From: guyeux Date: Wed, 30 Nov 2011 15:21:19 +0000 (+0100) Subject: Ménage X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/prng_gpu.git/commitdiff_plain/81be58cc120f6a94c5f98c292f65a74fc8df5973?ds=inline;hp=e443cbfb0433e0a7e1551c9841b0173698151f96 Ménage --- diff --git a/prng_gpu.tex b/prng_gpu.tex index dc3d3cb..1c7c9fe 100644 --- a/prng_gpu.tex +++ b/prng_gpu.tex @@ -1078,542 +1078,12 @@ obtain approximately 1.8GSample/s and on the GTX 280 about 1.6GSample/s. \label{fig:time_bbs_gpu} \end{figure} -Both these experimentations allows us to conclude that it is possible to +Both these experiments 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} - - -%\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. -% -