The Itinerary of the N-Type Map and Its Application

Abstract

The itinerary computation approach for the N-type map is proposed. Based on this itinerary approach, the topological conjugation between the N-type map and the shift map on the three-symbol space is proven. Additionally, the calculation of periodic points and the demonstration of the chaotic property of the N-type map are presented. As a practical application in the field of image security, two image scrambling encryption algorithms are proposed and implemented.

1. Introduction

Chaos is a crucial concept for characterizing the complex dynamical properties of dynamical systems. Chaotic systems exhibit complex dynamical properties, including topological transition, a dense set of periodic points, good pseudo-randomness, and strong sensitivity to initial values [1,2]. Li and Yorke first introduced the concept of chaos in their work “Period three implies chaos” [3]; Devaney extended the Li–Yorke chaos concept and put forward a widely adopted definition of chaos, known as Devaney chaos [4]. The chaos phenomena of some classical dynamical systems, such as the shift map, the tent map, and the Logistic map, along with the counting of their periodic points, have been thoroughly investigated, and the corresponding conclusions are well-established in the field of nonlinear science [5,6,7,8].

Topological conjugation establishes an equivalence relation among two dynamical systems, which is of great significance in the study of their properties. Two dynamical systems are topologically conjugate if there exists a homeomorphism (a continuous bijection with a continuous inverse) that maps the orbits of one system to the orbits of the other while preserving the dynamical structure. This equivalence relation reveals that the two dynamical systems possess common key topological properties, such as the identical number and configuration of periodic points, and similar long-term behaviors encompassing chaos or stability. Topological conjugacy allows us to study complex systems by relating them to simpler, well-understood systems. It filters out specific geometric and analytic details, thereby concentrating on the essential and intrinsic dynamical characteristics. As a matter of fact, one of the central questions in dynamical systems is to decide whether two nonlinear maps are topologically conjugate [9,10]. If two nonlinear maps are topologically conjugate, the homeomorphism connecting them plays a crucial role, particularly the existence of explicit analytical expressions. When explicit analytical expressions are available, upon knowing the orbits of a simple map, one can readily deduce the orbits of another map by means of this homeomorphic transformation. This linkage not only simplifies the calculation of periodic points remarkably but also enhances its computational viability. For example, Crampin and Heal provided a clear formula for computing the periodic points of the classical tent map, along with a directly illuminating proof of its Devaney chaos [8]. Since the Logistic map and the tent map are topologically conjugate via a simple homeomorphism [11], one can calculate the periodic points of the Logistic map via those of the classical tent map. Regarding the N-type map, Shi formulated the precise semi-conjugate expression between the N-type map and the classical tent map, thereby demonstrating that the semi-conjugate homeomorphic mapping is a continuous function exhibiting fractal features. Consequently, the Devaney chaos of the N-type map was indirectly proven [12]. Regrettably, the semi-conjugate expression established between the N-type map and the classical tent map turns out to be inapplicable in certain scenarios. The problem of computing periodic points in the N-type map remains to be further investigated. Generally, both the tent map and the N-shaped map satisfy the Devaney chaos criteria. The N-shaped map demonstrates significantly higher dynamical complexity due to its higher topological entropy $\ln 3$ and ternary symbolic encoding, compared to the tent map’s entropy $\ln 2$ and binary encoding. The structural intricacy of the N-shaped map enables richer symbolic dynamics.

Since the N-shaped map is defined on the interval [0, 1] and consists of three linear segments with slopes 3, −3, and 3, respectively, directly proving its chaotic properties on this interval is rather complicated. If it can be shown to be topologically conjugate to the shift map of a three-symbol space through a homeomorphism, the chaotic nature of the N-shaped map could be indirectly established by leveraging the straightforward proof of chaos in symbolic shift systems. However, the crux of such indirect proof lies in constructing a homeomorphism between these two dynamical systems. Generally, obtaining explicit analytical expressions for such homeomorphisms proves challenging. This paper addresses this by employing an itinerary approach to constructively derive an explicit analytical expression for the homeomorphism between these maps. This achievement resolves computational issues related to determining periodic points and chaotic orbits in the N-shaped map, representing one of the key contributions of this research. Moreover, the proposed itinerary methodology offers an alternative encoding scheme for natural numbers, which can be applied to pixel position rearrangement in digital images. This enables spatial scrambling of image pixels, disrupting both the natural spatial ordering of pixels and correlations between adjacent pixels, thereby achieving effective image encryption.

