+
+
+
+\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.
+
+
+
+
+\section{Efficient prng based on chaotic iterations}
+
+In order to implement efficiently a PRNG based on chaotic iterations it is
+possible to improve previous works [ref]. One solution consists in considering
+that the strategy used contains all the bits for which the negation is
+achieved out. Then in order to apply the negation on these bits we can simply
+apply the xor operator between the current number and the strategy. In
+order to obtain the strategy we also use a classical PRNG.
+
+Here is an example with 16-bits numbers showing how the bit operations are
+applied. Suppose that $x$ and the strategy $S^i$ are defined in binary mode.
+Then the following table shows the result of $x$ xor $S^i$.
+$$
+\begin{array}{|cc|cccccccccccccccc|}
+\hline
+x &=&1&0&1&1&1&0&1&0&1&0&0&1&0&0&1&0\\
+\hline
+S^i &=&0&1&1&0&0&1&1&0&1&1&1&0&0&1&1&1\\
+\hline
+x \oplus S^i&=&1&1&0&1&1&1&0&0&0&1&1&1&0&1&0&1\\
+\hline
+
+\hline
+ \end{array}
+$$
+
+%% \begin{figure}[htbp]
+%% \begin{center}
+%% \fbox{
+%% \begin{minipage}{14cm}
+%% unsigned int CIprng() \{\\
+%% static unsigned int x = 123123123;\\
+%% unsigned long t1 = xorshift();\\
+%% unsigned long t2 = xor128();\\
+%% unsigned long t3 = xorwow();\\
+%% x = x\textasciicircum (unsigned int)t1;\\
+%% x = x\textasciicircum (unsigned int)(t2$>>$32);\\
+%% x = x\textasciicircum (unsigned int)(t3$>>$32);\\
+%% x = x\textasciicircum (unsigned int)t2;\\
+%% x = x\textasciicircum (unsigned int)(t1$>>$32);\\
+%% x = x\textasciicircum (unsigned int)t3;\\
+%% return x;\\
+%% \}
+%% \end{minipage}
+%% }
+%% \end{center}
+%% \caption{sequential Chaotic Iteration PRNG}
+%% \label{algo:seqCIprng}
+%% \end{figure}
+
+
+
+\lstset{language=C,caption={C code of the sequential chaotic iterations based PRNG},label=algo:seqCIprng}
+\begin{lstlisting}
+unsigned int CIprng() {
+ static unsigned int x = 123123123;
+ unsigned long t1 = xorshift();
+ unsigned long t2 = xor128();
+ unsigned long t3 = xorwow();
+ x = x^(unsigned int)t1;
+ x = x^(unsigned int)(t2>>32);
+ x = x^(unsigned int)(t3>>32);
+ x = x^(unsigned int)t2;
+ x = x^(unsigned int)(t1>>32);
+ x = x^(unsigned int)t3;
+ return x;
+}
+\end{lstlisting}
+
+
+
+
+
+In listing~\ref{algo:seqCIprng} a sequential version of our chaotic iterations
+based PRNG is presented. The xor operator is represented by \textasciicircum. This function uses three classical 64-bits PRNG: the
+\texttt{xorshift}, the \texttt{xor128} and the \texttt{xorwow}. In the
+following, we call them xor-like PRNGSs. These three PRNGs are presented
+in~\cite{Marsaglia2003}. As each xor-like PRNG used works with 64-bits and as our PRNG
+works with 32-bits, the use of \texttt{(unsigned int)} selects the 32 least
+significant bits whereas \texttt{(unsigned int)(t3$>>$32)} selects the 32 most
+significants bits of the variable \texttt{t}. So to produce a random number
+realizes 6 xor operations with 6 32-bits numbers produced by 3 64-bits PRNG.
+This version successes the BigCrush of the TestU01 battery [P. L’ecuyer and
+ R. Simard. Testu01].
