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	<title>cp&#039;s mathem-o-blog</title>
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		<title>Newcastle MathsJam April 2012 Recap</title>
		<link>http://checkmyworking.com/2012/04/newcastle-mathsjam-april-2012-recap/</link>
		<comments>http://checkmyworking.com/2012/04/newcastle-mathsjam-april-2012-recap/#comments</comments>
		<pubDate>Sat, 21 Apr 2012 21:51:03 +0000</pubDate>
		<dc:creator>Christian Perfect</dc:creator>
				<category><![CDATA[MathsJam]]></category>
		<category><![CDATA[MathsJam Recaps]]></category>

		<guid isPermaLink="false">http://checkmyworking.com/?p=824</guid>
		<description><![CDATA[<p>April&#8217;s MathsJam was very enjoyable. We did a bit of arts and crafts, a bit of playing games, and if it had been <a href="http://www.youtube.com/watch?v=X7dFMbubxr4">NBA Jam</a> instead of Maths Jam I would have been entirely on fire because I used up &#8230;</p>]]></description>
			<content:encoded><![CDATA[<p>April&#8217;s MathsJam was very enjoyable. We did a bit of arts and crafts, a bit of playing games, and if it had been <a href="http://www.youtube.com/watch?v=X7dFMbubxr4">NBA Jam</a> instead of Maths Jam I would have been entirely on fire because I used up all my IQ points solving some very fun puzzles. Durham were still on their Easter holidays so the attendance was a modest six people. That was just enough for everyone to be doing the same thing at the same time, so we had a good time.</p>
<p><span id="more-824"></span></p>
<h2 style="text-align: center;">Sprouts</h2>
<p>Eamonn brought in the rules for <a href="http://en.wikipedia.org/wiki/Sprouts_(game)">Sprouts</a>. He and Sarah played a few games and quickly realised that games can&#8217;t go on forever, so they set about trying to work out the perfect strategy.</p>
<p><a title="sprouts by christianp, on Flickr" href="http://www.flickr.com/photos/christianp/7098976347/"><img class="aligncenter" src="http://farm8.staticflickr.com/7135/7098976347_9df8f4e960_n.jpg" alt="sprouts" width="320" height="249" /></a></p>
<p>Somebody asked what it means to &#8220;solve&#8221; a game, because I say that quite a lot. It means describing what happens when both players are playing perfectly, in particular which player wins.</p>
<p>I thought that Sprouts had been completely solved, but according to the wikipedia page and <a href="http://arxiv.org/abs/1008.2320">this paper</a> it hasn&#8217;t! It also seems that the link between Sprouts and <a href="http://en.wikipedia.org/wiki/Nim">Nim</a> isn&#8217;t as straightforward as I thought. Good job I didn&#8217;t mention it, then.</p>
<h2 style="text-align: -webkit-auto;"></h2>
<p>Matthew Taylor showed us something to do with integer partitions. I didn&#8217;t really understand how it worked, and he didn&#8217;t have a proof that it does what he says it does, so maybe he can explain it in the comments.</p>
<p>We answered Katie Steckles&#8217; question about whether we considered ourselves mathematicians. She&#8217;s written a very interesting article based on the responses she received, which should be on the Aperiodical soon.</p>
<p>&nbsp;</p>
<h2 style="text-align: center;">Hyperbolic spaaaaaaaace!</h2>
<p>Prompted by last month&#8217;s discussion about games in hyperbolic space, I&#8217;ve been collecting all sorts of interesting hyperbolic things. I <a href="http://kurage.nimh.nih.gov/tomh/public_html/hyperbolic/index.html">printed out some heptagons</a> which we glued together to make a model of the $\{7,3\}$ tiling of the hyperbolic plane.</p>
<p style="text-align: center;"><a title="IMAG0400 by christianp, on Flickr" href="http://www.flickr.com/photos/christianp/7099003049/"><img src="http://farm6.staticflickr.com/5469/7099003049_2dff3e726e_n.jpg" alt="IMAG0400" width="191" height="320" /></a><a title="IMAG0401 by christianp, on Flickr" href="http://www.flickr.com/photos/christianp/6952935436/"><img src="http://farm8.staticflickr.com/7052/6952935436_4166a3ea69_n.jpg" alt="IMAG0401" width="320" height="191" /></a></p>
<p style="text-align: left;">I had also seen <a href="http://www.youtube.com/watch?v=vyi1W717Bgg">a video</a> showing how to make the same model using <a href="http://www.getbuckyballs.com/">buckyballs</a> so we had a go at that as well. Steven seemed to be the only person capable of reliably making heptagons, but we got enough together in the end to make this small model. It doesn&#8217;t look like much in the photo but it was fantastic to hold and flexed in rather pleasant ways.</p>
<p><a title="IMAG0402 by christianp, on Flickr" href="http://www.flickr.com/photos/christianp/6952981312/"><img class="aligncenter" src="http://farm8.staticflickr.com/7191/6952981312_fb056814fb_n.jpg" alt="IMAG0402" width="320" height="191" /></a></p>
<p>I tried to talk about how fibonacci trees have something to do with this tiling but I forgot most of the details. I think I managed to just about get across how every positive integer can be represented uniquely as a sum of non-adjacent fibonacci numbers, aka <a href="http://en.wikipedia.org/wiki/Zeckendorf's_theorem">Zeckendorf&#8217;s theorem</a>.</p>
<p>Matthew took the opportunity to make Nessie.</p>
<p><a title="IMAG0403 by christianp, on Flickr" href="http://www.flickr.com/photos/christianp/6952983204/"><img class="aligncenter" src="http://farm6.staticflickr.com/5457/6952983204_2f1f2dc9a2_n.jpg" alt="IMAG0403" width="320" height="191" /></a></p>
<p>&nbsp;</p>
<h2 style="text-align: center;">Wine tasters on a plane</h2>
<p>We eventually got bored of playing with models of weird space and looked on twitter for some puzzles from the other MathsJams.</p>
<blockquote class="twitter-tweet"><p>MAN: We&#8217;re attempting to prove @<a href="https://twitter.com/aPaulTaylor">aPaulTaylor</a>&#8216;s claim that 219…978, with any number of 9s in the middle, reverses its digits when quadrupled.</p>
<p>— Maths Jam (@MathsJam) <a href="https://twitter.com/MathsJam/status/192327785648357376" data-datetime="2012-04-17T19:04:52+00:00">April 17, 2012</a></p></blockquote>
<p><a href="#" class="peekaboo_link peekaboo-quadruple peekaboo_onhide">Click here for the proof</a>. As usual, please try to have a think about it yourself before checking the solution, it&#8217;s much better that way.</p>
<div class="peekaboo_content peekaboo-quadruple peekaboo_onhide" style="display:none;"></p>
<p>Think about the numbers being carried forward when multiplying the number by $4$. The $78$ bit gives $312$, so the $3$ is carried forward to the first $9$ digit.</p>
<p>$(4 \times 9) + 3 = 39$, so each 9 digit produces $9$ with another $3$ carried forward. Because the same number is carried forward at each $9$ digit, you can have as many as you like.</p>
<p>Finally, $(4 \times 21) + 3 = 87$, so we&#8217;ve shown that $4 \times 219 \dots 978 = 879 \dots 912$.</p>
<p></div>
<p>&nbsp;</p>
<blockquote class="twitter-tweet"><p>GLA: 1000 barrels of wine and 10 tasters.1 barrel is poisoned and kills the drinker in 30 days.Can you single out the bad barrel in 31 days?</p>
<p>— Maths Jam (@MathsJam) <a href="https://twitter.com/MathsJam/status/192331833290137600" data-datetime="2012-04-17T19:20:57+00:00">April 17, 2012</a></p></blockquote>
<p>I&#8217;ll add another rule that Glasgow didn&#8217;t mention, but I think is needed for it to be an interesting puzzle: the poison kills a taster at some point on the 30th day after being drunk, not after an exact number of seconds. So you can&#8217;t get a single taster to drink from each barrel in turn and then use a stopwatch on the 30th day to work out which barrel killed him.</p>
<a href="#" class="peekaboo_link peekaboo-wine_hint peekaboo_onhide">Click here for a hint</a>
<div class="peekaboo_content peekaboo-wine_hint peekaboo_onhide" style="display:none;"></p>
<p>You might be thinking that the solution has something to do with the fact that $1000 = 10 \times 100$. Instead, notice that since you want to find the bad barrel in 31 days, and it takes 30 days for the poison to take effect, the tasters only have two days to do their drinking. With that in mind, $1000$ is suspiciously close to $1024 = 2^{10}$.</p>
<p><a href="#" class="peekaboo_link peekaboo-wine_solution peekaboo_onhide">Click here to see the solution</a>.</p>
<p></div>
<div class="peekaboo_content peekaboo-wine_solution peekaboo_onhide" style="display:none;"></p>
