A Powerful Zero-Watermarking Algorithm for Copyright Protection of Color Images Based on Quaternion Radial Fractional Hahn Moments and Artificial Bee Colony Algorithm

Abstract

In this paper, we propose a new zero-watermarking algorithm for copyright protection of color image based on a new type of fractional-order quaternion moments called quaternion radial fractional Hahn moments (QRFrHMs), the artificial bee colony (ABC) algorithm, and a chaotic system that uses a mixed linear–nonlinear coupling based on two-dimensional CML (2DCML). The proposed QRFrHMs are defined by the projection of the color image on an orthogonal basis formed by a quaternion circular function and the fractional version of discrete orthogonal Hahn polynomials which depends on an additional fractional parameter $\alpha \in {\mathbb{R}}$. The proposed zero-watermarking algorithm computes the QRFrHMs of the original image and then uses the ABC algorithm to choose the optimal parameters of the QRFrHMs. After that, the QRFrHMs with optimal parameters are used to construct a feature image called zero-watermark. The zero-watermark is used later to verify the copyright of the protected color image in a blind way. The combination of QRFrHMs and ABC algorithm gives the proposed zero-watermarking algorithm a high robustness with a bit error rate (BER) less than 0.03 for various attacks. In addition, the proposed zero-watermarking algorithm uses the chaotic 2DCML system to improve the security requirement so that only authorized persons can verify the copyright of the protected color image. The experimental results indicate the superiority of the proposed zero-watermarking algorithm.

Appendix

In this appendix, we establish the relationship between the proposed QRFrHMs and RFrHMs. We interest in this development only in the right-side QRFrHMs where the results found can also be obtained in the same way for left-side QRFrHMs.

Let $f_{{\text{R}}} (x,y)$, $f_{{\text{G}}} (x,y)$ and $f_{{\text{B}}} (x,y)$ be the red, green and blue channels of the color image $f(x,y)$, respectively, and $M_{n,m}^{\alpha } (f_{{\text{R}}} )$, $M_{n,m}^{\alpha } (f_{{\text{G}}} )$ and $M_{n,m}^{\alpha } (f_{{\text{B}}} )$ its RFrHMs, respectively, which can be calculated from Eq. (23). Let $\Re (x)$ and $\Im (x)$ represent the real and imaginary parts of number $x$, respectively.

Using Eqs. (13), (14) and (23), we have:

