# Image Encryption Technology Based on Fractional Two-Dimensional Triangle Function Combination Discrete Chaotic Map Coupled with Menezes-Vanstone Elliptic Curve Cryptosystem

Authors: Zeyu Liu; Tiecheng Xia; Jinbo Wang
Journal: Discrete Dynamics in Nature and Society
 (2018)

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## Abstract

A new fractional two-dimensional triangle function combination discrete chaotic map (2D-TFCDM) with the discrete fractional difference is proposed. We observe the bifurcation behaviors and draw the bifurcation diagrams, the largest Lyapunov exponent plot, and the phase portraits of the proposed map, respectively. On the application side, we apply the proposed discrete fractional map into image encryption with the secret keys ciphered by Menezes-Vanstone Elliptic Curve Cryptosystem (MVECC). Finally, the image encryption algorithm is analysed in four main aspects that indicate the proposed algorithm is better than others.


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## Body

## 1. Introduction

Nowadays, image encryption plays a significant role with the development of security technology in the areas of network, communication, and cloud service. Multifarious chaos-based image encryption algorithms have been developed up to now, such as in [1–6]; however a few of them have referred to the image encryption algorithm based on fractional discrete chaotic map accompanied with Elliptic Curve Cryptography (ECC).The theory of the fractional difference has been developed for decades [7–13]. Recently, Wu et al. [14–16] made a contribution to the application of the discrete fractional calculus (DFC) on an arbitrary time scale, and the theories of delta difference equations were utilized to reveal the discrete chaos behavior.ECC is a widely used technology in data security and communication security; it can achieve the same level of security with smaller key sizes and higher computational efficiency [17]. Many famous public-key algorithms, such as Diffie-Hellman, EIGamal, and Schnorr, can be implemented by means of elliptic curves over finite fields. MVECC is one of the popular elliptic curve public-key cryptosystems [18] and we adopt it in our cryptosystem.Many encryption methods based on fractional derivatives have been proposed in recent time, like fractional logistic maps [19], fractional-order chaos systems [20], and fractional form of hyperchaotic system [21].In [22], a new image encryption algorithm based on one-dimensional fractional chaotic time series within fractional-order difference has been proposed; however, the two-dimensional discrete chaotic map has seldom been used in image encryption except [23, 24].Our main purpose is to introduce a new two-dimensional discrete chaotic map based on fractional-order difference and apply it in image encryption. The rest of this paper is organized as follows. In Section2, the definitions and the properties of the DFC are introduced. After that, the definitions and operation of ECC are given. Then, the working principle of MVECC is described in the next section. In Section 5, we give the fractional 2D-TFCDM on time scales from the discrete integral expression. The bifurcation diagrams and the phase portraits of the map are presented while the difference orders and the coefficients are changing; the largest Lyapunov exponent plots are also displayed. Afterwards, we apply the proposed map into image encryption and show several examples. In Section 7, the performance of the proposed image encryption method is analysed systematically. Finally, we have come to some conclusions.


## 2. Preliminaries

The definitions of the fractional sum and difference are given as follows. LetNa denote the isolated time scale and Na=a,a+1,a+2,… (a∈R fixed). Within the DFC, the function f(t) is changed as a sequence f(n). The difference operator Δ is defined as Δf(n)=f(n+1)-f(n).Definition 1 (see [25]).
Letu: Na→R and 0<ν be given. The νth fractional sum is defined by(1)Δa-νut≔1Γν∑s=at-νt-s-1ν-1us,t∈Na+ν.Note that a is the starting point; t(ν) is the falling function defined as(2)tν=Γt+1Γt+1-ν.Definition 2 (see [26]).
For0<ν, ν∉N, and u(t) defined on Na, the ν-order Caputo fractional difference is defined by(3)ΔaνCut≔Δa-m-νΔmut=1Γm-ν∑s=at-m-νt-s-1m-ν-1Δmus,t∈Na+m-ν,m=ν+1.Theorem 3 (see [27]).
For the delta fractional difference equation(4)ΔaνCut=ft+ν-1,ut+ν-1,Δkua=uk,m=ν+1,k=0,…,m-1the equivalent discrete integral equation is(5)xn=u0t+1Γν∑s=a+m-νt-νt-s-1ν-1×fs+ν-1,us+ν-1,t∈Na+m,where(6)u0t=∑k=0m-1t-akk!Δkua.The complex difference equation with long-term memory is obtained here. It can reduce to the integer order one with the difference order ν=1 but the integer one does not hold the discrete memory. From (3) to (5), the domain is shifted from Na+m-ν to Na+m and the function u(t) is preserved to be defined on the isolated time scale Na in the fractional sums.


## 3. Introduction to Elliptic Curve

Definition 4.
An elliptic curve (EC)E over a prime field Fp denoted by E(Fp) refers to the set of all points (x,y) that satisfy the equation(7)E:y2≡x3+ax+bmodp,together with a special point O at infinity, where a,b∈Fp, p≠2,3 and 4a3+27b2≠0 [28, 29].


### 3.1. Elliptic Curve Operations

IfP=(x1,y1), Q=(x2,y2)∈E(Fp); then if x1=x2 but y1≠y2, P+Q=O; that is, Q=-P=(x1,-y1) [29].(8)P+Q=R=x3,y3,P≠-Q,O,P=-Q, where(9)x3≡λ2-2x1modp,y3≡λx1-x3-y1modp,λ=y2-y1x2-x1,P≠Q,3x12+a2y1,P=Q.The scalar multiplication overE(Fp) is defined by(10)kP=P+P+⋯+P︸ktimes, where k is an integer.Definition 5.
The order of an EC is defined by the number of points that lie on the EC denoted by#E [29].Definition 6.
SetP∈E(Fp), and then P is called a generator point if ord(P)=#E (ord(P) is the smallest positive integer n that makes nP=O) [29].


## 3.1. Elliptic Curve Operations

IfP=(x1,y1), Q=(x2,y2)∈E(Fp); then if x1=x2 but y1≠y2, P+Q=O; that is, Q=-P=(x1,-y1) [29].(8)P+Q=R=x3,y3,P≠-Q,O,P=-Q, where(9)x3≡λ2-2x1modp,y3≡λx1-x3-y1modp,λ=y2-y1x2-x1,P≠Q,3x12+a2y1,P=Q.The scalar multiplication overE(Fp) is defined by(10)kP=P+P+⋯+P︸ktimes, where k is an integer.Definition 5.
The order of an EC is defined by the number of points that lie on the EC denoted by#E [29].Definition 6.
SetP∈E(Fp), and then P is called a generator point if ord(P)=#E (ord(P) is the smallest positive integer n that makes nP=O) [29].


## 4. Menezes-Vanstone Elliptic Curve Cryptosystem (MVECC)

MVECC is one of most significant extensions of ECC; the working principle of MVECC is as follows.If Andy wants to encrypt and send the messageM to Bob, they should do the step as mentioned hereunder:( 1 ) Andy and Bob make an agreement on an elliptic curve E(Fp) and the base point α.( 2 ) Bob firstly selects a private key k to compute the public key γ=k·α (0≦k<ord(α)).( 3 ) If Andy wants to send a message M=(x1,x2)∈Zp∗×Zp∗ to Bob, he firstly chooses a random private key d (0≦d<ord(α)) and then computes his public key β=d·α. On the other hand, Andy calculates the secret key (c1,c2) by(11)c1,c2=d·γ=d·k·α=k·β.He should compute the ciphered message afterwards by(12)y1=x1∗c1modp,y2=x2∗c2modp.( 4 ) Then the ciphertext {γ,(y1,y2)} is sent to Bob. When Bob wants to get the plaintext (x1,x2), firstly, he computes the secret key (c1,c2)=k·β=k·d·α, and then he computes M=(x1,x2) by(13)x1=y1∗c1-1modp,x2=y2∗c2-1modp, to get the plaintext [18].Any adversary that only hasβ and γ without the private keys d and k very difficultly breaks the MVECC to get the plaintext M. What is more, if #E have only one big prime divisor, the EC is called a safe EC [29]; then, the MVECC can become an more efficient and secure cryptosystem.


## 5. Fractional 2D-TFCDM

From [14–16], we notice the application of the DFC in fractional generalizations of the discrete chaotic maps. Recently [30], the following 2D-TFCDM was proposed:(14)xn+1=k1cos⁡xn+yn,k1=8,yn+1=k2sin⁡xn-yn,k2=0.5. Now, consider the fractional generalization of x(n); the map was also studied in [31]:(15)ΔανCxt=k1cos⁡xt+ν+yt+ν-xt+ν,0<ν<1,t∈Na+1-ν,yn+1=k2sin⁡xn-yn,k2=0.5.From Theorem 3, we have the following equivalent discrete numerical formula for the variable k1: (k2=0.5) with 0<ν<1:(16)xn=x0+1Γν∑j=1nΓn-j+νΓn-j+1k1cos⁡xj-1+yj-1-xj-1,yn=k2sin⁡xn-1-yn-1,k2=0.5.Letν = 1, x0=0.19, y(0)=0.06, n=200, and k1 be fixed. In what follows, Figure 1 is the bifurcation diagram where the step size of k1 is 0.002. Figure 2 is the same bifurcation diagram except for ν = 0.8.Figure 1
The bifurcation diagram of the 2D-TFCDM of variablek1 for ν=1.Figure 2
The bifurcation diagram of the fractional 2D-TFCDM of variablek1 for ν=0.8.In Figures3 and 4, the largest Lyapunov exponent plots are drawn by use of the Jacobian matrix algorithm proposed in [32]. The largest Lyapunov exponent LE is positive somewhere; it is corresponding to the chaotic intervals in Figures 1 and 2.Figure 3
The largest Lyapunov exponent of the 2D-TFCDM of the variablek1.Figure 4
The largest Lyapunov exponent of the fractional 2D-TFCDM of the variablek1.By choosing 101 different initial values we can ploty(n) versus x(n) in one figure. The phase portraits of the integer map are derived from Figure 5. The cases of ν=0.8 and ν=0.6 are plotted in Figures 6 and 7, respectively.Figure 5
The phase portraits of the 2D-TFCDM fork1=8, k2=0.5, and ν=1.Figure 6
The phase portraits of the fractional 2D-TFCDM fork1=8, k2=0.5, and ν=0.8.Figure 7
The phase portraits of the fractional 2D-TFCDM fork1=8, k2=0.5, and ν=0.6.


## 6. Applications

The fractionalized chaotic map can be applied in image encryption. Exploit (16) into an algorithm, and set the initial values x0, y0, the order ν, and the coefficients k1, k2 of chaotic system as keys. In this paper, we propose the encryption algorithm and divide it into 3 parts.


### 6.1. Generation of New Keys Based on Elliptic Curve in a Finite Field

Settinga=1, b=6, and p=9996887 in (7), we can get E(F9996887). Since #E=10000721 is a prime number, according to [29], E(F9996887) is a safe EC. Let α=(2,4), randomly select d=9134417, k=1269960∈[1,#E]; then β=dα=(6020909,7282175), γ=kα=(7495358,7052635), and (c1,c2)=kβ=(3049362,3915118)=dγ. The initial key ν=0.6026331, x0=4.107532, ν01=ν×107=6026331, and x01=x0×106=4107532.Calculate(17)ν01′=c1∗ν01modp=3049362·6026331mod9996887=7123456mod9996887,x01′=c2·x01modp=3915118·4107532mod9996887=190000mod9996887. Then, the ciphertext is ((7495358, 7052635), 7123456, 190000), the enciphered key is ν′=ν01′/107=0.7123456, and x0′=x01′/106=0.19.Makey0=3.650991, k1=0.897029, and k2=0.434264, and compute y01=y0×106, k01=k1×106, and k02=k2×106; then(18)y01′=c1·y01modp=3049362·3650991mod9996887=60000mod9996887,k01′=c2·k01modp=3915118·897029mod9996887=8000000mod9996887,k02′=c1·k02modp=3049362·434264mod9996887=500000mod9996887.Sety0′=y01′/106=0.06, k1′=k01′/106=8, k2′=k02′/106=0.5, and then x0′, y0′, ν′, k1′, k2′ are taken as the keys of Section 6.2.


### 6.2. Permutation Procedure Based on Fractional 2D-TFCDM

Taking advantage of (16) with the initial values x0′, y0′, ν′, k1′, and k2′ generated in the last section, we can encrypt the image. The next step of encryption is permutation; it is subdivided into 4 steps:1 Set x0′ as x(1); iterate (16) for MN-1 times to generate the one-dimensional real number chaotic sequence x(i), i=1,2,…,MN; here M and N denote the length and width of the original image V, respectively.2 Reorder x(k) by the bubble sort and get x′(k), and record the change of the subscript of x(k) as z(k).3 Change M×N original image V into 1×MN sequence v(k), and rearrange v(k) according to z(k) to get the new sequence v′(k).4 Reshape v′(k) into M×N image as V′; V′ is the permutated image we needed.Reversing the above 4 steps, we can remove the effect of permutation to get the original image.


### 6.3. Encryption Method Based on Fractional 2D-TFCDM

1 In Section 6.2 we get the chaotic sequence x(i) and image V′. Reshape M×N image V′ into 1×MN sequence u(i); that is i=N(m-1)+n, (m=1,2,…,M, n=1,2,…,N). Another M×N image is used as a key image (K-image). Change the K-image also into 1×MN sequence w(i).2 Set i=0.3 Round x(i)×108 as x1(i), do modulus operation to x1(i) in (19), and get x2(i):(19)x2i=modx1i,256.4 Do the following operation and get u′(i):(20)u′i=ui⊕modwi+x2i,256,where ⊕ refers to the Xor operation, and u′(i) is the encrypted pixel value.The inverse form of (20) is(21)ui=u′i⊕modwi+x2i,256.5 Compute the iteration times k(i) according to (22)ki=1+modu′i,256.Then, iterate (16) for k(i) times to get x(i+1), circle from step 3 to step 5, until getting x(MN).6 Change u′(i) into M×N image as V′′, which is the finally encrypted figure we need.The decryption procedure is including 2 parts:1 Do all steps in encryption process except (20) which is replaced by (21).2 Reverse the procedure in Section 6.2. Then the decryption procedure is done.Figure8 shows the encryption process described in Sections 6.2 and 6.3 in a flow chart, and Figure 9 illustrates the iteration procedure of S box.Figure 8
The proposed encryption method.Figure 9
The S box.The original, encrypted, and decrypted images are shown in Figures10–18. The proposed algorithm can encrypt any rectangular image.Figure 10
Cameraman.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 11
Lena.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 12
Peppers.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 13
Lake.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 14
Dollar.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 15
Columbia.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 16
Lax.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 17
Boat.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 18
Aerial.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureThe adopted cryptosystem in Section6.1 is asymmetric; however, the ones in Sections 6.2 and 6.3 are symmetric.


## 6.1. Generation of New Keys Based on Elliptic Curve in a Finite Field

Settinga=1, b=6, and p=9996887 in (7), we can get E(F9996887). Since #E=10000721 is a prime number, according to [29], E(F9996887) is a safe EC. Let α=(2,4), randomly select d=9134417, k=1269960∈[1,#E]; then β=dα=(6020909,7282175), γ=kα=(7495358,7052635), and (c1,c2)=kβ=(3049362,3915118)=dγ. The initial key ν=0.6026331, x0=4.107532, ν01=ν×107=6026331, and x01=x0×106=4107532.Calculate(17)ν01′=c1∗ν01modp=3049362·6026331mod9996887=7123456mod9996887,x01′=c2·x01modp=3915118·4107532mod9996887=190000mod9996887. Then, the ciphertext is ((7495358, 7052635), 7123456, 190000), the enciphered key is ν′=ν01′/107=0.7123456, and x0′=x01′/106=0.19.Makey0=3.650991, k1=0.897029, and k2=0.434264, and compute y01=y0×106, k01=k1×106, and k02=k2×106; then(18)y01′=c1·y01modp=3049362·3650991mod9996887=60000mod9996887,k01′=c2·k01modp=3915118·897029mod9996887=8000000mod9996887,k02′=c1·k02modp=3049362·434264mod9996887=500000mod9996887.Sety0′=y01′/106=0.06, k1′=k01′/106=8, k2′=k02′/106=0.5, and then x0′, y0′, ν′, k1′, k2′ are taken as the keys of Section 6.2.


## 6.2. Permutation Procedure Based on Fractional 2D-TFCDM

Taking advantage of (16) with the initial values x0′, y0′, ν′, k1′, and k2′ generated in the last section, we can encrypt the image. The next step of encryption is permutation; it is subdivided into 4 steps:1 Set x0′ as x(1); iterate (16) for MN-1 times to generate the one-dimensional real number chaotic sequence x(i), i=1,2,…,MN; here M and N denote the length and width of the original image V, respectively.2 Reorder x(k) by the bubble sort and get x′(k), and record the change of the subscript of x(k) as z(k).3 Change M×N original image V into 1×MN sequence v(k), and rearrange v(k) according to z(k) to get the new sequence v′(k).4 Reshape v′(k) into M×N image as V′; V′ is the permutated image we needed.Reversing the above 4 steps, we can remove the effect of permutation to get the original image.


## 6.3. Encryption Method Based on Fractional 2D-TFCDM

1 In Section 6.2 we get the chaotic sequence x(i) and image V′. Reshape M×N image V′ into 1×MN sequence u(i); that is i=N(m-1)+n, (m=1,2,…,M, n=1,2,…,N). Another M×N image is used as a key image (K-image). Change the K-image also into 1×MN sequence w(i).2 Set i=0.3 Round x(i)×108 as x1(i), do modulus operation to x1(i) in (19), and get x2(i):(19)x2i=modx1i,256.4 Do the following operation and get u′(i):(20)u′i=ui⊕modwi+x2i,256,where ⊕ refers to the Xor operation, and u′(i) is the encrypted pixel value.The inverse form of (20) is(21)ui=u′i⊕modwi+x2i,256.5 Compute the iteration times k(i) according to (22)ki=1+modu′i,256.Then, iterate (16) for k(i) times to get x(i+1), circle from step 3 to step 5, until getting x(MN).6 Change u′(i) into M×N image as V′′, which is the finally encrypted figure we need.The decryption procedure is including 2 parts:1 Do all steps in encryption process except (20) which is replaced by (21).2 Reverse the procedure in Section 6.2. Then the decryption procedure is done.Figure8 shows the encryption process described in Sections 6.2 and 6.3 in a flow chart, and Figure 9 illustrates the iteration procedure of S box.Figure 8
The proposed encryption method.Figure 9
The S box.The original, encrypted, and decrypted images are shown in Figures10–18. The proposed algorithm can encrypt any rectangular image.Figure 10
Cameraman.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 11
Lena.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 12
Peppers.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 13
Lake.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 14
Dollar.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 15
Columbia.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 16
Lax.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 17
Boat.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 18
Aerial.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureThe adopted cryptosystem in Section6.1 is asymmetric; however, the ones in Sections 6.2 and 6.3 are symmetric.


## 7. Analysis of Results in Applications

### 7.1. Key Space

In the proposed algorithm, the initial valuesx0, y0, the order ν, and the coefficients k1, k2 are taken as the secret keys; consequently there are 5 keys. Assume the precision of x0, y0, ν, k1, and k2 are 10-16,3×10-17,10-16,10-15, and 10-16, respectively; then the key’s space is 1/3×1080≈1.12×2264. If the size of the plaintext is 512×512, then the key space of K-image is also 512×512×28=226. The total key space of the proposed algorithm is 1.12×2290.


### 7.2. Statistics Analysis

The quality against any statistical attack is important for a well-designed encryption method; it include 3 aspects as follows.


#### 7.2.1. Correlation of the Plain- and Cipher-Images

In an ordinary image, the adjacent pixels are related; therefore the correlation coefficient of adjacent pixels is usually high. A good encryption algorithm should make the correlation coefficients of encrypted image nearly equal to zero. The closer to zero the correlation coefficients is, the better the encryption algorithm is. Formulas (23) calculate the correlation coefficient. The correlations along the x direction of both original and encrypted images are displayed in Figures 19–27 from Cameraman to Aerial. The correlation coefficients are displayed in Table 1.(23)rxy=cov⁡x,yDxDycov⁡x,y=1N∑i=1Nxi-Exyi-EyEx=1N∑i=1NxiDx=1N∑i=1Nxi-Ex2.Table 1
Correlation coefficients of image.
Image Original image Encrypted image Horizontal Diagonal Vertical Horizontal Diagonal Vertical Cameraman 0.9276 0.9120 0.9597 0.0119 −0.0021 −0.0025 Lena 0.9722 0.9527 0.9860 −0.0140 −0.0086 −0.0034 Peppers 0.9667 0.9382 0.9694 −0.0088 0.0080 −0.0054 Lake 0.9768 0.9544 0.9748 −0.0155 0.0101 −0.0088 Dollar 0.8035 0.6952 0.6938 0.0131 −0.0183 0.0263 Columbia 0.9727 0.9403 0.9705 0.0060 −0.0104 −0.0093 Lax 0.7889 0.7151 0.8483 −0.0107 0.0147 0.0107 Boat 0.9407 0.9158 0.9545 0.0169 −0.0074 −0.0077 Aerial 0.9135 0.7952 0.8677 0.0084 −0.0123 −0.0133Figure 19
Cameraman.
(a)
The original figure (b)
The encrypted figureFigure 20
Lena.
(a)
The original figure (b)
The encrypted figureFigure 21
Peppers.
(a)
The original figure (b)
The encrypted figureFigure 22
Lake.
(a)
The original figure (b)
The encrypted figureFigure 23
Dollar.
(a)
The original figure (b)
The encrypted figureFigure 24
Columbia.
(a)
The original figure (b)
The encrypted figureFigure 25
Lax.
(a)
The original figure (b)
The encrypted figureFigure 26
Boat.
(a)
The original figure (b)
The encrypted figureFigure 27
Aerial.
(a)
The original figure (b)
The encrypted figureWith the sharp contrast of data between original image and encrypted image, Table1 indicates that the encryption process make pixels of the encrypted image almost independent with each other. Consequently, the encryption algorithm is good at pixel value randomization.Compared with other algorithm, we can observe that most correlation coefficients of encrypted image are nearer to 0 in Table2. As a consequence of this, the proposed encryption algorithm is superior to others.Table 2
Comparison of correlation coefficients of image.
Algorithm Image Original image Encrypted image Horizontal Vertical Diagonal Horizontal Vertical Diagonal Proposed Lena 0.9722 0.9527 0.9860 −0.0140 −0.0086 −0.0034 [1] Lena 0.9503 0.9755 0.9275 −0.0226 0.0041 0.0368 [2] Lena 0.927970 0.926331 0.839072 −0.010889 −0.018110 −0.006104 [5] Lena 0.946 0.973 0.921 −0.0055 −0.0075 0.0026 [6] Lena 0.9569 0.9236 0.9019 0.0042 −0.0043 0.0163


