$M$ est doublement stochastique.
\end{Proof}
+
+
+Montrons que
+\begin{lemma}
+$d$ est une distance sur $\mathcal{X}_{\mathsf{N},\mathcal{P}}$.
+\end{lemma}
+
+
+\begin{proof}
+ $d_{\mathds{B}^\mathsf{N}}$ est la distance de Hamming.
+ Prouvons que
+ $d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}$ est aussi une distance;
+$d$ sera ainsi une distance comme somme de deux distances.
+ \begin{itemize}
+\item De manière évidente, $d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}(s,\check{s})\geqslant 0$, et si $s=\check{s}$, alors
+$d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}(s,\check{s})=0$.
+Réciproquement si $d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}(s,\check{s})=0$, alors
+$\forall k \in \mathds{N}, v^k=\check{v}^k$ d'après la définition de $d$.
+Or les éléments entre les positions $p+1$ et $p+n$
+sont nules et correspondent à $|u^0-\check{u}^0|$,
+on peut conclure que $u^0=\check{u}^0$.
+On peut étendre ce résultat aux $n \times \max{(\mathcal{P})}$ premiers
+bloc engendrant $u^i=\check{u}^i$, $\forall i \leqslant v^0=\check{v}^0$,
+et en vérifiant tous les $n \times \max{(\mathcal{P})}$ blocs, $u=\check{u}$.
+ \item $d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}$ est évidemment symétrique
+($d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}(s,\check{s})=d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}(\check{s},s)$).
+\item l'inégalité triangulaire est établie puisque la valeur absolue la vérifie
+aussi.
+ \end{itemize}
+\end{proof}
+
+
+
+\begin{theorem}
+La fonction $G_{f_u,\mathcal{P}}$ est chaotique sur
+ $(\mathcal{X}_{\mathsf{N},\mathcal{P}},d)$ si et seulement si
+graphe d'itération $\textsc{giu}_{\mathcal{P}}(f)$
+est fortement connexe.
+\end{theorem}
+
+\begin{proof}
+Suppose that $\Gamma_{\mathcal{P}}(f)$ is strongly connected.
+Let $x=(e,(u,v)),\check{x}=(\check{e},(\check{u},\check{v}))
+\in \mathcal{X}_{\mathsf{N},\mathcal{P}}$ and $\varepsilon >0$.
+We will find a point $y$ in the open ball $\mathcal{B}(x,\varepsilon )$ and
+$n_0 \in \mathds{N}$ such that $G_f^{n_0}(y)=\check{x}$: this strong transitivity
+will imply the transitivity property.
+We can suppose that $\varepsilon <1$ without loss of generality.
+
+Let us denote by $(E,(U,V))$ the elements of $y$. As
+$y$ must be in $\mathcal{B}(x,\varepsilon)$ and $\varepsilon < 1$,
+$E$ must be equal to $e$. Let $k=\lfloor \log_{10} (\varepsilon) \rfloor +1$.
+$d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}((u,v),(U,V))$ must be lower than
+$\varepsilon$, so the $k$ first digits of the fractional part of
+$d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}((u,v),(U,V))$ are null.
+Let $k_1$ the smallest integer such that, if $V^0=v^0$, ..., $V^{k_1}=v^{k_1}$,
+ $U^0=u^0$, ..., $U^{\sum_{l=0}^{k_1}V^l-1} = u^{\sum_{l=0}^{k_1}v^l-1}$.
+Then $d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}((u,v),(U,V))<\varepsilon$.
+In other words, any $y$ of the form $(e,((u^0, ..., u^{\sum_{l=0}^{k_1}v^l-1}),
+(v^0, ..., v^{k_1}))$ is in $\mathcal{B}(x,\varepsilon)$.
+
+Let $y^0$ such a point and $z=G_f^{k_1}(y^0) = (e',(u',v'))$. $\Gamma_{\mathcal{P}}(f)$
+being strongly connected, there is a path between $e'$ and $\check{e}$. Denote
+by $a_0, \hdots, a_{k_2}$ the edges visited by this path. We denote by
+$V^{k_1}=|a_0|$ (number of terms in the finite sequence $a_1$),
+$V^{k_1+1}=|a_1|$, ..., $V^{k_1+k_2}=|a_{k_2}|$, and by
+$U^{k_1}=a_0^0$, $U^{k_1+1}=a_0^1$, ..., $U^{k_1+V_{k_1}-1}=a_0^{V_{k_1}-1}$,
+$U^{k_1+V_{k_1}}=a_1^{0}$, $U^{k_1+V_{k_1}+1}=a_1^{1}$,...
