Penrose Tilings

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Table of contents
1 What is a Penrose Tiling?

1.1 Adjacency Rules
1.2 Properties of Penrose Tilings

2 Determining the Index Sequence

2.1 Main Idea
2.2 Details

3 Forming a Groupoid

3.1 Defining the Equivalenece Relation, Rp
3.2 Defining the related groupoid, G

4 Works Cited
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What is a Penrose Tiling?

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Penrose tilings tile the plane with just two types of triangular tiles, one with angles of 36, 36, 108 degrees, and one with angles of 72, 72, 36 degrees. To form a Penrose tiling, the two types of isosceles triangles must be joined according to a set of adjacency rules.

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Adjacency Rules

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Description
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Description

Two different sets of adjacency rules can be used to build a Penrose tiling with the aforementioned two triangles. The first is the kite and dart method, and the second is the rhombi method. Remember that these adjacency rules ensure an aperiodic tiling but do not determine the tiling; there are uncountably many distinct Penrose tilings.

1. Kite and Dart

The second picture on the right illustrates the adjacency rules for a kite and dart tiling. The vertices of the individual tiles must be matched up according to color (pink to pink and purple to purple) so that in the infinite tiling each vertex is a single color. The top picture illustrates another set of matching rules that will yield this tiling.

2. Rhombi

The third picture illustrartes the adjacency rules for a tiling by rhombi. Just as in the kite and dart tiling, the vertices of the individual tiles must be matched up according to color so that in the infinite tiling each vertex is a single color.

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Properties of Penrose Tilings

1. If arranged properly, a Penrose tiling is an aperiodic tiling, meaning that the tiling does not possess translational symmetry. (If the triangular tiles are arranged without following the adjacency rules, a periodic tiling may result. This is not a Penrose tiling.)

2. The ratio of the edges is φ (the golden ratio) for both triangles used to construct a Penrose tiling. (One triangle is composed of two long edges and one short one, and the other triangle is composed of two short edges and one long one.)

3. Penrose tilings display instances of five-fold symmetry.

4. Any finite pattern in a Penrose tiling is repeated infinitely many times in EVERY infinite Penrose tiling.

5. We may associate with a tiling a sequence of zeros and ones where every one is followed by a zero, called the index sequence, {xn}. While the index sequence is not unique, one can recover a Penrose tiling from the index sequence that will be identical to the original tiling, except perhaps rotated.

6. There are uncountably many Penrose tilings. (Further down this page you can find information about the index sequence that will help you see why.)

7. These sequences can be compared using an equivalence relation, Rp.

8. This equivalence relation yields a groupoid, G.

Information from:Groupoids, Inverse Semigroups, and their Operator Algebras, Alan L.T. Patereson.

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Determining the Index Sequence

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Main Idea

To form an index sequence of a Penrose Tiling, T, we will alter T by combining adjacent tiles to make larger tiles of the same shape. This is done by combining one acute triangle and one obtuse triangle to form a larger obtuse triangle, and combining two acute triangles and one obtuse triangle to form a larger acute triangle. The new triangles will be similar to the original ones, and the result will be another infinite Penrose Tiling with the tiles scaled by a factor of φ. (The details of this process will be stated later.) As this process is iterated to infinity, the sequence, XP, is determined at each step by the kind of triangle that contains P. A tiling can then be reconstructed from XP that will be identical to the original tiling, except that it will be rotated based on the orientation of the starting triangle.

For more information, see: Groupoids, Inverse Semigroups, and their Operator Algebras, by Alan L.T. Paterson.

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Details

(This is the method presented in Paterson, 165.) Consider a tiling that is made out of kites and darts. In this tiling, the long edges have length φ, and the short edges have length 1. Divide the two types of tiles down the middle, splitting each into two triangles. The dart will decompose into two obtuse triangles (with long edge of length φ, and short edges of length 1), and the dart will decompose into two acute triangles (with long edges of length φ, and short edge of length 1). Note here that φ denotes the "golden number": \phi=(1+\sqrt{5})/2=1.618\dots.


Now, we are going to add some labels. Let this tiling be T1, let the smaller triangles (the obtuse ones, here) be S1, and let the larger triangles (the acute ones) be L1. Let P be a point in the plane tiled by T1. If P is inside an element of L1, the first number of the sequence is 0, and if it lies in an element of S1 the first number of the sequence is 1.


