for the body of the comics, but I believe there are aspects of the title I'm still missing, so I left the incomplete tag in place. [[User:Shachar|Shachar]] ([[User talk:Shachar|talk]]) 07:52, 18 September 2013 (UTC)<br />
<br />
Isn't google already running applications designed to continue running even if some of nodes they run on have a fatal hardware failure? Also, even if the claim would be true in "practical" sense, it would not solve the problem, because as you said, the stopping would be because of reasons external to the actual program. In other words, program running on turing machine will never stop by hardware failure, because turing machine BY DEFINITION doesn't have any. -- [[User:Hkmaly|Hkmaly]] ([[User talk:Hkmaly|talk]]) 08:57, 18 September 2013 (UTC)<br />
:Remembered this is wiki and added it to the actual explanation :-) -- [[User:Hkmaly|Hkmaly]] ([[User talk:Hkmaly|talk]]) 09:10, 18 September 2013 (UTC)<br />
:Several systems are running with redundant nodes. They will not run forever. They are in for example extremely unlikely to outlive the sun. [[Special:Contributions/85.19.71.131|85.19.71.131]] 11:29, 18 September 2013 (UTC)<br />
::System with ability to replace nodes can be deployed on nodes physically as distant as needed. Application which is currently starting on a multi-node system on earth can be later expanded to system with nodes in different star system. Although unless the nodes have FTL connection it would have inpractically large lag ... -- [[User:Hkmaly|Hkmaly]] ([[User talk:Hkmaly|talk]]) 09:39, 20 September 2013 (UTC)<br />
<br />
"For all practical purposes, this is the correct solution"<br />
:No, it's not. A very practical purpose would be "have my OS kill processes that won't stop". Other one would be "reject installing apps that contain algorithms that don't halt". If the OS assumes "every app will eventually halt" it would kill every process and reject every app. [[User:Osias|Osias]] ([[User talk:Osias|talk]]) 12:15, 18 September 2013 (UTC)<br />
::Changing the paragraph to say "a physical perspective" instead of "all practical purposes" was a good solution. [[User:Osias|Osias]] ([[User talk:Osias|talk]]) 14:16, 18 September 2013 (UTC)<br />
::It would, in fact, kill/reject none since it would find no nonhalters.[[Special:Contributions/178.0.89.106|178.0.89.106]] 20:51, 18 September 2013 (UTC)<br />
<br />
<br />
Google "halting problem" and do a little reeding so you are in the same mindset as Randall. This is a famous computer science problem. You aren't talking about the same thing in comments above. ''&mdash; [[User:Tbc|tbc]] ([[User talk:Tbc|talk]]) 12:30, 18 September 2013 (UTC)''<br />
<br />
What is the joke here? What does "big picture" mean? [[Special:Contributions/62.209.198.2|62.209.198.2]] 16:33, 18 September 2013 (UTC)<br />
:I believe it's related to the quote " In the long run we are all dead." by John Maynard Keynes. [[User:Osias|Osias]] ([[User talk:Osias|talk]]) 18:46, 18 September 2013 (UTC)<br />
<br />
Same kind of humor as in http://www.explainxkcd.com/wiki/index.php?title=221 [[Special:Contributions/176.67.13.14|176.67.13.14]] 18:47, 18 September 2013 (UTC)<br />
<br />
A program with its given input can be seen, as a whole, as a specific program. Therefore the halting function need not take two inputs and is equivalent to a function that takes the two described inputs. Therefore I feel the comment about the number of inputs in the explanation can be removed. {{unsigned ip|66.69.243.27}}<br />
:Yeah, the halting problem on the empty word/input is known to be equivalent to the general halting problem. I think that's the form used in this comic. {{unsigned ip|85.218.82.16}}<br />
<br />
Might there be a reference here to Isaac Asimov's famous story "The Last Question"? (The titular question was: 'Can entropy be reversed?' Throughout the lifetime of the universe, the computer only said: 'THERE IS INSUFFICIENT DATA FOR A MEANINGFUL ANSWER.') [[Special:Contributions/174.239.229.68|174.239.229.68]] 04:18, 20 September 2013 (UTC)<br />
<br />
;Missing the obvious?<br />
<br />