+
+Let $y=(e,((u^0, ..., u^{\sum_{l=0}^{k_1}v^l-1}, a_0^0, ..., a_0^{|a_0|}, a_1^0, ..., a_1^{|a_1|},...,
+ a_{k_2}^0, ..., a_{k_2}^{|a_{k_2}|},$ \linebreak
+ $\check{u}^0, \check{u}^1, ...),(v^0, ..., v^{k_1},|a_0|, ...,
+ |a_{k_2}|,\check{v}^0, \check{v}^1, ...)))$. So $y\in \mathcal{B}(x,\varepsilon)$
+ and $G_{f}^{k_1+k_2}(y)=\check{x}$.
+
+
+Conversely, if $\Gamma_{\mathcal{P}}(f)$ is not strongly connected, then there are
+2 vertices $e_1$ and $e_2$ such that there is no path between $e_1$ and $e_2$.
+That is, it is impossible to find $(u,v)\in \mathds{S}_{\mathsf{N},\mathcal{P}}$
+and $n \mathds{N}$ such that $G_f^n(e,(u,v))_1=e_2$. The open ball $\mathcal{B}(e_2, 1/2)$
+cannot be reached from any neighborhood of $e_1$, and thus $G_f$ is not transitive.
+\end{proof}
+
+
+We show now that,
+\begin{prpstn}
+If $\Gamma_{\mathcal{P}}(f)$ is strongly connected, then $G_f$ is
+regular on $(\mathcal{X}_{\mathsf{N},\mathcal{P}}, d)$.
+\end{prpstn}
+
+\begin{proof}
+Let $x=(e,(u,v)) \in \mathcal{X}_{\mathsf{N},\mathcal{P}}$ and $\varepsilon >0$.
+As in the proofs of Prop.~\ref{prop:trans}, let $k_1 \in \mathds{N}$ such
+that
+$$\left\{(e, ((u^0, ..., u^{v^{k_1-1}},U^0, U^1, ...),(v^0, ..., v^{k_1},V^0, V^1, ...)) \mid \right.$$
+$$\left.\forall i,j \in \mathds{N}, U^i \in \llbracket 1, \mathsf{N} \rrbracket, V^j \in \mathcal{P}\right\}
+\subset \mathcal{B}(x,\varepsilon),$$
+and $y=G_f^{k_1}(e,(u,v))$. $\Gamma_{\mathcal{P}}(f)$ being strongly connected,
+there is at least a path from the Boolean state $y_1$ of $y$ and $e$.
+Denote by $a_0, \hdots, a_{k_2}$ the edges of such a path.
+Then the point:
+$$(e,((u^0, ..., u^{v^{k_1-1}},a_0^0, ..., a_0^{|a_0|}, a_1^0, ..., a_1^{|a_1|},...,
+ a_{k_2}^0, ..., a_{k_2}^{|a_{k_2}|},u^0, ..., u^{v^{k_1-1}},$$
+$$a_0^0, ...,a_{k_2}^{|a_{k_2}|}...),(v^0, ..., v^{k_1}, |a_0|, ..., |a_{k_2}|,v^0, ..., v^{k_1}, |a_0|, ..., |a_{k_2}|,...))$$
+is a periodic point in the neighborhood $\mathcal{B}(x,\varepsilon)$ of $x$.
+\end{proof}
+
+$G_f$ being topologically transitive and regular, we can thus conclude that
+\begin{thrm}
+The function $G_f$ is chaotic on $(\mathcal{X}_{\mathsf{N},\mathcal{P}},d)$ if
+and only if its iteration graph $\Gamma_{\mathcal{P}}(f)$ is strongly connected.
+\end{thrm}
+
+\begin{crllr}
+ The pseudorandom number generator $\chi_{\textit{14Secrypt}}$ is not chaotic
+ on $(\mathcal{X}_{\mathsf{N},\{b\}},d)$ for the negation function.
+\end{crllr}
+\begin{proof}
+ In this context, $\mathcal{P}$ is the singleton $\{b\}$.
+ If $b$ is even, any vertex $e$ of $\Gamma_{\{b\}}(f_0)$ cannot reach
+ its neighborhood and thus $\Gamma_{\{b\}}(f_0)$ is not strongly connected.
+ If $b$ is even, any vertex $e$ of $\Gamma_{\{b\}}(f_0)$ cannot reach itself
+ and thus $\Gamma_{\{b\}}(f_0)$ is not strongly connected.
+\end{proof}
