2n-fold reflectively-symmetric tilings with unit length sides and inflation factor less than 2

This page describes a “Filling the Gaps” way to generate substitution tilings that have the following characteristics:

· Tiles have sides with unit length.

· Angles at vertices are multiples of pi/n, including pi (180 degrees).

· A unit thin rhomb is part of the set, with unit length sides and angles pi/n and (n-1)*pi/n.

· The inflation factor is thin rhomb’s long diagonal length = 2*cos(pi/2n) < 2.

· The inflation rules are reflectively and rotationally (180 degrees or in some cases 90 degrees) symmetric.

The goal is to find a tiling for each value of n.

First I show the results of the method.

Then I show early examples of such tilings. (I found these to be beautiful, and this motivated me to try and find a general method for other values of n.) Then the general method is presented that will work for any value of n (assuming it terminates in a finite set of tiles).

Results

Tilings for n=6 through n=10

	Filling-the-Gaps 12-fold tiling (n=6) 7 prototiles
	Filling-the-Gaps 14-fold tiling (n=7) 10 prototiles
	Filling-the-Gaps 16-fold tiling (n=8) 15 prototiles
	Filling-the-Gaps 18-fold tiling (n=9) 22 prototiles
	Filling-the-Gaps 20-fold tiling (n=10) 35 prototiles

Earlier Results

Other n-fold substitution tilings by others

Penrose tiling: The penrose tiling’s inflation factor is even smaller, the long diagonal of fat rhomb 2*cos(2pi/2*5). I tried looking at diagonals of other fat rhombs 2*cos(2pi/2n) for n odd but I don’t think larger n are possible, with symmetry constraints. Penrose is unique for 2*cos(2pi/2n) for odd n. I don’t think any others are possible for 2*cos(k*pi/2n) for k and n relatively prime.

Sub Rosa tiling by Kari and Rissanen: this creates rotationally and reflectively symmetric tilings with rhombs for any value of n, but the inflation factor is much larger than 2.

Harriss tiling, Binary (Lancon Billard) tiling (as described by Hibma): Hibma shows how each is derived with inflation factor < 2 for any value of n. But Harriss tiling is not rotationally symmetric, and Binary tiling is neither rotationally nor reflectively symmetric.

Stefan Pautze recently posted a paper that included the example for n=4 below, describing the technique in this method as “filling the gaps”. He documents other tilings with gaps filled for other values of n, but none of the others have inflation factor less than 2.

Earlier examples

	8-fold (n=4) mentioned by Stefan Pautze
	10-fold (n=5)
	12-fold (n=6). Dieter K. Steemann’s pinteret site attributes the n=6 tiling to Walton, though the original work is gone with the demise of yahoo groups. It was original at Yahoo true_tile group, Message #3286, 12-fold tiling, Dale Walton, 2016/Apr/01. Dieter mentioned that Peter Stampfli also described the tiling and also calls it “filling the gaps”.
	18-fold (n=9)

Goals of this particular method

Look for a method that works for any value of n.

While some differences between n even and n odd, I want a method that works similarly for all odd values of n, especially primes.

Thus this method excludes triangles as primitive shape where n is a multiple of 3 (replacing 12-fold and 18-fold tilings mentioned above with new tilings).

I’ll first present the method, then show the tilings for n=6 through n=10. But this order may be backwards. You may first want to see the tilings below before going through the method.

Method

Start with thin rhomb.

Repeat, until no new prototiles are generated:

A. Replace each edge with a thin rhomb. This creates an inner region with as many as twice as many sides, and corresponding inner angles less than the original by an amount pi/n.

In the picture above, the thin rhomb on the top left inflates to 4 rhombs plus interior yellow region on top right. Here the angle pi/7 goes to 0, shown by the adjacent red rhombs in the top right. The top angle 6pi/7 becomes 5pi/7 on the top right.

On the bottom left, the rhomb has angles 2pi/7 and 5 pi/7. Its inflation shows an orange region with angles pi/7 and 4pi/7. The orange region cannot be tiled by rhombs.

B. When inflating a shape, look at angles along the edges of the inner region.

a. If angle not gone to 0, leave it.