<p>Number the barrels from $0$ to $999$. Then each barrel&#8217;s number can be written down as a 10-digit binary string. Assign a digit to each taster.</p>
<p>On the first day, the $N^{\textrm{th}}$ taster drinks from each barrel whose $N^{\textrm{th}}$ digit is $0$. On the second day, they drink from the barrels with a $1$ in that digit.</p>
<p>On the 31st day, all the tasters are dead. A tragic yet unavoidable consequence of our brilliant logic.</p>
<p>Each taster can tell you one digit of the binary number of the poisoned barrel: if it was $0$, then they drank the poisoned wine on the first day and so died on the 30th. If it was $1$, then they drank the poisoned wine on the second day and died on the 31st. So if the guy who was assigned the first digit died on the 30th day, then the first digit of the poisoned barrel&#8217;s number is $0$. By applying that logic to each digit, we can work out the label of the poisoned barrel, and throw a heck of a wake for the dead tasters with the remaining safe wine.</p>
<p></div>
<p>&nbsp;</p>
<p>I thought this problem came from twitter but I can&#8217;t find the originating tweet. Whoever brought it up, can you say in the comments where it came from?</p>
<p><em>100 passengers are waiting to get on a plane with 100 seats. The first dude gets on, but has forgotten which seat is his, so he picks a seat at random. The rest of the passengers board one by one. If their seat is free, they sit in it, otherwise they pick another free seat at random. </em></p>
<p><em>What is the probability that the last passenger sits in his assigned seat?</em></p>
<p><a href="#" class="peekaboo_link peekaboo-plane peekaboo_onhide">Click here for the solution</a>, but give this one a proper thinking about first because it&#8217;s a good question.</p>
<div class="peekaboo_content peekaboo-plane peekaboo_onhide" style="display:none;"></p>
<p>Opinion was divided on this one. Some thought the probability was $\frac{1}{2}$, while others thought it was $\frac{1}{100}$.</p>
<p>Eamonn tried drawing probability trees and came up with an answer of $\frac{1}{100}$, but he made a mistake with conditional probabilities. Joseph reduced the problem to a smaller number of passengers and tried writing out all possible seating arrangements, also deciding that the answer for the original puzzle is $\frac{1}{100}$. He had missed that some arrangements can&#8217;t be produced by the passengers&#8217; behaviour.</p>
<p>My solution took a while to formulate in precise words but is quite wonderfully concise: the first passenger comes in and either sits in his own seat, in which case everyone else sits in their own seat, or he sits in someone else&#8217;s seat. Everyone else files in until the guy whose seat the first passenger took boards the plane. You can then say that the usurped passenger agrees to swap seats with the first guy, who has forgotten which seat is his. So this new passenger is in the same position as the first one was, but with fewer available seats. That might be enough to convince you the answer&#8217;s $\frac{1}{2}$, or you might think there&#8217;s some weirdness lurking somewhere, unacknowledged.</p>
<p>We can prove by induction on the number of seats that the probability the last passenger sits in his own seat is $\frac{1}{2}$. Let $P(n)$ be the probability that the last passenger sits in his own seat in a plane with $n$ seats.</p>
<p>If there are two seats, the probability the last passenger sits in his own seat is $\frac{1}{2}$, clearly. So $P(2) = \frac{1}{2}$.</p>
<p>If there are $N$ seats, then the following are true:</p>
<ul>
<li>with probability $\frac{1}{N}$, the first passenger sits in his own seat and hence everyone else sits in the right seats.</li>
<li>with probability $\frac{1}{N}$, the first passenger sits in the last passenger&#8217;s seat, so the last passenger has to sit somewhere else.</li>
<li>If the first passenger sits in seat $i$, $2 \leq i \lt N$, then the last passenger sits in his own seat with probability $P(i)$.</li>
</ul>
<p>So by summing up all those eventualities, we can get $P(N)$,</p>
<p>\[ P(N) = \frac{1}{N} \times 1 + \sum_{i=2}^{N-1} \left( \frac{1}{N} \times P(i) \right) + \frac{1}{N} \times 0. \]</p>
<p>By using the inductive hypothesis that $P(i) = \frac{1}{2}$ for $2 \leq i \lt N$, the previous equation means that</p>
<p>\[\begin{align} P(N) &amp;= \frac{1}{N} + \left( \frac{N-2}{N} \right) \times \frac{1}{2} \\ &amp;= \frac{2 + N -2}{2N} \\ &amp;= \frac{1}{2}. \end{align}\]</p>
<p>So for any number of seats greater than 1, the probability that the last passenger sits in his own seat is $\frac{1}{2}$.</p>
<p></div>
<p>&nbsp;</p>
<p>Finally, we played with some <a href="http://www.mathpuzzle.com/RollingCube.pdf">rolling cube puzzles from mathpuzzle.com</a>.</p>
<p><a title="IMAG0404 by christianp, on Flickr" href="http://www.flickr.com/photos/christianp/7099050997/"><img class="aligncenter" src="http://farm6.staticflickr.com/5453/7099050997_cb94bfdf16_n.jpg" alt="IMAG0404" width="320" height="191" /></a></p>
<p>The aim is to roll the cube so that you can read the word &#8220;Finish&#8221; the right way up. You can&#8217;t spin the cube on its face, and you can only roll the cube about green edges in contact with the table. They&#8217;re sort of like mazes in a bit of the symmetry group of the cube. I had to add some marks to distinguish the green edges because of how bad my eyes are at looking at colours.</p>
<p>&nbsp;</p>
<p>That&#8217;s it for April&#8217;s MathsJam. I thoroughly enjoyed myself yet again and I think everyone else did too. If I&#8217;ve missed anything, please mention it in the comments, and I hope I&#8217;ll see you next month!</p>
]]></content:encoded>
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		</item>
		<item>
		<title>Slides &#8211; &#8220;Putting maths notation online&#8221;</title>
		<link>http://checkmyworking.com/2012/04/slides-putting-maths-notation-online/</link>
		<comments>http://checkmyworking.com/2012/04/slides-putting-maths-notation-online/#comments</comments>
		<pubDate>Fri, 20 Apr 2012 15:41:31 +0000</pubDate>
		<dc:creator>Christian Perfect</dc:creator>
				<category><![CDATA[presentations]]></category>
		<category><![CDATA[Web Stuff]]></category>

		<guid isPermaLink="false">http://checkmyworking.com/?p=827</guid>
		<description><![CDATA[<p>On Monday I gave a talk at Birmingham at a workshop titled, <a href="http://mathstore.ac.uk/node/2096">Using social media to engage students in mathematical sciences</a>. I have no experience of doing that, but I was invited to talk a bit about <a href="http://checkmyworking.com/misc/hestemsoc%20talk/">putting maths </a>&#8230;</p>]]></description>
			<content:encoded><![CDATA[<p>On Monday I gave a talk at Birmingham at a workshop titled, <a href="http://mathstore.ac.uk/node/2096">Using social media to engage students in mathematical sciences</a>. I have no experience of doing that, but I was invited to talk a bit about <a href="http://checkmyworking.com/misc/hestemsoc%20talk/">putting maths notation online</a>. It&#8217;s basically just a collection of links to the posts I&#8217;ve written on the subject previously, but maybe big text in small slides will be more accessible.</p>
]]></content:encoded>
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		<slash:comments>0</slash:comments>
		</item>
		<item>
		<title>Newcastle MathsJam March 2012 Recap</title>
		<link>http://checkmyworking.com/2012/04/newcastle-mathsjam-aprilmarch-2012-recap/</link>
		<comments>http://checkmyworking.com/2012/04/newcastle-mathsjam-aprilmarch-2012-recap/#comments</comments>
		<pubDate>Fri, 20 Apr 2012 11:32:31 +0000</pubDate>
		<dc:creator>Christian Perfect</dc:creator>
				<category><![CDATA[MathsJam]]></category>
		<category><![CDATA[MathsJam Recaps]]></category>

		<guid isPermaLink="false">http://checkmyworking.com/?p=802</guid>
		<description><![CDATA[<p>It&#8217;s been two months since I last wrote one of these!  March was a haze of overwork and stress for me, so I didn&#8217;t write a recap for March&#8217;s MathsJam while it was still March. <a href="http://peterrowlett.net/">Peter Rowlett</a>, who was &#8230;</p>]]></description>
