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de Bruijn’s pentagrids

In a Rhombic tiling (aka a Penrose P3 tiling) we can identify five ribbons.

Opposite sides of a rhomb are parallel. We may form a ribbon by attaching rhombs along opposite sides. There are five directions taken by sides, so there are five families of ribbons that do not intersect, determined by the side directions.

Every ribbon determines a skeleton curve through the midpoints of opposite sides of the rhombs. If we straighten these skeleton curves we get a de Bruijn’s pentagrid.



A pentagrid is a grid of the plane consisting of five families of parallel lines, at angles multiples of 72o=360o/5, and with each family consisting of parallel lines with a spacing given by a number γi for 0i4, satisfying
γ0+γ1+γ2+γ3+γ4=0

The points of the plane in the j-th family of the pentagrid are
{xR2 | x.vj+γjZ}
where vj=ζj=(cos(2πj/5),sin(2πj/5)).

A pentagrid is regular if no point in R2 belongs to more than two of the five grids. For almost all choices γ0,,γ4 the corresponding pentagrid is regular. The pentagrid coordinates of a point xR2 are the five integers (K0(x),,K4(x)Z5 defined by
Kj(x)=x.vj+γj
with r the smallest integer greater of equal to r, and clearly the function Kj(x) is constant in regions between the j-th grid lines.



With these pentagrid-coordinates one can associate a vertex to any point xR2
V(x)=j=04Kj(x)vj
which is constant in regions between grid lines.

Here’s de Bruijn‘s result:

For x running over R2, the vertices V(x) coming from a pentagrid determine a Rhombic tiling of the plane. Even better, every Rhombic tiling comes from a pentagrid.



We’ll prove this for regular pentagrids (the singular case follows from a small deformation).

Any intersection point x0 of the pentagrid belongs to exactly two families of parallel lines, so we have two integers r and s with 0r<s4 and integers kr,ksZ such that x0 is determined by the equations {x.vr+γr=krx.vs+γs=ks In a small neighborhood of x0, V(x) takes the values of four vertices of a rhomb. These vertices are associated to the 5-tuples in Z5 given by (K0(x0),,K4(x0))+ϵ1(δ0r,,δ4r)+ϵ2(δ0s,,δ4s) with ϵ1,ϵ2{0,1}. So, the intersection points of the regular pentagrid lines correspond to rhombs, and the regions between the grid lines (which are called meshes) correspond to vertices, whose positions are given by V(x).

Observe that these four vertices give a thin rhombus if the angle between vr and vs is 144o and determine a thick rhombus if the angle is 72o.

Not all pentagrid coordinates (k0,,k4) occur in the tiling though. For xR2 we have
Kj(x)=x.vj+γj+λj(x)
with 0λj(x)<1. In a regular pentagrid at most two of the λj(x) can be zero and so we have 0<λ0(x)++λ4(x)<5 and we have assumed that γ0++γ4=0, giving us j=04Kj(x)=x.(j=04vj)+j=04γj+j=04λj(x)=j=04λj(x) The left hand side must be an integers and the right hand side a number strictly between 0 and 5. This defines the index of a vertex
ind(x)=j=04Kj(x){1,2,3,4}
Therefore, every vertex in the tiling may be represented as
k0v0++k4v4
with (k0,,k4)Z5 satisfying j=04kj{1,2,3,4}. If we move a point along the edges of a rhombus, we note that the index increases by 1 in the directions of v0,,v4 and decreases by 1 in the directions v0,,v4.



It follows that the index-values of the vertices of a thick rhombus are either 1 and 3 at the 72o angles and 2 at the 108o angles, or they are 2 and 4 at the 72o angles, and 3 at the 108o angles. For a thin rhombus the index-values must be either 1 and 3 at the 144o angles and 2 at the 36o angles, or 2 and 4 at the 144o angles and 3 at the 36o angles.

We still have to show that this gives a legal Rhombic tiling, that is, that the gluing restrictions of Penrose’s rhombs are satisfied.



We do this by orienting and colouring the edges of the thin and thick rhombus towards the vertex defined by the semi-circles on the rhombus-pieces. Edges connecting a point of index 3 to a point of index 2 are coloured red, edges connecting a point of index 1 to a point of index 2, or connecting a point of index 3 to one of index 4 are coloured blue.

We orient green edges pointing from index 2 to index 1, or from index 3 to index 4. As this orientation depends only on the index-values of the vertices, two rhombs sharing a common green edge also have the same orientation on that edge.

On each individual rhomb, knowing the green edges and their orientation forced the red edges and their orientation, but we still have to show that if two rhombs share a common red edge, this edge has the same orientation on both (note in the picture above that a red edge can both flow from 2 to 3 as well as from 3 to 2).

From the gluing conditions of Penrose’s rhombs we see that if PQ be a red edge, in common to two rhombs, and these rhombs have angle α resp. β in P, then α and β must be both less that 90o or both bigger than 90o. We translate this in terms of the pentragrid.



The two rhombs correspond to two intersection points A and B, and we may assume that they both lie on a line l of family 0 of parallel lines (other cases are treated similarly by cyclic permutation), and that A is an intersection with a family p-line and B with a family q-line, with p,q{1,2,3,4}. The interval AB crosses the unique edge common to the two rhombs determined by A and B, and we call the interval AB red if jKj(x) is 2 on one side of AB and 3 on the other side. The claim about the angles α and β above now translates to: if AB is red, then p+q is odd (in the picture above p=1 and q=2).

By a transformation we may assume that γ0=0 and that l is the Y-axis. For every yR we then get
{K1(0,y)=y.sin(2π/5)+γ1K2(0,y)=y.sin(4π/5)+γ2K3(0,y)=y.sin(4π/5)+γ3K4(0,y)=y.sin(2π/5)+γ4
and γ1+γ4 and γ2+γ3 are not integers (otherwise the pentagrid is not regular). If y runs from to + we find that
K1(0,y)+K4(0,y)γ1+γ4={01
with jumps from 0 to 1 at places where (γ1+γ4γ1)/sin(2π/5)Z and from 1 to 0 when γ4/sin(2π/5)Z. (A similar result holds replacing K1,K4,γ1,γ4,sin(2π/5) by K2,K3,γ2,γ3,sin(4π/5).

Because the points on l intersecting with 1-family and 4-family gridlines alternate (and similarly for 2- and 3-family gridlines) we know already that pq. If we assume that p+q is even, we have two possibilities, either {p,g}={1,3} or {2,4}. As γ0=0 we have γ1+γ2+γ3+γ4=0 and therefore
γ1+γ4+γ2+γ3=1
It then follows that K1(0,y)+K2(0,y)+K3(0,y)+K4(0,y)=1 or 3 between the points A and B. Therefore K0(y,0)++K4(0,y) is either 1 on the left side and 2 on the right side, or is 3 on the left side and 4 on the right side, so AB must be green, a contradiction. Therefore, p+q is odd, and the orientation of the red edge in both rhombs is the same. Done!

An alternative way to see this correspondence between regular pentagrids and Rhombic tilings as as follows. To every intersection of two gridlines we assign a rhombus, a thin one if they meet at an acute angle of 36o and a thich one if this angle is 72o, with the long diagonal dissecting the obtuse angle.



Do this for all intersections, surrounding a given mesh.



Attach the rhombs by translating them towards the mesh, and finally draw the colours of the edges.



Another time, we will connect this to the cut-and-project method, using the representation theory of D5.

Published in geometry GoV groups

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