A new image/video encryption scheme based on fractional discrete Tchebichef transform and singular value decomposition

Abstract

One of the major application areas of the fractional-order discrete transform (FrDTs) is in signal and information security, particularly in signal and image/video encryption. Recently, many researchers proposed techniques that implemented not only the fractional transforms, but also various randomized versions of the FrDTs, which add more security features to signal’s encryption. In this paper, we propose a new image/video encryption scheme based on fractional-order discrete Tchebichef transform (FrDTT) using singular value decomposition. The FrDTTs are derived algebraically using the spectral decomposition of discrete Tchebichef polynomials, then the singular value decomposition technique in order to build a basic set of orthonormal eigenvectors which help to develop FrDTTs. Finally, we implement and apply the scheme proposed in this paper for encrypting test images and video sequences. Moreover, we methodically perform the security evaluation in terms of brute force and statistical attacks as well as comparisons with the existing methods in terms of secret key sensitivity and space. The promising experiment results demonstrate the effectiveness and efficiency of our proposed FrDTTs based image encryption techniques.

Appendices

Appendix A

Proof of property (v)

Let λ be an eigenvalue of Tchebichef polynomial matrix and u the corresponding eigenvector, then Cu = λu, using the properties (iii) and (iv), we have:

$$ u=\mathrm{CC}u=\lambda \mathrm{C}u={\lambda}^2u $$

(A1)

thus. {Key1, Key2}

$$ \left({\lambda}^2-1\right)u=0 $$

(A2)

The matrix C has only two eigenvalues {1, −1}, The proof of Eq. (19) has been completed

Appendix B

Proof of properties (a)-(b)

Let C ∈ ℂ^N × Nbe Tchebichef polynomial matrix, with their eigenvalues on the diagonal of a diagonal matrix Λ =  diag (λ₁, ……λ_N) ∈ ℂ^N × N and the corresponding eigenvectors forming the columns of a matrix V = [u₁, ……, u_N] ∈ ℂ^N × N, we have:

$$ \mathrm{C}=V\Lambda {V}^{-1} $$

(B.1)

where, the orthonormal vectors u₁, ……u_N are eigenvectors of C, corresponding to eigenvalues λ₁, ……λ_N.

$$ \mathrm{C}=\left[{u}_1\cdots \cdots {u}_N\right]\left(\begin{array}{ccc}{\lambda}_1& & \\ {}& \ddots & \\ {}& & {\lambda}_N\end{array}\right)\left[\begin{array}{c}{u}_1^T\\ {}\vdots \\ {}{u}_N^T\end{array}\right]=\left[{\lambda}_1{u}_1\cdots \cdots {\lambda}_N{u}_N\right]\left[\begin{array}{c}{u}_1^T\\ {}\vdots \\ {}{u}_N^T\end{array}\right] $$

(B.2)

$$ \mathrm{C}={\lambda}_1{u}_1{u}_1^T\cdots \cdots {\lambda}_N{u}_N{u}_N^T $$

(B.3)

Notice that the matrices

$$ {\mathrm{P}}_j:= {u}_j{u}_j^T\in {\mathbb{C}}^{N\times N} $$

(B.4)

are orthogonal projectors, since ${\mathrm{P}}_j^T={\mathrm{P}}_j$and ${\mathrm{P}}_j^2:= {u}_j\left({u}_j^T{u}_j\right){u}_j^T={u}_j{u}_j^T={\mathrm{P}}_j$

$$ \mathrm{C}=\sum \limits_{j=1}^N{\lambda}_j{\mathrm{P}}_j $$

(B.5)

if j ≠ k, then the orthogonality of the eigenvectors implies

$$ {\mathrm{P}}_j{\mathrm{P}}_k={u}_j{u}_j^T{u}_k{u}_k^T=0 $$

(B.6)

The proof of properties (a) and (b) has been completed.

Proof of propriety (c)

Let γ, η and λ be respectively the eigenvalues of the matrices P₀,P₁ and C of size N × N, using Eq. (18), we have:

$$ {\displaystyle \begin{array}{c}\mid \gamma I-{\mathrm{P}}_0\mid =\mid \gamma I-0.5\left(\mathrm{C}+I\right)\mid =\mid \left(\gamma -0.5\right)I-0.5\mathrm{C}\mid \\ {}=\mid \left(\gamma -0.5\right)I-0.5\mathrm{C}\mid ={0.5}^N\mid \left(2\gamma -1\right)I-\mathrm{C}\mid \\ {}=0\end{array}} $$

(B.7)

Similarly, we have

$$ \mid \eta I-{\mathrm{P}}_1\mid ={0.5}^N\mid \left(2\eta -1\right)I-\mathrm{C}\mid =0,\boldsymbol{and}\mid \lambda I-\mathrm{C}\mid =0 $$

(B.8)

From (B.7) and (B.8), we have

$$ 2\gamma -1=\lambda, \boldsymbol{and}-\left(2\eta -1\right)=\lambda $$

(B.9)

Hence, if λ = 1, there is γ = 1, η = 0, and if λ =  − 1, then γ = 0, η = 1.

The proof of propriety (c) has been completed.

Appendix C

Let P₀ and P₁ the spectral projection matrices of Tchebichef polynomial matrix C ∈ ℂ^N × N, and u,v be their eigenvectors corresponding to λ = 1, respectively.

Proof of Lemma 1

From Table 1 and Property (c), we have:

$$ {\mathrm{P}}_0u=u,\boldsymbol{and}\ {\mathrm{P}}_1v=v $$

(C.1)

using (C.1) and Property (b), we have:

$$ {u}^Tv={\left({\mathrm{P}}_0u\right)}^T\left({\mathrm{P}}_1v\right)={u}^T{{\mathrm{P}}_0}^T{\mathrm{P}}_1v=0 $$

(C.2)

The proof of Lemma 1 has been completed.

Proof of Lemma 2

From Eq. (15), Lemma 1 and property (b) we have:

$$ {\displaystyle \begin{array}{c}\mathrm{C}u=\left({\lambda}_0{\mathrm{P}}_0+{\lambda}_1{\mathrm{P}}_1\right)u={\lambda}_0{\mathrm{P}}_0u+{\lambda}_1{\mathrm{P}}_1u\\ {}={\lambda}_0{\mathrm{P}}_0u+{\lambda}_1{\mathrm{P}}_1{\mathrm{P}}_0u={\lambda}_0{\mathrm{P}}_0u+{\lambda}_1{\mathrm{P}}_1^T{\mathrm{P}}_0u\\ {}={\lambda}_0{\mathrm{P}}_0u={\lambda}_0u\end{array}} $$

(C.3)

$$ {\displaystyle \begin{array}{c}\mathrm{C}v=\left({\lambda}_0{\mathrm{P}}_0+{\lambda}_1{\mathrm{P}}_1\right)v={\lambda}_0{\mathrm{P}}_0v+{\lambda}_1{\mathrm{P}}_1v\\ {}={\lambda}_0{\mathrm{P}}_0{\mathrm{P}}_1v+{\lambda}_1{\mathrm{P}}_1v={\lambda}_1{\mathrm{P}}_1v={\lambda}_1v\end{array}} $$

(C.4)

The proof of Lemma 2 has been completed.