			<content:encoded><![CDATA[<p>It&#8217;s been two months since I last wrote one of these!  March was a haze of overwork and stress for me, so I didn&#8217;t write a recap for March&#8217;s MathsJam while it was still March. <a href="http://peterrowlett.net/">Peter Rowlett</a>, who was visiting Newcastle as part of his mission to avoid having to think up new puzzles for MathsJams by always attending different ones (and also to give a talk at the university) has kindly sent me his notes, so here&#8217;s what I&#8217;ve reconstructed:<span id="more-802"></span></p>
<h2 style="text-align: center;">Big noughts and crosses</h2>
<p>I have this picture of Naomi playing noughts and crosses on a big board. I can&#8217;t remember what the rules were. A different version by David Cushing swept round the Newcastle maths department for a while until we proved it had ambiguously-won games. I won&#8217;t bother writing up the rules for that version.</p>
<p><a href="http://checkmyworking.com/wp-content/uploads/2012/04/tumblr_m25w2uDYp61qcaz8ho1_1280.jpg"><img class="aligncenter size-medium wp-image-813" title="tumblr_m25w2uDYp61qcaz8ho1_1280" src="http://checkmyworking.com/wp-content/uploads/2012/04/tumblr_m25w2uDYp61qcaz8ho1_1280-179x300.jpg" alt="" width="179" height="300" /></a></p>
<p>&nbsp;</p>
<h2 style="text-align: center;">Dice</h2>
<p>Somebody asked if you can make two dice, with different sets of numbers on their faces, so that when you roll them both the sum will have the same probability distribution as two normal dice? An additional constraint is that only positive numbers are allowed.Somebody at another MathsJam tweeted <a href="https://twitter.com/#!/MathsJam/status/182198580554117120/photo/1">a picture of just such a pair of dice</a>, and @outofthenorm2 tweeted <a href="https://twitter.com/#!/outofthenorm2/status/161997394240737280/photo/1">a nicely-rendered diagram of the solutions to the puzzle</a>.</p>
<p>&nbsp;</p>
<h2 style="text-align: center;">Mad Abel</h2>
<p>We played <a href="http://www.pagat.com/docs/madabel.pdf">Mad Abel</a> again. Here&#8217;s a picture of Peter losing.</p>
<p><a href="http://checkmyworking.com/wp-content/uploads/2012/04/tumblr_m17b8hkuh01rq44y0o1_1280.png"><img class="aligncenter size-medium wp-image-809" title="tumblr_m17b8hkuh01rq44y0o1_1280" src="http://checkmyworking.com/wp-content/uploads/2012/04/tumblr_m17b8hkuh01rq44y0o1_1280-224x300.png" alt="" width="224" height="300" /></a></p>
<h2 style="text-align: center;">Poker</h2>
<p>Matthew Taylor outed himself as a former president of Durham&#8217;s Poker society and suggested a two-player game involving poker hands. You deal out four piles of five cards, face-up. Players take turns to take the top card of one of the piles into their hand. When all the cards have been taken, the players&#8217; hands are judged as poker hands. I can&#8217;t remember if anyone came up with a strategy for this.</p>
<p>&nbsp;</p>
<h2 style="text-align: center;">An almost-Collatz game</h2>
<p>David Cushing made a hero&#8217;s return to MathsJam with a fun number theory game that I don&#8217;t think anyone quite got round to solving. I enjoyed it regardless.</p>
<ul>
<li>Both players agree on an upper bound, for example 100.</li>
<li>The first player writes down a whole number lower than the upper bound.</li>
<li>Players take turns writing down either a factor or a multiple of the previous number. Numbers can only be picked once per game.</li>
<li>The first person not able to pick a number (because all of the previous number&#8217;s factors and multiples have been picked) loses<a href="http://checkmyworking.com/wp-content/uploads/2012/04/tumblr_m17b2qqiZk1rq44y0o1_1280.jpg"><img class="aligncenter size-medium wp-image-808" title="tumblr_m17b2qqiZk1rq44y0o1_1280" src="http://checkmyworking.com/wp-content/uploads/2012/04/tumblr_m17b2qqiZk1rq44y0o1_1280-300x179.jpg" alt="" width="300" height="179" /></a>.</li>
</ul>
<p>Matthew was doing something with graphs of possible moves but I don&#8217;t think he managed to solve the game.</p>
<p><a href="http://checkmyworking.com/wp-content/uploads/2012/04/tumblr_m25w4hRyYa1qcaz8ho1_1280.jpg"><img class="aligncenter size-medium wp-image-814" title="tumblr_m25w4hRyYa1qcaz8ho1_1280" src="http://checkmyworking.com/wp-content/uploads/2012/04/tumblr_m25w4hRyYa1qcaz8ho1_1280-300x179.jpg" alt="" width="300" height="179" /></a></p>
<p>&nbsp;</p>
<h1 style="text-align: center;">Baffling graph theory</h1>
<p>(or graffling bath feary)</p>
<p>Peter has sent me a photo of a game I must have missed because it&#8217;s completely baffling. Apparently you draw $N$ points and label them with the numbers $1 \dots N$. The challenge is to draw edges such that each point is joined to as many edges as its label. An edge can start in the middle of another edge, but only once per edge.</p>
<p><a href="http://checkmyworking.com/wp-content/uploads/2012/04/photo.jpg"><img class="aligncenter size-medium wp-image-805" title="photo" src="http://checkmyworking.com/wp-content/uploads/2012/04/photo-224x300.jpg" alt="" width="224" height="300" /></a></p>
<p>A variation is to draw $N$ points, all with the label $N$. $1$ is a loop, $2$ is a triangle, and $3,4,5$ are also possible. Peter asks if any larger numbers are possible.</p>
<p>We played with the zero-magic graphs from <a href="http://digitalcommons.library.unlv.edu/cgi/viewcontent.cgi?article=2011&amp;context=thesesdissertations">Samuel Hansen&#8217;s thesis</a>. Peter had a sketchpad full of festively shaped graphs produced while trying to make last year&#8217;s Christmas cards.</p>
<p><a href="http://checkmyworking.com/wp-content/uploads/2012/04/tumblr_m17du7WA0o1rq44y0o1_1280.png"><img class="aligncenter size-medium wp-image-811" title="tumblr_m17du7WA0o1rq44y0o1_1280" src="http://checkmyworking.com/wp-content/uploads/2012/04/tumblr_m17du7WA0o1rq44y0o1_1280-224x300.png" alt="" width="224" height="300" /></a></p>
<p>&nbsp;</p>
<h2 style="text-align: center;">Bulletin of the British Society of Hysterical Mathematicians</h2>
<p>Peter brought an issue of the <a href="http://www.tandfonline.com/doi/abs/10.1080/17498430.2011.560511?journalCode=tbsh20">BSHM Bulletin</a> which prompted a discussion of the 15-game and <a href="http://peterrowlett.net/files/15puzzle.pdf">when it is solvable</a>.</p>
<p>&nbsp;</p>
<h2 style="text-align: center;">Jeff Weeks&#8217;s Torus Games</h2>
<p>David brought in his iPad and showed us <a href="http://www.geometrygames.org/TorusGames/index.html">a game which lets you play pool on a torus</a>. There was much confusion about shot placement.</p>
<p><a href="http://checkmyworking.com/wp-content/uploads/2012/04/tumblr_m17cj9UQWh1rq44y0o1_1280.png"><img class="aligncenter size-medium wp-image-810" title="tumblr_m17cj9UQWh1rq44y0o1_1280" src="http://checkmyworking.com/wp-content/uploads/2012/04/tumblr_m17cj9UQWh1rq44y0o1_1280-224x300.png" alt="" width="224" height="300" /></a></p>
<p>&nbsp;</p>
<p>I had a bit of a discussion with John about <a href="http://aperiodical.com/2012/03/hyperrogue-ii/">HyperRogue</a>, a roguelike game played in hyperbolic space. We both wanted to make something like a hyperbolic orienteering game, where the negative curvature of space messes with your ability to follow directions.</p>
<p>That&#8217;s all I&#8217;ve got for March. Sorry this wasn&#8217;t a proper post. April&#8217;s is really good though, so hopefully that makes up for it.</p>
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		<title>A little applet to make maths for screengrabbing</title>
		<link>http://checkmyworking.com/2012/04/a-little-applet-to-make-maths-for-screengrabbing/</link>
		<comments>http://checkmyworking.com/2012/04/a-little-applet-to-make-maths-for-screengrabbing/#comments</comments>
		<pubDate>Thu, 19 Apr 2012 20:50:20 +0000</pubDate>
		<dc:creator>Christian Perfect</dc:creator>
				<category><![CDATA[LaTeX]]></category>
		<category><![CDATA[Web Stuff]]></category>

		<guid isPermaLink="false">http://checkmyworking.com/?p=799</guid>
		<description><![CDATA[<p>Paul Taylor wanted an easy way to write some maths he could take a screengrab of, for use as an icon. Before I intervened he was doing something unnatural with wikipedia, so I wrote a little applet using MathJax: &#8220;&#8230;</p>]]></description>
