-\section{The relativity of disorder}
-\label{sec:de la relativité du désordre}
+\section{Cryptanalysis of the Proposed PRNG}
-In the next two sections, we investigate the impact of the choices that have
-lead to the definitions of measures in Sections \ref{sec:chaotic iterations} and \ref{deuxième def}.
-\subsection{Impact of the topology's finenesse}
+Mettre ici la preuve de PCH
-Let us firstly introduce the following notations.
+%\section{The relativity of disorder}
+%\label{sec:de la relativité du désordre}
-\begin{notation}
-$\mathcal{X}_\tau$ will denote the topological space
-$\left(\mathcal{X},\tau\right)$, whereas $\mathcal{V}_\tau (x)$ will be the set
-of all the neighborhoods of $x$ when considering the topology $\tau$ (or simply
-$\mathcal{V} (x)$, if there is no ambiguity).
-\end{notation}
+%In the next two sections, we investigate the impact of the choices that have
+%lead to the definitions of measures in Sections \ref{sec:chaotic iterations} and \ref{deuxième def}.
+%\subsection{Impact of the topology's finenesse}
+%Let us firstly introduce the following notations.
-\begin{theorem}
-\label{Th:chaos et finesse}
-Let $\mathcal{X}$ a set and $\tau, \tau'$ two topologies on $\mathcal{X}$ s.t.
-$\tau'$ is finer than $\tau$. Let $f:\mathcal{X} \to \mathcal{X}$, continuous
-both for $\tau$ and $\tau'$.
-
-If $(\mathcal{X}_{\tau'},f)$ is chaotic according to Devaney, then
-$(\mathcal{X}_\tau,f)$ is chaotic too.
-\end{theorem}
-
-\begin{proof}
-Let us firstly establish the transitivity of $(\mathcal{X}_\tau,f)$.
-
-Let $\omega_1, \omega_2$ two open sets of $\tau$. Then $\omega_1, \omega_2 \in
-\tau'$, becaus $\tau'$ is finer than $\tau$. As $f$ is $\tau'-$transitive, we
-can deduce that $\exists n \in \mathds{N}, \omega_1 \cap f^{(n)}(\omega_2) =
-\varnothing$. Consequently, $f$ is $\tau-$transitive.
-
-Let us now consider the regularity of $(\mathcal{X}_\tau,f)$, \emph{i.e.}, for
-all $x \in \mathcal{X}$, and for all $\tau-$neighborhood $V$ of $x$, there is a
-periodic point for $f$ into $V$.
-
-Let $x \in \mathcal{X}$ and $V \in \mathcal{V}_\tau (x)$ a $\tau-$neighborhood
-of $x$. By definition, $\exists \omega \in \tau, x \in \omega \subset V$.
-
-But $\tau \subset \tau'$, so $\omega \in \tau'$, and then $V \in
-\mathcal{V}_{\tau'} (x)$. As $(\mathcal{X}_{\tau'},f)$ is regular, there is a
-periodic point for $f$ into $V$, and the regularity of $(\mathcal{X}_\tau,f)$ is
-proven.
-\end{proof}
-
-\subsection{A given system can always be claimed as chaotic}
-
-Let $f$ an iteration function on $\mathcal{X}$ having at least a fixed point.
-Then this function is chaotic (in a certain way):
-
-\begin{theorem}
-Let $\mathcal{X}$ a nonempty set and $f: \mathcal{X} \to \X$ a function having
-at least a fixed point.
-Then $f$ is $\tau_0-$chaotic, where $\tau_0$ is the trivial (indiscrete)
-topology on $\X$.
-\end{theorem}
-
-
-\begin{proof}
-$f$ is transitive when $\forall \omega, \omega' \in \tau_0 \setminus
-\{\varnothing\}, \exists n \in \mathds{N}, f^{(n)}(\omega) \cap \omega' \neq
-\varnothing$.
-As $\tau_0 = \left\{ \varnothing, \X \right\}$, this is equivalent to look for
-an integer $n$ s.t. $f^{(n)}\left( \X \right) \cap \X \neq \varnothing$. For
-instance, $n=0$ is appropriate.
-
-Let us now consider $x \in \X$ and $V \in \mathcal{V}_{\tau_0} (x)$. Then $V =
-\mathcal{X}$, so $V$ has at least a fixed point for $f$. Consequently $f$ is
-regular, and the result is established.
-\end{proof}
-
-
+%\begin{notation}
+%$\mathcal{X}_\tau$ will denote the topological space
+%$\left(\mathcal{X},\tau\right)$, whereas $\mathcal{V}_\tau (x)$ will be the set
+%of all the neighborhoods of $x$ when considering the topology $\tau$ (or simply
+%$\mathcal{V} (x)$, if there is no ambiguity).
+%\end{notation}
-\subsection{A given system can always be claimed as non-chaotic}
-\begin{theorem}
-Let $\mathcal{X}$ be a set and $f: \mathcal{X} \to \X$.
-If $\X$ is infinite, then $\left( \X_{\tau_\infty}, f\right)$ is not chaotic
-(for the Devaney's formulation), where $\tau_\infty$ is the discrete topology.
-\end{theorem}
-
-\begin{proof}
-Let us prove it by contradiction, assuming that $\left(\X_{\tau_\infty},
-f\right)$ is both transitive and regular.
+%\begin{theorem}
+%\label{Th:chaos et finesse}
+%Let $\mathcal{X}$ a set and $\tau, \tau'$ two topologies on $\mathcal{X}$ s.t.
+%$\tau'$ is finer than $\tau$. Let $f:\mathcal{X} \to \mathcal{X}$, continuous
+%both for $\tau$ and $\tau'$.
-Let $x \in \X$ and $\{x\}$ one of its neighborhood. This neighborhood must
-contain a periodic point for $f$, if we want that $\left(\X_{\tau_\infty},
-f\right)$ is regular. Then $x$ must be a periodic point of $f$.
+%If $(\mathcal{X}_{\tau'},f)$ is chaotic according to Devaney, then
+%$(\mathcal{X}_\tau,f)$ is chaotic too.
+%\end{theorem}
-Let $I_x = \left\{ f^{(n)}(x), n \in \mathds{N}\right\}$. This set is finite
-because $x$ is periodic, and $\mathcal{X}$ is infinite, then $\exists y \in
-\mathcal{X}, y \notin I_x$.