The main contributions of the paper include three aspects. (1) An explicit analytic expression of the homomorphism between the N-type map and the shift map on the three-symbol space is proposed. Through this homomorphism, the periodic points of the shift map on the three-symbol space can be mapped to obtain the periodic points of the N-type map. (2) With the help of the shift map, the Devaney chaos of the N-type map is directly proved in ternary representation, which has a good teaching demonstration effect. (3) The truncated finite itinerary set of the N-type map is applied to the research of image information security. Two itinerary-based image scrambling algorithms are designed to encrypt images of arbitrary height and width.

The rest of the article is organized as follows. Section 2 describes some preliminaries, covering some definitions and conclusions of dynamical systems. Section 3 provides the itinerary computation approach for the N-type map. The explicit formulae for calculating the itineraries and the periodic points of the N-type map are presented. Section 4 proves the Devaney chaos of the N-type map using the itinerary approach. Section 5 provides an application of the itinerary for the N-type map, which realizes image scrambling encryption for images of arbitrary size. Section 6 draws some conclusions.

2. Preliminaries

Some notations and definitions related to dynamical systems are introduced briefly in this section. Let $f$ be a dynamical system. The set composed of all periodic points of $f$ is denoted by $Per(f)$. The number of $k$-periodic orbits of $f$ is denoted by $P(k)$. The complement operation of $b\in \{0,1,2\}$ is denoted by $\overline{b}$,$\overline{b}=2-b$. Let $0.\widetilde{b_1{b}_2\cdots b_k}$ denote the infinitely repeating number $0.b_1{b}_2\cdots b_k{b}_1{b}_2\cdots b_k\cdots$. For two integers $m,k$, $m|k$ means that $m$ can divide $k$.

Definition 1

 [4]. Assume $(X,\rho )$ is metric space, $f$ maps $X$ to itself. For arbitrary $x_0\in X$, the forward orbit of $x_0$ is denoted by $O_f^{+}(x_0)=\{x_k=f^k(x_0),k=0,1,\cdots \}$. $f$ is then called a discrete dynamical system, denoted by $\{X;f\}$.

Definition 2

 [4]. Let $({\Sigma }_N,d_c)$ denote the code space of $N$ symbols $T=\{0,1,\cdots ,N-1\}$, ${\Sigma }_N=\{s=s_0{s}_1{s}_2\cdots |s_i\in T\}$. The metric $d_c$ is given by Equation (1).

$$d_c(s,t)={\displaystyle \sum _{i=0}^{\infty }\frac{|s_i-t_i|}{N^i}},s=s_0{s}_1{s}_2\cdots ,t=t_0{t}_1{t}_2\cdots ,$$

(1)

Definition 3

[4]. If there exists positive integer $n$ such that $x_0=f^n(x_0)$, $x_0\ne f^k(x_0)$, $0<k<n$, then $x_0$ is called a $n$-periodic point of $f$, and $n$ is its period. A 1-periodic point is also called a fixed point. The orbit of an $n$-periodic point $O_f^{+}(x_0)=\{x_0,f(x_0),\cdots ,f^{n-1}(x_0)\}$ is called a cycle of $f$.

Definition 4

[4]. A dynamical system $\{X;f\}$ on metric space $(X,\rho )$ is chaotic if it satisfies the following conditions (a–c):

(a) $f$ is transitive, i.e., there exists $x_0\in X$, such that the orbit of $x_0$ is dense in $X$;

(b) $f$ is sensitive to initial conditions, i.e., there exists $\epsilon >0$ such that, for any $x\in X$ and any ball $B(x,\delta )$ with arbitrary radius $\delta >0$, there is $y\in B(x,\delta )$ and an integer $n\ge 0$ such that $\rho (f^n(x),f^n(y))>\epsilon$;

(c) the set of periodic points of $f$ is dense in $X$.

Definition 5

[13]. Two dynamical systems $\{X;f\}$ and $\{Y;g\}$ are said to be equivalent, or topologically conjugate, if there is a homeomorphism $\theta :X\to Y$ such that $\theta f(x)=g\theta (x),\forall x\in X$.

