-This section considers functions $f: \Bool^n \rightarrow \Bool^n $
+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
-First of all, let $\pi$, $\mu$ be two distributions on $\Bool^n$. The total
+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
defined by
-$$\tv{\pi-\mu}=\max_{A\subset \Bool^n} |\pi(A)-\mu(A)|.$$ It is known that
-$$\tv{\pi-\mu}=\frac{1}{2}\sum_{X\in\Bool^n}|\pi(X)-\mu(X)|.$$ Moreover, if
-$\nu$ is a distribution on $\Bool^n$, one has
+$$\tv{\pi-\mu}=\max_{A\subset \Bool^{\mathsf{N}}} |\pi(A)-\mu(A)|.$$ It is known that
+$$\tv{\pi-\mu}=\frac{1}{2}\sum_{X\in\Bool^{\mathsf{N}}}|\pi(X)-\mu(X)|.$$ Moreover, if
+$\nu$ is a distribution on $\Bool^{\mathsf{N}}$, one has
$$\tv{\pi-\mu}\leq \tv{\pi-\nu}+\tv{\nu-\mu}$$
-Let $P$ be the matrix of a Markov chain on $\Bool^n$. $P(X,\cdot)$ is the
+Let $P$ be the matrix of a Markov chain on $\Bool^{\mathsf{N}}$. $P(X,\cdot)$ 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
-$$d(t)=\max_{X\in\Bool^n}\tv{P^t(X,\cdot)-\pi}.$$
+$$d(t)=\max_{X\in\Bool^{\mathsf{N}}}\tv{P^t(X,\cdot)-\pi}.$$
and
-Let $(X_t)_{t\in \mathbb{N}}$ be a sequence of $\Bool^n$ valued random
+Let $(X_t)_{t\in \mathbb{N}}$ be a sequence of $\Bool^{\mathsf{N}}$ 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 $B_t\subseteq
-(\Bool^n)^{t+1}$ such that $\{\tau=t\}=\{(X_0,X_1,\ldots,X_t)\in B_t\}$.
+(\Bool^{\mathsf{N}})^{t+1}$ such that $\{\tau=t\}=\{(X_0,X_1,\ldots,X_t)\in B_t\}$.
In other words, the event $\{\tau = t \}$ only depends on the values of
$(X_0,X_1,\ldots,X_t)$, not on $X_k$ with $k > t$.
\begin{Theo}
-If $\tau$ is a strong stationary time, then $d(t)\leq \max_{X\in\Bool^n}
+If $\tau$ is a strong stationary time, then $d(t)\leq \max_{X\in\Bool^{\mathsf{N}}}
\P_X(\tau > t)$.
\end{Theo}
%Let $\Bool^n$ be the set of words of length $n$.
Let $E=\{(X,Y)\mid
-X\in \Bool^n, Y\in \Bool^n,\ X=Y \text{ or } X\oplus Y \in 0^*10^*\}$.
+X\in \Bool^{\mathsf{N}}, Y\in \Bool^{\mathsf{N}},\ X=Y \text{ or } X\oplus Y \in 0^*10^*\}$.
In other words, $E$ is the set of all the edges in the classical
-$n$-cube.
-Let $h$ be a function from $\Bool^n$ into $\llbracket 1, n \rrbracket$.
+${\mathsf{N}}$-cube.
+Let $h$ be a function from $\Bool^{\mathsf{N}}$ into $\llbracket 1, {\mathsf{N}} \rrbracket$.
Intuitively speaking $h$ aims at memorizing for each node
-$X \in \Bool^n$ which edge is removed in the Hamiltonian cycle,
-\textit{i.e.} which bit in $\llbracket 1, n \rrbracket$
+$X \in \Bool^{\mathsf{N}}$ which edge is removed in the Hamiltonian cycle,
+\textit{i.e.} which bit in $\llbracket 1, {\mathsf{N}} \rrbracket$
cannot be switched.
We denote by $E_h$ the set $E\setminus\{(X,Y)\mid X\oplus Y =
-0^{n-h(X)}10^{h(X)-1}\}$. This is the set of the modified hypercube,
-\textit{i.e.}, the $n$-cube where the Hamiltonian cycle $h$
+0^{{\mathsf{N}}-h(X)}10^{h(X)-1}\}$. This is the set of the modified hypercube,
+\textit{i.e.}, the ${\mathsf{N}}$-cube where the Hamiltonian cycle $h$
has been removed.
