Research Article: On the Penrose and Taylor–Socolar hexagonal tilings

Date Published: May 01, 2017

Publisher: International Union of Crystallography

Author(s): Jeong-Yup Lee, Robert V. Moody.

http://doi.org/10.1107/S2053273317003576

Abstract

A uniform geometric/algebraic approach to the Penrose and Taylor–Socolar hexagonal tilings is given, which clarifies their construction and the way in which they are inter-related.

Partial Text

From the very beginning, aperiodic tilings have played a significant role in unraveling the mysteries of aperiodic crystals. Knowing what is mathematically possible has often turned out to be a crucial element in conceiving what might be physically realizable. In this paper we discuss two remarkable aperiodic tilings of the plane that are built out of one of the most basic of all crystallographic structures: the standard periodic hexagonal lattice.

An arrowed hexagon is a regular hexagon in which each side has been given a direction, indicated by an arrowhead. An arrowed hexagon is called well arrowed if, up to rotation, the arrows form the pattern shown on the right side of Fig. 1 ▸. In fact, all three hexagons in this figure are well arrowed. The structure of the well arrowed hexagon gives it a well defined orientation in the plane, namely that provided by the two parallel arrows facing in the same direction.

There are other decorations of well arrowed hexagons and double hexagon tiles that are equivalent representations of the arrowing but help to make the underlying geometry of the tilings more transparent. The first of these is the marking of well arrowed hexagons shown in Fig. 3 ▸, which replaces the arrows of a well arrowed hexagon with a black stripe and two black corner markings (see Socolar & Taylor, 2011 ▸). Initially, we will use this representation of the arrowing with the small hexagons, and later for the outer hexagons.

Let us continue with a non-singular legal double hexagon tiling , in which we have completed its small hexagons to a well matched hexagonal tiling and then resolved everything into triangles by decorating each of the small hexagons.

If we start with a non-singular legal double hexagon tiling then we obtain a tiling of the plane with the small hexagons. The centers of these hexagons form a triangular lattice of the plane composed of level 0 (side length 1) equilateral triangles, as we have seen. For definiteness we now specify this lattice as a set of points in , namely the set of points , where , (Fig. 14 ▸). Joining nearest neighbors of Q produces the triangular lattice of level 0 triangles, indicated by the thin lines in Fig. 15 ▸.

At this point it is rather clear that given any triangulation and any choice of one coset leading to it ought to lead to a legal double hexagon tiling. Here are the details. We will assume that both triangulations are non-singular. This guarantees that there are no infinite edges, we get proper triangulations, and they are nested. This nesting can be geometrically manifested by laterally shifting the edges as indicated by the nesting. The Voronoi cells of the lattices (nearest neighbor cells) are hexagons centered, respectively, on the points of and . We know that every hexagon from has a triangle edge passing through it and this edge will be shifted laterally in nesting. This is shown in Fig. 18 ▸. The hexagon is made into a well arrowed hexagon by placing the pair of parallel arrows in the direction of the shift. The small hexagons now make a well arrowed and well matched hexagon tiling.

By definition, a Penrose tiling is precisely a legal double hexagon tiling. Taylor–Socolar tilings (T–S tilings) are usually defined by the T–S tiles shown in Fig. 20 ▸ and they are assembled as regular hexagonal tilings, but under the matching rules:

The purpose of this paper has been to clarify the unity that exists between the Taylor–Socolar tilings and the Penrose hexagonal tilings – a unity that can be expressed both geometrically and algebraically in terms of double hexagon tiles. Each non-singular legal double hexagon tiling encompasses both a Penrose tiling and a T–S tiling, and this pairing can be interpreted algebraically in terms of (2). Each of the two hexagonal tilings leads to a nested triangulation, and these two are bound together by the simple rule that triangle edges of each right-bisect edges of the other.

 

Source:

http://doi.org/10.1107/S2053273317003576

 

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