1. Introduction
 (1)
 The generation of Latin cubes in this algorithm is independent of plain image.
 (2)
 When any one bit in the plain image changes, the corresponding number of bits in the cipher image follows the change with obvious regularity.
2. Description of an Image Chaotic Encryption Algorithm
2.1. A Brief View of an Image Chaotic Encryption Algorithm
2.2. Logistic Map
2.3. Generation of Latin Cubes
Algorithm 1 Steps for Generation of Latin Cubes. 
Input: Secret keys ${key}_{0},\phantom{\rule{4.pt}{0ex}}{\mu}_{0},\phantom{\rule{4.pt}{0ex}}\alpha ,\phantom{\rule{4.pt}{0ex}}\beta ,\phantom{\rule{4.pt}{0ex}}\gamma $; Side length $q=\sqrt[3]{8\times w\times h}$; Output: Three orthogonal Latin cubes ${L}_{1}$, ${L}_{2}$ and ${L}_{3}$;

2.4. Steps for Image Chaotic Encryption
Algorithm 2 Steps for Image Chaotic Encryption. 
Input: Secret keys ${key}_{0},\phantom{\rule{4.pt}{0ex}}{\mu}_{0},\phantom{\rule{4.pt}{0ex}}\alpha ,\phantom{\rule{4.pt}{0ex}}\beta ,\phantom{\rule{4.pt}{0ex}}\gamma $; Plaintext image P; Output: Ciphertxet image E;

3. Security Analysis
3.1. Analysis of Chaotic Index Sequence $lx$
3.1.1. Relation between the FirstScrambling Image ${S}_{1}$ and the Plain Image M
3.1.2. The First Case for Analysis of Chaotic Index Sequence $lx$
3.1.3. The Second Case for Analysis of Chaotic Index Sequence $lx$
 (1)
 If ${t}_{1}^{\prime}<t<{t}_{2}^{\prime}$, then one has$$\left\{\begin{array}{c}{t}_{2}^{\prime}=({m}_{{k}_{1}^{\prime}{k}_{2}^{\prime},0}+{m}_{{k}_{1}^{\prime}{k}_{2}^{\prime}k,0}+t)/2={A}_{1},\hfill \\ {t}_{1}^{\prime}=({m}_{{k}_{1}^{\prime}{k}_{2}^{\prime},0}+{m}_{{k}_{1}^{\prime}{k}_{2}^{\prime}k,0}+t)/2={B}_{1}.\hfill \end{array}\right.$$
 (2)
 If ${t}_{1}^{\prime}>t>{t}_{2}^{\prime}$, then one has$$\left\{\begin{array}{c}{t}_{1}^{\prime}=({m}_{{k}_{1}^{\prime}{k}_{2}^{\prime},0}+{m}_{{k}_{1}^{\prime}{k}_{2}^{\prime}k,0}+t)/2={A}_{1},\hfill \\ {t}_{2}^{\prime}=({m}_{{k}_{1}^{\prime}{k}_{2}^{\prime},0}+{m}_{{k}_{1}^{\prime}{k}_{2}^{\prime}k,0}+t)/2={B}_{1}.\hfill \end{array}\right.$$
 (3)
 If ${t}_{2}^{\prime}<t<{t}_{3}^{\prime}$, then one has$$\left\{\begin{array}{c}{t}_{3}^{\prime}=({m}_{{k}_{2}^{\prime}{k}_{3}^{\prime},0}+{m}_{{k}_{2}^{\prime}{k}_{3}^{\prime}k,0}+t)/2={A}_{2},\hfill \\ {t}_{2}^{\prime}=({m}_{{k}_{2}^{\prime}{k}_{3}^{\prime},0}+{m}_{{k}_{2}^{\prime}{k}_{3}^{\prime}k,0}+t)/2={B}_{2}.\hfill \end{array}\right.$$
 (4)
 If ${t}_{2}^{\prime}>t>{t}_{3}^{\prime}$, then one has$$\left\{\begin{array}{c}{t}_{2}^{\prime}=({m}_{{k}_{2}^{\prime}{k}_{3}^{\prime},0}+{m}_{{k}_{2}^{\prime}{k}_{3}^{\prime}k,0}+t)/2={A}_{2},\hfill \\ {t}_{3}^{\prime}=({m}_{{k}_{2}^{\prime}{k}_{3}^{\prime},0}+{m}_{{k}_{2}^{\prime}{k}_{3}^{\prime}k,0}+t)/2={B}_{2}.\hfill \end{array}\right.$$
3.2. Analysis of Secret Keys $\alpha $, $\beta $, $\gamma $
3.3. Flowchart of Security Analysis
4. Steps for Deciphering the Image Chaotic Encryption Algorithm
Algorithm 3 Steps for Deciphering Image Chaotic Encryption Algorithm. 
Output: The equivalent secret keys $lx,\phantom{\rule{4.pt}{0ex}}\alpha ,\phantom{\rule{4.pt}{0ex}}\beta ,\phantom{\rule{4.pt}{0ex}}\gamma $;

5. Numerical Simulation Experiments
6. Suggestions for Improvement
7. Conclusions
Author Contributions
Funding
Conflicts of Interest
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