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
V ij for recovering the corresponding secret pixel as shown in Figure 16.4.
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.
C ij in Figure 16.6 is determined.
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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