# Posts Tagged ‘flexible polyrhombs’

## 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.

## Some Contributed Solutions

October 2nd, 2013

I’ve had a few solutions sent in recently, so I wanted to share them with you all.

First, Abaroth noticed that my rhombic-cell pentomino tiling had just enough space to fill out into a five pointed star if the tetrominoes were also included:

But that was just the beginning! He then proceeded to produce an entire collection of tilings with these pieces, which he calls flexominoes. One problem that can come up in tilings of this sort is that if there is a vertex with three rhombi around it, a polyomino containing all three rhombi has an ambiguous identity, since there is more than one way to “unglue” the polyomino at that point. I contributed an ambiguity-free solution to one of the patterns Abaroth found:

Speaking of rhombuses, Abaroth has been investigating color-matching puzzles using rhombic tiles. His puzzle page has more interesting material on color matching puzzles and symmetrical polyhex tilings.

Next up, George Sicherman sent in a symmetrical tiling for the flexible tetrarhombs:

What’s interesting here is that although the outline of the tiling is symmetrical, the pattern of the cells isn’t. The lesson here is that being able to trade off some cell-level symmetry for more pattern-outline symmetry can give us a little variety in our choices of what we can tile.

Finally, Bryce Herdt provided a de Bruijn sequence of invertible length 5 binary words. (That is, a cyclic sequence where each word occurs once as a substring.) Since he did so in text format, I made a visualization:

## 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.)