-\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}
+Montrons que:
+\begin{lemma}
+Le graphe d'itération $\textsc{giu}_{\mathcal{P}}(f)$
+est fortement connexe si et seulement si
+la fonction $G_{f_u,\mathcal{P}}$ est topologiquement transitive sur
+$(\mathcal{X}_{\mathsf{N},\mathcal{P}},d)$.
+\end{lemma}\label{prop:trans}
\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
+Supposons tout d'abord que $G_{f_u,\mathcal{P}}$ fortement connexe.
+Soit $x=(e,(u,v)),\check{x}=(\check{e},(\check{u},\check{v}))
+\in \mathcal{X}_{\mathsf{N},\mathcal{P}}$ et $\varepsilon >0$.
+On cherche un point $y$ dans une boule ouverte $\mathcal{B}(x,\varepsilon )$
+et un nombre
+$n_0 \in \mathds{N}$ tels que $G_{f_u,\mathcal{P}}^{n_0}(y)=\check{x}$:
+Cette transitivité forte entrainera la propriété de transitivité classique.
+On peut supposer que $\varepsilon <1$ sans perte de généralité.
+
+Soit $(E,(U,V))$ les éléments de $y$. Comme
+$y$ doit appartenir à $\mathcal{B}(x,\varepsilon)$ et $\varepsilon < 1$,
+$E$ est égal à $e$.
+Soit $k=\lfloor \log_{10} (\varepsilon) \rfloor +1$.
+La distance $d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}((u,v),(U,V))$ est inférieure à
+$\varepsilon$: les $k$ premiers éléments de la partie décimale de
+$d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}((u,v),(U,V))$ sont nuls.
+Soit $k_1$ le plus petit entier tel que, si $V^0=v^0$, ..., $V^{k_1}=v^{k_1}$,
+alors $U^0=u^0$, ..., $U^{\sum_{l=0}^{k_1}V^l-1} = u^{\sum_{l=0}^{k_1}v^l-1}$.
+Alors $d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}((u,v),(U,V))<\varepsilon$.
+En d'autres mots, chaque $y$ de la forme $(e,((u^0, ..., u^{\sum_{l=0}^{k_1}v^l-1}),
+(v^0, ..., v^{k_1}))$ est dans $\mathcal{B}(x,\varepsilon)$.
+
+Soit $y^0$ un tel point et $z=G_{f_u,\mathcal{P}}^{k_1}(y^0) = (e',(u',v'))$.
+$G_{f_u,\mathcal{P}}$ étant fortement connexe,
+il existe un chemin entre $e'$ et $\check{e}$.
+Soit $a_0, \hdots, a_{k_2}$ les arêtes visitées le long de ce chemin.
+On fixe $V^{k_1}=|a_0|$
+(le nombre de termes dans la séquence finie $a_1$),
+$V^{k_1+1}=|a_1|$, ..., $V^{k_1+k_2}=|a_{k_2}|$, et
$U^{k_1}=a_0^0$, $U^{k_1+1}=a_0^1$, ..., $U^{k_1+V_{k_1}-1}=a_0^{V_{k_1}-1}$,
-$U^{k_1+V_{k_1}}=a_1^{0}$, $U^{k_1+V_{k_1}+1}=a_1^{1}$,...
+$U^{k_1+V_{k_1}}=a_1^{0}$, $U^{k_1+V_{k_1}+1}=a_1^{1}$,\ldots
-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|},...,
+Soit $y=(e,((u^0, ..., u^{\sum_{l=0}^{k_1}v^l-1}, a_0^0, ..., a_0^{|a_0|}, a_1^0, ..., a_1^{|a_1|},...,
a_{k_2}^0, ..., a_{k_2}^{|a_{k_2}|},$ \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 $G_{f}^{k_1+k_2}(y)=\check{x}$.
-
+ |a_{k_2}|,\check{v}^0, \check{v}^1, ...)))$.
+Ainsi
+ $y\in \mathcal{B}(x,\varepsilon)$
+ et $G_{f}^{k_1+k_2}(y)=\check{x}$.
-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.
+Réciproquement, si $G_{f_u,\mathcal{P}}$ n'est pas fortement connexe,
+il y a donc deux n{\oe}uds
+$e_1$ et $e_2$ sans chemins entre eux.
+Il n'est ainsi pas possible de trouver un couple $(u,v)\in \mathds{S}_{\mathsf{N},\mathcal{P}}$
+et $n \mathds{N}$ tel que $G_{f_u,\mathcal{P}}^n(e,(u,v))_1=e_2$.
+La boule ouverte $\mathcal{B}(e_2, 1/2)$ ne peut ainsi pas être atteinte
+depuis n'importe quel voisins de $e_1$:
+$G_{f_u,\mathcal{P}}$ n'est pas 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}
+Montrons maintenant que
+\begin{lemma}
+Si $G_{f_u,\mathcal{P}}$ est fortement connexe, alors $G_{f_u,\mathcal{P}}$ est
+régulière sur $(\mathcal{X}_{\mathsf{N},\mathcal{P}}, d)$.
+\end{lemma}
\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
+Soit $x=(e,(u,v)) \in \mathcal{X}_{\mathsf{N},\mathcal{P}}$ et $\varepsilon >0$.
+Comme dans la preuve du lemme~\ref{prop:trans}, soit $k_1 \in \mathds{N}$ tel
+que
$$\left\{(e, ((u^0, ..., u^{v^{k_1-1}},U^0, U^1, ...),(v^0, ..., v^{k_1},V^0, V^1, ...)) \mid \right.$$
$$\left.\forall i,j \in \mathds{N}, U^i \in \llbracket 1, \mathsf{N} \rrbracket, V^j \in \mathcal{P}\right\}
\subset \mathcal{B}(x,\varepsilon),$$
-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:
+et $y=G_{f_u,\mathcal{P}}^{k_1}(e,(u,v))$.
+$G_{f_u,\mathcal{P}}$ étant fortement connexe,
+il existe au moins un chemin entre l'état booléen $y_1$ de $y$ et $e$.
+Nommons $a_0, \hdots, a_{k_2}$ les arêtes d'un tel chemin.
+Le 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 $x$.
+est un point périodique dans le voisinage $\mathcal{B}(x,\varepsilon)$ de $x$.
\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}
+$G_{f_u,\mathcal{P}}$ étant topologiquement transitive and regulière,
+on peut conclure le théorème:
+
+
+\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{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}
