# Posts Tagged ‘flexible polyforms’

## Flexible pentominoes on rhombic polyhedra

February 26th, 2018

If you subdivide the faces of a rhombic triacontahedron into 2×2 grids, you can tile the polyhedron with two copies of each pentomino. One way of looking at this figure is as a tiling of the projective hemi-rhombic triacontahedron. The projective (also known as abstract) polyhedra can be formed by identifying the opposite faces of certain polyhedra with each other. So the projective hemi-cube has three square faces, and the projective hemi-rhombic triacontahedron has 15 rhombic faces. Stitching together the opposite sides of the unshaded area in the figure is a way to form this 15 face “polyhedron”.

I came up with that one a couple of years ago, but I neglected to put up a blog post because I didn’t like the graphic enough. I suspect that it’d look really cool if the lines of the rhombic triacontahedron were properly projected onto a flat disk, but I don’t have the expertise to make that happen. I finally decided that it was worth sharing even if it doesn’t look as cool as it could.

Below is another tiling of subdivided rhombi. The significance of this figure is that four copies could be used to cover a rhombic hexecontahedron. ## Stop! In the name of Octagons!

March 29th, 2014

In the spirit of the flexible rhombic cell polyominoes that I posted about previously, here’s a hexiamond tiling of eight triangular segments squashed into an octagon: Of course, octagons can also be used as base cells for polyforms. In fact, any regular polygon (and quite a lot of other things) can be used in this way, but octagons are special. They don’t tessellate the plane by themselves like equilateral triangles, squares, and hexagons, but they do form a semi-regular tessellation of the plane along with squares. This makes polyocts behave fairly well; you won’t be able to tile something convex and hole-free with them, but you can tile something that’s reasonably symmetrical at least. For example, here’s a tiling of the 1-, 2-, 3-, and 4-octs: That’s not the most symmetry that polyocts are capable of, (full octagonal symmetry is possible) but it’s the most we’re going to get with this set of pieces. See this page by George Sicherman for some figures with full symmetry that can be tiled with various individual polyocts.

## Flexible polyrhombs

September 11th, 2013

From time to time, a pentomino tiling still manages to surprise me. Normally, the largest number of symmetries you can make a pentomino tiling have is eight, the number of symmetries of the square. For example, we can tile an 8×8 square with the corner cells removed. (If we leave the plane for other topologies like cylinders and tori we can get more.) However, it’s a basic principle of flexible polyform tilings that we can generally try to squeeze one more repeated unit around the center of a rotationally symmetric tiling. So I did.

But my starting point for this was not the pentominoes. In my Hinged Polyform post, I discussed polyforms where the connections between pieces could occur at arbitrary angles. Conversely, we could look at polyforms where the shapes of the individual cells could contain arbitrary angles. If we assume that all of the edges are the same length, there is only one possible triangle, so polyiamonds in this scheme aren’t interesting. Rhombuses are the simplest example of cells where the angles can vary. Since they have only one degree of freedom per cell, they are reasonably tractable. The flexible tetrarhombs include flexible versions of the five tetrominoes in addition to the three shapes in blue below: The flexible pentarhombs include the flexible pentominoes, the 18 forms that can be derived by adding a single red cell to one of the blue forms above, and the six additional forms in green. (I may well have missed some.)

It may be possible to find a good flexible tetrarhomb tiling, but I haven’t yet managed it. And 36 pentarhombs is too many for me to handle. If only there were some subset with a better number of pieces for a puzzle, something like the 12 pentominoes.

Oh. Right. (And that was more or less the thought process that led to the tiling above.)

## Hinged Polyforms

August 30th, 2013

Here’s a tiling of the nine hinged tetriamonds: Hinged polyforms meet at corners rather than edges as in regular polyforms. The corner connections, like hinges, are flexible: two hinged polyforms are equivalent if it is possible to turn one into the other by swinging the hinges at the vertices, in addition to rotating and reflecting the whole pieces. Hinge angles that cause two cells to lie flat against each other are disallowed, as it isn’t possible to visually distinguish which side of the edge has the hinge. In some cases, hinges may be “locked”, with angles that are completely determined by the geometry of the piece. (For instance, when three cells meet around an equilateral triangle.)

With the above piece set, it is possible to realize all of the pieces using a small set of angles for the individual triangles. Other sets may be trickier to work with.