			<content:encoded><![CDATA[<p>Paul Taylor wanted an easy way to write some maths he could take a screengrab of, for use as an icon. Before I intervened he was doing something unnatural with wikipedia, so I wrote a little applet using MathJax: &#8220;<a href="http://checkmyworking.com/misc/makebigmaths/">make big maths</a>&#8220;.</p>
<p>Quite a few tools like this exist, using mimetex or some other CGI tool to run LaTeX on a server and produce an image file. That&#8217;s far too slow and rubbish-looking for my liking, so I made my own with MathJax.</p>
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		<title>Interesting Esoterica Summation, volume 3</title>
		<link>http://checkmyworking.com/2012/04/interesting-esoterica-summation-volume-3/</link>
		<comments>http://checkmyworking.com/2012/04/interesting-esoterica-summation-volume-3/#comments</comments>
		<pubDate>Sun, 08 Apr 2012 12:17:06 +0000</pubDate>
		<dc:creator>Christian Perfect</dc:creator>
				<category><![CDATA[Interesting Esoterica]]></category>

		<guid isPermaLink="false">http://checkmyworking.com/?p=789</guid>
		<description><![CDATA[<p>Summing up some more interesting esoterica seems like the right thing to do at the moment, so here&#8217;s that.</p>
<p>A reminder: every now and then I encounter a paper or a book or an article that grabs my interest but isn’t &#8230;</p>]]></description>
			<content:encoded><![CDATA[<p>Summing up some more interesting esoterica seems like the right thing to do at the moment, so here&#8217;s that.</p>
<p>A reminder: every now and then I encounter a paper or a book or an article that grabs my interest but isn’t directly useful for anything. It might be about some niche sub-sub-subtopic I’ve never heard of, or it might talk about something old from a new angle, or it might just have a funny title. I put these things in my <a href="http://www.mendeley.com/groups/519971/interesting-esoterica/">Interesting Esoterica collection</a> on <a href="http://www.mendeley.com/" target="_blank">Mendeley</a>.</p>
<p>In this post the titles are links to the original sources, and I try to add some interpretation or explanation of why I think each thing is interesting below the abstract.</p>
<p>Some things might not be freely available, or even available for a reasonable price. <a href="http://www.thecostofknowledge.com/" target="_blank">Sorry</a>.</p>
<p><span id="more-789"></span></p>
<h3><a href="http://www.sciencedirect.com/science/article/pii/S0315086002000083">Passage to the limit in Proposition I, Book I of Newton&#8217;s <em>Principia</em></a></h3>
<blockquote><p>The purpose of this paper is to analyze the way in which Newton uses his polygon model and passes to the limit in Proposition I, Book I of his <em>Principia</em>. It will be evident from his method that the limit of the polygon is indeed the orbital arc of the body and that his approximation of the actual continuous force situation by a series of impulses passes correctly in the limit into the continuous centripetal force situation. The analysis of the polygon model is done in two ways: (1) using the modern concepts of force, linear momentum, linear impulse, and velocity, and (2) using Newton&#8217;s concepts of motive force and quantity of motion. It should be clearly understood that the term “force” without the adjective “motive,” is used in the modern sense, which is that force is a vector which is the time rate of change of the linear momentum. Newton did not use the word “force” in this modern sense. The symbol <em>$F$</em> denotes modern force. For Newton “force” was “motive force,” which is measured by the change in the quantity of motion of a body. Newton&#8217;s “quantity of motion” is proportional to the magnitude of the modern vector momentum. Motive force is a scalar and the symbol $F_m$ is used for motive force.</p></blockquote>
<p><em></em>A short history of maths article about how Newton approximated arcs using polygons whose boundaries tend to the actual arc as the number of sides increases. All this infinitesimal /passage to the limit stuff is very fiddly and sort of at the core of sorting out the calculus, but this article isn&#8217;t particularly interesting.</p>
<p>&nbsp;</p>
<h3><a href="http://arxiv.org/abs/1202.3033">Orange peels and Fresnel integrals</a></h3>
<blockquote><p>There are two standard ways of peeling an orange: either cut the skin along meridians, or cut it along a spiral. We consider here the second method, and study the shape of the spiral strip, when unfolded on a table. We derive a formula that describes the corresponding flattened-out spiral. Cutting the peel with progressively thinner strip widths, we obtain a sequence of increasingly long spirals. We show that, after rescaling, these spirals tends to a definite shape, known as the Euler spiral. The Euler spiral has applications in many fields of science. In optics, the illumination intensity at a point behind a slit is computed from the distance between two points on the Euler spiral. The Euler spiral also provides optimal curvature for train tracks between a straight run and an upcoming bend. It is striking that it can be also obtained with an orange and a kitchen knife.</p></blockquote>
<p>I&#8217;ve always found Euler spirals, also known as Cornu curves, interesting. They&#8217;re curves with a constant change in radius of curvature, so if you were driving around one you&#8217;d turn your steering wheel at a constant rate. They first found an application in train tracks, but have a variety of applications in optics and computer graphics.</p>
<p>This paper is about a cool fact that the authors noticed, which is that if you peel an orange in a spiral and lay it flat, it makes an Euler spiral.</p>
<p>Here&#8217;s a paper with some historical information about the Euler spiral: <a href="http://levien.com/phd/euler_hist.pdf">The Euler spiral: a mathematical history</a>.</p>
<p>&nbsp;</p>
<h3><a href="http://www.springerlink.com/content/q08n44457x236ln6/">Further evidence for addition and numerical competence by a Grey parrot (Psittacus erithacus)</a></h3>
<blockquote><p>A Grey parrot (<em>Psittacus erithacus</em>), able to quantify sets of eight or fewer items (including heterogeneous subsets), to sum two sequentially presented sets of 0–6 items (up to 6), and to identify and serially order Arabic numerals (1–8), all by using English labels, was tested on addition of two Arabic numerals or three sequentially presented collections (e.g., of variously sized jelly beans or nuts). He was, without explicit training and in the absence of the previously viewed addends, asked, “How many total?” and required to answer with a vocal English number label. In a few trials on the Arabic numeral addition, he was also shown variously colored Arabic numerals while the addends were hidden and asked “What color number (is the) total?” Although his death precluded testing on all possible arrays, his accuracy was statistically significant and suggested addition abilities comparable with those of nonhuman primates.</p></blockquote>
<p>The late parrot <a href="http://en.wikipedia.org/wiki/Alex_(parrot)">Alex</a> was quite famous. His owners were conducting an experiment to see, first of all, how much language he could understand and use and, later, whether he could understand quantity and number. This paper describes how Alex could add up collections of objects and say the answer out loud.</p>
<p>&nbsp;</p>
<h3><a href="http://www.ams.org/journals/bull/2008-45-01/S0273-0979-07-01191-3/">Barcodes: the persistent topology of data</a></h3>
<blockquote><p>This article surveys recent work of Carlsson and collaborators on applications of computational algebraic topology to problems of feature detection and shape recognition in high-dimensional data. The primary mathematical tool considered is a homology theory for point-cloud data sets&#8211;<strong>persistent homology</strong>&#8211;and a novel representation of this algebraic characterization&#8211;<strong>barcodes</strong>. We sketch an application of these techniques to the classification of natural images.</p></blockquote>
<p>This is really fascinating. Suppose you have a cloud of points in a space with a lot of dimensions. We can only perceive two dimensions directly, so you have to project the points down onto the plane. Can you work out the shape of the points from the projection? The author compares this to seeing what a connect-the-dots drawing is of before drawing the lines.</p>
