X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/16dcc.git/blobdiff_plain/3fc31015143d2bab7226a54390f3e1c5eba8f4d5..8dfce25f59a57787b53e1ba55971a5c4f8d5993f:/generating.tex?ds=sidebyside

diff --git a/generating.tex b/generating.tex
index 377c811..b4839db 100644
--- a/generating.tex
+++ b/generating.tex
@@ -21,7 +21,7 @@ $000,100,101,001,011,111,$ $110,010,000$
 has been removed.
 \end{xpl}
 
-We first have proven the following result, which 
+We  have first proven the following result, which 
 states that the ${\mathsf{N}}$-cube without one
 Hamiltonian cycle 
 has the awaited property with regard to the connectivity.
@@ -32,7 +32,7 @@ the ${\mathsf{N}}$-cube where an Hamiltonian
 cycle is removed, is strongly connected.
 \end{thrm}
 
-Moreover, if all the transitions have the same probability ($\frac{1}{n}$),
+Moreover, when all the transitions have the same probability ($\frac{1}{n}$),
 we have proven the following results:
 \begin{thrm}
 The Markov Matrix $M$ resulting from the ${\mathsf{N}}$-cube in
@@ -43,11 +43,11 @@ 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.
-The question which remains to solve is:
+The question which remains to be solved is:
 \emph{can we always find $b$ such that $\Gamma_{\{b\}}(f)$ is strongly connected?}
 
-The answer is indeed positive. We furthermore have the following strongest 
-result.
+The answer is indeed positive. Furthermore, we have the following results which are stronger
+than previous ones. 
 \begin{thrm}
 There exists $b \in \Nats$ such that $\Gamma_{\{b\}}(f)$ is complete.
 \end{thrm}
@@ -65,10 +65,10 @@ In such a context, the Hamiltonian cycle is equivalent to a Gray code.
 Many approaches have been proposed as a way to build such codes, for instance 
 the Reflected Binary Code. In this one and 
 for a $\mathsf{N}$-length cycle, one of the bits is exactly switched 
-$2^{\mathsf{N}-1}$ times whereas the others bits are modified at most 
+$2^{\mathsf{N}-1}$ times whereas the other bits are modified at most 
 $\left\lfloor \dfrac{2^{\mathsf{N-1}}}{\mathsf{N}-1} \right\rfloor$ times.
 It is clear that the function that is built from such a code would
-not provide an uniform output.  
+not provide a uniform output.  
 
 The next section presents how to build balanced Hamiltonian cycles in the 
 $\mathsf{N}$-cube with the objective to embed them into the