A. 2×2 integer edge sequence solution.

The smallest first generation supertile is one based on either the (1, -1) or the (\frac{1}{2}, -\frac{1}{2}) edge sequence.  In fact, the latter may be considered to be a special case of the first and will be discussed later. The (1,-1) edge sequence means that there is only one \pm pair of edge angles, or m_1=1. It follows that n_0=2 and n_2=1, i.e. the supertile S_s^n has to be constructed from the set of prototiles {2T_s^n,T_{s+2}^n, T_{s-2}^n}. In our general model the arrangement of the tiles is given by the supertile matrix

(1)   \begin{equation*}S_s^n=\begin{bmatrix}T_s^n &T_{s+2}^n \\T_{s-2}^n &T_s^n \\\end{bmatrix}\end{equation*}

In order to tile the plane, the inflation factor should be larger than one. Because L = 2\cos(\pi{}/n), n should be larger than three. Below, supertiles are shown for n=4,5 and 6. Rhomb prototiles T_s^n and T_{n-s}^n are identical objects, but supertiles S_s^n and S_{n-s}^n are not. The supertiles also loose the twofold rotation axis perpendicular to the plane of the prototiles. Consequently, after the first generation supertiles there are four different substitution rules for the next supertile generation, which we have denoted by a), b), c) and d). a) is the basic supertile, in b) the tiles of a) have been rotated over \pi{}, in c) the s-2 and s+2 tiles of case a) have been replaced by the n-s+2 and the n-s-2 tiles respectively, and in d) the tiles of case c) have been rotated over \pi{}. In matrix notation:

(2)   \begin{equation*}\begin{bmatrix}T_s^n &T_{s+2}^n \\T_{s-2}^n &T_s^n \\\end{bmatrix}  ,  \begin{bmatrix}\underline{T_s^n} &\underline{T_{s+2}^n} \\\underline{T_{s-2}^n }&\underline{T_s^n }\\\end{bmatrix}  ,  \begin{bmatrix}\underline{T_s^n} &T_{n-s-2}^n \\T_{n-s+2}^n &\underline{T_s^n} \\\end{bmatrix}  ,  \begin{bmatrix}T_s^n &\underline{T_{n-s-2}^n} \\\underline{T_{n-s+2}^n} &T_s^n \\\end{bmatrix}\end{equation*}

Underlined entries denote prototiles rotated over \pi.

  Fig. S1-S3 Substitution tiles for n=4,5,6.

The supertiles consist only of single positive and negative parts, i.e. the tile overlaps which arise during the tiling proces are exactly compensated by negative tiles or tile parts. The first generation supertiles consist of two prototiles T_s^n, one rotated over \pi{}/n and the other over -\pi{}/n and a T_{s+2}^n and T_{s-2}^n prototiles in the original position as drawn in the pictures. Negative tiles play a part for s=0, 1, n-1 and n. For s=1 the s=-1 negative tile cuts of pieces of the other three tiles, as is illustrated in the picture for n=5 below. For the s=n-1 supertile, on the other hand, the two n-1 prototiles partly overlap. This overlap is removed by the n+1 prototile.  The s=0 supertile is just a zigzag line because the s=-2 annihilates the s=2 tile. This is best seen by starting with s=\delta, 2-\delta and 2+\delta prototiles, \delta being a small value  approaching zero. The negative n+1 and the positive n-1 prototile do not overlap in the s=n supertile, although the total area is zero as it should.

build of n=5 1st supertiles
Fig.S4 Illustration of the construction of n=5 first generation supertiles. White tiles are negative tiles.

 B. 2×2 half-integer edge sequence solution.

Looking at the first three figures it becomes  clear that for substitution rules a) and b) there are two independent prototile sets for even and odd s. This is also true for substitution rules c) and d) if n is even. In fact, the half-integer edge sequence solution is identical to the integer edge sequence solution for a doubled n-value and even s. We will discuss these solutions in more detail below.

The (\frac{1}{2}, -\frac{1}{2}) edge sequence means that m_\frac{1}{2}=1. It follows that n_0=2 and n_1=1, i.e. the supertile S_s^n has to be constructed from the set of prototiles {2T_s^n,T_{s+1}^n, T_{s-1}^n}. In our general model the arrangement of the tiles is given by the supertile matrix

(3)   \begin{equation*}S_s^n=\begin{bmatrix}T_s^n &T_{s+1}^n \\T_{s-1}^n &T_s^n \\\end{bmatrix}\end{equation*}

Below the first two generations of 2\times2 supertiles for n=5 with a halfinteger edge sequence are shown. If one compares these tiles with the integer case it is clear that negative tiles play a smaller role. Only for substitution rule a) a negative part appears in the s=n-1 tile. For substitution rule c) and d) the tiles are not even cut up. The only tile having a negative part is the s=n tile. This tile is only involved in the construction of next generation s=n and s=1 tiles. In the latter case, its role is to move a patch of complete tiles from one end of the supertile to another, also for higher generations.

halfinteger 2x2 supertiles n=5
Fig. S5 n=5 half integer edge supertiles.

The tilings resulting from the application of substitution rules a), b) and c) will be called:

a) Basic (1, -1) rhomb tiling, S_s^n(a)=\begin{bmatrix}T_s^n &T_{s+1}^n \\T_{s-1}^n &T_s^n \\\end{bmatrix}

b) Koch (1, -1) rhomb tiling,  S_s^n(b)=\begin{bmatrix}\underline{T_s^n} &\underline{T_{s+1}^n} \\\underline{T_{s-1}^n} &\underline{T_s^n} \\\end{bmatrix}. The supertile edges are related to the so called Koch curves.

c) Lancon Billard (1,-1) rhomb tiling,  S_s^n(c)=\begin{bmatrix}\underline{T_s^n }&T_{n-s-1}^n \\T_{n-s+1}^n &\underline{T_s^n }\\\end{bmatrix} and its mirror image S_s^n(d)=\begin{bmatrix}T_s^n &\underline{T_{n-s-1}^n }\\\underline{T_{n-s+1}^n }&T_s^n \\\end{bmatrix}.  For n=5 the even s first generation substitution tiles are the Lancon Billard substitution tiles.

Larger supertiles for all three cases are shown on a separate page called Large Supertiles. Special pages have been devoted to the Koch Supertiles and the Lancon Billard Supertiles.

C. Harriss-like Supertiles.