+\label{sec:experiments}
+
+Different experiments have been performed in order to measure the generation
+speed. We have used a computer equiped with Tesla C1060 NVidia GPU card and an
+Intel Xeon E5530 cadenced at 2.40 GHz for our experiments and we have used
+another one equipped with a less performant CPU and a GeForce GTX 280. Both
+cards have 240 cores.
+
+In Figure~\ref{fig:time_xorlike_gpu} we compare the number of random numbers
+generated per second with the xor-like based PRNG. In this figure, the optimized
+version use the {\it xor64} described in~\cite{Marsaglia2003}. The naive version
+use the three xor-like PRNGs described in Listing~\ref{algo:seqCIprng}. In
+order to obtain the optimal performance we removed the storage of random numbers
+in the GPU memory. This step is time consuming and slows down the random numbers
+generation. Moreover, if one is interested by applications that consume random
+numbers directly when they are generated, their storage are completely
+useless. In this figure we can see that when the number of threads is greater
+than approximately 30,000 upto 5 millions the number of random numbers generated
+per second is almost constant. With the naive version, it is between 2.5 and
+3GSample/s. With the optimized version, it is approximately equals to
+20GSample/s. Finally we can remark that both GPU cards are quite similar. In
+practice, the Tesla C1060 has more memory than the GTX 280 and this memory
+should be of better quality.
+
+\begin{figure}[htbp]
+\begin{center}
+ \includegraphics[scale=.7]{curve_time_xorlike_gpu.pdf}
+\end{center}
+\caption{Number of random numbers generated per second with the xorlike based PRNG}
+\label{fig:time_xorlike_gpu}
+\end{figure}
+
+
+In comparison, Listing~\ref{algo:seqCIprng} allows us to generate about
+138MSample/s with only one core of the Xeon E5530.
+
+
+In Figure~\ref{fig:time_bbs_gpu} we highlight the performance of the optimized
+BBS based PRNG on GPU. Performances are less important. On the Tesla C1060 we
+obtain approximately 1.8GSample/s and on the GTX 280 about 1.6GSample/s.
+
+\begin{figure}[htbp]
+\begin{center}
+ \includegraphics[scale=.7]{curve_time_bbs_gpu.pdf}
+\end{center}
+\caption{Number of random numbers generated per second with the BBS based PRNG}
+\label{fig:time_bbs_gpu}
+\end{figure}
+
+Both these experimentations allows us to conclude that it is possible to
+generate a huge number of pseudorandom numbers with the xor-like version and
+about tens times less with the BBS based version. The former version has only
+chaotic properties whereas the latter also has cryptographically properties.
+
+
+%% \section{Cryptanalysis of the Proposed PRNG}
+
+
+%% Mettre ici la preuve de PCH
+
+%\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}
+%\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{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}
