Substitution scheme in Ammann–Beenker tiling On wiki page about Ammann–Beenker tiling is described the substitution scheme $R → R r R ; r → R$ that introduces the ratio as a scaling factor: its matrix is the Pell substitution matrix, and the series of words produced by the substitution have the property that the number of $r$s and $R$s are equal to successive Pell numbers.
Q: the order of the $r$s and $R$s in the assignation $R → R r R ; r → R$ I don't understand. applying recursive definition of Pell numbers we see that in every recursion step if we associate $R_{old}:= P_n, r_{old}:=P_{n-1}$ then recursively we step forward by $R_{new}:=P_{n+1} = 2R_{old} +r_{old}$ and $r_{new}=R_{old}$. what I don't understand is the order of $R$ and $r$ in the assignation $R → R r R$.  
 A: In this particular case, the order doesn't matter because the tilings associated with the substitutions $$\phi_1\colon R \mapsto RrR,\: r \mapsto R,\quad \phi_2\colon R \mapsto RRr,\: r \mapsto R, \quad \phi_3 \colon R \mapsto rRR,\: r \mapsto R$$ are all identical via the conjugation map $\rho \colon u \mapsto R u R^{-1}$ or its inverse $\rho^{-1} \colon u \mapsto R^{-1} u R$, where these maps are well defined.
For example, $\rho \circ \phi_1 = \phi_2$ and $\rho^{-1} \circ \phi_1 = \phi_3$.
In general, the order that letters appear in a substituted word can affect the resulting tiling. For example, the substitutions $\psi \colon a \mapsto abba,\: b \mapsto baab$ and $\psi' \colon a \mapsto aabb, b \mapsto abab$ really do have different associated tilings, as it can be shown that they are not mutually locally derivable (for example by looking at their resulting cohomology groups).
If one only cares about statistical properties of the letters in a tiling, as is the case here, then the order of the letters does not matter, as that is dependent only on the associated abelianisation matrix (substitution matrix). Here, that is
$$
\begin{pmatrix}
2 & 1\\
1 & 0
\end{pmatrix}.
$$
A: The substitution rule is to be understood as a combination of inflating (scaling) and replacement of patches.
r ->
R ->
RrR ->
(RrR)(R)(RrR) = RrRRRrR ->
(RrR)(R)(RrR)(RrR)(RrR)(R)(RrR) = RrRRRrRRrRRrRRRrR ->
etc.
Moreover you should be aware that this substitution only produces a 1D chain, while the Ammann-Beenker tiling clearly is 2D. 
But clearly there is a 2D substitution too, which produces the required tiling directly. Again you have patches which are to be scaled first and then substituted by the original patches again. This then is described by the following Wikipedia picture:

--- rk
