## Abstract

We study a family of substitution tilings with similar right triangles of two sizes which is obtained using the substitution rule introduced in Danzer and van Ophuysen (Res. Bull. Panjab Univ. Sci. **50**(1–4), 137–175 (2000)). In that paper, it is proved this family of tilings can be obtained from a local rule using decorated tiles. That is, that this family is *sofic*. In the present paper, we provide an alternative proof of this fact. We use more decorated tiles than Danzer and van Ophuysen (22 in place of 10). However, our decoration of supertiles is more intuitive and our local rule is simpler.

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## Notes

Here are more details. Consider the tree whose vertices are correct decorations of fragments of the form \(\{F_1,F_2,\dots , F_i\}\). Edges connect a decoration of a fragment \(\{F_1,F_2,\dots , F_i\}\) to a decoration of the fragment \(\{F_1,F_2,\dots , F_i, F_{i+1}\}\) whenever the decorations are consistent. This tree has arbitrary long branches. Any vertex of the tree has finitely many neighbors. By Kőnig’s lemma [9], the tree has an infinite branch, which provides a correct decoration of the entire tiling.

Goodman-Strauss formulates property P2, as “the local rule enforces the hierarchical structure associated with \(\sigma \)”, which means that every tiling satisfying the local rule can be uniquely partitioned into supertiles of order

*n*for each*n*.The name “golden” in a similar context was used to call isosceles triangles whose all angles are integer multiples of \(36^{\circ }\) (Robinson triangles). To avoid confusion, we add the attribute “right”.

This ignoring is important, as there are colored large tiles which possess stars that cannot be extended to legal stars.

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## Acknowledgements

The author is grateful to Daria Pchelina and Alexander Kozachinskii for verifying the proofs and reading the preliminary version of the paper, and to anonymous referees for valuable comments.

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The article is supported by the Russian Science Foundation (20-11-20203).

## Appendices

### Appendix A: Proof of Claim 5.1

(a) For the sake of the contradiction assume that the patch in Fig. 24 (a) occurs in an L-tiling. Look at the vertex colored black. A quick look at the list of legal stars reveals that only the star \(C_{3}\) (shown in Fig. 24 (b)) fits for the patch in that vertex. Adding that star and its neighborhood, we obtain the patch shown in Fig. 24 (d), where the added tiles are colored light-gray.

Now look at the blue vertex on the bottom left. In that vertex only the stars \(C_3\) and \(C_4\) fit. Those stars are shown in Fig. 24 (c). The star \(C_3\) has non-matching orientation of the green arrow, hence \(C_3\) cannot be the star within *T* in the blue vertex. Therefore, it is the star \(C_4\). Adding the star \(C_4\) and its neighborhood, we obtain the tiling shown in Fig. 25 (a).

A small search reveals that only the star \(C_3\) fits for that patch in the brown vertex on the top left. Figure 25 (b) shows the tiling that is obtained by adding the star \(C_3\) and its neighborhood. Thus we conclude that the axis of the initial star and its extension to the right must be directed from the right to the left (yellow arrows in Fig. 25 (c)). Now look at the beginning of this sequence of arrows (the red point on the top right in Fig. 25 (c)). Only the stars shown in Fig. 25 (d) fit for the resulting patch in the red vertex. However none of them can be there, since all they have non-matching orientation of the horizontal yellow arrow.

(b) For the sake of contradiction assume that an L-tiling *T* includes the patch shown in Fig. 26 (a). Our plan is the following. We first show that, in addition to all tiles in the patch in Fig. 26 (a), the tiling *T* must contain all the tiles shown in Fig. 26 (b).

Assume that this is done. It is easy to verify that only the stars \(C_5,C_6,C_7\) (shown in Fig. 26 (c)) fit for the patch in the rightmost brown vertex. In all three cases the axis of the initial star (the horizontal black line) must be directed rightwards. However, similar arguments applied to the leftmost brown vertex show that that axis must be directed leftwards, which is a contradiction. So we have to show that the tiling *T* must contain all the tiles shown in Fig. 26 (b). To this end, let us go back to Fig. 26 (a). We copied that patch in Fig. 27 (a).

Only the star \(C_3\) fits for this patch in the blue vertex. Figure 27 (b) shows the patch that is obtained by adding the star \(C_3\) and its neighborhood. Look now at the yellow vertex on the bottom. Only the stars \(C_4,C_5,C_6,C_7\) fit for the patch in that vertex, they are shown in Fig. 27 (c). Note that the stars \(C_4,C_6\) (from the left column) have non-matching orientation of the vertical blue arrow, which must direct downwards, hence cannot be there. We will consider the remaining two cases separately.

*Case 1: The star * \(C_7\) *is in the yellow vertex in Fig.* 27 (b). In this case we are able to derive a contradiction quite easily. Adding the star \(C_7\) and its neighborhood we obtain the patch shown in Fig. 28 (a).

Only the stars \(C_3,C_4\) (shown in Fig. 28 (b)) fit in the blue vertex. Adding the star \(C_4\), we obtain the patch shown in Fig. 28 (c). We can see that no legal star fits in the green vertex on the right. Hence only the star \(C_3\) can be in the blue vertex. Adding that star and its neighborhood, we obtain the patch shown in Fig. 29 (b).

