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A new fast and efficient active steganalysis based on combined geometrical blind source separation

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Abstract

We have presented a new active steganalysis method in order to break the block-discrete cosine transform (DCT) coefficients steganography. Our method is based on a combination of the Blind Source Separation (BSS) technique and Maximum A posteriori (MAP) estimator. We have additionally introduced a new geometrical BSS method based on the minimum range of mixed sources which reduces the computational cost of the proposed steganalysis. The high efficiency of this new combined method has been confirmed by enough experiments. These experiments show that, compared to the previous active steganalysis methods our active steganalysis method not only reduces the error rate but also causes a low computational cost.

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Authors and Affiliations

  1. Department of Electrical Engineering, Ferdowsi University of Mashhad, Mashhad, Iran

    Hamed Modaghegh & Seyed Alireza Seyedin

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  1. Hamed Modaghegh
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  2. Seyed Alireza Seyedin
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Correspondence to Hamed Modaghegh.

Appendix

Appendix

Proof of Propositions which was stated in paper.

Proof of Proposition 1:

We have by (7):

$$ \begin{array}{cc}\hfill \theta =0,\pi \Rightarrow \left\{\begin{array}{c}\hfill {z}_1=\pm {s}_1\hfill \\ {}\hfill {z}_2=\pm {s}_2\hfill \end{array}\right.,\hfill & \hfill \theta =\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.,\raisebox{1ex}{$3\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\Rightarrow \left\{\begin{array}{c}\hfill {z}_1=\pm {s}_2\hfill \\ {}\hfill {z}_2=\pm {s}_1\hfill \end{array}\right.\hfill \end{array} $$
(25)

So it is clear that:

$$ \min \left( \max {z}_1, \max {z}_2\right)= \min \left( \max {s}_1, \max {s}_2\right) $$
(26)

Now let us proof (8) for other θ. Without loss of generality assume:

$$ \max {s}_1=K \max {s}_2 $$
(27)

Then according to (1–1), we have:

$$ \begin{array}{l} \max {z}_1= \max {s}_2\times \left(\left| \cos \theta \right|K+\left| \sin \theta \right|\right)\\ {} \max {z}_2= \max {s}_2\times \left(\left| \sin \theta \right|K+\left| \cos \theta \right|\right)\end{array} $$
(28)
$$ \begin{array}{l} \max {z}_1= \max {s}_1\times \left(\left| \sin \theta \right|/K+\left| \cos \theta \right|\right)\\ {} \max {z}_2= \max {s}_1\times \left(\left| \cos \theta \right|/K+\left| \sin \theta \right|\right)\end{array} $$
(29)

Using (28) gives us:

$$ \begin{array}{l}K\ge 1\Rightarrow \min \left( \max {s}_1, \max {s}_2\right)= \max {s}_2\\ {}\kern2em , \max {z}_1\ge \max {s}_2\times \left(\left| \sin \theta \right|+\left| \cos \theta \right|\right)\\ {}\kern2em , \max {z}_2\ge \max {s}_2\times \left(\left| \sin \theta \right|+\left| \cos \theta \right|\right)\\ {}\kern2em \Rightarrow \max {z}_1, \max {z}_2\ge \min \left( \max {s}_1, \max {s}_2\right)\times \left(\left| \sin \theta \right|+\left| \cos \theta \right|\right)\ \left(1-6\right)\end{array} $$
(30)

And for other value of K we use (29):

$$ \begin{array}{l}K<1\Rightarrow \min \left( \max {s}_1, \max {s}_2\right)= \max {s}_1\\ {}\kern2em , \max {z}_1\ge \max {s}_1\times \left(\left| \sin \theta \right|+\left| \cos \theta \right|\right)\\ {}\kern2em , \max {z}_2\ge \max {s}_1\times \left(\left| \sin \theta \right|+\left| \cos \theta \right|\right)\\ {}\kern2em \Rightarrow \max {z}_1, \max {z}_2\ge \min \left( \max {s}_1, \max {s}_2\right)\times \left(\left| \sin \theta \right|+\left| \cos \theta \right|\right)\end{array} $$
(31)

