Image Steganography Using an Eight-Element Neighborhood Gaussian Markov Random Field Model

Abstract

Currently, the design of cost functions which measure the embedding distortion becomes the main task left in steganography with the emergence of Syndrome-Trellis Codes. Whether heuristically designed, e.g., Hill, WOW, or statistical model-based, e.g., HUGO, MiPOD, the embedding distortion is almost a sum of each element’s distortion. This paper proposes a new non-additive cost function designed by incorporating an eight-element neighborhood Gaussian Markov Random Field Model (8-GMRF). This proposed scheme, which could be viewed as an extension to MiPOD, derives change probabilities from minimizing the total KL divergence with a given payload and then implements adaptive steganography. Experimental results demonstrate that the proposed 8-GMRF performs superior or comparable to some of the state-of-the-art schemes in resisting steganalysis detectors.

Appendix

1.1 A. Joint probability of one 8-element neighborhood GMRF block

Recall Fig. 1, which illustrates 9 pixels of one block with index $\mathrm{{\{0}} \sim \mathrm{{8\}}}$. Every two pixels are adjacent if connected or independent if disconnected. For those independent (e.g. pixels, i, j), we can easily obtain (14). Accordingly, we proceed with a group of pixels according to Markovianity shown in (15), which will be repeatedly used in this proof.

$$\begin{aligned} p(i,j) = p(i|j) \cdot p(j) = p(i) \cdot p(j) \end{aligned}$$

(14)

$$\begin{aligned} p(i,{j_1}...{j_m}) = p(i|{j_1}...{j_m}) \cdot p({j_1}...{j_m}) = p(i|{j_k},i \leftrightarrow {j_k}) \cdot p({j_1}...{j_m}) \end{aligned}$$

(15)

Firstly, it is obvious to get (16) using (15) and then take the second term of former formula repeatedly we can get (17). (18) is derived from (17).

$$\begin{aligned} p(0 \sim 8) = p(5|0 \sim 4,6 \sim 8) \cdot p(0 \sim 4,6 \sim 8) = p(5|0,1,2) \cdot p(0 \sim 4,6 \sim 8) \end{aligned}$$

(16)

$$\begin{aligned} p(0 \sim 4,6 \sim 8) = p(6|0,2,3) \cdot p(0 \sim 4,7,8), p(0 \sim 4,7,8) = p(8|0,1,4) \cdot p(0 \sim 4,7) \end{aligned}$$

$$\begin{aligned} p(0 \sim 4,7) = \frac{{p(0,1,3,4)}}{{p(0,1,3)}} \cdot p(0,1,2,3), p(0,1,2,3) = \frac{{p(0,1,2)}}{{p(0,2)}} \cdot p(0,2,3) \end{aligned}$$

(17)

$$\begin{aligned} p(0,1,3,4) = \frac{{p(0,3,4)}}{{p(0,4)}} \cdot p(0,1,4), p(0,1,3) = \frac{{p(0,1)}}{{p(0)}} \cdot p(0,3) \end{aligned}$$

(18)

Substitute (18) into (17), (17) into (16), $p(0 \sim 8)$ can be formulated as:

$$\begin{aligned} p(0 \sim 8) = \frac{{p(1,0,2,5) \cdot p(3,0,2,6) \cdot p(3,0,4,7) \cdot p(1,0,4,8) \cdot p(0)}}{{p(0,1) \cdot p(0,2) \cdot p(0,3) \cdot p(0,4)}} \end{aligned}$$

(19)

1.2 B. $2 \times 2$ steganographic Fisher Information matrix

Stimulated by one-dimensional steganographic Fisher Information (2) in MG, we construct $2 \times 2$ steganographic Fisher Information matrix as following:

$$\begin{aligned} I_{ij}(0) = \left[ {\begin{array}{*{20}{c}} {I_{ij}^{(11)}(0)}&{}{I_{ij}^{(12)}(0)}\\ {I_{ij}^{(21)}(0)}&{}{I_{ij}^{(22)}(0)} \end{array}} \right] , \end{aligned}$$

(20)

where $I_{ij}^{(12)}(0) = I_{ij}^{(21)}(0) = {\sum _{k,l}}\frac{1}{{{p_{k,l}}}}(\frac{{\partial {q_{k,l}}({\beta _i},{\beta _j})}}{{\partial {\beta _i}}}{|_{{\beta _i},{\beta _j} = 0}}) \cdot (\frac{{\partial {q_{k,l}}({\beta _i},{\beta _j})}}{{\partial {\beta _j}}}{|_{{\beta _i},{\beta _j} = 0}})$,

$$\begin{aligned}I_{ij}^{(11)}(0) = {\sum _{k,l}}\frac{1}{{{p_{k,l}}}}{(\frac{{\partial {q_{k,l}}({\beta _i},{\beta _j})}}{{\partial {\beta _i}}}{|_{{\beta _i},{\beta _j} = 0}})^2}, I_{ij}^{(22)}(0) = {\sum _{k,l}}\frac{1}{{{p_{k,l}}}}{(\frac{{\partial {q_{k,l}}({\beta _i},{\beta _j})}}{{\partial {\beta _j}}}{|_{{\beta _i},{\beta _j} = 0}})^2}.\end{aligned}$$

