Jekyll2022-06-30T23:36:21+00:00https://11011110.github.io/blog/feed.xml11011110Geometry, graphs, algorithms, and moreDavid EppsteinLinkage2022-06-30T16:23:00+00:002022-06-30T16:23:00+00:00https://11011110.github.io/blog/2022/06/30/linkage<ul> <li> <p><a href="https://njas.blog/">Neil Sloane has a new blog</a> <span style="white-space:nowrap">(<a href="https://mathstodon.xyz/@christianp/108487272201741871">$$\mathbb{M}$$</a>),</span> subtitled “interesting sequences I need help with”. <a href="https://njas.blog/2022/06/03/the-two-up-sequence-a090252/">His first post concerns the two-up sequence</a>, formed in steps where the $k$th step adds two numbers that are not already in the sequence and are relatively prime to the preceding $k$. Most of the terms appear to be the primes (in order). The remaining terms appear to be prime powers or semiprimes but this has not been proven.</p> </li> <li> <p><a href="https://en.wikipedia.org/wiki/Schwarz_lantern">Schwarz lantern</a> <span style="white-space:nowrap">(<a href="https://mathstodon.xyz/@11011110/108497292976963088">$$\mathbb{M}$$</a>),</span> a polyhedral cylindrical surface that you can fold from paper. As you make its triangles smaller relative to the cylinder size, it approximates the cylinder in distance, but not necessarily in surface area. This example became one of the original motivations for research on non-obtuse mesh generation algorithms, because its non-convergence to the correct area happens when its triangles are very obtuse. Now a Good Article on Wikipedia.</p> </li> <li> <p><a href="https://rjlipton.wpcomstaging.com/2022/06/19/the-graph-of-ancestors/">The graph of ancestors</a> <span style="white-space:nowrap">(<a href="https://mathstodon.xyz/@11011110/108507916081000038">$$\mathbb{M}$$</a>).</span> For Father’s Day, Ken Regan tries to find a more principled way of quantifying pedigree collapse (the unavoidable existence of inbreeding among one’s ancestors) than “implex”, the difference between $$2^k$$ and the number of distinct $$k$$-step ancestors.</p> </li> <li> <p><a href="https://mathstodon.xyz/@johncarlosbaez/108504865028259313">John Baez describes the work of Hoàng Xuân Sính</a> (<a href="https://en.wikipedia.org/wiki/Ho%C3%A0ng_Xu%C3%A2n_S%C3%ADnh">see also</a>), a student of Grothendieck who became the first female mathematician in Vietnam.</p> </li> <li> <p><a href="https://www.torontomu.ca/canadian-conference-computational-geometry-2022/program/">Accepted papers to the 34th Canadian Conference on Computational Geometry</a> <span style="white-space:nowrap">(<a href="https://mathstodon.xyz/@11011110/108519879503501792">$$\mathbb{M}$$</a>),</span> to be held in August in Toronto. For mine, see recent blog posts on <a href="/blog/2022/06/22/dehn-rank-revisited.html">Dehn rank</a>, <a href="/blog/2022/06/24/reflections-octagonal-mirror.html">octagonal reflections</a>, and <a href="/blog/2022/06/28/motion-bend-lines.html">flattening paper surfaces</a>. In other news, from looking at the web site, I see that the host university, formerly Ryerson, has changed its name: now it’s Toronto Metropolitan University.</p> </li> <li> <p><a href="https://www.sligocki.com/2022/06/21/bb-6-2-t15.html">Tetration in busy-beaver Turing machines</a> and <a href="https://cp4space.hatsya.com/2022/06/23/tetrational-machines/">in Conway’s Game of Life</a> <span style="white-space:nowrap">(<a href="https://mathstodon.xyz/@11011110/108533292912244512">$$\mathbb{M}$$</a>,</span> <a href="https://btm.qva.mybluehost.me/telling-the-tale-of-two-tetrations/">see also</a>).</p> </li> <li> <p><a href="https://www.thisiscolossal.com/2022/06/paper-show-heron-arts/">Paper show</a> <span style="white-space:nowrap">(<a href="https://mathstodon.xyz/@11011110/108542378459285958">$$\mathbb{M}$$</a>):</span> Group art exhibit featuring 14 different artists who work in papercraft. Coming soon to a gallery in the Mission in San Francisco. Maybe the most geometric are the paper geodes of <a href="https://www.huntzliu.com/">Huntz Liu</a>.</p> </li> <li> <p>Two 3-regular penny graphs <span style="white-space:nowrap">(<a href="https://mathstodon.xyz/@11011110/108546890027577211">$$\mathbb{M}$$</a>).</span> You can’t make these avoid triangles altogether (see <a href="https://doi.org/10.7155/jgaa.00463">my paper on triangle-free penny graphs</a>) but the larger one of these two avoids pairs of triangles that share an edge. It’s more difficult than it seems like it should be to get all of the pennies that should be touching to actually touch, and I didn’t entirely succeed, but I think the pattern is clear.</p> <p style="text-align:center"><img src="https://www.ics.uci.edu/~eppstein/pix/3regpen/3regular-penny-m.jpg" alt="Two three-regular penny graphs, with vertices represented by pennies and edges represented by touching pennies" style="border-style:solid;border-color:black" /></p> </li> <li> <p><a href="https://en.wikipedia.org/wiki/Double_bubble_theorem">The double bubble theorem</a> <span style="white-space:nowrap">(<a href="https://mathstodon.xyz/@11011110/108551879168137037">$$\mathbb{M}$$</a>)</span> states that the minimum area enclosing two volumes looks like a double soap bubble, with three spherical patches meeting at $$120^\circ$$ angles on a circle. Just after getting Wikipedia’s article to Good Article, I learned of significant progress: Milman and Neeman announced a proof of the triple bubble conjecture in all dimensions and related results. See <a href="https://arxiv.org/abs/2205.09102">their preprint</a> or <a href="https://amathr.org/milman-and-neeman/">Frank Morgan’s review</a>.</p> </li> <li> <p><a href="https://www.sixthtone.com/news/1010653/she-spent-a-decade-writing-fake-russian-history.-wikipedia-just-noticed.-">An elaborate hoax history of medieval Russian history is uncovered on the Chinese-language Wikipedia</a> <span style="white-space:nowrap">(<a href="https://mathstodon.xyz/@11011110/108563040999391348">$$\mathbb{M}$$</a>,</span> <a href="https://news.ycombinator.com/item?id=31915937">via</a>, <a href="https://www.metafilter.com/195824/crumples-up-thesis-on-Kashin-silver-mine">mf</a>). Link goes to Chinese state media, but is in English; see also the <a href="https://en.wikipedia.org/wiki/Wikipedia:Fabricated_articles_and_hoaxes_of_Russia_in_2022">English Wikipedia internal report on the situation</a>.</p> </li> <li> <p><a href="http://www.ag.jku.at/geometrikum.shtml">Geometrikum</a> <span style="white-space:nowrap">(<a href="https://mathstodon.xyz/@11011110/108568817200153580">$$\mathbb{M}$$</a>):</span> geometric exhibits at the Institute for Applied Geometry of the University of Linz. Or, if you’re not near Linz, you can see them online. Includes string-art surfaces, Lego Stanford bunny and trefoil knot, some regular 4-polytope skeletons, papercraft polycubes and modular polyhedra, matchstick tetrastix, and more.</p> </li> </ul>David EppsteinNeil Sloane has a new blog ($$\mathbb{M}$$), subtitled “interesting sequences I need help with”. His first post concerns the two-up sequence, formed in steps where the kth step adds two numbers that are not already in the sequence and are relatively prime to the preceding k. Most of the terms appear to be the primes (in order). The remaining terms appear to be prime powers or semiprimes but this has not been proven.The motion of bend lines on smooth surfaces2022-06-28T16:25:00+00:002022-06-28T16:25:00+00:00https://11011110.github.io/blog/2022/06/28/motion-bend-lines<p>You may have played with a paper yoyo, a strip of paper wrapped around a stick so that when you flick it with your wrist, it extends outward into a long tube. Here’s one, <a href="https://commons.wikimedia.org/wiki/File:Arcade_Paper_Laser,_August_8th_2016.jpeg">the only example I could find on Wikimedia Commons</a>:</p> <p style="text-align:center"><img src="/blog/assets/2022/paper-yoyo.jpg" alt="Arcade Paper Laser, public domain image by Zhonghua88, 8 August 2016, from Wikimedia Commons" style="width:100%;max-width:720px" /></p> <p>If you’ve used one, then the way they work is simple and intuitive. But if you know a little bit about the <a href="https://en.wikipedia.org/wiki/Mathematics_of_paper_folding">mathematics of paper folding</a> and <a href="https://en.wikipedia.org/wiki/Developable_surface">developable surfaces</a>, and think about it, it starts to seem a little strange.</p> <p>Paper can bend or fold, but not stretch. When a flat sheet of paper is bent into a smooth but not flat surface (as it is in all states of the paper yoyo), it bends along straight “bend lines” that extend all the way across the surface, and that are forced to remain straight. You may be familiar with the way that a slice of pizza will droop if you try to hold it flat from its crust, but will remain straight if you bend it lengthwise: the bend lines are holding it in a rigid shape. Or, if you roll a poster into a cylinder, you can wave it around like a light saber, again making it much more rigid than the unrolled poster (at least until you hit it into something hard enough to crumple it).