We define the Markov matrix $P_h$ for each line $X$ and
each column $Y$ as follows:
$$\left\{
\begin{array}{ll}
-P_h(X,X)=\frac{1}{2}+\frac{1}{2n} & \\
+P_h(X,X)=\frac{1}{2}+\frac{1}{2{\mathsf{N}}} & \\
P_h(X,Y)=0 & \textrm{if $(X,Y)\notin E_h$}\\
-P_h(X,Y)=\frac{1}{2n} & \textrm{if $X\neq Y$ and $(X,Y) \in E_h$}
+P_h(X,Y)=\frac{1}{2{\mathsf{N}}} & \textrm{if $X\neq Y$ and $(X,Y) \in E_h$}
\end{array}
\right.
$$
-We denote by $\ov{h} : \Bool^n \rightarrow \Bool^n$ the function
-such that for any $X \in \Bool^n $,
-$(X,\ov{h}(X)) \in E$ and $X\oplus\ov{h}(X)=0^{n-h(X)}10^{h(X)-1}$.
-The function $\ov{h}$ is said {\it square-free} if for every $X\in \Bool^n$,
+We denote by $\ov{h} : \Bool^{\mathsf{N}} \rightarrow \Bool^{\mathsf{N}}$ the function
+such that for any $X \in \Bool^{\mathsf{N}} $,
+$(X,\ov{h}(X)) \in E$ and $X\oplus\ov{h}(X)=0^{{\mathsf{N}}-h(X)}10^{h(X)-1}$.
+The function $\ov{h}$ is said {\it square-free} if for every $X\in \Bool^{\mathsf{N}}$,
$\ov{h}(\ov{h}(X))\neq X$.
\begin{Lemma}\label{lm:h}
\begin{proof}
Let $\ov{h}$ be bijective.
-Let $k\in \llbracket 1, n \rrbracket$ s.t. $h(\ov{h}^{-1}(X))=k$.
+Let $k\in \llbracket 1, {\mathsf{N}} \rrbracket$ s.t. $h(\ov{h}^{-1}(X))=k$.
Then $(\ov{h}^{-1}(X),X)$ belongs to $E$ and
-$\ov{h}^{-1}(X)\oplus X = 0^{n-k}10^{k-1}$.
+$\ov{h}^{-1}(X)\oplus X = 0^{{\mathsf{N}}-k}10^{k-1}$.
Let us suppose $h(X) = h(\ov{h}^{-1}(X))$. In such a case, $h(X) =k$.
By definition of $\ov{h}$, $(X, \ov{h}(X)) \in E $ and
-$X\oplus\ov{h}(X)=0^{n-h(X)}10^{h(X)-1} = 0^{n-k}10^{k-1}$.
+$X\oplus\ov{h}(X)=0^{{\mathsf{N}}-h(X)}10^{h(X)-1} = 0^{{\mathsf{N}}-k}10^{k-1}$.
Thus $\ov{h}(X)= \ov{h}^{-1}(X)$, which leads to $\ov{h}(\ov{h}(X))= X$.
This contradicts the square-freeness of $\ov{h}$.
\end{proof}
Let $Z$ be a random variable that is uniformly distributed over
-$\llbracket 1, n \rrbracket \times \Bool$.
-For $X\in \Bool^n$, we
+$\llbracket 1, {\mathsf{N}} \rrbracket \times \Bool$.
+For $X\in \Bool^{\mathsf{N}}$, we
define, with $Z=(i,b)$,
$$
\left\{
\begin{array}{ll}
-f(X,Z)=X\oplus (0^{n-i}10^{i-1}) & \text{if } b=1 \text{ and } i\neq h(X),\\
+f(X,Z)=X\oplus (0^{{\mathsf{N}}-i}10^{i-1}) & \text{if } b=1 \text{ and } i\neq h(X),\\
f(X,Z)=X& \text{otherwise.}
\end{array}\right.
$$
%%%%%%%%%%%%%%%%%%%%%%%%%%%ù
%\section{Stopping time}
-An integer $\ell\in \llbracket 1,n \rrbracket$ is said {\it fair}
+An integer $\ell\in \llbracket 1,{\mathsf{N}} \rrbracket$ is said {\it fair}
at time $t$ if there
exists $0\leq j <t$ such that $Z_{j+1}=(\ell,\cdot)$ and $h(X_j)\neq \ell$.