<p>The barcodes mentioned in the title are representations of how connected the data are as you try to join them into simplices. From these you can make a guess at the homology of the space the points come from, which associates a group with its topology. The homology is a quick way of identifying the shape of the cloud. I&#8217;m not doing a great job of explaining all this but the paper does, and it gets bonus points for a completely gratuitous figure of a Klein bottle in front of a barcode.</p>
<h3></h3>
<h3><a href="http://www.cadaeic.net/naraven.htm">Poe, E. &#8211; Near A Raven</a></h3>
<p>The world&#8217;s most heroic mnemonic for $\pi$. The lengths of the words represent the (base-10) digits of $\pi$. If I ever felt the need to memorise a lot of digits of $\pi$, I think I&#8217;d use this. I won&#8217;t, though.</p>
<h3></h3>
<h3><a href="http://people.mpim-bonn.mpg.de/zagier/files/exp-math-2/fulltext.pdf">Quotients Homophones des Groupes Libres / Homophonic Quotients of Free Groups</a></h3>
<p>This paper has two titles. It has an excellent slogan &#8211; &#8220;<em>Ah, la recherche! Du temps perdu.</em>&#8221; It&#8217;s about what happens if you identify homophones in the free group generated by the letters of the alphabet. It has results in both French and English. I love this paper.</p>
<h3></h3>
<h3><a href="http://arxiv.org/abs/1104.3034">Random walks reaching against all odds the other side of the quarter plane</a></h3>
<blockquote><p>For a homogeneous random walk in the quarter plane with nearest-neighbor transitions, starting from some state $(i_0,j_0)$, we study the event that the walk reaches the vertical axis, before reaching the horizontal axis. We derive an exact expression for the probability of this event, and derive an asymptotic expression for the case when $i_0$ becomes large, a situation in which the event becomes highly unlikely. The exact expression follows from the solution of a boundary value problem and is in terms of an integral that involves a conformal gluing function. The asymptotic expression follows from the asymptotic evaluation of this integral. Our results find applications in a model for nucleosome shifting, the voter model and the asymmetric exclusion process.</p></blockquote>
<p>I wish I could remember why I saved this! The title is quite fun, but I have a feeling the result has an interesting application. It might be something to do with the application to the voter model they mention, but I can&#8217;t remember what.</p>
<p>&nbsp;</p>
<h3><a href="http://www-math.mit.edu/~poonen/papers/sampler.pdf">Undecidable Problems: A Sampler</a></h3>
<blockquote><p>After discussing two senses in which the notion of undecidability is used, we present a survey of undecidable decision problems arising in various branches of mathematics.</p></blockquote>
<p>Just a good reference of undecidable problems.</p>
<p>&nbsp;</p>
<h3><a href="http://www.myreckonings.com/modernnomogramsTMP/Doc_What_Are_My_Chances_UMAP_32-4-2011.pdf">Doc, what are my chances?</a></h3>
<blockquote><p>When a physician has to deliver bad news to a patient, the bottom-line question is, “Doc, what are my chances?” The doctor’s answers are then usually couched in the language of probability.</p>
<p>Recently, both patients and insurance companies have become more questioning, because they want to understand the doctor’s reasoning. What persists is an explanation gap, the distance between the doctor’s best assessment and the patient’s comprehension of both the probability and how it was obtained.</p>
<p>We seek to reveal the logic behind the medical diagnostic decision making process and arm the patient with the appropriate vocabulary. Because decisions hinge on probabilities—“we’ll proceed to this treatment if we can ascertain that there is at least a 50% chance you have the disease” — we demonstrate how the numbers are determined. This approach removes the fuzziness from the process, and the required arithmetic is not intimidating when taken one step at a time. To demonstrate the method, we work a sample problem from start to ﬁnish.</p>
<p>We illustrate how a graphical solution called a nomogram [Doerﬂer 2009] delivers numerical results easily. Moreover, nomograms can offer advantages over other computational methods; an appendix gives a comparison.</p></blockquote>
<p>I&#8217;ve seen a few nomograms in my various encounters with serious doctors. They&#8217;re a nice simple way of visualising how parameters affect some measurement. The authors of this paper have created a fantastic nomogram for using Bayesian reckoning to decide whether a test is worth doing. They also <a href="http://www.myreckonings.com/modernnomograms/">sell posters</a>, which I&#8217;ve considered buying for the department&#8217;s Bayesians.</p>
<p>&nbsp;</p>
<h3><a href="http://www.jstor.org/stable/2002657?seq=1">A new approximation to $\pi$ (conclusion)</a></h3>
<p>The March 16, 1946 edition of <em>Nature</em> contained a very short announcement that William Shanks (<a href="http://www.houghtonlespring.org.uk/articles/william_shanks.htm">resident of Houghton-le-Spring</a>, just down the road from where I grew up) had <a href="http://www.nature.com/nature/journal/v157/n3985/pdf/157342c0.pdf">got his calculation of $\pi$ wrong after the 528th digit</a>.</p>
<p>In July 1946, <em>Mathematical Tables and Computation</em> published <a href="http://www.jstor.org/stable/2002627?seq=3">a note with the digits </a><a href="http://www.jstor.org/stable/2002627?seq=3">corrected </a><a href="http://www.jstor.org/stable/2002627?seq=3">up to the 620th</a>.</p>
<p>In April1947, an article titled <em><a href="http://www.jstor.org/stable/2002296?seq=1">A new approximation to $\pi$</a></em> gave D.F. Ferguson and Dr. John Wrench&#8217;s independent calculations up to 808 digits. That wasn&#8217;t quite the end of the story. In the article I linked to in the headline above, both men published a few corrections to Dr Wrench&#8217;s longer calculation. And that was the end of the matter! <a href="http://www-history.mcs.st-andrews.ac.uk/HistTopics/Pi_chronology.html">Definitely</a>!</p>
<p>Several sources note that Ferguson was the first to use a mechanical desk calculator to do his computations. I&#8217;d really like to know what it was.</p>
<p>&nbsp;</p>
<h3><a href="http://udoo.uni-muenster.de/downloads/publications/1433.pdf">Topic-based Vector Space Model</a></h3>
<blockquote><p>This paper motivates and presents the Topic-based Vector Space Model (TVSM), a new vector-based approach for document comparison. The approach does not assume independence between terms and it is flexible regarding the specification of term-similarities. Stopword-list, stemming and thesaurus can be fully integrated into the model. This paper shows further how the TVSM can be fully implemented within the context of relational databases. This facilitates the use of this approach by generic applications. At the end short comparisons with other vector-based approaches namely the Vector Space Model (VSM) and the Generalized Vector Space Model (GVSM) are presented.</p></blockquote>
<p>I found this while trying to find references for a story we linked to on The Aperiodical about <a href="http://aperiodical.com/2012/03/automatically-tagging-the-world-service-archive/#comment-37">BBC R&amp;D&#8217;s efforts to automatically tag transcripts of radio programmes</a>. The paper itself isn&#8217;t enormously interesting but it&#8217;s good to know it&#8217;s been done.</p>
<p>&nbsp;</p>
<h3><a href="http://mathoverflow.net/questions/91423/gaussian-prime-spirals">Gaussian prime spirals</a></h3>
<blockquote><p>Imagine a particle in the complex plane, starting at $c_0$, a <a href="http://en.wikipedia.org/wiki/Gaussian_integer" rel="nofollow">Gaussian integer</a>, moving initially $\pm$ in the horizontal or vertical directions. When it hits a <a href="http://en.wikipedia.org/wiki/Gaussian_prime#As_a_unique_factorization_domain" rel="nofollow">Gaussian prime</a>, it turns left $90^{\circ}$. <em>Does the spiral always form a cycle?</em></p></blockquote>
<p>One of that kind of MathOverflow question that makes you forget the incomprehensible self-esteem-destroying horror of the rest of the site. A nice and simply stated question about an automaton walking between the Gaussian integers, with some clever insights and lots of nice pictures.</p>
