X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/rairo15.git/blobdiff_plain/6537ad6b39c8648e4c3597b469e79e505193e358..ac6bb3190276a11caaed94b7443f95f697328bcd:/stopping.tex

diff --git a/stopping.tex b/stopping.tex
index 9664549..0ad3e8a 100644
--- a/stopping.tex
+++ b/stopping.tex
@@ -1,3 +1,80 @@
+
+
+
+Let thus be given such kind of map.
+This article focuses 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 
+${\mathsf{N}}$-cube augmented with all the self-loop, \textit{i.e.}, all the 
+edges $(v,v)$ for any $v \in \Bool^{\mathsf{N}}$. 
+Next, if we add probabilities on the transition graph, iterations can be 
+interpreted as Markov chains.
+
+\begin{xpl}
+Let us consider for instance  
+the graph $\Gamma(f)$ defined 
+in \textsc{Figure~\ref{fig:iteration:f*}.} and 
+the probability function $p$ defined on the set of edges as follows:
+$$
+p(e) \left\{
+\begin{array}{ll}
+= \frac{2}{3} \textrm{ if $e=(v,v)$ with $v \in \Bool^3$,}\\
+= \frac{1}{6} \textrm{ otherwise.}
+\end{array}
+\right.  
+$$
+The matrix $P$ of the Markov chain associated to the function $f^*$ and to its probability function $p$ is 
+\[
+P=\dfrac{1}{6} \left(
+\begin{array}{llllllll}
+4&1&1&0&0&0&0&0 \\
+1&4&0&0&0&1&0&0 \\
+0&0&4&1&0&0&1&0 \\
+0&1&1&4&0&0&0&0 \\
+1&0&0&0&4&0&1&0 \\
+0&0&0&0&1&4&0&1 \\
+0&0&0&0&1&0&4&1 \\
+0&0&0&1&0&1&0&4 
+\end{array}
+\right)
+\]
+\end{xpl}
+
+
+% % 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 
+% % $\Bool^n$  by:
+% % $$\tv{\pi-\mu}=\max_{A\subset \Bool^n} |\pi(A)-\mu(A)|.$$ 
+% % It is known that
+% % $$\tv{\pi-\mu}=\frac{1}{2}\sum_{x\in\Bool^n}|\pi(x)-\mu(x)|.$$
+
+% % 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
+% % $$d(t)=\max_{x\in\Bool^n}\tv{M^t(x,\cdot)-\pi}.$$
+% 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.
+% Finally, let $\varepsilon$ be a positive number, the \emph{mixing time} 
+% with respect to $\varepsilon$ is given by
+% $$t_{\rm mix}(\varepsilon)=\min\{t \mid d(t)\leq \varepsilon\}.$$
+% It defines the smallest iteration number 
+% that is sufficient to obtain a deviation lesser than $\varepsilon$.
+% Notice that the upper and lower bounds of mixing times cannot    
+% directly be computed with eigenvalues formulae as expressed 
+% in~\cite[Chap. 12]{LevinPeresWilmer2006}. The authors of this latter work  
+% only consider reversible Markov matrices whereas we do no restrict our 
+% matrices to such a form.
+
+
+
+
+
+
+
 This section considers functions $f: \Bool^n \rightarrow \Bool^n $ 
 issued from an hypercube where an Hamiltonian path has been removed.
 A specific random walk in this modified hypercube is first 
@@ -268,4 +345,10 @@ Theorem~\ref{prop:stop} is a direct application of
 lemma~\ref{prop:lambda} and~\ref{lm:stopprime}.
 \end{proof}
 
-
+Notice that the calculus of the stationary time upper bound is obtained
+under the following constraint: for each vertex in the $\mathsf{N}$-cube 
+there are one ongoing arc and one outgoing arc that are removed. 
+The calculus does not consider (balanced) hamiltonian cycles, which 
+are more regular and more binding than this constraint.
+In this later context, we claim that the upper bound for the stopping time 
+should be reduced.