X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/16dcc.git/blobdiff_plain/c9d6a79d174af327059607a3e0a1f23ab7921a49..151dc81946eee7c7fb5fbc327ada099f2af6a2c3:/chaos.tex diff --git a/chaos.tex b/chaos.tex index acf42e3..69eedb0 100644 --- a/chaos.tex +++ b/chaos.tex @@ -1,3 +1,5 @@ + +\subsection{Motivations} Let us us first recall the chaos theoretical context presented in~\cite{bcgr11:ip}. In this article, the space of interest is $\Bool^{{\mathsf{N}}} \times \llbracket1;{\mathsf{N}}\rrbracket^{\Nats}$ @@ -22,13 +24,13 @@ We have proven~\cite[Theorem 1]{bcgr11:ip} that $\mathcal{H}_f$ is chaotic in $\Bool^{{\mathsf{N}}} \times \llbracket1;{\mathsf{N}}\rrbracket^{\Nats}$ if and only if $\Gamma(f)$ is strongly connected. -However, the corrolary which would say that $\chi_{\textit{14Secrypt}}$ is chaotic +However, the corollary which would say that $\chi_{\textit{14Secrypt}}$ is chaotic cannot be directly deduced since we do not output all the successive -values of iterating $F_f$. Only a a few of them is concerned and +values of iterating $F_f$. Only a few of them are concerned and any subsequence of a chaotic sequence is not necessarily -a chaotic sequence too. +a chaotic sequence as well. This necessitates a rigorous proof, which is the aim of this section. - +Let us firstly recall the theoretical framework in which this research takes place. @@ -38,7 +40,7 @@ This necessitates a rigorous proof, which is the aim of this section. Consider a topological space $(\mathcal{X},\tau)$ and a continuous function $f : -\mathcal{X} \rightarrow \mathcal{X}$. +\mathcal{X} \rightarrow \mathcal{X}$~\cite{Devaney}. \begin{definition} The function $f$ is said to be \emph{topologically transitive} if, for any pair of open sets @@ -153,13 +155,12 @@ Let us first introduce $\mathcal{P} \subset \mathds{N}$ a finite nonempty set having the cardinality $\mathsf{p} \in \mathds{N}^\ast$. Intuitively, this is the set of authorized numbers of iterations. Denote by $p_1, p_2, \hdots, p_\mathsf{p}$ the ordered elements of $\mathcal{P}$: $\mathcal{P} = \{ p_1, p_2, \hdots, p_\mathsf{p}\}$ -and $p_1< p_2< \hdots < p_\mathsf{p}$. In our algorithm, -$\mathsf{p}$ is 1 and $p_1$ is $b$. - +and $p_1< p_2< \hdots < p_\mathsf{p}$. -The Algorithm~\ref{CI Algorithm} -may be seen as $b$ functional composition of $F_f$. -However, it can be generalized with $p_i$, $p_i \in \mathcal{P}$, +In our Algorithm~\ref{CI Algorithm}, +$\mathsf{p}$ is 1 and $p_1$ is $b$. +But this algorithm can be seen as $b$ functional compositions of $F_f$. +Obviously, it can be generalized with $p_i$, $p_i \in \mathcal{P}$, functional compositions of $F_f$. Thus, for any $p_i \in \mathcal{P}$ we introduce the function $F_{f,p_i} : \mathds{B}^\mathsf{N} \times \llbracket 1, \mathsf{N} \rrbracket^{p_i} \rightarrow \mathds{B}^\mathsf{N}$ defined by @@ -179,10 +180,10 @@ $\mathds{S}_{\mathsf{N},\mathcal{P}}= Each element in this space is a pair where the first element is $\mathsf{N}$-uple in $\Bool^{\mathsf{N}}$, as in the previous space. The second element is a pair $((u^k)_{k \in \Nats},(v^k)_{k \in \Nats})$ of infinite sequences. -The sequence $(v^k)_{k \in \Nats}$ defines how many iterations are executed at time $k$ between two outputs. -The sequence $(u^k)_{k \in \Nats}$ defines which elements is modified. +The sequence $(v^k)_{k \in \Nats}$ defines how many iterations are executed at time $k$ before the next output, +while $(u^k)_{k \in \Nats}$ details which elements are modified. -Let us define the shift function $\Sigma$ for any element of $\mathds{S}_{\mathsf{N},\mathcal{P}}$. +Let us introduce the shift function $\Sigma$ for any element of $\mathds{S}_{\mathsf{N},\mathcal{P}}$. \[ \begin{array}{cccc} @@ -199,17 +200,18 @@ Let us define the shift function $\Sigma$ for any element of $\mathds{S}_{\maths In other words, $\Sigma$ receives two sequences $u$ and $v$, and it operates $v^0$ shifts on the first sequence and a single shift on the second one. -Let +Let us consider \begin{equation} \begin{array}{cccc} G_f :& \mathcal{X}_{\mathsf{N},\mathcal{P}} & \rightarrow & \mathcal{X}_{\mathsf{N},\mathcal{P}}\\ & (e,(u,v)) & \mapsto & \left( F_{f,v^0}\left( e, (u^0, \hdots, u^{v^0-1}\right), \Sigma (u,v) \right) . \end{array} \end{equation} -Then the outputs $(y^0, y^1, \hdots )$ produced by the $\textit{CIPRNG}_f^2(u,v)$ generator -are the first components of the iterations $X^0 = (x^0, (u,v))$ and $\forall n \in \mathds{N}, +Then the outputs $(y^0, y^1, \hdots )$ produced by the $\textit{CIPRNG}_f^2(u,v)$ generator~\cite{wbg10:ip} +are by definition the first components of the iterations $X^0 = (x^0, (u,v))$ and $\forall n \in \mathds{N}, X^{n+1} = G_f(X^n)$ on $\mathcal{X}_{\mathsf{N},\mathcal{P}}$. - +The new obtained generator can be shown as either a post-treatment over generators $u$ and $v$, or a +discrete dynamical system on a set constituted by binary vectors and couple of integer sequences. @@ -233,8 +235,8 @@ between finite sequences $u^0, u^1, \hdots, u^{v^0-1}$ and $\check{u}^0, \check between $u^{v^0}, u^{v^0+1}, \hdots, u^{v^1-1}$ and $\check{u}^{\check{v}^0}, \check{u}^{\check{v}^0+1}, \hdots, \check{u}^{\check{v}^1-1}$, etc. More precisely, let $p = \lfloor \log_{10}{(\max{\mathcal{P}})}\rfloor +1$ and $n = \lfloor \log_{10}{(\mathsf{N})}\rfloor +1$. \begin{itemize} -\item The $p$ first digits of $d(x,\check{x})$ is $|v^0-\check{v}^0|$ written in decimal numeration (and with $p$ digits). -\item The next $n\times \max{(\mathcal{P})}$ digits aim at measuring how much $u^0, u^1, \hdots, u^{v^0-1}$ differs from $\check{u}^0, \check{u}^1, \hdots, \check{u}^{\check{v}^0-1}$. The $n$ first +\item The $p$ first digits of $d(x,\check{x})$ are $|v^0-\check{v}^0|$ written in decimal numeration (and with $p$ digits: zeros are added on the left if needed). +\item The next $n\times \max{(\mathcal{P})}$ digits aim at measuring how much $u^0, u^1, \hdots, u^{v^0-1}$ differ from $\check{u}^0, \check{u}^1, \hdots, \check{u}^{\check{v}^0-1}$. The $n$ first digits are $|u^0-\check{u}^0|$. They are followed by $|u^1-\check{u}^1|$ written with $n$ digits, etc. \begin{itemize} @@ -251,12 +253,20 @@ $\check{u}^{v^0}$ (on $n$ digits), ..., $\check{u}^{\check{v}^0-1}$ (on $n$ digi \end{itemize} \end{itemize} - - -\newcommand{\ns}{$\hspace{.1em}$} - +This distance has been defined to capture all aspects of divergences between two sequences generated +by the $\textit{CIPRNG}_f^2$ method, when setting respectively $(u,v)$ and $(\check{u},\check{v})$ as inputted +couples of generators. The integral part measures the bitwise Hamming distance between +the two $\mathsf{N}$-length binary vectors chosen as seeds. The fractional part must decrease +when the number of identical iterations applied by the $\textit{CIPRNG}_f^2$ discrete dynamical system on these seeds, in both cases (that is, when inputting either $(u,v)$ or $(\check{u},\check{v})$), increases. +More precisely, the fractional part will alternately measure the following elements: +\begin{itemize} + \item Do we iterate the same number of times between the next two outputs, when considering either $(u,v)$ or $(\check{u},\check{v})$? + \item Then, do we iterate the same components between the next two outputs of $\textit{CIPRNG}_f^2$ ? + \item etc. +\end{itemize} +Finally, zeros are put to be able to recover what occurred at a given iteration. Such aims are illustrated in the two following examples. \begin{xpl} -Consider for instance that $\mathsf{N}=13$, $\mathcal{P}=\{1,2,11\}$ (so $\mathsf{p}=2$), and that +Consider for instance that $\mathsf{N}=13$, $\mathcal{P}=\{1,2,11\}$ (so $\mathsf{p}=3$, $p=\lfloor \log_{10}{(\max{\mathcal{P}})}\rfloor +1 = 2$, while $n=2$), and that $s=\left\{ \begin{array}{l} u=\underline{6,} ~ \underline{11,5}, ...\\ @@ -271,28 +281,29 @@ $\check{s}=\left\{ \end{array} \right.$. -So $d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}(s,\check{s}) = 0.01\ns00\ns04\ns00\ns00\ns00\ns00\ns00\ns00\ns00\ns00\ns00\ns01\ns10\ns05 ...$ +So +$$d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}(s,\check{s}) = 0.01~0004000000000000000000~01~1005...$$ Indeed, the $p=2$ first digits are 01, as $|v^0-\check{v}^0|=1$, and we use $p$ digits to code this difference ($\mathcal{P}$ being $\{1,2,11\}$, this difference can be equal to 10). We then take the $v^0=1$ first terms of $u$, each term being coded in $n=2$ digits, that is, 06. As we can iterate at most $\max{(\mathcal{P})}$ times, we must complete this value by some 0's in such a way that the obtained result has $n\times \max{(\mathcal{P})}=22$ digits, that is: -0600000000000000000000. Similarly, the $\check{v}^0=2$ first -terms in $\check{u}$ are represented by 0604000000000000000000, and the absolute value of their -difference is equal to 0004000000000000000000. These digits are concatenated to 01, and +0600000000000000000000. Similarly, the first $\check{v}^0=2$ +terms in $\check{u}$ are represented by 0604000000000000000000, and the value of their +digit per digit absolute difference is equal to 0004000000000000000000. These digits are concatenated to 01, and we start again with the remainder of the sequences. \end{xpl} \begin{xpl} -Consider now that $\mathsf{N}=9$, and $\mathcal{P}=\{2,7\}$, and that +Consider now that $\mathsf{N}=9$ ($n=1$), $\mathcal{P}=\{2,7\}$ ($\mathsf{p}=2, p=1$), and that $s=\left\{ \begin{array}{l} u=\underline{6,7,} ~ \underline{4,2,} ...\\ v=2,2,... \end{array} -\right.$ +\right.$\\ while $\check{s}=\left\{ \begin{array}{l} @@ -301,8 +312,10 @@ $\check{s}=\left\{ \end{array} \right.$ -So $d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}(s,\check{s}) = 0.5173633305600000...$, as $|v^0-\check{v}^0|=5$, $|4963667-6700000| = 1736333$, $|v^1-\check{v}^1|=0$, -and $|9800000-4200000| = 5600000$. +So: +$d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}(s,\check{s}) = 0.5~2263667~1~5600000...