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		<title>Fibonacci multiples</title>
		<link>http://mathlesstraveled.com/2012/05/15/fibonacci-multiples/</link>
		<comments>http://mathlesstraveled.com/2012/05/15/fibonacci-multiples/#comments</comments>
		<pubDate>Tue, 15 May 2012 15:26:05 +0000</pubDate>
		<dc:creator>Brent</dc:creator>
				<category><![CDATA[arithmetic]]></category>
		<category><![CDATA[challenges]]></category>
		<category><![CDATA[fibonacci]]></category>
		<category><![CDATA[number theory]]></category>
		<category><![CDATA[pattern]]></category>
		<category><![CDATA[divisibility]]></category>

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		<description><![CDATA[I haven&#8217;t written anything here in a while, but hope to write more regularly now that the semester is over&#8212;I have a series on combinatorial proofs to finish up, some books to review, and a few other things planned. But &#8230; <a href="http://mathlesstraveled.com/2012/05/15/fibonacci-multiples/">Continue reading <span class="meta-nav">&#8594;</span></a><img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=mathlesstraveled.com&#038;blog=21246689&#038;post=1639&#038;subd=mathlesstraveled&#038;ref=&#038;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p>I haven&#8217;t written anything here in a while, but hope to write more regularly now that the semester is over&#8212;I have a <a href="http://mathlesstraveled.com/2012/01/29/differences-of-powers-of-consecutive-integers/" title="Differences of powers of consecutive integers">series</a> <a href="http://mathlesstraveled.com/2012/02/16/differences-of-powers-of-consecutive-integers-part-ii/" title="Differences of powers of consecutive integers, part II">on</a> <a href="http://mathlesstraveled.com/2012/02/17/combinatorial-proofs/" title="Combinatorial proofs">combinatorial</a> <a href="http://mathlesstraveled.com/2012/03/03/making-our-equation-count/" title="Making our equation count">proofs</a> to finish up, some books to review, and a few other things planned.  But to ease back into things, here&#8217;s a little puzzle for you.  Recall that the Fibonacci numbers are defined by </p>
<p><img src='http://s0.wp.com/latex.php?latex=F_0+%3D+0%3B+F_1+%3D+1%3B+F_%7Bn%2B2%7D+%3D+F_%7Bn%2B1%7D+%2B+F_n&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='F_0 = 0; F_1 = 1; F_{n+2} = F_{n+1} + F_n' title='F_0 = 0; F_1 = 1; F_{n+2} = F_{n+1} + F_n' class='latex' />.</p>
<p>Can you figure out a way to prove the following cute theorem?</p>
<blockquote><p>
If <img src='http://s0.wp.com/latex.php?latex=m&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='m' title='m' class='latex' /> evenly divides <img src='http://s0.wp.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n' title='n' class='latex' />, then <img src='http://s0.wp.com/latex.php?latex=F_m&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='F_m' title='F_m' class='latex' /> evenly divides <img src='http://s0.wp.com/latex.php?latex=F_n&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='F_n' title='F_n' class='latex' /></i>.
</p></blockquote>
<p>(Incidentally, the existence of this theorem constitutes good evidence that the &#8220;correct&#8221; definition of <img src='http://s0.wp.com/latex.php?latex=F_0&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='F_0' title='F_0' class='latex' /> is <img src='http://s0.wp.com/latex.php?latex=0&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='0' title='0' class='latex' />, not <img src='http://s0.wp.com/latex.php?latex=1&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='1' title='1' class='latex' />.)</p>
<p>For example, <img src='http://s0.wp.com/latex.php?latex=5&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='5' title='5' class='latex' /> evenly divides <img src='http://s0.wp.com/latex.php?latex=10&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='10' title='10' class='latex' />, and sure enough, <img src='http://s0.wp.com/latex.php?latex=F_5+%3D+5&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='F_5 = 5' title='F_5 = 5' class='latex' /> evenly divides <img src='http://s0.wp.com/latex.php?latex=F_%7B10%7D+%3D+55&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='F_{10} = 55' title='F_{10} = 55' class='latex' />.  <img src='http://s0.wp.com/latex.php?latex=13&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='13' title='13' class='latex' /> evenly divides <img src='http://s0.wp.com/latex.php?latex=91&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='91' title='91' class='latex' />, and sure enough, <img src='http://s0.wp.com/latex.php?latex=F_%7B13%7D+%3D+233&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='F_{13} = 233' title='F_{13} = 233' class='latex' /> evenly divides <img src='http://s0.wp.com/latex.php?latex=F_%7B91%7D+%3D+4660046610375530309&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='F_{91} = 4660046610375530309' title='F_{91} = 4660046610375530309' class='latex' /> (in particular, <img src='http://s0.wp.com/latex.php?latex=4660046610375530309+%3D+233+%5Ctimes+20000200044530173&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='4660046610375530309 = 233 &#92;times 20000200044530173' title='4660046610375530309 = 233 &#92;times 20000200044530173' class='latex' />).</p>
<p>I know of two different ways to prove it; there are probably more!  Neither of the proofs I know is particularly obvious, but they do not require any difficult concepts.</p>
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			<media:title type="html">Brent</media:title>
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		<item>
		<title>Carnival of Mathematics 86</title>
		<link>http://mathlesstraveled.com/2012/05/08/carnival-of-mathematics-86/</link>
		<comments>http://mathlesstraveled.com/2012/05/08/carnival-of-mathematics-86/#comments</comments>
		<pubDate>Tue, 08 May 2012 17:05:56 +0000</pubDate>
		<dc:creator>Brent</dc:creator>
				<category><![CDATA[links]]></category>
		<category><![CDATA[Carnival of Mathematics]]></category>

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		<description><![CDATA[Welcome to the 86th Carnival of Mathematics! is semiprime, nontotient, and noncototient. It is also happy since and . In fact, it is the smallest happy, nontotient semiprime (the only smaller happy nontotient is 68&#8212;which is, of course, 86 in &#8230; <a href="http://mathlesstraveled.com/2012/05/08/carnival-of-mathematics-86/">Continue reading <span class="meta-nav">&#8594;</span></a><img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=mathlesstraveled.com&#038;blog=21246689&#038;post=1589&#038;subd=mathlesstraveled&#038;ref=&#038;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<div style="text-align:center;">
<img src="http://aperiodical.com/wp-content/uploads/2012/04/carnivalofmathematics_white-background.png" alt="Carnival of Mathematics logo" />
</div>
<p>Welcome to the 86th <a href="http://aperiodical.com/carnival-of-mathematics" title="Carnival of Mathematics">Carnival of Mathematics</a>!  <img src='http://s0.wp.com/latex.php?latex=86+%3D+2+%5Ccdot+43+%3D+222_6+%3D+20%2B21%2B22%2B23+%3D+3%5E2+%2B+4%5E2+%2B+5%5E2+%2B+6%5E2&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='86 = 2 &#92;cdot 43 = 222_6 = 20+21+22+23 = 3^2 + 4^2 + 5^2 + 6^2' title='86 = 2 &#92;cdot 43 = 222_6 = 20+21+22+23 = 3^2 + 4^2 + 5^2 + 6^2' class='latex' /> is <a href="http://mathworld.wolfram.com/Semiprime.html" title="Semiprime">semiprime</a>, <a href="http://mathworld.wolfram.com/Nontotient.html">nontotient</a>, and <a href="http://mathworld.wolfram.com/Noncototient.html">noncototient</a>. It is also <a href="http://mathworld.wolfram.com/HappyNumber.html">happy</a> since <img src='http://s0.wp.com/latex.php?latex=8%5E2+%2B+6%5E2+%3D+64+%2B+36+%3D+100&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='8^2 + 6^2 = 64 + 36 = 100' title='8^2 + 6^2 = 64 + 36 = 100' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=1%5E2+%2B+0%5E2+%2B+0%5E2+%3D+1&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='1^2 + 0^2 + 0^2 = 1' title='1^2 + 0^2 + 0^2 = 1' class='latex' />.  In fact, it is the smallest happy, nontotient semiprime (the only smaller happy nontotient is 68&#8212;which is, of course, 86 in reverse&#8212;but 68 is not semiprime).</p>
<p>However, the most interesting mathematical fact about 86 (in my opinion) is that it is the largest known integer <img src='http://s0.wp.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n' title='n' class='latex' /> for which the decimal expansion of <img src='http://s0.wp.com/latex.php?latex=2%5En&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='2^n' title='2^n' class='latex' /> contains no zeros!  In particular, <img src='http://s0.wp.com/latex.php?latex=2%5E%7B86%7D+%3D+77371252455336267181195264&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='2^{86} = 77371252455336267181195264' title='2^{86} = 77371252455336267181195264' class='latex' />.  Although no one has <i>proved</i> it is the largest such <img src='http://s0.wp.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n' title='n' class='latex' />, every <img src='http://s0.wp.com/latex.php?latex=2%5En&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='2^n' title='2^n' class='latex' /> up to <img src='http://s0.wp.com/latex.php?latex=n+%3D+4.6+%5Ctimes+10%5E7&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n = 4.6 &#92;times 10^7' title='n = 4.6 &#92;times 10^7' class='latex' /> (which is quite a lot, although still slightly less than the total number of integers) has been checked to contain at least one zero.  The probability that any larger power of 2 contains no zeros is vanishingly small, given some reasonable assumptions about the distribution of digits in base-ten expansions of powers of two.</p>
<p>&#8220;Eighty-six&#8221; is also apparently some sort of slang term in American English, but it really has nothing to do with math, so who cares?  Onward to the carnival!  I had a lot of fun reading all the submissions, and have decided to organize them somewhat thematically&#8212;though they don&#8217;t always fit perfectly, so don&#8217;t assume you won&#8217;t be interested in a post just because of my categorization!</p>
<h3>Art</h3>
<div style="text-align:center;">
<img width="300" src="http://somethingorotherwhatever.com/tumblr/money-1.jpg" alt="Money polyhedron" />
</div>
<p><a href="http://aperiodical.com/author/christianp/">Christian Perfect</a> has started a series of posts on the theme of <a href="http://aperiodical.com/category/columns/arty-maths/">&#8220;Arty Maths&#8221;</a>, with links to artistic images and videos with a mathematical bent.  Above is a cool example, some sort of stellated polyhedron made out of money by <a href="http://www.kristimalakoff.com/" title="Kristi Malakoff">Kristi Malakoff</a> (you can find <a href="http://www.kristimalakoff.com/Money%20Pieces.html" title="Money Pieces">more here</a>).</p>
<p>Katie Steckles submitted a link to Robby Ingebretsen&#8217;s blog post <a href="http://nerdplusart.com/first-3d-rendered-film-from-1972-and-my-visit-to-pixar">First Digital 3D Rendered Film (from 1972) and My Visit to Pixar</a>.  Katie says,</p>
<blockquote><p>
This is possibly the earliest example of a computer animation, and one of its two creators, Edwin Catmull, who went on to found Pixar, is credited with &#8220;having work[ed] out [the] math to handle things like texture mapping, 3D anti-aliasing and z-buffering&#8221;. Fascinating to think he had to invent all of that in order to do this!
