bla bla:

[16dcc.git] / stopping.tex

diff --git a/stopping.tex b/stopping.tex
index fb0b9e0dec27c2dc2215ea2e80b7304e0a034f91..1ac699934cc93a9f5bd6be19b60b634d87c42596 100644 (file)
--- a/stopping.tex
+++ b/stopping.tex
@@ -33,7 +33,7 @@ P=\dfrac{1}{6} \left(
 0&0&0&0&1&0&4&1 \\
 0&0&0&1&0&1&0&4 
 \end{array}
 0&0&0&0&1&0&4&1 \\
 0&0&0&1&0&1&0&4 
 \end{array}

-\right)
+\right).

 \]
 \end{xpl}
 
 \]
 \end{xpl}
 


@@ -69,9 +69,13 @@ and
 
 $$t_{\rm mix}(\varepsilon)=\min\{t \mid d(t)\leq \varepsilon\}.$$
 
 
 $$t_{\rm mix}(\varepsilon)=\min\{t \mid d(t)\leq \varepsilon\}.$$
 

-Intuitively speaking, $t_{\rm mix}$ is a mixing time 
-\textit{i.e.}, is the time until the matrix $X$ of a Markov chain  
-is $\epsilon$-close to a stationary distribution.
+%% Intuitively speaking, $t_{\rm mix}$ is a mixing time 
+%% \textit{i.e.}, is the time until the matrix $X$ of a Markov chain  
+%% is $\epsilon$-close to a stationary distribution.
+
+Intutively speaking,  $t_{\rm mix}(\varepsilon)$ is the time/steps required
+to be sure to be $\varepsilon$-close to the staionary distribution, wherever
+the chain starts. 

 
 
 
 
 
 


@@ -115,7 +119,7 @@ $$\P_X(X_\tau=Y)=\pi(Y).$$
 
 
 A stopping time $\tau$ is a {\emph strong stationary time} if $X_{\tau}$ is
 
 
 A stopping time $\tau$ is a {\emph strong stationary time} if $X_{\tau}$ is

-independent of $\tau$. 
+independent of $\tau$. The following result will be useful~\cite[Proposition~6.10]{LevinPeresWilmer2006},

 
 
 \begin{thrm}\label{thm-sst}
 
 
 \begin{thrm}\label{thm-sst}


@@ -231,7 +235,8 @@ This probability is independent of the value of the other bits.
 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
 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
+$\ell$-th bit of $X_t$ is $0$ or $1$ with the same probability,  and
+independently of the value of the other bits, proving the

 lemma.\end{proof}
 
 \begin{thrm} \label{prop:stop}
 lemma.\end{proof}
 
 \begin{thrm} \label{prop:stop}


@@ -345,7 +350,7 @@ direct application of lemma~\ref{prop:lambda} and~\ref{lm:stopprime}.
 \end{proof}
 
 Now using Markov Inequality, one has $\P_X(\tau > t)\leq \frac{E[\tau]}{t}$.
 \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:  $\P_X(\tau > t)\leq \frac{1}{4}$. 
+With $t_n=32N^2+16N\ln (N+1)$, one obtains:  $\P_X(\tau > t_n)\leq \frac{1}{4}$. 

 Therefore, using the defintion of $t_{\rm mix)}$ and
 Theorem~\ref{thm-sst}, it follows that
 $t_{\rm mix}\leq 32N^2+16N\ln (N+1)=O(N^2)$.
 Therefore, using the defintion of $t_{\rm mix)}$ and
 Theorem~\ref{thm-sst}, it follows that
 $t_{\rm mix}\leq 32N^2+16N\ln (N+1)=O(N^2)$.


@@ -354,11 +359,11 @@ $t_{\rm mix}\leq 32N^2+16N\ln (N+1)=O(N^2)$.
 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. 
 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 
+The calculus doesn't consider (balanced) Hamiltonian cycles, which 

 are more regular and more binding than this constraint.
 Moreover, the bound
 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
+is obtained using the coarse Markov Inequality. For the
+classical (lazzy) random walk 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 $\Theta(N\ln
 N)$.
 Hamiltonian cylce, the mixing time is in $\Theta(N\ln N)$. 
 We conjecture that in our context, the mixing time is also in $\Theta(N\ln
 N)$.


@@ -376,12 +381,6 @@ number of iterations such that all elements $\ell\in \llbracket 1,{\mathsf{N}} \
 by calling this code many times with many instances of function and many 
 seeds.
 
 by calling this code many times with many instances of function and many 
 seeds.
 

-Practically speaking, for each number $\mathsf{N}$,$ 3 \le \mathsf{N} \le 16$, 
-10 functions have been generaed 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 table~\ref{table:stopping:moy}
-summarizes results. It can be observed that the approximation is largely
-wœsmaller than the upper bound given in theorem~\ref{prop:stop}.
-

 \begin{algorithm}[ht]
 %\begin{scriptsize}
 \KwIn{a function $f$, an initial configuration $x^0$ ($\mathsf{N}$ bits)}
 \begin{algorithm}[ht]
 %\begin{scriptsize}
 \KwIn{a function $f$, an initial configuration $x^0$ ($\mathsf{N}$ bits)}


@@ -389,39 +388,59 @@ wœsmaller than the upper bound given in theorem~\ref{prop:stop}.
 
 $\textit{nbit} \leftarrow 0$\;
 $x\leftarrow x^0$\;
 
 $\textit{nbit} \leftarrow 0$\;
 $x\leftarrow x^0$\;

-$\textit{visited}\leftarrow\emptyset$\;
-
-\While{$\left\vert{\textit{visited}}\right\vert < \mathsf{N} $}
+$\textit{fair}\leftarrow\emptyset$\;
+\While{$\left\vert{\textit{fair}}\right\vert < \mathsf{N} $}

 {
 {

-        $ s \leftarrow \textit{Random}(n)$ \;
+        $ s \leftarrow \textit{Random}(\mathsf{N})$ \;

         $\textit{image} \leftarrow f(x) $\;
         $\textit{image} \leftarrow f(x) $\;

-        \If{$x[s] \neq \textit{image}[s]$}{
-            $\textit{visited} \leftarrow \textit{visited} \cup \{s\}$
+        \If{$\textit{Random}(1) \neq 0$ and $x[s] \neq \textit{image}[s]$}{
+            $\textit{fair} \leftarrow \textit{fair} \cup \{s\}$\;
+            $x[s] \leftarrow \textit{image}[s]$\;

           }
           }

-        $x[s] \leftarrow \textit{image}[s]$\;

         $\textit{nbit} \leftarrow \textit{nbit}+1$\;
 }
 \Return{$\textit{nbit}$}\;
 %\end{scriptsize}
         $\textit{nbit} \leftarrow \textit{nbit}+1$\;
 }
 \Return{$\textit{nbit}$}\;
 %\end{scriptsize}

-\caption{Pseudo Code of the stoping time calculus}
+\caption{Pseudo Code of stoping time calculus }

 \label{algo:stop}
 \end{algorithm}
 
 \label{algo:stop}
 \end{algorithm}
 

+Practically speaking, for each number $\mathsf{N}$, $ 3 \le \mathsf{N} \le 16$, 
+10 functions have been generaed 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}
-$$
-\begin{array}{|*{15}{l|}}
-\hline
-\mathsf{N}  & 3 & 4 & 5 & 6 & 7& 8 & 9 & 10& 11 & 12 & 13 & 14 & 15 & 16 \\
-\hline
-\mathsf{N}  & 3 & 10.9 & 5 & 17.7 & 7& 25 & 9 & 32.7& 11 & 40.8 & 13 & 49.2 & 15 & 16 \\
-\hline
-\end{array}
-$$
-\caption{Average Stopping Time}\label{table:stopping:moy}
-\end{table}
+
+
+% \begin{table}
+% $$
+% \begin{array}{|*{14}{l|}}
+% \hline
+% \mathsf{N}  & 4 & 5 & 6 & 7& 8 & 9 & 10& 11 & 12 & 13 & 14 & 15 & 16 \\
+% \hline
+% \mathsf{N}  & 21.8 & 28.4 & 35.4 & 42.5 & 50 & 57.7 & 65.6& 73.5 & 81.6 & 90 & 98.3 & 107.1 & 115.7 \\
+% \hline
+% \end{array}
+% $$
+% \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}
+
+

 
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