Cryptanalysis of a plaintext-related chaotic RGB image encryption scheme using total plain image characteristics

Abstract

Recently, a novel plaintext-related RGB image encryption scheme has been proposed, where the security is strengthened by associating the initial condition and control parameter of one logistic map with total plain image characteristics. This paper points out its weakness that the secret location storing the total characteristics cannot adapt to different plain images. Accordingly, we propose a strategy to break this encryption scheme by applying chosen/known plaintext attacks, where the data complexity of attack is reduced to the minimum. The experimental results prove that all the secret matrices can be revealed effectively. Meanwhile, three corresponding improvements are given to guarantee its security, which help to design more secure plaintext-related cryptosystems in the future.

Appendix

Inequality that the equivalent pixel sum of the plain image P ₀ must satisfy is deduced as follows.

If V ₀ > V ₃, then V _0h  > V _3h > V _3l and V _0l ≥ V _3h > V _3l. And inequality (50) can be deduced by subtracting inequality (30) from (31) as follows:

$$ {V}_{0\mathrm{l}}-{V}_{3\mathrm{h}}<\left(F+{S}_0\times 1000\right)\kern0.5em \operatorname{mod}1-\left(F+{S}_3\times 1000\right)\kern0.5em \operatorname{mod}1<{V}_{0\mathrm{h}}-{V}_{3\mathrm{l}}, $$

(50)

where V _0h − V _3l ∈ (0, 1) and V _0l − V _3h ∈ [0, 1). As the lower and upper bounds range between 0 and 1, inequality (50) can be converted into the following:

$$ {V}_{0\mathrm{l}}-{V}_{3\mathrm{h}}<\left(\left(F+{S}_0\times 1000\right)\kern0.5em \operatorname{mod}1-\left(F+{S}_3\times 1000\right)\kern0.5em \operatorname{mod}1\right)\operatorname{mod}1<{V}_{0\mathrm{h}}-{V}_{3\mathrm{l}}. $$

(51)

In turn, we obtain

$$ {V}_{0\mathrm{l}}-{V}_{3\mathrm{h}}<\left(F+{S}_0\times 1000-F-{S}_3\times 1000\right)\kern0.5em \operatorname{mod}1<{V}_{0\mathrm{h}}-{V}_{3\mathrm{l}}. $$

(52)

Further, we have the following:

$$ {V}_{0\mathrm{l}}-{V}_{3\mathrm{h}}<\left({S}_0\times 1000-{S}_3\times 1000\right)\kern0.5em \operatorname{mod}1<{V}_{0\mathrm{h}}-{V}_{3\mathrm{l}}, $$

$$ \left({V}_0-{V}_3-1\right)/253<\left({S}_0\times 1000-{S}_3\times 1000\right)\kern0.5em \operatorname{mod}1<\left({V}_0-{V}_3+1\right)/253. $$

(53)

If V ₀ < V ₃, we have the following:

$$ \left({V}_3-{V}_0-1\right)/253<\left({S}_3\times 1000-{S}_0\times 1000\right)\kern0.5em \operatorname{mod}1<\left({V}_3-{V}_0+1\right)/253. $$

(54)

From Eq. (21), we can replace the pixel sum at 10⁻¹⁵ decimal precision with that of the uint8 format. Let ${\tilde{S}}_0$ and ${\tilde{S}}_3$ be the sums of P ₀ ∈ [0, 255] and the known plain image C ₃ ∈ [0, 255], respectively.

In (53),

$$ {\displaystyle \begin{array}{l}\left({S}_0\times 1000-{S}_3\times 1000\right)\kern0.5em \operatorname{mod}1\\ {}=\left(\left({S}_0-{S}_3\right)\times 1000\right)\operatorname{mod}1\\ {}=\left(\left({\tilde{S}}_0-{\tilde{S}}_3\right)\times 1000/255\right)\operatorname{mod}1\\ {}=\left({\tilde{S}}_{0\_3}\times 200/51\right)\operatorname{mod}1\\ {}=\left(\left({\tilde{S}}_{0\_3}-51\times n\right)\times 200/51\right)\operatorname{mod}1\\ {}=\left({\tilde{\tilde{S}}}_{0\_3}\times 200/51\right) \operatorname {mod}1\kern1em \end{array}}, $$

(55)

where n is a non-negative integer, ${\tilde{S}}_{0\_3}={\tilde{S}}_0-{\tilde{S}}_3$.

$$ {\tilde{\tilde{S}}}_{0\_3}={\tilde{S}}_{0\_3}\operatorname{mod}51=\left({\tilde{S}}_0-{\tilde{S}}_3\right)\operatorname{mod}51=\left({\tilde{\tilde{S}}}_0-{\tilde{\tilde{S}}}_3\right)\operatorname{mod}51, $$

(56)

where ${\tilde{\tilde{S}}}_0={\tilde{S}}_0\operatorname{mod}51$ and ${\tilde{\tilde{S}}}_3={\tilde{S}}_3\operatorname{mod}51$ are the equivalent pixel sums of plain image P ₀ and the known plain image C ₃, respectively.

Then, (53) and (54) can be transformed into the following two inequalities:

$$ \left({V}_0-{V}_3-1\right)/253<\left(\left(\left({\tilde{\tilde{S}}}_0-{\tilde{\tilde{S}}}_3\right)\operatorname{mod}51\right)\times 200/51\right)\kern0.5em \operatorname{mod}1<\left({V}_0-{V}_3+1\right)/253. $$

(57)

$$ \left({V}_3-{V}_0-1\right)/253<\left(\left(\left({\tilde{\tilde{S}}}_3-{\tilde{\tilde{S}}}_0\right)\operatorname{mod}51\right)\times 200/51\right)\kern0.5em \operatorname{mod}1<\left({V}_3-{V}_0+1\right)/253. $$

(58)