Steganographic algorithm based on randomization of DCT kernel

Abstract

Calibration attack is one of the most important attacks specially on JPEG domain steganography. Prediction of cover image statistics from its stego image is an important requirement in calibration based attacks. Domain separation techniques are used as a counter measures against such attacks because they make the cover image prediction process rather difficult. Most of the algorithms in the past are based on randomization of embedding locations. In this paper, we propose a new domain separation technique which is based on randomization of DCT kernel matrix. A comprehensive set of experiments are done to validate that the proposed domain separation scheme performs better than the related existing schemes.

Appendices

Appendix I: example of PDCT

Consider a matrix A

$$ \left(\begin{array}{ccc}\hfill 12\hfill & \hfill 9\hfill & \hfill 5\hfill \\ {}\hfill 24\hfill & \hfill 7\hfill & \hfill 15\hfill \\ {}\hfill 31\hfill & \hfill 2\hfill & \hfill 19\hfill \end{array}\right) $$

The 3 × 3 DCT kernel is given by D

$$ \left(\begin{array}{ccc}\hfill 0.5774\hfill & \hfill 0.5774\hfill & \hfill 0.5774\hfill \\ {}\hfill 0.7071\hfill & \hfill 0.0000\hfill & \hfill -0.7071\hfill \\ {}\hfill 0.4082\hfill & \hfill -0.8165\hfill & \hfill 0.4082\hfill \end{array}\right) $$

DCT(A) = D ∗ A ∗ D′

$$ \left(\begin{array}{ccc}\hfill 41.3333\hfill & \hfill 11.4310\hfill & \hfill 16.4992\hfill \\ {}\hfill -10.6145\hfill & \hfill -2.5000\hfill & \hfill -13.5677\hfill \\ {}\hfill -3.2998\hfill & \hfill 0.2887\hfill & \hfill -0.8333\hfill \end{array}\right) $$

The 3 × 3 PDCT kernel when α = 1.25 is given by $\overline{D}$

$$ \left(\begin{array}{ccc}\hfill 0.6352-0.1474 i\hfill & \hfill 0.4655+0.2279 i\hfill & \hfill 0.5249+0.1749 i\hfill \\ {}\hfill 0.6275+0.2279 i\hfill & \hfill 0.1439-0.3523 i\hfill & \hfill -0.5800-0.2703 i\hfill \\ {}\hfill 0.3138+0.1749 i\hfill & \hfill -0.7166-0.2703 i\hfill & \hfill 0.4913-0.2074 i\hfill \end{array}\right) $$

PDCT(A) = $\overline{D}\ast A\ast {\overline{D}}^{\prime }$

$$ \left(\begin{array}{ccc}\hfill 37.3570+8.5973 i\hfill & \hfill 12.0092+3.3128 i\hfill & \hfill 13.8951+6.1228 i\hfill \\ {}\hfill -4.6532-15.5554 i\hfill & \hfill -0.6476-1.6196 i\hfill & \hfill -9.2278-6.7782 i\hfill \\ {}\hfill 0.8440-10.4383 i\hfill & \hfill 0.5825-3.5580 i\hfill & \hfill 1.2906-6.9777 i\hfill \end{array}\right) $$

Appendix II: Reversibility of randomized DCT (RDCT)

Proof for ${\overline{D}}^{\prime}\overline{D}= I$, where $\overline{D}$ is the RDCT kernel matrix

Let D ₁, D ₂, D ₃, D ₄, D ₅, D ₆, D ₇, D ₈ be the column vectors of DCT matrix D. We know that the DCT matrix D is a orthonormal matrix where the columns are unit vectors perpendicular to each other.

i.e. D _i . D _i  = 1 and D _i . D _j  = 0. This gives D′D = I.

Now consider the product ${\overline{D}}^{\prime}\overline{D}$. The diagonal entries are of the form D _i . D _i and other entries are of the form D _i . D _j . Hence it is clear that ${\overline{D}}^{\prime}\overline{D}= I$

Proof for Inverserdct(A _rdct ) = A

$$ Inverserdct\left({A}_{rdct}\right)={\overline{D}}^{\prime }{A}_{rdct}\overline{D} $$

(10)

From Eq. 3 we get,

$$ {\overline{D}}^{\prime }{A}_{rdct}\overline{D}={\overline{D}}^{\prime}\left(\overline{D} A{\overline{D}}^{\prime}\right)\overline{D}={\overline{D}}^{\prime}\overline{D} A{\overline{D}}^{\prime}\overline{D} $$

(11)

As we have proved ${\overline{D}}^{\prime}\overline{D}= I$, the above reduces to A. Therefore

$$ Inverserdct\left({A}_{rdct}\right)= A $$

Hence the reversibility is proved.

Appendix III: reversibility example of RDCT transform matrix

Consider a matrix A

$$ \left(\begin{array}{ccc}\hfill 12\hfill & \hfill 9\hfill & \hfill 5\hfill \\ {}\hfill 24\hfill & \hfill 7\hfill & \hfill 15\hfill \\ {}\hfill 31\hfill & \hfill 2\hfill & \hfill 19\hfill \end{array}\right) $$

The 3 × 3 DCT kernel is given by D

$$ \left(\begin{array}{ccc}\hfill 0.5774\hfill & \hfill 0.5774\hfill & \hfill 0.5774\hfill \\ {}\hfill 0.7071\hfill & \hfill 0.0000\hfill & \hfill -0.7071\hfill \\ {}\hfill 0.4082\hfill & \hfill -0.8165\hfill & \hfill 0.4082\hfill \end{array}\right) $$

Let the random matrix K

$$ \left(\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ {}\hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ {}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right) $$

The 3 × 3 RDCT kernel is given by $\overline{D}$ = KD

$$ \left(\begin{array}{ccc}\hfill 0.7071\hfill & \hfill 0.0000\hfill & \hfill -0.7071\hfill \\ {}\hfill 0.4082\hfill & \hfill -0.8165\hfill & \hfill 0.4082\hfill \\ {}\hfill 0.5774\hfill & \hfill 0.5774\hfill & \hfill 0.5774\hfill \end{array}\right) $$

RDCT(A) = A _rdct = $\overline{D}\ast A\ast {\overline{D}}^{\prime }$

$$ \left(\begin{array}{ccc}\hfill -2.5000\hfill & \hfill -13.5677\hfill & \hfill -10.6145\hfill \\ {}\hfill 0.2887\hfill & \hfill -0.8333\hfill & \hfill -3.2998\hfill \\ {}\hfill 11.4310\hfill & \hfill 16.4992\hfill & \hfill 41.3333\hfill \end{array}\right) $$

$Inverserdct\left({A}_{rdct}\right)={\overline{D}}^{\prime }{A}_{rdct}\overline{D}$

$$ \left(\begin{array}{ccc}\hfill 12\hfill & \hfill 9\hfill & \hfill 5\hfill \\ {}\hfill 24\hfill & \hfill 7\hfill & \hfill 15\hfill \\ {}\hfill 31\hfill & \hfill 2\hfill & \hfill 19\hfill \end{array}\right) $$