Maybe it is just me, but I interpreted this to be the "DoesItHalt" function actually *running* "program", then when "program" completes (halts) it returns true. This would be the "big picture" solution and does not actually deal with the "details". --[[User:Bpothier|B. P.]] ([[User talk:Bpothier|talk]]) 23:52, 20 September 2013 (UTC)<br />
<br />
What you thought you saw was the most obvious "implementation" of a solution to the halting problem.<br />
define DoesItHalt (program):<br />
{<br />
program();<br />
return true;<br />
}<br />
This solution returns <code>true</code> if <code>program</code> stops, and loops forever is <code>program</code> loops forever. [[User:Xhfz|Xhfz]] ([[User talk:Xhfz|talk]]) 20:44, 23 September 2013 (UTC)<br />
<br />
: It won't work if your program is exit() or shutdownYourComputer(). :) --[[Special:Contributions/61.223.87.164|61.223.87.164]] 00:49, 28 September 2013 (UTC)<br />
<br />
::It will. We are talking about Turing machines. A Turing machine can stop itself, but it cannot stop the calling Turing machine. [[User:Xhfz|Xhfz]] ([[User talk:Xhfz|talk]]) 12:43, 7 October 2013 (UTC)<br />
<br />
:::Turing machines are known to be really poor at I/O. If you trace the shutdownYourComputer function, it actually instructs your {{w|Power_supply_unit_(computer)|power supply unit}} (ATX required, AT lacked such capability) to turn power down. Similarly, rebootYourComputer function instruct outside hardware - usually {{w|Northbridge_(computing)|north bridge}} - to send a reset signal on the {{w|Peripheral_Component_Interconnect|PCI bus}} (and presumably other busses), which will reset all devices and start {{w|Power-on_self-test|POST}}. Unlike Turing machines, virtualized OSs may be able to reboot host computer if the hypervisor is not coded correctly and allows I/O for hardware acceleration. -- [[User:Hkmaly|Hkmaly]] ([[User talk:Hkmaly|talk]]) 09:22, 16 October 2013 (UTC)<br />
; The halting problem for every input<br />
The sentence in bold is false<br />
:It should be noted that Randall's solution, barring its unsoundness, solves more than the halting problem in the form it is usually stated. The halting problem requires two parameters (a program and its parameters), while Randall's function only accepts one (the program). The question of whether a program halts for every input '''can be shown to be even harder to solve''' than the halting problem, meaning that even if a Turing machine had an additional instruction allowing it to check whether a program halts with given parameters, it still could not always confirm that a given program that halts for all parameters does so.<br />
In fact, let A be a Turing machine that solves the halting problem for one input, and B a Turing machine that solves the halting problem for every input. B can be built using A as a subroutine. First B builds machine Y that takes its input, X, and halts if X loops with at least one input; Y loops if X stops for every input.<br />
:Turing machine B (Turing machine X) {<br />
:// manipulate X in order to build Y that calls X for every input and stops when the first input does not halt<br />
: Y = <code>"s = <nowiki>''</nowiki>; repeat { if (A(" + X + "," s + ") == 'false' then halt; s = next(s);}"</code><br />
: return A(Y, "") // second parameter is ignored<br />
:}<br />
The function next returns the next valid string. For example, if our alphabet is A..Z0..9, then next("AJ38YT") = "AJ38YU"<br />
<br />
Machine B determines if Y halts. And Y halts if machine X does not halt. If we had Turing machine A then we could build B. <br />
[[User:Xhfz|Xhfz]] ([[User talk:Xhfz|talk]]) 17:53, 18 October 2013 (UTC)<br />
<br />
; Karl Popper<br />
I think that the title text is a direct reference to Karl Popper's falsifiability argument, since this is one of the most common example of a non-falsifiable statement. [[User:Bonob|Bonob]] ([[User talk:Bonob|talk]]) 19:01, 18 October 2013 (UTC)</div>Bonobhttps://www.explainxkcd.com/wiki/index.php?title=Talk:1266:_Halting_Problem&diff=50879Talk:1266: Halting Problem2013-10-18T19:04:20Z<p>Bonob: </p>
<hr />
<div>I wrote an explanation for the body of the comics, but I believe there are aspects of the title I'm still missing, so I left the incomplete tag in place. [[User:Shachar|Shachar]] ([[User talk:Shachar|talk]]) 07:52, 18 September 2013 (UTC)<br />