$. +%as $|v^0-\check{v}^0|=5$, $|4963667-6700000| = 1736333$, $|v^1-\check{v}^1|=0$, +%and $|9800000-4200000| = 5600000$. \end{xpl} @@ -342,10 +355,10 @@ too, thus $d$ will also be a distance, being the sum of two distances. \item Obviously, $d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}(s,\check{s})\geqslant 0$, and if $s=\check{s}$, then $d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}(s,\check{s})=0$. Conversely, if $d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}(s,\check{s})=0$, then $\forall k \in \mathds{N}, v^k=\check{v}^k$ due to the -definition of $d$. Then, as digits between positions $p+1$ and $p+n$ are null and correspond to $|u^0-\check{u}^0|$, we can conclude that $u^0=\check{u}^0$. An extension of this result to the whole first $n \times \max{(\mathcal{P})}$ bloc leads to $u^i=\check{u}^i$, $\forall i \leqslant v^0=\check{v}^0$, and by checking all the $n \times \max{(\mathcal{P})}$ blocs, $u=\check{u}$. +definition of $d$. Then, as digits between positions $p+1$ and $p+n$ are null and correspond to $|u^0-\check{u}^0|$, we can conclude that $u^0=\check{u}^0$. An extension of this result to the whole first $n \times \max{(\mathcal{P})}$ blocs leads to $u^i=\check{u}^i$, $\forall i \leqslant v^0=\check{v}^0$, and by checking all the $n \times \max{(\mathcal{P})}$ blocs, $u=\check{u}$. \item $d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}$ is clearly symmetric ($d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}(s,\check{s})=d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}(\check{s},s)$). -\item The triangle inequality is obtained because the absolute value satisfies it too. +\item The triangle inequality is obtained because the absolute value satisfies it as well. \end{itemize} \end{proof} @@ -383,7 +396,7 @@ This implies that: \item $G_f(x^n)_1=G_f(x)_1$: they have the same Boolean vector as first coordinate. \item $d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}(\Sigma (u^n,v^n); \Sigma(u,v)) = 10^{p+n \max{(\mathcal{P})}} d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}((u^n,v^n); (u,v))$. As the right part of the equality tends -to 0, we can deduce that it is the case too for the left part of the equality, and so +to 0, we can deduce that it is also the case for the left part of the equality, and so $G_f(x^n)_2$ is convergent to $G_f(x)_2$. \end{itemize} \end{proof} @@ -403,26 +416,25 @@ $y=F_{f,p_i} (x, (u_0, \hdots, u_{p_i-1})) $. \end{itemize} It is not hard to see that the graph $\Gamma_{\{1\}}(f)$ is -$\Gamma(f)$. +$\Gamma(f)$ formerly introduced in~\cite{bcgr11:ip} for the $\textit{CIPRNG}_f^1(u)$ generator, +which is indeed $\textit{CIPRNG}_f^2(u,(1)_{n \in \mathds{N}})$. \begin{figure}[ht] \centering - \begin{subfigure}[b]{0.45\textwidth} - \centering - \includegraphics[scale=0.85]{graphe1.pdf} - \caption{$\Gamma(f_0)$} + \subfigure[$\Gamma(f_0)$]{ + \begin{minipage}{0.45\textwidth} + \centering + \includegraphics[scale=0.85]{graphe1.pdf} + \end{minipage} \label{graphe1} - \end{subfigure}% - -~ %add desired spacing between images, e. g. ~, \quad, \qquad, \hfill etc. - % (or a blank line to force the subfigure onto a new line) - \begin{subfigure}[b]{0.3\textwidth} - \centering - \includegraphics[scale=0.85]{graphe2.pdf} - \caption{$\Gamma_{\{2,3\}}(f_0)$} + } + \subfigure[$\Gamma_{\{2,3\}}(f_0)$]{ + \begin{minipage}{0.3\textwidth} + \centering + \includegraphics[scale=0.85]{graphe2.pdf} + \end{minipage} \label{graphe2} - \end{subfigure} - ~ %add desired spacing between images, e. g. ~, \quad, \qquad, \hfill etc. + } \caption{Iterating $f_0:(x_1,x_2) \mapsto (\overline{x_1}, \overline{x_2})$} \label{fig:itg} \end{figure} @@ -433,12 +445,13 @@ Consider for instance $\mathsf{N}=2$, Let $f_0:\mathds{B}^2 \longrightarrow \mathds{B}^2$ be the negation function, \textit{i.e.