If two dynamical systems $\{X;f\}$ and $\{Y;g\}$ are topologically conjugate, then they have the same dynamical properties. For example, $\{X;f\}$ has periodic orbits $\{x_0,x_1,\cdots ,x_{n-1}\}$, if and only if $\{Y;g\}$ has periodic orbits $\{h(x_0),h(x_1),\cdots ,h(x_{n-1})\}$; the two dynamical systems also have the same chaotic properties, and so on [13]. This property serves as the foothold for calculating the periodic points of the N-type map in Section 3, and it is also the basis for proving the chaotic characteristics of the N-type map in Section 4. The N-type map $T(x)$ defined in [0, 1] is shown in Equation (2). It is a piecewise continuous function connected by three-line segments. Figure 1 shows the graphs of $T(x)$ and its iteration $T^2(x)$.

$$T(x)=\left\{\begin{array}{l}3x,x\in [0,1/3),\hfill \\ -3x+2,x\in [1/3,2/3),\\ 3x-2,x\in [2/3,1].\hfill \end{array}\right.$$

(2)

Figure 1. The N-type map $T(x)$ (a) and its iteration $T^2(x)$ (b).

3. The Itineraries and the Periodic Points of the N-Type Map

The operation of multiplying a real number by 3 is a simple shift operation in ternary representation, implying that an intuitive explanation can be given for the iteration of the N-type map. Now rewrite any number in [0, 1] in ternary form:

$$x=b_1\times 3^{-1}+b_2\times 3^{-2}+b_3\times 3^{-3}+\cdots =(0.b_1{b}_2{b}_2\cdots {)}_3, b_i\in \{0,1,2\},i=1,2,\cdots$$

For the sake of simplicity, abbreviate $x=(0.b_1{b}_2{b}_2\cdots {)}_3$ in ternary expression as $x=0.b_1{b}_2{b}_2\cdots$.

Case 1. $x\in [0,1/3)$, $b_1=0$,

$$T(x)=3x=3\times (b_1\times 3^{-1}+b_2\times 3^{-2}+b_3\times 3^{-3}+\cdots )=b_1.b_2{b}_3\cdots =0.b_2{b}_3\cdots$$

Case 2. $x\in [1/3,2/3)$, $b_1=1$, $\overline{b_1}=1$,

$$\begin{array}{l}T(x)=-3x+2=3\times (1-x)-1=3\times (0.222\cdots -0.b_1{b}_2{b}_2\cdots )-1\\ \hspace{1em}\hspace{1em}=3\times 0.\overline{b_1}\overline{b_2}\overline{b_3}\cdots -1=\overline{b_1}.\overline{b_2}\overline{b_3}\cdots -1=0.\overline{b_2}\overline{b_3}\cdots .\end{array}$$

Case 3. $x\in [2/3,1]$, $b_1=2$,

$$T(x)=3x-2=3\times (b_1\times 3^{-1}+b_2\times 3^{-2}+b_3\times 3^{-3}+\cdots )-2=0.b_2{b}_3\cdots$$

From the above classification calculation, it can be observed that when $x\in [0,1/3)$ and $x\in [2/3,1]$, the operation $T(x)$ represents a shift operation in the ternary system. Meanwhile, the operation $T(x)$ is the compound operation of the shift operation and the complement operation if $x\in [1/3,2/3)$. Evidently, if $x_0$ is a periodic point of $T(x)$, then its ternary representation must be an infinite repetition of a finite sequence of ternary digits, similar to the representation of repeating decimals in the decimal system. In this paper, we present a convenient method for calculating the periodic points. This method is implemented through the itinerary calculation of the N-type map and its equivalent dynamical system, namely, the shift map on the three-symbol space.

3.1. The Itinerary Computation for the N-Type Map

Definition 6

[11]. For dynamical system $\{[0,1];f\}$, assume that the orbit of $x_0\in [0,1]$ is $O_f^{+}(x_0)=\{f^k(x_0),k=0,1,\cdots \}$, $[0,1]$ is evenly divided into three intervals:

$$I_0=[0,1/3],I_1=[1/3,2/3],I_2=[2/3,1]$$

If $x_k=f^k(x_0)\in I_{{\sigma }_{k+1}}$, then the itinerary of $x_0$ is defined by an element of ${\Sigma }_3$: $\sigma ={\sigma }_1{\sigma }_2\cdots {\sigma }_n\cdots ,{\sigma }_i\in \{0,1,2\}$. Its corresponding mapping rule is denoted by $\sigma =h(x_0)$, and $h:[0,1]\to \Sigma$ is called the itinerary map.

