such that
$x=(x_1,\dots,x_n)$ maps to $f(x)=(f_1(x),\dots,f_n(x))$.
Functions are iterated as follows.
-At the $t^{th}$ iteration, only the $s_{t}-$th component is
-``iterated'', where $s = \left(s_t\right)_{t \in \mathds{N}}$ is a sequence of indices taken in $\llbracket 1;n \rrbracket$ called ``strategy''. Formally,
+At the $t^{th}$ iteration, only the $s_{t}-$th component is said to be
+``iterated'', where $s = \left(s_t\right)_{t \in \mathds{N}}$ is a sequence of indices taken in $\llbracket 1;n \rrbracket$ called ``strategy''.
+Formally,
let $F_f: \llbracket1;n\rrbracket\times \Bool^{n}$ to $\Bool^n$ be defined by
\[
F_f(i,x)=(x_1,\dots,x_{i-1},f_i(x),x_{i+1},\dots,x_n).
Let us consider for instance $n=3$.
Let
$f^*: \Bool^3 \rightarrow \Bool^3$ be defined by
-
$f^*(x_1,x_2,x_3) =
(x_2 \oplus x_3, \overline{x_1}\overline{x_3} + x_1\overline{x_2},
-\overline{x_1}\overline{x_3} + x_1x_2)$
-
-
+\overline{x_1}\overline{x_3} + x_1x_2)$.
The iteration graph $\Gamma(f^*)$ of this function is given in
Figure~\ref{fig:iteration:f*}.
\vspace{-1em}
\begin{figure}[ht]
\begin{center}
-\includegraphics[scale=0.5]{images/iter_f0c.eps}
+\includegraphics[scale=0.5]{images/iter_f0c}
\end{center}
\vspace{-0.5em}
\caption{Iteration Graph $\Gamma(f^*)$ of the function $f^*$}\label{fig:iteration:f*}
\end{figure}
\end{xpl}
-\vspace{-0.5em}
-It is easy to associate a Markov Matrix $M$ to such a graph $G(f)$
-as follows:
+% \vspace{-0.5em}
+% It is easy to associate a Markov Matrix $M$ to such a graph $G(f)$
+% as follows:
+
+% $M_{ij} = \frac{1}{n}$ if there is an edge from $i$ to $j$ in $\Gamma(f)$ and $i \neq j$; $M_{ii} = 1 - \sum\limits_{j=1, j\neq i}^n M_{ij}$; and $M_{ij} = 0$ otherwise.
+
+% \begin{xpl}
+% The Markov matrix associated to the function $f^*$ is
+
+% \[
+% M=\dfrac{1}{3} \left(
+% \begin{array}{llllllll}
+% 1&1&1&0&0&0&0&0 \\
+% 1&1&0&0&0&1&0&0 \\
+% 0&0&1&1&0&0&1&0 \\
+% 0&1&1&1&0&0&0&0 \\
+% 1&0&0&0&1&0&1&0 \\
+% 0&0&0&0&1&1&0&1 \\
+% 0&0&0&0&1&0&1&1 \\
+% 0&0&0&1&0&1&0&1
+% \end{array}
+% \right)
+% \]
+%\end{xpl}
+
+Let thus be given such kind of map.
+This article focusses on studying its iterations according to
+the equation~(\ref{eq:asyn}) with a given strategy.
+First of all, this can be interpreted as walking into its iteration graph
+where the choice of the edge to follow is decided by the strategy.
+Notice that the iteration graph is always a subgraph of
+$n$-cube augemented with all the self-loop, \textit{i.e.}, all the
+edges $(v,v)$ for any $v \in \Bool^n$.
+Next, if we add probabilities on the transition graph, iterations can be
+interpreted as Markov chains.
+
+
+
+
+Let $\pi$, $\mu$ be two distribution on a same set $\Omega$. The total
+variation distance between $\pi$ and $\mu$ is denoted $\tv{\pi-\mu}$ and is
+defined by
+$$\tv{\pi-\mu}=\max_{A\subset \Omega} |\pi(A)-\mu(A)|.$$ It is known that
+$$\tv{\pi-\mu}=\frac{1}{2}\sum_{x\in\Omega}|\pi(x)-\mu(x)|.$$ Moreover, if
+$\nu$ is a distribution on $\Omega$, one has
+$$\tv{\pi-\mu}\leq \tv{\pi-\nu}+\tv{\nu-\mu}$$
+
+Let $P$ be the matrix of a markov chain on $\Omega$. $P(x,\cdot)$ is the
+distribution induced by the $x$-th row of $P$. If the markov chain induced by
+$P$ has a stationary distribution $\pi$, then we define
+$$d(t)=\max_{x\in\Omega}\tv{P^t(x,\cdot)-\pi},$$
+and
+
+$$t_{\rm mix}(\varepsilon)=\min\{t \mid d(t)\leq \varepsilon\}.$$
+One can prove that
-$M_{ij} = \frac{1}{n}$ if there is an edge from $i$ to $j$ in $\Gamma(f)$ and $i \neq j$; $M_{ii} = 1 - \sum\limits_{j=1, j\neq i}^n M_{ij}$; and $M_{ij} = 0$ otherwise.