+%\begin{proof}
+%Let us firstly establish the transitivity of $(\mathcal{X}_\tau,f)$.
-As $\left(\X_{\tau_\infty}, f\right)$ must be transitive, for all open nonempty
-sets $A$ and $B$, an integer $n$ must satisfy $f^{(n)}(A) \cap B \neq
-\varnothing$. However $\{x\}$ and $\{y\}$ are open sets and $y \notin I_x
-\Rightarrow \forall n, f^{(n)}\left( \{x\} \right) \cap \{y\} = \varnothing$.
-\end{proof}
+%Let $\omega_1, \omega_2$ two open sets of $\tau$. Then $\omega_1, \omega_2 \in
+%\tau'$, becaus $\tau'$ is finer than $\tau$. As $f$ is $\tau'-$transitive, we
+%can deduce that $\exists n \in \mathds{N}, \omega_1 \cap f^{(n)}(\omega_2) =
+%\varnothing$. Consequently, $f$ is $\tau-$transitive.
+%Let us now consider the regularity of $(\mathcal{X}_\tau,f)$, \emph{i.e.}, for
+%all $x \in \mathcal{X}$, and for all $\tau-$neighborhood $V$ of $x$, there is a
+%periodic point for $f$ into $V$.
+%Let $x \in \mathcal{X}$ and $V \in \mathcal{V}_\tau (x)$ a $\tau-$neighborhood
+%of $x$. By definition, $\exists \omega \in \tau, x \in \omega \subset V$.
+%But $\tau \subset \tau'$, so $\omega \in \tau'$, and then $V \in
+%\mathcal{V}_{\tau'} (x)$. As $(\mathcal{X}_{\tau'},f)$ is regular, there is a
+%periodic point for $f$ into $V$, and the regularity of $(\mathcal{X}_\tau,f)$ is
+%proven.
+%\end{proof}
+%\subsection{A given system can always be claimed as chaotic}
+%Let $f$ an iteration function on $\mathcal{X}$ having at least a fixed point.
+%Then this function is chaotic (in a certain way):
-\section{Chaos on the order topology}
-\label{sec: chaos order topology}
-\subsection{The phase space is an interval of the real line}
+%\begin{theorem}
+%Let $\mathcal{X}$ a nonempty set and $f: \mathcal{X} \to \X$ a function having
+%at least a fixed point.
+%Then $f$ is $\tau_0-$chaotic, where $\tau_0$ is the trivial (indiscrete)
+%topology on $\X$.
+%\end{theorem}
-\subsubsection{Toward a topological semiconjugacy}
-In what follows, our intention is to establish, by using a topological
-semiconjugacy, that chaotic iterations over $\mathcal{X}$ can be described as
-iterations on a real interval. To do so, we must firstly introduce some
-notations and terminologies.
+%\begin{proof}
+%$f$ is transitive when $\forall \omega, \omega' \in \tau_0 \setminus
+%\{\varnothing\}, \exists n \in \mathds{N}, f^{(n)}(\omega) \cap \omega' \neq
+%\varnothing$.
+%As $\tau_0 = \left\{ \varnothing, \X \right\}$, this is equivalent to look for
+%an integer $n$ s.t. $f^{(n)}\left( \X \right) \cap \X \neq \varnothing$. For
+%instance, $n=0$ is appropriate.
-Let $\mathcal{S}_\mathsf{N}$ be the set of sequences belonging into $\llbracket
-1; \mathsf{N}\rrbracket$ and $\mathcal{X}_{\mathsf{N}} = \mathcal{S}_\mathsf{N}
-\times \B^\mathsf{N}$.
+%Let us now consider $x \in \X$ and $V \in \mathcal{V}_{\tau_0} (x)$. Then $V =
+%\mathcal{X}$, so $V$ has at least a fixed point for $f$. Consequently $f$ is
+%regular, and the result is established.
+%\end{proof}
-\begin{definition}
-The function $\varphi: \mathcal{S}_{10} \times\mathds{B}^{10} \rightarrow \big[
-0, 2^{10} \big[$ is defined by:
-\begin{equation}
- \begin{array}{cccl}
-\varphi: & \mathcal{X}_{10} = \mathcal{S}_{10} \times\mathds{B}^{10}&
-\longrightarrow & \big[ 0, 2^{10} \big[ \\
- & (S,E) = \left((S^0, S^1, \hdots ); (E_0, \hdots, E_9)\right) & \longmapsto &
-\varphi \left((S,E)\right)
-\end{array}
-\end{equation}
-where $\varphi\left((S,E)\right)$ is the real number:
-\begin{itemize}
-\item whose integral part $e$ is $\displaystyle{\sum_{k=0}^9 2^{9-k} E_k}$, that
-is, the binary digits of $e$ are $E_0 ~ E_1 ~ \hdots ~ E_9$.
-\item whose decimal part $s$ is equal to $s = 0,S^0~ S^1~ S^2~ \hdots =
-\sum_{k=1}^{+\infty} 10^{-k} S^{k-1}.$
-\end{itemize}
-\end{definition}
+%\subsection{A given system can always be claimed as non-chaotic}
-$\varphi$ realizes the association between a point of $\mathcal{X}_{10}$ and a
-real number into $\big[ 0, 2^{10} \big[$. We must now translate the chaotic
-iterations $\Go$ on this real interval. To do so, two intermediate functions
-over $\big[ 0, 2^{10} \big[$ must be introduced:
+%\begin{theorem}
+%Let $\mathcal{X}$ be a set and $f: \mathcal{X} \to \X$.
+%If $\X$ is infinite, then $\left( \X_{\tau_\infty}, f\right)$ is not chaotic
+%(for the Devaney's formulation), where $\tau_\infty$ is the discrete topology.
+%\end{theorem}
+%\begin{proof}
+%Let us prove it by contradiction, assuming that $\left(\X_{\tau_\infty},
+%f\right)$ is both transitive and regular.
-\begin{definition}
-\label{def:e et s}
-Let $x \in \big[ 0, 2^{10} \big[$ and:
-\begin{itemize}
-\item $e_0, \hdots, e_9$ the binary digits of the integral part of $x$:
-$\displaystyle{\lfloor x \rfloor = \sum_{k=0}^{9} 2^{9-k} e_k}$.
-\item $(s^k)_{k\in \mathds{N}}$ the digits of $x$, where the chosen decimal
-decomposition of $x$ is the one that does not have an infinite number of 9:
-$\displaystyle{x = \lfloor x \rfloor + \sum_{k=0}^{+\infty} s^k 10^{-k-1}}$.