In other words, there exist a date $j$ before $t$ where
(\textit{i.e.}, $l$ is the strategy at date $j$)
and where the configuration $X_j$ allows to traverse the edge $l$.
-Let $\ts$ be the first time all the elements of $\llbracket 1, n \rrbracket$
+Let $\ts$ be the first time all the elements of $\llbracket 1, {\mathsf{N}} \rrbracket$
are fair. The integer $\ts$ is a randomized stopping time for
the Markov chain $(X_t)$.
\begin{Theo} \label{prop:stop}
If $\ov{h}$ is bijective and square-free, then
-$E[\ts]\leq 8n^2+ n\ln (n+1)$.
+$E[\ts]\leq 8{\mathsf{N}}^2+ {\mathsf{N}}\ln ({\mathsf{N}}+1)$.
\end{Theo}
-For each $X\in \Bool^n$ and $\ell\in\llbracket 1,n\rrbracket$,
+For each $X\in \Bool^{\mathsf{N}}$ and $\ell\in\llbracket 1,{\mathsf{N}}\rrbracket$,
let $S_{X,\ell}$ be the
random variable that counts the number of steps
from $X$ until we reach a configuration where
\begin{Lemma}\label{prop:lambda}
If $\ov{h}$ is a square-free bijective function, then the inequality
-$E[\lambda_h]\leq 8n^2$ is established.
+$E[\lambda_h]\leq 8{\mathsf{N}}^2$ is established.
\end{Lemma}
\begin{proof}
For every $X$, every $\ell$, one has $\P(S_{X,\ell})\leq 2)\geq
-\frac{1}{4n^2}$.
+\frac{1}{4{\mathsf{N}}^2}$.
Let $X_0= X$.
Indeed,
\begin{itemize}
\item if $h(X)\neq \ell$, then
-$\P(S_{X,\ell}=1)=\frac{1}{2n}\geq \frac{1}{4n^2}$.
+$\P(S_{X,\ell}=1)=\frac{1}{2{\mathsf{N}}}\geq \frac{1}{4{\mathsf{N}}^2}$.
\item otherwise, $h(X)=\ell$, then
$\P(S_{X,\ell}=1)=0$.
But in this case, intutively, it is possible to move
since $\ov{h}$ is square-free,
$\ov{h}(X)=\ov{h}(\ov{h}(\ov{h}^{-1}(X)))\neq \ov{h}^{-1}(X)$. It follows
that $(X,\ov{h}^{-1}(X))\in E_h$. We thus have
-$P(X_1=\ov{h}^{-1}(X))=\frac{1}{2N}$. Now, by Lemma~\ref{lm:h},
+$P(X_1=\ov{h}^{-1}(X))=\frac{1}{2{\MATHSF{N}}}$. Now, by Lemma~\ref{lm:h},
$h(\ov{h}^{-1}(X))\neq h(X)$. Therefore $\P(S_{x,\ell}=2\mid
-X_1=\ov{h}^{-1}(X))=\frac{1}{2N}$, proving that $\P(S_{x,\ell}\leq 2)\geq
-\frac{1}{4N^2}$.
+X_1=\ov{h}^{-1}(X))=\frac{1}{2{\MATHSF{N}}}$, proving that $\P(S_{x,\ell}\leq 2)\geq
+\frac{1}{4{\MATHSF{N}}^2}$.
\end{itemize}
-Therefore, $\P(S_{X,\ell}\geq 3)\leq 1-\frac{1}{4n^2}$. By induction, one
+Therefore, $\P(S_{X,\ell}\geq 3)\leq 1-\frac{1}{4{\mathsf{N}}^2}$. By induction, one
has, for every $i$, $\P(S_{X,\ell}\geq 2i)\leq
-\left(1-\frac{1}{4n^2}\right)^i$.
+\left(1-\frac{1}{4{\mathsf{N}}^2}\right)^i$.