#### 7.2.2. Histogram

Histogram reflects the distribution of colors inside the image. The adversary can get some effective information from the regularity of histogram. Therefore, a well-designed image encryption method should make the pixel value of encrypted image distribute uniformly. Figure28 shows the histogram of Cameraman. Similarly, the histograms of the other 8 cases are drawn in Figures 29–36. It is illustrated that the proposed encryption method has a good effect on pixel value distribution uniformization.Figure 28
Cameraman.
(a)
The original image (b)
The encrypted image (c)
The decrypted imageFigure 29
Lena.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 30
Peppers.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 31
Lake.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 32
Dollar.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 33
Columbia.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 34
Lax.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 35
Boat.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 36
Aerial.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figure


#### 7.2.3. Information Entropy

Information entropy defines the randomness and the unpredictability of information in an image. It is defined by(24)Hm=∑i=02n-1pmilog2⁡1pmi.Here p(mi) is the probability of mi; n is the number of bits that is required to represent the symbol mi. For the pixels values of the image are 0~255, according to (24) the information entropy is 8 bits for an ideally random image. Therefore, the closer to 8 bits the information entropy is, the better the encryption algorithm is. The information entropy of the 9 cases is gotten in Table 3; it indicates that the encrypted images are very close to the random images.Table 3
Information entropy.
Image Original image Encrypted image Cameraman 7.0097 7.9974 Peppers 7.5739 7.9976 Dollar 6.9785 7.9992 Lax 6.8272 7.9993 Aerial 6.9940 7.9992 Lena 7.2185 7.9993 Lake 7.4845 7.9993 Columbia 7.2736 7.9992 Boat 6.9391 7.9972From Table4, we can observe that the information entropy of proposed algorithm is nearer to 8 bits than other algorithms.Table 4
Comparison of information entropy.
Algorithm Image Original image Encrypted image Proposed Lena 7.2185 7.9993 [1] Lena 7.2072 7.9973 [4] Lena Undefined 7.9972 [19] Lena Undefined 7.987918 [20] Lena 7.447144 7.988847


### 7.3. Sensitivity Analysis

The different range between two images is measured by two criteria: number of pixels change rate (NPCR) and unified average changing intensity (UACI). They are defined as follows:(25)Di,j=0,T1i,j=T2i,j,1,T1i,j≠T2i,j,NPCR=∑i=1W∑j=1HDi,jW×H×100%UACI=∑i=1W∑j=1HT1i,j-T2i,j255W×H×100%.Here W and H are the width and the height of T1 and T2.


#### 7.3.1. Key Sensitivity

We encrypt the image by the keysx0=0.19, y0=0.06, ν=0.7123456, k1=8, and k2=0.5. Figure 37(a) is the decrypted image with the correct keys. Figure 37(b) represents the decrypted image under 10-16 adding to x0 with other keys unchanged. Similarly, the secret keys y0,ν,k1,k2 are added as 3×10-17,10-16,10-15 and 10-16 to decrypt the images separately with other keys unchanged. The results are shown in Figures 37(c)–37(f). The comparison of key space is shown in Table 5 and the NPCR and UACI between Figures 37(a) and 37(b)–37(f) are calculated in Table 6.Table 5
Comparison of key spaces.
Algorithm Proposed [2] [4] [6] Key spaces 2.23 × 1 0 87 1.12 × 2 290 2 128 ≈2273 2276Table 6
NPCR and UACI between Figures37(a) and 37(b)–37(f).
Image NPCR and UACI NPCR (%) UACI (%) Figure37(b) 99.61 31.26 Figure37(c) 97.02 30.23 Figure37(d) 99.60 31.03 Figure37(e) 99.61 31.01 Figure37(f) 99.62 31.27Figure 37
The test of key sensitivity.
(a)
The correct keys (b)
x 0 + 10 - 16 (c)
y 0 + 3 × 10 - 17 (d)
v + 10 - 16 (e)
k 1 + 10 - 15 (f)
k 2 + 10 - 16In contrast with other algorithm, the key space of proposed algorithm is larger than others.Most NPCR are near to 99.61% and most of UACI are higher than 30% in Table6. We cannot recognize the man inside from Figures 37(b)–37(f); therefore the encryption method is sensitive to the keys.


#### 7.3.2. Plaintext Sensitivity

By encrypting two same images with only one pixel difference, the attackers can obtain effective information by comparing the two encrypted images. Therefore an encryption method designed against differential attack should ensure that the two encrypted images are completely different even if there is only a pixel difference in the original image.In Table7, Figure 10(a)(x,y) is the same as Figure 10(a) except for a pixel locating (x,y). After that, the 2 images are encrypted with the same keys and the NPCR and UACI between the 2 ciphertext images are calculated. Similarly, the data of other 8 cases are obtained in Tables 8–15.Table 7
NPCR and UACI between cipher-images with slightly different plain-images.
Image NPCR and UACI of Cameraman NPCR (1-round %) UACI (1-round %) NPCR (2-round %) UACI (2-round %) Figure10(a)(30, 30) 4.84 1.64 99.57 33.61 Figure10(a)(50, 50) 81.43 27.39 99.62 33.56 Figure10(a)(80, 80) 80.87 27.19 99.59 33.51 Figure10(a)(100, 100) 6.82 2.28 99.59 33.46Table 8
NPCR and UACI between cipher-images with slightly different plain-images.
Image NPCR and UACI of Lena NPCR (1-round %) UACI (1-round %) NPCR (2-round %) UACI (2-round %) Figure11(a)(30, 30) 1.21 0.41 99.59 33.39 Figure11(a)(50, 50) 95.06 31.93 99.59 33.53 Figure11(a)(80, 80) 94.90 31.93 99.60 33.48 Figure11(a)(100, 100) 1.71 0.58 99.63 33.40Table 9
NPCR and UACI between cipher-images with slightly different plain-images.
Image NPCR and UACI of Peppers NPCR (1-round %) UACI (1-round %) NPCR (2-round %) UACI (2-round %) Figure12(a)(30, 30) 4.83 1.64 99.56 33.44 Figure12(a)(50, 50) 81.44 27.35 99.62 33.50 Figure12(a)(80, 80) 6.12 2.03 99.57 33.42 Figure12(a)(100, 100) 6.83 2.32 99.60 33.50Table 10
NPCR and UACI between cipher-images with slightly different plain-images.
Image NPCR and UACI of Lake NPCR (1-round %) UACI (1-round %) NPCR (2-round %) UACI (2-round %) Figure13(a)(30, 30) 1.21 0.41 99.61 33.46 Figure13(a)(50, 50) 95.09 31.88 99.61 33.42 Figure13(a)(80, 80) 94.92 31.89 99.60 33.46 Figure13(a)(100, 100) 1.71 0.58 99.59 33.47Table 11
NPCR and UACI between cipher-images with slightly different plain-images.
Image NPCR and UACI of Dollar NPCR (1-round %) UACI (1-round %) NPCR (2-round %) UACI (2-round %) Figure14(a)(30, 30) 1.21 0.41 99.64 33.42 Figure14(a)(50, 50) 95.08 32.00 99.60 33.49 Figure14(a)(80, 80) 94.90 31.93 99.61 33.48 Figure14(a)(100, 100) 1.71 0.57 99.61 33.41Table 12
NPCR and UACI between cipher-images with slightly different plain-images.
Image NPCR and UACI of Columbia NPCR (1-round %) UACI (1-round %) NPCR (2-round %) UACI (2-round %) Figure15(a)(30, 30) 94.83 31.96 99.61 33.47 Figure15(a)(50, 50) 93.48 31.40 99.48 33.36 Figure15(a)(80, 80) 0.96 0.32 99.60 33.45 Figure15(a)(100, 100) 1.11 0.38 99.61 33.51Table 13
NPCR and UACI between cipher-images with slightly different plain-images.
Image NPCR and UACI of Lax NPCR (1-round %) UACI (1-round %) NPCR (2-round %) UACI (2-round %) Figure16(a)(30, 30) 1.21 0.40 99.62 33.39 Figure16(a)(50, 50) 95.06 31.99 99.61 33.49 Figure16(a)(80, 80) 94.92 31.87 99.58 33.41 Figure16(a)(100, 100) 1.70 0.58 99.62 33.48Table 14
NPCR and UACI between cipher-images with slightly different plain-images.
Image NPCR and UACI of Boat NPCR (1-round %) UACI (1-round %) NPCR (2-round %) UACI (2-round %) Figure17(a)(30, 30) 4.83 1.60 99.59 33.47 Figure17(a)(50, 50) 81.46 27.45 99.62 33.59 Figure17(a)(80, 80) 80.82 27.24 99.58 33.48 Figure17(a)(100, 100) 6.82 2.32 99.62 33.61Table 15
NPCR and UACI between cipher-images with slightly different plain-images.
Image NPCR and UACI of Aerial NPCR (1-round %) UACI (1-round %) NPCR (2-round %) UACI (2-round %) Figure18(a)(30, 30) 1.21 0.41 99.61 33.49 Figure18(a)(50, 50) 95.06 31.93 99.62 33.43 Figure18(a)(80, 80) 94.91 31.88 99.61 33.53 Figure18(a)(100, 100) 1.71 0.57 99.61 33.52From Table16, the NPCR and UACI of proposed algorithm after 2-round encryption are nearer to the ideal values 99.61% and 33.46% [33] than others. Therefore the proposed method is better.Table 16
Comparison of NPCR and UACI of image.
Algorithm Image NPCR (%) UACI (%) Proposed Lena 99.60 33.48 [1] Lena 99.61 33.53 [2] Lena 99.6429 33.3935 [3] Lena 99.6304 33.5989 [5] Lena 99.932 39.520 [19] Lena 75.62561 34.84288 [20] Lena 99.6091 33.5038 [21] Lena 99.6330 34.1319


