+This former generator as successively passed various batteries of statistical tests, as the NIST tests~\cite{bcgr11:ip}.
+
+\subsection{Improving the Speed of the Former Generator}
+
+Instead of updating only one cell at each iteration, we can try to choose a
+subset of components and to update them together. Such an attempt leads
+to a kind of merger of the two sequences used in Algorithm
+\ref{CI Algorithm}. When the updating function is the vectorial negation,
+this algorithm can be rewritten as follows:
+
+\begin{equation}
+\left\{
+\begin{array}{l}
+x^0 \in \llbracket 0, 2^\mathsf{N}-1 \rrbracket, S \in \llbracket 0, 2^\mathsf{N}-1 \rrbracket^\mathds{N} \\
+\forall n \in \mathds{N}^*, x^n = x^{n-1} \oplus S^n,
+\end{array}
+\right.
+\label{equation Oplus}
+\end{equation}
+where $\oplus$ is for the bitwise exclusive or between two integers.
+This rewritten can be understood as follows. The $n-$th term $S^n$ of the
+sequence $S$, which is an integer of $\mathsf{N}$ binary digits, presents
+the list of cells to update in the state $x^n$ of the system (represented
+as an integer having $\mathsf{N}$ bits too). More precisely, the $k-$th
+component of this state (a binary digit) changes if and only if the $k-$th
+digit in the binary decomposition of $S^n$ is 1.
+
+The single basic component presented in Eq.~\ref{equation Oplus} is of
+ordinary use as a good elementary brick in various PRNGs. It corresponds
+to the following discrete dynamical system in chaotic iterations:
+
+\begin{equation}
+\forall n\in \mathds{N}^{\ast }, \forall i\in
+\llbracket1;\mathsf{N}\rrbracket ,x_i^n=\left\{
+\begin{array}{ll}
+ 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{eq:generalIC}
+\end{equation}
+where $f$ is the vectorial negation and $\forall n \in \mathds{N}$,
+$\mathcal{S}^n \subset \llbracket 1, \mathsf{N} \rrbracket$ is such that
+$k \in \mathcal{S}^n$ if and only if the $k-$th digit in the binary
+decomposition of $S^n$ is 1. Such chaotic iterations are more general
+than the ones presented in Definition \ref{Def:chaotic iterations} for
+the fact that, instead of updating only one term at each iteration,
+we select a subset of components to change.
+
+
+Obviously, replacing Algorithm~\ref{CI Algorithm} by
+Equation~\ref{equation Oplus}, possible when the iteration function is
+the vectorial negation, leads to a speed improvement. However, proofs
+of chaos obtained in~\cite{bg10:ij} have been established
+only for chaotic iterations of the form presented in Definition
+\ref{Def:chaotic iterations}. The question is now to determine whether the
+use of more general chaotic iterations to generate pseudo-random numbers
+faster, does not deflate their topological chaos properties.
+
+\subsection{Proofs of Chaos of the General Formulation of the Chaotic Iterations}
+\label{deuxième def}
+Let us consider the discrete dynamical systems in chaotic iterations having
+the general form:
+
+\begin{equation}
+\forall n\in \mathds{N}^{\ast }, \forall i\in
+\llbracket1;\mathsf{N}\rrbracket ,x_i^n=\left\{
+\begin{array}{ll}
+ 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
+contained into the set $S^{n}$ are iterated.
+
+Let us now rewrite these general chaotic iterations as usual discrete dynamical
+system of the form $X^{n+1}=f(X^n)$ on an ad hoc metric space. Such a formulation
+is required in order to study the topological behavior of the system.
+
+Let us introduce the following function:
+\begin{equation}
+\begin{array}{cccc}
+ \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}
+where $\mathcal{P}\left(X\right)$ is for the powerset of the set $X$, that is, $Y \in \mathcal{P}\left(X\right) \Longleftrightarrow Y \subset X$.
+
+Given a function $f:\mathds{B}^\mathsf{N} \longrightarrow \mathds{B}^\mathsf{N} $, define the function:
+\begin{equation}
+\begin{array}{lrll}
+F_{f}: & \mathcal{P}\left(\llbracket1;\mathsf{N}\rrbracket \right) \times \mathds{B}^{\mathsf{N}} &
+\longrightarrow & \mathds{B}^{\mathsf{N}} \\
+& (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}%
+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} = \mathcal{P}\left(\llbracket 1 ; \mathsf{N} \rrbracket\right)^\mathds{N} \times
+\mathds{B}^\mathsf{N},
+\end{equation}
+\noindent and the map defined on $\mathcal{X}$:
+\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 $\sigma
+(S^{n})_{n\in \mathds{N}}\in \mathcal{P}\left(\llbracket 1 ; \mathsf{N} \rrbracket\right)^\mathds{N}\longrightarrow (S^{n+1})_{n\in
+\mathds{N}}\in \mathcal{P}\left(\llbracket 1 ; \mathsf{N} \rrbracket\right)^\mathds{N}$ 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}
+X^0 \in \mathcal{X} \\
+X^{k+1}=G_{f}(X^k).%
+\end{array}%
+\right.
+\end{equation}%
+
+Another time, a shift function appears as a component of these general chaotic
+iterations.
+
+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 defined.
+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})}\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\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}]
+Firstly, strong transitivity implies transitivity.