Consider $x_0=0.b_1{b}_2{b}_3\cdots ,b_i\in \{0,1,2\}$ with itinerary $\sigma =h(x_0)={\sigma }_1{\sigma }_2\cdots {\sigma }_n\cdots$, ${\sigma }_i\in \{0,1,2\}$. If $x_0$ is expressed in ternary representation and $b_1=0,2$, the iteration function value $T(x_0)$ can be simply obtained by removing $b_1$. If $b_1=1$, $T(x_0)$ can be yielded by removing $b_1$ firstly and then carrying out the complement operations of all the subsequent digit values. Since the complement operation is a switch operation, an even number of complement operations is equivalent to a no complement operation, while an odd number of complement operations is equivalent to one complement operation. Instead of judging whether the value $x_{k+1}$ in the next iteration $x_{k+1}=T(x_k)$ needs to undergo a complement operation, one can directly make the judgment based on the ternary representation of the initial value $x_0$. And one can also determine the value ${\sigma }_k$ in the itinerary $\sigma$ of $x_0$.

In the ternary representation, when there is an odd number of 1s in the preceding digit segment $b_1{b}_2\cdots b_k$, the iterated function value $T^k(x_0)$ needs to complement the digit segment $b_{k+1}{b}_{k+2}\cdots$ in the ternary representation of $x_0$; otherwise, no operation is needed. Rewrite the iteration process to be equivalent to Equation (3).

$$\begin{array}{l}{x}_k=T^k(x_0)=T^k(0.b_1{b}_2{b}_3\cdots )\\ \hspace{1em}=\left\{\begin{array}{l}0.b_{k+1}{b}_{k+2}\cdots , \#\{b_i=1,i=1,\cdots ,k\} \mathrm{is} \mathrm{even};\\ 0.{\overline{b}}_{k+1}{\overline{b}}_{k+2}\cdots , \#\{b_i=1,i=1,\cdots ,k\} \mathrm{is} \mathrm{odd},\end{array}\right.\end{array}$$

(3)

where $\#\{\cdots \}$ denotes the number of elements in set $\{\cdots \}$. As stated in Equation (3), if there is an even number of 1s in $b_1{b}_2\cdots b_k$, $x_k=T^k(x_0)$ can be simply derived by removing $b_1{b}_2\cdots b_k$. Otherwise, $x_k=T^k(x_0)$ can be simply derived by removing $b_1{b}_2\cdots b_k$ and then complementing the subsequent digit values. As a result, the itinerary $\sigma =h(x_0)$$={\sigma }_1{\sigma }_2\cdots {\sigma }_n\cdots$ can be calculated by Equation (4).

$${\sigma }_1=b_1,{\sigma }_{k+1}=\left\{\begin{array}{l}{b}_{k+1}, \#\{b_i=1,i=1,\cdots ,k\} \mathrm{is} \mathrm{even};\\ {\overline{b}}_{k+1}, \#\{b_i=1,i=1,\cdots ,k\} \mathrm{is} \mathrm{odd}, k=1,2,\cdots .\end{array}\right.$$

(4)

For example, let $x_0=0.0120112012210\cdots$. It follows from Equation (4) that ${\sigma }_1=b_1=0$, ${\sigma }_2=b_2=1$, ${\sigma }_3=\overline{b_3}=\overline{2}=0$, ${\sigma }_4=\overline{b_4}=\overline{0}=2$, ${\sigma }_5=\overline{b_5}=\overline{1}=1$, and so on. The itinerary of $x_0$ is $\sigma =h(x_0)=0102110212212\cdots$.

