 $\pi \colon \ensuremath{I_q} \smallsetminus \partial\ensuremath{\mathcal{M}}(f_q) \to F_{\ensuremath{\mathbb{A}}}^{\ensuremath{\mathbb{N}}}$ in analogy to the case $q=3$.
The same arguments as there show that this map is bijective and conjugates $f_q$ to the shift map $\tau$ with $\pi\circ f_3= \tau \circ \pi$ on $\ensuremath{I_q} \smallsetminus \partial\ensuremath{\mathcal{M}}(f_q)$.

\smallskip

For $q=2h_q+3=\kappa_q+1$ finally define the alphabet $F$ as $F=\{\ensuremath{\varepsilon} 1_i,\, \ensuremath{\varepsilon}=+,-,\, 1\leq i\leq \kappa_q-1\}\cup \{\ensuremath{\varepsilon} 2_i,\, \ensuremath{\varepsilon}=+,-,\, \kappa_q-1\leq i\leq \kappa_q\} \cup \ensuremath{\mathbb{Z}}\smallsetminus\{0,\pm 1,\pm 2\} $.
The transition matrix $\ensuremath{\mathbb{A}}=\left(\ensuremath{\mathbb{A}}_{i,j}\right)_{i,j\in F}$ is given in Table~\ref{D2.3}.
\begin{table}
\begin{align*}
\ensuremath{\mathbb{A}}_{\ensuremath{\varepsilon} 1_{2i},\ensuremath{\varepsilon} 1_{2i+1}}           &= 1\quad 1\leq i\leq h_q-2,\\
\ensuremath{\mathbb{A}}_{\ensuremath{\varepsilon} 1_{2h_q-2},\ensuremath{\varepsilon} 1_{2h_q}}       &= 1, \\
\ensuremath{\mathbb{A}}_{\ensuremath{\varepsilon} 1_{2h_q-2},\ensuremath{\varepsilon} 2_{\kappa_q}}   &= 1, \\
\ensuremath{\mathbb{A}}_{\ensuremath{\varepsilon} 1_{2h_q},-\ensuremath{\varepsilon} 1_{i}}           &= 1 \quad 1\leq i\leq \kappa_q-1,\\
\ensuremath{\mathbb{A}}_{\ensuremath{\varepsilon} 1_{2h_q},-\ensuremath{\varepsilon} 2_{i}}           &= 1 \quad \kappa_q\leq i\leq \kappa_q+1,\\
\ensuremath{\mathbb{A}}_{\ensuremath{\varepsilon} 1_{2h_q},-\ensuremath{\varepsilon} m}               &= 1 \quad m\geq 3,\\
\ensuremath{\mathbb{A}}_{\ensuremath{\varepsilon} 1_{2i-1},\ensuremath{\varepsilon} 1_{2i+1}}         &= 1,\quad 1\leq i\leq h_q-1,\\
\ensuremath{\mathbb{A}}_{\ensuremath{\varepsilon} 1_{2h_q-1},\ensuremath{\varepsilon} 2_{\kappa_q+1}} &= 1,\\
\ensuremath{\mathbb{A}}_{\ensuremath{\varepsilon} 1_{2h_q-1},-\ensuremath{\varepsilon} m}             &= 1,\quad m\geq 3, \\
\ensuremath{\mathbb{A}}_{\ensuremath{\varepsilon} 2_{\kappa_q},-\ensuremath{\varepsilon} 1_1}         &= 1,\\
\ensuremath{\mathbb{A}}_{\ensuremath{\varepsilon} 2_{\kappa_q+1},\delta 1_i}      &= 1,\quad \delta = +,-, \quad 2\leq i \leq \kappa_q-1,\\
\ensuremath{\mathbb{A}}_{\ensuremath{\varepsilon} 2_{\kappa_q+1},\delta 2_i}      &= 1,\quad \delta = +,-,\quad 2\leq i \leq \kappa_q-1,\\
\ensuremath{\mathbb{A}}_{\ensuremath{\varepsilon} 2_{\kappa_q+1},\delta n}        &= 1,\quad \delta = +,-,\quad n\geq 3,\\
\ensuremath{\mathbb{A}}_{\ensuremath{\varepsilon} 2_{\kappa_q+1},-\ensuremath{\varepsilon} 1_1}       &= 1, \\
\ensuremath{\mathbb{A}}_{m,n}                                 &= 1, \quad m\in\ensuremath{\mathbb{Z}}\smallsetminus\{0,\pm 1,\pm 2\},\quad n\in F,
\end{align*}
\caption{The transition matrix $\ensuremath{\mathbb{A}}=(\ensuremath{\mathbb{A}}_{i,j})_{i,j\in F}$ with $\ensuremath{\varepsilon} = +,-$ for $q$-regular sequences and $q$ odd, $q \geq 5$. All other matrix elements vanish.}
\label{D2.3}
\end{table}