<br />
Isn't google already running applications designed to continue running even if some of nodes they run on have a fatal hardware failure? Also, even if the claim would be true in "practical" sense, it would not solve the problem, because as you said, the stopping would be because of reasons external to the actual program. In other words, program running on turing machine will never stop by hardware failure, because turing machine BY DEFINITION doesn't have any. -- [[User:Hkmaly|Hkmaly]] ([[User talk:Hkmaly|talk]]) 08:57, 18 September 2013 (UTC)<br />
:Remembered this is wiki and added it to the actual explanation :-) -- [[User:Hkmaly|Hkmaly]] ([[User talk:Hkmaly|talk]]) 09:10, 18 September 2013 (UTC)<br />
:Several systems are running with redundant nodes. They will not run forever. They are in for example extremely unlikely to outlive the sun. [[Special:Contributions/85.19.71.131|85.19.71.131]] 11:29, 18 September 2013 (UTC)<br />
::System with ability to replace nodes can be deployed on nodes physically as distant as needed. Application which is currently starting on a multi-node system on earth can be later expanded to system with nodes in different star system. Although unless the nodes have FTL connection it would have inpractically large lag ... -- [[User:Hkmaly|Hkmaly]] ([[User talk:Hkmaly|talk]]) 09:39, 20 September 2013 (UTC)<br />
<br />
"For all practical purposes, this is the correct solution"<br />
:No, it's not. A very practical purpose would be "have my OS kill processes that won't stop". Other one would be "reject installing apps that contain algorithms that don't halt". If the OS assumes "every app will eventually halt" it would kill every process and reject every app. [[User:Osias|Osias]] ([[User talk:Osias|talk]]) 12:15, 18 September 2013 (UTC)<br />
::Changing the paragraph to say "a physical perspective" instead of "all practical purposes" was a good solution. [[User:Osias|Osias]] ([[User talk:Osias|talk]]) 14:16, 18 September 2013 (UTC)<br />
::It would, in fact, kill/reject none since it would find no nonhalters.[[Special:Contributions/178.0.89.106|178.0.89.106]] 20:51, 18 September 2013 (UTC)<br />
<br />
<br />
Google "halting problem" and do a little reeding so you are in the same mindset as Randall. This is a famous computer science problem. You aren't talking about the same thing in comments above. ''&mdash; [[User:Tbc|tbc]] ([[User talk:Tbc|talk]]) 12:30, 18 September 2013 (UTC)''<br />
<br />
What is the joke here? What does "big picture" mean? [[Special:Contributions/62.209.198.2|62.209.198.2]] 16:33, 18 September 2013 (UTC)<br />
:I believe it's related to the quote " In the long run we are all dead." by John Maynard Keynes. [[User:Osias|Osias]] ([[User talk:Osias|talk]]) 18:46, 18 September 2013 (UTC)<br />
<br />
Same kind of humor as in http://www.explainxkcd.com/wiki/index.php?title=221 [[Special:Contributions/176.67.13.14|176.67.13.14]] 18:47, 18 September 2013 (UTC)<br />
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A program with its given input can be seen, as a whole, as a specific program. Therefore the halting function need not take two inputs and is equivalent to a function that takes the two described inputs. Therefore I feel the comment about the number of inputs in the explanation can be removed. {{unsigned ip|66.69.243.27}}<br />
:Yeah, the halting problem on the empty word/input is known to be equivalent to the general halting problem. I think that's the form used in this comic. {{unsigned ip|85.218.82.16}}<br />
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Might there be a reference here to Isaac Asimov's famous story "The Last Question"? (The titular question was: 'Can entropy be reversed?' Throughout the lifetime of the universe, the computer only said: 'THERE IS INSUFFICIENT DATA FOR A MEANINGFUL ANSWER.') [[Special:Contributions/174.239.229.68|174.239.229.68]] 04:18, 20 September 2013 (UTC)<br />
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;Missing the obvious?<br />
<br />
