# Anticontrol of a Fractional-Order Chaotic System and Its Application in Color Image Encryption

Authors: Yujun Niu; Xuming Sun; Cheng Zhang; Hongjun Liu
Journal: Mathematical Problems in Engineering
 (2020)

---

## Abstract

This paper investigates the anticontrol of the fractional-order chaotic system. The necessary condition of the anticontrol of the fractional-order chaotic system is proposed, and based on this necessary condition, a 3D fractional-order chaotic system is driven to two new 4D fractional-order hyperchaotic systems, respectively, without changing the parameters and fractional order. Hyperchaotic properties of these new fractional dynamic systems are confirmed by Lyapunov exponents and bifurcation diagrams. Furthermore, a color image encryption algorithm is designed based on these fractional hyperchaotic systems. The effectiveness of their application in image encryption is verified.


---

## Body

## 1. Introduction

In recent years, how to make a dynamic system chaotic or enhance the chaos of the system, that is to say, study on the anticontrol of the chaotic system, has become a very hot research topic [1–7]. A feedback control design method is proposed to make all the Lyapunov exponents of the discrete-time dynamical system strictly positive by Chen and Lai [1]. Based on time-delay feedback, a systematic design approach is developed for anticontrol of chaos in a continuous-time system [2, 3]. Li et al. present a simple parameter perturbation control technique to drive a unified chaotic system to hyperchaotic [4]. In [5], a state feedback control is used to design a hyperchaotic Chua system with piecewise-linear nonlinearity. A systematic methodology is proposed to construct the continuous-time autonomous hyperchaotic system with multiple positive Lyapunov exponents, and a 6-dimensional hyperchaotic circuit is implemented [6, 7].In the aforementioned works, the dynamical systems are all chaotic systems with integer order. However, compared with the integer-order chaotic system, the fractional-order system has higher nonlinearity. Moreover, the derivative orders can be used as secret keys in the encryption algorithm based on the chaotic system. At the same time, because high-dimensional chaotic systems have multiple positive Lyapunov exponents and control parameters, it can display more complex dynamical behaviors. Therefore, study on the fractional-order hyperchaotic systems has attracted interest of many scholars [8–19]. Some new high-dimensional fractional-order chaotic systems have been proposed and studied, including dynamic analysis [8–11], control [12, 13], synchronization [14, 15], circuit implementation [16], and application in information encryption [17–19]. But, these fractional-order hyperchaotic systems are obtained by directly modifying the order of integer-order hyperchaotic systems, instead of getting from the anticontrol of the fractional-order system.Inspired by the above discussions, in this paper, based on linear feedback and nonlinear feedback, we directly drive the 3D fractional-order chaotic system to two new 4D fractional-order hyperchaotic systems, respectively, without changing the parameters and fractional order. We propose the necessary conditions of the anticontrol for the fractional-order chaotic system and to calculate the Lyapunov exponents and bifurcation diagrams of the new fractional hyperchaotic dynamic systems. Furthermore, based on these two fractional-order hyperchaotic systems, a color image encryption algorithm is designed. The security analysis verifies that these hyperchaotic systems are effective for image encryption.


## 2. Problem Formulation

A 3D autonomous chaotic system is proposed by Qi et al. [20], which can be described as(1)x˙=ay−x+yz,y˙=cx−y−xz,z˙=xy−bz,where x, y, and z are state variables. When the parameters a=35, b=8/3, and c=55, system (1) shows a chaotic behavior. The fractional-order equation of system (1) can be expressed as(2)dq1xdtq1=ay−x+yz,dq2ydtq2=cx−y−xz,dq3zdtq3=xy−bz,where qi is the fractional order, 0<qi≤1i=1,2,3. According to the algorithm presented by Wolf et al. [21], we calculate the largest Lyapunov exponent of fractional-order system (2). When q1=q2=q3=0.96, system (2) exhibits a chaotic behavior with the largest Lyapunov exponent 2.168. In this paper, the numerical simulation of fractional-order systems is derived according to Caputo derivative. More detailed introduction about Caputo derivative definition can be seen in [22].In the following, fractional-order chaotic system (2) is driven to two new 4D fractional-order hyperchaotic systems, respectively, with the same parameters and fractional order, that is, a=35, b=8/3, c=55, and q1=q2=q3=0.96.


## 3. New Fractional-Order Hyperchaotic Systems

About the anticontrol of the fractional-order chaotic systems, we give the two necessary conditions as follows:(1)
The new dynamic system must be dissipative(2)
None of equilibriums of the new fractional-order system is stableThe stable and unstable region division at the zero equilibrium of the fractional-order system is shown in Figure1. According to the stability theory of the fractional-order system [23], it can be proved that, for n-dimensional fractional system, if all the eigenvalues (λ1,λ2,...,λn) of the Jacobian matrix of some equilibrium point satisfy(3)argλi>απ2,α=maxq1,q2,…,qn,i=1,2,…,n,then the fractional-order system is asymptotically steady at the equilibrium. From Figure 1, it can be seen that, for the fractional-order system, as long as there is a stable equilibrium, it will be steady at one point; it is in chaos only when none equilibrium is stable. Moreover, a dynamic system with chaotic characteristics must be dissipative. Therefore, the above necessary conditions can be obtained.Figure 1
Stability region of the fractional-order system.In the following, we analyze the dynamical behaviors of fractional-order system (2). We can calculate that system (2) has three equilibriums S0−19.3091,−7.5418,54.6094, S119.3091,7.5418,54.6094, and S20,0,0.For the equilibriumS0−19.3091,−7.5418,54.6094, it can be calculated that the eigenvalues of the Jacobian matrix are λ1=−43.0978, λ2=2.2156+24.4545i, and λ3=2.2156−24.4545i. Further, it can be obtained that argλ1=π, argλ2=1.4797, and argλ3=−1.4797. From formula (3), when qi>0.94201.4797×2/π, i=1,2,3, S0−19.3091,−7.5418,54.6094 is unstable equilibrium. In the same way, it can be obtained that the equilibrium S119.3091,7.5418,54.6094 is unstable when qi>0.94201.4797×2/π,i=1,2,3, and S20,0,0 is unstable equilibrium of system (2).To sum up, whenqi>0.9420,i=1,2,3, S0, S1, and S2 are all unstable equilibriums of system (2). So, the value of fractional order of system (2) must be between 0.9420 and 1. Finally, we select 0.96 as fractional order of system (2).