<p>&nbsp;</p>
<h3><a href="http://www.jstor.org/stable/10.2307/2323998">The bitangent sphere problem</a></h3>
<blockquote><p>Here is the problem: choose a point po on a (smooth, closed, embedded) surface $M$ in $\mathbb{R}_3$. Does there always exist a sphere which is tangent to $M$ at $p_o$ and at some other point $p \in M$?<br />
Here is the answer: no. But the answer is very nearly yes, for there is always a sphere or plane having the required property&#8230;</p></blockquote>
<p>A nice bit of geometry.</p>
<p>&nbsp;</p>
<h3><a href="http://arxiv.org/abs/1107.3707">Statistical Laws Governing Fluctuations in Word Use from Word Birth to Word Death</a></h3>
<blockquote><p>We analyze the dynamic properties of $10^7$ words recorded in English, Spanish and Hebrew over the period 1800&#8211;2008 in order to gain insight into the coevolution of language and culture. We report language independent patterns useful as benchmarks for theoretical models of language evolution. A significantly decreasing (increasing) trend in the birth (death) rate of words indicates a recent shift in the selection laws governing word use. For new words, we observe a peak in the growth-rate fluctuations around 40 years after introduction, consistent with the typical entry time into standard dictionaries and the human generational timescale. Pronounced changes in the dynamics of language during periods of war shows that word correlations, occurring across time and between words, are largely influenced by coevolutionary social, technological, and political factors. We quantify cultural memory by analyzing the long-term correlations in the use of individual words using detrended fluctuation analysis.</p></blockquote>
<p>&nbsp;</p>
<h3><a href="http://arxiv.org/abs/1201.5597">The mate-in-n problem of infinite chess is decidable</a></h3>
<blockquote><p>Infinite chess is chess played on an infinite edgeless chessboard. The familiar chess pieces move about according to their usual chess rules, and each player strives to place the opposing king into checkmate. The mate-in-n problem of infinite chess is the problem of determining whether a designated player can force a win from a given finite position in at most n moves. A naive formulation of this problem leads to assertions of high arithmetic complexity with 2n alternating quantifiers&#8212;there is a move for white, such that for every black reply, there is a counter-move for white, and so on. In such a formulation, the problem does not appear to be decidable; and one cannot expect to search an infinitely branching game tree even to finite depth. Nevertheless, the main theorem of this article, confirming a conjecture of the first author and C. D. A. Evans, establishes that the mate-in-n problem of infinite chess is computably decidable, uniformly in the position and in n. Furthermore, there is a computable strategy for optimal play from such mate-in-n positions. The proof proceeds by showing that the mate-in-n problem is expressible in what we call the first-order structure of chess, which we prove (in the relevant fragment) is an automatic structure, whose theory is therefore decidable. Indeed, it is definable in Presburger arithmetic. Unfortunately, this resolution of the mate-in-n problem does not appear to settle the decidability of the more general winning-position problem, the problem of determining whether a designated player has a winning strategy from a given position, since a position may admit a winning strategy without any bound on the number of moves required. This issue is connected with transfinite game values in infinite chess, and the exact value of the omega one of chess is not known.</p></blockquote>
<p>&#8220;Infinite chess&#8221; and &#8220;mental poker&#8221; are just wonderful things to say you&#8217;re researching. I need to think of an unsolved problem in something like &#8220;chaotic tiddlywinks&#8221;.</p>
<p>&nbsp;</p>
<h3><a href="http://arxiv.org/abs/1203.0100">Cake Cutting Mechanisms</a></h3>
<blockquote><p>We examine the history of cake cutting mechanisms and discuss the efficiency of their allocations. In the case of piecewise uniform preferences, we define a game that in the presence of strategic agents has equilibria that are not dominated by the allocations of any mechanism. We identify that the equilibria of this game coincide with the allocations of an existing cake cutting mechanism.</p></blockquote>
<p>An undergrad&#8217;s dissertation giving an exhaustive  survey of ways of dividing up cake between several people, who might want different amounts or pieces. I lost interest by the time I got to the game theory (nothing is more tedious!) but I&#8217;m always forgetting cake-slicing mechanisms so this will be good to have as a reference. It&#8217;s got a lot of good quotes too. Keen undergrads are what make life bearable.</p>
<p>&nbsp;</p>
<h3><a href="http://moniker.name/worldmaking/?p=409">Navigating Hyperbolic Space with Fibonacci Trees</a></h3>
<p>I recently found a joyful game called <a href="http://www.roguetemple.com/z/hyper.php">HyperRogue II</a>: a roguelike played in hyperbolic space. I&#8217;ve always wanted to do something similar, so seeing that it&#8217;s possible has prompted me to look up the maths behind tiling and navigating hyperbolic space. This post sums up, in a very handwavey way, how Maurice Margenstern does it, using Fibonacci trees. Bonus points for a justifiable application of the fibonacci sequence; deeply satisfying.</p>
<p>It took me a while to find an accessible copy of the paper the author refers to, <a href="https://sites.google.com/site/benoitmartin57/partage/acri-2004.pdf?attredirects=0&amp;d=1">Cellular automata in the hyperbolic plane: proposal for a new environment.</a></p>
<p>&nbsp;</p>
<p>That&#8217;s it for now!</p>
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		<title>Slides about the princess in a castle puzzle</title>
		<link>http://checkmyworking.com/2012/04/slides-about-the-princess-on-a-castle-puzzle/</link>
		<comments>http://checkmyworking.com/2012/04/slides-about-the-princess-on-a-castle-puzzle/#comments</comments>
		<pubDate>Mon, 02 Apr 2012 14:37:53 +0000</pubDate>
		<dc:creator>Christian Perfect</dc:creator>
				<category><![CDATA[MathsJam]]></category>
		<category><![CDATA[presentations]]></category>

		<guid isPermaLink="false">http://checkmyworking.com/?p=779</guid>
		<description><![CDATA[<p>I gave a talk to our internal postgrad forum last week about the <a href="http://checkmyworking.com/2011/12/solving-the-princess-on-a-graph-puzzle/">princess in a castle puzzle</a>. I made some slides for it using <a href="http://imakewebthings.com/deck.js/">deck.js</a>. They looked quite nice and I could just about get what I &#8230;</p>]]></description>
			<content:encoded><![CDATA[<p>I gave a talk to our internal postgrad forum last week about the <a href="http://checkmyworking.com/2011/12/solving-the-princess-on-a-graph-puzzle/">princess in a castle puzzle</a>. I made some slides for it using <a href="http://imakewebthings.com/deck.js/">deck.js</a>. They looked quite nice and I could just about get what I wanted in them, but I now know that using SVG in HTML is still an enormous faff if you want it to scale nicely, which is basically the only reason you would use SVG.</p>
<p><a href="http://checkmyworking.com/misc/princess%20pgf/">Click here to see the slides</a>.</p>
<p>I&#8217;m not sure if you can follow along with the slides without me talking; maybe I&#8217;ll do a transcript with <a href="https://github.com/dseif/slide-drive">slide drive</a> later.</p>
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		<title>Visualising the wrong data on the Guardian data blog</title>
		<link>http://checkmyworking.com/2012/03/visualising-the-wrong-data-on-the-guardian-data-blog/</link>
		<comments>http://checkmyworking.com/2012/03/visualising-the-wrong-data-on-the-guardian-data-blog/#comments</comments>
		<pubDate>Mon, 19 Mar 2012 21:31:29 +0000</pubDate>
		<dc:creator>Christian Perfect</dc:creator>
				<category><![CDATA[Musings]]></category>

		<guid isPermaLink="false">http://checkmyworking.com/?p=771</guid>