-\end{itemize}
-$e$ and $s$ are thus defined as follows:
-\begin{equation}
-\begin{array}{cccl}
-e: & \big[ 0, 2^{10} \big[ & \longrightarrow & \mathds{B}^{10} \\
- & x & \longmapsto & (e_0, \hdots, e_9)
-\end{array}
-\end{equation}
-and
-\begin{equation}
- \begin{array}{cccc}
-s: & \big[ 0, 2^{10} \big[ & \longrightarrow & \llbracket 0, 9
-\rrbracket^{\mathds{N}} \\
- & x & \longmapsto & (s^k)_{k \in \mathds{N}}
-\end{array}
-\end{equation}
-\end{definition}
-
-We are now able to define the function $g$, whose goal is to translate the
-chaotic iterations $\Go$ on an interval of $\mathds{R}$.
+%Let $x \in \X$ and $\{x\}$ one of its neighborhood. This neighborhood must
+%contain a periodic point for $f$, if we want that $\left(\X_{\tau_\infty},
+%f\right)$ is regular. Then $x$ must be a periodic point of $f$.
+
+%Let $I_x = \left\{ f^{(n)}(x), n \in \mathds{N}\right\}$. This set is finite
+%because $x$ is periodic, and $\mathcal{X}$ is infinite, then $\exists y \in
+%\mathcal{X}, y \notin I_x$.
+
+%As $\left(\X_{\tau_\infty}, f\right)$ must be transitive, for all open nonempty
+%sets $A$ and $B$, an integer $n$ must satisfy $f^{(n)}(A) \cap B \neq
+%\varnothing$. However $\{x\}$ and $\{y\}$ are open sets and $y \notin I_x
+%\Rightarrow \forall n, f^{(n)}\left( \{x\} \right) \cap \{y\} = \varnothing$.
+%\end{proof}
+
+
+
+
+
+
+%\section{Chaos on the order topology}
+%\label{sec: chaos order topology}
+%\subsection{The phase space is an interval of the real line}
+
+%\subsubsection{Toward a topological semiconjugacy}
+
+%In what follows, our intention is to establish, by using a topological
+%semiconjugacy, that chaotic iterations over $\mathcal{X}$ can be described as
+%iterations on a real interval. To do so, we must firstly introduce some
+%notations and terminologies.
+
+%Let $\mathcal{S}_\mathsf{N}$ be the set of sequences belonging into $\llbracket
+%1; \mathsf{N}\rrbracket$ and $\mathcal{X}_{\mathsf{N}} = \mathcal{S}_\mathsf{N}
+%\times \B^\mathsf{N}$.
+
+
+%\begin{definition}
+%The function $\varphi: \mathcal{S}_{10} \times\mathds{B}^{10} \rightarrow \big[
+%0, 2^{10} \big[$ is defined by:
+%\begin{equation}
+% \begin{array}{cccl}
+%\varphi: & \mathcal{X}_{10} = \mathcal{S}_{10} \times\mathds{B}^{10}&
+%\longrightarrow & \big[ 0, 2^{10} \big[ \\
+% & (S,E) = \left((S^0, S^1, \hdots ); (E_0, \hdots, E_9)\right) & \longmapsto &
+%\varphi \left((S,E)\right)
+%\end{array}
+%\end{equation}
+%where $\varphi\left((S,E)\right)$ is the real number:
+%\begin{itemize}
+%\item whose integral part $e$ is $\displaystyle{\sum_{k=0}^9 2^{9-k} E_k}$, that
+%is, the binary digits of $e$ are $E_0 ~ E_1 ~ \hdots ~ E_9$.
+%\item whose decimal part $s$ is equal to $s = 0,S^0~ S^1~ S^2~ \hdots =
+%\sum_{k=1}^{+\infty} 10^{-k} S^{k-1}.$
+%\end{itemize}
+%\end{definition}
+
+
+
+%$\varphi$ realizes the association between a point of $\mathcal{X}_{10}$ and a
+%real number into $\big[ 0, 2^{10} \big[$. We must now translate the chaotic
+%iterations $\Go$ on this real interval. To do so, two intermediate functions
+%over $\big[ 0, 2^{10} \big[$ must be introduced:
+
+
+%\begin{definition}
+%\label{def:e et s}
+%Let $x \in \big[ 0, 2^{10} \big[$ and:
+%\begin{itemize}
+%\item $e_0, \hdots, e_9$ the binary digits of the integral part of $x$:
+%$\displaystyle{\lfloor x \rfloor = \sum_{k=0}^{9} 2^{9-k} e_k}$.
+%\item $(s^k)_{k\in \mathds{N}}$ the digits of $x$, where the chosen decimal
+%decomposition of $x$ is the one that does not have an infinite number of 9:
+%$\displaystyle{x = \lfloor x \rfloor + \sum_{k=0}^{+\infty} s^k 10^{-k-1}}$.
+%\end{itemize}
+%$e$ and $s$ are thus defined as follows:
+%\begin{equation}
+%\begin{array}{cccl}
+%e: & \big[ 0, 2^{10} \big[ & \longrightarrow & \mathds{B}^{10} \\
+% & x & \longmapsto & (e_0, \hdots, e_9)
+%\end{array}
+%\end{equation}
+%and
+%\begin{equation}
+% \begin{array}{cccc}
+%s: & \big[ 0, 2^{10} \big[ & \longrightarrow & \llbracket 0, 9
+%\rrbracket^{\mathds{N}} \\
+% & x & \longmapsto & (s^k)_{k \in \mathds{N}}
+%\end{array}
+%\end{equation}
+%\end{definition}
+
+%We are now able to define the function $g$, whose goal is to translate the
+%chaotic iterations $\Go$ on an interval of $\mathds{R}$.
+
+%\begin{definition}
+%$g:\big[ 0, 2^{10} \big[ \longrightarrow \big[ 0, 2^{10} \big[$ is defined by:
+%\begin{equation}
+%\begin{array}{cccc}
+%g: & \big[ 0, 2^{10} \big[ & \longrightarrow & \big[ 0, 2^{10} \big[ \\
+% & x & \longmapsto & g(x)
+%\end{array}
+%\end{equation}
+%where g(x) is the real number of $\big[ 0, 2^{10} \big[$ defined bellow:
+%\begin{itemize}
+%\item its integral part has a binary decomposition equal to $e_0', \hdots,
+%e_9'$, with:
+% \begin{equation}
+%e_i' = \left\{
+%\begin{array}{ll}
+%e(x)_i & \textrm{ if } i \neq s^0\\
+%e(x)_i + 1 \textrm{ (mod 2)} & \textrm{ if } i = s^0\\
+%\end{array}
+%\right.