The set $\partial\ensuremath{\mathcal{M}}(f_q)$ and the map $\pi \colon \ensuremath{I_q} \smallsetminus \partial\ensuremath{\mathcal{M}}(f_q)\to F_{\ensuremath{\mathbb{A}}}^{\ensuremath{\mathbb{N}}}$  are defined similarly as in the foregoing cases $q=3$ and $q$ even and have the same properties.
The inverse $(\pi)^{-1} \colon F_{\ensuremath{\mathbb{A}}}^{\ensuremath{\mathbb{N}}}\to  \ensuremath{I_q} \smallsetminus \partial\ensuremath{\mathcal{M}}(f_q)$ is given by $\pi^{-1}(\underline{\xi})=x$ with $x \in J_{\xi_1}\cap \bigcap_{l=1}^\infty (f_{\xi_l}\circ\ldots
\circ f_{\xi_1})^{-1} J_{\xi_{l+1}}$ where $f_{\xi_i}=f\big|_{J_{\ensuremath{\varepsilon} 1}}$ for $\xi_i=\ensuremath{\varepsilon} 1_l$, $1 \leq l \leq \kappa_q-1$ respectively $f_{\xi_i}=f\big|_{J_{\ensuremath{\varepsilon} 2}}$ for $\xi_i=\ensuremath{\varepsilon} 2_l$, $\kappa_q\leq l\leq
\kappa_q+1$.
Hence also in this case the map $f_q$ gets conjugated by $\pi$ to the shift map $\tau$ on the space $ F_{\ensuremath{\mathbb{A}}}^{\ensuremath{\mathbb{N}}}$ of symbol sequences and therefore is itself a subshift of infinite type.




\subsection{A sofic system related to the map $f_q$ and the regular $\lambda_q$-CF}
\label{D2a}
The transition matrix $\ensuremath{\mathbb{A}}$ in \eqref{D2.1} for the subshift $f_3:I_3\to I_3$ shows that a symbol sequence $\underline{a}=(a_i)_{i\in \ensuremath{\mathbb{N}}}\in F_{\ensuremath{\mathbb{A}}}^{\ensuremath{\mathbb{N}}} = \pi\big(I_3\smallsetminus\partial\ensuremath{\mathcal{M}}(f_3)\big)$ if and only if $(a_i,a_{i+1}) \neq(\ensuremath{\varepsilon} 2, \ensuremath{\varepsilon} m),\, m\geq 2$ for all $i\in \ensuremath{\mathbb{N}}$.
Hence this sequence is $q$-regular for $q=3$ and $F_{\ensuremath{\mathbb{A}}}^{\ensuremath{\mathbb{N}}}=\ensuremath{\mathcal{A}}_3^{reg}$.
The inverse map $\pi^{-1} \colon F_{\ensuremath{\mathbb{A}}}^{\ensuremath{\mathbb{N}}}\to  \ensuremath{I_q} \smallsetminus \partial\ensuremath{\mathcal{M}}(f_q)$ therefore has the form $\pi^{-1}(\underline{a}) = \lb 0;a_1,a_2,\ldots\rb$.
This follows from

\begin{lemma}
\label{convergence3}
For $\underline{a}=(a_i)_{i\in \ensuremath{\mathbb{N}}} \in \ensuremath{\mathcal{A}}_3^\text{reg}$ a $3$-regular sequence the limit \linebreak $\lim_{n\to\infty} \lb 0;a_1,a_2,\ldots,a_n\rb$ exists and defines a point $x\in \ensuremath{\mathbb{R}}$.
\end{lemma}

\begin{proof}
Set $x_n:=\lb 0;a_1,a_2,\ldots,a_n\rb$ and denote by $J_{(a_1,\ldots,a_n)}$ the closed interval $J_{(a_1,\ldots,a_n)} := J_{a_1} \cap \bigcap_{l=1}^{n-1}\left(f_{a_l}\circ\ldots f_{a_1} \right)^{-1} J_{a_{l+1}}$ with $f_{a_i} := f_q\big|_{J_{a_i}}$.
Obviously $x_n \in J_{(a_1,\ldots,a_n)}$.
All these intervals are nonempty and $J_{(a_1,\ldots,a_{n+1})}\subset J_{(a_1,\ldots,a_{n})}$ for all $n$.
Hence $\bigcap_{n=1}^\infty J_{(a_1,\ldots,a_n)}$ is not empty.
Because the map $f_3 $ is strictly expanding this set contains exactly one