ajout de fichiers

[16dcc.git] / chaos.tex

diff --git a/chaos.tex b/chaos.tex
index ec2723aef86663af75c4d60192d7e8fab1104041..69eedb0b3b55c4504a2e8b8f0a88c29556172ca4 100644 (file)
--- 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 $\Bool^{{\mathsf{N}}} \times \llbracket1;{\mathsf{N}}\rrbracket^{\Nats}$ 
 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.
 $\mathcal{H}_f$ 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
 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  
 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.
 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.

 
 
 
 
 
 


@@ -38,43 +40,43 @@ This necessitates a rigorous proof, which is the aim of this section.
 
 
 Consider a topological space $(\mathcal{X},\tau)$ and a continuous function $f :
 
 
 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 $f^k(U) \cap V \neq
 \varnothing$.
 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 $Per(f).$
 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).
 $f$ 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.
 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:
 
 
 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 $f$.
 \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
 
 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


@@ -153,21 +155,21 @@ 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 $\mathcal{P}$: $\mathcal{P} = \{ p_1, p_2, \hdots, p_\mathsf{p}\}$
 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 
 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 
 
 
 The considered space is 


@@ -178,30 +180,38 @@ $\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.
 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}} \\
 &\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}
 \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. 
 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}
 \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}
 \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 $\mathcal{X}_{\mathsf{N},\mathcal{P}}$.
 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.

 
 
 
 
 
 


@@ -225,8 +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}
 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}
 digits are $|u^0-\check{u}^0|$. They are followed by 
 $|u^1-\check{u}^1|$ written with $n$ digits, etc.
 \begin{itemize}


@@ -243,12 +253,20 @@ $\check{u}^{v^0}$ (on $n$ digits), ..., $\check{u}^{\check{v}^0-1}$ (on $n$ digi
 \end{itemize}
 \end{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}
 \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}, ...\\
 $s=\left\{
 \begin{array}{l}
 u=\underline{6,} ~ \underline{11,5}, ...\\


@@ -263,28 +281,29 @@ $\check{s}=\left\{
 \end{array}
 \right.$.
 
 \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: 
 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}
 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}
 
 $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}
 while
 $\check{s}=\left\{
 \begin{array}{l}


@@ -293,8 +312,10 @@ $\check{s}=\left\{
 \end{array}
 \right.$
 
 \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}
 
 
 \end{xpl}
 
 


@@ -305,16 +326,18 @@ where: % $p=\max \mathcal{P}$ and:
 \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{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|  \\
    \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}
        \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}
 
 
 \end{itemize}
 
 


@@ -332,10 +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 
 \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 $d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}$ 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}
 
  \end{itemize}
 \end{proof}
 


@@ -373,7 +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
 \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}
 $G_f(x^n)_2$ is convergent to $G_f(x)_2$.
 \end{itemize}
 \end{proof}


@@ -393,25 +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 
 \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{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}
     \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}
     \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}
   \caption{Iterating $f_0:(x_1,x_2) \mapsto (\overline{x_1}, \overline{x_2})$}
   \label{fig:itg}
 \end{figure}


@@ -422,12 +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 
 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}
 
 since the function $f_0$ is equal to its inverse $f_0^{-1}$. 
 \end{xpl}
 


@@ -456,7 +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.
 $d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}((u,v),(U,V))$ 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}),
  $U^0=u^0$, ..., $U^{\sum_{l=0}^{k_1}V^l-1} = u^{\sum_{l=0}^{k_1}v^l-1}$.
 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}),


@@ -470,22 +494,29 @@ $V^{k_1+1}=|a_1|$, ..., $V^{k_1+k_2}=|a_{k_2}|$, and by
 $U^{k_1}=a_0^0$, $U^{k_1+1}=a_0^1$, ..., $U^{k_1+V_{k_1}-1}=a_0^{V_{k_1}-1}$,
 $U^{k_1+V_{k_1}}=a_1^{0}$, $U^{k_1+V_{k_1}+1}=a_1^{1}$,...
 
 $U^{k_1}=a_0^0$, $U^{k_1+1}=a_0^1$, ..., $U^{k_1+V_{k_1}-1}=a_0^{V_{k_1}-1}$,
 $U^{k_1+V_{k_1}}=a_1^{0}$, $U^{k_1+V_{k_1}+1}=a_1^{1}$,...
 

-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 $e_2$.
  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}
 
 
 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 $(\mathcal{X}_{\mathsf{N},\mathcal{P}}, d)$.
 \begin{prpstn}
 If $\Gamma_{\mathcal{P}}(f)$ is strongly connected, then $G_f$ is 
 regular on $(\mathcal{X}_{\mathsf{N},\mathcal{P}}, d)$.


@@ -495,22 +526,24 @@ 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 
 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,
 $$\left.\forall i,j \in \mathds{N}, U^i \in \llbracket 1, \mathsf{N} \rrbracket, V^j \in \mathcal{P}\right\}
 \subset \mathcal{B}(x,\varepsilon),$$
 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.
 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}
 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}
 
 and only if its iteration graph $\Gamma_{\mathcal{P}}(f)$ is strongly connected.
 \end{thrm}
 


@@ -520,15 +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 $\{b\}$.
 \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. 
   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}
 
   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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