From: cguyeux <cguyeux@iut-bm.univ-fcomte.fr>
Date: Mon, 4 Feb 2013 09:29:00 +0000 (+0100)
Subject: Relecture
X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/canny.git/commitdiff_plain/f9feb4cb5a39609ffdf39b086e8d796592a71da5?ds=sidebyside;hp=-c

Relecture
---

f9feb4cb5a39609ffdf39b086e8d796592a71da5
diff --git a/stc.tex b/stc.tex
index 1ed8b44..e98589f 100644
--- a/stc.tex
+++ b/stc.tex
@@ -11,14 +11,14 @@ 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. 
-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  
 $$
@@ -53,13 +53,13 @@ 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.
-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 $n$. 
+that minimizes $D_X(x,y)$, has an exponential complexity with respect to $n$. 
 The Syndrome-Trellis Codes  (STC) 
 presented by Filler \emph{et al.} in~\cite{DBLP:conf/mediaforensics/FillerJF10} 
 is a practical solution to this complexity. Thanks to this contribution,
@@ -73,7 +73,7 @@ every 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.
 \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}