Quantitative steganalysis of spatial ±1 steganography in JPEG decompressed images

Abstract

On the basis of the analysis of JPEG error and stegonoise, we propose a novel quantitative steganalyzer for spatial ±1 steganography in JPEG decompressed images. First, we present a particular theoretical argument that the cover images, which are originally stored in JPEG format, can be approximately estimated through JPEG recompression with the detected quantization table. Then, on the basis of the relationship between the message embedding rate and the variance of the stegonoise in the discrete cosine transformation (DCT) domain, we construct a polynomial regression model to estimate the secret message length. The extensive experimental results show that the proposed scheme is computationally feasible and that it significantly outperforms the existing state-of-the-art estimators, especially for the images with high quality factors and embedding rates. The order of magnitude of the prediction error using the proposed scheme can remain in the 10⁻⁴ range, as measured by the median absolute difference. Moreover, our estimator is stable and robust with respect to the embedding rate and quality factor.

Appendix

In the appendix, we show the proof of Theorem I. The process is as follows.

According to 2-D DCT and the values of u and v, three cases are discussed:

• I. When u ² + v ² = 0,

  $$ F\left(0,0\right)=\frac{1}{8}{\displaystyle \sum_{i=0}^7{\displaystyle \sum_{j=0}^7f\left(i,j\right)}} $$

  (18)

  As known, the i.i.d. random variable f(i, j) possesses a mean of 0 and variance σ ². According to the property of the mean and variance, we can easily obtain the mean of F(0, 0) is 0, and its variance is σ ². According to the Central Limit Theorem, F(0, 0) approximately follows the Gaussian distribution with a mean of 0 and variance σ ², which is denoted by F(0, 0) ~ N(0, σ ²).

• II. When uv = 0 and u ² + v ² ≠ 0, without loss of generality, we set u = 0, v ≠ 0. Then, we have

  $$ F\left(0,v\right)=\frac{1}{4\sqrt{2}}{\displaystyle \sum_{i=0}^7{\displaystyle \sum_{j=0}^7f\left(i,j\right) \cos \frac{\left(2j+1\right) v\pi}{16}}},1\le v\le 7 $$

  (19)

  The mean of F(0, v) is 0. Setting its variance to B ²₁ , then

  $$ \begin{array}{c}\hfill {B}_1{}^2=\frac{1}{32}{\displaystyle \sum_{i=0}^7{\displaystyle \sum_{j=0}^7{\sigma}^2{ \cos}^2\frac{\left(2j+1\right) v\pi}{16}}}\hfill \\ {}\hfill =\frac{\sigma^2}{32}{\displaystyle \sum_{i=0}^7\left(4+\frac{1}{2}{\displaystyle \sum_{j=0}^7 \cos \frac{\left(2j+1\right) v\pi}{8}}\right)}\hfill \\ {}\hfill =\frac{\sigma^2}{32}{\displaystyle \sum_{i=0}^7\left(4+{\displaystyle \sum_{j=0}^3 \cos \frac{\left(2j+1\right) v\pi}{8}}\right)}\hfill \end{array} $$

  (20)

  Let $S={\displaystyle \sum_{j=0}^3 \cos \frac{\left(2j+1\right) v\pi}{8}}= \cos \frac{ v\pi}{8}+ \cos \frac{3 v\pi}{8}+ \cos \frac{5 v\pi}{8}+ \cos \frac{7 v\pi}{8}$. When v is an odd number, $\cos \frac{7 v\pi}{8}= \cos \left( v\pi -\frac{ v\pi}{8}\right)=- \cos \frac{ v\pi}{8}$ and $\cos \frac{5 v\pi}{8}=- \cos \frac{3 v\pi}{8}$, then S = 0; when v is an even number, $\cos \frac{7 v\pi}{8}= \cos \frac{ v\pi}{8}$, $\cos \frac{5 v\pi}{8}= \cos \frac{3 v\pi}{8}$, then $S=2\left( \cos \frac{ v\pi}{8}+ \cos \frac{3 v\pi}{8}\right)=0$. In sum, $\begin{array}{cc}\hfill {\displaystyle \sum_{j=0}^3 \cos \frac{\left(2j+1\right) v\pi}{8}}=0,\hfill & \hfill 1\le v\le 7\hfill \end{array}$. Substituting it into Eq. (20), we obtain B ²₁  = σ ².

  According to the Central Limit Theorem, we have F(0, v) ~ N(0, σ ²), 1 ≤ v ≤ 7. Similarly, we can get F(u, 0) ~ N(0, σ ²) when v = 0 and 1 ≤ u ≤ 7.

• III. When uv ≠ 0,

  $$ F\left(u,v\right)=\frac{1}{4}{\displaystyle \sum_{i=0}^7{\displaystyle \sum_{j=0}^7f\left(i,j\right) \cos \frac{\left(2i+1\right) u\pi}{16} \cos \frac{\left(2j+1\right) v\pi}{16}}},1\le u,v\le 7 $$

  (21)

  The mean of F(u,v) is 0, which can be easily obtained. Setting its variance to B ²₂ , we have

  $$ \begin{array}{c}\hfill {B}_2{}^2=\frac{1}{16}{\displaystyle \sum_{i=0}^7{\displaystyle \sum_{j=0}^7{\sigma}^2{ \cos}^2\frac{\left(2i+1\right) u\pi}{16}{ \cos}^2\frac{\left(2j+1\right) v\pi}{16}}}\hfill \\ {}\hfill =\frac{\sigma^2}{16}\left({\displaystyle \sum_{i=0}^7{ \cos}^2\frac{\left(2i+1\right) u\pi}{16}}\right)\left({\displaystyle \sum_{j=0}^7{ \cos}^2\frac{\left(2j+1\right) v\pi}{16}}\right)\hfill \\ {}\hfill =\frac{\sigma^2}{16}\left(4+\frac{1}{2}{\displaystyle \sum_{i=0}^7 \cos \frac{\left(2i+1\right) u\pi}{8}}\right)\left(4+\frac{1}{2}{\displaystyle \sum_{j=0}^7 \cos \frac{\left(2j+1\right) v\pi}{8}}\right)\hfill \\ {}\hfill =\frac{\sigma^2}{16}\left(4+{\displaystyle \sum_{i=0}^3 \cos \frac{\left(2i+1\right) u\pi}{8}}\right)\left(4+{\displaystyle \sum_{j=0}^3 \cos \frac{\left(2j+1\right) v\pi}{8}}\right)\hfill \\ {}\hfill ={\sigma}^2\hfill \end{array} $$

  (22)

  According to the Central Limit Theorem, F(u, v) ~ N(0, σ ²), 1 ≤ u, v ≤ 7.

In conclusion, {F(u,v)|0 ≤ u, v ≤ 7} approximately follows a Gaussian distribution with a mean of 0 and a variance of σ ², namely, F(u, v) ~ N(0, σ ²).