at most by 2.
This uniformity is a global property on the cycle, \textit{i.e.},
a property that is established while traversing the whole cycle.
-On the opposite side, when the objective is to follow a subpart
+On the other hand, when the objective is to follow a subpart
of the Gray code and to switch each element approximately the
same amount of times,
local properties are wished.
\textit{i.e.}, a vertex in the
$\mathsf{N}$-cube.
Obviously, the number of iterations $b$ has to be sufficiently large
-to provide an uniform output distribution.
+to provide a uniform output distribution.
To reduce the number of iterations, it can be claimed
that the provided Gray code
should ideally possess both balanced and locally balanced properties.
-However, none of the two algorithms is compatible with the second one:
+However, both algorithms are incompatible with the second one:
balanced Gray codes that are generated by state of the art works~\cite{ZanSup04,DBLP:journals/combinatorics/BhatS96} are not locally balanced. Conversely,
locally balanced Gray codes yielded by Igor Bykov approach~\cite{Bykov2016}
are not globally balanced.
and~\cite{ZanSup04} have addressed the problem of providing an approach
to produce balanced gray code.
The authors of~\cite{Robinson:1981:CS} introduced an inductive approach
-aiming at producing balanced Gray codes, provided the user gives
+aiming at producing balanced Gray codes, assuming the user gives
a special subsequence of the transition sequence at each induction step.
-This work have been strengthened in~\cite{DBLP:journals/combinatorics/BhatS96}
-where the authors have explicitly shown how to construct such a subsequence.
+This work has been strengthened in~\cite{DBLP:journals/combinatorics/BhatS96}
+where the authors have explicitly shown how to build such a subsequence.
Finally the authors of~\cite{ZanSup04} have presented
the \emph{Robinson-Cohn extension}
-algorithm. Their rigorous presentation of this one
+algorithm. Their rigorous presentation of this algorithm
has mainly allowed them to prove two properties.
The former states that if
$\mathsf{N}$ is a 2-power, a balanced Gray code is always totally balanced.
are 2-powers whose exponents are either equal
or differ from each other by 1.
However, the authors do not prove that the approach allows to build
-(totally balanced) Gray code.
+(totally balanced) Gray codes.
What follows shows that this fact is established and first recalls the approach.
It has been proven in~\cite{ZanSup04} that
$S_{\mathsf{N}}$ is the transition sequence of a cyclic $\mathsf{N}$-bits Gray code
if $S_{\mathsf{N}-2}$ is.
-However, the step~(\ref{item:nondet}) is not a constructive
-step that precises how to select the subsequences which ensures that
+However, step~(\ref{item:nondet}) is not a constructive
+step that precises how to select the subsequences which ensure that
yielded Gray code is balanced.
-Next section shows how to choose the sequence $l$ to have the balance property.
+Following sections show how to choose the sequence $l$ to have the balance property.
\subsection{Balanced Codes}
Let us first recall how to formalize the balance property of a Gray code.
Its transition sequence is $S=3,1,3,2,3,1,3,2$ and its transition count function is
$\textit{TC}_3(1)= \textit{TC}_3(2)=2$ and $\textit{TC}_3(3)=4$. Such a Gray code is balanced.
-Let now
-$L^4=0000, 0010, 0110, 1110, 1111, 0111, 0011,$ $0001, 0101,0100,$ $1100, 1101, 1001, 1011, 1010, 1000$
+Let $L^4$ $=0000,0010,0110,1110,1111,0111,0011,0001,0101,0100,1100,1101,1001,1011,1010,1000$
be a cyclic Gray code. Since $S=2,3,4,1,4,$ $3,2,3,1,4,1,3,2,1,2,4$, $\textit{TC}_4$ is equal to 4 everywhere, this code
is thus totally balanced.
On the contrary, for the standard $4$-bits Gray code
-$L^{\textit{st}}=0000,0001,0011,$
-$0010,0110,0111,0101,0100,1100,1101,1111,1110,1010,1011,1001,1000$,
+$L^{\textit{st}}=0000,0001,0011,
+0010,0110,0111,0101,0100,$ \newline $1100,1101,1111,1110,1010,1011,1001,1000$,
we have $\textit{TC}_4(1)=8$ $\textit{TC}_4(2)=4$ $\textit{TC}_4(3)=\textit{TC}_4(4)=2$ and
the code is neither balanced nor totally balanced.
\end{xpl}
For the inductive case, let us first define some variables.
Let $c_{\mathsf{N}}$ (resp. $d_{\mathsf{N}}$) be the number of elements
whose transition count is exactly $a_{\mathsf{N}}$ (resp $a_{\mathsf{N}} +2$).
-These two variables are defined by the system
+Both of these variables are defined by the system
\[
\left\{
\textit{TC}_{4} & = & [4,4,4,4] \\
\textit{TC}_{6} & = & [10,10,10,10,12,12] \\
\end{eqnarray*}
-It is not hard to verify that all these instanciations verify the aformentioned contraints.
+It is not difficult to check that all these instanciations verify the aforementioned constraints.
When $N \ge 8$, $\textit{TC}_{\mathsf{N}}(i)$ is defined as follows:
\begin{equation}
