Cryptography Reference
In-Depth Information
c
i
, where i∈[0,L−1]. For watermarking, two bits can be embedded into one
block, one from W
1
and the other from W
2
.
Assuming that W
1,k
=0andW
2,k
= 1 are the two watermark bits to be
embedded into X
k
. With the algorithm given in Sec. 12.5.1 and Eq. (12.6),
and after searching for the nearest codeword in C
′
, we conclude that the
resulting codeword containing the first watermark is i
S
= 101. According to
Eq. (12.7), we embed the second watermark and obtain i
B
= 1011. Because
the binary form of 1011 has a decimal form of 19, we conclude that j
19
is
the codeword to be transmitted in Fig 12.6. Referring to Fig. 12.7, the two
descriptions for transmission are i
1
= 0110 and i
2
= 0111.
In extracting the watermark, for both descriptions, or K = 2 in Fig. 12.1,
there will be four possible cases that can occur.
(1) If both descriptions are received, then both i
1
= 0110 and i
2
= 0111 can be
used for extracting the watermark bit W
2,k
and reconstructing the image.
By performing the inverse operation of index assignment l
−1
in Eq. (12.1)
and computing the conditional probability, we can exactly determine that
the codeword j
19
is transmitted with probability 1, thus, the watermark
bit is W
2,k
= 1. In reconstructing the block from the received descriptions,
we use the look-up table, and determine X
(0)
= j
19
. We employ the spatial
domain representations of j
19
to represent the reconstructed block X
′
k
.
(2) If Channel 1 breaks down, then only i
2
= 0111 is received. By performing
the inverse operation of index assignment l
−1
, we find that j
19
, j
21
,and
j
23
are the three possible candidates transmitted. Because 19, 21, and 23
are odd numbers, and by using a majority vote, the embedded bit can be
estimated to be W
2,k
= 1. In reconstructing the received image, we need
to calculate the conditional probability using the given conditions:
1
3
,
if t =19or21or23;
p (j
t
i
2
= 0111) =
(12.11)
0,
otherwise.
Thus, with Eq. (12.11), the reconstructed vector is the conditional ex-
pectation
X
(2)
1
3
(j
19
+ j
21
+ j
23
) . We use the spatial domain repre-
sentations of j
19
, j
21
,andj
23
to calculate their average to complete the
reconstruction of X
(2)
.
(3) If Channel 2 breaks down, then only i
1
= 0110 is received. By doing the
inverse operation of index assignment, we find that j
17
, j
18
,andj
19
are
the three possible candidates transmitted. Using a majority vote, the em-
bedded bit can be estimated to be W
2,k
= 1. In some situations where
there are two candidates, we use random selection to determine the wa-
termark bit. In reconstructing the received image, we need to calculate
the conditional probability at the given conditions:
=
1
3
,
if t =17or18or19;
p (j
t
i
1
= 0110) =
(12.12)
0,
otherwise.