### 7.4. Resistance to Known-Plaintext and Chosen-Plaintext Attacks

In Section6.3, the iteration times of the next round are decided by the encrypted pixel value of present round. In (20), x2(i), generated from the fractional 2D-TFCDM, is dependent on k(i-1) and determines k(i). Therefore, the corresponding keystream is different when different plaintext is encrypted. For the resultant information is related to the chosen-images, the attacker cannot get useful information after encrypting some special images. As a result, the attacks proposed in [34–41] become ineffective for our scheme. In a word, the proposed scheme can primely resist the known-plaintext and the chosen-plaintext attacks.


## 7.1. Key Space

In the proposed algorithm, the initial valuesx0, y0, the order ν, and the coefficients k1, k2 are taken as the secret keys; consequently there are 5 keys. Assume the precision of x0, y0, ν, k1, and k2 are 10-16,3×10-17,10-16,10-15, and 10-16, respectively; then the key’s space is 1/3×1080≈1.12×2264. If the size of the plaintext is 512×512, then the key space of K-image is also 512×512×28=226. The total key space of the proposed algorithm is 1.12×2290.


## 7.2. Statistics Analysis

The quality against any statistical attack is important for a well-designed encryption method; it include 3 aspects as follows.


### 7.2.1. Correlation of the Plain- and Cipher-Images

In an ordinary image, the adjacent pixels are related; therefore the correlation coefficient of adjacent pixels is usually high. A good encryption algorithm should make the correlation coefficients of encrypted image nearly equal to zero. The closer to zero the correlation coefficients is, the better the encryption algorithm is. Formulas (23) calculate the correlation coefficient. The correlations along the x direction of both original and encrypted images are displayed in Figures 19–27 from Cameraman to Aerial. The correlation coefficients are displayed in Table 1.(23)rxy=cov⁡x,yDxDycov⁡x,y=1N∑i=1Nxi-Exyi-EyEx=1N∑i=1NxiDx=1N∑i=1Nxi-Ex2.Table 1
Correlation coefficients of image.
Image Original image Encrypted image Horizontal Diagonal Vertical Horizontal Diagonal Vertical Cameraman 0.9276 0.9120 0.9597 0.0119 −0.0021 −0.0025 Lena 0.9722 0.9527 0.9860 −0.0140 −0.0086 −0.0034 Peppers 0.9667 0.9382 0.9694 −0.0088 0.0080 −0.0054 Lake 0.9768 0.9544 0.9748 −0.0155 0.0101 −0.0088 Dollar 0.8035 0.6952 0.6938 0.0131 −0.0183 0.0263 Columbia 0.9727 0.9403 0.9705 0.0060 −0.0104 −0.0093 Lax 0.7889 0.7151 0.8483 −0.0107 0.0147 0.0107 Boat 0.9407 0.9158 0.9545 0.0169 −0.0074 −0.0077 Aerial 0.9135 0.7952 0.8677 0.0084 −0.0123 −0.0133Figure 19
Cameraman.
(a)
The original figure (b)
The encrypted figureFigure 20
Lena.
(a)
The original figure (b)
The encrypted figureFigure 21
Peppers.
(a)
The original figure (b)
The encrypted figureFigure 22
Lake.
(a)
The original figure (b)
The encrypted figureFigure 23
Dollar.
(a)
The original figure (b)
The encrypted figureFigure 24
Columbia.
(a)
The original figure (b)
The encrypted figureFigure 25
Lax.
(a)
The original figure (b)
The encrypted figureFigure 26
Boat.
(a)
The original figure (b)
The encrypted figureFigure 27
Aerial.
(a)
The original figure (b)
The encrypted figureWith the sharp contrast of data between original image and encrypted image, Table1 indicates that the encryption process make pixels of the encrypted image almost independent with each other. Consequently, the encryption algorithm is good at pixel value randomization.Compared with other algorithm, we can observe that most correlation coefficients of encrypted image are nearer to 0 in Table2. As a consequence of this, the proposed encryption algorithm is superior to others.Table 2
Comparison of correlation coefficients of image.
Algorithm Image Original image Encrypted image Horizontal Vertical Diagonal Horizontal Vertical Diagonal Proposed Lena 0.9722 0.9527 0.9860 −0.0140 −0.0086 −0.0034 [1] Lena 0.9503 0.9755 0.9275 −0.0226 0.0041 0.0368 [2] Lena 0.927970 0.926331 0.839072 −0.010889 −0.018110 −0.006104 [5] Lena 0.946 0.973 0.921 −0.0055 −0.0075 0.0026 [6] Lena 0.9569 0.9236 0.9019 0.0042 −0.0043 0.0163


### 7.2.2. Histogram

Histogram reflects the distribution of colors inside the image. The adversary can get some effective information from the regularity of histogram. Therefore, a well-designed image encryption method should make the pixel value of encrypted image distribute uniformly. Figure28 shows the histogram of Cameraman. Similarly, the histograms of the other 8 cases are drawn in Figures 29–36. It is illustrated that the proposed encryption method has a good effect on pixel value distribution uniformization.Figure 28
Cameraman.
(a)
The original image (b)
The encrypted image (c)
The decrypted imageFigure 29
Lena.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 30
Peppers.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 31
Lake.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 32
Dollar.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 33
Columbia.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 34
Lax.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 35
Boat.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 36
Aerial.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figure