</p></blockquote>
<p>Robby&#8217;s blog post (and the extensive comments on it) give a lot more context and fascinating details.  And, of course, you can watch the video itself!</p>
<div style="text-align:center;">
<img width="300" src="http://www.walkingrandomly.com/images/random/great_hall_sackville.png" alt="" />
</div>
<p>Mike Croucher of <a href="http://www.walkingrandomly.com/" title="Walking Randomly">Walking Randomly</a> writes about some cool <a href="http://www.walkingrandomly.com/?p=4203">mathematically-themed stained glass windows</a>, and wonders whether anyone knows of any others.</p>
<h3>Statistics/data analysis</h3>
<p>Arthur Charpentier of <a href="http://freakonometrics.blog.free.fr/" title="Freakonometrics">Freakonometrics</a> writes about <a href="http://freakonometrics.blog.free.fr/index.php?post/2012/04/05/Playing-indoor-paintball%2C-with-R">Nonconvexity, and playing indoor paintball</a>: if a bunch of people in a nonconvex playing area are all holding water pistols and shoot at the closest person, who doesn&#8217;t get wet?</p>
<p>Katie Steckles submitted a link to <a href="http://lightyears.blogs.cnn.com/2012/04/20/data-its-how-stores-know-youre-pregnant/">Data: it&#8217;s how stores know you&#8217;re pregnant</a>, an article by Matthew Lane of <a href="http://www.mathgoespop.com/">Math Goes Pop!</a>  Ever wonder how companies can predict various things about you (such as whether you are pregnant!) based on your browsing habits and other publicly available data?  This article explains some of the basic math underlying this sort of &#8220;data mining&#8221;.</p>
<p>John Cook of <a href="http://www.johndcook.com/blog/">The Endeavour</a> answers the question: <i>What is randomness?</i> in <a href="http://www.johndcook.com/blog/2012/04/19/random-is-as-random-does/">Random is as random does</a>.  It turns out that the best answer might just be to avoid answering at all!</p>
<h3>Geometry</h3>
<p>Augustus Van Dusen of <a href="http://thinkingmachineblog.wordpress.com/">thinkingmachineblog</a>  submitted a post titled <a href="http://thinkingmachineblog.wordpress.com/2012/04/16/superellipse/" title="Superellipse">Superellipse</a>, saying</p>
<blockquote><p>I read an article about Sergels torg, a plaza in Stockholm, being an example of a superellipse. When I looked up superellipse on Wolfram math world, I noticed that the area formula involved gamma functions. I then decided to derive the result myself to see if it could be simplified and how it would reduce to the familiar formula for the area of an ellipse.</p></blockquote>
<p>Frederick Koh of <a href="http://www.whitegroupmaths.com/">White Group Mathematics</a> shares a <a href="http://www.whitegroupmaths.com/2011/10/problem-7-optimisation-via-geometry.html">geometric solution to an optimization problem</a> that doesn&#8217;t initially seem like it has anything to do with geometry.</p>
<div style="text-align:center;">
<img src="http://blog.zacharyabel.com/wp-content/uploads/2012/03/Morley-puzzle.gif" alt="" />
</div>
<p>Zachary Abel of <a href="http://blog.zacharyabel.com/">Three-Cornered Things</a> has written a series of three &#8220;excursions into the miraculous and interconnected workings of the humble triangle&#8221;: <a href="http://blog.zacharyabel.com/2012/03/many-morley-triangles/">Many Morley Triangles</a>, <a href="http://blog.zacharyabel.com/2012/03/several-sneaky-circles/">Several Sneaky Circles</a>, and <a href="http://blog.zacharyabel.com/2012/04/three-cornered-deltoids/">Three-Cornered Deltoids</a>.  These are some of my personal favorites from this month&#8217;s Carnival: chock-full of surprising mathematics and beautiful illustrations and animations!</p>
<h3>Teaching</h3>
<p><a href="http://www.solipsys.co.uk/new/ColinWright.html">Colin Wright</a> writes <a href="http://www.solipsys.co.uk/new/TheTrapeziumConundrum.html?ColinsBlog">The Trapezium Conundrum</a>: how should a trapezium (aka trapezoid if you&#8217;re from the US) be defined&#8212;with <i>exactly</i> one pair of parallel sides or <i>at least</i> one pair of parallel sides?  More generally, how are definitions arrived at and agreed upon?  The answer may depend on the audience.</p>
<p>Karen G. of <a href="http://schoolarama.blogspot.com/">school&#8226;a&#8226;rama</a> muses upon the relationship between language and learning place value in her post <a href="http://schoolarama.blogspot.com/2012/04/looking-to-asia.html">Looking to Asia</a>.</p>
<p>On her blog <a href="http://mathmamawrites.blogspot.com/">Math Mama Writes&#8230;</a>, Sue VanHattum writes about <a href="http://mathmamawrites.blogspot.com/2012/04/linear-algebra-leading-into-eigen-stuff.html">Linear Algebra: Leading into the Eigen Stuff</a>.  Sue says, &#8220;I&#8217;m teaching linear algebra for the first time in over a decade. That has meant relearning it&#8212;a delightful experience.&#8221;</p>
<p>Paul Salomon of <a href="http://lostinrecursion.wordpress.com/">Lost In Recursion</a> writes <a href="http://lostinrecursion.wordpress.com/2012/03/24/exponents-and-the-scale-of-the-universe-a-21st-century-math-lesson/">Exponents and the Scale of the Universe – a 21st Century math lesson</a>, a fun story about how an initially dry lesson on exponents turned into a remarkable learning experience.</p>
<h3>Fun</h3>
<p>Alistair Bird submitted a link to <a href="http://johncarlosbaez.wordpress.com/2012/04/24/enormous-integers/">Enormous Integers</a>, saying,</p>
<blockquote><p>
It&#8217;s still a common enough misconception that pure mathematics research must be about larger and larger numbers, but it&#8217;s still nice to sometimes play up to this stereotype, as John Baez&#8217;s blogpost on Azimuth about &#8216;Enormous Integers&#8217; does. Comments are worth a look too.
</p></blockquote>
<p>Pat Ballew writes on <a href="http://pballew.blogspot.com/">Pat&#8217;sBlog</a> about <a href="http://pballew.blogspot.com/2012/04/pandigital-primes.html">Pandigital Primes</a>: &#8220;exploring pandigital primes and finding out how handy Computer programming skills might be&#8221;.</p>
<p>Quick, what comes next in the series <img src='http://s0.wp.com/latex.php?latex=%5Cpi%2F2%2C+%5Cpi%2F2%2C+%5Cpi%2F2%2C+%5Cpi%2F2%2C+%5Cpi%2F2%2C+%5Cpi%2F2%2C+%5Cpi%2F2%2C+%5Cdots&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;pi/2, &#92;pi/2, &#92;pi/2, &#92;pi/2, &#92;pi/2, &#92;pi/2, &#92;pi/2, &#92;dots' title='&#92;pi/2, &#92;pi/2, &#92;pi/2, &#92;pi/2, &#92;pi/2, &#92;pi/2, &#92;pi/2, &#92;dots' class='latex' />?  <a href="http://www.thebigquestions.com/2012/03/26/loose-ends/">The answer</a>, as explained by Steven Landsburg on his blog, <a href="http://www.thebigquestions.com/">The Big Questions</a>, may surprise you! (Thanks to Katie Steckles for the submission, <a href="http://micromath.wordpress.com/2012/04/11/recommend/">via Alexandre Borovik</a>.)</p>
<div style="text-align:center;">
<img src="http://lostinrecursion.files.wordpress.com/2012/03/20120312-113512.jpg?w=200" alt="" />
</div>
<p>Paul Salomon of <a href="http://lostinrecursion.wordpress.com/">Lost In Recursion</a> displays <a href="http://lostinrecursion.wordpress.com/2012/03/12/the-lost-in-recursion-recursion/">The &#8220;Lost in Recursion&#8221; Recursion</a>.  Can you figure out what&#8217;s going on without getting lost in the &#8220;The &#8216;Lost in Recursion&#8217; Recursion&#8221; recursion?</p>
<h3>Stuff That Did Not Fit In Any Other Category But Is Still Awesome</h3>
<p>Colin Beveridge of <a href="http://blog.flyingcoloursmaths.co.uk/" title="Flying Colours Maths">Flying Colours Maths</a> submitted <a href="http://blog.flyingcoloursmaths.co.uk/secrets-of-the-mathematical-ninja-the-surprising-integration-rule-you-dont-get-taught-in-school">Secrets of the Mathematical Ninja: The surprising integration rule you don’t get taught in school</a>, and writes,</p>
<blockquote><p>
When I stumbled across this rule, my reaction was &#8216;whoa.&#8217; It&#8217;s quick, it&#8217;s extremely dirty, and it&#8217;s surprisingly accurate. The kind of thing the mathematical ninja dreams of.
</p></blockquote>
<p>Andrew Taylor writes a guest post, <a href="http://aperiodical.com/2012/04/electoral-reforms-and-non-transitive-dice/">Electoral Reforms and Non-Transitive Dice</a>, on <a href="http://aperiodical.com/">The Aperiodical</a>, explaining Why Choosing a Voting System is Hard in terms of a set of nontransitive dice.</p>
<p>Peter Rowlett, of <a href="http://travels.aperiodical.com/">Travels in a Mathematical World</a>, opines in his post, <a href="http://travels.aperiodical.com/2012/04/what-nice-job-you-have.html">What a nice job you have</a>, that a popular ranking listing &#8220;mathematician&#8221; as one of the top ten best jobs shouldn&#8217;t just be accepted and repeated uncritically.</p>
<p>In her article <a href="http://esciencecommons.blogspot.com/2012/03/how-culture-shaped-mathematician.html">How culture shaped a mathematician</a>, Carol Clark gives a glimpse into the life and background of mathematician Skip Garibaldi. She writes:</p>
<blockquote><p>
Mathematicians see the world differently than me. I interviewed a mathematician to get a glimpse of that view, and learned how everything from fine art to popular films and books played a role in shaping that view.
