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.
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
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}
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