### 3.1. Fractional Hyperchaotic System Obtained by Linear Feedback

For convenience of expression, the variables of system (2) are replaced with xii=1,…,3. With the same parameters and fractional order, a new 4-dimensional fractional-order dynamic system is obtained by introducing a linear feedback control term dq4x4/dtq4=−63x2+r1x4 to the equation of system (2) as follows:(4)dq1x1dtq1=ax2−x1+x2x3,dq2x2dtq2=cx1−x2−x1x3+x4,dq3x3dtq3=x1x2−bx3,dq4x4dtq4=−63x1+r1x4,where r1 is a control parameter, and system (4) is possible to be chaotic only when its value satisfies the above necessary condition of the anticontrol.In order to ensure the dissipative structure of system (4), the requirement is that(5)∇V=∂x˙1∂x1+∂x˙2∂x2+∂x˙3∂x3+∂x˙4∂x4=−a−1−b+r1=r1−38.7<0.It is concluded that the value of control parameterr1 must be less than 38.7. In order to maintain the dissipative structure better, we choose r1=0.6 near zero.Whena=35, b=8/3, c=55, and r1=0.6, we calculate that system (4) has three equilibriums S00,0,0,0, S10.0487,0.0088,0.1598,5.1141, and S2−0.0487,−0.0088,0.1598,−5.1141.First, let us study whether equilibriumS00,0,0,0 is stable. The Jacobian matrix of system (4) at equilibrium point S0 is as follows:(6)J=−aa+x3x20c−x3−1−x11x2x1−b0−6300r1=−aa00c−10100−b0−6300r1.It is calculated that the eigenvalues of the Jacobian matrix areλ1=−65.4068, λ2=28.1963, λ3=1.8105, and λ4=−2.6667. Furthermore, we can get that argλ1=π, argλ2=0, argλ3=0, and argλ4=π, without satisfying that argλi>0.96×π/2 (i=1,2,…,4). Therefore, it can be concluded that S0 is unstable equilibrium.Next,S10.0487,0.0088,0.1598,5.1141 is chosen to study. We can compute that the eigenvalues of the Jacobian matrix λ1=−65.4416, λ2=28.2276, λ3=1.8140, and λ4=−2.6666, and we can obtain that argλ1=π, argλ2=0, argλ3=0, and argλ4=π. So, the equilibrium S1 is not stable. In the same way, it can be obtained that S2−0.0487,−0.0088,0.1598,−5.1141 is also unstable.To sum up, whenr1=0.6, S0, S1, and S2 are all unstable equilibriums of system (4). We obtain the Lyapunov exponents of system (4): λ1=1.4272, λ2=0.3705, λ3=−0.0028, and λ4=−41.3635 when r1=0.6 and q1=q2=q3=q4=0.96. Therefore, it is proved that system (4) shows a hyperchaotic behavior. The part of projections of the hyperchaotic attractor is shown in Figure 2.Figure 2
Whenr1=0.6, the part of projections of the attractor of system (4).
(a)
(b)
(c)According to the method presented by Ramasubramanian and Sriram [24], when −1.5≤r1≤2, the Lyapunov exponent spectrum of fractional-order system (4) is calculated, and it is shown in Figure 3(a). The corresponding bifurcation diagram of system (4) is shown in Figure 3(b). From Figure 3, it is easy to observe that when −1.05≤r1≤1.7, fractional-order system (4) is hyperchaotic with satisfying that λ1>0, λ2>0, λ3=0, and λ4<0.Figure 3
Lyapunov exponents and bifurcation diagram of system (4) for −1.5≤r1≤2. (a) Lyapunov exponents. (b) Bifurcation diagram.
(a)
(b)


### 3.2. Fractional Hyperchaotic System Obtained by Nonlinear Feedback

The variables of system (2) are taken the place of yii=1,…,3, and a nonlinear feedback control term is added to system (2). A new 4D fractional-order dynamic system is obtained as follows:(7)dq1y1dtq1=ay2−y1+y2y3,dq2y2dtq2=cy1−y2−y1y3+y4,dq3y3dtq3=y1y2−by3,dq4y4dtq4=−y1y3+r2y4,where r2 is a control parameter, and we choose r2=1.2 in order to ensure system (7) be dissipative.Whenr2=1.2, it can be computed that system (7) has only one equilibrium S00,0,0,0. Furthermore, we can calculate that argλ1=π, argλ2=0, argλ3=π, and argλ4=0. So, it is not satisfied that argλi>0.96×π/2. Therefore, S0 is not stable equilibrium.Whenr2=1.2 and q1=q2=q3=q4=0.96, fractional-order system (7) displays a hyperchaotic behavior with Lyapunov exponents λ1=1.3551, λ2=0.2182, λ3=0.0039, and λ4=−30.2820. The part of projections of the hyperchaotic attractor is shown in Figure 4.Figure 4
Whenr2=1.2, the part of projections of the hyperchaotic attractor of system (7).
(a)
(b)
(c)When0.5≤r2≤2.5, the Lyapunov exponent spectrum of system (7) is computed according to Ramasubramanian and Sriram method [24]. It is shown in Figure 5(a), and the corresponding bifurcation diagram of system (7) is shown in Figure 5(b). From Figure 5, it can be seen that when 1.1≤r2≤1.95, system (7) is hyperchaotic with two positive, a zero, and a negative Lyapunov exponents.Figure 5
Lyapunov exponents and bifurcation diagram of system (7) for 0.5≤r2≤2.5. (a) Lyapunov exponents. (b) Bifurcation diagram.
(a)
(b)


## 3.1. Fractional Hyperchaotic System Obtained by Linear Feedback

For convenience of expression, the variables of system (2) are replaced with xii=1,…,3. With the same parameters and fractional order, a new 4-dimensional fractional-order dynamic system is obtained by introducing a linear feedback control term dq4x4/dtq4=−63x2+r1x4 to the equation of system (2) as follows:(4)dq1x1dtq1=ax2−x1+x2x3,dq2x2dtq2=cx1−x2−x1x3+x4,dq3x3dtq3=x1x2−bx3,dq4x4dtq4=−63x1+r1x4,where r1 is a control parameter, and system (4) is possible to be chaotic only when its value satisfies the above necessary condition of the anticontrol.In order to ensure the dissipative structure of system (4), the requirement is that(5)∇V=∂x˙1∂x1+∂x˙2∂x2+∂x˙3∂x3+∂x˙4∂x4=−a−1−b+r1=r1−38.7<0.It is concluded that the value of control parameterr1 must be less than 38.7. In order to maintain the dissipative structure better, we choose r1=0.6 near zero.Whena=35, b=8/3, c=55, and r1=0.6, we calculate that system (4) has three equilibriums S00,0,0,0, S10.0487,0.0088,0.1598,5.1141, and S2−0.0487,−0.0088,0.1598,−5.1141.First, let us study whether equilibriumS00,0,0,0 is stable. The Jacobian matrix of system (4) at equilibrium point S0 is as follows:(6)J=−aa+x3x20c−x3−1−x11x2x1−b0−6300r1=−aa00c−10100−b0−6300r1.It is calculated that the eigenvalues of the Jacobian matrix areλ1=−65.4068, λ2=28.1963, λ3=1.8105, and λ4=−2.6667. Furthermore, we can get that argλ1=π, argλ2=0, argλ3=0, and argλ4=π, without satisfying that argλi>0.96×π/2 (i=1,2,…,4). Therefore, it can be concluded that S0 is unstable equilibrium.Next,S10.0487,0.0088,0.1598,5.1141 is chosen to study. We can compute that the eigenvalues of the Jacobian matrix λ1=−65.4416, λ2=28.2276, λ3=1.8140, and λ4=−2.6666, and we can obtain that argλ1=π, argλ2=0, argλ3=0, and argλ4=π. So, the equilibrium S1 is not stable. In the same way, it can be obtained that S2−0.0487,−0.0088,0.1598,−5.1141 is also unstable.To sum up, whenr1=0.6, S0, S1, and S2 are all unstable equilibriums of system (4). We obtain the Lyapunov exponents of system (4): λ1=1.4272, λ2=0.3705, λ3=−0.0028, and λ4=−41.3635 when r1=0.6 and q1=q2=q3=q4=0.96. Therefore, it is proved that system (4) shows a hyperchaotic behavior. The part of projections of the hyperchaotic attractor is shown in Figure 2.Figure 2
Whenr1=0.6, the part of projections of the attractor of system (4).
(a)
(b)
(c)According to the method presented by Ramasubramanian and Sriram [24], when −1.5≤r1≤2, the Lyapunov exponent spectrum of fractional-order system (4) is calculated, and it is shown in Figure 3(a). The corresponding bifurcation diagram of system (4) is shown in Figure 3(b). From Figure 3, it is easy to observe that when −1.05≤r1≤1.7, fractional-order system (4) is hyperchaotic with satisfying that λ1>0, λ2>0, λ3=0, and λ4<0.Figure 3
Lyapunov exponents and bifurcation diagram of system (4) for −1.5≤r1≤2. (a) Lyapunov exponents. (b) Bifurcation diagram.
(a)
(b)


