+To make this article self-contained, this section recalls
+the basis of the Syndrome Treillis Codes (STC).
Let
$x=(x_1,\ldots,x_n)$ be the $n$-bits cover vector of the image $X$,
$m$ be the message to embed, and
The objective is thus to find $y$ that minimizes $D_X(x,y)$.
Hamming embedding proposes a solution to this problem.
-Some steganographic
+This is why some steganographic
schemes~\cite{DBLP:conf/ih/Westfeld01,DBLP:conf/ih/KimDR06,DBLP:conf/mmsec/FridrichPK07} are based on this binary embedding.
Furthermore this code provides a vector $y$ s.t. $Hy$ is equal to
$m$ for a given binary matrix $H$.
Let us explain this embedding on a small illustrative example where
-$\rho_X(i,x,y)$ is identically equal to 1,
-$m$ and $x$ are respectively a 3 bits column
+$\rho_X(i,x,y)$ is equal to 1,
+whereas $m$ and $x$ are respectively a 3 bits column
vector and a 7 bits column vector.
Let then $H$ be the binary Hamming matrix
$$
$m = Hy$.
It is then possible to embed 3 bits in only 7 LSBs of pixels by modifying
at most 1 bit.
-In the general case, when communicating $n$ message bits in
-$2^n-1$ pixels needs $1-1/2^n$ average changes. \CG{Phrase pas claire}.
+In the general case, communicating $n$ message bits in
+$2^n-1$ pixels needs $1-1/2^n$ average changes.
Unfortunately, for any given $H$, finding $y$ that solves $Hy=m$ and
-that minimizes $D_X(x,y)$ has an exponential complexity with respect to $n$.
-The Syndrome-Trellis Codes (STC)
+that minimizes $D_X(x,y)$, has an exponential complexity with respect to $n$.
+The Syndrome-Trellis Codes
presented by Filler \emph{et al.} in~\cite{DBLP:conf/mediaforensics/FillerJF10}
is a practical solution to this complexity. Thanks to this contribution,
the solving algorithm has a linear complexity with respect to $n$.
-First of all, Filler et al. compute the matrix $H$
+First of all, Filler \emph{et al.} compute the matrix $H$
by placing a small sub-matrix $\hat{H}$ of size $h × w$ next
-to each other and shifted down by one row.
+to each other and by shifting down by one row.
Thanks to this special form of $H$, one can represent
-every solution of $m=Hy$ as a path through a trellis.
+any solution of $m=Hy$ as a path through a trellis.
-Next, the process of finding $y$ consists of two stages: a forward and a backward part.
+Next, the process of finding $y$ consists in two stages: a forward and a backward part.
\begin{enumerate}
-\item Forward construction of the trellis that depends on $\hat{H}$, on $x$, on $m$, and on $\rho$;
+\item Forward construction of the trellis that depends on $\hat{H}$, on $x$, on $m$, and on $\rho$.
\item Backward determination of $y$ that minimizes $D$, starting with
the complete path having the minimal weight.
\end{enumerate}
