-
-
-
-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.
+This section considers functions $f: \Bool^{\mathsf{N}} \rightarrow \Bool^{\mathsf{N}} $
+issued from an hypercube where an Hamiltonian path has been removed
+as described in previous section.
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 $v \in \Bool^{\mathsf{N}}$.
\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^{\mathsf{N}} \rightarrow \Bool^{\mathsf{N}} $
-issued from an hypercube where an Hamiltonian path has been removed.
A specific random walk in this modified hypercube is first
-introduced. We further detail
-a theoretical study on the length of the path
-which is sufficient to follow to get a uniform distribution.
+introduced (See section~\ref{sub:stop:formal}). We further
+ study this random walk in a theoretical way to
+provide an upper bound of fair sequences
+(See section~\ref{sub:stop:bound}).
+We finally complete these study with experimental
+results that reduce this bound (Sec.~\ref{sub:stop:exp}).
Notice that for a general references on Markov chains
-see~\cite{LevinPeresWilmer2006}
-, and particularly Chapter~5 on stopping times.
-
+see~\cite{LevinPeresWilmer2006},
+and particularly Chapter~5 on stopping times.
+\subsection{Formalizing the Random Walk}\label{sub:stop:formal}
First of all, let $\pi$, $\mu$ be two distributions on $\Bool^{\mathsf{N}}$. The total
variation distance between $\pi$ and $\mu$ is denoted $\tv{\pi-\mu}$ and is
$P$ has a stationary distribution $\pi$, then we define
$$d(t)=\max_{X\in\Bool^{\mathsf{N}}}\tv{P^t(X,\cdot)-\pi}.$$
+\ANNOT{incohérence de notation $X$ : entier ou dans $B^N$ ?}
and
$$t_{\rm mix}(\varepsilon)=\min\{t \mid d(t)\leq \varepsilon\}.$$
+
+Intuitively speaking, $t_{\rm mix}$ is a mixing time
+\textit{i.e.}, is the time until the matrix $X$ \ANNOT{pas plutôt $P$ ?} of a Markov chain
+is $\epsilon$-close to a stationary distribution.
+
+
+
One can prove that
$$t_{\rm mix}(\varepsilon)\leq \lceil\log_2(\varepsilon^{-1})\rceil t_{\rm mix}(\frac{1}{4})$$
% It is known that $d(t+1)\leq d(t)$. \JFC{references ? Cela a-t-il
-% un intérêt dans la preuve ensuite.}
+% un intérêt dans la preuve ensuite.}
%and
% $$t_{\rm mix}(\varepsilon)=\min\{t \mid d(t)\leq \varepsilon\}.$$
-% One can prove that \JFc{Ou cela a-t-il été fait?}
+% One can prove that \JFc{Ou cela a-t-il été fait?}
% $$t_{\rm mix}(\varepsilon)\leq \lceil\log_2(\varepsilon^{-1})\rceil t_{\rm mix}(\frac{1}{4})$$
such that the distribution of $X_\tau$ is $\pi$:
$$\P_X(X_\tau=Y)=\pi(Y).$$
-A stopping time $\tau$ is a {\emph strong stationary time} if $X_{\tau}$ is
+\subsection{Upper bound of Stopping Time}\label{sub:stop:bound}
+
+
+A stopping time $\tau$ is a \emph{strong stationary time} if $X_{\tau}$ is
independent of $\tau$.
-\begin{thrm}
+\begin{thrm}\label{thm-sst}
If $\tau$ is a strong stationary time, then $d(t)\leq \max_{X\in\Bool^{\mathsf{N}}}
\P_X(\tau > t)$.
\end{thrm}
-%%%%%%%%%%%%%%%%%%%%%%%%%%%ù
+%%%%%%%%%%%%%%%%%%%%%%%%%%%ù
%\section{Stopping time}
An integer $\ell\in \llbracket 1,{\mathsf{N}} \rrbracket$ is said {\it fair}
$\frac{1}{2}$. Since $h(X_{\tau_\ell-1})\neq\ell$ the value of the $\ell$-th
bit of $X_{\tau_\ell}$
is $0$ or $1$ with the same probability ($\frac{1}{2}$).
+This probability is independent of the value of the other bits.
+
- Moving next in the chain, at each step,
+
+Moving next in the chain, at each step,
the $l$-th bit is switched from $0$ to $1$ or from $1$ to $0$ each time with
the same probability. Therefore, for $t\geq \tau_\ell$, the
$\ell$-th bit of $X_t$ is $0$ or $1$ with the same probability, proving the
\begin{proof}
Since $\ts^\prime$ is the time used to obtain $\mathsf{N}-1$ fair bits.
Assume that the last unfair bit is $\ell$. One has
-$\ts=\ts^\prime+S_{X_\tau,\ell}$, and therefore
-$E[\ts] = E[\ts^\prime]+E[S_{X_\tau,\ell}]$. Therefore,
-Theorem~\ref{prop:stop} is a direct application of
-lemma~\ref{prop:lambda} and~\ref{lm:stopprime}.
+$\ts=\ts^\prime+S_{X_\tau,\ell}$, and therefore $E[\ts] =
+E[\ts^\prime]+E[S_{X_\tau,\ell}]$. Therefore, Theorem~\ref{prop:stop} is a
+direct application of lemma~\ref{prop:lambda} and~\ref{lm:stopprime}.
\end{proof}
+Now using Markov Inequality, one has $\P_X(\tau > t)\leq \frac{E[\tau]}{t}$.
+With $t=32N^2+16N\ln (N+1)$, one obtains: $\P_X(\tau > t)\leq \frac{1}{4}$.
+Therefore, using the defintion of $t_{\rm mix)}$ and
+Theorem~\ref{thm-sst}, it follows that
+$t_{\rm mix}\leq 32N^2+16N\ln (N+1)=O(N^2)$.
+
+
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.
+Moreover, the bound
+is obtained using Markov Inequality which is frequently coarse. For the
+classical random walkin the $\mathsf{N}$-cube, without removing any
+Hamiltonian cylce, the mixing time is in $\Theta(N\ln N)$.
+We conjecture that in our context, the mixing time is also in $\Theta(N\ln
+N)$.
+
+
In this later context, we claim that the upper bound for the stopping time
-should be reduced.
+should be reduced. This fact is studied in the next section.
+
+\subsection{Practical Evaluation of Stopping Times}\label{sub:stop:exp}
+
+Let be given a function $f: \Bool^{\mathsf{N}} \rightarrow \Bool^{\mathsf{N}}$
+and an initial seed $x^0$.
+The pseudo code given in algorithm~\ref{algo:stop} returns the smallest
+number of iterations such that all elements $\ell\in \llbracket 1,{\mathsf{N}} \rrbracket$ are fair. It allows to deduce an approximation of $E[\ts]$
+by calling this code many times with many instances of function and many
+seeds.
+
+\begin{algorithm}[ht]
+%\begin{scriptsize}
+\KwIn{a function $f$, an initial configuration $x^0$ ($\mathsf{N}$ bits)}
+\KwOut{a number of iterations $\textit{nbit}$}
+
+$\textit{nbit} \leftarrow 0$\;
+$x\leftarrow x^0$\;
+$\textit{fair}\leftarrow\emptyset$\;
+\While{$\left\vert{\textit{fair}}\right\vert < \mathsf{N} $}
+{
+ $ s \leftarrow \textit{Random}(\mathsf{N})$ \;
+ $\textit{image} \leftarrow f(x) $\;
+ \If{$\textit{Random}(1) \neq 0$ and $x[s] \neq \textit{image}[s]$}{
+ $\textit{fair} \leftarrow \textit{fair} \cup \{s\}$\;
+ $x[s] \leftarrow \textit{image}[s]$\;
+ }
+ $\textit{nbit} \leftarrow \textit{nbit}+1$\;
+}
+\Return{$\textit{nbit}$}\;
+%\end{scriptsize}
+\caption{Pseudo Code of stoping time calculus }
+\label{algo:stop}
+\end{algorithm}
+
+Practically speaking, for each number $\mathsf{N}$, $ 3 \le \mathsf{N} \le 16$,
+10 functions have been generated according to method presented in section~\ref{sec:hamilton}. For each of them, the calculus of the approximation of $E[\ts]$
+is executed 10000 times with a random seed. The Figure~\ref{fig:stopping:moy}
+summarizes these results. In this one, a circle represents the
+approximation of $E[\ts]$ for a given $\mathsf{N}$.
+The line is the graph of the function $x \mapsto 2x\ln(2x+8)$.
+It can firstly
+be observed that the approximation is largely
+smaller than the upper bound given in theorem~\ref{prop:stop}.
+It can be further deduced that the conjecture of the previous section
+is realistic according the graph of $x \mapsto 2x\ln(2x+8)$.
+
+
+
+
+
+% \begin{table}
+% $$
+% \begin{array}{|*{14}{l|}}
+% \hline
+% \mathsf{N} & 4 & 5 & 6 & 7& 8 & 9 & 10& 11 & 12 & 13 & 14 & 15 & 16 \\
+% \hline
+% \mathsf{N} & 21.8 & 28.4 & 35.4 & 42.5 & 50 & 57.7 & 65.6& 73.5 & 81.6 & 90 & 98.3 & 107.1 & 16 \\
+% \hline
+% \end{array}
+% $$
+% \caption{Average Stopping Time}\label{table:stopping:moy}
+% \end{table}
+
+\begin{figure}
+\centering
+\includegraphics[scale=0.5]{complexity}
+\caption{Average Stopping Time Approximation}\label{fig:stopping:moy}
+\end{figure}
+
+
+
+%%% Local Variables:
+%%% mode: latex
+%%% TeX-master: "main"
+%%% ispell-dictionary: "american"
+%%% mode: flyspell
+%%% End:
