# A Novel Image-Encryption Scheme Based on a Non-Linear Cross-Coupled Hyperchaotic System with the Dynamic Correlation of Plaintext Pixels

^{*}

## Abstract

**:**

## 1. Introduction

## 2. Non-Linear Cross-Coupled Chaotic Map

#### 2.1. Performance Evaluation of LF-NCHM Model

#### 2.1.1. Trajectory

#### 2.1.2. Bifurcation Diagram

#### 2.1.3. Lyapunov Exponent

#### 2.1.4. Permutation Entropy

## 3. The Image-Encryption Algorithm Related to Plaintext Pixels in the Scrambling Process

#### 3.1. The Generation of Keys and Hyperchaotic Sequences

- Use the image file to generate hash values, Obtain the K = k${}_{1}$, k${}_{2}$, ……, k${}_{32}$. They are composed of 8 bits of data.
- Calculate the intermediate variables h${}_{1}$, h${}_{2}$, h${}_{3}$, h${}_{4}$ through k${}_{1}$, k${}_{2}$, ……, k${}_{32}$. The calculation rules are shown in Formula (3).$$\left\{\begin{array}{c}{h}_{1}=\frac{\left({k}_{25}\oplus {k}_{26}\oplus \dots \oplus {k}_{32}\right)}{256}\hfill \\ {h}_{2}={h}_{1}+\frac{\left({k}_{1}\oplus {k}_{2}\oplus \dots \oplus {k}_{8}\right)}{256}\hfill \\ {h}_{3}={h}_{2}+\frac{\left({k}_{17}\oplus {k}_{18}\oplus \dots \oplus {k}_{24}\right)}{256}\hfill \\ h4={h}_{3}+\frac{\left({k}_{9}\oplus {k}_{10}\oplus \dots \oplus {k}_{16}\right)}{256}.\hfill \end{array}\right.$$
- Calculate the initial values $\mu $, $\lambda $, x${}_{0}$ and y${}_{0}$ according to h${}_{1}$, h${}_{2}$, h${}_{3}$ and h${}_{4}$. The calculation rules are shown in Formula (4).$$\left\{\begin{array}{c}\mu =mod\left({\mu}_{0}^{{}^{\prime}}+\frac{\left({h}_{1}+{h}_{2}\right)}{256},4\right)\hfill \\ \lambda =mod\left({\lambda}_{0}^{{}^{\prime}}+\frac{\left({h}_{2}+{h}_{3}\right)}{256},6\right)\hfill \\ {x}_{0}=mod\left({x}_{0}^{{}^{\prime}}+\frac{\left({h}_{3}+{h}_{4}\right)\ast {10}^{14}}{256},1\right)\hfill \\ {y}_{0}=mod\left({y}_{0}^{{}^{\prime}}+\frac{\left({h}_{4}+{h}_{1}\right)\ast {10}^{14}}{256},1\right)\hfill \end{array}\right.$$
- Bring the initial conditions generated by the hash algorithm into LF-NCHM to produce a sequence of the required length.

#### 3.2. A Scrambling Method Based on Pixel Values

- Convert the plaintext image Fig with the size of M*N into the one-dimensional vector P(i), and the size of the one-dimensional vector is M*N.
- According to the dynamic secret key K related to the plaintext and hyperchaotic system LF-NCHM mentioned above, generate a one-dimensional key stream E(i) with the size of M*N.
- Sort E(i), and a one-dimensional vector of position sequence number with the size of M*N is obtained. The vector of the position sequence number is named S(i). An example of this rule is shown in Figure 8.
- Calculate the sum of the pixel values and index for the plaintext image. index = sum mod 2.
- Use the MATLAB library function zeros to produce a one-dimensional vector C(i) of size M*N, which is used to save the pixel scrambling results.
- Judge the index. If the index is equal to zero, select the location information from the left side of $S\left(i\right)$, $C\left(1\right)=P\left(S\right(1\left)\right)$. If the index is equal to one, choose to take the location information from the right side of $S\left(i\right)$, $C\left(1\right)=P(S$(M*N)). Figure 9 shows this operation.
- Calculate index. $index=C(i-1)$ mod 2, i ≥ 2. If the index is equal to zero, select the location information from the left side of $S\left(i\right)$. $C\left(i\right)=P\left(S\right(left\left)\right)$, and left represents the latest sequence number of location sequence from the left. If index is equal to one, select the location information from the right side of $S\left(i\right)$. $C\left(i\right)=P\left(S\right(right\left)\right)$, right represents the latest sequence number of the position sequence starting from the right side. Repeat the operation until the assignment of C(M*N) is completed.
- Restore the one-dimensional vector $C\left(i\right)$ to the two-dimensional matrix $C(i,j)$ with the size of M*N, and the scrambling process is completed.

