Consider a topological space $(\mathcal{X},\tau)$ and a continuous function $f :
\mathcal{X} \rightarrow \mathcal{X}$.
-\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
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 sequence $(u^k)_{k \in \Nats}$ defines which elements is modified.
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
\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
+\newcommand{\ns}{$\hspace{.1em}$}
\begin{xpl}
-Consider for instance that $\mathsf{N}=13$, $\mathcal{P}=\{1,2,11\}$ (so $\mathsf{p}=3$), and that
+Consider for instance that $\mathsf{N}=13$, $\mathcal{P}=\{1,2,11\}$ (so $\mathsf{p}=2$), and that
$s=\left\{
\begin{array}{l}
u=\underline{6,} ~ \underline{11,5}, ...\\
\end{array}
\right.$.
-So $d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}(s,\check{s}) = 0.010004000000000000000000011005 ...$
+So $d_{\mathds{S}_{\mathsf{N},\mathcal{P}}}(s,\check{s}) = 0.01\ns00\ns04\ns00\ns00\ns00\ns00\ns00\ns00\ns00\ns00\ns00\ns01\ns10\ns05 ...$
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
\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}
\caption{$\Gamma(f_0)$}
\label{graphe1}
\end{subfigure}%
- ~ %add desired spacing between images, e. g. ~, \quad, \qquad, \hfill etc.
+
+~ %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
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 2 or 3 modification and next outputing results.
+when always applying either 2 or 3 modifications before generating results.
Notice that here, orientations of arcs are not necessary
since the function $f_0$ is equal to its inverse $f_0^{-1}$.
\end{xpl}
$U^{k_1}=a_0^0$, $U^{k_1+1}=a_0^1$, ..., $U^{k_1+V_{k_1}-1}=a_0^{V_{k_1}-1}$,
$U^{k_1+V_{k_1}}=a_1^{0}$, $U^{k_1+V_{k_1}+1}=a_1^{1}$,...
-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 $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 $y\in \mathcal{B}(x,\varepsilon)$
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 $(u,v)\in \mathds{S}_{\mathsf{N},\mathcal{P}}$
-and $n \mathds{N}$ such that $G_f^n(e,(u,v))_1=e_2$. The open ball $\mathcal{B}(e_2, 1/2)$
+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}
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 $e$.
+there is at least a path from the Boolean state $y_1$ of $y$ and $e$ \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}
In this context, $\mathcal{P}$ is the singleton $\{b\}$.
If $b$ is even, any vertex $e$ of $\Gamma_{\{b\}}(f_0)$ cannot reach
its neighborhood and thus $\Gamma_{\{b\}}(f_0)$ is not strongly connected.
- If $b$ is even, any vertex $e$ of $\Gamma_{\{b\}}(f_0)$ cannot reach itself
+ If $b$ is odd, any vertex $e$ of $\Gamma_{\{b\}}(f_0)$ cannot reach itself
and thus $\Gamma_{\{b\}}(f_0)$ is not strongly connected.
\end{proof}
such that $\Gamma_{\{b\}}$ is strongly connected.
-
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