		<description><![CDATA[<p style="text-align: center;"><a href="http://www.guardian.co.uk/news/datablog/2012/mar/19/minimum-wage-housing"><img class="aligncenter size-full wp-image-772" title="Housing minimum wage map" src="http://checkmyworking.com/wp-content/uploads/2012/03/Housing-minimum-wage-map-007.jpg" alt="" width="460" height="276" /></a></p>
<p><a href="http://www.guardian.co.uk/news/datablog/2012/mar/19/minimum-wage-housing">This visualisation</a> shows for each council or unitary authority how many hours a week you&#8217;d need to work, earning minimum wage, in order to pay the median rent for a one-bed flat. The minimum wage is a national constant.</p>
<p>No &#8230;</p>]]></description>
			<content:encoded><![CDATA[<p style="text-align: center;"><a href="http://www.guardian.co.uk/news/datablog/2012/mar/19/minimum-wage-housing"><img class="aligncenter size-full wp-image-772" title="Housing minimum wage map" src="http://checkmyworking.com/wp-content/uploads/2012/03/Housing-minimum-wage-map-007.jpg" alt="" width="460" height="276" /></a></p>
<p><a href="http://www.guardian.co.uk/news/datablog/2012/mar/19/minimum-wage-housing">This visualisation</a> shows for each council or unitary authority how many hours a week you&#8217;d need to work, earning minimum wage, in order to pay the median rent for a one-bed flat. The minimum wage is a national constant.</p>
<p>No justification is given for using the median rent. In a fair world, the median rent should be paid by someone on the median income. Assuming that people earning the minimum wage are the lowest earners<sup><a href="http://checkmyworking.com/2012/03/visualising-the-wrong-data-on-the-guardian-data-blog/#footnote_0_771" id="identifier_0_771" class="footnote-link footnote-identifier-link" title="this isn&amp;#8217;t true: people earning cash in hand, for example, might earn less than the minimum wage but they still have an effect on house prices">1</a></sup> and make up X% of the population, then an upper bound for rent paid by people earning minimum wage should be the Xth percentile, if housing is provided fairly<sup><a href="http://checkmyworking.com/2012/03/visualising-the-wrong-data-on-the-guardian-data-blog/#footnote_1_771" id="identifier_1_771" class="footnote-link footnote-identifier-link" title="another large assumption.">2</a></sup>. If you&#8217;re not paying attention to this kind of thing, you might as well just say &#8220;average&#8221; instead of &#8220;median&#8221;.</p>
<p>Finally, there&#8217;s this:</p>
<blockquote><p>Note that these figures are dramatic simplifications, used for illustrative purposes only: in reality tax payments, tax credits, housing benefit, council tax, utility bills and more substantially alter these figures.</p></blockquote>
<p>So the figures are meaningless anyway &#8211; benefits might be so generous that very little of the lowest earners&#8217; pay goes towards housing, or taxes might be so high that people really need to work many more hours than this statistic claims in order to keep a house.</p>
<p>Supposing that all of these problems are sorted out and we calculate a reasonable-sounding number of hours to work to earn rent each week, what should that number be? What if house prices plummeted but food became very expensive? I can&#8217;t think what these numbers are telling us that raw rent data aren&#8217;t already.</p>
<p>A question has occurred to me: would I be making life worse for those worse off me by living in a flat cheaper than I should be or, on the other hand, would spending a higher than average proportion of my income on rent drive prices up, having an effect on the bottom end of the market? I suppose taking a cheaper flat reduces the prices of nicer ones. This is why I don&#8217;t like economics.</p>
<ol class="footnotes"><li id="footnote_0_771" class="footnote">this isn&#8217;t true: people earning cash in hand, for example, might earn less than the minimum wage but they still have an effect on house prices</li><li id="footnote_1_771" class="footnote">another large assumption.</li></ol>]]></content:encoded>
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		<title>Gatterdam&#8217;s proof that free products with amalgamation are $\mathcal{E}^{n+1}$-computable</title>
		<link>http://checkmyworking.com/2012/03/gatterdams-proof-that-amalgamated-free-products-are-mathcalen1-computable/</link>
		<comments>http://checkmyworking.com/2012/03/gatterdams-proof-that-amalgamated-free-products-are-mathcalen1-computable/#comments</comments>
		<pubDate>Sat, 10 Mar 2012 17:24:25 +0000</pubDate>
		<dc:creator>Christian Perfect</dc:creator>
				<category><![CDATA[Research]]></category>

		<guid isPermaLink="false">http://checkmyworking.com/?p=749</guid>
		<description><![CDATA[<p>I haven&#8217;t posted any research in ages! I haven&#8217;t done any research in ages! Here&#8217;s a rewriting of a theorem from Ronald Gatterdam&#8217;s doctoral thesis, in more sensible language. I&#8217;ve proved the same theorem for fundamental groups of graphs of &#8230;</p>]]></description>
			<content:encoded><![CDATA[<p>I haven&#8217;t posted any research in ages! I haven&#8217;t done any research in ages! Here&#8217;s a rewriting of a theorem from Ronald Gatterdam&#8217;s doctoral thesis, in more sensible language. I&#8217;ve proved the same theorem for fundamental groups of graphs of groups.</p>
<p>The main problem with Gatterdam&#8217;s proof was that he didn&#8217;t invent any new notation to make life simpler, so he wrote out pages and pages of very repetitive function applications.</p>
<p>I&#8217;ve made up a lot of notation to make dealing with functions in the Grzegorczyk hierarchy a lot easier, so you&#8217;ll probably need to refer to <a href="https://github.com/christianp/PHD-Notes/raw/master/bass_serre_computability.pdf">my draft paper</a> for a few things. I keep all my research notes in <a href="https://github.com/christianp/PHD-Notes">a repository on github</a>, by the way.</p>
<h2><span id="more-749"></span>Theorem</h2>
<p>Let $G_1, G_2$ be $\mathcal{E}^n$-computable groups. Let $H_1, H_2$ be $\mathcal{E}^n$-decidable subgroups of the latter. Let $\phi&#8217;: H_1 \to H_2$ be an isomorphism, with $\phi&#8217;, \phi&#8217;^{-1}$ both $\mathcal{E}^n$-computable.</p>
<p>Then the free product of $G_1$ and $G_2$ with $H_1$ and $H_2$ amalgamated, $G = G_1 \underset{\phi&#8217;}{\ast} G_2$, is $\mathcal{E}^{n+1}$-computable.</p>
<h2>Proof</h2>
<p>Let $(i&#8217;_1,m&#8217;_1,j&#8217;_1)$ be the index of $G_1$ and $(i&#8217;_2,m&#8217;_2,j&#8217;_2)$ be the index of $G_2$. The dashes will be explained later.</p>
<p>We can assume, without loss of generality, that $0 \notin i&#8217;_a(G_a)$ and $i&#8217;_a(1) = 1$ for $a=1,2$. <em>(Might not even need this.)</em></p>
<p>By Magnus, Karrass, Solitar, etc., all elements $g \in G$ have normal form</p>
<p>\[ g = h g_1 \dots g_r \]</p>
<p>where $h \in H_1$, and the $g_i$ are coset representatives of $G_a / H_a$, $a=1,2$, such that $g_{i+1} \in G_1 \Leftrightarrow g_i \in G_2$.</p>
<p>The following proof becomes a lot easier if we redefine the factor group indices as follows:</p>
<p>\[ \begin{align} i_1(x) &amp;:= 2i'_1(x), \\ m_1(x,y) &amp;:= 2m'_1 \left( \frac{x}{2},\frac{y}{2} \right) \\ j_1(x) &amp;:= 2j'_1 \left( \frac{x}{2} \right). \end{align} \]</p>
<p>\[ \begin{align} i_2(x) &amp;:= 2i'_2(x) - 1, \\ m_2(x,y) &amp;:= 2m'_2 \left( \frac{x+1}{2},\frac{y+1}{2} \right) \\ j_2(x) &amp;:= 2j'_2 \left( \frac{x+1}{2} \right). \end{align} \]</p>
<p>Now,</p>
<p>\[ \begin{align} x \in i_1(G_1) &amp;\Leftrightarrow 2 \mid x \wedge \frac{x}{2} \in i'_1(G_1), \\ x \in i_1(H_1) &amp;\Leftrightarrow x \in i_1(G_1) \wedge \frac{x}{2} \in i'_1(H_1), \\ x \in i_2(G_2) &amp;\Leftrightarrow 2\not \mid x \wedge \frac{x+1}{2} \in i'_2(G_2), \\ x \in i_2(H_2) &amp;\Leftrightarrow x \in i_2(G_2) \wedge \frac{x+1}{2} \in i'_2(H_2). \end{align} \]</p>
<p>The subgroup isomorphism also needs to be redefined:</p>
<p>\[ \begin{align} \phi(x) &amp;:= 2\phi' \left( \frac{x}{2} \right)-1 \\ \phi^{-1}(x) &amp;:= 2 \phi'^{-1} \left( \frac{x+1}{2} \right) \end{align} \]</p>
<p>In order to do multiplication, we need to be able to split every $g_a \in G_a$, $a=1,2$, into a word of the form $h_ak_a$, where $h_a \in H_a$ and $k_a$ is a coset representative of $g_a$ in $G_a / H_a$.</p>
<p>Define:</p>
<p>\[ \begin{align} k_a(x) &amp;:= \min_{y \leq x} \left( m_a(x,j_a(y)) \in i_a(H_a) \right), \\ h_a(x) &amp;:= m_a(x,j_a(k_a(x))). \end{align} \]</p>