Moreover,
since $S_{X,\ell}$ is positive, it is known~\cite[lemma 2.9]{proba}, that
$$E[S_{X,\ell}]=\sum_{i=1}^{+\infty}\P(S_{X,\ell}\geq i).$$
\P(S_{X,\ell}\geq 1)+\P(S_{X,\ell}\geq 2)+2 \sum_{i=1}^{+\infty}\P(S_{X,\ell}\geq 2i).$$
Consequently,
$$E[S_{X,\ell}]\leq 1+1+2
-\sum_{i=1}^{+\infty}\left(1-\frac{1}{4n^2}\right)^i=2+2(4n^2-1)=8n^2,$$
+\sum_{i=1}^{+\infty}\left(1-\frac{1}{4{\mathsf{N}}^2}\right)^i=2+2(4{\mathsf{N}}^2-1)=8{\mathsf{N}}^2,$$
which concludes the proof.
\end{proof}
-Let $\ts^\prime$ be the first time that there are exactly $n-1$ fair
+Let $\ts^\prime$ be the first time that there are exactly ${\mathsf{N}}-1$ fair
elements.
\begin{Lemma}\label{lm:stopprime}
-One has $E[\ts^\prime]\leq n \ln (n+1).$
+One has $E[\ts^\prime]\leq {\mathsf{N}} \ln ({\mathsf{N}}+1).$
\end{Lemma}
\begin{proof}
This is a classical Coupon Collector's like problem. Let $W_i$ be the
random variable counting the number of moves done in the Markov chain while
-we had exactly $i-1$ fair bits. One has $\ts^\prime=\sum_{i=1}^{n-1}W_i$.
+we had exactly $i-1$ fair bits. One has $\ts^\prime=\sum_{i=1}^{{\mathsf{N}}-1}W_i$.
But when we are at position $X$ with $i-1$ fair bits, the probability of
- obtaining a new fair bit is either $1-\frac{i-1}{n}$ if $h(X)$ is fair,
- or $1-\frac{i-2}{n}$ if $h(X)$ is not fair. It follows that
-$E[W_i]\leq \frac{n}{n-i+2}$. Therefore
-$$E[\ts^\prime]=\sum_{i=1}^{n-1}E[W_i]\leq n\sum_{i=1}^{n-1}
- \frac{1}{n-i+2}=n\sum_{i=3}^{n+1}\frac{1}{i}.$$
-
-But $\sum_{i=1}^{n+1}\frac{1}{i}\leq 1+\ln(n+1)$. It follows that
-$1+\frac{1}{2}+\sum_{i=3}^{n+1}\frac{1}{i}\leq 1+\ln(n+1).$
+ obtaining a new fair bit is either $1-\frac{i-1}{{\mathsf{N}}}$ if $h(X)$ is fair,
+ or $1-\frac{i-2}{{\mathsf{N}}}$ if $h(X)$ is not fair. It follows that
+$E[W_i]\leq \frac{{\mathsf{N}}}{{\mathsf{N}}-i+2}$. Therefore
+$$E[\ts^\prime]=\sum_{i=1}^{{\mathsf{N}}-1}E[W_i]\leq {\mathsf{N}}\sum_{i=1}^{{\mathsf{N}}-1}
+ \frac{1}{{\mathsf{N}}-i+2}={\mathsf{N}}\sum_{i=3}^{{\mathsf{N}}+1}\frac{1}{i}.$$
+
+But $\sum_{i=1}^{{\mathsf{N}}+1}\frac{1}{i}\leq 1+\ln({\mathsf{N}}+1)$. It follows that
+$1+\frac{1}{2}+\sum_{i=3}^{{\mathsf{N}}+1}\frac{1}{i}\leq 1+\ln({\mathsf{N}}+1).$
Consequently,
-$E[\ts^\prime]\leq n (-\frac{1}{2}+\ln(n+1))\leq n\ln(n+1)$.
+$E[\ts^\prime]\leq {\mathsf{N}} (-\frac{1}{2}+\ln({\mathsf{N}}+1))\leq {\mathsf{N}}\ln({\mathsf{N}}+1)$.
\end{proof}
One can now prove Theorem~\ref{prop:stop}.
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.
