Cryptography Reference
In-Depth Information
its corresponding cover image block C
ij
by replacing the eight bits v
1
v
0
y
1
y
0
z
1
z
0
u
1
u
0
with s
1
s
2
...s
8
, and at the same time, the bits z
3
z
2
of pixel
Z
ij
are replaced by v
1
v
0
of pixel V
ij
in order to keep the complete pixel value
However, Yang et al.'s scheme cannot avoid the fact that four least sig-
nificant bits (LSBs) of the pixel value Z
ij
in each block must be modified.
Such modification may make the stego-images different from the cover images
visually and the risk of being observed by attackers is increased.
To further improve Lin-Tsai's authentication ability, Yang et al. used the
hash-based message authentication code (HMAC) in Equation 16.2.
H
k
(( C
ij
b)kid);
(16.2)
where id is the block ID; C
ij
is 31 bits except the check bit b; H( ) and k denote
a one-way hash function and the secret key, respectively. Next, as shown in
Equation 16.3, the 160-bit HMAC output and executes the XOR operation to
obtain the authentication bit b.
b = H
k
(( C
ij
b)kid)
(16.3)
However, the average probability of detecting any malicious modification is
just 50% because of the characteristic of the parity check itself. In other words,
the fake stego-images still have a very high probability of being authenticated
successfully.
16.2.3 Chang et al.'s Scheme
In 2008, Chang et al. proposed an enhanced secret sharing scheme based on
Lin-Tsai's scheme and Yang et al.'s scheme. They used the concept of Thien
and Lin's secret image sharing [23] to ensure that no distortion is introduced
into the secret image and applied the concept of the Chinese Remainder The-
orem (CRT) to improve authentication ability. As a result, their scheme can
achieve high authentication ability. The owchart of Chang et al.'s scheme [9]
is shown in
Figure 16.5.
Four check bits p
1
, p
2
, p
3
, p
4
can be calculated and embedded into the
stego-block (shown in Figure 16.5). The details are shown in the steps below.
Step 1:
The residues set of the block
R
ij;1
=(x
7
x
6
x
5
x
4
x
3
s
1
s
2
);
R
ij;2
=(v
7
v
6
v
5
v
4
s
3
s
4
);
R
ij;3
=(w
7
w
6
w
5
w
4
w
3
s
5
s
6
);
R
ij;4
=(z
7
z
6
z
5
z
4
z
3
s
7
s
8
);
R
ij;5
= i;
R
ij;6
= j.
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