### 7.2.3. Information Entropy

Information entropy defines the randomness and the unpredictability of information in an image. It is defined by(24)Hm=∑i=02n-1pmilog2⁡1pmi.Here p(mi) is the probability of mi; n is the number of bits that is required to represent the symbol mi. For the pixels values of the image are 0~255, according to (24) the information entropy is 8 bits for an ideally random image. Therefore, the closer to 8 bits the information entropy is, the better the encryption algorithm is. The information entropy of the 9 cases is gotten in Table 3; it indicates that the encrypted images are very close to the random images.Table 3
Information entropy.
Image Original image Encrypted image Cameraman 7.0097 7.9974 Peppers 7.5739 7.9976 Dollar 6.9785 7.9992 Lax 6.8272 7.9993 Aerial 6.9940 7.9992 Lena 7.2185 7.9993 Lake 7.4845 7.9993 Columbia 7.2736 7.9992 Boat 6.9391 7.9972From Table4, we can observe that the information entropy of proposed algorithm is nearer to 8 bits than other algorithms.Table 4
Comparison of information entropy.
Algorithm Image Original image Encrypted image Proposed Lena 7.2185 7.9993 [1] Lena 7.2072 7.9973 [4] Lena Undefined 7.9972 [19] Lena Undefined 7.987918 [20] Lena 7.447144 7.988847


## 7.2.1. Correlation of the Plain- and Cipher-Images

In an ordinary image, the adjacent pixels are related; therefore the correlation coefficient of adjacent pixels is usually high. A good encryption algorithm should make the correlation coefficients of encrypted image nearly equal to zero. The closer to zero the correlation coefficients is, the better the encryption algorithm is. Formulas (23) calculate the correlation coefficient. The correlations along the x direction of both original and encrypted images are displayed in Figures 19–27 from Cameraman to Aerial. The correlation coefficients are displayed in Table 1.(23)rxy=cov⁡x,yDxDycov⁡x,y=1N∑i=1Nxi-Exyi-EyEx=1N∑i=1NxiDx=1N∑i=1Nxi-Ex2.Table 1
Correlation coefficients of image.
Image Original image Encrypted image Horizontal Diagonal Vertical Horizontal Diagonal Vertical Cameraman 0.9276 0.9120 0.9597 0.0119 −0.0021 −0.0025 Lena 0.9722 0.9527 0.9860 −0.0140 −0.0086 −0.0034 Peppers 0.9667 0.9382 0.9694 −0.0088 0.0080 −0.0054 Lake 0.9768 0.9544 0.9748 −0.0155 0.0101 −0.0088 Dollar 0.8035 0.6952 0.6938 0.0131 −0.0183 0.0263 Columbia 0.9727 0.9403 0.9705 0.0060 −0.0104 −0.0093 Lax 0.7889 0.7151 0.8483 −0.0107 0.0147 0.0107 Boat 0.9407 0.9158 0.9545 0.0169 −0.0074 −0.0077 Aerial 0.9135 0.7952 0.8677 0.0084 −0.0123 −0.0133Figure 19
Cameraman.
(a)
The original figure (b)
The encrypted figureFigure 20
Lena.
(a)
The original figure (b)
The encrypted figureFigure 21
Peppers.
(a)
The original figure (b)
The encrypted figureFigure 22
Lake.
(a)
The original figure (b)
The encrypted figureFigure 23
Dollar.
(a)
The original figure (b)
The encrypted figureFigure 24
Columbia.
(a)
The original figure (b)
The encrypted figureFigure 25
Lax.
(a)
The original figure (b)
The encrypted figureFigure 26
Boat.
(a)
The original figure (b)
The encrypted figureFigure 27
Aerial.
(a)
The original figure (b)
The encrypted figureWith the sharp contrast of data between original image and encrypted image, Table1 indicates that the encryption process make pixels of the encrypted image almost independent with each other. Consequently, the encryption algorithm is good at pixel value randomization.Compared with other algorithm, we can observe that most correlation coefficients of encrypted image are nearer to 0 in Table2. As a consequence of this, the proposed encryption algorithm is superior to others.Table 2
Comparison of correlation coefficients of image.
Algorithm Image Original image Encrypted image Horizontal Vertical Diagonal Horizontal Vertical Diagonal Proposed Lena 0.9722 0.9527 0.9860 −0.0140 −0.0086 −0.0034 [1] Lena 0.9503 0.9755 0.9275 −0.0226 0.0041 0.0368 [2] Lena 0.927970 0.926331 0.839072 −0.010889 −0.018110 −0.006104 [5] Lena 0.946 0.973 0.921 −0.0055 −0.0075 0.0026 [6] Lena 0.9569 0.9236 0.9019 0.0042 −0.0043 0.0163


## 7.2.2. Histogram

Histogram reflects the distribution of colors inside the image. The adversary can get some effective information from the regularity of histogram. Therefore, a well-designed image encryption method should make the pixel value of encrypted image distribute uniformly. Figure28 shows the histogram of Cameraman. Similarly, the histograms of the other 8 cases are drawn in Figures 29–36. It is illustrated that the proposed encryption method has a good effect on pixel value distribution uniformization.Figure 28
Cameraman.
(a)
The original image (b)
The encrypted image (c)
The decrypted imageFigure 29
Lena.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 30
Peppers.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 31
Lake.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 32
Dollar.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 33
Columbia.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 34
Lax.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 35
Boat.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figureFigure 36
Aerial.
(a)
The original figure (b)
The encrypted figure (c)
The decrypted figure


## 7.2.3. Information Entropy

Information entropy defines the randomness and the unpredictability of information in an image. It is defined by(24)Hm=∑i=02n-1pmilog2⁡1pmi.Here p(mi) is the probability of mi; n is the number of bits that is required to represent the symbol mi. For the pixels values of the image are 0~255, according to (24) the information entropy is 8 bits for an ideally random image. Therefore, the closer to 8 bits the information entropy is, the better the encryption algorithm is. The information entropy of the 9 cases is gotten in Table 3; it indicates that the encrypted images are very close to the random images.Table 3
Information entropy.
Image Original image Encrypted image Cameraman 7.0097 7.9974 Peppers 7.5739 7.9976 Dollar 6.9785 7.9992 Lax 6.8272 7.9993 Aerial 6.9940 7.9992 Lena 7.2185 7.9993 Lake 7.4845 7.9993 Columbia 7.2736 7.9992 Boat 6.9391 7.9972From Table4, we can observe that the information entropy of proposed algorithm is nearer to 8 bits than other algorithms.Table 4
Comparison of information entropy.
Algorithm Image Original image Encrypted image Proposed Lena 7.2185 7.9993 [1] Lena 7.2072 7.9973 [4] Lena Undefined 7.9972 [19] Lena Undefined 7.987918 [20] Lena 7.447144 7.988847


## 7.3. Sensitivity Analysis

The different range between two images is measured by two criteria: number of pixels change rate (NPCR) and unified average changing intensity (UACI). They are defined as follows:(25)Di,j=0,T1i,j=T2i,j,1,T1i,j≠T2i,j,NPCR=∑i=1W∑j=1HDi,jW×H×100%UACI=∑i=1W∑j=1HT1i,j-T2i,j255W×H×100%.Here W and H are the width and the height of T1 and T2.