## 3.2. Fractional Hyperchaotic System Obtained by Nonlinear Feedback

The variables of system (2) are taken the place of yii=1,…,3, and a nonlinear feedback control term is added to system (2). A new 4D fractional-order dynamic system is obtained as follows:(7)dq1y1dtq1=ay2−y1+y2y3,dq2y2dtq2=cy1−y2−y1y3+y4,dq3y3dtq3=y1y2−by3,dq4y4dtq4=−y1y3+r2y4,where r2 is a control parameter, and we choose r2=1.2 in order to ensure system (7) be dissipative.Whenr2=1.2, it can be computed that system (7) has only one equilibrium S00,0,0,0. Furthermore, we can calculate that argλ1=π, argλ2=0, argλ3=π, and argλ4=0. So, it is not satisfied that argλi>0.96×π/2. Therefore, S0 is not stable equilibrium.Whenr2=1.2 and q1=q2=q3=q4=0.96, fractional-order system (7) displays a hyperchaotic behavior with Lyapunov exponents λ1=1.3551, λ2=0.2182, λ3=0.0039, and λ4=−30.2820. The part of projections of the hyperchaotic attractor is shown in Figure 4.Figure 4
Whenr2=1.2, the part of projections of the hyperchaotic attractor of system (7).
(a)
(b)
(c)When0.5≤r2≤2.5, the Lyapunov exponent spectrum of system (7) is computed according to Ramasubramanian and Sriram method [24]. It is shown in Figure 5(a), and the corresponding bifurcation diagram of system (7) is shown in Figure 5(b). From Figure 5, it can be seen that when 1.1≤r2≤1.95, system (7) is hyperchaotic with two positive, a zero, and a negative Lyapunov exponents.Figure 5
Lyapunov exponents and bifurcation diagram of system (7) for 0.5≤r2≤2.5. (a) Lyapunov exponents. (b) Bifurcation diagram.
(a)
(b)


## 4. Image Encryption Based on the Fractional-Order Hyperchaotic Systems

In this section, a color image encryption algorithm is designed based on fractional-order hyperchaotic system (4) and system (7).


### 4.1. Design of Image Encryption Algorithm

#### 4.1.1. Permutation Process

In order to break the correlation of the neighboring pixels in the plaintext, the plain image is permutated in bit-level. A color image with size ofm×n is chosen as the plain image P.Step 1. First, the plain imageP is converted into a grayscale image whose size is m×3n according to the red, green, and blue components of the color image. Then, each pixel of the grayscale image is transformed into an 8-bit array. So, the whole plain image P is changed into a binary matrix Pb with a size of m×24n.Step 2. The sequencesx1i,x2i,x3i,x4i are generated by fractional-order hyperchaotic system (4) with initial values x10,x20,x30,x40. The key vectors keyh and keyv are produced by equations (8) and (9), respectively:(8)keyhi=absx1i+x2i×g−floorabsx1i+x2i×g,i=1,2,…,m,(9)keyvj=absx3j×g+x4j−floorabsx3j×g+x4j,j=1,2,…,24n,whereg is a disturbance item related to the plain image, which can be obtained by equation (10) as follows:(10)g=sumPbmn,wheresumPb is the sum of all the elements with value of 1 in the matrix Pb. Therefore, the key vectors keyh and keyv are related to the plain image P. Because different plain images are encrypted by different keys, this algorithm can resist the chosen plaintext attack.Step 3. Set two auxiliary vectorshi,i=1,2,…,m and vj,j=1,2,…,24n. They represent the line numbers and the column numbers of Pb in the ascending order, respectively. Next, the vectors h′ and v′ are generated by equations (11) and (12), separately. Then, a binary matrix is obtained by permuting the rows and columns of image Pb, respectively, according to the vectors h′ and v′. Finally, we transform this binary matrix into color image Pc.(11)h′i=hfloorkeyhi×mmodi,i=1,2,…,m,(12)v′j=vfloorkeyvj×24nmodj,j=1,2,…,24n.


#### 4.1.2. Encryption Algorithm

In this stage, the permutated color imagePc is encrypted in pixel-level, and the detailed steps are as follows:Step 1. Separate the permuted color imagePc into red, green, and blue grayscale images, and the i-th pixel value of these three grayscale images is represented by ri, gi, and bii=1,2,...,mn, respectively.Step 2. With initial valuesy10,y20,y30,y40, the sequences y1i,y2i,y3i,y4i are produced by fractional-order hyperchaotic system (7). Then, an integer sequence skji between 0 and 255 is obtained:(13)skji=absyui+yvi−floorabsyui+yvi×1014mod256,j=1,…,6,whereu,v∈1,2,3,4 and u≠v. The encryption key sequences keyri,keygi,keybi can be gained by the key selection table, and it is shown in Table 1.Table 1
The key selection table.
Key
0
1
2
key
r
i
s
k
1
i
⊕
s
k
2
i
s
k
1
i
⊕
s
k
3
i
s
k
2
i
⊕
s
k
5
i
key
g
i
s
k
3
i
⊕
s
k
5
i
s
k
4
i
⊕
s
k
5
i
s
k
1
i
⊕
s
k
4
i
key
b
i
s
k
4
i
⊕
s
k
6
i
s
k
2
i
⊕
s
k
6
i
s
k
3
i
⊕
s
k
6
iThere are three groups of keys in Table1. si is used to decide which group is selected to encrypt ri, gi, and bi of the i-th pixel. So, it is realized that different plain images are encrypted by different key streams. The sequence si is generated by equations (14) and (15):(14)keysi=abs∑j=14xji−floorabs∑j=14xji×1014,i=1,2,…,mn,(15)si=floor∑j=2mnrj+gj+bj×keysimod3,i=1,floorspi−1−ri+gi+bi×keysimod3,i=2,…,mn,wherexji,j=1,2,…,4 are produced by fractional-order hyperchaotic system (4).
Step 3. Afterri, gi, and bi of three grayscale images are encrypted by equation (16), respectively, we can get the encrypted ri′, gi′, and bi′:(16)
r
i
′
=
r
i
+
r
i
−
1
′
mod
256
⊕
key
r
i
,
g
i
′
=
g
i
+
g
i
−
1
′
mod
256
⊕
key
g
i
,
i
=
1
,
…
,
m
n
,
b
i
′
=
b
i
+
b
i
−
1
′
mod
256
⊕
key
b
i
,where(17)
r
0
′
=
∑
j
=
1
m
n
r
j
+
s
k
1
1
mod
256
,
g
0
′
=
∑
j
=
1
m
n
g
j
+
s
k
2
1
mod
256
,
b
0
′
=
∑
j
=
1
m
n
b
j
+
s
k
3
1
mod
256
.Step 4. Repeat the aforementioned steps appropriately. In the end, the encrypted color imageIE is created by the composition of the three encrypted grayscale images.


