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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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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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Every canonical projection tiling (this means, tilings built with the cut-and-project method where the window is taken to be the projection of the unit square) would produce a tiling by parallelograms.

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

canonical projection tiling from dimensions 6,7,8,10,12,14

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

Paralellogram tilings

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  • $\begingroup$ 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. $\endgroup$
    – Adam Hughes
    Sep 1, 2014 at 15:07
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    $\begingroup$ raiz(y) means sqrt(y) $\endgroup$
    – Melquíades Ochoa
    May 14, 2016 at 18:45

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