Next, we are going to form larger obtuse triangles by combining one obtuse triangle and one acute triangle using the following procedure. In each case where an acute triangle and an obtuse triangle share a short edge (as in the picture to the left), delete that edge. This will give a new tiling, T2 with a set S2 that consists of all of the acute triangles (the small ones in this new tiling) and a set L2 consisting of all obtuse triangles in T2. This new tiling will be a tiling by rhombi. Look at the diagram to the right to convince yourself of this. To find the next number in the index sequence for T, consider which type of triangle P is inside of in tiling T2. If P is in an element of L2, then the next number in the sequence is 0, and if it lies in an element of S2 the next number of the sequence is 1.


Now, we will do a similar process to form larger acute triangles. Whenever the short edge of an obtuse triangle (the larger kind in the previous tiling) corresponds to the long edge of an acute triangle (the smaller kind in the previous tiling), delete the edge. Now, we have a new tiling T3 with large (acute) triangles forming the set L3, and small (obtuse) triangles forming the set S3. T3 is a kite and dart tiling.


We can generalize this process as follows. If T is a tiling by rhombi, follow the above process to form a kite and dart tiling. Instead of labeling this tiling T2, label it T1. If the original tiling is a kite and dart tiling, label it T1. Now, we produce tiling T2n from tiling T2n − 1 by combining triangles of S2n − 1 and triangles of L2n − 1 whenever they share a short edge. Similarly, we produce tiling T2n + 1 from tiling T2n by combining a triangle of S2n with a triangle of L2n whenever they share an edge that is a long edge of the triangle in S2n and the short edge of the triangle in L2n.

(Paterson, 165)

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Forming a Groupoid

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Defining the Equivalenece Relation, Rp

We define an equivalence relation, Rp on the set Xp of index sequences of Penrose tilings. (Note that Xp is the set of all sequences of zeros and ones, where a one is always followed by a zero. This must be the case, because in any tiling, the small triangles will be combined with large triangles to form the next tiling.) Let \{x_{n}\} \sim \{y_{n}\} \Leftrightarrow \exists m \mid x_{n} = y_{n} \,\forall n \geq m. More simply, two sequences are in the same equivalence class if they eventually coincide. This makes sense intuitively, because the choice of P is arbitrary, but as the sequence of tilings progresses, two different choices of P will eventually fall within the same triangle, and the sequences will be exactly the same from that point on. So, we have that the quotient space Xp / Rp is the set of all possible Penrose tilings, thereby making the choice of P insignificant. An interesting fact is that there are uncountably many distinct Penrose tilings. (You can figure out why from the information already presented here.)

(Paterson, 35, 165)

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Defining the related groupoid, G

From the equivalence relation Rp, mentioned above, we use the standard method of defining a groupoid, which I will demonstrate:

G=\{(x,y)\mid x\in X_{p},y\in X_{p},x\sim y\}.

We also define the set of composable pairs: G^{(2)}=\{ ((x,y),(y,z)) \mid (x,y) \in G, (y,z)\in G\}.

  • Multiplication: For (x,y),(y,z)\in G, we define (x,y)\circ (y,z) = (x,z).

(x,y),(y,z)\in G \Rightarrow x\sim y and y\sim z. Since Rp is transitive, x\sim z, so (x,z)\in G.

  • Associativity:

((x,y)\circ (y,z)) \circ (z,w)=(x,z)\circ (z,w)= (x,w)

(x,y)\circ((y,z)\circ(z,w))=(x,y)\circ (y,w)=(x,w)

  • Identity Section:

Let \alpha (x,y)=x, \beta (x,y)=y \,\forall (x,y)\in G.

Define \epsilon: X_{p} \rightarrow G \mid \epsilon (x)=(x,x).

We have that \epsilon (\alpha (x,y))\circ (x,y)=\epsilon (x)\circ(x,y)= (x,x)\circ (x,y)=(x,y),

and (x,y)\circ \epsilon (\beta (x,y))=(x,y)\circ\epsilon(y,y)=(x,y)\circ (y,y)=(x,y), so ε is the identity section for G.

  • Inversion:

Let (x,y)^{-1}= (y,x) \,\forall (x,y) \in G.

(x,y)\circ (x,y)^{-1}=(x,y)\circ (y,x)= (x,x)=\epsilon (\alpha (x,y))

and (x,y)^{-1}\circ (x,y)=(y,x)\circ(x,y)=(y,y)=\epsilon( \beta (x,y))

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Works Cited

The main source for work on this page was Groupoids, Inverse Semigroups, and their Operator Algebras by Alan L.T.Paterson.

The information on adjacency rules was taken from Tilings and Patterns by Grunbaum/Shephard.

BJ Kahng offered guidance on working with groupoids.


Also, in case you're interested, here's the link the Wikipedia article on Penrose Tilings: [1] (http://en.wikipedia.org/wiki/Penrose_tiling).


Related Pages: Tilings, Ammann Tilings

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