</p> <p>In the rolled-up state of the paper yoyo, it’s again rolled into a cylinder, with bend lines parallel to the cylinder axis. If we mark some of those bend lines and unroll it flat, it might look something like this:</p> <p style="text-align:center"><img src="/blog/assets/2022/paper-yoyo-rolled.svg" alt="Bend lines in a rolled paper yoyo" style="width:100%;max-width:720px" /></p> <p>But then, what happens when you extend it? The paper cannot stretch from its rectangular shape into a parallelogram. Instead, as it extends, the bend lines continuously rotate on the surface of the paper. Near the point where they attach to the stick, the non-perpendicular bend lines cause the paper to flare out a little from its cylindrical shape, but it is surrounded by more wrapped paper constraining it from flaring very widely, so the bend lines must still stay near their original orientations. Farther along the yoyo, the bend lines can twist to bigger angles:</p> <p style="text-align:center"><img src="/blog/assets/2022/paper-yoyo-extended.svg" alt="Bend lines in a rolled paper yoyo" style="width:100%;max-width:720px" /></p> <p>So although the bend lines of a smoothly bent sheet of paper are rigid along their length, they can slide and twist continuously within the sheet, as the sheet flexes continuously while remaining smooth. This twisting motion of bend lines is essential for understanding the seemingly-knotted surface below, two round disks connected by a thin band tied in an overhand knot:</p> <p style="text-align:center"><img src="/blog/assets/2022/knotted-dumbbell.svg" alt="Knotted paper dumbbell shape" /></p> <p>If you want to unknot it while keeping it smooth rather than crumpling it, you will obviously have to make the disks smaller by rolling them up somehow, perhaps into a cylinder around one of their diameters. But if you do that, you will put a bend line through or near the diameter, which will act like a rigid rod. And if you attach the knotted center band to two rigid rods, of length equal to the diameter of the disks, you definitely cannot untie it. The rods are too long to poke all the way through the knot before getting stuck.</p> <p>On the other hand, this surface can be untied while keeping it smooth! The trick is that, after you have rolled up one of the disks to make a rigid cylinder, with the tied band at one end, you can twist the roll, so that its bend lines rotate around the disk. As you do, the point where the band is tied will slide from one end of the cylinder to the other. And you can make this sliding motion coincide with the motion of the cylinder through a hole in the tied band, untying it.</p> <p>This and similar examples are the focus of my third new CCCG preprint, “Locked and unlocked smooth embeddings of surfaces”, <a href="https://arxiv.org/abs/2206.12989">arXiv:2206.12989</a>. In the phrasing of the title, the two disks with a tied center band, above, are unlocked: they can be reconfigured to a flat state while remaining smooth. My paper also shows that any compact shape with a continuous shrinking motion into itself, like the polygon below, is unlocked.</p> <p style="text-align:center"><img src="/blog/assets/2022/generalized-star.svg" alt="Polygon with a continuous shrinking motion" style="width:100%;max-width:600px" /></p> <p>On the other hand, there exist surfaces that, while topologically equivalent to their flat state, cannot be made flat through a continuous sequence of smooth embeddings. The simplest one that I found consists of a central square mat, rolled up into a cylinder and held into its cylindrical shape by two loops, entangled with each other in a pattern based on the Borromean rings:</p> <p style="text-align:center"><img src="/blog/assets/2022/tied-roll.svg" alt="Locked shape consisting of a rolled square mat with two entangled loops" /></p> <p>There is actually quite a lot of freedom in the pattern of bend lines of the central square. They can twist until the bend line through the center of the square gets halfway to the corners of the square:</p> <p style="text-align:center"><img src="/blog/assets/2022/bent-roll.svg" alt="Twisted bend lines in the rolled square mat" /></p> <p>But after that point, the entangled loops block the twisting from going any farther. They cannot be disentangled from each other without being pulled around the rigid central bend line, and they are too short for that to be possible, so the surface is locked.</p> <p>It’s not clear to me whether all simple polygons are unlocked, or whether it’s possible to make locked polygons without holes. I also don’t have any algorithmic results in this area; it seems likely that testing whether a surface is locked or unlocked is hard, but I don’t know how to prove it.</p> <p>(<a href="https://mathstodon.xyz/@11011110/108557742866716208">Discuss on Mastodon</a>)</p>David EppsteinYou may have played with a paper yoyo, a strip of paper wrapped around a stick so that when you flick it with your wrist, it extends outward into a long tube. Here’s one, the only example I could find on Wikimedia Commons:Reflections in an octagonal mirror maze2022-06-24T23:29:00+00:002022-06-24T23:29:00+00:00https://11011110.github.io/blog/2022/06/24/reflections-octagonal-mirror<p>The second preprint from my CCCG papers is now online. It is “Reflections in an octagonal mirror maze”, <a href="https://arxiv.org/abs/2206.11413">arXiv:2206.11413</a>. The title is quite literal: suppose you find yourself in a <a href="https://en.wikipedia.org/wiki/House_of_mirrors">mirror maze</a>, where the mirrors are aligned with the sides of an octagon, and have integer coordinates (meaning that, on a floorplan of the maze, the mirrors become line segments between points of an integer grid). What would you see if you looked in any given direction? It might be many reflections eventually leading to the back of your own head, to the exit, or some other non-reflective part of the maze. <a href="https://commons.wikimedia.org/wiki/File:Mirror_Maze_in_the_Museum_of_Science%2BIndustry_of_Chicago.jpg">The example below, from the Museum of Science &amp; Industry in Chicago</a>, is hexagonal rather than octagonal, but otherwise has much the same effect:</p> <p style="text-align:center"><img src="/blog/assets/2022/Chicago-mirror-maze.jpg" alt="Mirror Maze in the Museum of Science+Industry of Chicago, CC-BY-SA image by Kevdog686, 27 May 2019, from Wikimedia Commons" /></p> <p>In this example, it looks like you might be seeing something else, not the exit or your own head: <a href="https://en.wikipedia.org/wiki/Infinity_mirror">a corridor of infinitely many reflections</a>. But when looking in a direction of rational slope, and without tricks involving <a href="https://en.wikipedia.org/wiki/One-way_mirror">one-way mirrors</a> (my guess at what’s happening in the photo), that’s not actually possible. Rationality implies that there are only finitely many possible points where your view could hit a mirror: in the photo above, all the reflections down the corridor look like they are actually on the center lines of the mirrors. Because there are only finitely many of these points (in reality, if not in what you see), such a view would have to eventually repeat, in a finite cycle. And finite cycles like that are definitely possible, but you can’t see them, because putting your eye onto them blocks the view.</p> <p>Even if you can’t see a line with infinitely many reflections, you can see a very large number of them. The number of reflections you see between your eye and whatever opaque thing eventually blocks the view cannot be bounded by any function of the number of mirrors. If you take a diagonal view into a long hallway with its two sides mirrored, the number of reflections will be proportional to the length of the hall, even though there are only two mirrors. And with larger numbers of mirrors, more complex patterns of reflection can occur.</p> <p style="text-align:center"><img src="/blog/assets/2022/8reflex.svg" alt="Complex patterns of reflection in an octagonal mirror maze" /></p> <p>Because the number of reflections can be large, finding the eventual fate of a reflected light path by directly simulating the sequence of reflections any ray might take could be very slow. Instead, my new paper shows that it can be done in polynomial time! More precisely the time is polynomial in the number of mirrors and in the number of bits needed to specify a mirror coordinates or viewpoint direction using binary numbers.