Here are the hinged tetrominoes: Can a symmetrical tiling be found for these? Problem #43: Find one.

## Constellations

August 22nd, 2013

I made a presentation on flexible polyforms at the last Gathering for Gardner, but there were some polyform types that I didn’t get to, since I hadn’t yet come up with any good problems for them. One odd sort of polyform, which I am fancifully calling a constellation, can be obtained from configurations of points on the plane. We can consider two sets of points on the plane to be distinct if the pattern of collinearity among the points is different. Because every pair of points defines a line, the lines with only two points are, in a sense, not interesting; only the lines with three or more points need to be considered when determining whether two constellations differ. It seems reasonable to consider the order of points on a line to be significant; this gives us three different 5-point constellations with a pair of three point lines that meet at a point. There are 7 5-point constellations in all. Here’s the first tiling puzzle solution I found for them: One rule for constellation tiling puzzles that I like is to disallow any point from one constellation from falling directly between two points in another constellation. This keeps the constellations more compact, and adds a little challenge to the puzzle. I like to get as much symmetry as possible in one of these flexible polyform tilings, so I decided to try for one with 7-fold symmetry. This was a little harder, but eventually I found the following tiling: April 2018: Edited to update the image to a proper solution. Thanks to Bryce Herdt for noticing that the old solution was incorrect.

Where can we go from here? If I’ve counted right, there are 21 6-constellations. Of these, 7 can be formed by adding one independent point (a point on no line of 3) to each of the 5-constellations. The full set seems a little too big to solve by hand, but if we exclude the ones with independent points, a puzzle with the remaining 14 seems more manageable. (We may also want to exclude the 6-constellation with two separate lines of three points. With that one excluded, the remaining 13 6-constellations all can be formed from connected groups of lines with 3 or more points.) The 14 6-constellations with no independent points.

Problem #42: Find a tiling of 6-constellations with 6-fold dihedral symmetry. Either the set of 13 or the set of 14 will do. Even more symmetry is even better.

## Polycircles

January 23rd, 2012

A while back, (before I started this blog) I was exploring polyforms using unit-radius circles as their base cell type, which I called “polypennies”. We can think of these as “flexible” polyforms: since connections between the circles can occur at arbitrary angles, we consider two polypennies to be equivalent if we can continuously move the circles around each other without changing which circles are adjacent. (As with other polyform types, rotations and reflections are also considered equivalent.) The pentapennies

I called these polyforms “polypennies” rather than “polycircles” because “pennies” captured the equal size of the cells. (ETA: I forgot that I raided the word from the term “penny graph,” which has been used as an alternative to “unit coin graph” to describe the adjacency graph associated with a particular configuration of non-overlapping unit radius circles.) I also knew that eventually I would want to get to polyforms made of circles of arbitrary size, for which I was reserving the term “polycircle”. Well, it happens that I’ve been invited to Gathering for Gardner 10, where I plan to give a talk on flexible polyforms, so eventually is now.

For polycircles with cells of arbitrary size, another dimension of flexibility is required. Two polycircles are equivalent if they can be made congruent by continuously expanding or shrinking the circles without changing adjacencies, in addition to applying the transformations allowed with polypennies. This extra flexibility means that, in addition to the polycircles that are equivalent to polypennies, there are some polycircles that could only be formed by placing circles into spaces where they wouldn’t fit if all of the circles were forced to be the same size.

As with other flexible polyforms, elegant tiling puzzles for the polycircles can be produced by attempting to maximize the symmetry of the configuration to be tiled. Here’s an example, with fourfold rotational symmetry, of a tiling puzzle containing all of the polycircles of order 1 through 4: This was not a hard puzzle to solve, once I came up with a configuration to tile that would work. Adding smaller pieces is a time-honored trick for making polyform puzzles easier; I put in the 1- through 3-circles because I was failing to make any headway with the 4-circles alone. The extra dimension of flexibility was helpful in that one can generally resize the circles to touch more neighbors than is possible in polypenny puzzles, which tend to end up with a number of cells with only two neighbors. On the other hand, the 4-circles with a circle inside the gap between three others in a triangle were trickier to deal with than any of the 4-circles that are equivalent to 4-pennies.

Can we do better than the above? I think fivefold symmetry may be possible.

Problem #26: Find and solve a tiling puzzle for the 1-, 2-, 3-, and 4-circles with fivefold rotational symmetry.