Note that ${p_{k,l}}$, i.e. $p_{k,l}^{(i,j)}$, equals $P\{ {x_i} = k,{x_j} = l\}$, which can seemed as a point in the two-dimensional Gaussian distribution $f({x_i},{x_j})$ (21). ${x_i},{x_j},k,l$ are used as continuous for differential and integration in this proof. Given ${p_{k,l}} = f(k,l),{p_{k \pm 1,l}} = f(k \pm 1,l),{p_{k,l \pm 1}} = f(k,l \pm 1)$, we have the Taylor approximation at points $(k \pm 1,l),(k,l \pm 1)$:

$$\begin{aligned} f({x_i},{x_j}) = \frac{1}{{2\pi {\sigma _i}{\sigma _j}\sqrt{1 - \rho _{ij}^2} }}\exp ( - \frac{1}{{2(1 - \rho _{ij}^2)}}(\frac{{x_i^2}}{{\sigma _i^2}} - \frac{{2{\rho _{ij}}{x_i}{x_j}}}{{{\sigma _i}{\sigma _j}}} + \frac{{x_j^2}}{{\sigma _j^2}})). \end{aligned}$$

(21)

$$\begin{aligned} - 2{p_{k,l}} + {p_{k - 1,l}} + {p_{k + 1,l}} \approx \frac{{{\partial ^2}f({x_i},{x_j})}}{{\partial {x_i}^2}}{|_{{x_i} = k,{x_j} = 1}}, \end{aligned}$$

(22)

$$\begin{aligned} - 2{p_{k,l}} + {p_{k,l - 1}} + {p_{k,l + 1}} \approx \frac{{{\partial ^2}f({x_i},{x_j})}}{{\partial {x_j}^2}}{|_{{x_i} = k,{x_j} = 1}}. \end{aligned}$$

(23)

After substituting (21)–(23) and (8) into (20), we can finally derive:

$$\begin{aligned} I_{ij}^{(11)}(0) = \sum \limits _{k,l} {\frac{{{{( - 2{p_{k,l}} + {p_{k - 1,l}} + {p_{k + 1,l}})}^2}}}{{{p_{k,l}}}} \approx \int \limits _k {\int \limits _l {{{(\frac{{{\partial ^2}f({x_i},{x_j})}}{{\partial {x_i}^2}}{|_{{x_i} = k,{x_j} = l}})}^2}\frac{{dkdl}}{{f({x_i},{x_j})}}} } } \end{aligned}$$

$$\begin{aligned} I_{ij}^{(22)}(0) = \sum \limits _{k,l} {\frac{{{{( - 2{p_{k,l}} + {p_{k,l - 1}} + {p_{k,l + 1}})}^2}}}{{{p_{k,l}}}} \approx \int \limits _k {\int \limits _l {{{(\frac{{{\partial ^2}f({x_i},{x_j})}}{{\partial {x_j}^2}}{|_{{x_i} = k,{x_j} = l}})}^2}\frac{{dkdl}}{{f({x_i},{x_j})}}} } } \end{aligned}$$

$$\begin{aligned} I_{ij}^{(12)}(0) = I_{ij}^{(21)}(0) = \sum \limits _{k,l} {\frac{{( - 2{p_{k,l}} + {p_{k - 1,l}} + {p_{k + 1,l}})( - 2{p_{k,l}} + {p_{k,l - 1}} + {p_{k,l + 1}})}}{{{p_{k,l}}}}} \end{aligned}$$

$$\begin{aligned} \approx \int \limits _k {\int \limits _l {(\frac{{{\partial ^2}f({x_i},{x_j})}}{{\partial {x_i}^2}}{|_{{x_i} = k,{x_j} = l}})(\frac{{{\partial ^2}f({x_i},{x_j})}}{{\partial {x_j}^2}}{|_{{x_i} = k,{x_j} = l}})\frac{{dkdl}}{{f({x_i},{x_j})}}} } \end{aligned}$$

$$\begin{aligned} I_{ij}^{(11)}(0)\approx \frac{2}{{{\sigma _i}^2{{(1 - {\rho _{ij}}^2)}^2}}},I_{ij}^{(22)}(0)\approx \frac{2}{{{\sigma _j}^2{{(1 - {\rho _{ij}}^2)}^2}}},I_{ij}^{(12)}(0) = I_{ij}^{(21)}(0)\approx \frac{2{\rho _{ij}}^2}{{{\sigma _i}{\sigma _j}{{(1 - {\rho _{ij}}^2)}^2}}} \end{aligned}$$