+%\end{equation}
+%\item whose decimal part is $s(x)^1, s(x)^2, \hdots$
+%\end{itemize}
+%\end{definition}
+
+%\bigskip
+
+
+%In other words, if $x = \displaystyle{\sum_{k=0}^{9} 2^{9-k} e_k +
+%\sum_{k=0}^{+\infty} s^{k} ~10^{-k-1}}$, then:
+%\begin{equation}
+%g(x) =
+%\displaystyle{\sum_{k=0}^{9} 2^{9-k} (e_k + \delta(k,s^0) \textrm{ (mod 2)}) +
+%\sum_{k=0}^{+\infty} s^{k+1} 10^{-k-1}}.
+%\end{equation}
+
+
+%\subsubsection{Defining a metric on $\big[ 0, 2^{10} \big[$}
+
+%Numerous metrics can be defined on the set $\big[ 0, 2^{10} \big[$, the most
+%usual one being the Euclidian distance recalled bellow:
+
+%\begin{notation}
+%\index{distance!euclidienne}
+%$\Delta$ is the Euclidian distance on $\big[ 0, 2^{10} \big[$, that is,
+%$\Delta(x,y) = |y-x|^2$.
+%\end{notation}
+
+%\medskip
+
+%This Euclidian distance does not reproduce exactly the notion of proximity
+%induced by our first distance $d$ on $\X$. Indeed $d$ is finer than $\Delta$.
+%This is the reason why we have to introduce the following metric:
+
+
+
+%\begin{definition}
+%Let $x,y \in \big[ 0, 2^{10} \big[$.
+%$D$ denotes the function from $\big[ 0, 2^{10} \big[^2$ to $\mathds{R}^+$
+%defined by: $D(x,y) = D_e\left(e(x),e(y)\right) + D_s\left(s(x),s(y)\right)$,
+%where:
+%\begin{center}
+%$\displaystyle{D_e(E,\check{E}) = \sum_{k=0}^\mathsf{9} \delta (E_k,
+%\check{E}_k)}$, ~~and~ $\displaystyle{D_s(S,\check{S}) = \sum_{k = 1}^\infty
+%\dfrac{|S^k-\check{S}^k|}{10^k}}$.
+%\end{center}
+%\end{definition}
+
+%\begin{proposition}
+%$D$ is a distance on $\big[ 0, 2^{10} \big[$.
+%\end{proposition}
+
+%\begin{proof}
+%The three axioms defining a distance must be checked.
+%\begin{itemize}
+%\item $D \geqslant 0$, because everything is positive in its definition. If
+%$D(x,y)=0$, then $D_e(x,y)=0$, so the integral parts of $x$ and $y$ are equal
+%(they have the same binary decomposition). Additionally, $D_s(x,y) = 0$, then
+%$\forall k \in \mathds{N}^*, s(x)^k = s(y)^k$. In other words, $x$ and $y$ have
+%the same $k-$th decimal digit, $\forall k \in \mathds{N}^*$. And so $x=y$.
+%\item $D(x,y)=D(y,x)$.
+%\item Finally, the triangular inequality is obtained due to the fact that both
+%$\delta$ and $\Delta(x,y)=|x-y|$ satisfy it.
+%\end{itemize}
+%\end{proof}
+
+
+%The convergence of sequences according to $D$ is not the same than the usual
+%convergence related to the Euclidian metric. For instance, if $x^n \to x$
+%according to $D$, then necessarily the integral part of each $x^n$ is equal to
+%the integral part of $x$ (at least after a given threshold), and the decimal
+%part of $x^n$ corresponds to the one of $x$ ``as far as required''.
+%To illustrate this fact, a comparison between $D$ and the Euclidian distance is
+%given Figure \ref{fig:comparaison de distances}. These illustrations show that
+%$D$ is richer and more refined than the Euclidian distance, and thus is more
+%precise.
+
+
+%\begin{figure}[t]
+%\begin{center}
+% \subfigure[Function $x \to dist(x;1,234) $ on the interval
+%$(0;5)$.]{\includegraphics[scale=.35]{DvsEuclidien.pdf}}\quad
+% \subfigure[Function $x \to dist(x;3) $ on the interval
+%$(0;5)$.]{\includegraphics[scale=.35]{DvsEuclidien2.pdf}}
+%\end{center}
+%\caption{Comparison between $D$ (in blue) and the Euclidian distane (in green).}
+%\label{fig:comparaison de distances}
+%\end{figure}
+
+
+
+
+%\subsubsection{The semiconjugacy}
+
+%It is now possible to define a topological semiconjugacy between $\mathcal{X}$
+%and an interval of $\mathds{R}$:
+
+%\begin{theorem}
+%Chaotic iterations on the phase space $\mathcal{X}$ are simple iterations on
+%$\mathds{R}$, which is illustrated by the semiconjugacy of the diagram bellow:
+%\begin{equation*}
+%\begin{CD}
+%\left(~\mathcal{S}_{10} \times\mathds{B}^{10}, d~\right) @>G_{f_0}>>
+%\left(~\mathcal{S}_{10} \times\mathds{B}^{10}, d~\right)\\
+% @V{\varphi}VV @VV{\varphi}V\\
+%\left( ~\big[ 0, 2^{10} \big[, D~\right) @>>g> \left(~\big[ 0, 2^{10} \big[,
+%D~\right)
+%\end{CD}
+%\end{equation*}
+%\end{theorem}
+
+%\begin{proof}
+%$\varphi$ has been constructed in order to be continuous and onto.
+%\end{proof}
+
+%In other words, $\mathcal{X}$ is approximately equal to $\big[ 0, 2^\mathsf{N}
+%\big[$.