<p>Now we can define $i(G)$ by</p>
<p>\[ i(hg_1 \dots g_r) := [h,g_1,\dots,g_r]. \]</p>
<p>(encode it as a Gödel number)</p>
<p>$i(G)$ is $\mathcal{E}^n$-decidable because $x \in i(G)$ if and only if:</p>
<ul>
<li>$x$ is a Gödel number</li>
<li>$(x)_0 \in i_1(H_1)$</li>
<li>$\forall 2 \lt i \lt |x|$, $(x)_i \in i_1(G_1) \Leftrightarrow (x)_{i-1} \in i_2(G_2)$</li>
<li>$\forall 1 \lt i \lt |x|$, $(x)_i \in i_a(G_a) \rightarrow k_a((x)_i) = (x)_i$.</li>
</ul>
<p>Now define a function $r: (i_1(G_1) \cup i_2(G_2)) \times i(G) \to i(G)$ which does multiplication of elements of $G$ on the left by elements of $G_1$ and $G_2$.</p>
<p>$r$ is defined by cases in a decision tree, so here’s some programming-ish syntax:</p>
<p>$r(x,y):=$</p>
<p><strong>If</strong> {$x \in i_1(G_1)$}</p>
<p style="padding-left: 30px;"><strong>If</strong> {$x \in i_1(H_1)$}</p>
<p style="padding-left: 60px;">$[m_1(x,(y)_0)] +\!\!\!\!+ y[1 \dots]$</p>
<p style="padding-left: 30px;"><strong>else</strong></p>
<p style="padding-left: 60px;"><strong></strong><strong>If</strong> {$(y)_1 \in i_1(G_1)$}</p>
<p style="padding-left: 90px;"><strong>If</strong> {$m_1(m_1(x,(y)_0),(y)_1) \in i_1(H_1)$ }</p>
<p style="padding-left: 120px;">$[m_1(m_1(x,(y)_0),(y)_1)] +\!\!\!\!+ y[2 \dots]$</p>
<p style="padding-left: 90px;"><strong>else</strong></p>
<p style="padding-left: 120px;">$[h_1m_1(m_1(x,(y)_0),(y)_1), k_1m_1(m_1(x,(y)_0),(y)_1)] +\!\!\!\!+ y[2 \dots]$</p>
<p style="padding-left: 90px;"><strong>endif</strong></p>
<p style="padding-left: 60px;"><strong></strong><strong>else</strong></p>
<p style="padding-left: 90px;"><strong></strong>$[h_1m_1(x,(y)_0),k_1m_2(x,(y)_0)] +\!\!\!\!+ y[1 \dots]$</p>
<p style="padding-left: 60px;"><strong>endif</strong></p>
<p style="padding-left: 30px;"><strong></strong><strong>endif</strong></p>
<p><strong></strong><strong>else</strong></p>
<p style="padding-left: 30px;"><strong></strong><strong>If</strong> {$x \in i_2(H_2)$ }</p>
<p style="padding-left: 60px;">$[m_1(\phi^{-1}(x),(y)_0)] +\!\!\!\!+ y[1 \dots]$</p>
<p style="padding-left: 30px;"><strong>else</strong></p>
<p style="padding-left: 60px;"><strong></strong><strong>If</strong> {$(y)_1 \in i_2(G_2)$ }</p>
<p style="padding-left: 90px;"><strong>If</strong> {$m_2(m_2(x,\phi((y)_0)),(y)_1) \in i_2(H_2)$ }</p>
<p style="padding-left: 120px;">$[\phi^{-1}m_2(m_2(x,\phi((y)_0)),(y)_1)] +\!\!\!\!+ y[2 \dots]$</p>
<p style="padding-left: 90px;"><strong>else</strong></p>
<p style="padding-left: 120px;"><strong></strong>$[\phi^{-1}h_2m_2(m_2(x,\phi((y)_0)),(y)_1), k_2m_2(m_2(x,\phi((y)_0)),(y)_1)] +\!\!\!\!+ y[2 \dots]$</p>
<p style="padding-left: 90px;"><strong>endif</strong></p>
<p style="padding-left: 60px;"><strong>else</strong></p>
<p style="padding-left: 90px;"><strong></strong>$[\phi^{-1}h_2m_2(x,\phi((y)_0)),k_2m_2(x,\phi((y)_0))] +\!\!\!\!+ y[1 \dots]$</p>
<p style="padding-left: 60px;"><strong>endif</strong></p>
<p style="padding-left: 30px;"><strong></strong><strong>endif</strong></p>
<p><strong></strong><strong>endif</strong></p>
<p>&nbsp;</p>
<p>Now we can do multiplication on $G$ in general:</p>
<p>\[ \begin{align} m([],y) &amp;:= y, \\ m( x:xs, y) &amp;:= r(x,m(xs,y)). \end{align} \]</p>
<p>And inversion:</p>
<p>\[ \begin{align} j([1]) &amp;:= [1], \\ j(x:xs) &amp;:= r(\hat{j}(x), j(xs)).\end{align} \]</p>
<p>\[ \hat{j}(x) := \begin{cases} j_1(x), &amp; x \in i_1(G_1), \\ j_2(x), &amp; x \in i_2(G_2). \end{cases} \]</p>
<p>($x:xs$ means a list of the form $[x,x_1,x_2, \dots]$. I borrowed this syntax from Haskell. I need to show that you can define recursive functions on lists like this, but I hope you agree it works)</p>
<p>The functions $m$ and $j$ are defined recursively. Because we don&#8217;t know anything about how quickly the original multiplication functions grow, the recursion is unbounded, so the best we can do is to say that the index of $G$ is at worst $\mathcal{E}^{n+1}$-computable.</p>
<p><strong>QED</strong></p>
<p>Gatterdam also shows that the embeddings $G_a \to G$ are natural, but I lost the will.</p>
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		<title>Using a zero-knowledge protocol to prove you can solve a sudoku</title>
		<link>http://checkmyworking.com/2012/03/using-a-zero-knowledge-protocol-to-prove-you-can-solve-a-sudoku/</link>
		<comments>http://checkmyworking.com/2012/03/using-a-zero-knowledge-protocol-to-prove-you-can-solve-a-sudoku/#comments</comments>
		<pubDate>Sat, 10 Mar 2012 10:31:02 +0000</pubDate>
		<dc:creator>Christian Perfect</dc:creator>
				<category><![CDATA[Musings]]></category>
		<category><![CDATA[videos]]></category>

		<guid isPermaLink="false">http://checkmyworking.com/?p=744</guid>
		<description><![CDATA[<p>I&#8217;ve just uploaded to youtube a video I made with <a href="http://twitter.com/#!/stecks">Katie Steckles</a> to demonstrate why zero-knowledge protocols exist and how one works.</p>
<p style="text-align: left;">
</p><p>Katie is a habitual liar, so we followed the zero-knowledge protocol described in the paper, &#8220;Cryptographic and Physical &#8230;</p>]]></description>
			<content:encoded><![CDATA[<p>I&#8217;ve just uploaded to youtube a video I made with <a href="http://twitter.com/#!/stecks">Katie Steckles</a> to demonstrate why zero-knowledge protocols exist and how one works.</p>
<p style="text-align: left;"><span style="text-align:center; display: block;"><a href="http://checkmyworking.com/2012/03/using-a-zero-knowledge-protocol-to-prove-you-can-solve-a-sudoku/"><img src="http://img.youtube.com/vi/i0sffXfDCUI/2.jpg" alt="" /></a></span></p>
<p>Katie is a habitual liar, so we followed the zero-knowledge protocol described in the paper, &#8220;Cryptographic and Physical Zero-Knowledge Proof Systems for Solutions of Sudoku Puzzles&#8221; which you can download from <a title="http://www.mit.edu/~rothblum/papers/sudoku.pdf" dir="ltr" href="http://www.mit.edu/~rothblum/papers/sudoku.pdf" rel="nofollow" target="_blank">http://www.mit.edu/~rothblum/papers/sudoku.pdf</a></p>
<p>By following this protocol, Katie can prove that she isn&#8217;t lying to me about being able to solve the puzzle, without revealing anything about how she solved it.</p>
<p>The paper I mentioned, &#8220;How to explain zero-knowledge protocols to your children&#8221; is an excellent explanation of the ideas behind zero-knowledge proof. You can get it from <a title="http://pages.cs.wisc.edu/~mkowalcz/628.pdf" dir="ltr" href="http://pages.cs.wisc.edu/~mkowalcz/628.pdf" rel="nofollow" target="_blank">http://pages.cs.wisc.edu/~mkowalcz/628.pdf</a></p>
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		<title>Fractal dimension in IKEA</title>
		<link>http://checkmyworking.com/2012/02/fractal-dimension-in-ikea/</link>
		<comments>http://checkmyworking.com/2012/02/fractal-dimension-in-ikea/#comments</comments>
		<pubDate>Fri, 24 Feb 2012 11:07:44 +0000</pubDate>
		<dc:creator>Christian Perfect</dc:creator>
				<category><![CDATA[Musings]]></category>

		<guid isPermaLink="false">http://checkmyworking.com/?p=730</guid>
		<description><![CDATA[<p>A long time ago, I realised that IKEA&#8217;s shopfitters must be experts in fractal dimension &#8211; they manage to lay out their shop so that you have to walk past every single thing they&#8217;re selling. You can&#8217;t just nip into &#8230;</p>]]></description>
			<content:encoded><![CDATA[<p>A long time ago, I realised that IKEA&#8217;s shopfitters must be experts in fractal dimension &#8211; they manage to lay out their shop so that you have to walk past every single thing they&#8217;re selling. You can&#8217;t just nip into IKEA &#8211; you have to go through the whole hour-long &#8220;It&#8217;s A Small World&#8221; of affordably wobbly furniture even if all you want is some kitchen utensils from the bit at the end.</p>
<p>I&#8217;d been meaning to add something about this to the Maths in the City site but it required going in to IKEA and taking a picture of their floor plan for illustration. I just remembered this morning that I&#8217;ve already taken one of those pictures, so I spent half an hour writing up this article &#8211; <a href="http://www.mathsinthecity.com/sites/fractal-dimension-ikea" target="_blank">Fractal Dimension in IKEA</a>.</p>
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