</p> <p>Although the problem is simple to state, the solution uses sophisticated methods from computational topology. It’s an application of <a href="/blog/2022/01/30/fast-iterated-exchange.html">an idea from an earlier post</a>, originally by Mark Bell, and included in an update to my earlier preprint “<a href="https://arxiv.org/abs/2112.11607">The Complexity of Iterated Reversible Computation</a>”. This idea concerns <em>iterated integer interval exchange transformations</em>: if you have a function on an interval of integers that acts by permuting subintervals, then how easy is it to compute the result of applying the same function many times? As Bell observed, this computation could be transformed into an equivalent problems of tracing normal curves on a topological surface, which had been solved in earlier work by Erickson and Nayyeri. The new paper generalizes these integer transformations to partial functions, extends the same method to compute iterated values of these partial functions, and shows that these partial functions can be used to model the way the mirrors in a mirror maze permute sightlines.</p> <p>(<a href="https://mathstodon.xyz/@11011110/108536811841948097">Discuss on Mastodon</a>)</p>David EppsteinThe second preprint from my CCCG papers is now online. It is “Reflections in an octagonal mirror maze”, arXiv:2206.11413. The title is quite literal: suppose you find yourself in a mirror maze, where the mirrors are aligned with the sides of an octagon, and have integer coordinates (meaning that, on a floorplan of the maze, the mirrors become line segments between points of an integer grid). What would you see if you looked in any given direction? It might be many reflections eventually leading to the back of your own head, to the exit, or some other non-reflective part of the maze. The example below, from the Museum of Science &amp; Industry in Chicago, is hexagonal rather than octagonal, but otherwise has much the same effect:Dehn rank revisited2022-06-22T18:22:00+00:002022-06-22T18:22:00+00:00https://11011110.github.io/blog/2022/06/22/dehn-rank-revisited<p>In <a href="/blog/2022/04/03/dissection-into-rectangles.html">a recent post</a>, I discussed dissection of orthogonal polygons into each other by axis-parallel cuts, translation, and gluing. Each polygon has a value associated with it, called its <a href="https://en.wikipedia.org/wiki/Dehn_invariant">Dehn invariant</a>, that cannot be changed by dissection, so two polygons that can be dissected into each other must have equal invariants. And for past usage of Dehn invariants, that was pretty much all we looked at: are they equal or not? But my post pointed out that these invariants actually have a lot of structure (you can think of them as matrices, after an arbitrary choice of basis) and this structure is geometrically meaningful. Matrices (or tensors) have a rank, and the rank of the Dehn invariant is a lower bound on the number of rectangles into which a polygon can be dissected. This in turn has implications on the ability of a polygon or its dissections to tile the plane.</p> <p>Now it’s a paper: “Orthogonal dissection into few rectangles”, <a href="https://arxiv.org/abs/2206.10675">arXiv:2206.10675</a>, to appear at CCCG. The results of the paper are stronger: instead of just using the Dehn rank as a lower bound, I proved that it always equals the minimum number of rectangles into which a given polygon can be dissected, and can be used to compute this number of rectangles efficiently. The two main steps of the proof are:</p> <ul> <li> <p>constructing a set of the right number of rectangles for any given tensor, by a direct geometric construction that turns an algebraic realization of the rank (a combination of positive and negative rectangles) into a geometric representation without the negativity, and</p> <p style="text-align:center"><img src="/blog/assets/2022/dehn-realizability.svg" alt="Construction of a set of rectangles realizing a given tensor as their Dehn invariant" style="width:100%;max-width:540px" /></p> </li> <li> <p>finding a dissection for any two polygons with equal invariants, by induction on the dimensions of the matrix describing their invariants.</p> <p style="text-align:center"><img src="/blog/assets/2022/dehn-dissectability.svg" alt="Induction step for proving the existence of a dissection between polygons with equal Dehn invariants" style="width:100%;max-width:540px" /></p> </li> </ul> <p>The images above are taken from the illustrations for the proofs of these two results from the paper, and maybe will provide a little insight into how they might be proven. For details, see the paper. Instead, here, I wanted to highlight a different, related problem, that I wasn’t able to prove as much about.</p> <p>Mostly when mathematicians talk about Dehn invariants, it’s for 3d dissections: cutting polyhedra up along planes, translating and rotating the pieces, and gluing them back together. The 3d Dehn invariant combines lengths and angles of polyhedron edges in the same way that the 2d invariant combines widths and heights of rectangles. Therefore, 3d Dehn rank is a lower bound on the number of edges you can dissect something into. It’s easy to construct polyhedra with arbitrarily large rank, and these polyhedra are forced to have arbitrarily large numbers of edges, no matter how you try to cut them up and reassemble them.</p> <p>However, the 3d Dehn rank is not exactly equal to the minimum number of edges after dissection. A cube has Dehn invariant zero (with rank zero), but the minimum number of edges of a polyhedron you can dissect it into is four, for a <a href="https://www.jstor.org/stable/2689983">space-filling tetrahedron</a> of the same volume. A regular tetrahedron is not space-filling, has a Dehn invariant of rank one, and already has the minimum number of edges among anything you can dissect it into. In both cases the rank is unequal to the minimum number of edges. Also, the rank is different in these cases but the minimum number of edges is the same.</p> <p>Nevertheless, I was hoping that the rank of the Dehn invariant would be usable as a constant-factor approximation to the minimum number of edges after dissection. To prove this, I’d need to find a polyhedron, having the same Dehn invariant as any given polyhedron, but with a number of edges proportional to the rank. The existence of a dissection would then follow from known results. But finding this few-edge polyhedron would have to use some knowledge of the starting polyhedron (unlike my set-of-rectangles construction), because not all tensors are realizable as Dehn invariants. So far, I haven’t been able to find any construction of these few-edge polyhedra.</p> <p>So: does every polyhedron, of Dehn rank $$r$$, have an equivalent polyhedron with $$O(r)$$ edges? Or are there some polyhedra that cannot be dissected into another polyhedron with few edges, even though their Dehn invariants have low rank?</p> <p>(<a href="https://mathstodon.xyz/@11011110/108524203085266737">Discuss on Mastodon</a>)</p>David EppsteinIn a recent post, I discussed dissection of orthogonal polygons into each other by axis-parallel cuts, translation, and gluing. Each polygon has a value associated with it, called its Dehn invariant, that cannot be changed by dissection, so two polygons that can be dissected into each other must have equal invariants. And for past usage of Dehn invariants, that was pretty much all we looked at: are they equal or not? But my post pointed out that these invariants actually have a lot of structure (you can think of them as matrices, after an arbitrary choice of basis) and this structure is geometrically meaningful. Matrices (or tensors) have a rank, and the rank of the Dehn invariant is a lower bound on the number of rectangles into which a polygon can be dissected. This in turn has implications on the ability of a polygon or its dissections to tile the plane.The shapes of triangular pencils2022-06-18T16:42:00+00:002022-06-18T16:42:00+00:00https://11011110.github.io/blog/2022/06/18/shapes-triangular-pencils<p><a href="https://ima.org.uk/19140/westward-ho-conway-concorde-and-curiously-curved-coins/">The Institute of Mathematics &amp; its Applications tells us</a> that applications of the <a href="https://en.wikipedia.org/wiki/Reuleaux_triangle">Reuleaux triangle</a> include “the cross-section of some pencils that are thought to be more ergonomic than traditional hexagonal ones”. It it true?</p> <p>It’s a bit difficult to tell, because there are many brands of triangular pencil and they don’t all come with easy-to-interpret cross-sections or dimensions. A true Reuleaux triangle pencil would have a cross section with rounded sides and 120° corners, making the same angle as the corners of a hexagonal pencil, as shown in the upper left cross-section of the figure below.