}, $f_0(x_1,x_2) = (\overline{x_1}, \overline{x_2})$, and consider $\mathcal{P}=\{2,3\}$. The graphs of iterations are given in -\textsc{Figure~\ref{fig:itg}}. -The \textsc{Figure~\ref{graphe1}} shows what happens when -displaying each iteration result. -On the contrary, the \textsc{Figure~\ref{graphe2}} explicits the behaviors -when always applying either 2 or 3 modifications before generating results. -Notice that here, orientations of arcs are not necessary +Figure~\ref{fig:itg}. +Figure~\ref{graphe1} shows what happens when each iteration result +is displayed . +On the contrary, Figure~\ref{graphe2} illustrates what happens when 2 or 3 modifications +are systematically applied +before results are generated. +Notice that here, the orientations of arcs are not necessary since the function $f_0$ is equal to its inverse $f_0^{-1}$. \end{xpl} @@ -467,7 +480,7 @@ $E$ must be equal to $e$. Let $k=\lfloor \log_{10} (\varepsilon) \rfloor +1$. $d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}((u,v),(U,V))$ must be lower than $\varepsilon$, so the $k$ first digits of the fractional part of $d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}((u,v),(U,V))$ are null. -Let $k_1$ the smallest integer such that, if $V^0=v^0$, ..., $V^{k_1}=v^{k_1}$, +Let $k_1$ be the smallest integer such that, if $V^0=v^0$, ..., $V^{k_1}=v^{k_1}$, $U^0=u^0$, ..., $U^{\sum_{l=0}^{k_1}V^l-1} = u^{\sum_{l=0}^{k_1}v^l-1}$. Then $d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}((u,v),(U,V))<\varepsilon$. In other words, any $y$ of the form $(e,((u^0, ..., u^{\sum_{l=0}^{k_1}v^l-1}), @@ -481,22 +494,29 @@ $V^{k_1+1}=|a_1|$, ..., $V^{k_1+k_2}=|a_{k_2}|$, and by $U^{k_1}=a_0^0$, $U^{k_1+1}=a_0^1$, ..., $U^{k_1+V_{k_1}-1}=a_0^{V_{k_1}-1}$, $U^{k_1+V_{k_1}}=a_1^{0}$, $U^{k_1+V_{k_1}+1}=a_1^{1}$,... -Let $y=(e,((u^0, \dots, u^{\sum_{l=0}^{k_1}v^l-1}, a_0^0, \dots, a_0^{|a_0|}, a_1^0, \dots, $ \linebreak -$a_1^{|a_1|},\dots, a_{k_2}^0, \dots, a_{k_2}^{|a_{k_2}|},$ - $\check{u}^0, \check{u}^1, \dots),(v^0, \dots, v^{k_1},|a_0|, \dots,$\linebreak - $|a_{k_2}|,\check{v}^0, \check{v}^1, \dots)))$. So $y\in \mathcal{B}(x,\varepsilon)$ +Let + +\begin{eqnarray*} +y&=&(e,( +(u^0, \dots, u^{\sum_{l=0}^{k_1}v^l-1}, a_0^0, \dots, a_0^{|a_0|}, a_1^0, \dots, +a_1^{|a_1|},\dots, a_{k_2}^0, \dots, a_{k_2}^{|a_{k_2}|}, + \check{u}^0, \check{u}^1, \dots), \\ +&&\qquad(v^0, \dots, v^{k_1},|a_0|, \dots, + |a_{k_2}|,\check{v}^0, \check{v}^1, \dots))). +\end{eqnarray*} +So $y\in \mathcal{B}(x,\varepsilon)$ and $G_{f}^{k_1+k_2}(y)=\check{x}$. Conversely, if $\Gamma_{\mathcal{P}}(f)$ is not strongly connected, then there are 2 vertices $e_1$ and $e_2$ such that there is no path between $e_1$ and $e_2$. -That is, it is impossible to find $(u,v)\in \mathds{S}_{\mathsf{N},\mathcal{P}}$ +Thus, it is impossible to find $(u,v)\in \mathds{S}_{\mathsf{N},\mathcal{P}}$ and $n\in \mathds{N}$ such that $G_f^n(e,(u,v))_1=e_2$. The open ball $\mathcal{B}(e_2, 1/2)$ cannot be reached from any neighborhood of $e_1$, and thus $G_f$ is not transitive. \end{proof} -We show now that, +We now show that, \begin{prpstn} If $\Gamma_{\mathcal{P}}(f)$ is strongly connected, then $G_f$ is regular on $(\mathcal{X}_{\mathsf{N},\mathcal{P}}, d)$. @@ -523,7 +543,7 @@ is a periodic point in the neighborhood $\mathcal{B}(x,\varepsilon)$ of $x$. $G_f$ being topologically transitive and regular, we can thus conclude that \begin{thrm} -The function $G_f$ is chaotic on $(\mathcal{X}_{\mathsf{N},\mathcal{P}},d)$ if +Function $G_f$ is chaotic on $(\mathcal{X}_{\mathsf{N},\mathcal{P}},d)$ if and only if its iteration graph $\Gamma_{\mathcal{P}}(f)$ is strongly connected. \end{thrm} @@ -533,15 +553,61 @@ and only if its iteration graph $\Gamma_{\mathcal{P}}(f)$ is strongly connected. \end{crllr} \begin{proof} In this context, $\mathcal{P}$ is the singleton $\{b\}$. - If $b$ is even, any vertex $e$ of $\Gamma_{\{b\}}(f_0)$ cannot reach + If $b$ is even, no vertex $e$ of $\Gamma_{\{b\}}(f_0)$ can reach its neighborhood and thus $\Gamma_{\{b\}}(f_0)$ is not strongly connected. - If $b$ is odd, any vertex $e$ of $\Gamma_{\{b\}}(f_0)$ cannot reach itself + If $b$ is odd, no vertex $e$ of $\Gamma_{\{b\}}(f_0)$ can reach itself and thus $\Gamma_{\{b\}}(f_0)$ is not strongly connected. \end{proof} -The next section recalls a general scheme to produce -functions and a iteration number $b$ -such that $\Gamma_{\{b\}}$ is strongly connected. + +\subsection{Comparison with other well-known generators} + +\begin{table} +\centering + \begin{tabular}{c|ccccccc} + PRNG & LCG & MRG & AWC & SWB & SWC & GFSR & INV\\ + \hline + NIST & 11 & 14 & 15 & 15 & 14 & 14 & 14\\ + DieHARD & 16 & 16 & 15 & 16 &18 & 16 & 16 + \end{tabular} + \caption{Statistical evaluation of known PRNGs: number of succeeded tests} + \label{table:comparisonWithout} +\end{table} + +\begin{table} +\centering + \begin{tabular}{c|ccccccc} + PRNG & LCG & MRG & AWC & SWB & SWC & GFSR & INV\\ + \hline + NIST & 15 & 15 & 15 & 15 & 15 & 15 & 15\\ + DieHARD & 18 & 18 & 18 & 18 & 18 & 18 & 18 + \end{tabular} + \caption{Statistical effects of CIPRNG on the succeeded tests} + \label{table:comparisonWith} +\end{table} +The objective of this section is to evaluate the statistical performance of the +proposed CIPRNG method, by comparing the effects of its application on well-known +but defective generators. We considered during the experiments the following PRNGs: +linear congruential generator (LCG), multiple recursive generators (MRG) +add-with-carry (AWC), subtract-with-borrow (SWB), shift-with-carry (SWC) +Generalized Feedback Shift Register (GFSR), and nonlinear inversive generator. +A general overview and a reminder of these generators +can be found, for instance, in the documentation of the TestU01 statistical +battery of tests~\cite{LEcuyerS07}. For each studied generator, we have compared +their scores according to both NIST~\cite{Nist10} and DieHARD~\cite{Marsaglia1996} +statistical batteries of tests, by launching them alone or inside the $\textit{CIPRNG}_f^2(v,v)$ +dynamical system, where $v$ is the considered PRNG set with most usual parameters, +and $f$ is the vectorial negation. + +Obtained results are reproduced in Tables~\ref{table:comparisonWithout} and \ref{table:comparisonWith}. +As can be seen, all these generators considered alone failed to pass either the 15 NIST tests or the +18 DieHARD ones, while both batteries of tests are always passed when applying the $\textit{CIPRNG}_f^2$ +post-treatment. Other results in the same direction, which can be found in~\cite{bfgw11:ip}, illustrate +the fact that operating a provable chaotic post-treatment on defective generators tends to improve +their statistical profile. + +Such post-treatment depending on the properties of the inputted function $f$, we need to recall a general scheme to produce +functions and an iteration number $b$ such that $\Gamma_{\{b\}}$ is strongly connected. %%% Local Variables: