cells leads to the definition of a particular \emph{state of the
system}. A sequence which elements belong to $\llbracket 1;\mathsf{N}
\rrbracket $ is called a \emph{strategy}. The set of all strategies is
-denoted by $\mathbb{S}.$
+denoted by $\llbracket 1, \mathsf{N} \rrbracket^\mathds{N}.$
\begin{definition}
\label{Def:chaotic iterations}
The set $\mathds{B}$ denoting $\{0,1\}$, let
$f:\mathds{B}^{\mathsf{N}}\longrightarrow \mathds{B}^{\mathsf{N}}$ be
-a function and $S\in \mathbb{S}$ be a strategy. The so-called
+a function and $S\in \llbracket 1, \mathsf{N} \rrbracket^\mathds{N}$ be a strategy. The so-called
\emph{chaotic iterations} are defined by $x^0\in
\mathds{B}^{\mathsf{N}}$ and
\begin{equation}
G_f\left(S,E\right) = \left(\sigma(S), F_f(i(S),E)\right), \label{Gf}
\end{equation}
\noindent where $\sigma$ is the \emph{shift} function defined by $\sigma
-(S^{n})_{n\in \mathds{N}}\in \mathbb{S}\longrightarrow (S^{n+1})_{n\in
-\mathds{N}}\in \mathbb{S}$ and $i$ is the \emph{initial function}
-$i:(S^{n})_{n\in \mathds{N}} \in \mathbb{S}\longrightarrow S^{0}\in \llbracket
+(S^{n})_{n\in \mathds{N}}\in \llbracket 1, \mathsf{N} \rrbracket^\mathds{N}\longrightarrow (S^{n+1})_{n\in
+\mathds{N}}\in \llbracket 1, \mathsf{N} \rrbracket^\mathds{N}$ and $i$ is the \emph{initial function}
+$i:(S^{n})_{n\in \mathds{N}} \in \llbracket 1, \mathsf{N} \rrbracket^\mathds{N}\longrightarrow S^{0}\in \llbracket
1;\mathsf{N}\rrbracket$. Then the chaotic iterations defined in
(\ref{sec:chaotic iterations}) can be described by the following iterations:
\begin{equation}
\end{itemize}
The distance presented above follows these recommendations. Indeed, if the floor
value $\lfloor d(X,Y)\rfloor $ is equal to $n$, then the systems $E, \check{E}$
-differ in $n$ cells. In addition, $d(X,Y) - \lfloor d(X,Y) \rfloor $ is a
+differ in $n$ cells ($d_e$ is indeed the Hamming distance). In addition, $d(X,Y) - \lfloor d(X,Y) \rfloor $ is a
measure of the differences between strategies $S$ and $\check{S}$. More
precisely, this floating part is less than $10^{-k}$ if and only if the first
$k$ terms of the two strategies are equal. Moreover, if the $k^{th}$ digit is
x_i^{n-1} & \text{ if } i \notin \mathcal{S}^n \\
\left(f(x^{n-1})\right)_{S^n} & \text{ if }i \in \mathcal{S}^n.
\end{array}\right.
+\label{general CIs}
\end{equation}
In other words, at the $n^{th}$ iteration, only the cells whose id is
Let us introduce the following function:
\begin{equation}
\begin{array}{cccc}
- \delta: & \llbracket 1; \mathsf{N} \rrbracket \times \mathcal{P}\left(\llbracket 1; \mathsf{N} \rrbracket\right) & \longrightarrow & \mathds{B}\\
+ \chi: & \llbracket 1; \mathsf{N} \rrbracket \times \mathcal{P}\left(\llbracket 1; \mathsf{N} \rrbracket\right) & \longrightarrow & \mathds{B}\\
& (i,X) & \longmapsto & \left\{ \begin{array}{ll} 0 & \textrm{if }i \notin X, \\ 1 & \textrm{if }i \in X, \end{array}\right.
\end{array}
\end{equation}
Given a function $f:\mathds{B}^\mathsf{N} \longrightarrow \mathds{B}^\mathsf{N} $, define the function:
\begin{equation}
\begin{array}{lrll}
-F_{f}: & \llbracket1;\mathsf{N}\rrbracket\times \mathds{B}^{\mathsf{N}} &
+F_{f}: & \mathcal{P}\left(\llbracket1;\mathsf{N}\rrbracket \right) \times \mathds{B}^{\mathsf{N}} &
\longrightarrow & \mathds{B}^{\mathsf{N}} \\
-& (k,E) & \longmapsto & \left( E_{j}.\delta (k,j)+f(E)_{k}.\overline{\delta
-(k,j)}\right) _{j\in \llbracket1;\mathsf{N}\rrbracket},%
+& (P,E) & \longmapsto & \left( E_{j}.\chi (j,P)+f(E)_{j}.\overline{\chi
+(j,P)}\right) _{j\in \llbracket1;\mathsf{N}\rrbracket},%
\end{array}%
\end{equation}%
-\noindent where + and . are the Boolean addition and product operations.
+where + and . are the Boolean addition and product operations, and $\overline{x}$
+is the negation of the Boolean $x$.
Consider the phase space:
\begin{equation}
-\mathcal{X} = \llbracket 1 ; \mathsf{N} \rrbracket^\mathds{N} \times
+\mathcal{X} = \mathcal{P}\left(\llbracket 1 ; \mathsf{N} \rrbracket\right)^\mathds{N} \times
\mathds{B}^\mathsf{N},
\end{equation}
\noindent and the map defined on $\mathcal{X}$:
G_f\left(S,E\right) = \left(\sigma(S), F_f(i(S),E)\right), \label{Gf}
\end{equation}
\noindent where $\sigma$ is the \emph{shift} function defined by $\sigma
-(S^{n})_{n\in \mathds{N}}\in \mathbb{S}\longrightarrow (S^{n+1})_{n\in
-\mathds{N}}\in \mathbb{S}$ and $i$ is the \emph{initial function}
-$i:(S^{n})_{n\in \mathds{N}} \in \mathbb{S}\longrightarrow S^{0}\in \llbracket
-1;\mathsf{N}\rrbracket$. Then the chaotic iterations defined in
-(\ref{sec:chaotic iterations}) can be described by the following iterations:
+(S^{n})_{n\in \mathds{N}}\in \mathcal{P}\left(\llbracket 1 ; \mathsf{N} \rrbracket\right)^\mathds{N}\longrightarrow (S^{n+1})_{n\in
+\mathds{N}}\in \mathcal{P}\left(\llbracket 1 ; \mathsf{N} \rrbracket\right)^\mathds{N}$ and $i$ is the \emph{initial function}
+$i:(S^{n})_{n\in \mathds{N}} \in \mathcal{P}\left(\llbracket 1 ; \mathsf{N} \rrbracket\right)^\mathds{N}\longrightarrow S^{0}\in \mathcal{P}\left(\llbracket 1 ; \mathsf{N} \rrbracket\right)$.
+Then the general chaotic iterations defined in Equation \ref{general CIs} can
+be described by the following discrete dynamical system:
\begin{equation}
\left\{
\begin{array}{l}
\right.
\end{equation}%
-With this formulation, a shift function appears as a component of chaotic
-iterations. The shift function is a famous example of a chaotic
-map~\cite{Devaney} but its presence is not sufficient enough to claim $G_f$ as
-chaotic.
+Another time, a shift function appears as a component of these general chaotic
+iterations.
-To study this claim, a new distance between two points $X = (S,E), Y =
-(\check{S},\check{E})\in
-\mathcal{X}$ has been introduced in \cite{guyeux10} as follows:
+To study the Devaney's chaos property, a distance between two points
+$X = (S,E), Y = (\check{S},\check{E})$ of $\mathcal{X}$ must be introduced.
+Let us introduce:
\begin{equation}
d(X,Y)=d_{e}(E,\check{E})+d_{s}(S,\check{S}),
+\label{nouveau d}
\end{equation}
\noindent where
\begin{equation}
\left\{
\begin{array}{lll}
\displaystyle{d_{e}(E,\check{E})} & = & \displaystyle{\sum_{k=1}^{\mathsf{N}%
-}\delta (E_{k},\check{E}_{k})}, \\
+}\delta (E_{k},\check{E}_{k})}\textrm{ is another time the Hamming distance}, \\
\displaystyle{d_{s}(S,\check{S})} & = & \displaystyle{\dfrac{9}{\mathsf{N}}%
-\sum_{k=1}^{\infty }\dfrac{|S^k-\check{S}^k|}{10^{k}}}.%
+\sum_{k=1}^{\infty }\dfrac{|S^k\Delta {S}^k|}{10^{k}}}.%
\end{array}%
\right.