3.2. The Computation of the N-Type Map’s Periodic Points

The shift dynamical system $\{{\Sigma }_3;S\}$ is defined by

$$S:{\Sigma }_3\to {\Sigma }_3,S(\sigma )=S({\sigma }_1{\sigma }_2\cdots {\sigma }_n\cdots )={\sigma }_2\cdots {\sigma }_n\cdots ,\forall \sigma ={\sigma }_1{\sigma }_2\cdots {\sigma }_n\cdots$$

It is well known that $\{{\Sigma }_3;S\}$ is chaotic, and the periodic points and chaotic nature can be simply proven [13]. Theorem 1 shows that the N-type map $\{[0,1];T\}$ is topologically conjugate to $\{{\Sigma }_3;S\}$ with homeomorphism $\sigma =h(x)$. The inverse map $x=h^{-1}(\sigma )=0.b_1{b}_2\cdots$, $\sigma ={\sigma }_1{\sigma }_2\cdots$ is given by Equation (5).

$$b_1={\sigma }_1,b_{k+1}=\left\{\begin{array}{l}{\sigma }_{k+1}, \#\{b_i=1,i=1,\cdots ,k\} \mathrm{is} \mathrm{even};\\ {\overline{\sigma }}_{k+1}, \#\{b_i=1,i=1,\cdots ,k\} \mathrm{is} \mathrm{odd},k=1,2,\cdots .\end{array}\right.$$

(5)

For example, let $\sigma =0102110212212\cdots$, and then it follows from Equation (5) that $b_1={\sigma }_1=0$, $b_2={\sigma }_2=1$, $b_3=\overline{{\sigma }_3}=\overline{0}=2$, $b_4=\overline{{\sigma }_4}=\overline{2}=0$, $b_5=\overline{{\sigma }_5}=\overline{1}=1$, and so on. $x_0=h^{-1}(\sigma )$ $=0.0120112012210\cdots$.

Theorem 1.

The N-type map $\{[0,1];T\}$ is topologically conjugate to $\{{\Sigma }_3;S\}$ with homeomorphism $\sigma =h(x)$, such that $h(T(x_0))=S(h(x_0)),\forall x_0\in [0,1]$.

Proof of Theorem 1.

For any $x_0\in [0,1]$ with $\sigma =h(x_0)={\sigma }_1{\sigma }_2\cdots {\sigma }_n\cdots ,{\sigma }_i\in \{0,1,2\}$, the itinerary of $T(x_0)$ is ${\sigma }_2\cdots {\sigma }_n\cdots$, that is, $h(T(x_0))={\sigma }_2\cdots {\sigma }_n\cdots =S(\sigma )=S(h(x_0))$. Therefore, the N-type map $\{[0,1];T\}$ is topologically conjugate to $\{{\Sigma }_3;S\}$ with homeomorphism $\sigma =h(x)$. □

Theorem 2.

Regarding the N-type map $\{[0,1];T\}$, $\#\{b_i=1,i=1,\cdots ,k\}=\#\{{\sigma }_i=1,i=1,$ $\cdots ,k\}$ for any $x_0=0.b_1{b}_2\cdots$ with $\sigma =h(x_0)={\sigma }_1{\sigma }_2\cdots {\sigma }_n\cdots ,{\sigma }_i\in \{0,1,2\}$.

Proof of Theorem 2.

For any $x_0\in [0,1]$ with $\sigma =h(x_0)={\sigma }_1{\sigma }_2\cdots {\sigma }_n\cdots ,{\sigma }_i\in \{0,1,2\}$, the statement can be proved by mathematical induction. (1) $k=1$, it follows from Equation (4) that ${\sigma }_1=b_1$, and the statement is correct. (2) Suppose $\#\{b_i=1,i=1,\cdots ,n\}=$ $\#\{{\sigma }_i=1,i=1,\cdots ,n\}$, it follows from Equation (4) that $b_{n+1}={\sigma }_{n+1}$ or $b_{n+1}={\overline{\sigma }}_{n+1}$. Thanks to $1=\overline{1}$, the number of 1s in $b_1{b}_2\cdots b_n{b}_{n+1}$ is the same as the number of 1s in ${\sigma }_1{\sigma }_2\cdots {\sigma }_n{\sigma }_{n+1}$, i.e., $\#\{b_i=1,i=1,\cdots ,n+1\}=\#\{{\sigma }_i=1,i=1,\cdots ,n+1\}$. □

