An efficient iterative method for the Ammann-Beenker tiling

Iterative methods are often very fast and easy to program. But the Amman-Beenker tiling (see my earlier post) seems at first sight to be too complicated as a rhomb or a square has to be divided into more than 20 smaller squares and rhombs. Their sides are smaller by a factor of 3+2*sqrt(2) = 5.8 , which is huge. Programming this would just be too tedious. Fortunately, I found a much more efficient iteration, which probably already has been discovered by others. To get something new and more interesting I am using two different colors for each tile.

Division of a yellow rhomb.

Division of a brown rhomb.

I am using smaller squares, which I divide into two triangles. These triangles and rhombs serve as tiles. A triangle always has to fit together with an other triangle to form a square. This determines

Division of a violet triangle.

Division of a blue triangle.

how the rhombs and squares have to be divided into smaller rhombs and squares. The new tiles are smaller by a factor of 1+sqrt(2) = 2.4. Note that this is the square root of 3+2*sqrt(2). We are free to try different colorings. I am just giving a particular example. Here, the triangles mix up the colors. A blue triangle gives violet triangles and a violet triangle gives blue ones in an iteration. I am amazed, that such a seemingly asymmetric scheme gives such a symmetric result:

Amman-Beenker tiling with two-fold color symmetry in rotation resulting from the iterative method.

In this particular case, all squares are diagonally split into two colors. The rhombs and triangles of the previous iteration are plainly visible.

This iteration methods shows clearly, how lines of rhombs arise. Their orientations form one-dimensional quasiperiodic sequences.

One more iterative division of a blue triangle.

One more iterative division of a yellow rhomb.

A note added later: The orientation of the isosceles triangles is not explicitly given in the iteration scheme above. In order to have matching pairs of triangles we can only use one specific orientation. You can see this from the pictures of one more iteration as shown here. This iteration method, but without special colors, is published in Wikipedia. However, I could not understand it because of a confusion with the correct orientation of the triangles.

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