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::I'd do it in Perl, but that's because I'm a bit partial to Perl.  I don't think it matters too much.  It could even be a semi-manual process.
 
::I'd do it in Perl, but that's because I'm a bit partial to Perl.  I don't think it matters too much.  It could even be a semi-manual process.
 
::However, whatever way it's done, if there was a loop (or a flip-flop state, i.e. more black pixels overall means less black pixels on a graph, which means less black pixels overall, without a point of stability) then I'd detect for that and work out which "immutable" parts (e.g. lengths of drawn axes) could be altered by an appropriate number of pixels to have another go at looking for stability.  In Perl, that'd be detected by something like a simple "$coverage{$no_of_black_pixels}++" for every state visited, with an "if (exists $coverage{$no_of_black_pixels}) { reject_and_renew() }"-style check before that, probably "die"ing the program to let me read the log of rejections that led there and let me choose a basic change (or other mutable element) that could lead us in the right direction.
 
::However, whatever way it's done, if there was a loop (or a flip-flop state, i.e. more black pixels overall means less black pixels on a graph, which means less black pixels overall, without a point of stability) then I'd detect for that and work out which "immutable" parts (e.g. lengths of drawn axes) could be altered by an appropriate number of pixels to have another go at looking for stability.  In Perl, that'd be detected by something like a simple "$coverage{$no_of_black_pixels}++" for every state visited, with an "if (exists $coverage{$no_of_black_pixels}) { reject_and_renew() }"-style check before that, probably "die"ing the program to let me read the log of rejections that led there and let me choose a basic change (or other mutable element) that could lead us in the right direction.
βˆ’
:::In an [http://www.maa.org/publications/periodicals/math-horizons/the-mathematics-behind-xkcd-a-conversation-with-randall-munroe interview], Randall Munroe explains that he did it completely by hand, counting the black pixels with Photoshop and iterating manually. Notice that, once you chose the radius of the disc, the width and scale for the rectangles of the second panel, and the text, decorations and legend, it is fairly easy to write the equations satisfied by the amount of black ink each panel. It turns out to be a set of linear equations, easily solved. This is for the continuous problem (say, if the comic strip were drawn with Bezier curves and a vector image). For the discrete problem, you have to iterate a little bit from this first insight, but not that much. [[Special:Contributions/138.96.199.247|138.96.199.247]] 10:15, 16 September 2013 (UTC)
 
 
::As a more simple example, if the original title-text hadn't turned out to be one where a certain stated number of characters made the text that same number of characters, I'd add, remove or change a word (or item of punctiation!) towards something that worked.  As a dumb example of the way I'd do it: "This sentence has <foo> characters." has 35, there, including the five of "<foo>", so "thirty-five" would be six too many, "forty-one" would be 39-long, "thirty-nine" makes it "forty-one", and we know that loops back.  I could be more intelligent and choose a number where own_length==(what it depicts, minus thirty), where the easy answer is "32" in digits.  But there's no obvious set of number words that obet that rule, so let's change the sentence to "There are <foo> characters in this sentence.", and see where ''that'' leads us.  Quick answer? 39+length of added number words.  If I'm right, that's "forty-nine".
 
::As a more simple example, if the original title-text hadn't turned out to be one where a certain stated number of characters made the text that same number of characters, I'd add, remove or change a word (or item of punctiation!) towards something that worked.  As a dumb example of the way I'd do it: "This sentence has <foo> characters." has 35, there, including the five of "<foo>", so "thirty-five" would be six too many, "forty-one" would be 39-long, "thirty-nine" makes it "forty-one", and we know that loops back.  I could be more intelligent and choose a number where own_length==(what it depicts, minus thirty), where the easy answer is "32" in digits.  But there's no obvious set of number words that obet that rule, so let's change the sentence to "There are <foo> characters in this sentence.", and see where ''that'' leads us.  Quick answer? 39+length of added number words.  If I'm right, that's "forty-nine".
 
::Of course, there are multiple loop-backs with a self-referential image.  But while it would be 'obvious' if extra spaces were inserted (or some removed!) to make a line of text fit itself in a self-referential way, an image has more "neutral space" (or 'fill') that can be changed with no effect on itself but (in a non-linear way and deminishing returns, especially with the multiple levels of recursion in the third panel) can shuffle values in the rest, perhaps to hit upon a self-consistent result overall.  Narrowing or widening the panes (thus making more/less white space, and only slightly different black space) could change the ratio enough to hit a solution.  Or altering the radius of the pie-chart by a pixel or three (while obviously also updating the angle filled in) could help.  And if it didn't work with a pie-chart, for some reason, a big block of text saying "x% black vs y% white", or similar, could have possibly set up a result.  The problem is not finding a method of solving the problem, but that there are way too many ways.  But once you hit one that doesn't look forced, that'd be good enough and you could roll with it.
 
::Of course, there are multiple loop-backs with a self-referential image.  But while it would be 'obvious' if extra spaces were inserted (or some removed!) to make a line of text fit itself in a self-referential way, an image has more "neutral space" (or 'fill') that can be changed with no effect on itself but (in a non-linear way and deminishing returns, especially with the multiple levels of recursion in the third panel) can shuffle values in the rest, perhaps to hit upon a self-consistent result overall.  Narrowing or widening the panes (thus making more/less white space, and only slightly different black space) could change the ratio enough to hit a solution.  Or altering the radius of the pie-chart by a pixel or three (while obviously also updating the angle filled in) could help.  And if it didn't work with a pie-chart, for some reason, a big block of text saying "x% black vs y% white", or similar, could have possibly set up a result.  The problem is not finding a method of solving the problem, but that there are way too many ways.  But once you hit one that doesn't look forced, that'd be good enough and you could roll with it.

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