+
+
+
+
+
+
+%\subsection{Study of the chaotic iterations described as a real function}
+
+
+%\begin{figure}[t]
+%\begin{center}
+% \subfigure[ICs on the interval
+%$(0,9;1)$.]{\includegraphics[scale=.35]{ICs09a1.pdf}}\quad
+% \subfigure[ICs on the interval
+%$(0,7;1)$.]{\includegraphics[scale=.35]{ICs07a95.pdf}}\\
+% \subfigure[ICs on the interval
+%$(0,5;1)$.]{\includegraphics[scale=.35]{ICs05a1.pdf}}\quad
+% \subfigure[ICs on the interval
+%$(0;1)$]{\includegraphics[scale=.35]{ICs0a1.pdf}}
+%\end{center}
+%\caption{Representation of the chaotic iterations.}
+%\label{fig:ICs}
+%\end{figure}
+
+
+
+
+%\begin{figure}[t]
+%\begin{center}
+% \subfigure[ICs on the interval
+%$(510;514)$.]{\includegraphics[scale=.35]{ICs510a514.pdf}}\quad
+% \subfigure[ICs on the interval
+%$(1000;1008)$]{\includegraphics[scale=.35]{ICs1000a1008.pdf}}
+%\end{center}
+%\caption{ICs on small intervals.}
+%\label{fig:ICs2}
+%\end{figure}
+
+%\begin{figure}[t]
+%\begin{center}
+% \subfigure[ICs on the interval
+%$(0;16)$.]{\includegraphics[scale=.3]{ICs0a16.pdf}}\quad
+% \subfigure[ICs on the interval
+%$(40;70)$.]{\includegraphics[scale=.45]{ICs40a70.pdf}}\quad
+%\end{center}
+%\caption{General aspect of the chaotic iterations.}
+%\label{fig:ICs3}
+%\end{figure}
+
+
+%We have written a Python program to represent the chaotic iterations with the
+%vectorial negation on the real line $\mathds{R}$. Various representations of
+%these CIs are given in Figures \ref{fig:ICs}, \ref{fig:ICs2} and \ref{fig:ICs3}.
+%It can be remarked that the function $g$ is a piecewise linear function: it is
+%linear on each interval having the form $\left[ \dfrac{n}{10},
+%\dfrac{n+1}{10}\right[$, $n \in \llbracket 0;2^{10}\times 10 \rrbracket$ and its
+%slope is equal to 10. Let us justify these claims:
+
+%\begin{proposition}
+%\label{Prop:derivabilite des ICs}
+%Chaotic iterations $g$ defined on $\mathds{R}$ have derivatives of all orders on
+%$\big[ 0, 2^{10} \big[$, except on the 10241 points in $I$ defined by $\left\{
+%\dfrac{n}{10} ~\big/~ n \in \llbracket 0;2^{10}\times 10\rrbracket \right\}$.
+
+%Furthermore, on each interval of the form $\left[ \dfrac{n}{10},
+%\dfrac{n+1}{10}\right[$, with $n \in \llbracket 0;2^{10}\times 10 \rrbracket$,
+%$g$ is a linear function, having a slope equal to 10: $\forall x \notin I,
+%g'(x)=10$.
+%\end{proposition}
-\begin{definition}
-$g:\big[ 0, 2^{10} \big[ \longrightarrow \big[ 0, 2^{10} \big[$ is defined by:
-\begin{equation}
-\begin{array}{cccc}
-g: & \big[ 0, 2^{10} \big[ & \longrightarrow & \big[ 0, 2^{10} \big[ \\
- & x & \longmapsto & g(x)
-\end{array}
-\end{equation}
-where g(x) is the real number of $\big[ 0, 2^{10} \big[$ defined bellow:
-\begin{itemize}
-\item its integral part has a binary decomposition equal to $e_0', \hdots,
-e_9'$, with:
- \begin{equation}
-e_i' = \left\{
-\begin{array}{ll}
-e(x)_i & \textrm{ if } i \neq s^0\\
-e(x)_i + 1 \textrm{ (mod 2)} & \textrm{ if } i = s^0\\
-\end{array}
-\right.
-\end{equation}
-\item whose decimal part is $s(x)^1, s(x)^2, \hdots$
-\end{itemize}
-\end{definition}
-
-\bigskip
-
-
-In other words, if $x = \displaystyle{\sum_{k=0}^{9} 2^{9-k} e_k +
-\sum_{k=0}^{+\infty} s^{k} ~10^{-k-1}}$, then:
-\begin{equation}
-g(x) =
-\displaystyle{\sum_{k=0}^{9} 2^{9-k} (e_k + \delta(k,s^0) \textrm{ (mod 2)}) +
-\sum_{k=0}^{+\infty} s^{k+1} 10^{-k-1}}.
-\end{equation}
-
-
-\subsubsection{Defining a metric on $\big[ 0, 2^{10} \big[$}
-
-Numerous metrics can be defined on the set $\big[ 0, 2^{10} \big[$, the most
-usual one being the Euclidian distance recalled bellow:
-
-\begin{notation}
-\index{distance!euclidienne}
-$\Delta$ is the Euclidian distance on $\big[ 0, 2^{10} \big[$, that is,
-$\Delta(x,y) = |y-x|^2$.
-\end{notation}
-
-\medskip
-
-This Euclidian distance does not reproduce exactly the notion of proximity
-induced by our first distance $d$ on $\X$. Indeed $d$ is finer than $\Delta$.
-This is the reason why we have to introduce the following metric:
-
-
-
-\begin{definition}
-Let $x,y \in \big[ 0, 2^{10} \big[$.
-$D$ denotes the function from $\big[ 0, 2^{10} \big[^2$ to $\mathds{R}^+$
-defined by: $D(x,y) = D_e\left(e(x),e(y)\right) + D_s\left(s(x),s(y)\right)$,
-where:
-\begin{center}
-$\displaystyle{D_e(E,\check{E}) = \sum_{k=0}^\mathsf{9} \delta (E_k,
-\check{E}_k)}$, ~~and~ $\displaystyle{D_s(S,\check{S}) = \sum_{k = 1}^\infty
-\dfrac{|S^k-\check{S}^k|}{10^k}}$.
-\end{center}
-\end{definition}
-
-\begin{proposition}
-$D$ is a distance on $\big[ 0, 2^{10} \big[$.
-\end{proposition}
-
-\begin{proof}
-The three axioms defining a distance must be checked.
-\begin{itemize}
-\item $D \geqslant 0$, because everything is positive in its definition. If
-$D(x,y)=0$, then $D_e(x,y)=0$, so the integral parts of $x$ and $y$ are equal
-(they have the same binary decomposition). Additionally, $D_s(x,y) = 0$, then
-$\forall k \in \mathds{N}^*, s(x)^k = s(y)^k$. In other words, $x$ and $y$ have
-the same $k-$th decimal digit, $\forall k \in \mathds{N}^*$. And so $x=y$.
-\item $D(x,y)=D(y,x)$.
-\item Finally, the triangular inequality is obtained due to the fact that both
-$\delta$ and $\Delta(x,y)=|x-y|$ satisfy it.
-\end{itemize}
-\end{proof}
-
-
-The convergence of sequences according to $D$ is not the same than the usual
-convergence related to the Euclidian metric. For instance, if $x^n \to x$
-according to $D$, then necessarily the integral part of each $x^n$ is equal to
-the integral part of $x$ (at least after a given threshold), and the decimal
-part of $x^n$ corresponds to the one of $x$ ``as far as required''.
-To illustrate this fact, a comparison between $D$ and the Euclidian distance is
-given Figure \ref{fig:comparaison de distances}. These illustrations show that
-$D$ is richer and more refined than the Euclidian distance, and thus is more
-precise.
-
-
-\begin{figure}[t]
-\begin{center}
- \subfigure[Function $x \to dist(x;1,234) $ on the interval
-$(0;5)$.]{\includegraphics[scale=.35]{DvsEuclidien.pdf}}\quad
- \subfigure[Function $x \to dist(x;3) $ on the interval
-$(0;5)$.]{\includegraphics[scale=.35]{DvsEuclidien2.pdf}}
-\end{center}
-\caption{Comparison between $D$ (in blue) and the Euclidian distane (in green).}
-\label{fig:comparaison de distances}
-\end{figure}
-
-
-
-
-\subsubsection{The semiconjugacy}
-
-It is now possible to define a topological semiconjugacy between $\mathcal{X}$
-and an interval of $\mathds{R}$:
-
-\begin{theorem}
-Chaotic iterations on the phase space $\mathcal{X}$ are simple iterations on
-$\mathds{R}$, which is illustrated by the semiconjugacy of the diagram bellow:
-\begin{equation*}
-\begin{CD}
-\left(~\mathcal{S}_{10} \times\mathds{B}^{10}, d~\right) @>G_{f_0}>>
-\left(~\mathcal{S}_{10} \times\mathds{B}^{10}, d~\right)\\
- @V{\varphi}VV @VV{\varphi}V\\
-\left( ~\big[ 0, 2^{10} \big[, D~\right) @>>g> \left(~\big[ 0, 2^{10} \big[,
-D~\right)
-\end{CD}
-\end{equation*}
-\end{theorem}
-
-\begin{proof}
-$\varphi$ has been constructed in order to be continuous and onto.
-\end{proof}
-
-In other words, $\mathcal{X}$ is approximately equal to $\big[ 0, 2^\mathsf{N}
-\big[$.
-
-
-
-
-
-
-\subsection{Study of the chaotic iterations described as a real function}
-
-
-\begin{figure}[t]
-\begin{center}
- \subfigure[ICs on the interval
-$(0,9;1)$.]{\includegraphics[scale=.35]{ICs09a1.pdf}}\quad
- \subfigure[ICs on the interval
-$(0,7;1)$.]{\includegraphics[scale=.35]{ICs07a95.pdf}}\\
- \subfigure[ICs on the interval
-$(0,5;1)$.]{\includegraphics[scale=.35]{ICs05a1.pdf}}\quad
- \subfigure[ICs on the interval
-$(0;1)$]{\includegraphics[scale=.35]{ICs0a1.pdf}}
-\end{center}
-\caption{Representation of the chaotic iterations.}
-\label{fig:ICs}
-\end{figure}
-
-
-
-
-\begin{figure}[t]
-\begin{center}
- \subfigure[ICs on the interval
-$(510;514)$.]{\includegraphics[scale=.35]{ICs510a514.pdf}}\quad
- \subfigure[ICs on the interval
-$(1000;1008)$]{\includegraphics[scale=.35]{ICs1000a1008.pdf}}
-\end{center}
-\caption{ICs on small intervals.}
-\label{fig:ICs2}
-\end{figure}
-
-\begin{figure}[t]
-\begin{center}
- \subfigure[ICs on the interval
-$(0;16)$.]{\includegraphics[scale=.3]{ICs0a16.pdf}}\quad
- \subfigure[ICs on the interval
-$(40;70)$.]{\includegraphics[scale=.45]{ICs40a70.pdf}}\quad
-\end{center}
-\caption{General aspect of the chaotic iterations.}
-\label{fig:ICs3}
-\end{figure}
-
-
-We have written a Python program to represent the chaotic iterations with the
-vectorial negation on the real line $\mathds{R}$. Various representations of
-these CIs are given in Figures \ref{fig:ICs}, \ref{fig:ICs2} and \ref{fig:ICs3}.
-It can be remarked that the function $g$ is a piecewise linear function: it is
-linear on each interval having the form $\left[ \dfrac{n}{10},
-\dfrac{n+1}{10}\right[$, $n \in \llbracket 0;2^{10}\times 10 \rrbracket$ and its
-slope is equal to 10. Let us justify these claims:
-
-\begin{proposition}
-\label{Prop:derivabilite des ICs}
-Chaotic iterations $g$ defined on $\mathds{R}$ have derivatives of all orders on
-$\big[ 0, 2^{10} \big[$, except on the 10241 points in $I$ defined by $\left\{
-\dfrac{n}{10} ~\big/~ n \in \llbracket 0;2^{10}\times 10\rrbracket \right\}$.
-
-Furthermore, on each interval of the form $\left[ \dfrac{n}{10},
-\dfrac{n+1}{10}\right[$, with $n \in \llbracket 0;2^{10}\times 10 \rrbracket$,
-$g$ is a linear function, having a slope equal to 10: $\forall x \notin I,
-g'(x)=10$.
-\end{proposition}
-
-
-\begin{proof}
-Let $I_n = \left[ \dfrac{n}{10}, \dfrac{n+1}{10}\right[$, with $n \in \llbracket
-0;2^{10}\times 10 \rrbracket$. All the points of $I_n$ have the same integral
-prat $e$ and the same decimal part $s^0$: on the set $I_n$, functions $e(x)$
-and $x \mapsto s(x)^0$ of Definition \ref{def:e et s} only depend on $n$. So all
-the images $g(x)$ of these points $x$:
-\begin{itemize}
-\item Have the same integral part, which is $e$, except probably the bit number
-$s^0$. In other words, this integer has approximately the same binary
-decomposition than $e$, the sole exception being the digit $s^0$ (this number is
-then either $e+2^{10-s^0}$ or $e-2^{10-s^0}$, depending on the parity of $s^0$,
-\emph{i.e.}, it is equal to $e+(-1)^{s^0}\times 2^{10-s^0}$).
-\item A shift to the left has been applied to the decimal part $y$, losing by
-doing so the common first digit $s^0$. In other words, $y$ has been mapped into
-$10\times y - s^0$.
-\end{itemize}
-To sum up, the action of $g$ on the points of $I$ is as follows: first, make a
-multiplication by 10, and second, add the same constant to each term, which is
-$\dfrac{1}{10}\left(e+(-1)^{s^0}\times 2^{10-s^0}\right)-s^0$.
-\end{proof}
-
-\begin{remark}
-Finally, chaotic iterations are elements of the large family of functions that
-are both chaotic and piecewise linear (like the tent map).
-\end{remark}
-
-
-
-\subsection{Comparison of the two metrics on $\big[ 0, 2^\mathsf{N} \big[$}
-
-The two propositions bellow allow to compare our two distances on $\big[ 0,
-2^\mathsf{N} \big[$:
-
-\begin{proposition}
-Id: $\left(~\big[ 0, 2^\mathsf{N} \big[,\Delta~\right) \to \left(~\big[ 0,
-2^\mathsf{N} \big[, D~\right)$ is not continuous.
-\end{proposition}
-
-\begin{proof}
-The sequence $x^n = 1,999\hdots 999$ constituted by $n$ 9 as decimal part, is
-such that:
-\begin{itemize}
-\item $\Delta (x^n,2) \to 0.$
-\item But $D(x^n,2) \geqslant 1$, then $D(x^n,2)$ does not converge to 0.
-\end{itemize}
-
-The sequential characterization of the continuity concludes the demonstration.
-\end{proof}
-
-
-
-A contrario:
-
-\begin{proposition}
-Id: $\left(~\big[ 0, 2^\mathsf{N} \big[,D~\right) \to \left(~\big[ 0,
-2^\mathsf{N} \big[, \Delta ~\right)$ is a continuous fonction.
-\end{proposition}
-
-\begin{proof}
-If $D(x^n,x) \to 0$, then $D_e(x^n,x) = 0$ at least for $n$ larger than a given
-threshold, because $D_e$ only returns integers. So, after this threshold, the
-integral parts of all the $x^n$ are equal to the integral part of $x$.
-
-Additionally, $D_s(x^n, x) \to 0$, then $\forall k \in \mathds{N}^*, \exists N_k
-\in \mathds{N}, n \geqslant N_k \Rightarrow D_s(x^n,x) \leqslant 10^{-k}$. This
-means that for all $k$, an index $N_k$ can be found such that, $\forall n
-\geqslant N_k$, all the $x^n$ have the same $k$ firsts digits, which are the
-digits of $x$. We can deduce the convergence $\Delta(x^n,x) \to 0$, and thus the
-result.
-\end{proof}
-
-The conclusion of these propositions is that the proposed metric is more precise
-than the Euclidian distance, that is:
-
-\begin{corollary}
-$D$ is finer than the Euclidian distance $\Delta$.
-\end{corollary}
-
-This corollary can be reformulated as follows:
-
-\begin{itemize}
-\item The topology produced by $\Delta$ is a subset of the topology produced by
-$D$.
-\item $D$ has more open sets than $\Delta$.
-\item It is harder to converge for the topology $\tau_D$ inherited by $D$, than
-to converge with the one inherited by $\Delta$, which is denoted here by
-$\tau_\Delta$.
-\end{itemize}
-
-
-\subsection{Chaos of the chaotic iterations on $\mathds{R}$}
-\label{chpt:Chaos des itérations chaotiques sur R}
-
-
-
-\subsubsection{Chaos according to Devaney}
-
-We have recalled previously that the chaotic iterations $\left(\Go,
-\mathcal{X}_d\right)$ are chaotic according to the formulation of Devaney. We
-can deduce that they are chaotic on $\mathds{R}$ too, when considering the order
-topology, because:
-\begin{itemize}
-\item $\left(\Go, \mathcal{X}_d\right)$ and $\left(g, \big[ 0, 2^{10}
-\big[_D\right)$ are semiconjugate by $\varphi$,
-\item Then $\left(g, \big[ 0, 2^{10} \big[_D\right)$ is a system chaotic
-according to Devaney, because the semiconjugacy preserve this character.
-\item But the topology generated by $D$ is finer than the topology generated by
-the Euclidian distance $\Delta$ -- which is the order topology.
-\item According to Theorem \ref{Th:chaos et finesse}, we can deduce that the
-chaotic iterations $g$ are indeed chaotic, as defined by Devaney, for the order
-topology on $\mathds{R}$.
-\end{itemize}
-
-This result can be formulated as follows.
-
-\begin{theorem}
-\label{th:IC et topologie de l'ordre}
-The chaotic iterations $g$ on $\mathds{R}$ are chaotic according to the
-Devaney's formulation, when $\mathds{R}$ has his usual topology, which is the
-order topology.
-\end{theorem}
-Indeed this result is weaker than the theorem establishing the chaos for the
-finer topology $d$. However the Theorem \ref{th:IC et topologie de l'ordre}
-still remains important. Indeed, we have studied in our previous works a set
-different from the usual set of study ($\mathcal{X}$ instead of $\mathds{R}$),
-in order to be as close as possible from the computer: the properties of
-disorder proved theoretically will then be preserved when computing. However, we
-could wonder whether this change does not lead to a disorder of a lower quality.
-In other words, have we replaced a situation of a good disorder lost when
-computing, to another situation of a disorder preserved but of bad quality.
-Theorem \ref{th:IC et topologie de l'ordre} prove exactly the contrary.
-
+%\begin{proof}
+%Let $I_n = \left[ \dfrac{n}{10}, \dfrac{n+1}{10}\right[$, with $n \in \llbracket
+%0;2^{10}\times 10 \rrbracket$. All the points of $I_n$ have the same integral
+%prat $e$ and the same decimal part $s^0$: on the set $I_n$, functions $e(x)$
+%and $x \mapsto s(x)^0$ of Definition \ref{def:e et s} only depend on $n$. So all
+%the images $g(x)$ of these points $x$:
+%\begin{itemize}
+%\item Have the same integral part, which is $e$, except probably the bit number
+%$s^0$. In other words, this integer has approximately the same binary
+%decomposition than $e$, the sole exception being the digit $s^0$ (this number is
+%then either $e+2^{10-s^0}$ or $e-2^{10-s^0}$, depending on the parity of $s^0$,
+%\emph{i.e.}, it is equal to $e+(-1)^{s^0}\times 2^{10-s^0}$).
+%\item A shift to the left has been applied to the decimal part $y$, losing by
+%doing so the common first digit $s^0$. In other words, $y$ has been mapped into
+%$10\times y - s^0$.
+%\end{itemize}
+%To sum up, the action of $g$ on the points of $I$ is as follows: first, make a
+%multiplication by 10, and second, add the same constant to each term, which is
+%$\dfrac{1}{10}\left(e+(-1)^{s^0}\times 2^{10-s^0}\right)-s^0$.
+%\end{proof}
+
+%\begin{remark}
+%Finally, chaotic iterations are elements of the large family of functions that
+%are both chaotic and piecewise linear (like the tent map).
+%\end{remark}
+
+
+
+%\subsection{Comparison of the two metrics on $\big[ 0, 2^\mathsf{N} \big[$}
+
+%The two propositions bellow allow to compare our two distances on $\big[ 0,
+%2^\mathsf{N} \big[$:
+
+%\begin{proposition}
+%Id: $\left(~\big[ 0, 2^\mathsf{N} \big[,\Delta~\right) \to \left(~\big[ 0,
+%2^\mathsf{N} \big[, D~\right)$ is not continuous.
+%\end{proposition}
+
+%\begin{proof}
+%The sequence $x^n = 1,999\hdots 999$ constituted by $n$ 9 as decimal part, is
+%such that:
+%\begin{itemize}
+%\item $\Delta (x^n,2) \to 0.$
+%\item But $D(x^n,2) \geqslant 1$, then $D(x^n,2)$ does not converge to 0.
+%\end{itemize}
+
+%The sequential characterization of the continuity concludes the demonstration.
+%\end{proof}
+
+
+
+%A contrario:
+
+%\begin{proposition}
+%Id: $\left(~\big[ 0, 2^\mathsf{N} \big[,D~\right) \to \left(~\big[ 0,
+%2^\mathsf{N} \big[, \Delta ~\right)$ is a continuous fonction.
+%\end{proposition}
+
+%\begin{proof}
+%If $D(x^n,x) \to 0$, then $D_e(x^n,x) = 0$ at least for $n$ larger than a given
+%threshold, because $D_e$ only returns integers. So, after this threshold, the
+%integral parts of all the $x^n$ are equal to the integral part of $x$.
+
+%Additionally, $D_s(x^n, x) \to 0$, then $\forall k \in \mathds{N}^*, \exists N_k
+%\in \mathds{N}, n \geqslant N_k \Rightarrow D_s(x^n,x) \leqslant 10^{-k}$. This
+%means that for all $k$, an index $N_k$ can be found such that, $\forall n
+%\geqslant N_k$, all the $x^n$ have the same $k$ firsts digits, which are the
+%digits of $x$. We can deduce the convergence $\Delta(x^n,x) \to 0$, and thus the
+%result.
+%\end{proof}
+
+%The conclusion of these propositions is that the proposed metric is more precise
+%than the Euclidian distance, that is:
+
+%\begin{corollary}
+%$D$ is finer than the Euclidian distance $\Delta$.
+%\end{corollary}
+
+%This corollary can be reformulated as follows:
+
+%\begin{itemize}
+%\item The topology produced by $\Delta$ is a subset of the topology produced by
+%$D$.
+%\item $D$ has more open sets than $\Delta$.
+%\item It is harder to converge for the topology $\tau_D$ inherited by $D$, than
+%to converge with the one inherited by $\Delta$, which is denoted here by
+%$\tau_\Delta$.
+%\end{itemize}
+
+
+%\subsection{Chaos of the chaotic iterations on $\mathds{R}$}
+%\label{chpt:Chaos des itérations chaotiques sur R}
+
+
+
+%\subsubsection{Chaos according to Devaney}
+
+%We have recalled previously that the chaotic iterations $\left(\Go,
+%\mathcal{X}_d\right)$ are chaotic according to the formulation of Devaney. We
+%can deduce that they are chaotic on $\mathds{R}$ too, when considering the order
+%topology, because:
+%\begin{itemize}
+%\item $\left(\Go, \mathcal{X}_d\right)$ and $\left(g, \big[ 0, 2^{10}
+%\big[_D\right)$ are semiconjugate by $\varphi$,
+%\item Then $\left(g, \big[ 0, 2^{10} \big[_D\right)$ is a system chaotic
+%according to Devaney, because the semiconjugacy preserve this character.
+%\item But the topology generated by $D$ is finer than the topology generated by
+%the Euclidian distance $\Delta$ -- which is the order topology.
+%\item According to Theorem \ref{Th:chaos et finesse}, we can deduce that the
+%chaotic iterations $g$ are indeed chaotic, as defined by Devaney, for the order
+%topology on $\mathds{R}$.
+%\end{itemize}
+
+%This result can be formulated as follows.
+
+%\begin{theorem}
+%\label{th:IC et topologie de l'ordre}
+%The chaotic iterations $g$ on $\mathds{R}$ are chaotic according to the
+%Devaney's formulation, when $\mathds{R}$ has his usual topology, which is the
+%order topology.
+%\end{theorem}
+
+%Indeed this result is weaker than the theorem establishing the chaos for the
+%finer topology $d$. However the Theorem \ref{th:IC et topologie de l'ordre}
+%still remains important. Indeed, we have studied in our previous works a set
+%different from the usual set of study ($\mathcal{X}$ instead of $\mathds{R}$),
+%in order to be as close as possible from the computer: the properties of
+%disorder proved theoretically will then be preserved when computing. However, we
+%could wonder whether this change does not lead to a disorder of a lower quality.
+%In other words, have we replaced a situation of a good disorder lost when
+%computing, to another situation of a disorder preserved but of bad quality.
+%Theorem \ref{th:IC et topologie de l'ordre} prove exactly the contrary.
+%
