X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/canny.git/blobdiff_plain/8290657e61738a93b7b6eb60966c49efbc4f0af8..f9feb4cb5a39609ffdf39b086e8d796592a71da5:/stc.tex?ds=sidebyside

diff --git a/stc.tex b/stc.tex
index 6570c73..e98589f 100644
--- a/stc.tex
+++ b/stc.tex
@@ -1,24 +1,24 @@
 Let 
 $x=(x_1,\ldots,x_n)$ be the $n$-bits cover vector of the image $X$, 
-$m$ be the message to embed and 
+$m$ be the message to embed, and 
 $y=(y_1,\ldots,y_n)$ be the $n$-bits stego vector.
-The usual additive embbeding impact of replacing $x$ by $y$ in $X$
+The usual additive embedding impact of replacing $x$ by $y$ in $X$
 is given by a distortion function 
-$D_X(x,y)= \Sigma_{i=1}^n \rho_X(i,x,y)$ where the function $\rho_X$
-expressed the cost of replacing $x_i$ by $y_i$ in $X$.
+$D_X(x,y)= \Sigma_{i=1}^n \rho_X(i,x,y)$, where the function $\rho_X$
+expresses the cost of replacing $x_i$ by $y_i$ in $X$.
 Let us consider that $x$ is fixed: 
 this is for instance the LSBs of the image edge bits. 
 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
+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  
 $$
@@ -34,7 +34,7 @@ The objective is to modify $x$ to get $y$ s.t. $m = Hy$.
 In this algebra, the sum and the product respectively correspond to
 the exclusive \emph{or} and to the \emph{and} Boolean operators.
 If $Hx$ is already equal to $m$, nothing has to be changed and $x$ can be sent.
-Otherwise we consider the difference $\delta = d(m,Hx)$ which is expressed 
+Otherwise we consider the difference $\delta = d(m,Hx)$, which is expressed 
 as a vector : 
 $$
 \delta = \left( \begin{array}{l}
@@ -45,28 +45,25 @@ $$
 \right)  
 \textrm{ where $\delta_i$ is 0 if $m_i = Hx_i$ and 1 otherwise.} 
 $$
-Let us thus consider the $j$th column of $H$ which is equal to $\delta$.   
-We denote by $\overline{x}^j$ the vector  we obtain by
-switching the $j$th component of $x$, 
+Let us thus consider the $j-$th column of $H$, which is equal to $\delta$.   
+We denote by $\overline{x}^j$ the vector  obtained by
+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 
 $m = Hy$.
-It is then possible to embed 3 bits in only 7 LSB of pixels by modifying
-1 bit at most.
-In the general case, when comunicating $n$ message bits in 
-$2^n-1$ pixels needs $1-1/2^n$ average changes. 
+It is then possible to embed 3 bits in only 7 LSBs of pixels by modifying
+at most 1 bit.
+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 
-that minimizes $D_X(x,y)$ has exponential complexity with respect to $n$. 
+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) 
-presented by Filler et al. in~\cite{DBLP:conf/mediaforensics/FillerJF10} 
+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 resspect to $n$.
+the solving algorithm has a linear complexity with respect to $n$.
 
 First of all, Filler et al. compute the matrix $H$
 by placing a small sub-matrix $\hat{H}$ of size $h × w$ next
@@ -74,12 +71,12 @@ to each other and shifted 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.
 
-Next, the  process of finding $y$ consists of a forward and a backward part:
+Next, the  process of finding $y$ consists of 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 Backward determinization of $y$ which minimizes $D$ starting with 
-the complete path with minimal weight
+\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} 
-Let us now give some details about these two parts.
+