+
+Let $(S,E) \in\mathcal{X}$ and $\varepsilon >0$. To
+prove that $G_f$ is regular, it is sufficient to prove that
+there exists a strategy $\tilde S$ such that the distance between
+$(\tilde S,E)$ and $(S,E)$ is less than $\varepsilon$, and such that
+$(\tilde S,E)$ is a periodic point.
+
+Let $t_1=\lfloor-\log_{10}(\varepsilon)\rfloor$, and let $E'$ be the
+configuration that we obtain from $(S,E)$ after $t_1$ iterations of
+$G_f$. As $G_f$ is strongly transitive, there exists a strategy $S'$
+and $t_2\in\mathds{N}$ such
+that $E$ is reached from $(S',E')$ after $t_2$ iterations of $G_f$.
+
+Consider the strategy $\tilde S$ that alternates the first $t_1$ terms
+of $S$ and the first $t_2$ terms of $S'$: $$\tilde
+S=(S_0,\dots,S_{t_1-1},S'_0,\dots,S'_{t_2-1},S_0,\dots,S_{t_1-1},S'_0,\dots,S'_{t_2-1},S_0,\dots).$$ It
+is clear that $(\tilde S,E)$ is obtained from $(\tilde S,E)$ after
+$t_1+t_2$ iterations of $G_f$. So $(\tilde S,E)$ is a periodic
+point. Since $\tilde S_t=S_t$ for $t<t_1$, by the choice of $t_1$, we
+have $d((S,E),(\tilde S,E))<\epsilon$.
+\end{proof}
+
+
+
+\section{Efficient PRNG based on Chaotic Iterations}
+
+In order to implement efficiently a PRNG based on chaotic iterations it is
+possible to improve previous works [ref]. One solution consists in considering
+that the strategy used contains all the bits for which the negation is
+achieved out. Then in order to apply the negation on these bits we can simply
+apply the xor operator between the current number and the strategy. In
+order to obtain the strategy we also use a classical PRNG.
+
+Here is an example with 16-bits numbers showing how the bitwise operations
+are
+applied. Suppose that $x$ and the strategy $S^i$ are defined in binary mode.
+Then the following table shows the result of $x$ xor $S^i$.
+$$
+\begin{array}{|cc|cccccccccccccccc|}
+\hline
+x &=&1&0&1&1&1&0&1&0&1&0&0&1&0&0&1&0\\
+\hline
+S^i &=&0&1&1&0&0&1&1&0&1&1&1&0&0&1&1&1\\
+\hline
+x \oplus S^i&=&1&1&0&1&1&1&0&0&0&1&1&1&0&1&0&1\\
+\hline
+
+\hline
+ \end{array}
+$$
+
+%% \begin{figure}[htbp]
+%% \begin{center}
+%% \fbox{
+%% \begin{minipage}{14cm}
+%% unsigned int CIprng() \{\\
+%% static unsigned int x = 123123123;\\
+%% unsigned long t1 = xorshift();\\
+%% unsigned long t2 = xor128();\\
+%% unsigned long t3 = xorwow();\\
+%% x = x\textasciicircum (unsigned int)t1;\\
+%% x = x\textasciicircum (unsigned int)(t2$>>$32);\\
+%% x = x\textasciicircum (unsigned int)(t3$>>$32);\\
+%% x = x\textasciicircum (unsigned int)t2;\\
+%% x = x\textasciicircum (unsigned int)(t1$>>$32);\\
+%% x = x\textasciicircum (unsigned int)t3;\\
+%% return x;\\
+%% \}
+%% \end{minipage}
+%% }
+%% \end{center}
+%% \caption{sequential Chaotic Iteration PRNG}
+%% \label{algo:seqCIprng}
+%% \end{figure}
+
+
+
+\lstset{language=C,caption={C code of the sequential chaotic iterations based
+PRNG},label=algo:seqCIprng}
+\begin{lstlisting}
+unsigned int CIprng() {
+ static unsigned int x = 123123123;
+ unsigned long t1 = xorshift();
+ unsigned long t2 = xor128();
+ unsigned long t3 = xorwow();
+ x = x^(unsigned int)t1;
+ x = x^(unsigned int)(t2>>32);
+ x = x^(unsigned int)(t3>>32);
+ x = x^(unsigned int)t2;
+ x = x^(unsigned int)(t1>>32);
+ x = x^(unsigned int)t3;
+ return x;
+}
+\end{lstlisting}
+
+
+
+
+
+In listing~\ref{algo:seqCIprng} a sequential version of our chaotic iterations
+based PRNG is presented. The xor operator is represented by \textasciicircum.
+This function uses three classical 64-bits PRNG: the \texttt{xorshift}, the
+\texttt{xor128} and the \texttt{xorwow}. In the following, we call them
+xor-like PRNGSs. These three PRNGs are presented in~\cite{Marsaglia2003}. As
+each xor-like PRNG used works with 64-bits and as our PRNG works with 32-bits,
+the use of \texttt{(unsigned int)} selects the 32 least significant bits whereas
+\texttt{(unsigned int)(t3$>>$32)} selects the 32 most significants bits of the
+variable \texttt{t}. So to produce a random number realizes 6 xor operations
+with 6 32-bits numbers produced by 3 64-bits PRNG. This version successes the
+BigCrush of the TestU01 battery~\cite{LEcuyerS07}.
+
+\section{Efficient prng based on chaotic iterations on GPU}
