+
+\subsection{Motivations}
Let us us first recall the chaos theoretical context presented
in~\cite{bcgr11:ip}. In this article, the space of interest
is $\Bool^{{\mathsf{N}}} \times \llbracket1;{\mathsf{N}}\rrbracket^{\Nats}$
$\mathcal{H}_f$ 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.
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{dfntn}
+\begin{definition}
The function $f$ is said to be \emph{topologically transitive} if, for any pair of open sets
$U,V \subset \mathcal{X}$, there exists $k>0$ such that $f^k(U) \cap V \neq
\varnothing$.
-\end{dfntn}
+\end{definition}
-\begin{dfntn}
+\begin{definition}
An element $x$ is a \emph{periodic point} for $f$ of period $n\in \mathds{N}^*$
if $f^{n}(x)=x$.% The set of periodic points of $f$ is denoted $Per(f).$
-\end{dfntn}
+\end{definition}
-\begin{dfntn}
+\begin{definition}
$f$ is said to be \emph{regular} on $(\mathcal{X}, \tau)$ if the set of periodic
points for $f$ is dense in $\mathcal{X}$: for any point $x$ in $\mathcal{X}$,
any neighborhood of $x$ contains at least one periodic point (without
necessarily the same period).
-\end{dfntn}
+\end{definition}
-\begin{dfntn}[Devaney's formulation of chaos~\cite{Devaney}]
+\begin{definition}[Devaney's formulation of chaos~\cite{Devaney}]
The function $f$ is said to be \emph{chaotic} on $(\mathcal{X},\tau)$ if $f$ is regular and
topologically transitive.
-\end{dfntn}
+\end{definition}
The chaos property is strongly linked to the notion of ``sensitivity'', defined
on a metric space $(\mathcal{X},d)$ by:
-\begin{dfntn}
+\begin{definition}
\label{sensitivity} The function $f$ has \emph{sensitive dependence on initial conditions}
if there exists $\delta >0$ such that, for any $x\in \mathcal{X}$ and any
neighborhood $V$ of $x$, there exist $y\in V$ and $n > 0$ such that
$d\left(f^{n}(x), f^{n}(y)\right) >\delta $.
The constant $\delta$ is called the \emph{constant of sensitivity} of $f$.
-\end{dfntn}
+\end{definition}
Indeed, Banks \emph{et al.} have proven in~\cite{Banks92} that when $f$ is
chaotic and $(\mathcal{X}, d)$ is a metric space, then $f$ has the property of
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
-
-$$
-F_{f,p_i} (x,(u^0, u^1, \hdots, u^{p_i-1})) \mapsto
-F_f(\hdots (F_f(F_f(x,u^0), u^1), \hdots), u^{p_i-1}).
-$$
+\[
+\begin{array}{l}
+F_{f,p_i} (x,(u^0, u^1, \hdots, u^{p_i-1})) \mapsto \\
+\qquad F_f(\hdots (F_f(F_f(x,u^0), u^1), \hdots), u^{p_i-1}).
+\end{array}
+\]
The considered space is
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 introduce the shift function $\Sigma$ for any element of $\mathds{S}_{\mathsf{N},\mathcal{P}}$.
-Let us define the shift function $\Sigma$ for any element of $\mathds{S}_{\mathsf{N},\mathcal{P}}$.
-$$\begin{array}{cccc}
-\Sigma:&\mathds{S}_{\mathsf{N},\mathcal{P}} &\longrightarrow
+\[
+\begin{array}{cccc}
+\Sigma:&\mathds{S}_{\mathsf{N},\mathcal{P}} &\rightarrow
&\mathds{S}_{\mathsf{N},\mathcal{P}} \\
-& \left((u^k)_{k \in \mathds{N}},(v^k)_{k \in \mathds{N}}\right) & \longmapsto & \left(\sigma^{v^0}\left((u^k)_{k \in \mathds{N}}\right),\sigma\left((v^k)_{k \in \mathds{N}}\right)\right).
+& \left((u^k)_{k \in \mathds{N}},(v^k)_{k \in \mathds{N}}\right) & \mapsto &
+\begin {array}{l}
+ \left(\sigma^{v^0}\left((u^k)_{k \in \mathds{N}}\right), \right. \\
+ \qquad \left. \sigma\left((v^k)_{k \in \mathds{N}}\right)\right).
+ \end{array}
\end{array}
-$$
+\]
+
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}} & \longrightarrow & \mathcal{X}_{\mathsf{N},\mathcal{P}}\\
- & (e,(u,v)) & \longmapsto & \left( F_{f,v^0}\left( e, (u^0, \hdots, u^{v^0-1}\right), \Sigma (u,v) \right) .
+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.
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}
\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
$s=\left\{
\begin{array}{l}
u=\underline{6,} ~ \underline{11,5}, ...\\
\end{array}
\right.$.
