Quantum image encryption algorithm based on generalized Arnold transform and Logistic map

Abstract

In the era of big data, image security and real-time processing become more and more important and increasingly difficult to satisfy. To improve the security and processing efficiency of image encryption algorithm, an enhanced quantum scheme is proposed for generalized novel enhanced quantum image representation. The proposed quantum encryption scheme mainly consists of two-stage operation in order, i.e., twice scrambling based generalized Arnold transform and pixel encryption based on the quantum key image (which are generated and prepared based on Logistic map). In the first stage, generalized Arnold transform are employed to simultaneously disturb the coordinate information and pixel gray value of quantum plain image. Following that, the scrambled image is further encrypted into a quantum cipher image based on quantum key image, which is divided into three sub-processes in detail, i.e., CNOT operations, bit-plane scrambling and controlled perfect shuffle permutations are executed orderly. The quantum image decryption process can be easily implemented in a reverse way. The complete quantum circuit implementation for above two stages operation is constructed and analyzed in terms of quantum cost and time complexity. Compared to classical image processing algorithm, the investigated quantum encryption algorithm demonstrates an exponential speedup with computational cost of ${\rm O}\left( n \right)$ for a $2^{n} \times 2^{n}$ quantum grayscale or color images. The proposed scheme is simulated and verified on a classical computer with MATLAB environments, i.e., not in a real quantum version that not considers the effects of quantum noise. Experimental results and numerical analysis indicate that the presented quantum algorithm has good visual effects and high security.

Appendix 1

For two n-qubit numbers A and B, $\left( {A - B} \right)\;\bmod 2^{n}$ can be expressed as:

$$\left( {A - B} \right)\;\bmod 2^{n} = \left\{ \begin{gathered} A - B,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;A \ge B \hfill \\ 2^{n} - (B - A),\;\;\;\;\;A < B \hfill \\ \end{gathered} \right..$$

Proof

1.1 It is obvious that $\left( {A - B} \right)\;\bmod 2^{n} = A - B$ when $A \ge B$.

1.2 When $A < B$, it can be verified as follows:

Assume that $\overline{B} = \overline{b}_{n - 1} \overline{b}_{n - 2} \cdot \cdot \cdot \overline{b}_{1} \overline{b}_{0}$ denotes the inverse code of B, then we can obtain that:

$$\begin{gathered} B + \overline{B} \hfill \\ = b_{n - 1} \cdot 2^{n - 1} + b_{n - 2} \cdot 2^{n - 2} + \cdot \cdot \cdot + b_{1} \cdot 2^{1} + b_{0} \cdot 2^{0} { + }\overline{{b_{n - 1} }} \cdot 2^{n - 1} \hfill \\ \;\; + \overline{{b_{n - 2} }} \cdot 2^{n - 2} + \cdot \cdot \cdot + \overline{{b_{1} }} \cdot 2^{1} + \overline{{b_{0} }} \cdot 2^{0} \hfill \\ { = }\left( {b_{n - 1} { + }\overline{{b_{n - 1} }} } \right) \cdot 2^{n - 1} + \left( {b_{n - 2} { + }\overline{{b_{n - 2} }} } \right) \cdot 2^{n - 2} + \cdot \cdot \cdot \hfill \\ \;\;\; + \left( {b_{1} { + }\overline{{b_{1} }} } \right) \cdot 2^{1} + \left( {b_{0} { + }\overline{{b_{0} }} } \right) \cdot 2^{0} \hfill \\ = 2^{n - 1} + 2^{n - 2} + \cdot \cdot \cdot + 2^{1} + 2^{0} \hfill \\ { = }2^{n} - 1. \hfill \\ \end{gathered}$$

Thus, $- B = (\overline{B} + 1) - 2^{n} ,\;\overline{B} + 1 = 2^{n} - B$, and $(A - B)\bmod 2^{n}$ can be deduced as follows:

$$\begin{gathered} \left( {A - B} \right)\bmod 2^{n} \hfill \\ = \left[ {A{ + }\left( { - B} \right)} \right]\bmod 2^{n} = \left[ {A + \left( {\overline{B} + 1} \right) - 2^{n} } \right]\bmod 2^{n} \hfill \\ = \left[ {A + \left( {\overline{B} + 1} \right)} \right]\bmod 2^{n} { = }\left[ {A + \left( {2^{n} - B} \right)} \right]\bmod 2^{n} \hfill \\ = \left[ {2^{n} - \left( {B - A} \right)} \right]\bmod 2^{n} . \hfill \\ \end{gathered}$$

Since $A < B \Rightarrow \left[ {2^{n} - \left( {B - A} \right)} \right] < 2^{n}$, we can obtain that:

$$\left[ {2^{n} - \left( {B - A} \right)} \right]\bmod 2^{n} = 2^{n} - \left( {B - A} \right),\;A < B.\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\square$$