Flat-foldability of crease patterns can be tested in time polynomial in the pattern size, exponential in the treewidth of a certain arrangement graph, and factorial in the ply of the folded pattern. The exponential dependence on treewidth is necessary under the strong exponential time hypothesis. This is intended as the journal version of my CCCG 2023 paper.
The nets of certain polyhedra, despite being easy to construct, are hard to fold: the polyhedron they fold into has coordinates that cannot be computed, in an an algebraic computation tree model, allowing either the extraction of \(n\)th roots of previously-computed values for arbitrary \(n\), or the extraction of roots of polynomials of bounded degree. I posted about this 2015 but didn’t otherwise publish it.
For the “flaps and flips” model of origami reconfiguration, in which sheets of origami paper are taped by one edge to a tabletop, and we can change the position of one sheet at a time as long as all sheets lie flat at the end of each move, getting from one flat state to another is \(\mathsf{PSPACE}\)-complete, and counting the number of flat states is \(\mathsf{\#P}\)-complete. I posted about its \(\mathsf{PSPACE}\)-completeness last August. The \(\mathsf{\#P}\)-completeness proof translates counting non-bipartite perfect matchings into nondeterministic constraint logic and from there into flaps and flips.
For infinite crease patterns, it is undecidable to determine whether a flat folding exists. This strengthens recent Turing-completeness results of Tom Hull and Inna Zakharevich in several ways:
The proof uses a finite square sheet of origami, with infinitely many creases on it, rather than the infinite half-plane of Hull and Zakharevich.
The crease patterns of Hull and Zakharevich always fold, so testing whether they fold is not hard (just say yes). Instead, their construction proves the undecidability of a more complicated problem: whether a finite perturbation of a periodic crease pattern can be folded so that the perturbation remains finite. The new preprint constructs repeating crease patterns for which foldability itself is hard.
Hull and Zakharevich use crease patterns in which the creases are labeled both by whether they are mountain or valley folds, and whether they are mandatory or optional. The new undecidability proofs do not need optional folds, and can work either with labeled or unlabeled crease patterns.
I haven’t posted in detail about the undecidability results here, so maybe I should say a little more about them. They are heavily based on the binary tilings of the hyperbolic plane, and on the fact that a binary tiling with tiles in the form of \(2\times 1\) rectangles fits very neatly into a square of origami paper:
The main idea of the proof is to form infinite crease patterns in which each tile of a binary tiling gets creased in the same way, with creases that simulate part of a logic circuit. The idea of using creases to simulate circuits was also used by Hull and Zakharevich, and before them in \(\mathsf{NP}\)-completeness proofs for finite flat-foldability by Bern and Hayes and by Akitaya et al. In fact, my preprint uses exactly the same circuit building blocks as Akitaya et al., with some extra care to show that they can produce arbitrary finite circuits within each tile that connect together across tile boundaries.
There are actually two variations of the proof. One variation shows that it is hard to test foldability for a fixed crease pattern or circuit within each tile, plus a partial folding of the pattern on finitely many creases. The second variation avoids the partial folding, at the expense of making the crease pattern within each tile vary.
The first variation is based on simulating Turing machines. In the binary tiling above, the successive states of a universal Turing machine and its tape can be recorded in the red vertical stripes, ordered from left to right. The circuit within each tile ensures that the successive red stripes carry out the correct behavior of the Turing machine, that the blue tiles convey the state of the machine across from one stripe to the next, and that if the Turing machine ever halts the pattern will fail to fold. The initial partial folding sets the initial state of the Turing machine in the leftmost stripe. Since the halting problem is hard, so is testing foldability of this crease pattern and partial folding.
When I spoke about this at JCDCG^{3}, I had only this variation. Someone in the audience (unfortunately I forget who) asked whether the finite set of initial folds were necessary. Later in the conference, Stefan Langerman spoke about certain problems involving Wang tiling. These two things come together in the second variant of my undecidability proof for flat foldability.
In the Wang tiling problem, you are given a finite set of prototiles (usually square, with colored edges), and you are asked to tile the plane by translating these prototiles without rotation. The tiles are required to fit together, edge-to-edge, with matching edge colors. This is, famously, undecidable. It’s easy to prove this if you’re allowed to force one special starting tile to be included. Then, you can design Wang tiles that, if they tile, will form the time-space diagram of a Turing machine, with the one special tile used to fix its initial state. It’s harder to prove, but still true, that Wang tiling is hard even without a fixed special tile. Stefan used this problem as the basis for a proof, with Erik Demaine, that it’s undecidable to test whether three given polygons can tile the plane (allowing rotations), improving a previous result that tiling for five polygons is hard. Their preprint, “Tiling with three polygons is undecidable”, was also recently uploaded as arXiv:2409.11582.
One can ask about Wang tiling problems for other shapes of tiles, and a 2007 paper of Jarkko Kari showed that Wang tiling is hard for the tiles of a binary tiling. A minor technicality is that a tiling of the entire hyperbolic plane and a tiling of an origami square are not quite the same thing, but one exists if the other does. An example of a binary tiling Wang tiling instance (with four prototiles) and a partial solution to it is shown above, with the matching edge colors shown as colored circles. It’s easy to design finite logic circuits, for any binary Wang tiling instance, that fit within each tile, nondeterministically guess a color for each edge, check that each tile is colored to match a prototile, and fail to fold if not. The crease pattern formed by using creases to simulate these circuits, in each tile of a binary tiling, is flat foldable if and only if the given Wang tiling instance is solvable. Therefore, it’s undecidable to test whether crease patterns that repeat in the pattern of a binary tiling can be folded flat, even when no partial folding is given.