As it can be seen (30) and (31) are equal so it can be written:

$$ \max {z}_1, \max {z}_2\ge \min \left( \max {s}_1, \max {s}_2\right)\times \left(\left| \cos \theta \right|+\left| \sin \theta \right|\right) $$
(32)

Since \( \theta \ne 0,\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.,\pi, \raisebox{1ex}{$3\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right. \) and triangle inequality theorem:

$$ \begin{array}{l}\left| \cos \theta \right|+\left| \sin \theta \right|>1\Rightarrow \max {z}_1, \max {z}_2> \min \left( \max {s}_1, \max {s}_2\right)\\ {}\Rightarrow \min \left( \max {z}_1, \max {z}_2\right)> \min \left( \max {s}_1, \max {s}_2\right)\end{array} $$
(33)

Proof of Proposition 2:

According to (7), we have:

$$ \begin{array}{c}\hfill \max {z}_1=\left| \cos \theta \right| \max {s}_1+\left| \sin \theta \right| \max {s}_2\hfill \\ {}\hfill \max {z}_2=\left| \sin \theta \right| \max {s}_1+\left| \cos \theta \right| \max {s}_2\hfill \\ {}\hfill \min \left({z}_1\right)=\left| \cos \theta \right| \min \left({s}_1\right)+\left| \sin \theta \right| \min \left({s}_2\right)\hfill \\ {}\hfill \min \left({Z}_2\right)=\left| \sin \theta \right| \min \left({S}_1\right)+\left| \cos \theta \right| \min \left({S}_2\right)\hfill \end{array} $$
(34)

Without loss of generality assume:

$$ Range\left({s}_1\right)=K\times Range\left({s}_2\right) $$
(35)

Using (34) gives us:

$$ \begin{array}{c}\hfill Range\left({z}_1\right)= \max {z}_1- \min {z}_1=\left(\left| \cos \theta \right| \max {s}_1+\left| \sin \theta \right| \max {s}_2\right)-\left(\left| \cos \theta \right| \min {s}_1+\left| \sin \theta \right| \min {s}_2\right)\hfill \\ {}\hfill =\left| \cos \theta \right|\left( \max {s}_1- \min {s}_1\right)+\left| \sin \theta \right|\left( \max {s}_2- \min {s}_2\right)=\left| \cos \theta \right| Range\left({s}_1\right)+\left| \sin \theta \right| Range\left({s}_2\right)\hfill \\ {}\hfill =\left| \cos \theta \right|K\times Range\left({s}_2\right)+\left| \sin \theta \right| Range\left({s}_2\right)= Rang\left({s}_2\right)\times \left(\left| \cos \theta \right|K+\left| \sin \theta \right|\right)\hfill \end{array} $$
(36)
$$ Range\left({z}_1\right)=\left| \cos \theta \right| Range\left({s}_1\right)+\left| \sin \theta \right|/ KRange\left({s}_1\right)= Range\left({s}_1\right)\times \left(\left| \cos \theta \right|+\left| \sin \theta \right|/K\right) $$
(37)

In a similar way, we write above equations for Range(z 2 ):

$$ Range\left({z}_2\right)=\left| \sin \theta \right|K\times Range\left({s}_2\right)+\left| \cos \theta \right| Range\left({s}_2\right)= Range\left({s}_2\right)\times \left(\left| \sin \theta \right|K+\left| \cos \theta \right|\right) $$
(38)
$$ Range\left({z}_2\right)=\left| \sin \theta \right| Range\left({s}_1\right)+\left| \cos \theta \right|/ KRange\left({s}_1\right)= Range\left({s}_1\right)\left(\left| \sin \theta \right|+\left| \cos \theta \right|/K\right) $$
(39)