</p> <p style="text-align:center"><img src="/blog/assets/2022/tri-pencils.svg" alt="Hypothetical cross-sections of triangular and hexagonal pencils: upper left, Reuleaux triangle; upper right, hexagon; lower left, flat triangle with rounded corners; lower-right, smooth three-lobed curve of constant width" /></p> <p>Instead, all the ones I found online appear to fall into two other classes of shapes: flat-sided triangles with rounded-off corners (lower left of the figure) or smooth three-lobed shapes with no corners at all (lower right). Here are some:</p> <ul> <li> <p>The best image I could find of the Dixon Tri-Conderoga is <a href="https://woodclinched.com/2010/09/15/tri-conderoga/">this brief post by Woodclinched</a>. It appears to be a smooth three-lobed shape. Dixon also produce a “Tri-Write” triangular pencil, aimed at a lower-level market than the Tri-Conderoga; the <a href="https://dixonticonderogacompany.com/wp-content/uploads/2021/05/2021_Dixon-Prod-Catalog_04.21.pdf">2021 Dixon product catalog</a> shows both it and the Tri-Conderoga as having a cross-section with slightly rounded sides (flatter than a Reuleaux triangle but not completely flat) and rounded corners.</p> </li> <li> <p>The Faber-Castell Grip 2001 is another three-sided pencil, with added bumps on its sides to make it grippier. If we ignore the bumps for the purpose of determining its shape, <a href="https://www.penciltalk.org/2008/10/faber-castell-grip-2001-and-jumbo-grip-in-alternate-finishes">Pencil Talk shows it as a flat triangle with rounded corners</a>.</p> </li> <li> <p>The Koh-I-Noor Triograph, as shown in <a href="https://www.penciltalk.org/2007/09/koh-i-noor-hardmuth-triograph-1830-pencil">a review by Pencil Talk</a>, has rounded corners, and flat-enough sides to clearly print the brand name in large letters on it.</p> </li> <li> <p>The Lyra Ferby is a German children’s brand with a smooth three-lobed shape.</p> </li> <li> <p>The Marco Grip-Rite appears, from <a href="https://polarpencilpusher.home.blog/2019/07/11/pencil-review-marco-grip-rite-2b-w-bonus-sharpener/">its review by Polar Pencil Pusher</a>, to have a smooth three-lobed cross-section, even though the review calls it a Reuleaux triangle.</p> </li> <li> <p>The <a href="https://micador.com.au/products/micador-jr-colourush-jumbo-pencils-fsc-100">Micador ColouRush Jumbo Triangular Pencils</a> (an Asutralian branch of children’s pencils) have a cross-section image on the manufacturer website clearly showing a flat triangle with rounded corners.</p> </li> <li> <p>PaperMate make several triangular writing tools under the “Handwriting” brand, including PaperMate Handwriting woodcase pencils, aimed at children. I couldn’t find a good image of a cross-section of one, but their mechanical pencil in the same branch is clearly a smooth three-lobed shape. <a href="https://www.pencilrevolution.com/2019/03/papermate-handwriting-pencil/">Pencil Revolution</a> appears to show a similar shape for the wood pencils.</p> </li> <li> <p>I don’t know what brand of pencils the IMA used for the photo of their example, but it appears to be one with flat sides and rounded corners.</p> </li> </ul> <p>I also didn’t find pencils with flat-sided triangles (and non-rounded or only very lightly rounded corners) as their cross-sections. But if you search for “triangle pencil” you will find plenty of colorful plastic things with that shape that you can slide over your pencil to make it easier to hold.</p> <p>I can’t tell you whether any of these are any good to write or draw with; for that, I prefer fountain pens. (I don’t have an <a href="https://www.fountainpennetwork.com/forum/topic/229621-modern-pens-the-omas-360/">Omas 360</a>, but that one really does appear to be based on a Reuleaux triangle.) Beyond writing, if you’re hoping for a pencil that doubles as a constant-width roller for rolling your books smoothly across your desk, and don’t want boring circular pencils, the smooth three-lobed ones may be your best bet. The one I drew above really is a curve of constant width. It’s just not a Reuleaux triangle.</p> <p>This has been yet another episode of “things that are not Reuleaux triangles”. See previously <a href="/blog/2020/11/27/study-triangular-bottle.html">a triangular bottle</a>, <a href="/blog/2020/11/15/linkage.html">the Kölntriangle</a>, <a href="/blog/2020/11/02/constant-width-involutes.html">rotors within rotors</a>, <a href="/blog/2020/08/30/linkage.html">a Polish football logo</a>, <a href="/blog/2020/07/05/shape-wankel-rotor.html">Wankel engines, and many more</a>.</p> <p>(<a href="https://mathstodon.xyz/@11011110/108501435837486329">Discuss on Mastodon</a>)</p>David EppsteinThe Institute of Mathematics &amp; its Applications tells us that applications of the Reuleaux triangle include “the cross-section of some pencils that are thought to be more ergonomic than traditional hexagonal ones”. It it true?Linkage2022-06-15T16:36:00+00:002022-06-15T16:36:00+00:00https://11011110.github.io/blog/2022/06/15/linkage<ul> <li> <p><a href="https://thegradient.pub/working-on-the-weekends-an-academic-necessity/">Claas Voelcker on academic work-life balance</a> <span style="white-space:nowrap">(<a href="https://mathstodon.xyz/@11011110/108406720822630002">$$\mathbb{M}$$</a>,</span> <a href="https://news.ycombinator.com/item?id=31562264">via</a>). I think we all know that many academics (myself included!) struggle to keep our weekend and evening time free of work-related distractions. Voelcker investigates where this pressure to work comes from (often internally) and suggests that overwork may block creativity; taking time off can make you more productive.</p> </li> <li> <p>From a face-up deck of cards, repeatedly deal off the number of cards showing on the top card (counting Jacks as 11, etc.). What’s the probability that you empty the deck by dealing out exactly the right number of cards in the last step? <a href="https://mathstodon.xyz/@christianp/108401164015456979">Christian Lawson-Perfect’s post</a> inspired a group discussion leading to the result that, for decks with large numbers of suits, the answer should tend to 1/7, and that for a standard 52-card deck it is approximately 0.1420342593977892.</p> </li> <li> <p>Another Wikipedia illustration <span style="white-space:nowrap">(<a href="https://mathstodon.xyz/@11011110/108415828596664345">$$\mathbb{M}$$</a>):</span> empty regions for the <a href="https://en.wikipedia.org/wiki/Euclidean_minimum_spanning_tree">Euclidean minimum spanning tree</a>. If the red vertical segment is to be an MST edge, the outer white lens needs to be empty of other points; this emptiness implies that the edge is part of the relative neighborhood graph. The emptiness of the light blue diameter circle inside the lens defines the Gabriel graph in the same way. The inner rhombus must not only be empty, but disjoint from the rhombi of other edges.</p> <p style="text-align:center"><img src="/blog/assets/2022/EMST-empty-regions.svg" alt="Empty regions for the Euclidean minimum spanning tree" style="width:100%;max-width:720px" /></p> </li> <li> <p><a href="https://somethingorotherwhatever.com/shunting-yard-animation/">Shunting yard animation</a> <span style="white-space:nowrap">(<a href="https://mathstodon.xyz/@christianp/108424166828836198">$$\mathbb{M}$$</a>).</span> Cutesy train animation of the <a href="https://en.wikipedia.org/wiki/Shunting_yard_algorithm">shunting yard algorithm</a> for parsing infix expressions.</p> </li> <li> <p><a href="https://beachpackagingdesign.com/boxvox/pseudo-cylindrical-concave-polyhedral-packaging">Pseudo-cylindrical concave polyhedral packaging</a> <span style="white-space:nowrap">(<a href="MLIhttps://mathstodon.xyz/@11011110/108434883781055479NK">$$\mathbb{M}$$</a>).</span> This post describes multiple examples of the <a href="https://en.wikipedia.org/wiki/Yoshimura_buckling">Yoshimura buckling pattern</a> or <a href="https://en.wikipedia.org/wiki/Schwarz_lantern">Schwarz lantern</a> in food/drink packaging, not in the obvious way (it happens when you crumple a can end-on) but deliberately by the manufacturer. It doesn’t say why they did, though. Maybe because it looks cool.</p> </li> <li> <p><a href="https://cp4space.hatsya.com/2022/05/25/threelds/">Threelds</a> <span style="white-space:nowrap">(<a href="https://mathstodon.xyz/@11011110/108440193859317506">$$\mathbb{M}$$</a>).</span> I have no idea whether it’s useful for anything, but a threeld is a pair of fields where the multiplication operation on the inner one forms the addition on the outer one. The finite ones have inner order 3 and outer order 2, or inner order a Mersenne prime and outer order the adjacent power of two, but there also exist infinite ones with inner field of characteristic 0 and outer of characteristic 2.