</p></blockquote>
<hr />
<p>The <a href="http://travels.aperiodical.com/2012/04/carnival-of-mathematics-85.html">previous Carnival of Mathematics</a> was hosted at <a href="http://travels.aperiodical.com/">Travels in a Mathematical World</a>; next month, the 87th Carnival will be hosted by Mr. Chase at <a href="http://mrchasemath.wordpress.com/">Random Walks</a>, so start getting your submissions ready now!  For lists of past and future carnivals, instructions on submitting, and answers to frequently asked questions, see the <a href="http://aperiodical.com/carnival-of-mathematics/" title="Carnival of Mathematics">main Carnival of Mathematics site</a>.  The next <a href="http://letsplaymath.net/mtap/">Math Teachers at Play</a> carnival is also coming up soon, with a <a href="http://letsplaymath.net/mtap/mtap-submission-form/">submission</a> deadline of this Friday.</p>
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		<title>Carnival of Mathematics submissions due in a week</title>
		<link>http://mathlesstraveled.com/2012/04/23/carnival-of-mathematics-submissions-due-in-a-week/</link>
		<comments>http://mathlesstraveled.com/2012/04/23/carnival-of-mathematics-submissions-due-in-a-week/#comments</comments>
		<pubDate>Tue, 24 Apr 2012 01:49:21 +0000</pubDate>
		<dc:creator>Brent</dc:creator>
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		<description><![CDATA[I&#8217;ll be hosting the Carnival of Mathematics, and the submission deadline is coming up soon&#8212;Tuesday, May 1. Please submit something! It could be something you wrote, or something someone else wrote that you enjoyed. All mathematics ranging from elementary to &#8230; <a href="http://mathlesstraveled.com/2012/04/23/carnival-of-mathematics-submissions-due-in-a-week/">Continue reading <span class="meta-nav">&#8594;</span></a><img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=mathlesstraveled.com&#038;blog=21246689&#038;post=1574&#038;subd=mathlesstraveled&#038;ref=&#038;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p>I&#8217;ll be hosting the <a href="http://aperiodical.com/carnival-of-mathematics/">Carnival of Mathematics</a>, and the submission deadline is coming up soon&#8212;Tuesday, May 1.  Please <a href="https://docs.google.com/spreadsheet/viewform?formkey=dC1IREpjV1lVRE1xcml2bXJDUzhqTXc6MQ#gid=0">submit something</a>!  It could be something you wrote, or something someone else wrote that you enjoyed.  All mathematics ranging from elementary to advanced is welcome.</p>
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		<title>Book review: In Pursuit of the Traveling Salesman</title>
		<link>http://mathlesstraveled.com/2012/04/07/book-review-in-pursuit-of-the-traveling-salesman-2/</link>
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		<pubDate>Sat, 07 Apr 2012 20:00:46 +0000</pubDate>
		<dc:creator>Brent</dc:creator>
				<category><![CDATA[books]]></category>
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		<category><![CDATA[TSP. traveling]]></category>

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		<description><![CDATA[As mathematical problems go, the &#8220;traveling salesman problem&#8221; (TSP) is a rare gem: it is simultaneously of great theoretical, historical, and practical interest. On the theoretical front, it is a well-known example of the class of &#8220;NP-complete&#8221; problems, which lie &#8230; <a href="http://mathlesstraveled.com/2012/04/07/book-review-in-pursuit-of-the-traveling-salesman-2/">Continue reading <span class="meta-nav">&#8594;</span></a><img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=mathlesstraveled.com&#038;blog=21246689&#038;post=1566&#038;subd=mathlesstraveled&#038;ref=&#038;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p><img alt="" src="http://press.princeton.edu/images/k9531.gif" class="alignright" width="300" height="456" /></p>
<p>As mathematical problems go, the &#8220;<a href="http://www.tsp.gatech.edu/">traveling salesman problem</a>&#8221; (TSP) is a rare gem: it is simultaneously of great theoretical, historical, <em>and</em> practical interest. On the theoretical front, it is a well-known example of the class of &#8220;NP-complete&#8221; problems, which lie at the heart of the <a href="http://www.claymath.org/millennium/P_vs_NP/">million-dollar &#8220;P vs NP&#8221; question</a> (which I still intend to explain at some point). Historically, it has been studied for almost 200 years (given a sufficiently inclusive definition of &#8220;study&#8221;), and has occupied a place in the public consciousness for at least the last 50. And this great historical interest is partly due to the problem&#8217;s practical significance.</p>
<p>So, what is it? Given a set of points in the plane (or, more generally, a set of points with &#8220;distances&#8221; specified somehow between each pair of points), the problem is to determine the <em>shortest</em> path which visits every point exactly once and then returns to the starting location. Of course, in one sense this is &#8220;easy&#8221;: just list all possible paths and compute the length of each. Note, however, that for a set of <img src='http://s0.wp.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n' title='n' class='latex' /> points, there are <img src='http://s0.wp.com/latex.php?latex=%28n-1%29%21&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='(n-1)!' title='(n-1)!' class='latex' /> (that is, <img src='http://s0.wp.com/latex.php?latex=1+%5Ccdot+2+%5Ccdot+3+%5Ccdot+%5Cdots+%5Ccdot+%28n-1%29&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='1 &#92;cdot 2 &#92;cdot 3 &#92;cdot &#92;dots &#92;cdot (n-1)' title='1 &#92;cdot 2 &#92;cdot 3 &#92;cdot &#92;dots &#92;cdot (n-1)' class='latex' />) possible paths that visit every point once and then return to the start. Even with only <img src='http://s0.wp.com/latex.php?latex=30&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='30' title='30' class='latex' /> points, that&#8217;s a whopping <img src='http://s0.wp.com/latex.php?latex=8841761993739701954543616000000&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='8841761993739701954543616000000' title='8841761993739701954543616000000' class='latex' /> possible paths&#8212;if you could compute the length of one trillion paths every second, it would still take <em>280 million millenia</em> (that&#8217;s <img src='http://s0.wp.com/latex.php?latex=2.8+%5Ctimes+10%5E%7B11%7D&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='2.8 &#92;times 10^{11}' title='2.8 &#92;times 10^{11}' class='latex' /> years, slightly longer than I&#8217;ve been alive) to check all of them! And that&#8217;s only <img src='http://s0.wp.com/latex.php?latex=30&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='30' title='30' class='latex' /> points&#8212;in practice, people want to solve the TSP for sets of points much larger than <img src='http://s0.wp.com/latex.php?latex=30&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='30' title='30' class='latex' />. So the point is not just to &#8220;solve&#8221; the problem; the real question is, can it be solved <em>efficiently</em>?</p>
<p>Amazingly, <em>no one knows</em>! But that hasn&#8217;t stopped people from coming up with extremely clever algorithms that seem to work well in practice, on very large sets of points (<em>i.e.</em> thousands, or even tens of thousands of points)&#8212;even though there are &#8220;pathological&#8221; inputs for which these algorithms do essentially no better than just trying every path. So these algorithms constitute a good solution to the TSP from a <em>practical</em> point of view but not a <em>theoretical</em> one!</p>
<p>William Cook&#8217;s new book, <em>In Pursuit of the Traveling Salesman: Mathematics at the Limits of Computation</em>, does a wonderful job presenting the history and significance of the TSP and an overview of cutting-edge research. It&#8217;s a beautiful, visually rich book, full of color photographs and diagrams that enliven both the narrative and mathematical presentation. And it includes a wealth of information (perhaps even a bit <em>too</em> much at times; I got lost in a few places). But it actually bills itself, partly, as an introduction to cutting-edge ideas in TSP research&#8212;and I think overall it succeeds admirably, explaining ideas in ways that are accessible but not patronizing. Read this book if you want a fun, beautifully illustrated introduction to (this one fascinating piece of) the edge of human knowledge!</p>
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		<title>New Carnival of Mathematics</title>
		<link>http://mathlesstraveled.com/2012/04/05/new-carnival-of-mathematics/</link>
		<comments>http://mathlesstraveled.com/2012/04/05/new-carnival-of-mathematics/#comments</comments>
		<pubDate>Thu, 05 Apr 2012 19:07:54 +0000</pubDate>
		<dc:creator>Brent</dc:creator>
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		<description><![CDATA[The Carnival of Mathematics has been revived! A big thanks to Mike Croucher of Walking Randomly for organizing it for the past few years, and to Katie Steckles, Christian Perfect, and Peter Rowlett for taking over. The latest edition, carnival &#8230; <a href="http://mathlesstraveled.com/2012/04/05/new-carnival-of-mathematics/">Continue reading <span class="meta-nav">&#8594;</span></a><img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=mathlesstraveled.com&#038;blog=21246689&#038;post=1558&#038;subd=mathlesstraveled&#038;ref=&#038;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p>The Carnival of Mathematics <a href="http://aperiodical.com/carnival-of-mathematics/" title="Carnival of Mathematics">has been revived</a>!  A big thanks to Mike Croucher of <a href="http://www.walkingrandomly.com/">Walking Randomly</a> for organizing it for the past few years, and to <a href="http://aperiodical.com/author/katie/">Katie Steckles</a>, <a href="http://aperiodical.com/author/christianp/">Christian Perfect</a>, and <a href="http://aperiodical.com/author/peter/">Peter Rowlett</a> for taking over.  The latest edition, <a href="http://travels.aperiodical.com/2012/04/carnival-of-mathematics-85.html" title="Carnival of Mathematics #85">carnival #85</a> (wow, has it really been going that long?) is now up at Peter Rowlett&#8217;s blog, <a href="http://travels.aperiodical.com/" title="Travels in a Mathematical World">Travels in a Mathematical World</a>.  Lots of cool stuff there, so <a href="http://travels.aperiodical.com/2012/04/carnival-of-mathematics-85.html">be sure to check it out</a> if you haven&#8217;t already.</p>