By using (35),(36) and (38) we obtain:

$$ \begin{array}{l}K\ge 1\kern0.48em \Rightarrow Range\left({s}_1\right)> Range\left({s}_2\right)\Rightarrow \min \left( Range\left({s}_1\right), Range\left({s}_2\right)\right)= Range\left({s}_2\right)\\ {}\kern0.72em Range\left({z}_1\right)> Range\left({s}_2\right)\\ {}\kern0.6em Range\left({z}_2\right)> Range\left({s}_2\right)\\ {}\kern0.6em Range\left({z}_2\right), Range\left({z}_1\right)> Range\left({s}_2\right)= \min \left( Range\left({s}_1\right), Range\left({s}_2\right)\right)\end{array} $$
(40)

Also (35), (37) and (39) give us:

$$ \begin{array}{l}K<1\kern0.48em \Rightarrow Range\left({s}_1\right)< Range\left({s}_2\right)\Rightarrow \min \left( Range\left({s}_1\right), Range\left({s}_2\right)\right)= Range\left({s}_1\right)\\ {}\kern0.72em Range\left({z}_1\right)> Range\left({s}_1\right)\\ {}\kern0.6em Range\left({z}_2\right)> Range\left({s}_1\right)\\ {}\kern0.6em Range\left({z}_1\right), Range\left({z}_2\right)> Range\left({s}_1\right)= \min \left( Range\left({s}_1\right), Range\left({s}_2\right)\right)\end{array} $$
(41)

So we can write for every K:

$$ \begin{array}{l} Range\left({z}_1\right), Range\left({z}_2\right)> \min \left( Range\left({s}_1\right), Range\left({s}_2\right)\right)\\ {} \min \left( Range\left({z}_1\right), Range\left({z}_2\right)\right)> \min \left( Range\left({s}_1\right), Range\left({s}_2\right)\right)\end{array} $$
(42)

Proof of Proposition 3:

According to (12) and (13) for θ i  = − θ we have:

$$ \left[\begin{array}{c}\hfill {z}_1^{-\theta}\hfill \\ {}\hfill {z}_2^{-\theta}\hfill \end{array}\right]= rotate\left(-\theta \right) rotate\left(\theta \right)\left[\begin{array}{c}\hfill {s}_1\hfill \\ {}\hfill {s}_2\hfill \end{array}\right]=\left[\begin{array}{c}\hfill {s}_1\hfill \\ {}\hfill {s}_2\hfill \end{array}\right] $$
(43)

The Eq. (14) is true for every θ i so we can substitute it for − θ:

$$ \min \left({z}_{{}_1}^{\theta_0},{z}_2^{\theta_0}\right)\le \min \left({z}_1^{-\theta },{z}_2^{-\theta}\right)= \min \left({s}_1,{s}_2\right) $$
(44)

Also preposition 1 gives us:

$$ \min \left({z}_{{}_1}^{\theta_0},{z}_2^{\theta_0}\right)\ge \min \left({s}_1,{s}_2\right) $$
(45)

From (44) and (45) and sandwich theorem we obtain:

$$ \min \left({z}_{{}_1}^{\theta_0},{z}_2^{\theta_0}\right)= \min \left({s}_1,{s}_2\right) $$
(46)

Finally according to preposition 1 and (46) we can conclude that \( {\theta}_0=0,\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.,\pi, \raisebox{1ex}{$3\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right. \) or:

$$ \left\{\begin{array}{c}\hfill {z}_{{}_1}^{\theta_0}=\pm {s}_1\hfill \\ {}\hfill {z}_{{}_2}^{\theta_0}=\pm {s}_2\hfill \end{array}\kern0.96em or\kern0.6em \right.\left\{\begin{array}{c}\hfill {z}_{{}_1}^{\theta_0}=\pm {s}_2\hfill \\ {}\hfill {z}_{{}_2}^{\theta_0}=\pm {s}_1\hfill \end{array}\right. $$
(47)

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Modaghegh, H., Seyedin, S.A. A new fast and efficient active steganalysis based on combined geometrical blind source separation. Multimed Tools Appl 74, 5825–5843 (2015). https://doi.org/10.1007/s11042-014-1890-9

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