#### 3.3. Diffusion Process

- According to the generation rules of the key and hyperchaotic sequences, generate sequences $x\left(i\right),y\left(i\right)$ with the size of M*N. The sequences $x\left(i\right),y\left(i\right)$ are transformed into two-dimensional matrixes $x(i,j),y(i,j)$ of size M*N.
- Diffuse the scrambled image $C(i,j)$. First, from the lower right corner of the image, from right to left, from bottom to top. The current pixel is associated with the adjacent pixels on the right and bottom, and the intermediate image P1 is obtained. The specific operations are as follows:$for\phantom{\rule{3.33333pt}{0ex}}i\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}M:\phantom{\rule{3.33333pt}{0ex}}-1:\phantom{\rule{3.33333pt}{0ex}}1$$for\phantom{\rule{3.33333pt}{0ex}}j\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}N:\phantom{\rule{3.33333pt}{0ex}}-1:\phantom{\rule{3.33333pt}{0ex}}1$$if\phantom{\rule{3.33333pt}{0ex}}i\phantom{\rule{3.33333pt}{0ex}}==\phantom{\rule{3.33333pt}{0ex}}M\phantom{\rule{3.33333pt}{0ex}}\&\&\phantom{\rule{3.33333pt}{0ex}}j\phantom{\rule{3.33333pt}{0ex}}==\phantom{\rule{3.33333pt}{0ex}}N$$P1(i,\phantom{\rule{3.33333pt}{0ex}}j)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}C(i,\phantom{\rule{3.33333pt}{0ex}}j)\oplus floor\left(y\right(i,\phantom{\rule{3.33333pt}{0ex}}j)\phantom{\rule{3.33333pt}{0ex}}\ast \phantom{\rule{3.33333pt}{0ex}}255)$$elseif\phantom{\rule{3.33333pt}{0ex}}i\phantom{\rule{3.33333pt}{0ex}}==\phantom{\rule{3.33333pt}{0ex}}M$$P1(i,\phantom{\rule{3.33333pt}{0ex}}j)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}C(i,\phantom{\rule{3.33333pt}{0ex}}j)\oplus P1(i,\phantom{\rule{3.33333pt}{0ex}}j+1)\oplus floor\left(y\right(i,\phantom{\rule{3.33333pt}{0ex}}j)\phantom{\rule{3.33333pt}{0ex}}\ast \phantom{\rule{3.33333pt}{0ex}}255)$$elseif\phantom{\rule{3.33333pt}{0ex}}j\phantom{\rule{3.33333pt}{0ex}}==\phantom{\rule{3.33333pt}{0ex}}N$$P1(i,\phantom{\rule{3.33333pt}{0ex}}j)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}C(i,\phantom{\rule{3.33333pt}{0ex}}j)\oplus P1(i+1,\phantom{\rule{3.33333pt}{0ex}}j)\oplus floor\left(y\right(i,\phantom{\rule{3.33333pt}{0ex}}j)\phantom{\rule{3.33333pt}{0ex}}\ast \phantom{\rule{3.33333pt}{0ex}}255)$$else$$P1(i,\phantom{\rule{3.33333pt}{0ex}}j)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}C(i,\phantom{\rule{3.33333pt}{0ex}}j)\oplus P1(i,\phantom{\rule{3.33333pt}{0ex}}j+1)\oplus P1(i+1,\phantom{\rule{3.33333pt}{0ex}}j)\oplus floor\left(y\right(i,\phantom{\rule{3.33333pt}{0ex}}j)\phantom{\rule{3.33333pt}{0ex}}\ast \phantom{\rule{3.33333pt}{0ex}}255)$ where ⊕ indicates that binary values corresponding to two numbers are exclusive or by bits, and $floor\left(x\right)$ indicates the largest integer not greater than x. P1 is the first round of diffusion of encrypted images.Here we use four pixels as an example to give the calculation method. Figure 10 shows the operation process.