<p>I&#8217;ll be hosting Carnival #86 here, so <a href="https://docs.google.com/spreadsheet/viewform?formkey=dC1IREpjV1lVRE1xcml2bXJDUzhqTXc6MQ#gid=0">please submit something</a>!  The deadline is May 1st, and I&#8217;ll post the carnival sometime the week after that.</p>
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			<media:title type="html">Brent</media:title>
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	</item>
		<item>
		<title>Making our equation count</title>
		<link>http://mathlesstraveled.com/2012/03/03/making-our-equation-count/</link>
		<comments>http://mathlesstraveled.com/2012/03/03/making-our-equation-count/#comments</comments>
		<pubDate>Sun, 04 Mar 2012 03:09:09 +0000</pubDate>
		<dc:creator>Brent</dc:creator>
				<category><![CDATA[combinatorics]]></category>
		<category><![CDATA[pictures]]></category>
		<category><![CDATA[binomial coefficients]]></category>
		<category><![CDATA[functions]]></category>
		<category><![CDATA[matching]]></category>
		<category><![CDATA[permutation]]></category>

		<guid isPermaLink="false">http://mathlesstraveled.com/?p=1539</guid>
		<description><![CDATA[[This is post #4 in a series; previous posts can be found here: Differences of powers of consecutive integers, Differences of powers of consecutive integers, part II, Combinatorial proofs.] We&#8217;re still trying to find a proof of the equation which &#8230; <a href="http://mathlesstraveled.com/2012/03/03/making-our-equation-count/">Continue reading <span class="meta-nav">&#8594;</span></a><img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=mathlesstraveled.com&#038;blog=21246689&#038;post=1539&#038;subd=mathlesstraveled&#038;ref=&#038;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p>[This is post #4 in a series; previous posts can be found here: <a href="http://mathlesstraveled.com/2012/01/29/differences-of-powers-of-consecutive-integers/" title="Differences of powers of consecutive integers">Differences of powers of consecutive integers</a>, <a href="http://mathlesstraveled.com/2012/02/16/differences-of-powers-of-consecutive-integers-part-ii/" title="Differences of powers of consecutive integers, part II">Differences of powers of consecutive integers, part II</a>, <a href="http://mathlesstraveled.com/2012/02/17/combinatorial-proofs/" title="Combinatorial proofs">Combinatorial proofs</a>.]</p>
<p>We&#8217;re still trying to find a proof of the equation</p>
<p>
<div align="center"><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Csum_%7Bi%3D0%7D%5En+%28-1%29%5E%7Bn-i%7D+%5Cbinom+n+i+%28k%2Bi%29%5En+%3D+n%21&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;displaystyle &#92;sum_{i=0}^n (-1)^{n-i} &#92;binom n i (k+i)^n = n!' title='&#92;displaystyle &#92;sum_{i=0}^n (-1)^{n-i} &#92;binom n i (k+i)^n = n!' class='latex' /></div>
</p>
<p>which expresses the fact that a <a href="http://mathlesstraveled.com/2012/01/29/differences-of-powers-of-consecutive-integers/">certain arithmetic procedure</a> always seems to result, strangely enough, in the factorial of <img src='http://s0.wp.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n' title='n' class='latex' />. <a href="http://mathlesstraveled.com/2012/02/17/combinatorial-proofs/">Last time</a> I introduced the idea of using a <em>combinatorial proof</em>, and gave a simple example involving a binomial coefficient identity.</p>
<p>In order for this idea to yield any fruit, we need a way to interpret the various pieces of the equation as <em>counting</em> something. Let&#8217;s go over the pieces one by one and discuss some ways to interpret them combinatorially.</p>
<div id="factorial-permutations-and-matchings">
<h2>Factorial, permutations, and matchings</h2>
<p>Let&#8217;s start with the right-hand side, <img src='http://s0.wp.com/latex.php?latex=n%21&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n!' title='n!' class='latex' />. This one is not too hard: <img src='http://s0.wp.com/latex.php?latex=n%21&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n!' title='n!' class='latex' /> counts the number of <em>permutations</em> of <img src='http://s0.wp.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n' title='n' class='latex' /> objects, that is, the number of different ways to take <img src='http://s0.wp.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n' title='n' class='latex' /> distinct objects and arrange them in an ordered list. Why is that? Well, there are <img src='http://s0.wp.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n' title='n' class='latex' /> objects we could choose to put first; once we&#8217;ve made that choice, there are <img src='http://s0.wp.com/latex.php?latex=n-1&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n-1' title='n-1' class='latex' /> remaining objects we could choose to go second; then <img src='http://s0.wp.com/latex.php?latex=n-2&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n-2' title='n-2' class='latex' /> choices for the third object, and so on, for a total of <img src='http://s0.wp.com/latex.php?latex=n+%28n-1%29+%28n-2%29+%5Cdots+1+%3D+n%21&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n (n-1) (n-2) &#92;dots 1 = n!' title='n (n-1) (n-2) &#92;dots 1 = n!' class='latex' /> choices. For example, here are the <img src='http://s0.wp.com/latex.php?latex=4%21+%3D+4+%5Ccdot+3+%5Ccdot+2+%5Ccdot+1+%3D+24&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='4! = 4 &#92;cdot 3 &#92;cdot 2 &#92;cdot 1 = 24' title='4! = 4 &#92;cdot 3 &#92;cdot 2 &#92;cdot 1 = 24' class='latex' /> different permutations of size <img src='http://s0.wp.com/latex.php?latex=4&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='4' title='4' class='latex' />:</p>
<p><a href="http://mathlesstraveled.files.wordpress.com/2012/03/permutations.png"><img src="http://mathlesstraveled.files.wordpress.com/2012/03/permutations.png?w=640" alt="" title="Permutations"   class="aligncenter size-full wp-image-1547" /></a></p>
<p>However, there&#8217;s another way to think about permutations which will come in handy later. Namely, we can think of a permutation as a <em>matching</em> between two sets of size <img src='http://s0.wp.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n' title='n' class='latex' />. You know, like those puzzles that give two side-by-side lists and say &#8220;draw a line matching each cartoon character with their favorite cheese!&#8221; (&#8230;or whatever). Like this:</p>
<p><a href="http://mathlesstraveled.files.wordpress.com/2012/03/matchingexample.png"><img src="http://mathlesstraveled.files.wordpress.com/2012/03/matchingexample.png?w=640" alt="" title="MatchingExample"   class="aligncenter size-full wp-image-1545" /></a></p>
<p>Here we have a matching between two sets of size <img src='http://s0.wp.com/latex.php?latex=6&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='6' title='6' class='latex' />. Each dot on the left is matched with exactly one dot on the right, and vice versa.</p>
<p>Why are matchings another way of thinking about permutations? First, it&#8217;s not too hard to see that there are also <img src='http://s0.wp.com/latex.php?latex=n%21&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n!' title='n!' class='latex' /> matchings between two sets of size <img src='http://s0.wp.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n' title='n' class='latex' />: we have <img src='http://s0.wp.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n' title='n' class='latex' /> possible choices of what to match the first element with; then there are <img src='http://s0.wp.com/latex.php?latex=n-1&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n-1' title='n-1' class='latex' /> choices left over for what to match the second element with, and so on.</p>
<p>But we can also see a correspondence between permutations and matchings more directly. Start by labeling the dots on the left of a matching with consecutive numbers:</p>
<p><a href="http://mathlesstraveled.files.wordpress.com/2012/03/matchingnumbered.png"><img src="http://mathlesstraveled.files.wordpress.com/2012/03/matchingnumbered.png?w=640" alt="" title="MatchingNumbered"   class="aligncenter size-full wp-image-1548" /></a></p>
<p>Now, imagine each number &#8220;traveling&#8221; along the corresponding red edge until it reaches the dot on the other side. Like this:</p>
<p><a href="http://mathlesstraveled.files.wordpress.com/2012/03/matchingnumbered2.png"><img src="http://mathlesstraveled.files.wordpress.com/2012/03/matchingnumbered2.png?w=640" alt="" title="MatchingNumbered2"   class="aligncenter size-full wp-image-1549" /></a></p>
<p>See how the <img src='http://s0.wp.com/latex.php?latex=1&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='1' title='1' class='latex' /> traveled down the steep edge to end up at the fourth dot from the top; the <img src='http://s0.wp.com/latex.php?latex=2&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='2' title='2' class='latex' /> traveled across the horizontal edge to stay in the same position; the <img src='http://s0.wp.com/latex.php?latex=3&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='3' title='3' class='latex' /> traveled up to the top; and so on.</p>
<p>What we get out is a list of the numbers from 1&#8211;6 in some order; in this example we get <img src='http://s0.wp.com/latex.php?latex=3%2C2%2C5%2C1%2C6%2C4&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='3,2,5,1,6,4' title='3,2,5,1,6,4' class='latex' />. In other words, we can view a matching as a little physical machine for taking a list of objects and putting them into some particular order.</p>
<p>Here are all the permutations of size <img src='http://s0.wp.com/latex.php?latex=4&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='4' title='4' class='latex' /> again, this time visualized as matchings.</p>
<p><a href="http://mathlesstraveled.files.wordpress.com/2012/03/fourperms.png"><img src="http://mathlesstraveled.files.wordpress.com/2012/03/fourperms.png?w=640" alt="" title="FourPerms"   class="aligncenter size-full wp-image-1550" /></a></p>
<p>Now, at this point I am very tempted to go off on a tangent exploring group theory, symmetry groups, and all sorts of other stuff, but I shall restrain myself (for now!).</p>
</div>
<div id="binomial-coefficients">
<h2>Binomial coefficients</h2>