+$$t_{\rm mix}(\varepsilon)\leq \lceil\log_2(\varepsilon^{-1})\rceil t_{\rm mix}(\frac{1}{4})$$
-\begin{xpl}
-The Markov matrix associated to the function $f^*$ is
+It is known that $d(t+1)\leq d(t)$.
-\[
-M=\dfrac{1}{3} \left(
-\begin{array}{llllllll}
-1&1&1&0&0&0&0&0 \\
-1&1&0&0&0&1&0&0 \\
-0&0&1&1&0&0&1&0 \\
-0&1&1&1&0&0&0&0 \\
-1&0&0&0&1&0&1&0 \\
-0&0&0&0&1&1&0&1 \\
-0&0&0&0&1&0&1&1 \\
-0&0&0&1&0&1&0&1
-\end{array}
-\right)
-\]
+Let $(X_t)_{t\in \mathbb{N}}$ be a sequence of $\Omega$ valued random
+variables. A $\mathbb{N}$-valued random variable $\tau$ is a {\it stopping
+ time} for the sequence $(X_i)$ if for each $t$ there exists $B_t\subseteq
+\omega^{t+1}$ such that $\{tau=t\}=\{(X_0,X_1,\ldots,X_t)\in B_t\}$.
-
+Let $(X_t)_{t\in \mathbb{N}}$ be a markov chain and $f(X_{t-1},Z_t)$ a
+random mapping representation of the markov chain. A {\it randomized
+ stopping time} for the markov chain is a stopping time for
+$(Z_t)_{t\in\mathbb{N}}$. It he markov chain is irreductible and has $\pi$
+as stationary distribution, then a {\it stationay time} $\tau$ is a
+randomized stopping time (possibily depending on the starting position $x$),
+such that the distribution of $X_\tau$ is $\pi$:
+$$\P_x(X_\tau=y)=\pi(y).$$
-\end{xpl}
+\JFC{Ou ceci a-t-il ete prouvé}
+\begin{Theo}
+If $\tau$ is a strong stationary time, then $d(t)\leq \max_{x\in\Omega}
+\P_x(\tau > t)$.
+\end{Theo}
-It is usual to check whether rows of such kind of matrices
-converge to a specific
-distribution.
-Let us first recall the \emph{Total Variation} distance $\tv{\pi-\mu}$,
-which is defined for two distributions $\pi$ and $\mu$ on the same set
-$\Omega$ by:
-$$\tv{\pi-\mu}=\max_{A\subset \Omega} |\pi(A)-\mu(A)|.$$
+% Let us first recall the \emph{Total Variation} distance $\tv{\pi-\mu}$,
+% which is defined for two distributions $\pi$ and $\mu$ on the same set
+% $\Omega$ by:
+% $$\tv{\pi-\mu}=\max_{A\subset \Omega} |\pi(A)-\mu(A)|.$$
% It is known that
% $$\tv{\pi-\mu}=\frac{1}{2}\sum_{x\in\Omega}|\pi(x)-\mu(x)|.$$
-Let then $M(x,\cdot)$ be the
-distribution induced by the $x$-th row of $M$. If the Markov chain
-induced by
-$M$ has a stationary distribution $\pi$, then we define
-$$d(t)=\max_{x\in\Omega}\tv{M^t(x,\cdot)-\pi}.$$
+% Let then $M(x,\cdot)$ be the
+% distribution induced by the $x$-th row of $M$. If the Markov chain
+% induced by
+% $M$ has a stationary distribution $\pi$, then we define
+% $$d(t)=\max_{x\in\Omega}\tv{M^t(x,\cdot)-\pi}.$$
Intuitively $d(t)$ is the largest deviation between
the distribution $\pi$ and $M^t(x,\cdot)$, which
is the result of iterating $t$ times the function.