]]>The right side of the illustration depicts a binary tiling, in one of its conventional views using the Poincaré half plane model of hyperbolic geometry. The vertical sides of the tiles lie on hyperbolic lines, but their horizontal sides are arcs of horocycles, special hyperbolic curves intermediate between a circle and a line. They are not straight: the tiles are concave on top and convex on the bottom. The black line at the bottom of the image is the boundary of the half plane model.
One can measure the sides of each tile using standard formulas for hyperbolic geometry in the half-plane model. The hyperbolic length of a vertical line segment is the logarithm of the ratio of its top and bottom \(x\)-coordinates. Therefore, the vertical sides of each tile have length \(\ln 2\approx 0.69315\), as they always do in binary tilings. The widths of tiles in binary tilings can vary arbitrarily; I chose this drawing to make the tiles look square but they could be rectangles of arbitrary aspect ratio. The hyperbolic length of a horizontal line segment is its Euclidean length divided by its height above the \(x\)-axis. Therefore, the top edge of each square has hyperbolic length \(\tfrac12\), while the bottom edge has hyperbolic length \(1\). That’s why it’s possible for two top edges of tiles to match up with a single bottom edge.
The vertical black line in the middle of the illustration passes only through edges of tiles, not their interiors. Not every binary tiling has such a line, but when it exists the tiling has one-dimensional symmetry: it can be translated upwards and downwards along this line, and reflected across the line. (Binary tilings cannot have full two-dimensional symmetry.) The tile edges divide this line into intervals, all the same hyperbolic length, \(\ln 2\).
That leaves the black semicircle to explain. It is a hyperbolic line, perpendicular to the vertical symmetry line of the binary tiling. The left side of the top illustration shows what happens if we reflect the binary tiling across this perpendicular. In Euclidean geometry, this reflection is modeled as an inversion through the semicircle. Its (Euclidean) radius is the geometric mean of the two endpoints of the uppermost visible interval on the vertical black line, \(\sqrt2\) times the height of the inner endpoint, causing these two endpoints to invert to each other. The tiles on the right and the inverted tiles on the left match up along the vertical symmetry line, giving us a tiling of the entire hyperbolic plane. The inversion reverses top and bottom, so the bottom sides of tiles on the right (touching two other tiles) become the sides farthest from the origin of tiles on the left (still touching two other tiles). The vertical black line continues to be a symmetry line for the half-flipped tiling, but with different symmetries: translations and \(180^\circ\) rotations through tile corners or tile edge midpoints.
The horizontal horocycles on the right are inverted on the left into circles that are mutually tangent to each other and to the \(x\)-axis at the origin. We see only semicircles, but a complete horocycle would invert to a complete circle. The top point of each semicircle lies on the vertical symmetry line, at a corner of four tiles. These semicircles are scaled by powers of two, just like the horizontal lines on the right, to make them equally spaced at hyperbolic distance \(\ln 2\) from each other. Each row of equal-size squares on the right, between two consecutive horocycles, continues in inverted form on the left as a horn-shaped region between two of these semicircles.
Within a single horn, the tiles are separated from each other by hyperbolic lines, modeled as semicircles that are mutually tangent to each other and to the \(y\)-axis at the origin. On the right, the vertical sides of tiles in a row are equally spaced in Euclidean distance. On the left, this translates into a harmonic sequence: the radii of the semicircles modeling these lines are proportional to \(1,\tfrac12,\tfrac13,\tfrac14,\dots\). The constant of proportionality, for the horn whose top tile is crossed by the inversion circle, is set by the observation that the inversion circle passes through the bottom right corner of the top right red tile, and therefore must also pass through a corner of the top left blue tile. The first semicircle in this harmonic sequence also passes through this corner (which turns out to be its apex). Once we have found the tiles of one horn in this way, all of the other horns are scaled from it by powers of two. The semicircles with proportions \(1,\tfrac12,\tfrac13,\tfrac14,\dots\) in one horn continue with proportions \(\tfrac12,\tfrac14,\tfrac16,\tfrac18,\dots\) in the next horn out, and so on.
With these calculations and measurements, we have all we need to construct the illustration of the half-flipped binary tiling!
]]>Shift networks (\(\mathbb{M}\)). Jeremy Kun asks how to permute vectors using few vector additions, elementwise multiplications, and rotation operations, for an application involving homomorphic encryption.
Our fractional universe (\(\mathbb{M}\)). Jim Propp and his guest columnist Jeff Glibb parody those breathless but vague pop-sci expositions of new mathematical discoveries.
Byrne’s Euclid (\(\mathbb{M}\)), the one with colored diagrams replacing all the symbols, newly digitized and available online from the Harvard library.
Noted algorithmic bias researcher Timnit Gebru had a recent application for a grant to study the risks of AI turned down because she wouldn’t put it under the thumb of the big corporate players.
The call for papers for the 41st International Symposium on Computational Geometry (SoCG 2025) in Kanazawa, Japan next June is now live (\(\mathbb{M}\)). Abstracts are due November 26, with full submissions due December 3.
Small volume bodies of constant width (\(\mathbb{M}\)). Here “small” means exponentially small, as a function of dimension, relative to the unit ball. See also breathless but vague pop-sci exposition in Quanta.
