X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/canny.git/blobdiff_plain/03f2db2f776786ee6e9435d3c0f603b248cdd2ae..d5631915e956b49c89f78d66d7b4d25682a7e405:/stc.tex?ds=sidebyside

diff --git a/stc.tex b/stc.tex
index db98585..5e5b66e 100644
--- a/stc.tex
+++ b/stc.tex
@@ -1,5 +1,6 @@
 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 
 $x=(x_1,\ldots,x_n)$ be the $n$-bits cover vector issued from an image $X$, 
@@ -19,13 +20,13 @@ $m$ for a given binary matrix $H$.
 
 Let us explain this embedding on a small illustrative example where
 $m$ and $x$ are respectively  a 3 bits column
-vector and a 7 bits column vector and where 
+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).  
+(\textit{i.e.}, $\rho_X(i,x,y) = 0$ if $x = y$ and $1$ otherwise).  
 
-Let  $H$ be the binary Hamming matrix  
+Let  $\dot{H}$ be the binary Hamming matrix  
 $$
-H = \left(
+\dot{H} = \left(
 \begin{array}{lllllll}
  0 & 0 & 0 & 1 & 1 & 1 & 1 \\
  0 & 1 & 1 & 0 & 0 & 1 & 1 \\
@@ -33,11 +34,11 @@ H = \left(
 \end{array}
 \right).
 $$
-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.
-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}
@@ -53,14 +54,14 @@ 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 
-$m = Hy$.
+$m = \dot{H}y$.
 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.
 
-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.
+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 $n$.