$$ \begin{aligned} {}^{R}M_{n,m}^{\alpha } & = \frac{1}{l}\sum\limits_{r = 0}^{{\left\lfloor {N/2} \right\rfloor - 1}} {\sum\limits_{k = 0}^{l - 1} {\left[ {f_{R} (r,\theta_{k} )i + f_{G} (r,\theta_{k} )j + f_{B} (r,\theta_{k} )k} \right]\gamma_{r,n} e^{{ - \mu m\theta_{k} }} } } \\ & = i\frac{1}{l}\sum\limits_{r = 0}^{{\left\lfloor {N/2} \right\rfloor - 1}} {\sum\limits_{k = 0}^{l - 1} {f_{R} (r,\theta_{k} )} } \gamma_{r,n} e^{{ - \mu m\theta_{k} }}\\&\quad + j\frac{1}{l}\sum\limits_{r = 0}^{{\left\lfloor {N/2} \right\rfloor - 1}} {\sum\limits_{k = 0}^{l - 1} {f_{G} (r,\theta_{k} )} } \gamma_{r,n} e^{{ - \mu m\theta_{k} }} \\&\quad+ k\frac{1}{l}\sum\limits_{r = 0}^{{\left\lfloor {N/2} \right\rfloor - 1}} {\sum\limits_{k = 0}^{l - 1} {f_{B} (r,\theta_{k} )} } \gamma_{r,n} e^{{ - \mu m\theta_{k} }} \\ {}^{R}M_{n,m}^{\alpha } & = i\left[ {\frac{1}{l}\sum\limits_{r = 0}^{{\left\lfloor {N/2} \right\rfloor - 1}} {\sum\limits_{k = 0}^{l - 1} {f_{R} (r,\theta_{k} )} } \gamma_{r,n} \cos (m\theta_{k} )}\right.\\&\quad\left. {- \mu \frac{1}{l}\sum\limits_{r = 0}^{{\left\lfloor {N/2} \right\rfloor - 1}} {\sum\limits_{k = 0}^{l - 1} {f_{R} (r,\theta_{k} )} } \gamma_{r,n} \sin (m\theta_{k} )} \right] \\ & \quad + \;j\left[ {\frac{1}{l}\sum\limits_{r = 0}^{{\left\lfloor {N/2} \right\rfloor - 1}} {\sum\limits_{k = 0}^{l - 1} {f_{G} (r,\theta_{k} )} } \gamma_{r,n} \cos (m\theta_{k} ) }\right.\\&\quad\left. {- \mu \frac{1}{l}\sum\limits_{r = 0}^{{\left\lfloor {N/2} \right\rfloor - 1}} {\sum\limits_{k = 0}^{l - 1} {f_{G} (r,\theta_{k} )} } \gamma_{r,n} \sin (m\theta_{k} )} \right] \\ & \quad + \;k\left[ {\frac{1}{l}\sum\limits_{r = 0}^{{\left\lfloor {N/2} \right\rfloor - 1}} {\sum\limits_{k = 0}^{l - 1} {f_{B} (r,\theta_{k} )} } \gamma_{r,n} \cos (m\theta_{k} )}\right.\\&\quad\left. { - \mu \frac{1}{l}\sum\limits_{r = 0}^{{\left\lfloor {N/2} \right\rfloor - 1}} {\sum\limits_{k = 0}^{l - 1} {f_{B} (r,\theta_{k} )} } \gamma_{r,n} \sin (m\theta_{k} )} \right] \\ {}^{R}M_{n,m}^{\alpha } & = i(\Re (M_{n,m}^{\alpha } (f_{R} )) \\&\quad+ \frac{i + j + k}{{\sqrt 3 }}\Im (M_{n,m}^{\alpha } (f_{R} )) + j(\Re (M_{n,m}^{\alpha } (f_{G} )) + \frac{i + j + k}{{\sqrt 3 }}\Im (M_{n,m}^{\alpha } (f_{G} )) \\ & \quad + \;k(\Re (M_{n,m}^{\alpha } (f_{B} ))\\&\quad + \frac{i + j + k}{{\sqrt 3 }}\Im (M_{n,m}^{\alpha } (f_{B} )) \\ {}^{R}M_{n,m}^{\alpha } & = A_{n,m} + iB_{n,m} + jC_{n,m} + kD_{n,m} \\ \end{aligned} $$

(A1)

where

$$ \begin{aligned} A_{n,m} & = \frac{ - 1}{{\sqrt 3 }}[\Im (M_{n,m}^{\alpha } (f_{R} ) + \Im (M_{n,m}^{\alpha } (f_{G} ) + \Im (M_{n,m}^{\alpha } (f_{B} )] \\ B_{n,m} & = (\Re (M_{n,m}^{\alpha } (f_{R} )) + \frac{1}{\sqrt 3 }[\Im (M_{n,m}^{\alpha } (f_{G} ) - \Im (M_{n,m}^{\alpha } (f_{B} )) \\ C_{n,m} & = (\Re (M_{n,m}^{\alpha } (f_{G} )) + \frac{1}{\sqrt 3 }[\Im (M_{n,m}^{\alpha } (f_{B} ) - \Im (M_{n,m}^{\alpha } (f_{R} )) \\ D_{n,m} & = (\Re (M_{n,m}^{\alpha } (f_{B} )) + \frac{1}{\sqrt 3 }[\Im (M_{n,m}^{\alpha } (f_{R} ) - \Im (M_{n,m}^{\alpha } (f_{G} )) \\ \end{aligned} $$

(A2)

Equation (A1) shows that right-side QRFrHMs $\left\{ {{}^{R}M_{n,m}^{\alpha } } \right\}$ can be derived directly from RFrHMs $\left\{ {M_{n,m}^{\alpha } } \right\}$ of each color image channel.