</p> </li> <li> <p>Roundup of recent <em>Quanta</em> popularizations and the research they come from <span style="white-space:nowrap">(<a href="https://mathstodon.xyz/@11011110/108442789631056740">$$\mathbb{M}$$</a>):</span></p> <ul> <li> <p><a href="https://www.quantamagazine.org/impossible-seeming-surfaces-confirmed-decades-after-conjecture-20220602/">Near-optimal expansion for 2d surfaces</a>, based on “<a href="https://arxiv.org/abs/2107.05292">Near optimal spectral gaps for hyperbolic surfaces</a>”, by Will Hide and Michael Magee.</p> </li> <li> <p><a href="https://www.quantamagazine.org/mathematicians-transcend-geometric-theory-of-motion-20211209/">Inequality between cohomology rank and number of Hamiltonian flow orbits</a>, based on “<a href="https://arxiv.org/abs/2103.01507">Arnold conjecture and Morava K-theory</a>”, by Mohammed Abouzaid and Andrew J. Blumberg.</p> </li> <li> <p><a href="https://www.quantamagazine.org/graduate-students-side-project-proves-prime-number-conjecture-20220606/">Among pairwise-coprime sequences, primes maximize $$\sum 1/n_i\log n_i$$</a>, based on “<a href="https://arxiv.org/abs/2202.02384">A proof of the Erdős primitive set conjecture</a>”, by Jared Duker Lichtman.</p> </li> <li> <p><a href="https://www.quantamagazine.org/researchers-achieve-absurdly-fast-algorithm-for-network-flow-20220608/">Fast maximum flow algorithms</a>, based on “<a href="https://arxiv.org/abs/2203.00671">Maximum flow and minimum-cost flow in almost-linear time</a>”, by Li Chen, Rasmus Kyng, Yang P. Liu, Richard Peng, Maximilian Probst Gutenberg, and Sushant Sachdeva.</p> </li> </ul> </li> <li> <p>Kirk Smith asks Mastodon: “<a href="https://scholar.social/@kirk/108439384108355311">Do people on here edit/write Wikipedia articles related to your field? and what’s your experience/motivation?</a>” There’s a conflict here between desiring academic credit for your work, and maintaining the protection of pseudonymity. But the real-world harassment that pseudonymity prevents is real, whereas I think the possibility of getting much academic credit for this sort of work is largely illusory.</p> </li> <li> <p>New Wikipedia article: <a href="https://en.wikipedia.org/wiki/Staircase_paradox">Staircase paradox</a> <span style="white-space:nowrap">(<a href="https://mathstodon.xyz/@11011110/108454961367151468">$$\mathbb{M}$$</a>),</span> on the familiar example of staircase curves in a unit square that uniformly converge to the diagonal of the square, while their lengths converge to the wrong number ($$2$$, rather than $$\sqrt2$$). Somehow we don’t seem to have already had an article on this example. I’m sure there must be many more published sources on this example than the ones I used; if you think I missed something important, please let me know.</p> </li> <li> <p><a href="https://www.math.columbia.edu/~woit/wordpress/?p=12936">Physicists discover never-before seen particle sitting on a tabletop</a> <span style="white-space:nowrap">(<a href="https://mathstodon.xyz/@11011110/108465478766177772">$$\mathbb{M}$$</a>).</span> Peter Woit’s headline for this <em>Not Even Wrong</em> post repeats the breathless hype from the churnalism on a new condensed-matter-physics preprint, which promises applications to dark matter and quantum computing and turns out to be much much less. From the comments, it seems that the condensed matter physicists have been guilty of misapplying the tag “Higgs field” for a long time.</p> </li> <li> <p><a href="https://blogs.ams.org/beyondreviews/2021/07/18/yoshimura-crush-patterns/">Yoshimura crushing patterns on the <em>Inside MathSciNet</em> blog</a> <span style="white-space:nowrap">(<a href="https://mathstodon.xyz/@11011110/108469088531022175">$$\mathbb{M}$$</a>).</span> I don’t think it’s accurate to say that a crush pattern and a crease pattern are synonyms, though. One is a description of the output of a crushing process; the other is an input to a folding process that guides you to put the folds into their intended places. The similarity of the crushing pattern and the Yoshimura fold is not coincidental but the purpose is different.</p> </li> <li> <p><a href="https://doi.org/10.1016/j.jclinepi.2022.05.019">Study of all open-access papers on BioMed Central from a month-long window</a> <span style="white-space:nowrap">(<a href="https://mathstodon.xyz/@11011110/108474483987356656">$$\mathbb{M}$$</a>,</span> <a href="https://news.ycombinator.com/item?id=31660239">via</a>) finds that although 42% claim their data to be available on reasonable request, only 7% actually responded and provided their data.</p> </li> <li> <p><a href="https://b-mehta.github.io/unit-fractions/">Formalization in Lean of Thomas Bloom’s proof</a> of the density version of the <a href="https://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Graham_problem">Erdős–Graham problem</a> <span style="white-space:nowrap">(<a href="https://mathstodon.xyz/@11011110/108484103209861017">$$\mathbb{M}$$</a>,</span> <a href="https://twitter.com/XenaProject/status/1536099892694859777">via</a>), according to which every set of integers with positive upper density includes the denominators of an Egyptian fraction representation of one. The blueprint appears to show Theorem 2 of <a href="https://arxiv.org/abs/2112.03726">Bloom’s preprint</a> as verified, but Theorem 3 (log density) still to go.</p> </li> </ul>David EppsteinClaas Voelcker on academic work-life balance ($$\mathbb{M}$$, via). I think we all know that many academics (myself included!) struggle to keep our weekend and evening time free of work-related distractions. Voelcker investigates where this pressure to work comes from (often internally) and suggests that overwork may block creativity; taking time off can make you more productive.The analyst’s minimum spanning tree2022-06-14T08:47:00+00:002022-06-14T08:47:00+00:00https://11011110.github.io/blog/2022/06/14/analysts-minimum-spanning<p>Infinite sets of points in the Euclidean plane, even discrete sets, do not always have <a href="https://en.wikipedia.org/wiki/Euclidean_minimum_spanning_tree">Euclidean minimum spanning trees</a>. For instance, consider the points with coordinates</p> $\left(i, \pm\left(1+\frac1i\right)\right),$ <p>for positive <span style="white-space:nowrap">integers $$i$$.</span> You can connect the <span style="white-space:nowrap">positive-$$y$$</span> points and the <span style="white-space:nowrap">negative-$$y$$</span> points into two chains with edges of length less than two, but then you have to pick one edge of length greater than two to span from one chain to the other. Whichever edge you choose, the next edge along would always be a better choice. So a tree that minimizes the multiset of its edge weights (as finite minimum spanning trees do) does not exist for this example. And as the same example shows, the sum of edge weights may be infinite, so how can we use minimization of this sum to define a tree?</p> <p style="text-align:center"><img src="/blog/assets/2022/ladder-no-mst.svg" alt="Discrete infinite set of points with no Euclidean minimum spanning tree" /></p> <p>Despite that, here’s a construction that works for any <a href="https://en.wikipedia.org/wiki/Compact_space">compact set</a>, even one with infinitely many components, and that generalizes easily to higher-dimensional Euclidean spaces. I think it deserves to be called the Euclidean minimum spanning tree. Given a compact <span style="white-space:nowrap">set $$C$$,</span> consider every partition $$C=A\cup (C\setminus A)$$ of $$C$$ into two disjoint nonempty compact subsets. For each such partition, find a line segment $$s_A$$ of minimum length with endpoints in $$A$$ <span style="white-space:nowrap">and $$C\setminus A$$,</span> breaking ties lexicographically by coordinates. By the assumed compactness of $$A$$ <span style="white-space:nowrap">and $$C\setminus A$$,</span> such a line segment exists. Let $$T_C$$ be the union of $$C$$ itself and of all line segments obtained in this way. For example, the union of a triangle, square, and circle shown below has three partitions into two nonempty compact subsets, separating one of these three shapes from the other two. Two of these partitions choose the diagonal pink segment as their shortest connection, and the third partition chooses the horizontal pink segment. So in this case, $$T_C$$ consists of the three blue given shapes and two pink segments.</p> <p style="text-align:center"><img src="/blog/assets/2022/trisquircle.svg" alt="Minimum spanning tree of a circle, square, and triangle" /></p> <p>When $$C$$ is a finite point set, $$T_C$$ is just a Euclidean minimum spanning tree. When $$C$$ has finitely many connected components, like the example above, $$T_C$$ is again a minimum spanning tree, for the component-component distances. In the general case, $$T_C$$ still has many of the familiar properties of Euclidean minimum spanning trees:</p> <ul> <li> <p>It consists of the input and a collection of line segments connecting pairs of input points, by construction.</p> </li> <li> <p>It is a <a href="https://en.wikipedia.org/wiki/Connected_space">connected set</a>. Topologically, this means that it cannot be covered by two disjoint open sets that both have a nonempty intersection with it. (This is different from being path-connected, a stronger property.) Any nontrivial open disjoint cover of $$C$$ would be spanned by a line segment from one set to the other, and no new disjoint covers can separate these line segments from their endpoints.</p> </li> <li> <p>For any added <span style="white-space:nowrap">segment $$s_A$$,</span> the intersection of two disks with that segment as radius (a “lune”) has no point of $$C$$ in its interior. Any interior point would form one end of a shorter connecting segment between $$A$$ <span style="white-space:nowrap">and $$C\setminus A$$,</span> with the other end at an endpoint <span style="white-space:nowrap">of $$s_A$$.</span> No two added segments can cross without violating the empty lune property.</p> <p style="text-align:center"><img src="/blog/assets/2022/vesica.svg" alt="The empty lune of an edge" /></p> </li> <li> <p>For any added <span style="white-space:nowrap">segment $$s_A$$,</span> the open rhombus with angles $$60^\circ$$ and $$120^\circ$$ having $$s_A$$ as its long diagonal is disjoint from the rhombi formed in the same way from the other segments. Any two overlapping rhombi would allow the longer of the two segments they come from to be replaced by a shorter segment crossing the same compact partition, on a three-segment path connecting its endpoints via the other segment endpoints. Because these non-overlapping rhombi cover a region of bounded area, the squared segment lengths have a bounded sum, and only finitely many segments can be longer than any given length threshold.</p> <p style="text-align:center"><img src="/blog/assets/2022/ivy-rhombs.svg" alt="An infinite minimum spanning tree and its empty rhombi" /></p> </li> <li> <p>The union of $$C$$ with any subset of added segments is compact. If $$p$$ is a limit point of a <span style="white-space:nowrap">sequence $$\sigma_i$$</span> of points in this union, it must either lie in the empty rhombus of a segment (in which case it can only be a point of the same segment), or it is a limit point of a sequence of points <span style="white-space:nowrap">in $$C$$,</span> obtained by replacing each point in $$\sigma_i$$ that is interior to a segment by the nearest segment endpoint. This replacement only increases the distance from the replaced point to $$p$$ by a constant factor, which does not affect convergence. By compactness the replaced sequence converges to a point <span style="white-space:nowrap">in $$C$$.</span></p> </li> <li> <p>For <span style="white-space:nowrap">any $$i$$,</span> the set $$T_i$$ of the largest $$i$$ added segments (with the same tie-breaking order) are edges of a minimum spanning tree for a family of $$i-1$$ sets. To construct these sets, find the components of the union of $$C$$ with all shorter segments, and intersect each component <span style="white-space:nowrap">with $$C$$.</span> None of these components can cross between $$A$$ and $$C\setminus A$$ for any <span style="white-space:nowrap">edge $$s_A\in T_i$$.</span> Because adding $$T_i$$ connects all these components, there can be at most $$i-1$$ components. Each edge in $$T_i$$ is shortest (with a consistent tie-breaking rule) across some partition of the components, one of the ways of determining the edges in a finite minimum spanning tree. In particular, $$T_C$$ is minimally connected: removing any edge $$s_A\in S_i$$ separates some of the components from each other.</p> </li> <li> <p>$$T_C$$ has the minimum sum of squared edge lengths of all collections of line segments between points of $$C$$ that <span style="white-space:nowrap">connect $$C$$.</span> To see this, consider any other connecting set $$X$$ of line segments with a finite sum of squared edge lengths. Truncate the sorted sequence of edges of $$T_C$$ to a finite initial <span style="white-space:nowrap">sequence $$T_i$$</span> such that the rest of the sequence has negligible sum of squares. Because $$T_i$$ is a minimum spanning tree of its components, and $$X$$ connects those same components (perhaps redundantly), the sequence of edge lengths in $$T_i$$ is, step for step, less than or equal to the sorted sequence of lengths <span style="white-space:nowrap">in $$X$$.</span></p> </li> </ul> <p>There may exist other sets of line segments that connect $$C$$ with the same sum of squared edge lengths but they all are minimally connected, with the same sequence of edge lengths, the same empty lune and empty rhombus properties, and the same property that their initial sequences form finite minimum spanning trees of their components.</p> <p>(<a href="https://mathstodon.xyz/@11011110/108476695929961897">Discuss on Mastodon</a>)</p>David EppsteinInfinite sets of points in the Euclidean plane, even discrete sets, do not always have Euclidean minimum spanning trees. For instance, consider the points with coordinatesMaybe powers of π don’t have unexpectedly good approximations?2022-06-04T21:03:00+00:002022-06-04T21:03:00+00:00https://11011110.github.io/blog/2022/06/04/maybe-powers-pi<p>After <a href="/blog/2022/05/31/linkage.html">I wrote recently</a> about Ramanujan’s approximation $$\pi^4\approx 2143/22$$, writing “why do powers of $$\pi$$ seem to have unusually good rational approximations?”, Timothy Chow emailed to challenge my assumption, asking what evidence I had that their approximations were unusually good. So that led me to do a little statistical experiment to test that hypothesis, and the experiment showed…that the approximations seem to be about as good as we would expect, no more, no less. Not unusually good. Chow was correct, and my earlier statement was overstated. So if Ramanujan’s approximation is not just random fluctuation (which for all I know it could be), it at least does not seem to be part of a pattern of many good rational approximations for small powers of $$\pi$$. Below are some details of how I came to this conclusion.</p> <p>First off, one gets good rational approximations by truncating the <a href="https://en.wikipedia.org/wiki/Continued_fraction">continued fraction representation</a> of a number. for instance, Ramanujan’s approximation can be obtained from the continued fraction</p> $\pi^4=97 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{3 + \cfrac{1}{1+\cfrac{1}{16539+\cdots}}}}}$ <p>by stopping just before the big number $$16539$$. You want to stop at those points, roughly speaking, because the accuracy of your approximation (as measured for instance by how many decimal digits you get correct) comes from combining all of the terms up to and including the big one you stopped before, but the complexity of your approximation (how many digits it takes to write it down) comes from only the terms before you stopped. When you stop just before a big term, you get a lot of accuracy for free. And so the question I was really asking was “why do the powers of $$\pi$$ have unusually big terms in their continued fractions?” But the question I should have asked first was “do the powers of $$\pi$$ have unusually big terms in their continued fractions?”</p> <p>For a random number, the terms of the continued fraction will be distributed according to the <a href="https://en.wikipedia.org/wiki/Gauss%E2%80%93Kuzmin_distribution">Gauss–Kuzmin distribution</a>, according to which the probability of seeing a term with value exactly $$k$$ is</p> $- \log_2 \left( 1 - \frac{1}{(1+k)^2}\right).$ <p>This is a <a href="https://en.wikipedia.org/wiki/Heavy-tailed_distribution">heavy-tailed distribution</a>, so one should expect to see some large numbers from time to time. What I wanted to test was a <a href="https://en.wikipedia.org/wiki/Null_hypothesis">null hypothesis</a> that the continued fraction for powers of $$\pi$$ are distributed in this way. If the tests revealed a low likelihood of this being true, it would suggest that they have some other distribution, confirming what I wrote in my earlier post. If they didn’t, it still might mean that there was something unusual about those terms, but it would have to be more subtle, subtle enough to be undetected by the statistical tests I used.