### 4.2. Design of the Decryption Algorithm

#### 4.2.1. Decryption Algorithm

Step 1. Separate the encrypted imageIE into red, green, and blue grayscale images, and set ri′, gi′, and bi′i=1,2,…,mn to represent the i-th pixel value of these grayscale images, respectively. The decryption begins from the back to the front, that is to say that the mn-th pixel is firstly decrypted.
Step 2. With the same initial values with the encryption process, the sequencesy1i,y2i,y3i,y4i are generated by fractional-order dynamic system (7). Next, it can be calculated that smn=0 by equation (15). Then, the keys keyrmn,keygmn,keybmn can be obtained by equation (13) and smn=0. Finally, we can get rmn, gmn, and bmn after the following equation:
(18)
r
i
=
r
i
′
⊕
key
r
i
−
r
i
−
1
′
mod
256
,
g
i
=
g
i
′
⊕
key
g
i
−
g
i
−
1
′
mod
256
,
b
i
=
b
i
′
⊕
key
b
i
−
b
i
−
1
′
mod
256.
Step 3. With values ofrmn, gmn, and bmn and smn=0, the value of smn−1 can be computed by equations (14) and (15). Then, the (mn−1)-th pixel can be decrypted. Similarly, it is finished until the values of r1, g1, and b1 are decrypted.
Step 4. After repeating the above steps for the same rounds with the encryption process, the decrypted imageI′ is obtained.


#### 4.2.2. Inverse Permutation Process

Step 1. According to the red, green, and blue components of the color image, the imageI′ is converted into a grayscale image with size of m×3n. Then, a binary matrix Ib′ with size of m×24n is obtained by each pixel of the grayscale image which is changed into an 8-bit array.
Step 2. Because the sum of all the elements with value of 1 in the binary matrixIb′ is equal to one of all the elements with value of 1 in the binary matrix Pb, the value of g is computed by the following equation:
(19)
g
=
sum
I
b
′
m
n
.
Step 3. With the same initial values as permutation process, the sequencesx1i,x2i,x3i,x4i are generated by system (4). Then, the keys keyh and keyv are computed by equations (8) and (9), respectively. Finally, the vectors h′ and v′ are obtained by equations (11) and (12) separately.
Step 4. According to the array vectorsh′ and v′, the rows and columns of binary image Ib′ are inversely permuted, respectively. The image Ib is obtained by this inverse permutation. We convert Ib with size of m×24n into the plain color image I whose size is of m×n. Thus, the decryption is achieved completely.


### 4.3. Experimental Result

The color image named pepper is selected as the plain images, whose size is197×206. The initial values of fractional-order hyperchaotic system (4) and system (7) are x10=0.67185367890218, x20=0.24566789543262, x30=0.15492289843576, x40=1.32854321678987, y10=0.98165678567657, y20=0.12345678901234,y30=1.65432109876543, and y40=0.57167689592916. The experimental results are shown in Figure 6.Figure 6
The encryption results of pepper image. (a) The original image. (b) The grayscale image of decomposing Figure6(a). (c) The image of permuting Figure 6(b). (d) The result of composing Figure 6(c). (e) The encrypted image.
(a)
(b)
(c)
(d)
(e)


### 4.4. Performance and Security Analysis

#### 4.4.1. Key Space and Sensitivity

In the encryption algorithm, the secret keys are the initial values of the two hyperchaotic systems, that is,x10,x20,x30,x40 and y10,y20,y30,y40. Because the maximum precision of these initial values is 10−14, the total key space is that 10148=10112, which is much larger than 2100 [25]. Therefore, our algorithm can resist all kinds of brute force attacks.For the key sensitive test, the tiny change10−14 is selected as the error of these initial values, and it is shown in Table 2. Due to space limitations, only the changes of x10, x20, y30, and y40 were made. The experimental results are shown in Figure 7. It can be seen that these decrypted images are extremely similar to noise and absolutely different from the plain image, and those pixel distribution histograms are uniform. Moreover, we calculate the NPCR and UACI of the original image and recovered images (peppers) with different keys; it is shown in Table 3. It is easy to observe that NPCR is over 99%, and the UACI is close to 33%. It demonstrates that the recovered images with different keys are greatly different from their original form. Therefore, our algorithm is sensitive to the secret key.Table 2
The tiny change of secret keys.
Key
Original value
New value
x
1
0
0.67185367890218
0.67185367890218 + 10−14
x
2
0
0.24566789543262
0.24566789543262 − 10−14
y
3
0
1.65432109876543
1.65432109876543 + 10−14
y
4
0
0.57167689592916
0.57167689592916 − 10−14Figure 7
The decrypted images in tiny change of keys and their histograms. (a)x1(0) = 0.67185367890219. (b) x2(0) = 0.24566789543261. (c) y3(0) = 1.65432109876544. (d) y4(0) = 0.57167689592915. (e) The pixel distribution histogram of Figure 7(a). (f) The pixel distribution histogram of Figure 7(b). (g) The pixel distribution histogram of Figure 7(c). (h) The pixel distribution histogram of Figure 7(d).
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)Table 3
The NPCR and UACI of the original image and recovered image (peppers) with different keys.
Image
NPCR (%)
UACI (%)
R
G
B
R
G
B
Recovered image withx10
99.4259
99.3569
99.4185
28.3860
32.9807
32.8930
Recovered image withx20
99.3421
99.3864
99.4061
28.4061
32.8305
32.7692
Recovered image withy30
99.3174
99.3618
99.4209
28.2771
32.8916
32.7201
Recovered image withy40
99.3544
99.2755
99.3544
28.1678
32.4912
32.7000


#### 4.4.2.χ2 Test

We compute theχ2 values of both plain image and encrypted image named Pepper, Flower, Yacht, and Baboon, whose sizes are all 197×206. It is shown in Table 4. Obviously, the χ2 values of the ciphered images are far lower than those of the plain images and similar to the algorithm in [26, 27]; they are all within the effective interval 210.7918,293.2478.Table 4
The results of theχ2 test.
χ
2
Plain image
Ciphered image
R
G
B
R
G
B
Pepper
35,300
32,679
64,067
258.8352
247.7227
208.7380
Flower
25,554
27,012
47,904
235.1820
250.1199
260.4022
Yacht
18,431
21,044
17,323
259.1406
271.4038
229.4794
Baboon
19,059
27,759
17,654
249.9811
237.2006
219.9287
Pepper in [26]
57,362
62,180
122,870
265.4625
269.3956
289.2321
Image 1 in [27]
—
—
—
230.8105
234.7070
250.3419
Image 2 in [27]
—
—
—
265.7558
286.3242
276.2754


#### 4.4.3. Statistical Analysis

From the plain image and encrypted image, 2500 pairs of adjacent pixels are randomly selected in horizontal, vertical, and diagonal directions, respectively. Then, we calculate the correlation coefficients of two adjacent pixels, which are shown in Table5. From Table 5, it can be observed that the correlations of two adjacent pixels are close to 1 in the plain images, while the ones of the encrypted images are round 0 and similar to the algorithm in [28, 29]. Therefore, the encryption algorithm can eliminate the correlation of the adjacent pixels.Table 5
Correlation coefficients of the plain image and ciphered image.
Correlation
Horizontal
Vertical
Diagonal
Pepper
0.9373
0.9716
0.9478
Ciphered pepper
0.0021
0.0018
−0.0195
Flower
0.9414
0.9792
0.9769
Ciphered flower
0.0009
0.0173
−0.0194
Lena in [28] (R)
0.95409435
0.97692808
0.92946084
Lena in [28] (G)
0.93859702
0.96847242
0.91318153
Lena in [28] (B)
0.92230178
0.95144503
0.89275171
Ciphered lena in [28] (R)
0.00268849
0.00113425
0.00526812
Ciphered lena in [28] (G)
0.00979889
0.00302981
0.00038029
Ciphered lena in [28] (B)
0.00098796
0.00056287
0.00111043
Lena in [29]
0.9494
0.9667
0.9336
Ciphered lena in [29]
0.0054
0.0035
0.0016