Maybe it is just me, but I interpreted this to be the "DoesItHalt" function actually *running* "program", then when "program" completes (halts) it returns true. This would be the "big picture" solution and does not actually deal with the "details". --[[User:Bpothier|B. P.]] ([[User talk:Bpothier|talk]]) 23:52, 20 September 2013 (UTC)<br />
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What you thought you saw was the most obvious "implementation" of a solution to the halting problem.<br />
define DoesItHalt (program):<br />
{<br />
program();<br />
return true;<br />
}<br />
This solution returns <code>true</code> if <code>program</code> stops, and loops forever is <code>program</code> loops forever. [[User:Xhfz|Xhfz]] ([[User talk:Xhfz|talk]]) 20:44, 23 September 2013 (UTC)<br />
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: It won't work if your program is exit() or shutdownYourComputer(). :) --[[Special:Contributions/61.223.87.164|61.223.87.164]] 00:49, 28 September 2013 (UTC)<br />
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::It will. We are talking about Turing machines. A Turing machine can stop itself, but it cannot stop the calling Turing machine. [[User:Xhfz|Xhfz]] ([[User talk:Xhfz|talk]]) 12:43, 7 October 2013 (UTC)<br />
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:::Turing machines are known to be really poor at I/O. If you trace the shutdownYourComputer function, it actually instructs your {{w|Power_supply_unit_(computer)|power supply unit}} (ATX required, AT lacked such capability) to turn power down. Similarly, rebootYourComputer function instruct outside hardware - usually {{w|Northbridge_(computing)|north bridge}} - to send a reset signal on the {{w|Peripheral_Component_Interconnect|PCI bus}} (and presumably other busses), which will reset all devices and start {{w|Power-on_self-test|POST}}. Unlike Turing machines, virtualized OSs may be able to reboot host computer if the hypervisor is not coded correctly and allows I/O for hardware acceleration. -- [[User:Hkmaly|Hkmaly]] ([[User talk:Hkmaly|talk]]) 09:22, 16 October 2013 (UTC)<br />
; The halting problem for every input<br />
The sentence in bold is false<br />
:It should be noted that Randall's solution, barring its unsoundness, solves more than the halting problem in the form it is usually stated. The halting problem requires two parameters (a program and its parameters), while Randall's function only accepts one (the program). The question of whether a program halts for every input '''can be shown to be even harder to solve''' than the halting problem, meaning that even if a Turing machine had an additional instruction allowing it to check whether a program halts with given parameters, it still could not always confirm that a given program that halts for all parameters does so.<br />
In fact, let A be a Turing machine that solves the halting problem for one input, and B a Turing machine that solves the halting problem for every input. B can be built using A as a subroutine. First B builds machine Y that takes its input, X, and halts if X loops with at least one input; Y loops if X stops for every input.<br />
:Turing machine B (Turing machine X) {<br />
:// manipulate X in order to build Y that calls X for every input and stops when the first input does not halt<br />
: Y = <code>"s = <nowiki>''</nowiki>; repeat { if (A(" + X + "," s + ") == 'false' then halt; s = next(s);}"</code><br />
: return A(Y, "") // second parameter is ignored<br />
:}<br />
The function next returns the next valid string. For example, if our alphabet is A..Z0..9, then next("AJ38YT") = "AJ38YU"<br />
<br />
Machine B determines if Y halts. And Y halts if machine X does not halt. If we had Turing machine A then we could build B. <br />
[[User:Xhfz|Xhfz]] ([[User talk:Xhfz|talk]]) 17:53, 18 October 2013 (UTC)<br />
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<br />
; Karl Popper<br />
I think that the title text is a direct reference to Karl Popper's falsifiability argument, since this is one of the most common example of a non-falsifiable statement.[[User:Bonob|Bonob]] ([[User talk:Bonob|talk]]) 19:01, 18 October 2013 (UTC)</div>Bonobhttps://www.explainxkcd.com/wiki/index.php?title=Talk:1266:_Halting_Problem&diff=50878Talk:1266: Halting Problem2013-10-18T19:01:39Z<p>Bonob: </p>
