modif abstract

[16dcc.git] / generating.tex

diff --git a/generating.tex b/generating.tex
index 3d22441cf0684057f949b1c90f68ed18474f7a97..d512a9803b59c27f0a95a6a6bff958b887a8f30d 100644 (file)
--- a/generating.tex
+++ b/generating.tex
@@ -45,10 +45,10 @@ cycle is removed, is doubly stochastic.
 Let us consider now a ${\mathsf{N}}$-cube where an Hamiltonian 
 cycle is removed.
 Let $f$ be the corresponding function.
 Let us consider now a ${\mathsf{N}}$-cube where an Hamiltonian 
 cycle is removed.
 Let $f$ be the corresponding function.

-The question which remains to solve is
-can we always find $b$ such that $\Gamma_{\{b\}}(f)$ is strongly connected.
+The question which remains to solve is:
+\emph{can we always find $b$ such that $\Gamma_{\{b\}}(f)$ is strongly connected?}

 
 

-The answer is indeed positive. We furtheremore have the following strongest 
+The answer is indeed positive. We furthermore have the following strongest 

 result.
 \begin{thrm}
 There exist $b \in \Nats$ such that $\Gamma_{\{b\}}(f)$ is complete.
 result.
 \begin{thrm}
 There exist $b \in \Nats$ such that $\Gamma_{\{b\}}(f)$ is complete.


@@ -58,7 +58,7 @@ There is an arc $(x,y)$ in the
 graph $\Gamma_{\{b\}}(f)$ if and only if $M^b_{xy}$ is positive
 where $M$ is the Markov matrix of $\Gamma(f)$.
 It has been shown in~\cite[Lemma 3]{bcgr11:ip}  that $M$ is regular.
 graph $\Gamma_{\{b\}}(f)$ if and only if $M^b_{xy}$ is positive
 where $M$ is the Markov matrix of $\Gamma(f)$.
 It has been shown in~\cite[Lemma 3]{bcgr11:ip}  that $M$ is regular.

-There exists thus $b$ such there is an arc between any $x$ and $y$.
+Thus, there exists $b$ such that there is an arc between any $x$ and $y$.

 \end{proof}
 
 This section ends with the idea of removing a Hamiltonian cycle in the 
 \end{proof}
 
 This section ends with the idea of removing a Hamiltonian cycle in the 


@@ -66,12 +66,12 @@ $\mathsf{N}$-cube.
 In such a context, the Hamiltonian cycle is equivalent to a Gray code.
 Many approaches have been proposed a way to build such codes, for instance 
 the Reflected Binary Code. In this one, one of the bits is switched 
 In such a context, the Hamiltonian cycle is equivalent to a Gray code.
 Many approaches have been proposed a way to build such codes, for instance 
 the Reflected Binary Code. In this one, one of the bits is switched 

-exactly $2^{\mathsf{N}-}$ for a $\mathsf{N}$-length cycle. 
+exactly $2^{\mathsf{N}-}$ \ANNOT{formule incomplète : $2^{\mathsf{N}-1}$ ??} for a $\mathsf{N}$-length cycle. 

 
 %%%%%%%%%%%%%%%%%%%%%
 
 The function that is built 
 
 %%%%%%%%%%%%%%%%%%%%%
 
 The function that is built 

-from the 
+from the \ANNOT{Phrase non terminée}

 
 The next section presents how to build balanced Hamiltonian cycles in the 
 $\mathsf{N}$-cube with the objective to embed them into the 
 
 The next section presents how to build balanced Hamiltonian cycles in the 
 $\mathsf{N}$-cube with the objective to embed them into the