# Survey on Data Hiding Based on Block Truncation Coding
## 3 Survey of Related Works
### Non-Reversible Data Hiding
#### 3.1.5 Least Significant Bits (LSB) Substitution  
[PERSON] et al. [43] use quotient value difference (QVD) and LSB substitution to hide secret data in AMBTC. This method includes three phases of embedding. In the first phase, QVD and LSB substitution are applied to embed secret digits into the low mean and high mean values. In the second phase, an additional bit is embedded by alternating the order of the low mean and high mean or not. Secret digits are embedded in the third phase by substituting the bitmap of smooth blocks. Assume that the decimal conversion value of \(n\) hidden bits to be hidden is \(D\). In order to hide \(D\) in two quantization values, we need to know the difference and range of the two quantization values. First, the difference between the quotients of the two quantized values is \(\delta=Q_{H}-Q_{L}\), where \(Q_{L}=\mu_{0}/4\) and \(Q_{H}=\mu_{1}/4\). Second, the range of \(|\delta|\) is \(r=[r_{j}^{L},r_{j}^{H}];r_{1}=[0,7],r_{2}=[8,15],r_{3}=[16,31],\text{ and }r_{4}=[32,63]\). Third, the new \(\delta^{\prime}\) is \(r_{j}^{L}+D\), and the DH is determined by Equation (8).  
\[(Q_{L}^{\prime},Q_{H}^{\prime})=\begin{cases}\left(Q_{L}-\left\lfloor\frac{d^ {\prime}-d}{2}\right\rfloor,Q_{H}+\left\lceil\frac{d^{\prime}-d}{2}\right\rceil \right),&\text{if $d$ is even },\\ \left(Q_{L}-\left\lceil\frac{d^{\prime}-d}{2}\right\rfloor,Q_{H}+\left\lfloor \frac{d^{\prime}-d}{2}\right\rfloor\right),&\text{if $d$ is odd }.\end{cases} \tag{8}\]  
[PERSON] et al. [44] present a DH method based on Adaptive BTC Edge Quantization (ABTC-EQ) using an optimal pixel adjustment process (OPAP) [45] to optimize two quantization levels. ABTC-EQ as a cover media is superior to AMBTC in maintaining a high-quality image after encoding is completed. ABTC-EQ is represented by a form of trio \((\mu_{0},\mu_{1},[\mu_{2}],BM)\) where \(\mu\) is quantization levels \((\mu_{0},\mu_{1},\mu_{2})\) and BM is a bitmap. LSB substitution for OPAP is performing the following procedure: First, suppose that the \(n\)-bit secret message \(D\) is to be embedded into the \(k\)-rightmost LSBs of two quantization levels, where \(D=\{d_{i}|0\leq i<n,d_{i}\in\{0,1\}\}\). Second, the secret message \(D\) is rearranged to form a \(k\)-bit as \(D^{\prime}=\{d_{i}^{\prime}|d_{i}^{\prime}\in\{0,1,\ldots,2^{k}-1\}\}\). The mapping between the \(n\)-bit secret message \(D\) and the embedded message \(D^{\prime}\) can be defined as \(d_{i}^{\prime}=\sum_{j=0}^{k-1}d_{i\times k+j}\times 2^{k-1-j}\). Third, the pixel value \(p\) (here, two quantization levels) is modified to form \(p^{\prime}\) as \(p^{\prime}=p-(p\text{ mod }2^{k})+d_{i}^{\prime}\).  
**Example 6**.: _The given trio, secret bits, and \(k\) are \((\mu_{0},\mu_{1},BM)=\) (146, 149, 0 0 1 1 0 1 0 1 1 1 0 0 0 ), \(d=[1\ 0\ 1\ 0]\), and \(k=2\), respectively; we embed the secret bit \(d^{\prime}=[2\ 2]\) using the LSB replacement. That is, \(\mu_{0}^{\prime}=\mu_{0}-(\mu_{0}\ mod\ 2^{k})+d_{i}^{\prime}=146-(146\ mod\ 2^{2})+2=146\) and \(\mu_{1}^{\prime}=\mu_{1}-(\mu_{1}\ mod\ 2^{k})+d_{i}^{\prime}=149-(149\ mod\ 2^{2})+2=150\). Here, 4 bits were hidden in 2 pixels, and an error of 1 bit occurred. Assuming that the transmitting side and the receiving side share \(k\), the remainders obtained by dividing each pixel by \(2^{k}\) become the hidden values._