\end{equation}
+where $|X|$ is the cardinality of a set $X$ and $A\Delta B$ is for the symmetric difference, defined for sets A, B as
+$A\,\Delta\,B = (A \setminus B) \cup (B \setminus A)$.
+
+
+\begin{proposition}
+The function $d$ defined in Eq.~\ref{nouveau d} is a metric on $\mathcal{X}$.
+\end{proposition}
+
+\begin{proof}
+ $d_e$ is the Hamming distance. We will prove that $d_s$ is a distance
+too, thus $d$ will be a distance as sum of two distances.
+ \begin{itemize}
+\item Obviously, $d_s(S,\check{S})\geqslant 0$, and if $S=\check{S}$, then
+$d_s(S,\check{S})=0$. Conversely, if $d_s(S,\check{S})=0$, then
+$\forall k \in \mathds{N}, |S^k\Delta {S}^k|=0$, and so $\forall k, S^k=\check{S}^k$.
+ \item $d_s$ is symmetric
+($d_s(S,\check{S})=d_s(\check{S},S)$) due to the commutative property
+of the symmetric difference.
+\item Finally, $|S \Delta S''| = |(S \Delta \varnothing) \Delta S''|= |S \Delta (S'\Delta S') \Delta S''|= |(S \Delta S') \Delta (S' \Delta S'')|\leqslant |S \Delta S'| + |S' \Delta S''|$,
+and so for all subsets $S,S',$ and $S''$ of $\llbracket 1, \mathsf{N} \rrbracket$,
+we have $d_s(S,S'') \leqslant d_e(S,S')+d_s(S',S'')$, and the triangle
+inequality is obtained.
+ \end{itemize}
+\end{proof}
+
+
+Before being able to study the topological behavior of the general
+chaotic iterations, we must firstly establish that:
+
+\begin{proposition}
+ For all $f:\mathds{B}^\mathsf{N} \longrightarrow \mathds{B}^\mathsf{N} $, the function $G_f$ is continuous on
+$\left( \mathcal{X},d\right)$.
+\end{proposition}
+
+
+\begin{proof}
+We use the sequential continuity.
+Let $(S^n,E^n)_{n\in \mathds{N}}$ be a sequence of the phase space $%
+\mathcal{X}$, which converges to $(S,E)$. We will prove that $\left(
+G_{f}(S^n,E^n)\right) _{n\in \mathds{N}}$ converges to $\left(
+G_{f}(S,E)\right) $. Let us remark that for all $n$, $S^n$ is a strategy,
+thus, we consider a sequence of strategies (\emph{i.e.}, a sequence of
+sequences).\newline
+As $d((S^n,E^n);(S,E))$ converges to 0, each distance $d_{e}(E^n,E)$ and $d_{s}(S^n,S)$ converges
+to 0. But $d_{e}(E^n,E)$ is an integer, so $\exists n_{0}\in \mathds{N},$ $%
+d_{e}(E^n,E)=0$ for any $n\geqslant n_{0}$.\newline
+In other words, there exists a threshold $n_{0}\in \mathds{N}$ after which no
+cell will change its state:
+$\exists n_{0}\in \mathds{N},n\geqslant n_{0}\Rightarrow E^n = E.$
+
+In addition, $d_{s}(S^n,S)\longrightarrow 0,$ so $\exists n_{1}\in %
+\mathds{N},d_{s}(S^n,S)<10^{-1}$ for all indexes greater than or equal to $%
+n_{1}$. This means that for $n\geqslant n_{1}$, all the $S^n$ have the same
+first term, which is $S^0$: $\forall n\geqslant n_{1},S_0^n=S_0.$
+
+Thus, after the $max(n_{0},n_{1})^{th}$ term, states of $E^n$ and $E$ are
+identical and strategies $S^n$ and $S$ start with the same first term.\newline
+Consequently, states of $G_{f}(S^n,E^n)$ and $G_{f}(S,E)$ are equal,
+so, after the $max(n_0, n_1)^{th}$ term, the distance $d$ between these two points is strictly less than 1.\newline
+\noindent We now prove that the distance between $\left(
+G_{f}(S^n,E^n)\right) $ and $\left( G_{f}(S,E)\right) $ is convergent to
+0. Let $\varepsilon >0$. \medskip
+\begin{itemize}
+\item If $\varepsilon \geqslant 1$, we see that distance
+between $\left( G_{f}(S^n,E^n)\right) $ and $\left( G_{f}(S,E)\right) $ is
+strictly less than 1 after the $max(n_{0},n_{1})^{th}$ term (same state).