#### 4.4.4. Information Entropy

We compute the information entropy of the encrypted images according to the red, green, and blue components of the color image, which is shown in Table6. It can be seen that the information entropy of these ciphered images is all close to the value 8. So, it proves that the unpredictability of the encrypted images is very high.Table 6
The information entropy of encrypted images.
Color image
Red
Green
Blue
Pepper
7.9875
7.9880
7.9876
Flower
7.9882
7.9881
7.9871
Yacht
7.9867
7.9883
7.9874
Baboon
7.9881
7.9882
7.9875


#### 4.4.5. Differential Attack

To test the differential attack, some color images are encrypted with altering a pixel in the plain images, respectively, and NPCR and UACI are calculated in Table7. It can be found that the NPCR is over 99%, the UACI is over 33%, and similar to the algorithm in [28–30]. It indicates that our algorithm can resist the differential attack.Table 7
The NPCR and UACI of ciphered images.
Image
NPCR (%)
UACI (%)
R
G
B
R
G
B
Pepper
99.5762
99.6205
99.5762
33.2831
33.4199
33.5400
Flower
99.6772
99.7142
99.6550
33.4547
33.7477
33.3845
Yacht
99.5269
99.5810
99.5146
33.3787
33.4017
33.4997
Baboon
99.6550
99.6723
99.6747
33.3987
33.3903
33.3644
Lena in [28]
99.98779
99.98779
99.98779
50.17915
50.18009
25.19263
Lena in [29] (avg.)
99.5723
33.3159
Lena in [30]
99.6231
99.6338
99.6170
33.4747
33.5683
33.3382


## 4.1. Design of Image Encryption Algorithm

### 4.1.1. Permutation Process

In order to break the correlation of the neighboring pixels in the plaintext, the plain image is permutated in bit-level. A color image with size ofm×n is chosen as the plain image P.Step 1. First, the plain imageP is converted into a grayscale image whose size is m×3n according to the red, green, and blue components of the color image. Then, each pixel of the grayscale image is transformed into an 8-bit array. So, the whole plain image P is changed into a binary matrix Pb with a size of m×24n.Step 2. The sequencesx1i,x2i,x3i,x4i are generated by fractional-order hyperchaotic system (4) with initial values x10,x20,x30,x40. The key vectors keyh and keyv are produced by equations (8) and (9), respectively:(8)keyhi=absx1i+x2i×g−floorabsx1i+x2i×g,i=1,2,…,m,(9)keyvj=absx3j×g+x4j−floorabsx3j×g+x4j,j=1,2,…,24n,whereg is a disturbance item related to the plain image, which can be obtained by equation (10) as follows:(10)g=sumPbmn,wheresumPb is the sum of all the elements with value of 1 in the matrix Pb. Therefore, the key vectors keyh and keyv are related to the plain image P. Because different plain images are encrypted by different keys, this algorithm can resist the chosen plaintext attack.Step 3. Set two auxiliary vectorshi,i=1,2,…,m and vj,j=1,2,…,24n. They represent the line numbers and the column numbers of Pb in the ascending order, respectively. Next, the vectors h′ and v′ are generated by equations (11) and (12), separately. Then, a binary matrix is obtained by permuting the rows and columns of image Pb, respectively, according to the vectors h′ and v′. Finally, we transform this binary matrix into color image Pc.(11)h′i=hfloorkeyhi×mmodi,i=1,2,…,m,(12)v′j=vfloorkeyvj×24nmodj,j=1,2,…,24n.


### 4.1.2. Encryption Algorithm

In this stage, the permutated color imagePc is encrypted in pixel-level, and the detailed steps are as follows:Step 1. Separate the permuted color imagePc into red, green, and blue grayscale images, and the i-th pixel value of these three grayscale images is represented by ri, gi, and bii=1,2,...,mn, respectively.Step 2. With initial valuesy10,y20,y30,y40, the sequences y1i,y2i,y3i,y4i are produced by fractional-order hyperchaotic system (7). Then, an integer sequence skji between 0 and 255 is obtained:(13)skji=absyui+yvi−floorabsyui+yvi×1014mod256,j=1,…,6,whereu,v∈1,2,3,4 and u≠v. The encryption key sequences keyri,keygi,keybi can be gained by the key selection table, and it is shown in Table 1.Table 1
The key selection table.
Key
0
1
2
key
r
i
s
k
1
i
⊕
s
k
2
i
s
k
1
i
⊕
s
k
3
i
s
k
2
i
⊕
s
k
5
i
key
g
i
s
k
3
i
⊕
s
k
5
i
s
k
4
i
⊕
s
k
5
i
s
k
1
i
⊕
s
k
4
i
key
b
i
s
k
4
i
⊕
s
k
6
i
s
k
2
i
⊕
s
k
6
i
s
k
3
i
⊕
s
k
6
iThere are three groups of keys in Table1. si is used to decide which group is selected to encrypt ri, gi, and bi of the i-th pixel. So, it is realized that different plain images are encrypted by different key streams. The sequence si is generated by equations (14) and (15):(14)keysi=abs∑j=14xji−floorabs∑j=14xji×1014,i=1,2,…,mn,(15)si=floor∑j=2mnrj+gj+bj×keysimod3,i=1,floorspi−1−ri+gi+bi×keysimod3,i=2,…,mn,wherexji,j=1,2,…,4 are produced by fractional-order hyperchaotic system (4).
Step 3. Afterri, gi, and bi of three grayscale images are encrypted by equation (16), respectively, we can get the encrypted ri′, gi′, and bi′:(16)
r
i
′
=
r
i
+
r
i
−
1
′
mod
256
⊕
key
r
i
,
g
i
′
=
g
i
+
g
i
−
1
′
mod
256
⊕
key
g
i
,
i
=
1
,
…
,
m
n
,
b
i
′
=
b
i
+
b
i
−
1
′
mod
256
⊕
key
b
i
,where(17)
r
0
′
=
∑
j
=
1
m
n
r
j
+
s
k
1
1
mod
256
,
g
0
′
=
∑
j
=
1
m
n
g
j
+
s
k
2
1
mod
256
,
b
0
′
=
∑
j
=
1
m
n
b
j
+
s
k
3
1
mod
256
.Step 4. Repeat the aforementioned steps appropriately. In the end, the encrypted color imageIE is created by the composition of the three encrypted grayscale images.


## 4.1.1. Permutation Process

In order to break the correlation of the neighboring pixels in the plaintext, the plain image is permutated in bit-level. A color image with size ofm×n is chosen as the plain image P.Step 1. First, the plain imageP is converted into a grayscale image whose size is m×3n according to the red, green, and blue components of the color image. Then, each pixel of the grayscale image is transformed into an 8-bit array. So, the whole plain image P is changed into a binary matrix Pb with a size of m×24n.Step 2. The sequencesx1i,x2i,x3i,x4i are generated by fractional-order hyperchaotic system (4) with initial values x10,x20,x30,x40. The key vectors keyh and keyv are produced by equations (8) and (9), respectively:(8)keyhi=absx1i+x2i×g−floorabsx1i+x2i×g,i=1,2,…,m,(9)keyvj=absx3j×g+x4j−floorabsx3j×g+x4j,j=1,2,…,24n,whereg is a disturbance item related to the plain image, which can be obtained by equation (10) as follows:(10)g=sumPbmn,wheresumPb is the sum of all the elements with value of 1 in the matrix Pb. Therefore, the key vectors keyh and keyv are related to the plain image P. Because different plain images are encrypted by different keys, this algorithm can resist the chosen plaintext attack.Step 3. Set two auxiliary vectorshi,i=1,2,…,m and vj,j=1,2,…,24n. They represent the line numbers and the column numbers of Pb in the ascending order, respectively. Next, the vectors h′ and v′ are generated by equations (11) and (12), separately. Then, a binary matrix is obtained by permuting the rows and columns of image Pb, respectively, according to the vectors h′ and v′. Finally, we transform this binary matrix into color image Pc.(11)h′i=hfloorkeyhi×mmodi,i=1,2,…,m,(12)v′j=vfloorkeyvj×24nmodj,j=1,2,…,24n.


