# Data Hiding Method for Color AMBTC Compressed Images Using Color Difference
## 4 Proposed Method
### AMBTC Compression and DH Example  
**Example 1**.: _Traditional color AMBTC method._  
It denotes \(\mathcal{OI}_{R},\mathcal{OI}_{G}\), and \(\mathcal{OI}_{B}\) as layers of R, G, and B of the original color image \(\mathcal{OI}\). We describe the traditional color AMBTC encoding procedure with an example of Figure 4, where a block size is \(4\times 4\).  
Traditional color AMBC is compressed into two quantization levels and one bitmap for each layer of R, G, and B as shown in Figure 4. The mean (144) and bitmap (Figure 4b) based on the \(4\times 4\) of \(\mathcal{OI}_{R}\) can be obtained by using Equations (1) and (2). For \(\mathcal{OI}_{G}\) and \(\mathcal{OI}_{B}\), the mean and a bitmap can be obtained in the same way, respectively. As for the quantization level, \((Q_{1}^{R},Q_{2}^{R})=(38,144)\) can be obtained as shown in Figure 4b as performing Equations (3) and (4).  
**Example 2**.: _The proposed color AMBC method._  
With the RGB blocks in Figure 4a in Example 1, the obtaining procedure of a common  
Unlike the quantization levels obtained in Example 1, there are three quantization levels for each color, thus it is possible to represent the medium value for the block as well as it has excellent compression performance.  
**Example 3**.: _DH procedure using an example._  
To explain the embedding procedure, we use the quantization levels and a bitmap obtained in Example 1. Assuming that secret bit \(\mathcal{S}=(111000)\) and threshold \(T=60\). Moreover, only \(Q^{R}\) is considered for avoiding repeated embedding process for describing this example. First, in order to hide 2-bit data in \(Q^{R}_{1}=136\), which is the first pixel of \(Q^{R}=(136,162,174)\) and then, transform to \(Q^{R^{\prime}}_{1}=\left\lfloor 136/2\right\rfloor=68\). Next, extract 2 LSB and LSB, i.e., (2 LSB LSB) = (0 0) and \(\mathcal{S}_{1}=1\). According to Equation (13), \(Q^{R^{\prime\prime}}_{1}=Q^{R^{\prime}}_{1}(68)-1=67\) is obtained. Since (2 LSB LSB LSB) = (1 1) and \(S_{2}=1\), we obtain \(Q^{R^{\prime\prime}}_{1}(67)-1=66\) by using Equation (14). \(Q^{R^{\prime\prime}}_{1}(66)\) is recovered by the calculation, \(Q^{R^{\prime\prime}}_{1}\times 2=132\). Next, for \(Q^{R}_{2}\), the way of embedding \(S_{34}\) in \(Q^{R}_{2}\) is the same procedure of \(Q^{R}_{1}\). After embedding \(S_{34}\) in \(Q^{R}_{2}\), it is \(Q^{R}_{2}\)=166. In case of \(Q^{R}_{3}\), when \(S_{5,6}=(00)\) is embedded in \(Q^{R}_{3}\), it become \(Q^{R}_{R}=176\).  
The condition for embedding secret bits in the common bitmap is that the difference between the two quantization levels (\(|Q^{G}_{1}-Q^{G}_{2}|\leq T\)) must be less than or equal to the threshold value \(T\). Here, it is less than \(T=60\), thus the criterion is enough. Therefore, it can conceal a given 16-bit secret bit. Assuming \(S=(0010100101001101)\), the 16 bits are hidden in bitmap \(M\) using Equation (15). \(S_{7,8}\) is (00), which is the same as \(M\), so there is no need to change \(M\). For the following \(S_{9,10}=(10)\), \(+1\) is performed according to the rule of Equation (15). For embedding the remaining secret bits, if the bitmap \(M\) is applied successively by using Equation (15), it becomes \(M=(00000100010000010000110100101)\).