Computing the Tile Frequencies

We will see that it is helpful to compute the frequencies of the 6 tiles of the Penrose P1 tiling first.

We abbreviate the name and frequencies of the corresponding tiles with S, B, D, PA, PB and PC

Star S , Boat B ,Diamond D ,Pentagon PA , Pentagon PB , Pentagon PC

From the picture above which shows the overlay of the tiling and an doubly inflated tiling we can see how the number of tiles changes by the inflation::

S   → 1*S + 5*B + 5*PC
B   → 1*S + 3*B + 3*PC
D   → 1*S + 1*B + 1*PC
PA → 2.5*D + 1*PA + 5*PB
PB → 1.5*D + 1*PA + 3*PB + 2*PC
PC → 1.0*D + 1*PA + 1*PB +4*PC

The relations above may not be totally obvious. You may wonder for example why the inflated star S contains 5 boats since parts of the boats are outside of S. But we just add these parts to the star and exclude them from the adjacent pentagons. Similar considerations hold for the inflated boat B and the inflated diamond D.

We define the Matrix A as:

1	1	1	0	0	0
5	3	1	0	0	0
0	0	0	5/2	3/2	1
0	0	0	1	1	1
0	0	0	5	3	1
5	3	1	0	2	4

With the frequency vector x=(S,B,D,PA,PB,PC), A*x then gives the frequencies for the doubly inflated tiling.

The largest eigenvalue of A is φ⁴, where φ is the golden ratio .The corresponding eigenvector - normed for a total probability of 1 - gives the relative frequencies for the tiles.

P_S = φ^-5/√5 ≈ 4.0%
P_B = φ^-5 ≈ 9.0%
      P_D = φ^-4 ≈14.6%
      P_PA = φ^-3/√5 ≈ 10.6%
     P_PB = φ^-3 ≈ 23.6%
      P_PC = φ^-2 ≈ 38.2%

The relative frequency of all 3 pentagons together is φ/√5 ≈ 72.4%

The key to retrieve the frequency distribution of our "girih"-tiling with hexagons, boats, stars and decagons (outlined in dark red in the picture below) is to see the connection with the Penrose P1 tiling. To avoid confusion about the tile names (there are boats and stars in both tilings) we call this tiling HBSD-tiling .

If we appropiately connect adjacent centers of all decagon-candidates we get the P1-tiling! In the center of a pentagon is exactly one HBSD-star tile. The three pentagon types PA, PB and BC of the P1-tiling correspond to the stars with 0, 1 or 2 spikes of a different colour of the HBSD-tiling, we call them star0, star1 and star2.

To compute the relative frequencies of the decagons we observe in the picture above that the centers of the decagons map one to one to the vertices of the P1-diamonds, P1-boats and P1-stars. Overlapping occurs only when the centers of the decagons are vertices of the same diamond/boat/star. Decagons with centers on different diamonds/boats/stars cannot overlap.

:The P1-diamond allows 3 HBSD-decagons. 1 HBSD-boat and 1 HBSD-hexagon are left over..

The P1-boat allows 5 HBSD-decagons. 2 HBSD-boats and 1 HBSD-hexagon are left over..

The star allows 7 HBSD-decagons. 3 HBSD-boats and 1 hexagon are left over..

In a P1-tiling patch with a large number N of black outlined P1-tiles we now have

N*P_PA stars of type S0,
N*P_PB stars of type S1,
N*P_PC stars of type S2,
N*P_S*7 + N*P_B*5 + N*P_D*3 decagons
N*P_S*3 + N*P_B*2 + N*P_D*1 boats
N*P_S*1 + N*P_B*1 + N*P_D*1 hexagons

If we add all these terms we get the total number of tiles in the underlying HBSD tiling. The sum magically simplifies to N*φ². From this we get the relative frequencies:

P_Star0 = P_PA / φ² = φ^-5/√5 ≈ 4.0%
P_Star1 = P_PB / φ² = φ^-5≈ 9.0%
P_Star2 = P_PC / φ² = φ^-4 ≈14.6%

P_Decagon = (P_S*7 + P_B*5 + P_D*3) / φ^{2 =}1 / √5 ≈ 44.7%

P_Boat = (P_S*3 + P_B*2 + P_D*1 / φ^{2 =}φ^-2 / √5 ≈ 17.1%

P_Hexagon =(P_S*1 + P_B*1 + P_D*1) / φ^{2 =}φ^-3 / √5 ≈ 10.6%

If we only want to know the relative frequencies of stars, decagons and the rest (boats and hexagons) we have

P_Decagon = 1 / √5 ≈ 44.7%
P_Star = P_Star0 + P_Star1 + P_Star2 = P_Decagon / φ ≈ 27.6%
P_BoatOrHexagon = P_Boat + P_Hexagon = P_Decagon / φ ≈ 27.6%

I really wonder about the simplicity and beauty of the resulting formulas.

Even nicer, the relative frequencies can be visualized as areas of golden triangles, golden gnomons and a pentagon within a large pentagon..

If we take into account the size of the different tiles and ask which area A in the resulting tiling is occupied by the different tile types it is possible to derive the following expressions:

A_Decagon = φ^-1 ≈ 61.8%

A_Star = φ^-3≈ 23.6%

A_Boat ≈ φ^-3 / √5 ≈ 10.6%

A_Hexagon ≈ φ^-5 / √5 ≈ 4.0%

So more than 85% of the area is covered by decagons and stars.

To derive these formulas we use the fact that the area of the boat-tile is φ times the area of the hexagon-tile, the area of the star-tile is √5 times the area of the hexagon-tile and the decagon can be dissected into a boat and two hexagons.

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