2 La condition de l'algorithme de marquage est nécressaire et suffisante
3 pour permettre l'extraction du message du média marqué.
7 For sufficiency, let $d_i$ be the last iteration (date) the element $i \in \Im(S_p)$
8 of $x$ has been modified:% is defined by
10 d_i = \max\{j | S^j_p = i \}.
12 Let $D=\{d_i|i \in \Im(S_p) \}$.
13 The set $\Im(S_c)_{|D}$ is thus
14 the restriction of the image of $S_c$ to $D$.
17 The host that results from this iteration scheme is thus
18 $(x^l_0,\ldots,x^l_{\mathsf{N}-1})$ where
19 $x^l_i$ is either $x^{d_i}_i$ if $i$ belongs to $\Im(S_p)$ or $x^0_i$ otherwise.
20 Moreover, for each $i \in \Im(S_p)$, the element $x^{d_i}_i$ is equal to
21 $m^{d_i-1}_{S^{d_i}_c}$.
22 Thanks to constraint \ref{itm2:Sc}, all the indexes
23 $j \in \llbracket 0 ;\mathsf{P} -1 \rrbracket$ belong to
25 Let then $j \in \llbracket 0 ;\mathsf{P} -1 \rrbracket$ s.t.
27 Thus we have all the elements $m^._j$ of the vector $m$.
28 Let us focus now on some $m^{d_i-1}_j$.
29 Thus the value of $m^0_j$ can be immediately
30 deduced by counting in $S_c$ how many
31 times the component $j$ has been switched
34 Let us focus now on necessity.
35 If $\Im(S_c)_{|D} \subsetneq
36 \llbracket 0 ;\mathsf{P} -1 \rrbracket$,
37 there exist some $j \in \llbracket 0 ;\mathsf{P} -1 \rrbracket$ that
38 do not belong to $\Im(S_c)_{|\Im(S_p)}$.
39 Thus $m_j$ is not present in $x^l$ and the message cannot be extracted.
42 When the constraint \ref{itm2:Sc} is satisfied, we obtain a scheme
43 that always finds the original message provided the watermarked media
44 has not been modified.
45 In that context, correctness and completeness are established.
48 Thanks to constraint~\ref{itm2:Sc}, the cardinality $k$ of
49 $\Im(S_p)$ is larger than $\mathsf{P}$.
50 Otherwise the cardinality of $D$ would be smaller than $\mathsf{P}$
51 and similar to the cardinality of $\Im(S_c)_{|D}$,
52 which is contradictory.
54 One bit of index $j$ of the original message $m^0$
55 is thus embedded at least twice in $x^l$.
56 By counting the number of times this bit has been switched in $S_m$, the value of
57 $m_j$ can be deduced in many places.
58 Without attack, all these values are equal and the message is immediately
60 After an attack, the value of $m_j$ is obtained as mean value of all
62 The scheme is thus complete.
63 Notice that if the cover is not attacked, the returned message is always equal to the original
64 due to the definition of the mean function.