<p>Another piece of the equation is the <em>binomial coefficient</em> <img src='http://s0.wp.com/latex.php?latex=%5Cbinom+n+i&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;binom n i' title='&#92;binom n i' class='latex' />. But of course we <a href="http://mathlesstraveled.com/2007/02/14/binomial-coefficients/">already know</a> what binomial coefficients count&#8212;<img src='http://s0.wp.com/latex.php?latex=%5Cbinom+n+i&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;binom n i' title='&#92;binom n i' class='latex' /> is the number of ways to choose <img src='http://s0.wp.com/latex.php?latex=i&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='i' title='i' class='latex' /> things out of <img src='http://s0.wp.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n' title='n' class='latex' />, that is, the number of size-<img src='http://s0.wp.com/latex.php?latex=i&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='i' title='i' class='latex' /> subsets of a size-<img src='http://s0.wp.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n' title='n' class='latex' /> set. (I also <a href="http://mathlesstraveled.com/2012/02/17/combinatorial-proofs/">talked about this last time</a>.)</p>
</div>
<div id="exponentiation-and-functions">
<h2>Exponentiation and functions</h2>
<p>What about <img src='http://s0.wp.com/latex.php?latex=%28k%2Bi%29%5En&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='(k+i)^n' title='(k+i)^n' class='latex' />? What does that count? It turns out that exponentiation corresponds to counting <em>functions</em>: in particular, <img src='http://s0.wp.com/latex.php?latex=b%5Ea&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='b^a' title='b^a' class='latex' /> is the number of functions from a set of size <img src='http://s0.wp.com/latex.php?latex=a&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='a' title='a' class='latex' /> to a set of size <img src='http://s0.wp.com/latex.php?latex=b&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='b' title='b' class='latex' />. Why is that? Well, for each of the <img src='http://s0.wp.com/latex.php?latex=a&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='a' title='a' class='latex' /> elements of the domain, we have <img src='http://s0.wp.com/latex.php?latex=b&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='b' title='b' class='latex' /> choices for where a function could send it, and each of these choices is independent&#8212;so the total number of choices is <img src='http://s0.wp.com/latex.php?latex=b+%5Ccdot+b+%5Ccdot+%5Cdots+%5Ccdot+b+%3D+b%5Ea&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='b &#92;cdot b &#92;cdot &#92;dots &#92;cdot b = b^a' title='b &#92;cdot b &#92;cdot &#92;dots &#92;cdot b = b^a' class='latex' />.</p>
<p>For example, here are all of the <img src='http://s0.wp.com/latex.php?latex=3%5E3+%3D+27&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='3^3 = 27' title='3^3 = 27' class='latex' /> functions from a size-<img src='http://s0.wp.com/latex.php?latex=3&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='3' title='3' class='latex' /> set to a size-<img src='http://s0.wp.com/latex.php?latex=3&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='3' title='3' class='latex' /> set:</p>
<p><a href="http://mathlesstraveled.files.wordpress.com/2012/03/threefuncs.png"><img src="http://mathlesstraveled.files.wordpress.com/2012/03/threefuncs.png?w=640" alt="" title="ThreeFuncs"   class="aligncenter size-full wp-image-1551" /></a></p>
<p>Hmmm&#8230; this looks familiar! Note that some of these functions are matchings, and some aren&#8217;t. Perhaps you&#8217;re starting to get an inkling now why I introduced the idea of permutations as matchings&#8230;</p>
<p>All the pieces are almost in place now. The one piece of the equation we still haven&#8217;t yet talked about is that mysterious <img src='http://s0.wp.com/latex.php?latex=%28-1%29%5E%7Bn-i%7D&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='(-1)^{n-i}' title='(-1)^{n-i}' class='latex' />. It certainly doesn&#8217;t make sense to interpret that as the number of functions from a set of size <img src='http://s0.wp.com/latex.php?latex=n-i&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n-i' title='n-i' class='latex' /> to a &#8220;set of size negative one&#8221;, because of course there is no such thing as a set with a negative size. So how <em>can</em> we interpret it combinatorially? The answer lies in something called the <em>Principle of Inclusion-Exclusion</em> (or PIE for short), which will be the subject of my next post!</p>
</div>
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			<media:title type="html">Brent</media:title>
		</media:content>

		<media:content url="http://mathlesstraveled.files.wordpress.com/2012/03/permutations.png" medium="image">
			<media:title type="html">Permutations</media:title>
		</media:content>

		<media:content url="http://mathlesstraveled.files.wordpress.com/2012/03/matchingexample.png" medium="image">
			<media:title type="html">MatchingExample</media:title>
		</media:content>

		<media:content url="http://mathlesstraveled.files.wordpress.com/2012/03/matchingnumbered.png" medium="image">
			<media:title type="html">MatchingNumbered</media:title>
		</media:content>

		<media:content url="http://mathlesstraveled.files.wordpress.com/2012/03/matchingnumbered2.png" medium="image">
			<media:title type="html">MatchingNumbered2</media:title>
		</media:content>

		<media:content url="http://mathlesstraveled.files.wordpress.com/2012/03/fourperms.png" medium="image">
			<media:title type="html">FourPerms</media:title>
		</media:content>

		<media:content url="http://mathlesstraveled.files.wordpress.com/2012/03/threefuncs.png" medium="image">
			<media:title type="html">ThreeFuncs</media:title>
		</media:content>
	</item>
		<item>
		<title>Combinatorial proofs</title>
		<link>http://mathlesstraveled.com/2012/02/17/combinatorial-proofs/</link>
		<comments>http://mathlesstraveled.com/2012/02/17/combinatorial-proofs/#comments</comments>
		<pubDate>Fri, 17 Feb 2012 13:49:08 +0000</pubDate>
		<dc:creator>Brent</dc:creator>
				<category><![CDATA[combinatorics]]></category>
		<category><![CDATA[pictures]]></category>
		<category><![CDATA[proof]]></category>
		<category><![CDATA[binomial coefficients]]></category>
		<category><![CDATA[combinatorial proof]]></category>
		<category><![CDATA[identity]]></category>

		<guid isPermaLink="false">http://mathlesstraveled.com/?p=1525</guid>
		<description><![CDATA[Continuing from a previous post, we found that if we begin with th powers of consecutive integers and then repeatedly take successive differences, it seems like we always end up with the factorial of , that is, . We then &#8230; <a href="http://mathlesstraveled.com/2012/02/17/combinatorial-proofs/">Continue reading <span class="meta-nav">&#8594;</span></a><img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=mathlesstraveled.com&#038;blog=21246689&#038;post=1525&#038;subd=mathlesstraveled&#038;ref=&#038;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p>Continuing from <a href="http://mathlesstraveled.com/2012/02/16/differences-of-powers-of-consecutive-integers-part-ii/">a previous post</a>, we found that if we begin with <img src='http://s0.wp.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n' title='n' class='latex' />th powers of <img src='http://s0.wp.com/latex.php?latex=%28n%2B1%29&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='(n+1)' title='(n+1)' class='latex' /> consecutive integers and then repeatedly take successive differences, it seems like we always end up with the factorial of <img src='http://s0.wp.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n' title='n' class='latex' />, that is, <img src='http://s0.wp.com/latex.php?latex=n%21+%3D+1+%5Ccdot+2+%5Ccdot+3+%5Ccdot+%5Cdots+%5Ccdot+n&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n! = 1 &#92;cdot 2 &#92;cdot 3 &#92;cdot &#92;dots &#92;cdot n' title='n! = 1 &#92;cdot 2 &#92;cdot 3 &#92;cdot &#92;dots &#92;cdot n' class='latex' />. We then derived an algebraic expression for the result of the iterative difference procedure. So the goal now is to prove that</p>
<p><div align="center"><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Csum_%7Bi%3D0%7D%5En+%28-1%29%5E%7Bn-i%7D+%5Cbinom+n+i+%28k%2Bi%29%5En+%3D+n%21&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;displaystyle &#92;sum_{i=0}^n (-1)^{n-i} &#92;binom n i (k+i)^n = n!' title='&#92;displaystyle &#92;sum_{i=0}^n (-1)^{n-i} &#92;binom n i (k+i)^n = n!' class='latex' /></div>
</p>
<p>that is,</p>
<p><div align="center"><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%28-1%29%5En+%5Cbinom+n+0+k%5En+%2B+%28-1%29%5E%7Bn-1%7D+%5Cbinom+n+1+%28k%2B1%29%5En+%2B+%5Cdots+%2B+%5Cbinom+n+n+%28k%2Bn%29%5En+%3D+n%21&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;displaystyle (-1)^n &#92;binom n 0 k^n + (-1)^{n-1} &#92;binom n 1 (k+1)^n + &#92;dots + &#92;binom n n (k+n)^n = n!' title='&#92;displaystyle (-1)^n &#92;binom n 0 k^n + (-1)^{n-1} &#92;binom n 1 (k+1)^n + &#92;dots + &#92;binom n n (k+n)^n = n!' class='latex' /></div>
</p>
<p>Now, it&#8217;s possible (probable, in fact) that this can be proved by purely algebraic means. If you come up with such a proof I would love to see it! But I must confess that I spent several hours banging my head against it (algebraically speaking) without making any progress. Eventually I turned to the idea of a <em>combinatorial</em> proof.</p>
<p>What do I mean by that? <em>Combinatorics</em> is the subfield of mathematics concerned with <em>counting</em>. The essence of a combinatorial proof is to show that two different expressions are just two different ways of counting the same set of objects&#8212;and must therefore be equal. I&#8217;ve <a href="http://mathlesstraveled.com/2009/04/14/distributing-cookies-solutions/">described some combinatorial proofs before</a>, in counting the number of ways to distribute cookies.</p>
<p>As another simple example, consider the <a href="http://mathlesstraveled.com/2007/02/14/binomial-coefficients/">binomial coefficient</a> identity</p>
<p><div align="center"><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cbinom+n+k+%3D+%5Cbinom+%7Bn-1%7D%7Bk%7D+%2B+%5Cbinom%7Bn-1%7D%7Bk-1%7D&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;displaystyle &#92;binom n k = &#92;binom {n-1}{k} + &#92;binom{n-1}{k-1}' title='&#92;displaystyle &#92;binom n k = &#92;binom {n-1}{k} + &#92;binom{n-1}{k-1}' class='latex' /></div>
</p>
<p>It&#8217;s certainly possible to prove this algebraically, by expanding out the binomial coefficients (using <img src='http://s0.wp.com/latex.php?latex=%5Cbinom+n+k+%3D+%5Cfrac%7Bn%21%7D%7Bk%21%28n-k%29%21%7D&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;binom n k = &#92;frac{n!}{k!(n-k)!}' title='&#92;binom n k = &#92;frac{n!}{k!(n-k)!}' class='latex' />), but we can give a more elegant proof, based on the fact that <img src='http://s0.wp.com/latex.php?latex=%5Cbinom+n+k&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;binom n k' title='&#92;binom n k' class='latex' /> is the number of ways to choose a subset of <img src='http://s0.wp.com/latex.php?latex=k&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='k' title='k' class='latex' /> things out of a set of <img src='http://s0.wp.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n' title='n' class='latex' /> things. For example, here are the <img src='http://s0.wp.com/latex.php?latex=%5Cbinom+5+3+%3D+10&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;binom 5 3 = 10' title='&#92;binom 5 3 = 10' class='latex' /> ways to choose three things out a set of five:</p>
