X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/16dcc.git/blobdiff_plain/3fc31015143d2bab7226a54390f3e1c5eba8f4d5..HEAD:/stopping.tex diff --git a/stopping.tex b/stopping.tex index 142da7f..bb95663 100644 --- a/stopping.tex +++ b/stopping.tex @@ -1,6 +1,6 @@ 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. +as described in the 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 $v \in \Bool^{\mathsf{N}}$. @@ -137,7 +137,7 @@ In other words, $E$ is the set of all the edges in the classical ${\mathsf{N}}$-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 -$X \in \Bool^{\mathsf{N}}$ which edge is removed in the Hamiltonian cycle, +$X \in \Bool^{\mathsf{N}}$ whose edge is removed in the Hamiltonian cycle, \textit{i.e.}, which bit in $\llbracket 1, {\mathsf{N}} \rrbracket$ cannot be switched. @@ -164,7 +164,7 @@ P_h(X,Y)=\frac{1}{2{\mathsf{N}}} & \textrm{if $X\neq Y$ and $(X,Y) \in E_h$} 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 $X\in \Bool^{\mathsf{N}}$, +The function $\ov{h}$ is said to be {\it square-free} if for every $X\in \Bool^{\mathsf{N}}$, $\ov{h}(\ov{h}(X))\neq X$. \begin{lmm}\label{lm:h} @@ -211,7 +211,7 @@ exists $0\leq j t)\leq \frac{E[\tau]}{t}$. With $t_n=32N^2+16N\ln (N+1)$, one obtains: $\P_X(\tau > t_n)\leq \frac{1}{4}$. Therefore, using the definition of $t_{\rm mix}$ and Theorem~\ref{thm-sst}, it follows that -$t_{\rm mix}\leq 32N^2+16N\ln (N+1)=O(N^2)$. +$t_{\rm mix}(\frac{1}{4})\leq 32N^2+16N\ln (N+1)=O(N^2)$ and that + Notice that the calculus of the stationary time upper bound is obtained @@ -376,7 +377,7 @@ The calculus doesn't consider (balanced) Hamiltonian cycles, which are more regular and more binding than this constraint. Moreover, the bound is obtained using the coarse Markov Inequality. For the -classical (lazzy) random walk the $\mathsf{N}$-cube, without removing any +classical (lazy) random walk the $\mathsf{N}$-cube, without removing any Hamiltonian cycle, the mixing time is in $\Theta(N\ln N)$. We conjecture that in our context, the mixing time is also in $\Theta(N\ln N)$. @@ -419,16 +420,16 @@ $\textit{fair}\leftarrow\emptyset$\; \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 $E[\ts]$ +10 functions have been generated according to the 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. Figure~\ref{fig:stopping:moy} -summarizes these results. In this one, a circle represents the +summarizes these results. 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 $x \mapsto 2x\ln(2x+8)$. +is realistic according to the graph of $x \mapsto 2x\ln(2x+8)$.