Let us explain this embedding on a small illustrative example where
$m$ and $x$ are respectively a 3 bits column
-vector and a 7 bits column vector and where
+vector and a 7 bits column vector, and where
$\rho_X(i,x,y)$ is equal to 1 for any $i$, $x$, $y$
-(\textit{i.e.}, $\rho_X(i,x,y) = 0$ if $x = y$ and $0$ otherwise).
+(\textit{i.e.}, $\rho_X(i,x,y) = 0$ if $x = y$ and $1$ otherwise).
-Let $H$ be the binary Hamming matrix
+Let $\dot{H}$ be the binary Hamming matrix
$$
-H = \left(
+\dot{H} = \left(
\begin{array}{lllllll}
0 & 0 & 0 & 1 & 1 & 1 & 1 \\
0 & 1 & 1 & 0 & 0 & 1 & 1 \\
\end{array}
\right).
$$
-The objective is to modify $x$ to get $y$ s.t. $m = Hy$.
+The objective is to modify $x$ to get $y$ s.t. $m = \dot{H}y$.
In this algebra, the sum and the product respectively correspond to
the exclusive \emph{or} and to the \emph{and} Boolean operators.
-If $Hx$ is already equal to $m$, nothing has to be changed and $x$ can be sent.
-Otherwise we consider the difference $\delta = d(m,Hx)$, which is expressed
+If $\dot{H}x$ is already equal to $m$, nothing has to be changed and $x$ can be sent.
+Otherwise we consider the difference $\delta = d(m,\dot{H}x)$, which is expressed
as a vector :
$$
\delta = \left( \begin{array}{l}
switching the $j-$th component of $x$,
that is, $\overline{x}^j = (x_1 , \ldots, \overline{x_j},\ldots, x_n )$.
It is not hard to see that if $y$ is $\overline{x}^j$, then
-$m = Hy$.
+$m = \dot{H}y$.
It is then possible to embed 3 bits in 7 LSBs of pixels by modifying
at most 1 bit.
In the general case, communicating a message of $p$ bits in a cover of
$n=2^p-1$ pixels needs $1-1/2^p$ average changes.
-This Hamming embeding is really efficient to very small payload and is
-not well suited when the size of the message is larger, as in real situation.
+This Hamming embedding is really efficient to very small payload and is
+not well suited when the size of the message is larger, as in real situations.
The matrix $H$ should be changed to deal with higher payload.
Moreover, for any given $H$, finding $y$ that solves $Hy=m$ and
that minimizes $D_X(x,y)$, has an exponential complexity with respect to $n$.