<p><a href="http://mathlesstraveled.files.wordpress.com/2012/02/binom-5-3.png"><img src="http://mathlesstraveled.files.wordpress.com/2012/02/binom-5-3.png?w=640" alt="Size-3 subsets of a set of size 5" title="Size-3 subsets of a set of size 5"   class="aligncenter size-full wp-image-1531" /></a>
<p>Consider the <em>first</em> element of the size-<img src='http://s0.wp.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n' title='n' class='latex' /> set. Every subset of size <img src='http://s0.wp.com/latex.php?latex=k&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='k' title='k' class='latex' /> either includes this first element, or it doesn&#8217;t. The number of size-<img src='http://s0.wp.com/latex.php?latex=k&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='k' title='k' class='latex' /> subsets which do <em>not</em> include the first element is <img src='http://s0.wp.com/latex.php?latex=%5Cbinom%7Bn-1%7D%7Bk%7D&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;binom{n-1}{k}' title='&#92;binom{n-1}{k}' class='latex' />, since that&#8217;s the number of ways to choose <img src='http://s0.wp.com/latex.php?latex=k&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='k' title='k' class='latex' /> things from the remaining <img src='http://s0.wp.com/latex.php?latex=n-1&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n-1' title='n-1' class='latex' /> elements. The number of size-<img src='http://s0.wp.com/latex.php?latex=k&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='k' title='k' class='latex' /> subsets which <em>do</em> include the first element is <img src='http://s0.wp.com/latex.php?latex=%5Cbinom%7Bn-1%7D%7Bk-1%7D&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;binom{n-1}{k-1}' title='&#92;binom{n-1}{k-1}' class='latex' />, because they correspond to choosing <img src='http://s0.wp.com/latex.php?latex=k-1&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='k-1' title='k-1' class='latex' /> of the remaining <img src='http://s0.wp.com/latex.php?latex=n-1&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n-1' title='n-1' class='latex' /> things. Therefore <img src='http://s0.wp.com/latex.php?latex=%5Cbinom+n+k+%3D+%5Cbinom%7Bn-1%7D%7Bk%7D+%2B+%5Cbinom%7Bn-1%7D%7Bk-1%7D&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;binom n k = &#92;binom{n-1}{k} + &#92;binom{n-1}{k-1}' title='&#92;binom n k = &#92;binom{n-1}{k} + &#92;binom{n-1}{k-1}' class='latex' />.</p>
<p>Here&#8217;s an illustration of how this works in the particular case when <img src='http://s0.wp.com/latex.php?latex=n+%3D+5&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n = 5' title='n = 5' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=k+%3D+3&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='k = 3' title='k = 3' class='latex' />:</p>
<p><a href="http://mathlesstraveled.files.wordpress.com/2012/02/binom-5-3-sm.png"><img src="http://mathlesstraveled.files.wordpress.com/2012/02/binom-5-3-sm.png?w=640" alt="Size-3 subsets of 5 elements, grouped by first element" title="Size-3 subsets of 5 elements, grouped by first element"   class="aligncenter size-full wp-image-1533" /></a>
<p>Notice how the ten subsets from above have been split into two groups: the first group, on the left, are those that don&#8217;t include the first element; you can see that each of them corresponds to one of the <img src='http://s0.wp.com/latex.php?latex=%5Cbinom+4+3+%3D+4&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;binom 4 3 = 4' title='&#92;binom 4 3 = 4' class='latex' /> size-<img src='http://s0.wp.com/latex.php?latex=3&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='3' title='3' class='latex' /> subsets of the remaining four elements. The second group, on the right, are those that do include the first element; each corresponds to one of the <img src='http://s0.wp.com/latex.php?latex=%5Cbinom+4+2+%3D+6&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;binom 4 2 = 6' title='&#92;binom 4 2 = 6' class='latex' /> size-<img src='http://s0.wp.com/latex.php?latex=2&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='2' title='2' class='latex' /> subsets of the remaining four elements.</p>
<p>So that&#8217;s the idea of a combinatorial proof. And we want to do something similar for the identity we are trying to prove&#8212;although, of course, it&#8217;s going to be a bit more difficult!</p>
<p>You might have fun trying to think about what a combinatorial proof of our target equation might look like; although if you don&#8217;t have much experience with combinatorics you may have trouble coming up with what sorts of things the two sides of the equation might be counting! That&#8217;s what I&#8217;ll talk about in my next post.</p>
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			<media:title type="html">Brent</media:title>
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			<media:title type="html">Size-3 subsets of a set of size 5</media:title>
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			<media:title type="html">Size-3 subsets of 5 elements, grouped by first element</media:title>
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		<title>Differences of powers of consecutive integers, part II</title>
		<link>http://mathlesstraveled.com/2012/02/16/differences-of-powers-of-consecutive-integers-part-ii/</link>
		<comments>http://mathlesstraveled.com/2012/02/16/differences-of-powers-of-consecutive-integers-part-ii/#comments</comments>
		<pubDate>Fri, 17 Feb 2012 00:52:20 +0000</pubDate>
		<dc:creator>Brent</dc:creator>
				<category><![CDATA[arithmetic]]></category>
		<category><![CDATA[iteration]]></category>
		<category><![CDATA[pascal's triangle]]></category>
		<category><![CDATA[binomial coefficients]]></category>
		<category><![CDATA[consecutive]]></category>
		<category><![CDATA[difference]]></category>
		<category><![CDATA[integers]]></category>
		<category><![CDATA[powers]]></category>

		<guid isPermaLink="false">http://mathlesstraveled.com/?p=1444</guid>
		<description><![CDATA[If you spent some time playing around with the procedure from Differences of powers of consecutive integers (namely, raise consecutive integers to the th power, and repeatedly take pairwise differences until reaching a single number) you probably noticed the curious &#8230; <a href="http://mathlesstraveled.com/2012/02/16/differences-of-powers-of-consecutive-integers-part-ii/">Continue reading <span class="meta-nav">&#8594;</span></a><img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=mathlesstraveled.com&#038;blog=21246689&#038;post=1444&#038;subd=mathlesstraveled&#038;ref=&#038;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p>If you spent some time playing around with the procedure from <a href="http://mathlesstraveled.com/2012/01/29/differences-of-powers-of-consecutive-integers/">Differences of powers of consecutive integers</a> (namely, raise <img src='http://s0.wp.com/latex.php?latex=n%2B1&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n+1' title='n+1' class='latex' /> consecutive integers to the <img src='http://s0.wp.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n' title='n' class='latex' />th power, and repeatedly take pairwise differences until reaching a single number) you probably noticed the curious fact that it always seems to result in a factorial&#8212;in the factorial of <img src='http://s0.wp.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n' title='n' class='latex' />, to be precise.</p>
<p>For example:</p>
<p><img src='http://s0.wp.com/latex.php?latex=3%5E2%2C+4%5E2%2C+5%5E2+%3D+9%2C16%2C25+%5Cto+7%2C9+%5Cto+2&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='3^2, 4^2, 5^2 = 9,16,25 &#92;to 7,9 &#92;to 2' title='3^2, 4^2, 5^2 = 9,16,25 &#92;to 7,9 &#92;to 2' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=196%5E2%2C+197%5E2%2C+198%5E2+%3D+38416%2C38809%2C39204+%5Cto+393%2C395+%5Cto+2&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='196^2, 197^2, 198^2 = 38416,38809,39204 &#92;to 393,395 &#92;to 2' title='196^2, 197^2, 198^2 = 38416,38809,39204 &#92;to 393,395 &#92;to 2' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=7%5E3%2C+8%5E3%2C+9%5E3%2C+10%5E3+%3D+343%2C512%2C729%2C1000+%5Cto+169%2C217%2C271+%5Cto+48%2C54+%5Cto+6&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='7^3, 8^3, 9^3, 10^3 = 343,512,729,1000 &#92;to 169,217,271 &#92;to 48,54 &#92;to 6' title='7^3, 8^3, 9^3, 10^3 = 343,512,729,1000 &#92;to 169,217,271 &#92;to 48,54 &#92;to 6' class='latex' /></p>
<p><img src='http://s0.wp.com/latex.php?latex=0%5E4%2C1%5E4%2C2%5E4%2C3%5E4%2C4%5E4+%3D+0%2C1%2C16%2C81%2C256+%5Cto+1%2C15%2C65%2C175+%5Cto+14%2C50%2C110+%5Cto+36%2C60+%5Cto+24&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='0^4,1^4,2^4,3^4,4^4 = 0,1,16,81,256 &#92;to 1,15,65,175 &#92;to 14,50,110 &#92;to 36,60 &#92;to 24' title='0^4,1^4,2^4,3^4,4^4 = 0,1,16,81,256 &#92;to 1,15,65,175 &#92;to 14,50,110 &#92;to 36,60 &#92;to 24' class='latex' /></p>
<p>Several commenters figured this out as well. Does this always happen? If so, can we prove it?</p>
<p>Let&#8217;s start by thinking about what happens when we do the successive-differencing procedure. If we start with the list <img src='http://s0.wp.com/latex.php?latex=a%2Cb&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='a,b' title='a,b' class='latex' />, then we get <img src='http://s0.wp.com/latex.php?latex=-a+%2B+b&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='-a + b' title='-a + b' class='latex' />. (I want to keep the letters in order, which is why I wrote <img src='http://s0.wp.com/latex.php?latex=-a+%2B+b&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='-a + b' title='-a + b' class='latex' /> instead of <img src='http://s0.wp.com/latex.php?latex=b+-+a&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='b - a' title='b - a' class='latex' />. Instead of subtracting the first value from the second, we can think of it as adding the negation of the first value to the second.) If we start with <img src='http://s0.wp.com/latex.php?latex=a%2Cb%2Cc&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='a,b,c' title='a,b,c' class='latex' />, we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=a%2Cb%2Cc+%5Cto+%28-a%2Bb%29%2C+%28-b%2Bc%29+%5Cto+%28a+-+2b+%2B+c%29&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='a,b,c &#92;to (-a+b), (-b+c) &#92;to (a - 2b + c)' title='a,b,c &#92;to (-a+b), (-b+c) &#92;to (a - 2b + c)' class='latex' />.</p>