$P1(2,2)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}C(2,2)\oplus floor\left(y\right(2,2)\phantom{\rule{3.33333pt}{0ex}}\ast \phantom{\rule{3.33333pt}{0ex}}255)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}60$$P1(2,1)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}C(2,1)\oplus P1(2,2)\oplus floor\left(y\right(2,1)\phantom{\rule{3.33333pt}{0ex}}\ast \phantom{\rule{3.33333pt}{0ex}}255)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}58$$P1(1,2)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}C(1,2)\oplus P1(2,2)\oplus floor\left(y\right(1,2)\phantom{\rule{3.33333pt}{0ex}}\ast \phantom{\rule{3.33333pt}{0ex}}255)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}35$$P1(1,1)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}C(1,1)\oplus P1(1,2)\oplus P1(2,1)\oplus floor\left(y\right(1,1)\phantom{\rule{3.33333pt}{0ex}}\ast \phantom{\rule{3.33333pt}{0ex}}255)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}4$
- Carry out the second round of diffusion on P1. Starting from the upper left corner of the image, from left to right, and from top to bottom, the current pixel is associated with the adjacent pixels on the left and upper sides to obtain the encrypted image P2. The specific operations are as follows:$for\phantom{\rule{3.33333pt}{0ex}}i\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}1:\phantom{\rule{3.33333pt}{0ex}}1:\phantom{\rule{3.33333pt}{0ex}}M$$for\phantom{\rule{3.33333pt}{0ex}}j\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}1:\phantom{\rule{3.33333pt}{0ex}}1:\phantom{\rule{3.33333pt}{0ex}}N$$if\phantom{\rule{3.33333pt}{0ex}}i\phantom{\rule{3.33333pt}{0ex}}==\phantom{\rule{3.33333pt}{0ex}}1\phantom{\rule{3.33333pt}{0ex}}\&\&\phantom{\rule{3.33333pt}{0ex}}j\phantom{\rule{3.33333pt}{0ex}}==\phantom{\rule{3.33333pt}{0ex}}1$$P2(i,\phantom{\rule{3.33333pt}{0ex}}j)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}P1(i,\phantom{\rule{3.33333pt}{0ex}}j)\oplus floor\left(x\right(i,\phantom{\rule{3.33333pt}{0ex}}j)\phantom{\rule{3.33333pt}{0ex}}\ast \phantom{\rule{3.33333pt}{0ex}}255)$$elseif\phantom{\rule{3.33333pt}{0ex}}i\phantom{\rule{3.33333pt}{0ex}}==\phantom{\rule{3.33333pt}{0ex}}1$$P2(i,\phantom{\rule{3.33333pt}{0ex}}j)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}P1(i,\phantom{\rule{3.33333pt}{0ex}}j)\oplus P2(i,\phantom{\rule{3.33333pt}{0ex}}j-1)\oplus floor\left(x\right(i,\phantom{\rule{3.33333pt}{0ex}}j)\phantom{\rule{3.33333pt}{0ex}}\ast \phantom{\rule{3.33333pt}{0ex}}255)$$elseif\phantom{\rule{3.33333pt}{0ex}}j\phantom{\rule{3.33333pt}{0ex}}==\phantom{\rule{3.33333pt}{0ex}}1$$P2(i,\phantom{\rule{3.33333pt}{0ex}}j)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}P1(i,\phantom{\rule{3.33333pt}{0ex}}j)\oplus P2(i-1,\phantom{\rule{3.33333pt}{0ex}}j)\oplus floor\left(x\right(i,\phantom{\rule{3.33333pt}{0ex}}j)\phantom{\rule{3.33333pt}{0ex}}\ast \phantom{\rule{3.33333pt}{0ex}}255)$$else$$P2(i,\phantom{\rule{3.33333pt}{0ex}}j)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}P1(i,\phantom{\rule{3.33333pt}{0ex}}j)\oplus P2(i,\phantom{\rule{3.33333pt}{0ex}}j-1)\oplus P2(i-1,\phantom{\rule{3.33333pt}{0ex}}j)\oplus floor\left(x\right(i,\phantom{\rule{3.33333pt}{0ex}}j)\phantom{\rule{3.33333pt}{0ex}}\ast \phantom{\rule{3.33333pt}{0ex}}255)$This process is the reverse of Step 2.

#### 3.4. Decryption Process

**Input.**Security key K ={μ${}_{01}$, λ${}_{01}$, x${}_{01}$, y${}_{01}$, μ${}_{02}$, λ${}_{02}$, x${}_{02}$, y${}_{02}$} and encrypted image P2.

- Using key K, the hyperchaotic sequences $x\left(i\right),y\left(i\right)$ for inverse diffusion and the hyperchaotic sequence $E\left(i\right)$ for the inverse scrambling are obtained.
- Through $x\left(i\right)$, the first round of reverse diffusion is carried out, and the result is P3. The method is as follows.$for\phantom{\rule{3.33333pt}{0ex}}i\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}M:\phantom{\rule{3.33333pt}{0ex}}-1:\phantom{\rule{3.33333pt}{0ex}}1$$for\phantom{\rule{3.33333pt}{0ex}}j\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}N:\phantom{\rule{3.33333pt}{0ex}}-1:\phantom{\rule{3.33333pt}{0ex}}1$$if\phantom{\rule{3.33333pt}{0ex}}i\phantom{\rule{3.33333pt}{0ex}}==\phantom{\rule{3.33333pt}{0ex}}1\phantom{\rule{3.33333pt}{0ex}}\&\&\phantom{\rule{3.33333pt}{0ex}}j\phantom{\rule{3.33333pt}{0ex}}==\phantom{\rule{3.33333pt}{0ex}}1$$P3(i,\phantom{\rule{3.33333pt}{0ex}}j)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}P2(i,\phantom{\rule{3.33333pt}{0ex}}j)\oplus floor\left(x\right(i,\phantom{\rule{3.33333pt}{0ex}}j)\phantom{\rule{3.33333pt}{0ex}}\ast \phantom{\rule{3.33333pt}{0ex}}255)$$elseif\phantom{\rule{3.33333pt}{0ex}}i\phantom{\rule{3.33333pt}{0ex}}==\phantom{\rule{3.33333pt}{0ex}}1$$P3(i,\phantom{\rule{3.33333pt}{0ex}}j)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}P2(i,\phantom{\rule{3.33333pt}{0ex}}j)\oplus P2(i,\phantom{\rule{3.33333pt}{0ex}}j-1)\oplus floor\left(x\right(i,\phantom{\rule{3.33333pt}{0ex}}j)\phantom{\rule{3.33333pt}{0ex}}\ast \phantom{\rule{3.33333pt}{0ex}}255)$$elseif\phantom{\rule{3.33333pt}{0ex}}j\phantom{\rule{3.33333pt}{0ex}}==\phantom{\rule{3.33333pt}{0ex}}1$$P3(i,\phantom{\rule{3.33333pt}{0ex}}j)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}P2(i,\phantom{\rule{3.33333pt}{0ex}}j)\oplus P2(i-1,\phantom{\rule{3.33333pt}{0ex}}j)\oplus floor\left(x\right(i,\phantom{\rule{3.33333pt}{0ex}}j)\phantom{\rule{3.33333pt}{0ex}}\ast \phantom{\rule{3.33333pt}{0ex}}255)$$else$$P3(i,\phantom{\rule{3.33333pt}{0ex}}j)\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}P2(i,\phantom{\rule{3.33333pt}{0ex}}j)\oplus P2(i,\phantom{\rule{3.33333pt}{0ex}}j-1)\oplus P2(i-1,\phantom{\rule{3.33333pt}{0ex}}j)\oplus floor\left(x\right(i,\phantom{\rule{3.33333pt}{0ex}}j)\phantom{\rule{3.33333pt}{0ex}}\ast \phantom{\rule{3.33333pt}{0ex}}255)$This process starts from i = M and j = N, ⊕ indicates that the binary values corresponding to the two numbers are bitwise exclusive or, $floor\left(x\right)$ is the largest integer not greater than x, and P3 is the transformed image.
- Carry out the second round of reverse diffusion through $y\left(i\right)$, and obtain P4. The method is similar to Step 2 and will not be repeated.
- According to the scrambling method, P4 is transformed into a one-dimensional matrix C${}^{{}^{\prime}}$ and the pixel sum of P4 is calculated.
- Sort $E\left(i\right)$ to obtain the position serial number ${S}^{{}^{\prime}}\left(i\right)$.
- Calculate the index’. $index{}^{\prime}=sum\phantom{\rule{3.33333pt}{0ex}}mod\phantom{\rule{3.33333pt}{0ex}}2,i=1$. If index’ equals zero, $P\left(S{}^{\prime}\right(1\left)\right)=C{}^{\prime}\left(1\right)$. Otherwise, P(S’(M*N))= C’(1).
- Calculate the index’. $index{}^{\prime}={C}^{\prime}(i-1)\phantom{\rule{3.33333pt}{0ex}}mod\phantom{\rule{3.33333pt}{0ex}}2,i>=2$. If index’ equals zero, P(S’(left))=C’(i), left = left +1. Otherwise, $P\left(S{}^{\prime}\right(right\left)\right)=C{}^{\prime}\left(i\right),right=right-1$. Until i is equal to M*N.
- Restore vector P(i) to a decrypted image P of size M*N.