How not to prove the four color theorem (\(\mathbb{M}\)). Why a failed 2022 proof based on finding a large fraction of planar triangulations that are 4-colorable could never have worked: any large fraction of these graphs contains all planar graphs as subgraphs, and cannot be easier to prove 4-colorable than all planar graphs.
I spent an afternoon at the local beach instead of just staying online all day (\(\mathbb{M}\)). We saw an osprey, hovering low over the shallow water near some skin divers, but I didn’t bring the long lens needed for a good photo of it. Instead, here’s a wide shot of the beach in late afternoon. More photos.
Incidentally this photo totally confused Adobe Lightroom’s level function. It was about \(0.5^\circ\) off-level but I had to correct it manually despite the obvious visible horizon line. The level function wanted to level the sand ripples in the foreground.
Ryan Williams attempts to refute the orthogonal vectors conjecture (\(\mathbb{M}\)), and with it the strong exponential time hypothesis, by proving that if we could find orthogonal vectors (or disjoint sets) quickly, we could prove strong lower bounds in circuit complexity on the simulation of read-once 2-DNFs. He also uses SAT solvers to develop new algorithms for disjointness with interesting time bounds (but not strong enough bounds to refute these conjectures).
See-through illustration of a 38-sided space-filling polyhedron (\(\mathbb{M}\)), with some of its neighbors. This is the most sides known but the best upper bound proven so far is 92.
Innovations in Graph Theory publishes its first issue (\(\mathbb{M}\)). This is a diamond-model open-access journal published by the Centre Mersenne in Grenoble; see also the journal’s home page.
A method from one of my older papers, “Algorithms for coloring quadtrees”, 3-colors the binary tiling of the hyperbolic plane (\(\mathbb{M}\)). The tiles of each color form three interwoven infinite binary trees under corner-to-corner adjacencies.
Tired: publishers shortcutting academic peer-review to profit from open-access publication fees. Wired: publishers shortcutting academic peer-review to generate AI training data as quickly as possible (\(\mathbb{M}\)). Their authors get no say in whether their content is taken for this purpose. The linked story targets Taylor & Francis / Routledge, but Wiley and the Oxford University Press are also complicit in similar deals. The “feed the beast” illustration below is stolen from John Baez’s Mastodon post, where he writes “Yes, this image is AI-generated. I thought that was appropriate, in a darkly humorous way.”
This is the subject of a recent paper by ten authors including myself: “Noncrossing longest paths and cycles”, by Greg Aloupis, Ahmad Biniaz, Jit Bose, Jean-Lou De Carufel, me, Anil Maheshwari, Saeed Odak, Michiel Smid, Csaba Tóth, and Pavel Valtr, most of whom worked together on this at the 10th Annual Workshop on Geometry and Graphs in Barbados last year. It was presented last week at Graph Drawing 2024, for which the program has links to preliminary versions of all papers, at least until the published proceedings is ready.
Three points, or four points in non-convex position, have no crossings, but for larger numbers of points in general position a crossing is always possible. In a recent paper in Graphs and Combinatorics, Jose Luis Álvarez-Rebollar, Jorge Cravioto-Lagos, Nestaly Marín, Oriol Andreu Solé-Pi, and Jorge Urrutia conjectured that sets of five or more points in general position always have crossings in their longest cycles (maximum traveling salesperson tours) and asked whether they also have crossings in their longest paths (maximum traveling salesperson paths). Our new work provides strong counterexamples to both questions: for all \(n\), there exist sets of \(n\) points in general position in the plane with a non-crossing longest cycle, and sets of \(n\) points with a non-crossing longest path.
The path construction for even numbers of points is simplest. Its basic ideas are:
All points will be placed close to the \(x\)-axis, with \(x\)-coordinate much larger than \(y\), so that the differences in \(x\)-coordinate dominate their distances.
If the \(y\)-coordinates are zero, and the points are placed with half of their \(x\)-coordinates positive and half negative, then the (many) equally longest paths all start and end at the two middle points, with their edges all crossing the \(y\)-axis. Edges that do not cross the axis can be paired off on both sides of the axis and replaced by longer pairs of edges that cross. Among paths where all edges cross the axis, the total length is the sum of distances to the axis, doubled for non-endpoints, so the endpoints should be the two middle points.
We can use equally spaced \(x\)-coordinates, half positive and half negative, perturbed by very small \(y\)-coordinates. The longest path will continue to have all edges crossing the \(y\)-axis, but the perturbation will force one such path to be longest. (The construction for an odd number of points uses a slightly uneven placement that is not important for this post.)
When we perturb two points, the length of their edge increases (compared to its length with zero \(y\)-coordinates) by an amount that can be calculated by the Pythagorean theorem to be
\[\frac{\Delta_y^2}{2\Delta_x}\bigl(1+o(1)\bigr),\]where \(\Delta_x\) and \(\Delta_y\) are the differences in \(x\)- and \(y\)-coordinates, assuming \(\Delta_x\gg\Delta_y\).
Our perturbation will assign exponentially increasing \(y\)-coordinates to the points, with the largest perturbations going to the points closest to the \(x\)-axis, except for leaving one of the two middle points unperturbed.
Because of this exponential growth, the terms \(\Delta_y\) in the length increase are up to a small error term the same as the \(y\)-coordinates themselves, regardless of the choice of path. But different paths can pair up different \(\Delta_y\) terms with different \(\Delta_x\) terms. To get the longest path, we should pair up the largest \(\Delta_y\) term (coming from the edge incident to the highest point) with the smallest \(\Delta_x\) term (coming from the non-middle vertex closest to the other side of the \(x\)-axis), and continue greedily choosing edges that at each step pair the highest remaining vertex with the non-middle vertex closest to the other side of the axis.
