-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
a theoretical study on the length of the path
which is sufficient to follow to get a uniform distribution.
-
+Notice that for a general references on Markov chains
+see~\cite{LevinPeresWilmer2006}
+, and particularly Chapter~5 on stopping times.
-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 $k > t$.
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
+independent of $\tau$.
+
-\begin{Theo}
-If $\tau$ is a strong stationary time, then $d(t)\leq \max_{X\in\Bool^n}
+\begin{thrm}
+If $\tau$ is a strong stationary time, then $d(t)\leq \max_{X\in\Bool^{\mathsf{N}}}
\P_X(\tau > t)$.
-\end{Theo}
+\end{thrm}
%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:
-$$\left\{
+\begin{equation}
+\left\{
\begin{array}{ll}
-P_h(X,X)=\frac{1}{2}+\frac{1}{2n} & \\
+P_h(X,X)=\frac{1}{2}+\frac{1}{2{\mathsf{N}}} & \\
P_h(X,Y)=0 & \textrm{if $(X,Y)\notin E_h$}\\
-P_h(X,Y)=\frac{1}{2n} & \textrm{if $X\neq Y$ and $(X,Y) \in E_h$}
+P_h(X,Y)=\frac{1}{2{\mathsf{N}}} & \textrm{if $X\neq Y$ and $(X,Y) \in E_h$}
\end{array}
\right.
-$$
+\label{eq:Markov:rairo}
+\end{equation}
-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{lmm}\label{lm:h}
If $\ov{h}$ is bijective and square-free, then $h(\ov{h}^{-1}(X))\neq h(X)$.
-\end{Lemma}
+\end{lmm}
\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{Lemma}
+\begin{lmm}
The integer $\ts$ is a strong stationary time.
-\end{Lemma}
+\end{lmm}
\begin{proof}
Let $\tau_\ell$ be the first time that $\ell$ is fair. The random variable
$\ell$-th bit of $X_t$ is $0$ or $1$ with the same probability, proving the
lemma.\end{proof}
-\begin{Theo} \label{prop:stop}
+\begin{thrm} \label{prop:stop}
If $\ov{h}$ is bijective and square-free, then
-$E[\ts]\leq 8n^2+ n\ln (n+1)$.
-\end{Theo}
+$E[\ts]\leq 8{\mathsf{N}}^2+ 4{\mathsf{N}}\ln ({\mathsf{N}}+1)$.
+\end{thrm}
-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
$\ell$ is fair. More formally
$$S_{X,\ell}=\min \{t \geq 1\mid h(X_{t-1})\neq \ell\text{ and }Z_t=(\ell,.)\text{ and } X_0=X\}.$$
- We denote by
-$$\lambda_h=\max_{X,\ell} S_{X,\ell}.$$
+% We denote by
+% $$\lambda_h=\max_{X,\ell} S_{X,\ell}.$$
-\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.
-
-\end{Lemma}
+\begin{lmm}\label{prop:lambda}
+Let $\ov{h}$ is a square-free bijective function. Then
+for all $X$ and
+all $\ell$,
+the inequality
+$E[S_{X,\ell}]\leq 8{\mathsf{N}}^2$ is established.
+\end{lmm}
\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
+But in this case, intuitively, it is possible to move
from $X$ to $\ov{h}^{-1}(X)$ (with probability $\frac{1}{2N}$). And in
$\ov{h}^{-1}(X)$ the $l$-th bit can be switched.
More formally,
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
-elements.
+Let $\ts^\prime$ be the time used to get all the bits but one fair.
-\begin{Lemma}\label{lm:stopprime}
-One has $E[\ts^\prime]\leq n \ln (n+1).$
-\end{Lemma}
+\begin{lmm}\label{lm:stopprime}
+One has $E[\ts^\prime]\leq 4{\mathsf{N}} \ln ({\mathsf{N}}+1).$
+\end{lmm}
\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.
+
+Therefore,
+$\P (W_i=k)\leq \left(\frac{i-1}{{\mathsf{N}}}\right)^{k-1} \frac{{\mathsf{N}}-i+2}{{\mathsf{N}}}.$
+Consequently, we have $\P(W_i\geq k)\leq \left(\frac{i-1}{{\mathsf{N}}}\right)^{k-1} \frac{{\mathsf{N}}-i+2}{{\mathsf{N}}-i+1}.$
+It follows that $E[W_i]=\sum_{k=1}^{+\infty} \P (W_i\geq k)\leq {\mathsf{N}} \frac{{\mathsf{N}}-i+2}{({\mathsf{N}}-i+1)^2}\leq \frac{4{\mathsf{N}}}{{\mathsf{N}}-i+2}$.
+
+
+
+It follows that
+$E[W_i]\leq \frac{4{\mathsf{N}}}{{\mathsf{N}}-i+2}$. Therefore
+$$E[\ts^\prime]=\sum_{i=1}^{{\mathsf{N}}-1}E[W_i]\leq
+4{\mathsf{N}}\sum_{i=1}^{{\mathsf{N}}-1} \frac{1}{{\mathsf{N}}-i+2}=
+4{\mathsf{N}}\sum_{i=3}^{{\mathsf{N}}+1}\frac{1}{i}.$$
+
+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,
-$E[\ts^\prime]\leq n (-\frac{1}{2}+\ln(n+1))\leq n\ln(n+1)$.
+$E[\ts^\prime]\leq
+4{\mathsf{N}} (-\frac{1}{2}+\ln({\mathsf{N}}+1))\leq
+4{\mathsf{N}}\ln({\mathsf{N}}+1)$.
\end{proof}
One can now prove Theorem~\ref{prop:stop}.
\begin{proof}
-One has $\ts\leq \ts^\prime+\lambda_h$. Therefore,
+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}.
\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
+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.
