$\mathsf{N}$-cube.
Obviously, the number of iterations $b$ has to be sufficiently large
to provide a uniform output distribution.
-To reduce the number of iterations, the provided Gray code
+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:
balanced Gray codes that are generated by state of the art works~\cite{ZanSup04,DBLP:journals/combinatorics/BhatS96} are not locally balanced. Conversely,
\begin{xpl}
-Let $L^*=000,100,101,001,011,111,110,010$ be the Gray code that corresponds to
+Let $L^*=000,100,101,001,011,111,$ $110,010$ be the Gray code that corresponds to
the Hamiltonian cycle that has been removed in $f^*$.
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
-$L^4=0000, 0010, 0110, 1110, 1111, 0111, 0011, 0001, 0101,$
+$L^4=0000, 0010, 0110, 1110, 1111, 0111, 0011,$ $0001, 0101,$
$0100, 1100, 1101, 1001, 1011, 1010, 1000$
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.
whose transition count is exactly $a_{\mathsf{N}}$ (resp $a_{\mathsf{N}} +2$).
These two variables are defined by the system
-$$
+\[
\left\{
\begin{array}{lcl}
c_{\mathsf{N}} + d_{\mathsf{N}} & = & \mathsf{N} \\
c_{\mathsf{N}}a_{\mathsf{N}} + d_{\mathsf{N}}(a_{\mathsf{N}}+2) & = & 2^{\mathsf{N}}
\end{array}
\right.
-\qquad
\Leftrightarrow
-\qquad
\left\{
\begin{array}{lcl}
d_{\mathsf{N}} & = & \dfrac{2^{\mathsf{N}} -\mathsf{N}.a_{\mathsf{N}}}{2} \\
c_{\mathsf{N}} &= &\mathsf{N} - d_{\mathsf{N}}
\end{array}
\right.
-$$
+\]
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
We thus have
-$$
+\[
\begin{array}{rcl}
\textit{TC}_{\mathsf{N}}(i) - 2.\textit{TC}_{\mathsf{N}-2}(i)
&\ge&
\frac{2^{\mathsf{N}}-2N}{\mathsf{N}}
-2 \left( \frac{2^{\mathsf{N-2}}}{\mathsf{N-2}}+2\right)\\
&\ge&
-\dfrac{(\mathsf{N} -2).2^{\mathsf{N}}-2N.2^{\mathsf{N-2}}-6N(N-2)}{\mathsf{N.(N-2)}}\\
+\frac{(\mathsf{N} -2).2^{\mathsf{N}}-2N.2^{\mathsf{N-2}}-6N(N-2)}{\mathsf{N.(N-2)}}\\
\end{array}
-$$
+\]
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