<p>(The negation of <img src='http://s0.wp.com/latex.php?latex=%28-a+%2B+b%29&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='(-a + b)' title='(-a + b)' class='latex' /> is <img src='http://s0.wp.com/latex.php?latex=%28a+-+b%29&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='(a - b)' title='(a - b)' class='latex' />; adding this to <img src='http://s0.wp.com/latex.php?latex=%28-b+%2B+c%29&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='(-b + c)' title='(-b + c)' class='latex' /> yields <img src='http://s0.wp.com/latex.php?latex=%28a+-+2b+%2B+c%29&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='(a - 2b + c)' title='(a - 2b + c)' class='latex' />.) From <img src='http://s0.wp.com/latex.php?latex=a%2Cb%2Cc%2Cd&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='a,b,c,d' title='a,b,c,d' class='latex' /> we get</p>
<p><img src='http://s0.wp.com/latex.php?latex=a%2Cb%2Cc%2Cd+%5Cto+%5C%5C+%28-a%2Bb%29%2C+%28-b%2Bc%29%2C+%28-c%2Bd%29+%5Cto+%5C%5C+%28a+-+2b+%2B+c%29%2C+%28b+-+2c+%2B+d%29+%5Cto+%5C%5C+%28-a+%2B+3b+-+3c+%2B+d%29.&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='a,b,c,d &#92;to &#92;&#92; (-a+b), (-b+c), (-c+d) &#92;to &#92;&#92; (a - 2b + c), (b - 2c + d) &#92;to &#92;&#92; (-a + 3b - 3c + d).' title='a,b,c,d &#92;to &#92;&#92; (-a+b), (-b+c), (-c+d) &#92;to &#92;&#92; (a - 2b + c), (b - 2c + d) &#92;to &#92;&#92; (-a + 3b - 3c + d).' class='latex' /></p>
<p>Do you see any patterns yet? Let&#8217;s do one more. From the above calculation we can already see that doing four iterations on <img src='http://s0.wp.com/latex.php?latex=a%2Cb%2Cc%2Cd%2Ce&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='a,b,c,d,e' title='a,b,c,d,e' class='latex' /> will result in <img src='http://s0.wp.com/latex.php?latex=%28-a+%2B+3b+-+3c+%2B+d%29%2C+%28-b+%2B+3c+-+3d+%2B+e%29&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='(-a + 3b - 3c + d), (-b + 3c - 3d + e)' title='(-a + 3b - 3c + d), (-b + 3c - 3d + e)' class='latex' /> (do you see why?). Doing one final iteration gives us</p>
<p><img src='http://s0.wp.com/latex.php?latex=a+-+4b+%2B+6c+-+4d+%2B+e&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='a - 4b + 6c - 4d + e' title='a - 4b + 6c - 4d + e' class='latex' />.</p>
<p>Hmm. Let&#8217;s make a table.</p>
<div align="center">
<table style="border:none;width:200px;">
<tr>
<td><img src='http://s0.wp.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n' title='n' class='latex' /></td>
<td>Result</td>
</tr>
<tr>
<td>1</td>
<td><img src='http://s0.wp.com/latex.php?latex=-a+%2B+b&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='-a + b' title='-a + b' class='latex' /></td>
</tr>
<tr>
<td>2</td>
<td><img src='http://s0.wp.com/latex.php?latex=a+-+2b+%2B+c&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='a - 2b + c' title='a - 2b + c' class='latex' /></td>
</tr>
<tr>
<td>3</td>
<td><img src='http://s0.wp.com/latex.php?latex=-a+%2B+3b+-+3c+%2B+d&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='-a + 3b - 3c + d' title='-a + 3b - 3c + d' class='latex' /></td>
</tr>
<tr>
<td>4</td>
<td><img src='http://s0.wp.com/latex.php?latex=a+-+4b+%2B+6c+-+4d+%2B+e&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='a - 4b + 6c - 4d + e' title='a - 4b + 6c - 4d + e' class='latex' /></td>
</tr>
<tr>
<td>5</td>
<td><img src='http://s0.wp.com/latex.php?latex=-a+%2B+5b+-+10c+%2B+10d+-+5e+%2B+f&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='-a + 5b - 10c + 10d - 5e + f' title='-a + 5b - 10c + 10d - 5e + f' class='latex' /></td>
</tr>
</table>
</div>
<p>I included one more row (which you can verify if you like). Now do you see a pattern? The coefficients seem to be taken from <a href="http://mathlesstraveled.com/2007/02/17/binomial-coefficients-and-pascals-triangle/">Pascal&#8217;s triangle</a>, but with alternating signs!</p>
<p>In fact, it&#8217;s actually not too hard to see why this happens. At each step we take two offset copies of the previous row (by &quot;offset&quot; I mean that the letters are shifted by one, like <img src='http://s0.wp.com/latex.php?latex=%28a+-+2b+%2B+c%29&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='(a - 2b + c)' title='(a - 2b + c)' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=%28b+-+2c+%2B+d%29&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='(b - 2c + d)' title='(b - 2c + d)' class='latex' />) and add the negation of the first to the second. Since the signs are alternating, we really end up <em>adding</em> absolute values of the coefficients. Probably the best way to really see this is through an example:</p>
<p><div align="center"><img src='http://s0.wp.com/latex.php?latex=%5Cbegin%7Barray%7D%7Bcccccccccc%7D+-%28+%26+-a+%26+%2B+%26+3b+%26+-+%26+3c+%26+%2B+%26+d+%26+%29+%26+%5C%5C+%26+%26+%26+-b+%26+%2B+%26+3c+%26+-+%26+3d+%26+%2B+%26+e+%5C%5C+%5Chline+%26+a+%26+-+%26+4b+%26+%2B+%26+6c+%26+-+%26+4d+%26+%2B+%26+e+%5Cend%7Barray%7D&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;begin{array}{cccccccccc} -( &amp; -a &amp; + &amp; 3b &amp; - &amp; 3c &amp; + &amp; d &amp; ) &amp; &#92;&#92; &amp; &amp; &amp; -b &amp; + &amp; 3c &amp; - &amp; 3d &amp; + &amp; e &#92;&#92; &#92;hline &amp; a &amp; - &amp; 4b &amp; + &amp; 6c &amp; - &amp; 4d &amp; + &amp; e &#92;end{array}' title='&#92;begin{array}{cccccccccc} -( &amp; -a &amp; + &amp; 3b &amp; - &amp; 3c &amp; + &amp; d &amp; ) &amp; &#92;&#92; &amp; &amp; &amp; -b &amp; + &amp; 3c &amp; - &amp; 3d &amp; + &amp; e &#92;&#92; &#92;hline &amp; a &amp; - &amp; 4b &amp; + &amp; 6c &amp; - &amp; 4d &amp; + &amp; e &#92;end{array}' class='latex' /></div>
</p>
<p>Notice how we flip all the signs in the first row, so that they match the signs in the second row. But this is exactly how Pascal&#8217;s triangle is generated&#8212;each row is the sum of the previous row with itself, offset by one.</p>
<p>Now, in the real problem, we don&#8217;t start with <img src='http://s0.wp.com/latex.php?latex=a%2Cb%2Cc%2Cd%2C%5Cdots&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='a,b,c,d,&#92;dots' title='a,b,c,d,&#92;dots' class='latex' />, but with the <img src='http://s0.wp.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n' title='n' class='latex' />th powers of <img src='http://s0.wp.com/latex.php?latex=n%2B1&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n+1' title='n+1' class='latex' /> consecutive integers. Let&#8217;s call the first integer <img src='http://s0.wp.com/latex.php?latex=k&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='k' title='k' class='latex' />, so the sequence of consecutive integers is <img src='http://s0.wp.com/latex.php?latex=k%2C+k%2B1%2C+k%2B2%2C+%5Cdots%2C+k%2Bn&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='k, k+1, k+2, &#92;dots, k+n' title='k, k+1, k+2, &#92;dots, k+n' class='latex' />. Given this, we can now write down an expression for what we end up with after doing the iterated difference procedure:</p>
<p><div align="center"><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Csum_%7Bi%3D0%7D%5En+%28-1%29%5E%7Bn-i%7D+%5Cbinom%7Bn%7D%7Bi%7D+%28k%2Bi%29%5En&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;displaystyle &#92;sum_{i=0}^n (-1)^{n-i} &#92;binom{n}{i} (k+i)^n' title='&#92;displaystyle &#92;sum_{i=0}^n (-1)^{n-i} &#92;binom{n}{i} (k+i)^n' class='latex' /></div>
</p>
<p>Let&#8217;s break this down a bit. We know that we get a sum of <img src='http://s0.wp.com/latex.php?latex=n%2B1&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n+1' title='n+1' class='latex' /> terms; that&#8217;s the <img src='http://s0.wp.com/latex.php?latex=%5Csum_%7Bi%3D0%7D%5En&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;sum_{i=0}^n' title='&#92;sum_{i=0}^n' class='latex' /> part (you can <a href="http://mathlesstraveled.com/appendices/sigma-notation/">read more about sigma notation here</a>). We&#8217;ll use <img src='http://s0.wp.com/latex.php?latex=i&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='i' title='i' class='latex' /> to index the terms. We also know that the terms alternate sign, so we need to include <img src='http://s0.wp.com/latex.php?latex=%28-1%29&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='(-1)' title='(-1)' class='latex' /> raised to some power involving <img src='http://s0.wp.com/latex.php?latex=i&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='i' title='i' class='latex' />; the <em>last</em> term is always positive, so we use <img src='http://s0.wp.com/latex.php?latex=%28-1%29%5E%7Bn-i%7D&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='(-1)^{n-i}' title='(-1)^{n-i}' class='latex' />, which is equal to <img src='http://s0.wp.com/latex.php?latex=1&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='1' title='1' class='latex' /> when <img src='http://s0.wp.com/latex.php?latex=i+%3D+n&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='i = n' title='i = n' class='latex' />. The <a href="http://mathlesstraveled.com/2007/02/14/binomial-coefficients/">binomial coefficient</a> <img src='http://s0.wp.com/latex.php?latex=%5Cbinom+n+i&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;binom n i' title='&#92;binom n i' class='latex' /> gives us the <img src='http://s0.wp.com/latex.php?latex=i&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='i' title='i' class='latex' />th entry on the <img src='http://s0.wp.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n' title='n' class='latex' />th row of <a href="http://mathlesstraveled.com/2007/02/17/binomial-coefficients-and-pascals-triangle/">Pascal&#8217;s triangle</a>. Finally, of course, <img src='http://s0.wp.com/latex.php?latex=%28k%2Bi%29%5En&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='(k+i)^n' title='(k+i)^n' class='latex' /> is the <img src='http://s0.wp.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='n' title='n' class='latex' />th power of one of the integers we started with.</p>
<p>The claim, therefore, is that</p>
<p><div align="center"><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Csum_%7Bi%3D0%7D%5En+%28-1%29%5E%7Bn-i%7D+%5Cbinom+n+i+%28k%2Bi%29%5En+%3D+n%21&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;displaystyle &#92;sum_{i=0}^n (-1)^{n-i} &#92;binom n i (k+i)^n = n!' title='&#92;displaystyle &#92;sum_{i=0}^n (-1)^{n-i} &#92;binom n i (k+i)^n = n!' class='latex' /></div>
</p>
<p>And I will prove it to you, with pretty pictures, as promised!</p>
