Flexible polyrhombs

September 11th, 2013 by munizao Leave a reply »

From time to time, a pentomino tiling still manages to surprise me.

pentomino tiling with 5-fold dihedral symmetry

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.)


  1. Bryce Herdt says:

    This got me thinking of other five-fold layouts. I didn’t get anywhere with it, though.

    The basic shape I was considering was five 3*3 rhombi, plus L-trirhombi stretching between the outer edges of neighboring nonorhombi. If I’ve described it well enough, a little thought will show that there are other tilings for this shape’s outline, achievable by inverting areas with symmetric outlines. (These can probably be only hexagons, O-trirhombi here, but it’s easier to skip steps and use larger polygons.) So you could probably tile a grid that has fivefold rotational symmetry and a tenfold reflectively-symmetric outline.

  2. Bryce Herdt says:

    …oh. I took a while to post that comment, and hadn’t seen the update.

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