## 4. Simulation Results and Attack Test

#### 4.1. Simulation Results

#### 4.2. Statistical Analysis

#### 4.2.1. Histogram Analysis

#### 4.2.2. Correlation Analysis

#### 4.3. Key Space Analysis

#### 4.4. Sensitivity Analysis

#### 4.4.1. Key Sensitivity

#### 4.4.2. Differential Attack

#### 4.5. Information Entropy Analysis

#### 4.6. Chosen-Plaintext Analysis and Known-Plaintext Analysis

#### 4.7. Encryption Efficiency

## 5. Conclusions

## Author Contributions

## Funding

## Acknowledgments

## Conflicts of Interest

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**Figure 3.**Trajectories of the (

**a**) logistic Feigenbaum non-linear cross-coupled hyperchaotic map (LF-NCHM), (

**b**) FL-NMLD, (

**c**) 2D-SLMM and (

**d**) 2D-SIMM.

**Figure 11.**The simulation test results. (

**a**–

**c**) are the original image, encrypted image and decrypted image of the Lena image, respectively. (

**d**–

**f**) are the original image, encrypted image and decrypted image of the Cameraman image, respectively.

**Figure 12.**Histograms. (

**a**,

**b**) are histograms of the Lena image and the encrypted Lena image. (

**c**,

**d**) are histograms of the Cameraman image and the encrypted Cameraman image.

**Figure 13.**Distributions of the adjacent pixels in the original images and encrypted images of Lena. (

**a**,

**c**,

**e**) are the distributions of the original image in the horizontal, vertical and diagonal directions, respectively. (

**b**,

**d**,

**f**) are the distributions of the encrypted images in the horizontal, vertical and diagonal directions, respectively.

**Figure 14.**The key sensitivity test results. (

**a**) Lena, (

**b**) the encrypted image of Lena, (

**c**) the decrypted image of Lena, (

**d**) the parameter $\mu $ is modified, (

**e**) the parameter $\lambda $ is modified, (

**f**) the parameter x${}_{0}$ is modified and (

**g**) the parameter y${}_{0}$ is modified.

Horizontal | Vertical | Diagonal | |
---|---|---|---|

Lena | 0.9874 | 0.9973 | 0.9662 |

Encrypted Lena | 0.0009 | −0.0006 | 0.0007 |

Cameraman | 0.9606 | 0.9355 | 0.9001 |

Encrypted Cameraman | 0.0181 | 0.0004 | −0.0024 |

Horizontal | Vertical | Diagonal | |
---|---|---|---|

Lena | 0.9874 | 0.9973 | 0.9662 |

Encrypted Lena | 0.0009 | −0.0006 | 0.0007 |

[34] | 0.0008 | 0.0015 | 0.0032 |

[19] | 0.0024 | −0.0086 | 0.0402 |

[35] | 0.0030 | −0.0024 | −0.0013 |

Index | Lena | Cameraman | 4.1.01 | Theoretical Value |
---|---|---|---|---|

NPCR | 99.6078 | 99.5575 | 99.6307 | 99.6094 |

UACI | 33.4404 | 33.4058 | 33.4920 | 33.4635 |

BACI | 26.7386 | 26.8189 | 26.6824 | 26.7712 |

Information Entropy | |
---|---|

Lena | 7.4486 |

Encrypted Lena | 7.9970 |

Cameraman | 7.1048 |

Encrypted Cameraman | 7.9974 |

4.1.01 | 6.8981 |

Encrypted 4.1.01 | 7.9973 |

© 2020 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http://creativecommons.org/licenses/by/4.0/).

## Share and Cite

**MDPI and ACS Style**

Hou, W.; Li, S.; He, J.; Ma, Y.
A Novel Image-Encryption Scheme Based on a Non-Linear Cross-Coupled Hyperchaotic System with the Dynamic Correlation of Plaintext Pixels. *Entropy* **2020**, *22*, 779.
https://doi.org/10.3390/e22070779

**AMA Style**

Hou W, Li S, He J, Ma Y.
A Novel Image-Encryption Scheme Based on a Non-Linear Cross-Coupled Hyperchaotic System with the Dynamic Correlation of Plaintext Pixels. *Entropy*. 2020; 22(7):779.
https://doi.org/10.3390/e22070779

**Chicago/Turabian Style**

Hou, Wenjin, Shouliang Li, Jiapeng He, and Yide Ma.
2020. "A Novel Image-Encryption Scheme Based on a Non-Linear Cross-Coupled Hyperchaotic System with the Dynamic Correlation of Plaintext Pixels" *Entropy* 22, no. 7: 779.
https://doi.org/10.3390/e22070779