balanced Gray codes. In the transition sequence of these codes,
the number of transitions of each element must differ
at most by 2.
-This uniformity is a global property on the cycle, \textit{i.e.}
+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.
and an algorithm that establishes locally balanced Gray codes is given.
The current context is to provide a function
-$f:\Bool^{\mathsf{N}} \rightarrow \Bool^{\mathsf{N}}$ by removing a Hamiltonian
+$f:\Bool^{\mathsf{N}} \rightarrow \Bool^{\mathsf{N}}$ by removing an Hamiltonian
cycle in the $\mathsf{N}$-cube. Such a function is going to be iterated
-$b$ times to produce a pseudo random number,
-\textit{i.e.} a vertex in the
+$b$ times to produce a pseudorandom number,
+\textit{i.e.}, a vertex in the
$\mathsf{N}$-cube.
Obviously, the number of iterations $b$ has to be sufficiently large
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. There rigorous presentation of this one
-have mainly allowed them to prove two properties.
+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.
The latter states that for every $\mathsf{N}$ there
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.
Let be given a $\mathsf{N}-2$-bit Gray code whose transition sequence is
$S_{\mathsf{N}-2}$. What follows is the
\emph{Robinson-Cohn extension} method~\cite{ZanSup04}
-which produces a $n$-bits Gray code.
+which produces a $\mathsf{N}$-bits Gray code.
\begin{enumerate}
\item \label{item:nondet}Let $l$ be an even positive integer. Find
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
-$L^4=0000, 0010, 0110, 1110, 1111, 0111, 0011,$ $0001, 0101,$
-$0100, 1100, 1101, 1001, 1011, 1010, 1000$
-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
+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}
The proof is done by induction on $\mathsf{N}$. Let us immediately verify
-that it is established for both odd and even smallest values, \textit{i.e.}
+that it is established for both odd and even smallest values, \textit{i.e.},
$3$ and $4$.
-For the initial case where $\mathsf{N}=3$, \textit{i.e.} $\mathsf{N-2}=1$ we successively have: $S_1=1,1$, $l=2$, $u_0 = \emptyset$, and $v=\emptyset$.
+For the initial case where $\mathsf{N}=3$, \textit{i.e.}, $\mathsf{N-2}=1$ we successively have: $S_1=1,1$, $l=2$, $u_0 = \emptyset$, and $v=\emptyset$.
Thus again the algorithm successively produces
$U= 1,2,1$, $V = 3$, $W= 2, 1, 1,3$, and $W' = 1,2,1,3$.
Finally, $S_3$ is $1,2,1,3,1,2,1,3$ which obviously verifies the theorem.
- For the initial case where $\mathsf{N}=4$, \textit{i.e.} $\mathsf{N-2}=2$
+ For the initial case where $\mathsf{N}=4$, \textit{i.e.}, $\mathsf{N-2}=2$
we successively have: $S_1=1,2,1,2$, $l=4$,
$u_0,u_1,u_2 = \emptyset,\emptyset,\emptyset$, and $v=\emptyset$.
Thus again the algorithm successively produces
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
\[
\left\{
\]
Since $a_{\mathsf{N}}$ is even, $d_{\mathsf{N}}$ is an integer.
-Let us first proove that both $c_{\mathsf{N}}$ and $d_{\mathsf{N}}$ are positive
+Let us first prove that both $c_{\mathsf{N}}$ and $d_{\mathsf{N}}$ are positive
integers.
Let $q_{\mathsf{N}}$ and $r_{\mathsf{N}}$, respectively, be
-the quotient and the remainder in the Euclidean disvision
-of $2^{\mathsf{N}}$ by $2\mathsf{N}$, \textit{i.e.}
+the quotient and the remainder in the Euclidean division
+of $2^{\mathsf{N}}$ by $2\mathsf{N}$, \textit{i.e.},
$2^{\mathsf{N}} = q_{\mathsf{N}}.2\mathsf{N} + r_{\mathsf{N}}$, with $0 \le r_{\mathsf{N}} <2\mathsf{N}$.
First of all, the integer $r$ is even since $r_{\mathsf{N}} = 2^{\mathsf{N}} - q_{\mathsf{N}}.2\mathsf{N}= 2(2^{\mathsf{N}-1} - q_{\mathsf{N}}.\mathsf{N})$.
Next, $a_{\mathsf{N}}$ is $\frac{2^{\mathsf{N}}-r_{\mathsf{N}}}{\mathsf{N}}$. Consequently
For any $i$, $1 \le i \le \mathsf{N}$, let $zi_{\mathsf{N}}$ (resp. $ti_{\mathsf{N}}$ and $bi_{\mathsf{N}}$)
-be the occurence number of element $i$ in the sequence $u_0, \dots, u_{l-2}$
+be the occurrence number of element $i$ in the sequence $u_0, \dots, u_{l-2}$
(resp. in the sequences $s_{i_1}, \dots , s_{i_l}$ and $v$)
in step (\ref{item:nondet}) of the algorithm.
Let us now prove that the resulting system has always positive integer
solutions $z_i$, $t_i$, $0 \le z_i, t_i \le \textit{TC}_{\mathsf{N}-2}(i)$
and s.t. their sum is equal to $\textit{TC}_{\mathsf{N}-2}(i)$.
-This latter consraint is obviously established if the system has a solution.
+This latter constraint is obviously established if the system has a solution.
We thus have the following system.
\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}
A simple variation study of the function $t:\R \rightarrow \R$ such that
$x \mapsto t(x) = (x -2).2^{x}-2x.2^{x-2}-6x(x-2)$ shows that
-its derivative is strictly postive if $x \ge 6$ and $t(8)=224$.
+its derivative is strictly positive if $x \ge 6$ and $t(8)=224$.
The integer $\textit{TC}_{\mathsf{N}}(i) - 2.\textit{TC}_{\mathsf{N}-2}(i)$ is thus positive
for any $\mathsf{N} \ge 8$ and the proof is established.
For each element $i$, we are then left to choose $zi_{\mathsf{N}}$ positions
among $\textit{TC}_{\mathsf{N}}(i)$, which leads to
${\textit{TC}_{\mathsf{N}}(i) \choose zi_{\mathsf{N}} }$ possibilities.
-Notice that all such choices lead to a hamiltonian path.
+Notice that all such choices lead to an Hamiltonian path.
