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}
\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.
-In this later context, we claim that the upper bound for the stopping time
-should be reduced.
+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.