## 4.1.2. Encryption Algorithm

In this stage, the permutated color imagePc is encrypted in pixel-level, and the detailed steps are as follows:Step 1. Separate the permuted color imagePc into red, green, and blue grayscale images, and the i-th pixel value of these three grayscale images is represented by ri, gi, and bii=1,2,...,mn, respectively.Step 2. With initial valuesy10,y20,y30,y40, the sequences y1i,y2i,y3i,y4i are produced by fractional-order hyperchaotic system (7). Then, an integer sequence skji between 0 and 255 is obtained:(13)skji=absyui+yvi−floorabsyui+yvi×1014mod256,j=1,…,6,whereu,v∈1,2,3,4 and u≠v. The encryption key sequences keyri,keygi,keybi can be gained by the key selection table, and it is shown in Table 1.Table 1
The key selection table.
Key
0
1
2
key
r
i
s
k
1
i
⊕
s
k
2
i
s
k
1
i
⊕
s
k
3
i
s
k
2
i
⊕
s
k
5
i
key
g
i
s
k
3
i
⊕
s
k
5
i
s
k
4
i
⊕
s
k
5
i
s
k
1
i
⊕
s
k
4
i
key
b
i
s
k
4
i
⊕
s
k
6
i
s
k
2
i
⊕
s
k
6
i
s
k
3
i
⊕
s
k
6
iThere are three groups of keys in Table1. si is used to decide which group is selected to encrypt ri, gi, and bi of the i-th pixel. So, it is realized that different plain images are encrypted by different key streams. The sequence si is generated by equations (14) and (15):(14)keysi=abs∑j=14xji−floorabs∑j=14xji×1014,i=1,2,…,mn,(15)si=floor∑j=2mnrj+gj+bj×keysimod3,i=1,floorspi−1−ri+gi+bi×keysimod3,i=2,…,mn,wherexji,j=1,2,…,4 are produced by fractional-order hyperchaotic system (4).
Step 3. Afterri, gi, and bi of three grayscale images are encrypted by equation (16), respectively, we can get the encrypted ri′, gi′, and bi′:(16)
r
i
′
=
r
i
+
r
i
−
1
′
mod
256
⊕
key
r
i
,
g
i
′
=
g
i
+
g
i
−
1
′
mod
256
⊕
key
g
i
,
i
=
1
,
…
,
m
n
,
b
i
′
=
b
i
+
b
i
−
1
′
mod
256
⊕
key
b
i
,where(17)
r
0
′
=
∑
j
=
1
m
n
r
j
+
s
k
1
1
mod
256
,
g
0
′
=
∑
j
=
1
m
n
g
j
+
s
k
2
1
mod
256
,
b
0
′
=
∑
j
=
1
m
n
b
j
+
s
k
3
1
mod
256
.Step 4. Repeat the aforementioned steps appropriately. In the end, the encrypted color imageIE is created by the composition of the three encrypted grayscale images.


## 4.2. Design of the Decryption Algorithm

### 4.2.1. Decryption Algorithm

Step 1. Separate the encrypted imageIE into red, green, and blue grayscale images, and set ri′, gi′, and bi′i=1,2,…,mn to represent the i-th pixel value of these grayscale images, respectively. The decryption begins from the back to the front, that is to say that the mn-th pixel is firstly decrypted.
Step 2. With the same initial values with the encryption process, the sequencesy1i,y2i,y3i,y4i are generated by fractional-order dynamic system (7). Next, it can be calculated that smn=0 by equation (15). Then, the keys keyrmn,keygmn,keybmn can be obtained by equation (13) and smn=0. Finally, we can get rmn, gmn, and bmn after the following equation:
(18)
r
i
=
r
i
′
⊕
key
r
i
−
r
i
−
1
′
mod
256
,
g
i
=
g
i
′
⊕
key
g
i
−
g
i
−
1
′
mod
256
,
b
i
=
b
i
′
⊕
key
b
i
−
b
i
−
1
′
mod
256.
Step 3. With values ofrmn, gmn, and bmn and smn=0, the value of smn−1 can be computed by equations (14) and (15). Then, the (mn−1)-th pixel can be decrypted. Similarly, it is finished until the values of r1, g1, and b1 are decrypted.
Step 4. After repeating the above steps for the same rounds with the encryption process, the decrypted imageI′ is obtained.


### 4.2.2. Inverse Permutation Process

Step 1. According to the red, green, and blue components of the color image, the imageI′ is converted into a grayscale image with size of m×3n. Then, a binary matrix Ib′ with size of m×24n is obtained by each pixel of the grayscale image which is changed into an 8-bit array.
Step 2. Because the sum of all the elements with value of 1 in the binary matrixIb′ is equal to one of all the elements with value of 1 in the binary matrix Pb, the value of g is computed by the following equation:
(19)
g
=
sum
I
b
′
m
n
.
Step 3. With the same initial values as permutation process, the sequencesx1i,x2i,x3i,x4i are generated by system (4). Then, the keys keyh and keyv are computed by equations (8) and (9), respectively. Finally, the vectors h′ and v′ are obtained by equations (11) and (12) separately.
Step 4. According to the array vectorsh′ and v′, the rows and columns of binary image Ib′ are inversely permuted, respectively. The image Ib is obtained by this inverse permutation. We convert Ib with size of m×24n into the plain color image I whose size is of m×n. Thus, the decryption is achieved completely.


## 4.2.1. Decryption Algorithm

Step 1. Separate the encrypted imageIE into red, green, and blue grayscale images, and set ri′, gi′, and bi′i=1,2,…,mn to represent the i-th pixel value of these grayscale images, respectively. The decryption begins from the back to the front, that is to say that the mn-th pixel is firstly decrypted.
Step 2. With the same initial values with the encryption process, the sequencesy1i,y2i,y3i,y4i are generated by fractional-order dynamic system (7). Next, it can be calculated that smn=0 by equation (15). Then, the keys keyrmn,keygmn,keybmn can be obtained by equation (13) and smn=0. Finally, we can get rmn, gmn, and bmn after the following equation:
(18)
r
i
=
r
i
′
⊕
key
r
i
−
r
i
−
1
′
mod
256
,
g
i
=
g
i
′
⊕
key
g
i
−
g
i
−
1
′
mod
256
,
b
i
=
b
i
′
⊕
key
b
i
−
b
i
−
1
′
mod
256.
Step 3. With values ofrmn, gmn, and bmn and smn=0, the value of smn−1 can be computed by equations (14) and (15). Then, the (mn−1)-th pixel can be decrypted. Similarly, it is finished until the values of r1, g1, and b1 are decrypted.
Step 4. After repeating the above steps for the same rounds with the encryption process, the decrypted imageI′ is obtained.


## 4.2.2. Inverse Permutation Process

Step 1. According to the red, green, and blue components of the color image, the imageI′ is converted into a grayscale image with size of m×3n. Then, a binary matrix Ib′ with size of m×24n is obtained by each pixel of the grayscale image which is changed into an 8-bit array.
Step 2. Because the sum of all the elements with value of 1 in the binary matrixIb′ is equal to one of all the elements with value of 1 in the binary matrix Pb, the value of g is computed by the following equation:
(19)
g
=
sum
I
b
′
m
n
.
Step 3. With the same initial values as permutation process, the sequencesx1i,x2i,x3i,x4i are generated by system (4). Then, the keys keyh and keyv are computed by equations (8) and (9), respectively. Finally, the vectors h′ and v′ are obtained by equations (11) and (12) separately.
Step 4. According to the array vectorsh′ and v′, the rows and columns of binary image Ib′ are inversely permuted, respectively. The image Ib is obtained by this inverse permutation. We convert Ib with size of m×24n into the plain color image I whose size is of m×n. Thus, the decryption is achieved completely.


