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[canny.git] / stc.tex

diff --git a/stc.tex b/stc.tex
index 1ed8b448e1d58ce9bb55edb583e55ea55c199b2a..09ae4363ed1e98775c4499b5cdd1c826212bd659 100644 (file)
--- a/stc.tex
+++ b/stc.tex
@@ -1,3 +1,5 @@
+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 
 Let 
 $x=(x_1,\ldots,x_n)$ be the $n$-bits cover vector of the image $X$, 
 $m$ be the message to embed, and 


@@ -11,14 +13,14 @@ 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. 
 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
 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  
 $$
 vector and a 7 bits column vector. 
 Let then $H$ be the binary Hamming matrix  
 $$


@@ -53,27 +55,27 @@ 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 LSBs of pixels by modifying
 at most 1 bit.
 $m = Hy$.
 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  
 
 
 
 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 $n$.
 
 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
 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
 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}
 \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} 
 \item Backward determination of $y$ that minimizes $D$, starting with 
 the complete path having the minimal weight.
 \end{enumerate}