+\medskip
+\item If $\varepsilon <1$, then $\exists k\in \mathds{N},10^{-k}\geqslant
+\varepsilon > 10^{-(k+1)}$. But $d_{s}(S^n,S)$ converges to 0, so
+\begin{equation*}
+\exists n_{2}\in \mathds{N},\forall n\geqslant
+n_{2},d_{s}(S^n,S)<10^{-(k+2)},
+\end{equation*}%
+thus after $n_{2}$, the $k+2$ first terms of $S^n$ and $S$ are equal.
+\end{itemize}
+\noindent As a consequence, the $k+1$ first entries of the strategies of $%
+G_{f}(S^n,E^n)$ and $G_{f}(S,E)$ are the same ($G_{f}$ is a shift of strategies) and due to the definition of $d_{s}$, the floating part of
+the distance between $(S^n,E^n)$ and $(S,E)$ is strictly less than $%
+10^{-(k+1)}\leqslant \varepsilon $.\bigskip \newline
+In conclusion,
+$$
+\forall \varepsilon >0,\exists N_{0}=max(n_{0},n_{1},n_{2})\in \mathds{N}%
+,\forall n\geqslant N_{0},
+ d\left( G_{f}(S^n,E^n);G_{f}(S,E)\right)
+\leqslant \varepsilon .
+$$
+$G_{f}$ is consequently continuous.
+\end{proof}
+
+
+It is now possible to study the topological behavior of the general chaotic
+iterations. We will prove that,
+
+\begin{theorem}
+\label{t:chaos des general}
+ The general chaotic iterations defined on Equation~\ref{general CIs} satisfy
+the Devaney's property of chaos.
+\end{theorem}
+
+Let us firstly prove the following lemma.
+
+\begin{lemma}[Strong transitivity]
+\label{strongTrans}
+ For all couples $X,Y \in \mathcal{X}$ and any neighborhood $V$ of $X$, we can
+find $n \in \mathds{N}^*$ and $X' \in V$ such that $G^n(X')=Y$.
+\end{lemma}
+
+\begin{proof}
+ Let $X=(S,E)$, $\varepsilon>0$, and $k_0 = \lfloor log_{10}(\varepsilon)+1 \rfloor$.
+Any point $X'=(S',E')$ such that $E'=E$ and $\forall k \leqslant k_0, S'^k=S^k$,
+are in the open ball $\mathcal{B}\left(X,\varepsilon\right)$. Let us define
+$\check{X} = \left(\check{S},\check{E}\right)$, where $\check{X}= G^{k_0}(X)$.
+We denote by $s\subset \llbracket 1; \mathsf{N} \rrbracket$ the set of coordinates
+that are different between $\check{E}$ and the state of $Y$. Thus each point $X'$ of
+the form $(S',E')$ where $E'=E$ and $S'$ starts with
+$(S^0, S^1, \hdots, S^{k_0},s,\hdots)$, verifies the following properties:
+\begin{itemize}
+ \item $X'$ is in $\mathcal{B}\left(X,\varepsilon\right)$,
+ \item the state of $G_f^{k_0+1}(X')$ is the state of $Y$.
+\end{itemize}
+Finally the point $\left(\left(S^0, S^1, \hdots, S^{k_0},s,s^0, s^1, \hdots\right); E\right)$,
+where $(s^0,s^1, \hdots)$ is the strategy of $Y$, satisfies the properties
+claimed in the lemma.
+\end{proof}
+
+We can now prove the Theorem~\ref{t:chaos des general}...
+
+\begin{proof}[Theorem~\ref{t:chaos des general}]
+ On the one hand, strong transitivity implies transitivity. On the other hand,
+the regularity is exactly Lemma~\ref{strongTrans} with $Y=X$. As the sensitivity
+to the initial condition is implied by these two properties, we thus have
+the theorem.
+\end{proof}
+
\section{Efficient PRNG based on Chaotic Iterations}
