Merge branch 'master' of ssh://info.iut-bm.univ-fcomte.fr/canny

[canny.git] / stc.tex

diff --git a/stc.tex b/stc.tex
index 26ac732cb1805fe186b11981e1fcfb57283d9d5a..5e5b66e6258df38d81475c859e714ec93b6223c1 100644 (file)
--- a/stc.tex
+++ b/stc.tex
@@ -1,15 +1,15 @@
 To make this article self-contained, this section recalls
 To make this article self-contained, this section recalls

-the basis of the Syndrome Treillis Codes  (STC).
+the basis of the Syndrome Treillis Codes  (STC). 
+\JFC{A reader who is familar with syndrome coding can skip it.}
+

 Let 
 Let 

-$x=(x_1,\ldots,x_n)$ be the $n$-bits cover vector of the image $X$, 
+$x=(x_1,\ldots,x_n)$ 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.
 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 $\rho_X$
 expresses the cost of replacing $x_i$ by $y_i$ in $X$.
 $m$ be the message to embed, and 
 $y=(y_1,\ldots,y_n)$ be the $n$-bits stego vector.
 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 $\rho_X$
 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. 
 The objective is thus to find $y$ that minimizes $D_X(x,y)$.
 
 Hamming embedding proposes a solution to this problem. 


@@ -19,12 +19,14 @@ 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
 $m$ for a given binary matrix $H$. 
 
 Let us explain this embedding on a small illustrative example where

-$\rho_X(i,x,y)$ 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  
+$m$ and $x$ are respectively  a 3 bits column
+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 $1$ otherwise).  
+
+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 \\
 \begin{array}{lllllll}
  0 & 0 & 0 & 1 & 1 & 1 & 1 \\
  0 & 1 & 1 & 0 & 0 & 1 & 1 \\


@@ -32,11 +34,11 @@ H = \left(
 \end{array}
 \right).
 $$
 \end{array}
 \right).
 $$

-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.
 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 : 
 $$
 \delta = \left( \begin{array}{l}
 as a vector : 
 $$
 \delta = \left( \begin{array}{l}


@@ -52,18 +54,19 @@ 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 
 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 LSBs of pixels by modifying
+$m = \dot{H}y$.
+It is then possible to embed 3 bits in 7 LSBs of pixels by modifying

 at most 1 bit.
 at most 1 bit.

-In the general case, communicating $n$ message bits in 
-$2^n-1$ pixels needs $1-1/2^n$ average changes.
-
+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  
+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 $n$. 
 The Syndrome-Trellis Codes  
 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} 
+presented by Filler \emph{et al.} in~\cite{FillerJF11}

 is a practical solution to this complexity. Thanks to this contribution,
 the solving algorithm has a linear complexity with respect to $n$.
 
 is a practical solution to this complexity. Thanks to this contribution,
 the solving algorithm has a linear complexity with respect to $n$.
 


@@ -75,10 +78,10 @@ any solution of  $m=Hy$ as a path through a trellis.
 
 Next, the  process of finding $y$ consists in two stages: a forward and a backward part.
 \begin{enumerate}
 
 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 $\rho$. This step is linear in $n$.

 \item Backward determination of $y$ that minimizes $D$, starting with 
 \item Backward determination of $y$ that minimizes $D$, starting with 

-the complete path having the minimal weight.
+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} 
 
 

-