## 4.3. Experimental Result

The color image named pepper is selected as the plain images, whose size is197×206. The initial values of fractional-order hyperchaotic system (4) and system (7) are x10=0.67185367890218, x20=0.24566789543262, x30=0.15492289843576, x40=1.32854321678987, y10=0.98165678567657, y20=0.12345678901234,y30=1.65432109876543, and y40=0.57167689592916. The experimental results are shown in Figure 6.Figure 6
The encryption results of pepper image. (a) The original image. (b) The grayscale image of decomposing Figure6(a). (c) The image of permuting Figure 6(b). (d) The result of composing Figure 6(c). (e) The encrypted image.
(a)
(b)
(c)
(d)
(e)


## 4.4. Performance and Security Analysis

### 4.4.1. Key Space and Sensitivity

In the encryption algorithm, the secret keys are the initial values of the two hyperchaotic systems, that is,x10,x20,x30,x40 and y10,y20,y30,y40. Because the maximum precision of these initial values is 10−14, the total key space is that 10148=10112, which is much larger than 2100 [25]. Therefore, our algorithm can resist all kinds of brute force attacks.For the key sensitive test, the tiny change10−14 is selected as the error of these initial values, and it is shown in Table 2. Due to space limitations, only the changes of x10, x20, y30, and y40 were made. The experimental results are shown in Figure 7. It can be seen that these decrypted images are extremely similar to noise and absolutely different from the plain image, and those pixel distribution histograms are uniform. Moreover, we calculate the NPCR and UACI of the original image and recovered images (peppers) with different keys; it is shown in Table 3. It is easy to observe that NPCR is over 99%, and the UACI is close to 33%. It demonstrates that the recovered images with different keys are greatly different from their original form. Therefore, our algorithm is sensitive to the secret key.Table 2
The tiny change of secret keys.
Key
Original value
New value
x
1
0
0.67185367890218
0.67185367890218 + 10−14
x
2
0
0.24566789543262
0.24566789543262 − 10−14
y
3
0
1.65432109876543
1.65432109876543 + 10−14
y
4
0
0.57167689592916
0.57167689592916 − 10−14Figure 7
The decrypted images in tiny change of keys and their histograms. (a)x1(0) = 0.67185367890219. (b) x2(0) = 0.24566789543261. (c) y3(0) = 1.65432109876544. (d) y4(0) = 0.57167689592915. (e) The pixel distribution histogram of Figure 7(a). (f) The pixel distribution histogram of Figure 7(b). (g) The pixel distribution histogram of Figure 7(c). (h) The pixel distribution histogram of Figure 7(d).
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)Table 3
The NPCR and UACI of the original image and recovered image (peppers) with different keys.
Image
NPCR (%)
UACI (%)
R
G
B
R
G
B
Recovered image withx10
99.4259
99.3569
99.4185
28.3860
32.9807
32.8930
Recovered image withx20
99.3421
99.3864
99.4061
28.4061
32.8305
32.7692
Recovered image withy30
99.3174
99.3618
99.4209
28.2771
32.8916
32.7201
Recovered image withy40
99.3544
99.2755
99.3544
28.1678
32.4912
32.7000


### 4.4.2.χ2 Test

We compute theχ2 values of both plain image and encrypted image named Pepper, Flower, Yacht, and Baboon, whose sizes are all 197×206. It is shown in Table 4. Obviously, the χ2 values of the ciphered images are far lower than those of the plain images and similar to the algorithm in [26, 27]; they are all within the effective interval 210.7918,293.2478.Table 4
The results of theχ2 test.
χ
2
Plain image
Ciphered image
R
G
B
R
G
B
Pepper
35,300
32,679
64,067
258.8352
247.7227
208.7380
Flower
25,554
27,012
47,904
235.1820
250.1199
260.4022
Yacht
18,431
21,044
17,323
259.1406
271.4038
229.4794
Baboon
19,059
27,759
17,654
249.9811
237.2006
219.9287
Pepper in [26]
57,362
62,180
122,870
265.4625
269.3956
289.2321
Image 1 in [27]
—
—
—
230.8105
234.7070
250.3419
Image 2 in [27]
—
—
—
265.7558
286.3242
276.2754


### 4.4.3. Statistical Analysis

From the plain image and encrypted image, 2500 pairs of adjacent pixels are randomly selected in horizontal, vertical, and diagonal directions, respectively. Then, we calculate the correlation coefficients of two adjacent pixels, which are shown in Table5. From Table 5, it can be observed that the correlations of two adjacent pixels are close to 1 in the plain images, while the ones of the encrypted images are round 0 and similar to the algorithm in [28, 29]. Therefore, the encryption algorithm can eliminate the correlation of the adjacent pixels.Table 5
Correlation coefficients of the plain image and ciphered image.
Correlation
Horizontal
Vertical
Diagonal
Pepper
0.9373
0.9716
0.9478
Ciphered pepper
0.0021
0.0018
−0.0195
Flower
0.9414
0.9792
0.9769
Ciphered flower
0.0009
0.0173
−0.0194
Lena in [28] (R)
0.95409435
0.97692808
0.92946084
Lena in [28] (G)
0.93859702
0.96847242
0.91318153
Lena in [28] (B)
0.92230178
0.95144503
0.89275171
Ciphered lena in [28] (R)
0.00268849
0.00113425
0.00526812
Ciphered lena in [28] (G)
0.00979889
0.00302981
0.00038029
Ciphered lena in [28] (B)
0.00098796
0.00056287
0.00111043
Lena in [29]
0.9494
0.9667
0.9336
Ciphered lena in [29]
0.0054
0.0035
0.0016


### 4.4.4. Information Entropy

We compute the information entropy of the encrypted images according to the red, green, and blue components of the color image, which is shown in Table6. It can be seen that the information entropy of these ciphered images is all close to the value 8. So, it proves that the unpredictability of the encrypted images is very high.Table 6
The information entropy of encrypted images.
Color image
Red
Green
Blue
Pepper
7.9875
7.9880
7.9876
Flower
7.9882
7.9881
7.9871
Yacht
7.9867
7.9883
7.9874
Baboon
7.9881
7.9882
7.9875


### 4.4.5. Differential Attack

To test the differential attack, some color images are encrypted with altering a pixel in the plain images, respectively, and NPCR and UACI are calculated in Table7. It can be found that the NPCR is over 99%, the UACI is over 33%, and similar to the algorithm in [28–30]. It indicates that our algorithm can resist the differential attack.Table 7
The NPCR and UACI of ciphered images.
Image
NPCR (%)
UACI (%)
R
G
B
R
G
B
Pepper
99.5762
99.6205
99.5762
33.2831
33.4199
33.5400
Flower
99.6772
99.7142
99.6550
33.4547
33.7477
33.3845
Yacht
99.5269
99.5810
99.5146
33.3787
33.4017
33.4997
Baboon
99.6550
99.6723
99.6747
33.3987
33.3903
33.3644
Lena in [28]
99.98779
99.98779
99.98779
50.17915
50.18009
25.19263
Lena in [29] (avg.)
99.5723
33.3159
Lena in [30]
99.6231
99.6338
99.6170
33.4747
33.5683
33.3382