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		<slash:comments>3</slash:comments>
		<georss:point>39.953605 -75.213937</georss:point>
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		<geo:long>-75.213937</geo:long>
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			<media:title type="html">Brent</media:title>
		</media:content>
	</item>
		<item>
		<title>17&#215;17 4-coloring with no monochromatic rectangles</title>
		<link>http://mathlesstraveled.com/2012/02/09/17x17-4-coloring-with-no-monochromatic-rectangles/</link>
		<comments>http://mathlesstraveled.com/2012/02/09/17x17-4-coloring-with-no-monochromatic-rectangles/#comments</comments>
		<pubDate>Thu, 09 Feb 2012 04:45:29 +0000</pubDate>
		<dc:creator>Brent</dc:creator>
				<category><![CDATA[open problems]]></category>
		<category><![CDATA[pattern]]></category>
		<category><![CDATA[people]]></category>
		<category><![CDATA[pictures]]></category>
		<category><![CDATA[17x17]]></category>
		<category><![CDATA[four-coloring]]></category>
		<category><![CDATA[graph]]></category>
		<category><![CDATA[grid]]></category>
		<category><![CDATA[monochromatic]]></category>
		<category><![CDATA[rectangles]]></category>

		<guid isPermaLink="false">http://mathlesstraveled.com/?p=1486</guid>
		<description><![CDATA[Quick, what&#8217;s special about the following picture? As just announced by Bill Gasarch, this is a grid which has been four-colored (that is, each point in the grid has been assigned one of four colors) in such a way that &#8230; <a href="http://mathlesstraveled.com/2012/02/09/17x17-4-coloring-with-no-monochromatic-rectangles/">Continue reading <span class="meta-nav">&#8594;</span></a><img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=mathlesstraveled.com&#038;blog=21246689&#038;post=1486&#038;subd=mathlesstraveled&#038;ref=&#038;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p>Quick, what&#8217;s special about the following picture?</p>
<p><a href="http://mathlesstraveled.files.wordpress.com/2012/02/rect17.png"><img src="http://mathlesstraveled.files.wordpress.com/2012/02/rect17.png?w=640" alt="" title="4-colored 17x17 grid with no monochromatic rectangles"   class="aligncenter size-full wp-image-1487" /></a></p>
<p>As <a href="http://blog.computationalcomplexity.org/2012/02/17x17-problem-solved-also-18x18.html">just announced by Bill Gasarch</a>, this is a <img src='http://s0.wp.com/latex.php?latex=17+%5Ctimes+17&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='17 &#92;times 17' title='17 &#92;times 17' class='latex' /> grid which has been four-colored (that is, each point in the grid has been assigned one of four colors) in such a way that there are no <i>monochromatic rectangles</i>, that is, no four grid points forming the corners of an axis-aligned rectangle are all of the same color.  For example, if we change the top-left grid point to red, we can see several monochromatic rectangles pop up:</p>
<p><a href="http://mathlesstraveled.files.wordpress.com/2012/02/rect17-alter1.png"><img src="http://mathlesstraveled.files.wordpress.com/2012/02/rect17-alter1.png?w=640" alt="" title="Altered four-coloring with monochromatic rectangles highlighted"   class="aligncenter size-full wp-image-1488" /></a></p>
<p>Or here&#8217;s another version where I randomly picked a grid point in the middle, changed its color, and sure enough, more monochromatic rectangles result:</p>
<p><a href="http://mathlesstraveled.files.wordpress.com/2012/02/rect17-alter2.png"><img src="http://mathlesstraveled.files.wordpress.com/2012/02/rect17-alter2.png?w=640" alt="" title="Altered four-coloring with monochromatic rectangles highlighted"   class="aligncenter size-full wp-image-1489" /></a></p>
<p>As you can try verifying for yourself (and as I also verified using a computer program), there are no such monochromatic rectangles in the four-coloring at the top of this post!  (If you want to play with the four-coloring yourself, <a href="http://www.cs.umd.edu/~gasarch/BLOGPAPERS/17.txt">here it is</a> in a simple data format.)</p>
<p>For several years <a href="http://blog.computationalcomplexity.org/2009/11/17x17-challenge-worth-28900-this-is-not.html">no one knew if this was possible</a>, and Bill had offered a prize of $289 (that&#8217;s <img src='http://s0.wp.com/latex.php?latex=%5C%2417%5E2&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='&#92;$17^2' title='&#92;$17^2' class='latex' />, of course) to anyone who could find such a four-coloring.  The prize will be collected by Bernd Steinbach and Christian Posthoff&#8212;you can find more details in <a href="http://blog.computationalcomplexity.org/2012/02/17x17-problem-solved-also-18x18.html">Bill&#8217;s post</a>.  No one yet knows exactly how they found their four-coloring, but apparently they will be presenting a paper about it in May.  I&#8217;ll try to write more about it then (if I understand it at all)!</p>
<p>If you&#8217;re interested in reading more about the history and math behind this problem (and to get some intuition for why it is difficult), take a look at these posts by Brian Hayes on his blog, bit-player: <a href="http://bit-player.org/2009/the-17x17-challenge">The 17&#215;17 challenge</a> and <a href="http://bit-player.org/2009/17-x-17-a-nonprogress-report">17 x 17: A nonprogress report</a>.  <s>I also remember seeing</s> <a href="http://www.martinschweitzer.com/squaregame.html">Here&#8217;s a fun interactive applet</a> where you can play around with the problem, created by Martin Schweitzer.</p>
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			<media:title type="html">4-colored 17x17 grid with no monochromatic rectangles</media:title>
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			<media:title type="html">Brent</media:title>
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			<media:title type="html">4-colored 17x17 grid with no monochromatic rectangles</media:title>
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			<media:title type="html">Altered four-coloring with monochromatic rectangles highlighted</media:title>
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			<media:title type="html">Altered four-coloring with monochromatic rectangles highlighted</media:title>
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		<title>Book review: Nine Algorithms that Changed the Future</title>
		<link>http://mathlesstraveled.com/2012/02/04/book-review-nine-algorithms-that-changed-the-future/</link>
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		<pubDate>Sat, 04 Feb 2012 17:13:00 +0000</pubDate>
		<dc:creator>Brent</dc:creator>
				<category><![CDATA[books]]></category>
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		<category><![CDATA[John MacCormick]]></category>

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		<description><![CDATA[Nine Algorithms that Changed the Future: the Ingenious Ideas that Drive Today&#8217;s Computers, by John MacCormick. Princeton University Press, 2012. I&#8217;m often wary of books written for general audiences on technical topics. It&#8217;s quite difficult to write in a way &#8230; <a href="http://mathlesstraveled.com/2012/02/04/book-review-nine-algorithms-that-changed-the-future/">Continue reading <span class="meta-nav">&#8594;</span></a><img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=mathlesstraveled.com&#038;blog=21246689&#038;post=1370&#038;subd=mathlesstraveled&#038;ref=&#038;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p><a href="http://www.amazon.com/Nine-Algorithms-That-Changed-Future/dp/0691147140"><img src="http://press.princeton.edu/images/k9528.gif" alt="Nine Algorithms that Changed the Future" style="float:right;margin:20px;" width="200" /></a></p>
<p><i>Nine Algorithms that Changed the Future: the Ingenious Ideas that Drive Today&#8217;s Computers</i>, by John MacCormick. Princeton University Press, 2012.</p>
<p>I&#8217;m often wary of books written for general audiences on technical topics.  It&#8217;s quite difficult to write in a way that is both accessible to a wide audience and technically accurate.  Many such books end up sacrificing accuracy in the name of accessibility, trying to convey just the &#8220;intuition&#8221; or &#8220;general sense&#8221; of some topic, but often end up giving people the wrong idea instead.</p>
<p>I was quite happy to find, therefore, that John MacCormick nails it: &#8220;9 Algorithms that Changed the Future&#8221; is technically right on the money, but manages to explain things in ways that are both understandable and fun.  Want to understand how Google ranks search results? Or how Amazon manages to never lose or mess up your order information, even though they get hundreds of thousands of orders each day and (as we all know) networks and hard drives are unreliable?  Ever wonder how you can order something over the internet without your credit card number being stolen?  Or how &#8220;zip&#8221; is able to make your files smaller, seemingly by magic?  Even if you have never wondered about these things, perhaps I have made you wonder about them just now.  And that&#8217;s exactly the point of this book: there are quite a few ingenious algorithmic ideas that most of us rely on <i>every day</i> that we rarely&#8212;or never&#8212;even stop to wonder about.  </p>
<p>For example, I actually learned something new: I knew about public-key cryptography but had never really known much about Diffie-Hellman key exchange, which is what allows your web browser to talk to, say, Amazon&#8217;s servers securely even though they have never communicated before. It&#8217;s like having a secret conversation in code with a pen-pal whom you&#8217;ve never met, even though lots of people are reading your mail.  How can you ever <i>agree</i> on a secret code in the first place without the people reading your mail finding out (and hence being able to read all your subsequent coded messages)?  Sounds impossible, doesn&#8217;t it?  But it turns out that it <i>is</i> possible, with some clever ideas, which MacCormick skillfully explains using a fun metaphor about mixing colors of paint.</p>
<p>Each chapter starts out very simply, gradually building up more complex examples until you reach a full understanding of the algorithm being explained.  Along the way MacCormick introduces the &#8220;tricks&#8221;&#8212;the clever, central ideas&#8212;that make each algorithm work.  The writing is excellent: clear, precise, and fun.  I highly recommend this book to anyone curious about the ingenious mathematical and algorithmic ideas underlying some of today&#8217;s most ubiquitous technology.</p>
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