X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/16dcc.git/blobdiff_plain/236f25b2f3a081b11c71bedad6d044d695ce2cca..28bad0743e67c1875d9e849082024377e3a66321:/chaos.tex?ds=sidebyside diff --git a/chaos.tex b/chaos.tex index 511a11a..dafc635 100644 --- a/chaos.tex +++ b/chaos.tex @@ -40,41 +40,41 @@ This necessitates a rigorous proof, which is the aim of this section. Consider a topological space $(\mathcal{X},\tau)$ and a continuous function $f : \mathcal{X} \rightarrow \mathcal{X}$. -\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 @@ -163,11 +163,12 @@ However, 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 @@ -182,20 +183,27 @@ The sequence $(v^k)_{k \in \Nats}$ defines how many iterations are executed at t 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 @@ -245,6 +253,7 @@ $\check{u}^{v^0}$ (on $n$ digits), ..., $\check{u}^{\check{v}^0-1}$ (on $n$ digi +\newcommand{\ns}{$\hspace{.1em}$} \begin{xpl} Consider for instance that $\mathsf{N}=13$, $\mathcal{P}=\{1,2,11\}$ (so $\mathsf{p}=2$), and that @@ -262,7 +271,7 @@ $\check{s}=\left\{ \end{array} \right.$. -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 @@ -304,16 +313,18 @@ 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{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} @@ -402,7 +413,8 @@ $\Gamma(f)$. \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 @@ -469,17 +481,17 @@ $V^{k_1+1}=|a_1|$, ..., $V^{k_1+k_2}=|a_{k_2}|$, 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 $y=(e,((u^0, ..., u^{\sum_{l=0}^{k_1}v^l-1}, a_0^0, ..., a_0^{|a_0|}, a_1^0, ..., a_1^{|a_1|},..., - a_{k_2}^0, ..., a_{k_2}^{|a_{k_2}|},$ \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} @@ -494,16 +506,17 @@ 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 $e$. +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}