The resulting unique longest path is uncrossed! Each edge connects two vertices that are consecutive when sorted by their \(y\)-coordinates. Because the edges span disjoint ranges of \(y\), they cannot cross. In a not-to-scale view with untuned parameters, it looks something like the following.
To construct uncrossed longest paths with odd numbers of points we choose \(x\)-coordinates a little more carefully, and then merely omit the topmost point. The cycle construction uses a similar idea, but in two variations depending on whether the number of points is supposed to be odd or even. For evenly many points we double each point except the first and last in the non-crossing path path and carefully place its two copies so that a thin polygon can zigzag in the same way as the path. For odd points we find a polygon that adds one vertex and two edges to the long path to connect it into a polygon. See the paper for details.
]]>Corentin’s work first came to my attention at SoCG 2023, where he had a paper on the treewidth of knots and links with his advisors, later incorporated into his thesis. Treewidth is a concept in graph theory but it can be applied to knots and links by projecting them onto a plane and considering the 4-regular planar multigraphs that result. The treewidth of a knot or link is the minimum treewidth, obtained in this way, of any of its projections. It was known that this number can be arbitrarily large but Corentin provides a simpler construction: the torus knots. The illustration below is torus knot (11,10) where the first number counts the lobes and the second counts the number of strands you would have to cross to get from the center to the outside. This drawing style for these knots looks like a large grid, having high treewidth, and Corentin shows that its treewidth is within a constant factor of optimal for these knots.
The tools he uses to prove this also seem quite interesting. He defines a notion of “sphere decompositions” for three-dimensional structures, obtained by forming a family of nested double bubbles that partition space into annular shells between an enclosing bubble and the double bubble that it encloses, and topologically sweeping each of these shells by a sphere. He defines the “spherewidth” of a link to be minmax number of crossing points one of the spheres in this sweep, minimized over sphere decompositions and maximized among the spheres of the decomposition. If the treewidth of a knot or link is low, it also has a sphere decomposition of low spherewidth. But the spherewidth of the torus knots \((p,q)\) turns out to be at least proportional to \(\min(p,q)\): any lower width would allow you to attach a disk across the hole of the torus whose boundary crosses few strands of the torus knot, known to be impossible. Therefore, the treewidth is also large.
A second part of Corentin’s thesis is also related to structural graph theory in a different way. He defines a class of links obtained by “plumbing” double-twisted bands together according to a certain tree structure; this means laying one band perpendicularly across another and gluing them together where they overlap. The links are the boundaries of the resulting surfaces. He then uses Kruskal’s tree theorem to show that when a link property behave nicely with respect to cutting the surface without separating it, it can be characterized by a finite set of forbidden subsurfaces. This leads to a proof of existence of an algorithm for testing whether one of these links has bounded “genus defect”, the difference between the minimum genus of a Seifert surface and of certain four-dimensional surfaces. But although we can prove the algorithm exists, we don’t know what it is! That’s because we don’t know how to explicitly list the forbidden subsurfaces. This is all from another paper in SoCG 2024.
The thesis concludes with a third result: there exist links that are split (there exists a sphere that separates some components of the link from the rest) but for which proving this by a sequence of Reidemeister moves may have to make arbitrarily large increases in the crossing number. Both the construction (from two linked torus knots, drawn as above, and a third unknot that is split from them but not drawn as split) and the proof that the crossing number must increase use the methods from the SoCG 2023 paper. This raises the strong hope that we can prove similar results on large increases in crossing number for the unknotting problem as well.
I think these are very interesting results based on powerful new machinery that could prove very useful in continued research in this area. My understanding is that Corentin is continuing as a postdoctoral researcher in Montpellier, but he will go on the job market again soon, and should be a very strong candidate when he does.
Congratulations, Corentin!
]]>The Pentium as a Navajo weaving (\(\mathbb{M}\)), a 1994 rug by Marilou Schultz in the the National Gallery of Art designed from an image of the Intel Pentium integrated circuit.
On matching algorithms to solve single rows, columns, and boxes of sudoku puzzles (\(\mathbb{M}\)). As mentioned in my first blog post, 19 years ago.
Planar point sets with forbidden four-point patterns and few distinct distances (\(\mathbb{M}\)), Terry Tao. The forbidden patterns (as befits the problem) involve distances rather than order types. They are sets of four points with fewer than five distinct distances (classified into eight types) and the result is that some sets of \(n\) points avoiding these patterns nevertheless have only \(o(n^2)\) distances, solving an old Erdős problem. The solution combines random sampling of a grid (which reduces all but one of the patterns to few enough instances that you can afford to delete one more point per remaining one) with the modular parabola idea used e.g. by Erdős in the no-3-in-line problem (which without randomization kills the remaining pattern but not all the others).
Two new Wikipedia Good Articles (\(\mathbb{M}\)):
Skolem’s paradox, the seeming contradiction between the Löwenheim–Skolem theorem (first order theories such as ZFC have countable models) and Cantor’s theorem (there exists an uncountable set). In a countable model of ZFC, only countably many elements can model members of an uncountable set like \(\mathbb{R}\), but their countability is not visible within the model. This idea led to the “Skolemite” philosophy that all sets are secretly countable from some external view of set theory. I reviewed this for GA but the edits to bring it up to GA standard were by another editor, Pagliaccious.
Universal vertex, a vertex in a graph that is adjacent to all others. Although seemingly very basic, this touches on some nontrivial ideas: for instance, almost all cop-win graphs are cop-win because they have a universal vertex. And for an even number of labeled vertices, the number of graphs that include a universal vertex is odd, a property that is central to proving the evasiveness of having a universal vertex (if you learn about the graph by testing pairs of vertices for adjacency, you may be forced to test all pairs before finding out whether it has a universal vertex).
An annoyance in graph algorithms (\(\mathbb{M}\)): if your graph API gives you only one chance to list the neighbors of each vertex, then remembering them blows up the space for depth-first search from linear in the number of vertices to linear in the number of edges. You can work around this if all you need is the depth-first search tree but it’s still a problem for DFS-based algorithms that do something for each non-tree edge.
Katrine Hildebrandt embraces symmetry in paper, wire, and reed (\(\mathbb{M}\)), with geometric art often featuring concentric circles and parallel or radiating lines.
A new high-rank elliptic curve and a new book that systematically catalogues Diophantine equations (\(\mathbb{M}\)).
Lance Fortnow’s favorite theorem of the month: quasipolynomial time for parity games (\(\mathbb{M}\)).
Updated slides for my talk “Computational Complexities of Folding” (\(\mathbb{M}\)), which I gave again at JCDCG\(^3\) 2024. Compared to the OSME version, there’s some new material (and some removals of unimportant slides to make room):
It’s \(\mathsf{\#P}\)-complete to count flat-folded states for the same model of folding for which I already proved \(\mathsf{PSPACE}\)-completeness of reconfiguration.
It’s undecidable to determine whether a finite square of paper with a fractal infinite crease pattern can fold flat. This strengthens recent results of Hull and Zakharevich by using a finite rather than infinite sheet, by avoiding the use of optional folds, by using rational coordinates rather than multiples of \(\sqrt{3}\), and by simplifying the undecidable problem statement to flat-foldability instead of whether a certain finite perturbation to a repeating pattern remains finite.
Dan Gardner rebuts the idea that Wikipedia has become “Wokepedia” (\(\mathbb{M}\), via). In related Wikipedia news, Nicole Venus studies “the representation of female economists on Wikipedia” (via Ipigott on WikiProject Women in Red). Although Venus’s study showed the women in her data set to be 53% less likely than men to have a Wikipedia biography, much of that gap comes from differences in their visible accomplishments (likely caused by biases elsewhere, but beyond the control of Wikipedia editors). When that is factored out, Venus found that women were still 9% less likely to be profiled than men of comparable accomplishment. Initiatives to create more articles on women (like Women in Red) have helped lower that number, and Venus didn’t really see any systematic anti-women editing efforts on Wikipedia; she attributes the remaining gap to a greater tendency for self-promotion among men than women.
What makes mathematicians believe unproved mathematical statements? (\(\mathbb{M}\)), Timothy Gowers in Ann. Math. Phil. (2023).
Ordinary graphs (\(\mathbb{M}\)). Ed Pegg constructs some well-known graphs (the Shrikhande Graph, Petersen Graph, and Cuboctahedral Graph) from the ordinary lines of certain symmetric configurations of points. He finds two more interesting regular graphs that appear not to be so famous from the 16-point orchard configuration and the Grünbaum–Rigby configuration, and asks what other interesting graphs might be formed in the same way.
Cheating at scale (\(\mathbb{M}\)), on the difference in what you gain from poring over textbooks searching for a solved problem resembling your assigned problem, versus just asking an LLM to solve it for you. See also Ted Chiang in The New Yorker on why he thinks AI cannot make art: “Using ChatGPT to complete assignments is like bringing a forklift into the weight room; you will never improve your cognitive fitness that way.” (Via.)
A too-cute trick for zero-based indexing in Python from the end of an array backwards (\(\mathbb{M}\)): A[~0]
, A[~1]
, A[~2]
, …
Congratulations to Rajeev Alur of the the University of Pennsylvania for winning the 2024 Knuth Prize (\(\mathbb{M}\)), “for his introduction of novel models of computation which provide the theoretical foundations for analysis, design, synthesis, and verification of computer systems”!
Boxed cat heads: (\(\mathbb{M}\)).
New OEIS sequence (\(\mathbb{M}\)): 1, 1, 2, 2, 5, 5, 10, 10, 26, 26, 52, 52, 130, 130, 260, 260, 677, 677, … If you want a puzzle you can try guessing what it means before clicking through. For a bigger puzzle (one I don’t have an answer to): is there a natural meaning (other than the trivial \(a(2n)\)) for the sequence 1, 2, 5, 10, 26, … that you get by removing the duplicates?
Can an equilateral convex polygon be combinatorially equivalent to the regular dodecahedron without being congruent to it (\(\mathbb{M}\))? Without convexity the answer is yes: the endododecahedron.
The Chronicle of Higher Education on the consequences of MIT dropping race as an admissions criterion (\(\mathbb{M}\), archived). It frames the story as “Black and Latino Enrollment Plunges”, which I think we can all agree is bad. But buried in the article text one finds “Asian American students account for almost half, or 47 percent, of the incoming class, an increase of six percentage points over their share from 2020 to 2023. White enrollment stayed almost unchanged.”
My conclusion is the same as when affirmative action was banned in California in 1996: affirmative action, as implemented, never took positions from white students to give to minorities. Instead, it was done at the expense of other minorities not considered disadvantaged, today and then Asian Americans. (In an earlier time it might have been Jews.) When race-based admission is banned, and with it both affirmative action for disadvantaged minorities and suppression of not-considered-disadvantaged minorities, the suppression is lifted and we see from the difference how much they were suppressed.
The structural oppression faced by the disadvantaged minorities, causing them to be less competitive as applicants, remains. If we can no longer counterbalance it with affirmative action, we need some other way to reduce its effects. A common proxy for race is to let applicants offer hard-luck stories and give spots to those whose story is the most convincing. This plausibly has the advantage of targeting the people most affected by structural oppression, but I don’t have data or links on how effectively it works.
Polysticks and polyominoes, together at last (\(\mathbb{M}\)). Alexandre Muñiz on tiling polyominoes, and then tiling their edges with polysticks.
MIT’s 2020 cancellation of all Elsevier subscriptions saves it approximately $2M/year (\(\mathbb{M}\), via) with most faculty facing few problems and supportive.
Benford’s law applied to software reliability metrics (\(\mathbb{M}\)). Seconds until first acknowledgement appears to follow the law; incident duration does not.
How to fold a bowtie (\(\mathbb{M}\)). This example shows that, unlike finite crease patterns, infinite origami crease patterns might not have a unique flat-folded shape.
Glass artist Jiyong Lee’s geometric abstractions of biological cell division (\(\mathbb{M}\)).
The polynomial-time hierarchy is infinite relative to a random oracle (\(\mathbb{M}\)), this month’s favorite theorem from Lance Fortnow.
Dave Richeson uses salt to sculpt a roofline in the shape of a hat tile, related to his construction of a fold-and-cut pattern from its straight skeleton (last time).
Convex Optimization in the Age of LLMs (\(\mathbb{M}\)). Ben Recht kicks off liveblogging a new course.
An infinite family of polyhedra with a single non-right angle, Michael Engen. For large \(n\) these look like ruffled disks, with increasingly high-degree trigonometric olynomials for their non-right angles.
Smooth manifolds of any dimension can be triangulated with twin-width bounded by a function of the dimension (\(\mathbb{M}\)), Édouard Bonnet & Kristóf Huszár. Treewidth doesn’t work: 2-manifolds have bounded width (just thicken a cactus graph) but for 3-manifolds (even Haken manifolds) it is unbounded; see Huszár and Spreer in SoCG 2023.
An instance of this problem consists of some sheets of square origami paper, all the same size, spread out flat on a tabletop. Each of the sheets of paper (a “flap”) is taped to the tabletop along one of its edges, making a flexible hinge that allows it to “flip” to a different position, as long as the hinge stays attached to the table along the same line segment. No other flap can come between this hinge and the table, but that is the only restriction on how the flaps can be placed. For instance, there can be cycles in their above-below relations, as with the central red, pink, and green flaps below.
What you’re allowed to do is to move one flap at a time to a different position, keeping its hinge in place. In the example above, the bottom blue flap can be flipped to the other side of its hinge. But the moving flap doesn’t have to stay rigid, like a sheet of glass. For instance, it would also be allowed to flip the green flap down, even though another flap partly covers it and its hinge. These are the only two flips possible in the state shown, because the other flaps all have part of their hinges covered, fixing them in place. However, flipping the blue or green flap frees up other flaps that can flip in turn. It would also be allowed to flip a flap by moving it above or below others, keeping its hinge fixed, but allowing or disallowing that possibility turns out not to be important.
Now, for a given system of flaps and hinges, can you get from every flat state to every other flat state by a sequence of flips? Or if I give you two different flat states of the same system of flaps and hinges, can you get from one to the other? If you can, how many moves will it take? It turns out to be very difficult to answer these questions (\(\mathsf{PSPACE}\)-complete), and the number of moves may be exponential!
As usual, the proof of hardness involves showing that this problem can simulate another problem known to be hard. That other problem is nondeterministic constraint logic. Instances of nondeterministic constraint logic take the form of 3-regular directed graphs, with the edges colored blue and red, and with an odd number of blue edges at each vertex. Every vertex must either have at least one blue edge directed into it, or both red edges directed into it. The moves you’re allowed to make, in this system, are to reverse one edge at a time, maintaining these constraints. For instance, in the illustration below, some of the red edges can be reversed, but none of the blue ones can, so there is no way to get from the orientation shown to its mirror image. Getting from one orientation of the edges to another, or testing whether all orientations are connected, is known to be \(\mathsf{PSPACE}\)-complete, even when the graph is planar and has bounded bandwidth.
So to prove \(\mathsf{PSPACE}\)-hardness (the more difficult part of \(\mathsf{PSPACE}\)-completeness) for the flaps and flips problem, all we need to do is to show how to represent directed edges, and the two types of vertex with one blue edge and three blue edges, using flaps. But this turns out to be easy, using the three gadgets depicted below for an arrow, blue-red-red vertex, and blue-blue-blue vertex.
For an arrow, just lay out a sequence of flaps, each of which will (when flipped to one side or the other) cover part of the hinge of its neighboring flap in the sequence. If you think of this sequence as forming a directed edge, the arrowhead points to the uncovered hinge at one end. To reverse the edge, you can flip its flaps in order, from the arrowhead to the tail. The intermediate states of these flips will leave the edge having two tails and no arrowheads, but that isn’t problematic.
