integration de remarques de reviewers

[canny.git] / stc.tex

diff --git a/stc.tex b/stc.tex
index e8c695cf1a3963369fa61f56df0c1abe0b4d369f..db985853fa8dd98b21b8c676d242e8f4d65e4f63 100644 (file)
--- a/stc.tex
+++ b/stc.tex
@@ -1,26 +1,29 @@
+To make this article self-contained, this section recalls
+the basis of the Syndrome Treillis Codes  (STC).
+

 Let 
 Let 

-$x=(x_1,\ldots,x_n)$ be the $n$-bits cover vector of the image $X$, 
-$m$ be the message to embed and 
+$x=(x_1,\ldots,x_n)$ be the $n$-bits cover vector issued from an image $X$, 
+$m$ be the message to embed, and 

 $y=(y_1,\ldots,y_n)$ be the $n$-bits stego vector.
 $y=(y_1,\ldots,y_n)$ 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 
 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 $X$.
-Let us consider that $x$ is fixed: 
-this is for instance the LSBs of the image edge bits. 
+$D_X(x,y)= \Sigma_{i=1}^n \rho_X(i,x,y)$, where the function $\rho_X$
+expresses the cost of replacing $x_i$ by $y_i$ in $X$.

 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
 $m$ 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 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).  
+
+Let  $H$ be the binary Hamming matrix  

 $$
 H = \left(
 \begin{array}{lllllll}
 $$
 H = \left(
 \begin{array}{lllllll}


@@ -34,7 +37,7 @@ 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.
 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 : 
 $$
 \delta = \left( \begin{array}{l}
 as a vector : 
 $$
 \delta = \left( \begin{array}{l}


@@ -45,38 +48,39 @@ $$
 \right)  
 \textrm{ where $\delta_i$ is 0 if $m_i = Hx_i$ and 1 otherwise.} 
 $$
 \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 
 $m = Hy$.
 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 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.

 
 

-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 $n$. 
-The Syndrome-Trellis Codes  (STC) 
-presented by Filler et al. in~\cite{DBLP:conf/mediaforensics/FillerJF10} 
+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.
+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 $n$. 
+The Syndrome-Trellis Codes  
+presented by Filler \emph{et al.} in~\cite{FillerJF11}

 is a practical solution to this complexity. Thanks to this contribution,
 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
-to each other and shifted down by one row. 
+First of all, Filler \emph{et al.} compute the matrix $H$
+by placing a small sub-matrix $\hat{H}$ next
+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 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 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 $\rho$. This step is linear in $n$.
+\item Backward determination of $y$ that minimizes $D$, starting with 
+the complete path having the minimal weight. This corresponds to traversing 
+a graph and has a complexity which is linear in $n$. 

 \end{enumerate} 
 \end{enumerate} 

-Let us now give some details about these two parts.