## 4.4.1. Key Space and Sensitivity

In the encryption algorithm, the secret keys are the initial values of the two hyperchaotic systems, that is,x10,x20,x30,x40 and y10,y20,y30,y40. Because the maximum precision of these initial values is 10−14, the total key space is that 10148=10112, which is much larger than 2100 [25]. Therefore, our algorithm can resist all kinds of brute force attacks.For the key sensitive test, the tiny change10−14 is selected as the error of these initial values, and it is shown in Table 2. Due to space limitations, only the changes of x10, x20, y30, and y40 were made. The experimental results are shown in Figure 7. It can be seen that these decrypted images are extremely similar to noise and absolutely different from the plain image, and those pixel distribution histograms are uniform. Moreover, we calculate the NPCR and UACI of the original image and recovered images (peppers) with different keys; it is shown in Table 3. It is easy to observe that NPCR is over 99%, and the UACI is close to 33%. It demonstrates that the recovered images with different keys are greatly different from their original form. Therefore, our algorithm is sensitive to the secret key.Table 2
The tiny change of secret keys.
Key
Original value
New value
x
1
0
0.67185367890218
0.67185367890218 + 10−14
x
2
0
0.24566789543262
0.24566789543262 − 10−14
y
3
0
1.65432109876543
1.65432109876543 + 10−14
y
4
0
0.57167689592916
0.57167689592916 − 10−14Figure 7
The decrypted images in tiny change of keys and their histograms. (a)x1(0) = 0.67185367890219. (b) x2(0) = 0.24566789543261. (c) y3(0) = 1.65432109876544. (d) y4(0) = 0.57167689592915. (e) The pixel distribution histogram of Figure 7(a). (f) The pixel distribution histogram of Figure 7(b). (g) The pixel distribution histogram of Figure 7(c). (h) The pixel distribution histogram of Figure 7(d).
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)Table 3
The NPCR and UACI of the original image and recovered image (peppers) with different keys.
Image
NPCR (%)
UACI (%)
R
G
B
R
G
B
Recovered image withx10
99.4259
99.3569
99.4185
28.3860
32.9807
32.8930
Recovered image withx20
99.3421
99.3864
99.4061
28.4061
32.8305
32.7692
Recovered image withy30
99.3174
99.3618
99.4209
28.2771
32.8916
32.7201
Recovered image withy40
99.3544
99.2755
99.3544
28.1678
32.4912
32.7000


## 4.4.2.χ2 Test

We compute theχ2 values of both plain image and encrypted image named Pepper, Flower, Yacht, and Baboon, whose sizes are all 197×206. It is shown in Table 4. Obviously, the χ2 values of the ciphered images are far lower than those of the plain images and similar to the algorithm in [26, 27]; they are all within the effective interval 210.7918,293.2478.Table 4
The results of theχ2 test.
χ
2
Plain image
Ciphered image
R
G
B
R
G
B
Pepper
35,300
32,679
64,067
258.8352
247.7227
208.7380
Flower
25,554
27,012
47,904
235.1820
250.1199
260.4022
Yacht
18,431
21,044
17,323
259.1406
271.4038
229.4794
Baboon
19,059
27,759
17,654
249.9811
237.2006
219.9287
Pepper in [26]
57,362
62,180
122,870
265.4625
269.3956
289.2321
Image 1 in [27]
—
—
—
230.8105
234.7070
250.3419
Image 2 in [27]
—
—
—
265.7558
286.3242
276.2754


## 4.4.3. Statistical Analysis

From the plain image and encrypted image, 2500 pairs of adjacent pixels are randomly selected in horizontal, vertical, and diagonal directions, respectively. Then, we calculate the correlation coefficients of two adjacent pixels, which are shown in Table5. From Table 5, it can be observed that the correlations of two adjacent pixels are close to 1 in the plain images, while the ones of the encrypted images are round 0 and similar to the algorithm in [28, 29]. Therefore, the encryption algorithm can eliminate the correlation of the adjacent pixels.Table 5
Correlation coefficients of the plain image and ciphered image.
Correlation
Horizontal
Vertical
Diagonal
Pepper
0.9373
0.9716
0.9478
Ciphered pepper
0.0021
0.0018
−0.0195
Flower
0.9414
0.9792
0.9769
Ciphered flower
0.0009
0.0173
−0.0194
Lena in [28] (R)
0.95409435
0.97692808
0.92946084
Lena in [28] (G)
0.93859702
0.96847242
0.91318153
Lena in [28] (B)
0.92230178
0.95144503
0.89275171
Ciphered lena in [28] (R)
0.00268849
0.00113425
0.00526812
Ciphered lena in [28] (G)
0.00979889
0.00302981
0.00038029
Ciphered lena in [28] (B)
0.00098796
0.00056287
0.00111043
Lena in [29]
0.9494
0.9667
0.9336
Ciphered lena in [29]
0.0054
0.0035
0.0016


## 4.4.4. Information Entropy

We compute the information entropy of the encrypted images according to the red, green, and blue components of the color image, which is shown in Table6. It can be seen that the information entropy of these ciphered images is all close to the value 8. So, it proves that the unpredictability of the encrypted images is very high.Table 6
The information entropy of encrypted images.
Color image
Red
Green
Blue
Pepper
7.9875
7.9880
7.9876
Flower
7.9882
7.9881
7.9871
Yacht
7.9867
7.9883
7.9874
Baboon
7.9881
7.9882
7.9875


## 4.4.5. Differential Attack

To test the differential attack, some color images are encrypted with altering a pixel in the plain images, respectively, and NPCR and UACI are calculated in Table7. It can be found that the NPCR is over 99%, the UACI is over 33%, and similar to the algorithm in [28–30]. It indicates that our algorithm can resist the differential attack.Table 7
The NPCR and UACI of ciphered images.
Image
NPCR (%)
UACI (%)
R
G
B
R
G
B
Pepper
99.5762
99.6205
99.5762
33.2831
33.4199
33.5400
Flower
99.6772
99.7142
99.6550
33.4547
33.7477
33.3845
Yacht
99.5269
99.5810
99.5146
33.3787
33.4017
33.4997
Baboon
99.6550
99.6723
99.6747
33.3987
33.3903
33.3644
Lena in [28]
99.98779
99.98779
99.98779
50.17915
50.18009
25.19263
Lena in [29] (avg.)
99.5723
33.3159
Lena in [30]
99.6231
99.6338
99.6170
33.4747
33.5683
33.3382


## 5. Conclusion

In this paper, the anticontrol of the fractional-order chaotic system is studied. We give the necessary conditions for the anticontrol of the fractional-order chaotic system. With the same parameters and fractional order, a 3D chaotic system is driven to two new 4D fractional-order hyperchaotic systems, respectively. We compute Lyapunov exponents and bifurcation diagrams of these new fractional dynamic systems.Based on these two fractional-order hyperchaotic systems, a color image encryption algorithm is designed. In permutation process, the key is related to the plain image, and the key in encryption process is dynamically changing with different plain images and encrypted images. Therefore, our algorithm can resist the chosen plaintext (ciphertext) attacks and overcome the difficulty of key management in the “one time and one secret (one time one key)” scheme. In addition, the security analysis shows that our algorithm has better security. Therefore, it verifies the effectiveness of these fractional dynamic systems for image encryption.In the future, we will further study on the anticontrol of the fractional-order chaotic system and identify the new chaotic system scientifically by the criteria from the work [31]. In image encryption, the fractional hyperchaotic system could be extended to a computational model for parallel image encryption algorithms.



---

*Source: 6795964-2020-03-12.xml*