</p> <p>This is exactly the sort of thing <a href="https://en.wikipedia.org/wiki/Pearson%27s_chi-squared_test">Pearson’s $$\chi^2$$ test</a> is good for. So I set up a $$\chi^2$$ test in R comparing the observed data (the frequencies of each value in the continued fraction terms for $$\pi$$, $$\pi^2$$, $$\pi^3$$, and $$\pi^4$$, as taken from OEIS) with the Gauss–Kuzmin distribution. I omitted the initial term from each continued fraction, because those are the integer parts of the powers of $$\pi$$, which are both nonrandom and uninteresting for approximation purposes. I also grouped the terms of the continued fractions into six exponentially-growing buckets, by mapping each term $$t_i$$ to $$\min\bigl(5,\lfloor\log_2 t_i\rfloor\bigr)$$, because the $$\chi^2$$ test works better when it has only a small number of well-populated frequency counts to compare, rather than $$16539$$ of them, many empty or with only one representative. This meant also computing the Gauss–Kuzmin probabilities for the same buckets. These buckets still don’t have uniform probabilities but I thought it better to stick to my first choice of bucketing rather than to repeatedly adjust the parameters of the test. Even with this bucketing the $$\chi^2$$ test still wasn’t strong enough to work well on the OEIS data taken separately for each individual power of $$\pi$$. Instead I had to concatenate the data for the four powers I had chosen into a single test to get it to work without R complaining.</p> <p>Because I am more fluent in Python than R, I did this as <a href="/blog/assets/2022/pipower.py">a Python script</a> to generate a very short R script, but it would have been as easy to do the whole thing directly in R. The results: a <a href="https://en.wikipedia.org/wiki/P-value">$$p$$-value</a> somewhat greater than $$\tfrac12$$ (not small). So the null hypothesis was <em>not</em> rejected and I do not have evidence that the powers of $$\pi$$ have unusually big terms in their continued fractions.</p> <p>(<a href="https://mathstodon.xyz/@11011110/108423157545534693">Discuss on Mastodon</a>)</p>David EppsteinAfter I wrote recently about Ramanujan’s approximation $$\pi^4\approx 2143/22$$, writing “why do powers of $$\pi$$ seem to have unusually good rational approximations?”, Timothy Chow emailed to challenge my assumption, asking what evidence I had that their approximations were unusually good. So that led me to do a little statistical experiment to test that hypothesis, and the experiment showed…that the approximations seem to be about as good as we would expect, no more, no less. Not unusually good. Chow was correct, and my earlier statement was overstated. So if Ramanujan’s approximation is not just random fluctuation (which for all I know it could be), it at least does not seem to be part of a pattern of many good rational approximations for small powers of $$\pi$$. Below are some details of how I came to this conclusion.Linkage2022-05-31T23:09:00+00:002022-05-31T23:09:00+00:00https://11011110.github.io/blog/2022/05/31/linkage<ul> <li> <p>Irritated by repeated claims that spiral galaxies are logarithmic spirals <span style="white-space:nowrap">(<a href="https://mathstodon.xyz/@11011110/108316217782130412">$$\mathbb{M}$$</a>),</span> sourced to pop science instead of expert research, I found <a href="https://doi.org/10.1093/mnras/stt1627">“Pitch angle variations in spiral galaxies” (Savchenko &amp; Reshetnikov, <em>MNRAS</em> 2013)</a> which tells a more complicated story. In logarithmic spirals, the pitch angle is constant, in Archimedean spirals it decreases with radius, and in hyperbolic spirals it increases. All three are used as models of galaxies. But real galaxies have more varied pitch angles that do not match these models.</p> </li> <li> <p><a href="https://mathenchant.wordpress.com/2022/05/17/good-shurik-grothendieck/">Good Shurik Grothendieck</a> <span style="white-space:nowrap">(<a href="https://mathstodon.xyz/@11011110/108321635569223835">$$\mathbb{M}$$</a>).</span> James Propp writes (another) account of Grothendieck, as a way of explaining of how a scene from <em>Good Will Hunting</em> on the nature of genius may not be as inaccurate as he once thought it was.</p> </li> <li> <p><a href="https://mathstodon.xyz/@Ianagol/108323524092080204">Ian Agol posts a link</a> to <a href="https://www.quantamagazine.org/how-complex-is-a-knot-new-proof-reveals-ranking-system-that-works-20220518/">a <em>Quanta</em> story</a> on <a href="https://arxiv.org/abs/2201.03626">his research showing that certain four-dimensional surfaces connecting pairs of knots can be used to partial order them</a>, asking “how well mathematicians that aren’t topologists can grok the concepts described”.</p> </li> <li> <p><a href="https://github.blog/changelog/2022-05-19-render-mathematical-expressions-in-markdown/">Github’s markdown rendering now supports MathJax, finally</a> <span style="white-space:nowrap">(<a href="https://mathstodon.xyz/@11011110/108330496497692414">$$\mathbb{M}$$</a>,</span> <a href="https://news.ycombinator.com/item?id=31438864">via</a>). Github uses this rendering e.g. for the home pages of github packages. For some reason their chosen syntax is a little different from Kramdown and from modern LateX. Kramdown (which is what my Jekyll-based blog uses) delimits both inline and display math with double-dollar. Modern LateX (and Mathstodon) uses backslash-paren and backslash-square-bracket. Github uses old-school-TeX single-dollar and double-dollar.</p> </li> <li> <p><a href="https://mathstodon.xyz/@nilesjohnson/108193220633795260">Niles Johnson posts an example of tensor products</a>. Over the integers, $$(\mathbb{Z}/a)\otimes(\mathbb{Z}/b)=\mathbb{Z}/\gcd(a,b)$$, maybe not what you might have expected for a product. In the ensuing thread, Jacob Siehler writes that tensor products are “one of the things where you can vacillate for years between ‘This is so simple I don’t understand what all the fuss is,’ and ‘Apparently I do not understand this at all.’ ” I agree.</p> </li> <li> <p><a href="http://troika.uk.com/work/troika-squaring-the-circle/">“Squaring the Circle”, Troika, 2013</a> <span style="white-space:nowrap">(<a href="https://mathstodon.xyz/@11011110/108349773535124980">$$\mathbb{M}$$</a>):</span> an artwork that, from the intended viewpoint, looks like a circle, but with a square reflection in the mirror behind it. See also <a href="https://doi.org/10.1080/10724117.2020.1714290">David Richeson’s “Squaring the circle in a mirror” in <em>Math. Horizons</em></a> about this piece.</p> </li> <li> <p><a href="https://twitter.com/Morphinart/status/1487200838070284292">Ship in a Klein bottle</a> <span style="white-space:nowrap">(<a href="https://mathstodon.xyz/web/@Bruce/108353214826961618">via</a>).</span> Nice rendered artwork by J. Miguel Medina.</p> </li> <li> <p>If you’re an academic, you’re probably unhappy with Chegg <span style="white-space:nowrap">(<a href="https://mathstodon.xyz/@11011110/108361284820125145">$$\mathbb{M}$$</a>).</span> I know I am. Not so much for promoting intellectual property theft, but because they profit from miseducating students by luring them into copying rather than getting the benefit of their own work, and because they greatly boost my workload by forcing me to make up new problems instead of reusing them. But did you know that <a href="https://www.chronicle.com/article/work-in-public-education-and-hate-chegg-you-might-be-an-investor">your university’s retirement fund investors may well be financial supporters of Chegg?</a></p> </li> <li> <p>You can make a room whose walls are entirely mirrored <span style="white-space:nowrap">(<a href="https://mathstodon.xyz/@11011110/108366902122043626">$$\mathbb{M}$$</a>),</span> with the property that a light source at any point will leave some parts of the room dark. I’d previously only seen this as a mathematical construct but <a href="https://www.youtube.com/watch?v=x3VluzZTReE">in this video Steve Mould plays with a physical one</a>.</p> </li> <li> <p>New Wikipedia Good Article: <a href="https://en.wikipedia.org/wiki/Squaring_the_circle">Squaring the circle</a> <span style="white-space:nowrap">(<a href="https://mathstodon.xyz/@11011110/108369499495102486">$$\mathbb{M}$$</a>).</span> It mentions an approximation to $$\pi$$ by Ramanujan based on truncating the continued fraction</p> $\pi^4=97 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{3 + \cfrac{1}{1+\cfrac{1}{16539+\cdots}}}}}$ <p>Does anyone understand why this continued fraction has such a big term? Equivalently, why do powers of $$\pi$$ seem to have unusually good rational approximations?</p> <p>Relatedly, I reviewed (but did not significantly edit) another new Good Article, <a href="https://en.wikipedia.org/wiki/Lonely_runner_conjecture">Lonely runner conjecture</a>, on the unproven conjecture that for $$n$$ runners moving at different linear speeds around a circular track, each runner will eventually be far from all other runners.</p> </li> <li> <p>When I was an undergrad, I had a job on campus as a DECsystem-20 systems programmer <span style="white-space:nowrap">(<a href="https://mathstodon.xyz/@11011110/108376805535251788">$$\mathbb{M}$$</a>).