For a blue-red-red vertex, arrange the flaps at the ends of the three edges so that the blue flap can overlap both red hinges, but each red flap can only overlap the blue hinge, as shown. If the blue flap hinge is covered, the blue flap itself must be flipped away from the vertex. In terms of the underlying nondeterministic constraint logic instance, this means that the blue edge points in, as required. If the blue flap is not flipped away, it covers the two red flaps, and the two red edges point in. So this arrangement of flaps has exactly the same behavior as a blue-red-red vertex is required to have: its allowed states are exactly the ones where the blue edge points in, the two red edges both point in, or both of those things happen.
The blue-blue-blue vertex is shown in two arrangements below, but only one of these can be made with paper and tape. In the left part of the illustration, three flaps at the ends of three blue edges are flipped towards the vertex, corresponding to the forbidden state where the three edges are all directed outward from the vertex. Each flap overlaps the hinge of a different flap, forming a cyclic system of above-below relations that cannot be realized in the central area where all three flaps overlap. However, if at least one flap is flipped away from the vertex (corresponding to the arrow pointing in), it is realizable, as shown on the right. And in this state, with one arrow pointing in, the other two arrows can flip freely, unobstructed by the flipped-out flap or each other.
So to get a hard instance for the flaps and flips problem, or one that takes many moves to solve, just take a hard instance of nondeterministic constraint logic and translate it in this way. Any move you make in the resulting flaps and flips problem can be translated back into moves for nondeterministic constraint logic, so if you could solve one problem you could solve the other.
What I’d like to prove is something a little closer to actual origami: that for a single folded sheet of paper, it can be hard to get from one folded state to another using moves that change only a small part of the folding at a time: maybe only a bounded number of creases, or a single above-below relation, or only the creases along a single line segment of the folded model. It is possible to make origami crease patterns that fold to something resembling a tabletop with flaps attached, so maybe this could be used to simulate flaps and flips, simulating nondeterministic constraint logic, simulating arbitrary polynomial-space computation.
The last two images are from the origami maze font generator by Erik Demaine, Martin Demaine, and Jason Ku, using just punctuation as text: ',''
. Check it out and design yourself crease patterns that fold to your favorite message!
Fractals from hinged hexagon and triangle tilings (\(\mathbb{M}\)), Helena Verrill, from Bridges 2024. See also Verrill’s interactive site for generating these patterns.
Nature on Ken Ono’s work modeling and tuning the performance of Olympic swimmers (\(\mathbb{M}\), via; not very heavy on details).
Regular and semi-regular tessellations of origami flashers (\(\mathbb{M}\)). A flasher is a pattern that folds into a prismatic shape with a spiral wrapping around the sides of the prism. In this work from 8OSME, Nachat Jatusripitak and Manan Arya show that tessellated fold patterns can fold to tessellations by triangular, square, and hexagonal flashers.
Somehow I missed this when Antoine Chambert-Loir noted it last June (\(\mathbb{M}\)), but zbMATH is the inaugural winner of the Jean-Pierre Demailly Prize for Open Science in Mathematics, sponsored by the French societies for mathematics, applied mathematics, and statisti]cs (SMF, SMAI, and SFdS).
David Wood encourages combinatorists to send their papers to broad-focus mathematics journals, and when reviewing such submissions, to affirm that they are of broad interest to mathematics, to counter the widespread bias against combinatorics in such journals. See also Wood’s list of combinatorists in editorial positions in these journals, and of major journals that have no representation of combinatorics on their editorial boards.
Dan Drake collects pilgrimage sites for the mathematical tourist. That is, places where something important in the history of mathematics happened, not merely places where mathematical content can be found now.
Catherine Pfaff shows off some notes from her mathematics research, writing “When one isn’t sure of what’s actually true, sometimes it helps to just compute it all, or at least it feels good to do so.”
Peter Rowlett in New Scientist on connections between mathematics and poetry (\(\mathbb{M}\), archived).
Better Living Through Algorithms, by Naomi Kritzer in ClarkesWorld, just won the Hugo for short SF story (\(\mathbb{M}\), via), which put just enough pressure on me to actually read it (a little ironically given the story content; mild spoiler) and I’m happy I did.
Evolution of the notation for the empty set, with a cute story about André Weil impressing his daughter as the inventor of the \(\emptyset\) notation.
A binary number system from 1617: Napier’s location numerals (\(\mathbb{M}\)). Chris Staecker explains why these numbers never caught on: like a binary version of Roman numerals, they were simultaneously far ahead of and far behind their time. See also Staecker’s previous video on Napier’s binary multiplication board using this system.
Fold-and-cut hat and spectre tiles (\(\mathbb{M}\)). Dave Richeson used the straight skeleton method to construct the downloadable folding patterns for these, which led me to wonder how the kite tiling construction of the hat, the diamond-kite circle packing, and the circle-packing based fold-and-cut construction might play together.
My guess (not having worked it out in detail) is that there is some simple origami molecule that can fold all the kites of the plane tiling by kites into a half-plane, with their boundaries all on the same line segment; that you can then pop the kites of the hat onto the opposite side of the line segment, and that cutting nearly along the line segment (on the side away from the hat’s kites) will cut out a hat with a little bit of boundary fringe. But one might not find that boundary fringe part satisfactory.
Welcome to the Fold (\(\mathbb{M}\)). American Mathematical Society feature column by Sara Chari and Adriana Salerno on the mathematical properties of point systems constructed by paper folding.