<hr />
<div>I wrote an explanation for the body of the comics, but I believe there are aspects of the title I'm still missing, so I left the incomplete tag in place. [[User:Shachar|Shachar]] ([[User talk:Shachar|talk]]) 07:52, 18 September 2013 (UTC)<br />
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Isn't google already running applications designed to continue running even if some of nodes they run on have a fatal hardware failure? Also, even if the claim would be true in "practical" sense, it would not solve the problem, because as you said, the stopping would be because of reasons external to the actual program. In other words, program running on turing machine will never stop by hardware failure, because turing machine BY DEFINITION doesn't have any. -- [[User:Hkmaly|Hkmaly]] ([[User talk:Hkmaly|talk]]) 08:57, 18 September 2013 (UTC)<br />
:Remembered this is wiki and added it to the actual explanation :-) -- [[User:Hkmaly|Hkmaly]] ([[User talk:Hkmaly|talk]]) 09:10, 18 September 2013 (UTC)<br />
:Several systems are running with redundant nodes. They will not run forever. They are in for example extremely unlikely to outlive the sun. [[Special:Contributions/85.19.71.131|85.19.71.131]] 11:29, 18 September 2013 (UTC)<br />
::System with ability to replace nodes can be deployed on nodes physically as distant as needed. Application which is currently starting on a multi-node system on earth can be later expanded to system with nodes in different star system. Although unless the nodes have FTL connection it would have inpractically large lag ... -- [[User:Hkmaly|Hkmaly]] ([[User talk:Hkmaly|talk]]) 09:39, 20 September 2013 (UTC)<br />
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"For all practical purposes, this is the correct solution"<br />
:No, it's not. A very practical purpose would be "have my OS kill processes that won't stop". Other one would be "reject installing apps that contain algorithms that don't halt". If the OS assumes "every app will eventually halt" it would kill every process and reject every app. [[User:Osias|Osias]] ([[User talk:Osias|talk]]) 12:15, 18 September 2013 (UTC)<br />
::Changing the paragraph to say "a physical perspective" instead of "all practical purposes" was a good solution. [[User:Osias|Osias]] ([[User talk:Osias|talk]]) 14:16, 18 September 2013 (UTC)<br />
::It would, in fact, kill/reject none since it would find no nonhalters.[[Special:Contributions/178.0.89.106|178.0.89.106]] 20:51, 18 September 2013 (UTC)<br />
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Google "halting problem" and do a little reeding so you are in the same mindset as Randall. This is a famous computer science problem. You aren't talking about the same thing in comments above. ''&mdash; [[User:Tbc|tbc]] ([[User talk:Tbc|talk]]) 12:30, 18 September 2013 (UTC)''<br />
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What is the joke here? What does "big picture" mean? [[Special:Contributions/62.209.198.2|62.209.198.2]] 16:33, 18 September 2013 (UTC)<br />
:I believe it's related to the quote " In the long run we are all dead." by John Maynard Keynes. [[User:Osias|Osias]] ([[User talk:Osias|talk]]) 18:46, 18 September 2013 (UTC)<br />
<br />
Same kind of humor as in http://www.explainxkcd.com/wiki/index.php?title=221 [[Special:Contributions/176.67.13.14|176.67.13.14]] 18:47, 18 September 2013 (UTC)<br />
<br />
A program with its given input can be seen, as a whole, as a specific program. Therefore the halting function need not take two inputs and is equivalent to a function that takes the two described inputs. Therefore I feel the comment about the number of inputs in the explanation can be removed. {{unsigned ip|66.69.243.27}}<br />
:Yeah, the halting problem on the empty word/input is known to be equivalent to the general halting problem. I think that's the form used in this comic. {{unsigned ip|85.218.82.16}}<br />
<br />