In the picture above, the tile on the left inflates as shown in the middle picture, with each edge replaced with the pi/10 thin rhomb. In this tiling, none of the angles went to 0. So the interior region, shown on the right in pink, becomes an additional prototile.

b. If angle goes to 0, then look to fill internally

i. Determine the midpoints of the long edges of the inflated pattern (could be considered corners). It may help to draw lines through the midpoints. See the second figure below. The top and bottom corner angles went to 0 in the inflation. The lines are drawn through the midpoints on the long corners.

ii. fill from the zero angle to the midpoints

1. To fill, start with single edge and place rhomb 2pi/n. Then next 2 edges, then next 4 edges, etc. until you get to the midpoint. In the third figure below, it starts with yellow, then green, then blue. Stop at blue, since it is placed in the middle, where the line is drawn.

2. For the regions to be symmetric, you need to fill the cracks. See the fourth figure above, where a thin rhomb is inserged between the green and blue.

There is a pattern to the cracks, as each is an iterated inflation of the thin rhomb without trim.

3. You need to pre-fill the crack at the inner region when getting to the midpoint. See the last figure above, where rhombs are added at the border of the inner region. The rhomb pattern must be symmetric.

4. For n odd, shown above, you will fill in only either the horizontal or the vertical direction.

5. For n even, you will fill in both the horizontal and the vertical directions.

In the inflation above, multiple points on each edge went to 0. In the second figure, lines are at the middle of the long corners. The third picture shows the inner region filled in all directions with yellow, then green, then red. The fourth figure shows cracks are filled, and center gap is prefilled.

iii. For each direction you are filling (one for n odd, two for n even)

1. Look further in the same direction for other sections that are as long as the half the long edges (to midpoint). Where those sections dip, it helps to draw lines. Continuing the previous example, the second picture below shows the additional lines drawn, and the long edges identified with dark rhombs.

2. At the lines, pre-fill the cracks in a way to create symmetric inner regions. This is show in the third picture below.

The following example shows the process for n even, where the same process is done from all 4 sides.

The second picture shows an interesting case of three points on each of 4 sides with angles going to 0. Tiles were filled up to the midpoints (identified by the lines) and cracks were filled. The third picture shows the additional lines included where edge dips from a peak to a valley. The fourth picture shows the cracks pre-filled from all 4 directions.

3. The remaining regions are prototiles to be added to the set, if not already. This is shown in the last pictures above. In the n odd case above, one of the regions is the original (purple) tile that was inflated. The new region in light purple is added to the set of prototiles.

4. A special case is when the drop happens at a point from both directions. This happens for example when one of the additional lines is in the middle of the inner region. In this case, pre-fill the cracks in BOTH directions to make symmetric regions.

The second picture above shows the inflation with pieces added, along with the cracks filled. The third picture shows a line in the middle where the edge has a valley, a drop from high to low from both directions. The fourth picture shows that the cracks are pre-filled to the middle line from both directions. The resulting regions are symmetric.

The last green region also shows how the larger prototiles tend towards a rectangular shape with jagged sides.

The trivial cases of tessellations arising from 4-fold (n=2) and 6-fold (n=3) are consistent with this method. Each has one prototile.

Tilings for larger values of n

I feel I am at the limits of what inkscape can do. Sometimes snapping does not work, and I don’t like placing by hand, especially when I try to reuse parts when creating similar inflations. (Also I suspect for n=11 there will be at least 46 shapes, and finding that many colors for the tiles is challenging, though not the biggest challenge.)

Conclusion

The method produces a single tiling for each value of n—assuming it terminates. I cannot prove yet that repeated tile generation with this method will find a finite set for every value of n.

When starting on this “filling the gaps” tiling, I was expecting, or at least hoping, that the prototiles would be smaller in size. Previous results show n-polygons, n-polygrams (stars), bowties/hourglasses, and the other rhombs (at most 2 rhombs are used for each value of n in this tiling). With this method, the tiles grow to be quite big. For example, the number of sides of the largest tile for n=8 is 96 sides. This number keeps growing for higher values of n.

I was also hoping that the number of prototiles would be smaller in number. It seems the number of tiles grows exponentially as 2^n/2 (at least to n=9, seems slightly more at n=10 and maybe beyond).

I think further work to develop other sets of tiles could improve on this set.

Go back to Gallery of substitution tilings.