Only the stars \(C_3,C_4\) (shown in Fig. 29 (c)) fit for the resulting patch in the green vertex on the right. However the star \(C_3\) has non-matching orientation of the green arrow, hence the star centered at the green vertex is \(C_4\). Adding it and its neighborhood, we obtain the patch in Fig. 30 (a).

Only the stars \(C_4,C_5, C_6, C_7\) (shown in Fig. 30 (b)) fit for the patch in the brown vertex on the top right. However all those stars have non-matching orientation of the yellow arrow in the lower half of the star. Thus we have derived a contradiction in the first case.

*Case 2: The star centered at yellow vertex in Fig.* 27 (b) *is* \(C_5\). In this case we need a more involved analysis. Let us go back to Fig. 27 (b) and add the star \(C_5\) and its neighborhood in the yellow vertex. We obtain the patch shown in Fig. 31 (b).

Which stars fit for the patch in the leftmost vertex (colored in red)? These are the stars \(C_6,C_7\), which are shown in Fig. 31(a). Assume first that it is \(C_6\). Adding the neighborhood of the star \(C_6\) centered at the red point, we get the patch shown in Fig. 31 (c). If it is \(C_7\), we get a patch that differs from this one in orientation and labels of some sides. This difference does not matter and therefore we will consider only the case of \(C_6\).

Look at the vertex colored black (on the top left). Only the star \(C_4\) fits for the patch in that vertex. Adding \(C_4\) and its neighborhood in the black vertex we obtain the patch shown in Fig. 32 (a).

We have shown that our tiling *T* includes all the tiles from Fig. 26 (b) except for both triangles labeled by letter *B* and one triangle labeled by letter *A*. We have also shown that the tiling *T* includes the image of the initial star under the inversion through the white point. Via central symmetrical arguments we can prove that *T* includes also the other triangle labeled by letter *A* in Fig. 26 (b). It remains to show that *T* includes both triangles labeled by *B*.

To this end look at the blue vertex on the right in Fig. 32 (a). Only the stars \(C_3\) and \(C_4\), shown in Fig. 32 (b), fit for the patch in that vertex. If the star centered at the blue vertex is \(C_3\), we obtain the patch shown in Fig. 32 (c), and no legal star fits for it in the green vertex on the right.

In the remaining case the star centered at the blue vertex is \(C_4\), and we get the patch shown in Fig. 33 (a).

Adding its neighborhood, we get the patch that includes the sought triangle *B* (see Fig. 33 (b)). Via central symmetrical arguments we can prove that *T* includes also the other triangle labeled by letter *B*. We have reached our goal: we have proved that the tiling *T* includes the patch shown in Fig. 26 (b).

(c) For the sake of contradiction assume that an L-tiling includes the patch shown in Fig. 34 (a). First add the neighborhood of the star \(C_2\) centered at the blue vertex, we obtain Fig. 34 (b). Only the stars \(C_6\) and \(C_7\), shown on Fig. 34 (c), fit for the resulting patch in the yellow vertex (on the bottom). However the star \(C_7\) cannot be there, since its vertical green arrow has the non-matching orientation. Hence the star of the yellow vertex is \(C_6\). Fig. 34 (d) shows the patch which is obtained by adding that star and its neighborhood. Now look at the red vertex on the right. Only the stars \(C_5,C_6,C_7\) (Fig. 34 (e)) fit for the patch in that vertex. In all the three cases there is a horizontal blue arrow that goes into the red vertex. That arrow lies on the axis of the initial star. Now we know the color and orientation of that axis (see Fig. 35 (a)).

We can find now the stars in the leftmost and the bottommost vertices (both are colored blue). Indeed, only the stars \(C_5,C_6,C_7\) fit for the patch in the bottommost vertex, and two latter stars fit in two ways (see Fig. 35 (b)). In four cases the vertical arrow has the red (and not green) color. Hence those cases are impossible and only the lower star \(C_6\) can stand in the bottommost blue vertex.

A similar situation occurs in the leftmost blue vertex: the stars \(C_3,C_4, C_5,C_6,C_7\) fit for the patch there (two of them fit in two ways, see Fig. 36 (a)). However only one star, the lower \(C_7\), can have the matching color (blue) of the horizontal arrow. Adding to the patch the stars in the blue vertices and adding then their neighborhoods, we obtain the patch shown in Fig. 37 (b).

Only the star \(C_4\) fits for that patch in the yellow vertex, that star is shown in Fig. 37 (a). Now it is obvious that neither of the legal stars can stand in the adjacent black vertex, since the yellow arrow that goes out the yellow vertex into the black one cannot change its color to blue in the black vertex. We have considered all the cases. The claim is proved.

### Appendix B: To Cut Out of Paper

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Vereshchagin, N. A Family of Non-Periodic Tilings of the Plane by Right Golden Triangles.
*Discrete Comput Geom* **68**, 188–217 (2022). https://doi.org/10.1007/s00454-021-00367-4

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DOI: https://doi.org/10.1007/s00454-021-00367-4