### 7.3.1. Key Sensitivity

We encrypt the image by the keysx0=0.19, y0=0.06, ν=0.7123456, k1=8, and k2=0.5. Figure 37(a) is the decrypted image with the correct keys. Figure 37(b) represents the decrypted image under 10-16 adding to x0 with other keys unchanged. Similarly, the secret keys y0,ν,k1,k2 are added as 3×10-17,10-16,10-15 and 10-16 to decrypt the images separately with other keys unchanged. The results are shown in Figures 37(c)–37(f). The comparison of key space is shown in Table 5 and the NPCR and UACI between Figures 37(a) and 37(b)–37(f) are calculated in Table 6.Table 5
Comparison of key spaces.
Algorithm Proposed [2] [4] [6] Key spaces 2.23 × 1 0 87 1.12 × 2 290 2 128 ≈2273 2276Table 6
NPCR and UACI between Figures37(a) and 37(b)–37(f).
Image NPCR and UACI NPCR (%) UACI (%) Figure37(b) 99.61 31.26 Figure37(c) 97.02 30.23 Figure37(d) 99.60 31.03 Figure37(e) 99.61 31.01 Figure37(f) 99.62 31.27Figure 37
The test of key sensitivity.
(a)
The correct keys (b)
x 0 + 10 - 16 (c)
y 0 + 3 × 10 - 17 (d)
v + 10 - 16 (e)
k 1 + 10 - 15 (f)
k 2 + 10 - 16In contrast with other algorithm, the key space of proposed algorithm is larger than others.Most NPCR are near to 99.61% and most of UACI are higher than 30% in Table6. We cannot recognize the man inside from Figures 37(b)–37(f); therefore the encryption method is sensitive to the keys.


### 7.3.2. Plaintext Sensitivity

By encrypting two same images with only one pixel difference, the attackers can obtain effective information by comparing the two encrypted images. Therefore an encryption method designed against differential attack should ensure that the two encrypted images are completely different even if there is only a pixel difference in the original image.In Table7, Figure 10(a)(x,y) is the same as Figure 10(a) except for a pixel locating (x,y). After that, the 2 images are encrypted with the same keys and the NPCR and UACI between the 2 ciphertext images are calculated. Similarly, the data of other 8 cases are obtained in Tables 8–15.Table 7
NPCR and UACI between cipher-images with slightly different plain-images.
Image NPCR and UACI of Cameraman NPCR (1-round %) UACI (1-round %) NPCR (2-round %) UACI (2-round %) Figure10(a)(30, 30) 4.84 1.64 99.57 33.61 Figure10(a)(50, 50) 81.43 27.39 99.62 33.56 Figure10(a)(80, 80) 80.87 27.19 99.59 33.51 Figure10(a)(100, 100) 6.82 2.28 99.59 33.46Table 8
NPCR and UACI between cipher-images with slightly different plain-images.
Image NPCR and UACI of Lena NPCR (1-round %) UACI (1-round %) NPCR (2-round %) UACI (2-round %) Figure11(a)(30, 30) 1.21 0.41 99.59 33.39 Figure11(a)(50, 50) 95.06 31.93 99.59 33.53 Figure11(a)(80, 80) 94.90 31.93 99.60 33.48 Figure11(a)(100, 100) 1.71 0.58 99.63 33.40Table 9
NPCR and UACI between cipher-images with slightly different plain-images.
Image NPCR and UACI of Peppers NPCR (1-round %) UACI (1-round %) NPCR (2-round %) UACI (2-round %) Figure12(a)(30, 30) 4.83 1.64 99.56 33.44 Figure12(a)(50, 50) 81.44 27.35 99.62 33.50 Figure12(a)(80, 80) 6.12 2.03 99.57 33.42 Figure12(a)(100, 100) 6.83 2.32 99.60 33.50Table 10
NPCR and UACI between cipher-images with slightly different plain-images.
Image NPCR and UACI of Lake NPCR (1-round %) UACI (1-round %) NPCR (2-round %) UACI (2-round %) Figure13(a)(30, 30) 1.21 0.41 99.61 33.46 Figure13(a)(50, 50) 95.09 31.88 99.61 33.42 Figure13(a)(80, 80) 94.92 31.89 99.60 33.46 Figure13(a)(100, 100) 1.71 0.58 99.59 33.47Table 11
NPCR and UACI between cipher-images with slightly different plain-images.
Image NPCR and UACI of Dollar NPCR (1-round %) UACI (1-round %) NPCR (2-round %) UACI (2-round %) Figure14(a)(30, 30) 1.21 0.41 99.64 33.42 Figure14(a)(50, 50) 95.08 32.00 99.60 33.49 Figure14(a)(80, 80) 94.90 31.93 99.61 33.48 Figure14(a)(100, 100) 1.71 0.57 99.61 33.41Table 12
NPCR and UACI between cipher-images with slightly different plain-images.
Image NPCR and UACI of Columbia NPCR (1-round %) UACI (1-round %) NPCR (2-round %) UACI (2-round %) Figure15(a)(30, 30) 94.83 31.96 99.61 33.47 Figure15(a)(50, 50) 93.48 31.40 99.48 33.36 Figure15(a)(80, 80) 0.96 0.32 99.60 33.45 Figure15(a)(100, 100) 1.11 0.38 99.61 33.51Table 13
NPCR and UACI between cipher-images with slightly different plain-images.
Image NPCR and UACI of Lax NPCR (1-round %) UACI (1-round %) NPCR (2-round %) UACI (2-round %) Figure16(a)(30, 30) 1.21 0.40 99.62 33.39 Figure16(a)(50, 50) 95.06 31.99 99.61 33.49 Figure16(a)(80, 80) 94.92 31.87 99.58 33.41 Figure16(a)(100, 100) 1.70 0.58 99.62 33.48Table 14
NPCR and UACI between cipher-images with slightly different plain-images.
Image NPCR and UACI of Boat NPCR (1-round %) UACI (1-round %) NPCR (2-round %) UACI (2-round %) Figure17(a)(30, 30) 4.83 1.60 99.59 33.47 Figure17(a)(50, 50) 81.46 27.45 99.62 33.59 Figure17(a)(80, 80) 80.82 27.24 99.58 33.48 Figure17(a)(100, 100) 6.82 2.32 99.62 33.61Table 15
NPCR and UACI between cipher-images with slightly different plain-images.
Image NPCR and UACI of Aerial NPCR (1-round %) UACI (1-round %) NPCR (2-round %) UACI (2-round %) Figure18(a)(30, 30) 1.21 0.41 99.61 33.49 Figure18(a)(50, 50) 95.06 31.93 99.62 33.43 Figure18(a)(80, 80) 94.91 31.88 99.61 33.53 Figure18(a)(100, 100) 1.71 0.57 99.61 33.52From Table16, the NPCR and UACI of proposed algorithm after 2-round encryption are nearer to the ideal values 99.61% and 33.46% [33] than others. Therefore the proposed method is better.Table 16
Comparison of NPCR and UACI of image.
Algorithm Image NPCR (%) UACI (%) Proposed Lena 99.60 33.48 [1] Lena 99.61 33.53 [2] Lena 99.6429 33.3935 [3] Lena 99.6304 33.5989 [5] Lena 99.932 39.520 [19] Lena 75.62561 34.84288 [20] Lena 99.6091 33.5038 [21] Lena 99.6330 34.1319


## 7.3.1. Key Sensitivity

We encrypt the image by the keysx0=0.19, y0=0.06, ν=0.7123456, k1=8, and k2=0.5. Figure 37(a) is the decrypted image with the correct keys. Figure 37(b) represents the decrypted image under 10-16 adding to x0 with other keys unchanged. Similarly, the secret keys y0,ν,k1,k2 are added as 3×10-17,10-16,10-15 and 10-16 to decrypt the images separately with other keys unchanged. The results are shown in Figures 37(c)–37(f). The comparison of key space is shown in Table 5 and the NPCR and UACI between Figures 37(a) and 37(b)–37(f) are calculated in Table 6.Table 5
Comparison of key spaces.
Algorithm Proposed [2] [4] [6] Key spaces 2.23 × 1 0 87 1.12 × 2 290 2 128 ≈2273 2276Table 6
NPCR and UACI between Figures37(a) and 37(b)–37(f).
Image NPCR and UACI NPCR (%) UACI (%) Figure37(b) 99.61 31.26 Figure37(c) 97.02 30.23 Figure37(d) 99.60 31.03 Figure37(e) 99.61 31.01 Figure37(f) 99.62 31.27Figure 37
The test of key sensitivity.
(a)
The correct keys (b)
x 0 + 10 - 16 (c)
y 0 + 3 × 10 - 17 (d)
v + 10 - 16 (e)
k 1 + 10 - 15 (f)
k 2 + 10 - 16In contrast with other algorithm, the key space of proposed algorithm is larger than others.Most NPCR are near to 99.61% and most of UACI are higher than 30% in Table6. We cannot recognize the man inside from Figures 37(b)–37(f); therefore the encryption method is sensitive to the keys.