</span> So when my colleague Rich Pattis announced that, in preparation for his impending retirement, he was clearing out his office and giving away, among other things, the PDP-10 front console he had rescued from the trash pile of another university, I jumped at the opportunity. Here it is, atop one of my office bookshelves. Note the 36 keys in the front row: its memory was organized into 36-bit words, not bytes.</p> <p style="text-align:center"><img src="https://www.ics.uci.edu/~eppstein/pix/pdp10/PDP10-m.jpg" alt="PDP-10 front console" style="border-style:solid;border-color:black;width:100%;max-width:720px" /></p> </li> <li> <p>In Poland, getting a full professor title requires sign-off by the president of the country, currently Andrzej Duda. As part of a pattern of holocaust denial from Duda’s party, <a href="https://web.archive.org/web/20220525204138/https://www.timeshighereducation.com/news/polish-president-stifles-genocide-researchers-professorship-bid">Duda has refused for three years to sign off on the application of Michał Bilewicz, head of the Center for Research on Prejudice at the University of Warsaw and a prominent researcher on, among other things, anti-semitism in Poland</a> <span style="white-space:nowrap">(<a href="https://mathstodon.xyz/@11011110/108380954223141014">$$\mathbb{M}$$</a>,</span> <a href="https://retractionwatch.com/2022/05/28/weekend-reads-female-driver-stereotypes-stealth-research-ai-comes-to-fake-scientific-images/">via</a>).</p> </li> <li> <p>Here’s a new illustration for <a href="https://en.wikipedia.org/wiki/Mrs._Miniver%27s_problem">the Wikipedia article on Mrs. Miniver’s problem</a> <span style="white-space:nowrap">(<a href="https://mathstodon.xyz/@11011110/108387115665183435">$$\mathbb{M}$$</a>).</span> The problem asks to arrange two circles so that their intersection (yellow) has the same area as the surrounding parts of the union (blue). It involves solving a transcendental equation, so this seemed like a good time to write <a href="https://commons.wikimedia.org/wiki/File:Mrs_Miniver%27s_Problem.svg">a script to find the geometry numerically</a> before switching to a graphics editor to color it in.</p> <p><img src="/blog/assets/2022/MrsMiniver.svg" alt="Three instances of Mrs. Miniver's problem" /></p> </li> <li> <p><a href="http://colinmorris.github.io/blog/unpopular-wiki-articles">In search of the least viewed article on Wikipedia</a> <span style="white-space:nowrap">(<a href="https://mathstodon.xyz/@11011110/108393231911048502">$$\mathbb{M}$$</a>,</span> <a href="https://news.ycombinator.com/item?id=31524943">via</a>). The trick is that the “random article” button isn’t uniformly random on articles. It tags articles with permanent random numbers in the unit interval, and when pressed picks another random number and looks for the next larger tag. For seldom-viewed articles, most views are random, and the probability of getting a random view is proportional to the length of the gap between an article’s permanent random number and the next smaller one. So the tail of least-viewed articles is dominated by unlucky articles whose gap is small.</p> </li> <li> <p>Any baker knows: <a href="https://www.epicurious.com/expert-advice/for-easier-parchment-paper-crumple-it-article">crumple your parchment paper to make it more flexible and fit neatly into any curved pie dish</a> <span style="white-space:nowrap">(<a href="https://mathstodon.xyz/@11011110/108399624501508690">$$\mathbb{M}$$</a>).</span> But one reason this works involves deep mathematics: cut or cleanly folded paper can only fit <a href="https://en.wikipedia.org/wiki/Developable_surface">piecewise-developable surfaces</a> such as patches of planes, cylinders, or cones. By <a href="https://en.wikipedia.org/wiki/Nash%E2%80%93Kuiper_theorem">the Nash–Kuiper theorem</a>, continuous but non-smoothly-differentiable (crumpled) paper surfaces can fit any surface.</p> </li> </ul>David EppsteinIrritated by repeated claims that spiral galaxies are logarithmic spirals ($$\mathbb{M}$$), sourced to pop science instead of expert research, I found “Pitch angle variations in spiral galaxies” (Savchenko &amp; Reshetnikov, MNRAS 2013) which tells a more complicated story. In logarithmic spirals, the pitch angle is constant, in Archimedean spirals it decreases with radius, and in hyperbolic spirals it increases. All three are used as models of galaxies. But real galaxies have more varied pitch angles that do not match these models.Congratulations, Dr. Osegueda!2022-05-21T16:49:00+00:002022-05-21T16:49:00+00:00https://11011110.github.io/blog/2022/05/21/congratulations-dr-osegueda<p>Martha Osegueda, one of Mike Goodrich’s students at UC Irvine, successfully defended her doctoral dissertation yesterday. Martha is originally Salvadorean, and came to UCI from an undergraduate degree at the University of Texas El Paso. Her research at UCI has spanned a wide variety of topics in graph drawing, phylogeny, computational geometry, and parallel query complexity:</p> <ul> <li> <p>“Minimum-width drawings of phylogenetic trees” from <em>COCOA</em> 2019 (<a href="https://doi.org/10.1007/978-3-030-36412-0_4">doi:10.1007/978-3-030-36412-0_4</a>) is about drawing evolutionary trees compactly. For the version of tree drawing they study, it turns out to be NP-hard when the order of the leaves is unconstrained but linear-time solvable for fixed leaf orderings.</p> </li> <li> <p>“Reconstructing binary trees in parallel” from <em>SPAA</em> 2020 (<a href="https://arxiv.org/abs/2006.15259">arXiv:2006.15259</a>) involved the reconstruction of evolutionary trees from queries asking which pair of a triple of species is closest.</p> </li> <li> <p>“Concatenation arguments and their applications to polyominoes and polycubes”, in <em>CGTA</em> 2021 (<a href="https://doi.org/10.1016/j.comgeo.2021.101790">doi:10.1016/j.comgeo.2021.101790</a>) and “On the number of compositions of two polycubes” from <em>EuroComb</em> 2021 (<a href="https://doi.org/10.1007/978-3-030-83823-2_12">doi:10.1007/978-3-030-83823-2_12</a>) strengthen the known lower bounds on the growth rates of polyominoes and polycubes, using arguments based on counting ways of gluing pairs of smaller polyforms into larger ones.</p> </li> <li> <p>“Angles of arc-polygons and Lombardi drawings of cacti”, from <em>CCCG</em> 2021 (<a href="https://arxiv.org/abs/2107.03615">arXiv:2107.03615</a>) is my only collaboration with Martha and one <a href="/blog/2021/07/10/angles-arc-triangles.html">I have posted about here</a>. It characterizes the vertex angles that are possible in triangles with circular-arc sides, and uses that characterization to prove the existence of a Lombardi drawing for any cactus graph.</p> </li> <li> <p>“Parallel network mapping algorithms” from <em>SPAA</em> 2021 (<a href="https://doi.org/10.1145/3409964.3461822">doi:10.1145/3409964.3461822</a>) and “Mapping Networks via parallel $$k$$th-hop traceroute queries” from <em>STACS</em> 2022 (<a href="https://doi.org/10.4230/LIPIcs.STACS.2022.4">doi:10.4230/LIPIcs.STACS.2022.4</a>) are on traceroute-like problems: mapping the structure of a computer communication network using a small number of parallel rounds of pings to determine distances from nodes on the network.</p> </li> <li> <p>“Taming the knight’s tour: Minimizing turns and crossings” from <em>FUN</em> 2021 and <em>TCS</em> 2022 (<a href="https://doi.org/10.1016/j.tcs.2021.12.002">doi:10.1016/j.tcs.2021.12.002</a>) shows how to move a chess knight around all of the squares of a chessboard, touching each square once, so that the number of times the knight’s path crosses itself or deviates from a straight path are both simultaneously approximately minimized.</p> </li> <li> <p>“Diamonds are forever in the blockchain: Geometric polyhedral point-set pattern matching” is her newest work, in submission. It studies problems of aligning a polyhedron to sample points from a similarly-shaped surface, motivated by an application in tracing the provenance of ethically sourced diamonds.</p> </li> </ul> <p>Despite all of these good publications, Mike tells me that it was difficult to persuade Martha that she had done enough for a doctorate. But she had, easily, and now she’s moving on to a position at Meta. Congratulations, Martha!</p> <p>(<a href="https://mathstodon.xyz/@11011110/108342653334960089">Discuss on Mastodon</a>)</p>David EppsteinMartha Osegueda, one of Mike Goodrich’s students at UC Irvine, successfully defended her doctoral dissertation yesterday. Martha is originally Salvadorean, and came to UCI from an undergraduate degree at the University of Texas El Paso. Her research at UCI has spanned a wide variety of topics in graph drawing, phylogeny, computational geometry, and parallel query complexity: