## 3 Proposed Method
(Due to FOBP, the secret data is not embedded for all cover image pixels.) If these two values are equal and boundary of the range \(\left[3\times 2^{(n-2)}\,\ 256-3\times 2^{(n-2)}\right]\), it means that the secret data in these pixels in the recording series is zero. If the pixel values in two stego images are outside this interval and are equal, it is understood that no secret data is hidden there. The flow chart of the data extraction process is given in Figure 5.  
In the data extraction process, the initial series cannot be directly acquired as the pixel values of two stego images change. For this reason, the original cover image must first be  
Figure 4: The state of change interval for \(n=4\). Figure 2:0 The flow chart of the data extraction process. obtained in the data extraction process. In other words, it is impossible to extract the secret data intact without the recovery of the original cover image. Figure 2:1  
To recover the original cover image, the two stego images are averaged and rounded up, as in Equation (13). The main reason for rounding up is to neutralize the rounding down process performed during data hiding. After the original cover image has been recovered, the initial series can be obtained using Equation (7). The extraction of the master record series is accomplished by subtracting the first stego image from the second stego image as shown in Equation (14). Figure 2:2  
Finally, the initial series and shift amount are added to the master record series and the secret data symbols expressed as s are obtained using Equation (15). Figure 2:3  
For an example, the pixel values of the cover image are \(X_{10}=\{164,\,63,120,135\}\). When we consider the value of \(n=3\), we can embed \(k=n+1=4\) for each pixel. The sum of that makes 16 bits secret data for 4 pixels. When we assume the secret data as \(S_{10}=\{232,34\}\) and get the binary system of it, we get \(S_{2}=\{11101000,\,00100010\}\). If this expression is converted to \(k\) = 4-bit data, \(S_{2}=\{1110,\,1000,\,0010,\,0010\}\) data are obtained. These data are equal to \(s=\{14,\,8,\,2,\,2\,\}\) in decimal system. At the next step, we need to compute initial series by Equation (8). Then, we find the initial series as \(d=\{5,\,1,\,3,\,4\}\). When we add secret data to initial series by Equation (9), we get \(d^{\prime}=\{9,\,7,-1,-2\}\). In the next step, the shifting operation is performed in an amount of \(2^{n-1}\) by Equation (10), and \(d^{\prime\prime}=\{5,\,3,\,-5,-6\}\) series is obtained. Using Equation (11), it is calculated as \(d_{1}=\{2,\,1,-3,-3\}\) and \(d_{2}=\{-3,-2,\,2,\,3\}\). These two series are added to the cover image pixel values according to Equation (12), \(X^{\prime}=\{166,\,64,\,117,\,132\}\) and \(X^{\prime\prime}=\{161,\,61,\,122,138\}\) stego images pixel values are obtained. For extraction of secret data, we need to first calculate to original cover image pixel values by Equation (13). After computing Equation (13), we get \(X=\{164,\,63,120,135\}\). In the second step, we must get difference of these two stego image pixel values by Equation (14). So, we obtain \(d^{\prime\prime}=\{5,\,3,\,-5,-6\}\). Then the initial series is computing as \(d=\{5,\,1,\,3,\,4\}\) by using Equation (8), by the same method with the data hiding process. For the last step, the secret data \(s=\{14,\,8,\,2,\,2\}\) are found by Equation (15). And these secret data are converted to binary series as \(S_{2}=\{1110,\,1000,\,0010,\,0010\,\,\}\) to obtain a binary system of secret data. Then, secret data (\(S_{10}=\{232,34\}\)) are obtained by converting them from a binary to a decimal system.