In a range voting system, one can obviously vote truthfully, by assigning scores that represent one’s best estimate of a preference for each candidate. But, like other voting systems, range voting presents an opportunity for strategic voting, in which voters choose their scores based on how it affects the outcome, with the knowledge that this may not accurately reflect their actual preferences. Being strategic is not in any way illegal and cannot be prevented, but different voters may choose to vote truthfully or strategically according to their own motivations or understanding of the situation:
The goal for an individualist voter is to have the outcome reflect their own preferences. I think this is likely the case for most voters in most political elections. For such a voter, the rational choice is to vote strategically.
Collectivist voters have preferences, but their goal is to have the outcome reflect the overall preferences of the whole electorate rather than their own. A collectivist voter, who believes that other voters are also collectivist, should vote truthfully in order to more accurately assess the preferences of the electorate.
Norm-following voters may believe that systems like range voting are associated with certain norms of social behavior, for instance that one should vote truthfully regardless of one’s goal for the overall election. For these voters, following those norms is more important than obtaining their preferred goal.
Naive voters may see descriptions of the voting system as collecting their preferences, and vote truthfully without realizing that there is any alternative.
Effort-avoiding voters may vote truthfully because they think it is easier than voting strategically. Alternatively, they might assign high scores to candidates they like and low scores to candidates they dislike, because that is even easier.
The problem with range voting is not so much that these options are possible; they are possible with all systems. But unlike other voting systems, range voting makes it so blatantly obvious how to vote strategically that strategic voting can become commonplace, leaving the truthful voters at a severe disadvantage. Many of the sources used to advocate range voting discuss it as if most or all voters are likely to be truthful, but there is no inherent reason for honesty (unlike other systems where honesty has the advantage of being much easier) and my experience is that expecting truthfulness from most or all range voters is naive to the point of disingenuousness.
The obvious way to vote strategically in range voting is to assign a top score to all candidates for whom you want to increase the probability of winning, and a bottom score to all other candidates, avoiding any of the middle score values. But rather than recounting anecdotes I thought it might be more interesting to formulate a simple model where this obvious strategic choice can be proven to be optimal, and moreover where the optimal choice of which candidates to score up or down can be calculated directly.
We have to make some assumptions to get going, and the following are somewhat artificial, but I think not too strange. The intuitive idea is that it is only the relative score between the top two candidates that would matter in an election, if only the voter knew who those two candidates were going to be. One can define the regret of a strategic voter to be how much the voter would prefer to have adjusted the scores of these two candidates, in retrospect, and then ask for an outcome that minimizes this regret. Essentially the same thing is formulated below in positive rather than negative terms.
To normalize things, all voter scores are real numbers in the interval [0,1]. (It does not affect the argument whether these scores are arbitrary in this interval or limited to some discrete values.)
There is a fixed pool of candidates to vote for, among whom we will pick a single winner; the rest are losers.
Each voter has either a known preference ordering among all candidates.
Each voter has some information about the likely election outcomes, which we formulate as a probability distribution over which pair of candidates is likely to be the top two.
When a given pair of pair of candidates is the top two, each voter would prefer to have assigned their favorite of these two candidates a score of 1 and the other candidate a score of 0. We define the regret of the voter to be the total difference between these ideal scores and the voter’s actual scores.
The goal of each voter is to vote in a way that minimizes their expected regret.
Really the only important aspect of these assumptions is that the value a strategic voter wants to optimize (the expected regret) is a linear function of their chosen scores. For linear optimization, each score should be an extreme: 0 or 1. The regret-minimizing scores for a given voter are not necessarily monotonic in the voter’s preferences, though; for instance, if a voter prefers \(p > q > r > s\) but there is an overwhelming probability of seeing either \(pq\) or \(rs\) as the top two, then the voter should score 1 for \(p\) and \(r\), and 0 for \(q\) and \(s\).
There’s an easy method for a voter to determine which scores should be 0 and which should be 1, to achieve the minimum expected regret. For each candidate, sum over pairs of candidates to determine the probabilities that the candidate will be the favorite of the top two candidates, or the less-favorite of the top two candidates. If a candidate is more likely to be favorite than less-favorite, assign score 1; otherwise assign score 0.
So at least with these assumptions, in range voting, the strategic voters will always choose 0 and 1 for all their votes. One could try to put all the other voters on a more equal footing by telling them all to use 0 and 1. That is exactly what a different system, approval voting, does. Alternatively, one could require the strategic voters (and everyone else) to spread out their scores more evenly, by providing a regularly-spaced set of score values and requiring each value to be used only for a single candidate. That is exactly what a different system, the Borda count, does. But if we use range voting, and all voters are strategic, we get approval voting anyway, regardless of what we call it, and the only effect of calling it range voting is to give voters options that they will not use. And if we have a mix of strategic and truthful voters, the truthful voters are put at a disadvantage. Either way, range voting promises something different to the voters than what it delivers. If you think it’s the system you should use, choose Borda or approval instead. If you don’t like approval voting, then exactly the same disadvantages accrue to strategic range voting, because it ends up being the same as approval voting.
To be fair, there is one situation where I think range voting can work: where all participants are collectivist. In that situation, range voting collects more nuanced information than approval voting, exactly what the voters want to do. But although the assumption of collectivism might be true of some academic program committees and grant panels, I don’t think it’s true of most single-winner political elections.
Or, if you are strategic and think that your faction is going to be clever enough to vote strategically and your opponents are more likely to vote honestly, giving you an unfair advantage, then continue to advocate for range voting.
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