Judebert.com (Entries tagged as math)
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http://judebert.com/progress/
100100The Alpha Geek Programs Math
http://judebert.com/progress/archives/351-The-Alpha-Geek-Programs-Math.html
Programminghttp://judebert.com/progress/archives/351-The-Alpha-Geek-Programs-Math.html#commentshttp://judebert.com/progress/wfwcomment.php?cid=3512http://judebert.com/progress/rss.php?version=2.0&type=comments&cid=351judebert@judebert.com (Judebert)
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I know I've been neglecting my blogs lately. And my online forums. And FaceBook. I apologize.
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I've been working on a neat little puzzle game. Not a "puzzle game" like Bejeweled or those other match-up toys. Something that actually requires a little bit of thought, like Sudoku or Griddlers, but more logic. I've got a prototype, but I'm not ready to release it just yet; I promise to let you know when it's ready.
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What's taken me so long is the mathematical whirlwind I've immersed myself in. It's been delightful. (Yes, the Alpha Geek enjoys math.) For a rather minor aspect of the coding, I needed a way to sum any pair of 36 possible numbers and guarantee that the result was unique for that pair. (In mathematical terms, I needed a set such that a<sub>i</sub> + a<sub>j</sub> is distinct for all i < j.) I searched online for days before finding what I need: weak Sidon sets, also known as "well-spread sets" or S<sub>2</sub> sets.
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It turns out that nobody has published any weak Sidon sequences of any appreciable length, though. I needed to write a program to generate them.
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My first try turned up a 36-member set and a 100-member set in short order. But I wanted smaller numbers, so I programmed more exhaustive search methods. One of them ran for 36 hours before Kayla decided to log me out, destroying all my progress.
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So I modified the program. It now finds all well-spread sequences of a particular length chosen from the set of integers smaller than a specified maximum. And it saves its progress, so if I'm ever interrupted again, I can simply resume. I can even run on multiple computers by generating new start states (although of course I'll have to quit one of the programs manually).
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When I've got my first set of sequences, I'll publish my results and my program. I just thought you should know why you haven't heard much from me.
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Thu, 11 Jun 2009 15:41:46 -0400http://judebert.com/progress/archives/351-guid.htmlalpha geekmathprogrammingMath Injury
http://judebert.com/progress/archives/317-Math-Injury.html
Tatianahttp://judebert.com/progress/archives/317-Math-Injury.html#commentshttp://judebert.com/progress/wfwcomment.php?cid=3170http://judebert.com/progress/rss.php?version=2.0&type=comments&cid=317judebert@judebert.com (Judebert)
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Tatiana went to another Mu Alpha Theta competition this weekend. She took 9th place in the individual geometry competition! <a href="http://judebert.com/progress/archives/317-Math-Injury.html#extended">(Read on for one of the interesting problems she couldn't get.)</a>
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Unfortunately, she suffered an injury. You don't normally associate injuries with math; it's not a rough sport or anything. But it turns out Tatiana inherited her mother's brittle feet. She severely sprained her ankle on the way to the stage.
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She's on crutches and ibuprofen.
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<strong>Update 2009-02-04:</strong> I will no longer take my family to the CentraCare on University Blvd. After 48 hours without a reduction in swelling, we took her to our pediatrician, who discovered that the leg is actually fractured. Tatiana is now in a walking cast for 4 weeks. <p>
Problem number 15 was interesting. Given a right triangle, such that the hypotenuse c = a + 1, what is b<sup>2</sup>?
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It's not really difficult. c<sup>2</sup> = a<sup>2</sup> + b<sup>2</sup> so b<sup>2</sup> = c<sup>2</sup> - a<sup>2</sup>. Find c<sup>2</sup> with the classic FOIL method: (a + 1)<sup>2</sup> = a<sup>2</sup> + 2a +1. Subtract a<sup>2</sup>, and you see that c = 2a + 1.
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2a + 1 wasn't among the answers.
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<em>But c + a was.</em>
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You have to do an extra substitution step to realize that 2a + 1 is the same as a + c. After all, 2a + 1 = a + (a + 1). And earlier, we said that c = a + 1. So 2a + 1 = a + c = c + a.
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I suppose you could also have realized that a 3,4,5 triangle could be said to satisfy c = a + 1 (if you swap the common definitions for a and b). Then b<sup>2</sup> is 9. And 9 = 4 + 5.
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If you wanted to do it the easy, sissy way.
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Mon, 02 Feb 2009 14:35:20 -0500http://judebert.com/progress/archives/317-guid.htmlinjuriesmathschool