-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
$\check{s}=\left\{
\begin{array}{l}
\end{array}
\right.$
-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}
\begin{itemize}
\item $d_{\mathds{B}^\mathsf{N}}$ is the Hamming distance,
\item $\forall s=(u,v), \check{s}=(\check{u},\check{v}) \in \mathcal{S}_{\mathsf{N},\mathcal{P}}$,\newline
-$$\begin{array}{rcl}
- d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}(s,\check{s}) &= &
- \sum_{k=0}^\infty \dfrac{1}{10^{(k+1)p+kn\max{(\mathcal{P})}}}
+\[
+\begin{array}{l}
+ d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}(s,\check{s}) = \\
+ \quad \sum_{k=0}^\infty \dfrac{1}{10^{(k+1)p+kn\max{(\mathcal{P})}}}
\bigg(|v^k - \check{v}^k| \\
- & & + \left| \sum_{l=0}^{v^k-1}
+ \quad\quad + \left| \sum_{l=0}^{v^k-1}
\dfrac{u^{\sum_{m=0}^{k-1} v^m +l}}{ 10^{(l+1)n}} -
\sum_{l=0}^{\check{v}^k-1}
\dfrac{\check{u}^{\sum_{m=0}^{k-1} \check{v}^m +l}}{ 10^{(l+1)n}} \right| \bigg)
\end{array}
-$$ %\left| \sum_{l=0}^{v^k-1} \dfrac{u^{\sum_{m=0}^{k-1} v^m +l}}{ 10^{l}} - \sum_{l=0}^{\check{v}^k-1} \dfrac{\check{u}^{\sum_{m=0}^{k-1} \check{v}^m +l}}{ 10^{l}}\right|\right)}.$$
+\]
+ %\left| \sum_{l=0}^{v^k-1} \dfrac{u^{\sum_{m=0}^{k-1} v^m +l}}{ 10^{l}} - \sum_{l=0}^{\check{v}^k-1} \dfrac{\check{u}^{\sum_{m=0}^{k-1} \check{v}^m +l}}{ 10^{l}}\right|\right)}.
\end{itemize}
\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}
\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}
\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}
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}
$d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}((u,v),(U,V))$ 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}),
$U^{k_1}=a_0^0$, $U^{k_1+1}=a_0^1$, ..., $U^{k_1+V_{k_1}-1}=a_0^{V_{k_1}-1}$,
$U^{k_1+V_{k_1}}=a_1^{0}$, $U^{k_1+V_{k_1}+1}=a_1^{1}$,...
-Let $y=(e,((u^0, ..., u^{\sum_{l=0}^{k_1}v^l-1}, a_0^0, ..., a_0^{|a_0|}, a_1^0, ..., a_1^{|a_1|},...,
- a_{k_2}^0, ..., a_{k_2}^{|a_{k_2}|},$ \linebreak
- $\check{u}^0, \check{u}^1, ...),(v^0, ..., v^{k_1},|a_0|, ...,
- |a_{k_2}|,\check{v}^0, \check{v}^1, ...)))$. 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 $(\mathcal{X}_{\mathsf{N},\mathcal{P}}, d)$.
Let $x=(e,(u,v)) \in \mathcal{X}_{\mathsf{N},\mathcal{P}}$ and $\varepsilon >0$.
As in the proofs of Prop.~\ref{prop:trans}, let $k_1 \in \mathds{N}$ such
that
-$$\left\{(e, ((u^0, ..., u^{v^{k_1-1}},U^0, U^1, ...),(v^0, ..., v^{k_1},V^0, V^1, ...)) \mid \right.$$
+$$\left\{(e, ((u^0, \dots, u^{v^{k_1-1}},U^0, U^1, \dots),(v^0, \dots, v^{k_1},V^0, V^1, \dots)) \mid \right.$$
$$\left.\forall i,j \in \mathds{N}, U^i \in \llbracket 1, \mathsf{N} \rrbracket, V^j \in \mathcal{P}\right\}
\subset \mathcal{B}(x,\varepsilon),$$
and $y=G_f^{k_1}(e,(u,v))$. $\Gamma_{\mathcal{P}}(f)$ being strongly connected,
-there is at least a path from the Boolean state $y_1$ of $y$ and $e$ \ANNOT{Phrase pas claire : "from ... " mais pas de "to ..."}.
+there is at least a path from the Boolean state $y_1$ of $y$ to $e$.
+%\ANNOT{Phrase pas claire : "from \dots " mais pas de "to \dots"}.
Denote by $a_0, \hdots, a_{k_2}$ the edges of such a path.
-Then the point:
-$$(e,((u^0, ..., u^{v^{k_1-1}},a_0^0, ..., a_0^{|a_0|}, a_1^0, ..., a_1^{|a_1|},...,
- a_{k_2}^0, ..., a_{k_2}^{|a_{k_2}|},u^0, ..., u^{v^{k_1-1}},$$
-$$a_0^0, ...,a_{k_2}^{|a_{k_2}|}...),(v^0, ..., v^{k_1}, |a_0|, ..., |a_{k_2}|,v^0, ..., v^{k_1}, |a_0|, ..., |a_{k_2}|,...))$$
+Then the point:\linebreak
+$(e,((u^0, \dots, u^{v^{k_1-1}},a_0^0, \dots, a_0^{|a_0|}, a_1^0, \dots, a_1^{|a_1|},\dots,
+ a_{k_2}^0, \dots,$ \linebreak
+$\,a_{k_2}^{|a_{k_2}|},u^0, \dots, u^{v^{k_1-1}},a_0^0, \dots,a_{k_2}^{|a_{k_2}|}\dots),$\linebreak
+$(v^0, \dots, v^{k_1}, |a_0|, \dots, |a_{k_2}|,v^0, \dots, v^{k_1}, |a_0|, \dots, |a_{k_2}|,\dots))$
is a periodic point in the neighborhood $\mathcal{B}(x,\varepsilon)$ of $x$.
\end{proof}
$G_f$ 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}
\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.
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