-\label{fig:time_naive_gpu}
-\end{figure}
-
-
-First of all we have compared the time to generate X random numbers with both
-the CPU version and the GPU version.
-
-Faire une courbe du nombre de random en fonction du nombre de threads,
-éventuellement en fonction du nombres de threads par bloc.
-
-
-
-\section{The relativity of disorder}
-\label{sec:de la relativité du désordre}
-
-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{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}
-
-
-
-\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}
-
-
-
-
-\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.
-
-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}
-
-\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}