## 7.3.2. Plaintext Sensitivity

By encrypting two same images with only one pixel difference, the attackers can obtain effective information by comparing the two encrypted images. Therefore an encryption method designed against differential attack should ensure that the two encrypted images are completely different even if there is only a pixel difference in the original image.In Table7, Figure 10(a)(x,y) is the same as Figure 10(a) except for a pixel locating (x,y). After that, the 2 images are encrypted with the same keys and the NPCR and UACI between the 2 ciphertext images are calculated. Similarly, the data of other 8 cases are obtained in Tables 8–15.Table 7
NPCR and UACI between cipher-images with slightly different plain-images.
Image NPCR and UACI of Cameraman NPCR (1-round %) UACI (1-round %) NPCR (2-round %) UACI (2-round %) Figure10(a)(30, 30) 4.84 1.64 99.57 33.61 Figure10(a)(50, 50) 81.43 27.39 99.62 33.56 Figure10(a)(80, 80) 80.87 27.19 99.59 33.51 Figure10(a)(100, 100) 6.82 2.28 99.59 33.46Table 8
NPCR and UACI between cipher-images with slightly different plain-images.
Image NPCR and UACI of Lena NPCR (1-round %) UACI (1-round %) NPCR (2-round %) UACI (2-round %) Figure11(a)(30, 30) 1.21 0.41 99.59 33.39 Figure11(a)(50, 50) 95.06 31.93 99.59 33.53 Figure11(a)(80, 80) 94.90 31.93 99.60 33.48 Figure11(a)(100, 100) 1.71 0.58 99.63 33.40Table 9
NPCR and UACI between cipher-images with slightly different plain-images.
Image NPCR and UACI of Peppers NPCR (1-round %) UACI (1-round %) NPCR (2-round %) UACI (2-round %) Figure12(a)(30, 30) 4.83 1.64 99.56 33.44 Figure12(a)(50, 50) 81.44 27.35 99.62 33.50 Figure12(a)(80, 80) 6.12 2.03 99.57 33.42 Figure12(a)(100, 100) 6.83 2.32 99.60 33.50Table 10
NPCR and UACI between cipher-images with slightly different plain-images.
Image NPCR and UACI of Lake NPCR (1-round %) UACI (1-round %) NPCR (2-round %) UACI (2-round %) Figure13(a)(30, 30) 1.21 0.41 99.61 33.46 Figure13(a)(50, 50) 95.09 31.88 99.61 33.42 Figure13(a)(80, 80) 94.92 31.89 99.60 33.46 Figure13(a)(100, 100) 1.71 0.58 99.59 33.47Table 11
NPCR and UACI between cipher-images with slightly different plain-images.
Image NPCR and UACI of Dollar NPCR (1-round %) UACI (1-round %) NPCR (2-round %) UACI (2-round %) Figure14(a)(30, 30) 1.21 0.41 99.64 33.42 Figure14(a)(50, 50) 95.08 32.00 99.60 33.49 Figure14(a)(80, 80) 94.90 31.93 99.61 33.48 Figure14(a)(100, 100) 1.71 0.57 99.61 33.41Table 12
NPCR and UACI between cipher-images with slightly different plain-images.
Image NPCR and UACI of Columbia NPCR (1-round %) UACI (1-round %) NPCR (2-round %) UACI (2-round %) Figure15(a)(30, 30) 94.83 31.96 99.61 33.47 Figure15(a)(50, 50) 93.48 31.40 99.48 33.36 Figure15(a)(80, 80) 0.96 0.32 99.60 33.45 Figure15(a)(100, 100) 1.11 0.38 99.61 33.51Table 13
NPCR and UACI between cipher-images with slightly different plain-images.
Image NPCR and UACI of Lax NPCR (1-round %) UACI (1-round %) NPCR (2-round %) UACI (2-round %) Figure16(a)(30, 30) 1.21 0.40 99.62 33.39 Figure16(a)(50, 50) 95.06 31.99 99.61 33.49 Figure16(a)(80, 80) 94.92 31.87 99.58 33.41 Figure16(a)(100, 100) 1.70 0.58 99.62 33.48Table 14
NPCR and UACI between cipher-images with slightly different plain-images.
Image NPCR and UACI of Boat NPCR (1-round %) UACI (1-round %) NPCR (2-round %) UACI (2-round %) Figure17(a)(30, 30) 4.83 1.60 99.59 33.47 Figure17(a)(50, 50) 81.46 27.45 99.62 33.59 Figure17(a)(80, 80) 80.82 27.24 99.58 33.48 Figure17(a)(100, 100) 6.82 2.32 99.62 33.61Table 15
NPCR and UACI between cipher-images with slightly different plain-images.
Image NPCR and UACI of Aerial NPCR (1-round %) UACI (1-round %) NPCR (2-round %) UACI (2-round %) Figure18(a)(30, 30) 1.21 0.41 99.61 33.49 Figure18(a)(50, 50) 95.06 31.93 99.62 33.43 Figure18(a)(80, 80) 94.91 31.88 99.61 33.53 Figure18(a)(100, 100) 1.71 0.57 99.61 33.52From Table16, the NPCR and UACI of proposed algorithm after 2-round encryption are nearer to the ideal values 99.61% and 33.46% [33] than others. Therefore the proposed method is better.Table 16
Comparison of NPCR and UACI of image.
Algorithm Image NPCR (%) UACI (%) Proposed Lena 99.60 33.48 [1] Lena 99.61 33.53 [2] Lena 99.6429 33.3935 [3] Lena 99.6304 33.5989 [5] Lena 99.932 39.520 [19] Lena 75.62561 34.84288 [20] Lena 99.6091 33.5038 [21] Lena 99.6330 34.1319


## 7.4. Resistance to Known-Plaintext and Chosen-Plaintext Attacks

In Section6.3, the iteration times of the next round are decided by the encrypted pixel value of present round. In (20), x2(i), generated from the fractional 2D-TFCDM, is dependent on k(i-1) and determines k(i). Therefore, the corresponding keystream is different when different plaintext is encrypted. For the resultant information is related to the chosen-images, the attacker cannot get useful information after encrypting some special images. As a result, the attacks proposed in [34–41] become ineffective for our scheme. In a word, the proposed scheme can primely resist the known-plaintext and the chosen-plaintext attacks.


## 8. Conclusions

Fractional 2D-TFCDM is obtained from the 2D-TFCDM. After that, we found new chaotic dynamics behaviors from the fractionalized map. Moreover, the map can be converted into image encryption algorithm as an application. Finally, the encryption effect is analysed in 4 main aspects; we find the proposed scheme is superior to others almost anywhere in comparison. As far as we know, the proposed image encryption algorithm has never been reported before.



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*Source: 4585083-2018-04-23.xml*