# Substitution tilings with parallelograms

I'm looking for a substitution tiling made with parallelograms, that is, a tiling of the plane with parallelograms (which do not have to be of the same shape) such that we can take one parallelogram and replicate the tiling inside it in a smaller scale.

I seek a decomposition of a given parallelogram into a set of smaller parallelograms that is not the obvious subdivision into four equal subparallelograms using midpoint subdivision.

I've searched the Tilings Encyclopedia with no luck. Does anyone know of such a tiling?

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if rhombus is ok you could skew a mathworld.wolfram.com/PerfectRectangle.html – ryu jin Jan 11 '13 at 1:14

You'll get something similar to a rectangle or square tiling. If the outer parallelogram leans to the left, then all the parallelograms will lean to the left, with sides parallel to the outer sides.

To see this, for a right-leaning parallelogram, fill in the small angle at the top left. That will create a new small angle at the top. Filling it is forced, and eventually the top area is filled.

So ... this is equivalent to rectangle tilings.

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By example this. This substitution tile the plane. I can't think in most interesting tilings.

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You should translate and explain a bit as well: not everyone on this site speaks Spanish. Always endeavor to make every post available to as many people as possible. – Adam Hughes Sep 1 '14 at 15:07
raiz(y) means sqrt(y) – Melquíades Ochoa May 14 at 18:45

Every canonical projection tiling would produce a tiling by parallelograms.

Harriss & Lamb studied substitution rules for such tilings. These includes both Penrose rhombs and Ammann-Beenker tilings

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