Might there be a reference here to Isaac Asimov's famous story "The Last Question"? (The titular question was: 'Can entropy be reversed?' Throughout the lifetime of the universe, the computer only said: 'THERE IS INSUFFICIENT DATA FOR A MEANINGFUL ANSWER.') [[Special:Contributions/174.239.229.68|174.239.229.68]] 04:18, 20 September 2013 (UTC)<br />
<br />
;Missing the obvious?<br />
<br />
Maybe it is just me, but I interpreted this to be the "DoesItHalt" function actually *running* "program", then when "program" completes (halts) it returns true. This would be the "big picture" solution and does not actually deal with the "details". --[[User:Bpothier|B. P.]] ([[User talk:Bpothier|talk]]) 23:52, 20 September 2013 (UTC)<br />
<br />
What you thought you saw was the most obvious "implementation" of a solution to the halting problem.<br />
define DoesItHalt (program):<br />
{<br />
program();<br />
return true;<br />
}<br />
This solution returns <code>true</code> if <code>program</code> stops, and loops forever is <code>program</code> loops forever. [[User:Xhfz|Xhfz]] ([[User talk:Xhfz|talk]]) 20:44, 23 September 2013 (UTC)<br />
<br />
: It won't work if your program is exit() or shutdownYourComputer(). :) --[[Special:Contributions/61.223.87.164|61.223.87.164]] 00:49, 28 September 2013 (UTC)<br />
<br />
::It will. We are talking about Turing machines. A Turing machine can stop itself, but it cannot stop the calling Turing machine. [[User:Xhfz|Xhfz]] ([[User talk:Xhfz|talk]]) 12:43, 7 October 2013 (UTC)<br />
<br />
:::Turing machines are known to be really poor at I/O. If you trace the shutdownYourComputer function, it actually instructs your {{w|Power_supply_unit_(computer)|power supply unit}} (ATX required, AT lacked such capability) to turn power down. Similarly, rebootYourComputer function instruct outside hardware - usually {{w|Northbridge_(computing)|north bridge}} - to send a reset signal on the {{w|Peripheral_Component_Interconnect|PCI bus}} (and presumably other busses), which will reset all devices and start {{w|Power-on_self-test|POST}}. Unlike Turing machines, virtualized OSs may be able to reboot host computer if the hypervisor is not coded correctly and allows I/O for hardware acceleration. -- [[User:Hkmaly|Hkmaly]] ([[User talk:Hkmaly|talk]]) 09:22, 16 October 2013 (UTC)<br />
; The halting problem for every input<br />
The sentence in bold is false<br />
:It should be noted that Randall's solution, barring its unsoundness, solves more than the halting problem in the form it is usually stated. The halting problem requires two parameters (a program and its parameters), while Randall's function only accepts one (the program). The question of whether a program halts for every input '''can be shown to be even harder to solve''' than the halting problem, meaning that even if a Turing machine had an additional instruction allowing it to check whether a program halts with given parameters, it still could not always confirm that a given program that halts for all parameters does so.<br />
In fact, let A be a Turing machine that solves the halting problem for one input, and B a Turing machine that solves the halting problem for every input. B can be built using A as a subroutine. First B builds machine Y that takes its input, X, and halts if X loops with at least one input; Y loops if X stops for every input.<br />
:Turing machine B (Turing machine X) {<br />
:// manipulate X in order to build Y that calls X for every input and stops when the first input does not halt<br />
: Y = <code>"s = <nowiki>''</nowiki>; repeat { if (A(" + X + "," s + ") == 'false' then halt; s = next(s);}"</code><br />
: return A(Y, "") // second parameter is ignored<br />
:}<br />
The function next returns the next valid string. For example, if our alphabet is A..Z0..9, then next("AJ38YT") = "AJ38YU"<br />
<br />
Machine B determines if Y halts. And Y halts if machine X does not halt. If we had Turing machine A then we could build B. <br />
[[User:Xhfz|Xhfz]] ([[User talk:Xhfz|talk]]) 17:53, 18 October 2013 (UTC)<br />
<br />
I think that the title text is a direct reference to Karl Popper's falsifiability argument, since this is the most common example of a non-falsifiable statement.[[User:Bonob|Bonob]] ([[User talk:Bonob|talk]]) 19:01, 18 October 2013 (UTC)</div>Bonob