Now, one can calculate the periodic points of the N-type map using the periodic points of $\{{\Sigma }_3;S\}$. Thanks to the easy computation of the periodic points of the dynamical system $\{{\Sigma }_3;S\}$, one can operate the inverse itinerary map $x=h^{-1}(\sigma )$ on the periodic points $\sigma$ of $\{{\Sigma }_3;S\}$ by Theorem 1 to obtain the periodic points of the N-type map. The $k$-periodic points $\sigma$ of $\{{\Sigma }_3;S\}$ such that $S^k(\sigma )=\sigma$, and thus $\sigma$ can be given by $\sigma ={\sigma }_1{\sigma }_2\cdots {\sigma }_k{\sigma }_1{\sigma }_2\cdots {\sigma }_k\cdots ={\displaystyle \widetilde{{\sigma }_1{\sigma }_2\cdots {\sigma }_k}}$.

(i) Let $\sigma =\widetilde{{\sigma }_1{\sigma }_2\cdots {\sigma }_k}$, $\#\{{\sigma }_i=1,i=1,\cdots ,k\}$ is even, then $S^k(\sigma )=\sigma$, $\sigma$ is a periodic point with period that can divide $k$. $x=h^{-1}(\sigma )$ is then a periodic point of the N-type map with the same period of $\sigma$. For example, let $k=3$, and then the following points are the periodic points of $S$, with period 1 or 3:

$$\tilde{0},\tilde{2},\widetilde{011},\widetilde{110},\widetilde{101}$$

where $\tilde{0},\tilde{2}$ are the fixed points of $S$, and $\widetilde{011},\widetilde{110},\widetilde{101}$ are 3-periodic points of $S$, consisting of a 3-periodic cycle. Using the inverse itinerary map Equation (5), one can calculate the corresponding periodic points of $T$:

$$0.\tilde{0},0.\tilde{2},0.\widetilde{011},0.\widetilde{110},0.\widetilde{121}$$

where $0.\tilde{0}=0,0.\tilde{2}=1$ are fixed points, and $0.\widetilde{011}={\displaystyle \frac{2}{13}},0.\widetilde{110}={\displaystyle \frac{6}{13}},0.\widetilde{121}={\displaystyle \frac{8}{13}}$ are 3-periodic points.

Since $\sigma$ satisfies $\sigma =\widetilde{{\sigma }_1{\sigma }_2\cdots {\sigma }_k}$ and $\#\{{\sigma }_i=1,i=1,\cdots ,k\}$ is even, during the $k$ times of operation procedure, an even number of complement operations are carried out. As a result, the periodic points $x=0.\widetilde{b_1{b}_2\cdots b_k}$ of $T$ are such that $\#\{b_i=1,i=1,\cdots ,k\}$ is also even. Consequently, one can determine the periodic points whose periods can divide $k$:

$$x=0.b_1{b}_2\cdots b_k{b}_1{b}_2\cdots b_k\cdots ,$$

$$3^{k}x=b_1{b}_2\cdots b_k+0.b_1{b}_2\cdots b_k\cdots =b_1{b}_2\cdots b_k+x.$$

Assume the ternary number $b_1{b}_2\cdots b_k$ corresponds to decimal number $p$, and then $x={\displaystyle \frac{p}{3^k-1}}$. For any $b_1{b}_2\cdots b_k$ such that $\#\{b_i=1,i=1,\cdots ,k\}$ is even, one can obtain one periodic point $x$ whose period can divide $k$. For instance, let $k=3$, the ternary number $011$ corresponds to decimal number $p=4$, and then $x={\displaystyle \frac{4}{3^3-1}}={\displaystyle \frac{2}{13}}$ is a 3-periodic point, and two other 3-periodic points are ${\displaystyle \frac{6}{13}},{\displaystyle \frac{8}{13}}$ derived from ternary numbers $110,121$, respectively.

(ii) Let $\sigma =\widetilde{{\sigma }_1{\sigma }_2\cdots {\sigma }_k}$, $\#\{{\sigma }_i=1,i=1,\cdots ,k\}$ is odd, $S^k(\sigma )=\sigma$. In this case, according to Equation (5), $x$ cannot be expressed as a $\mathit{k-}$digit cycle form due to $T^k(x)=0\widetilde{.\overline{b_1}\overline{b_2}\cdots \overline{b_k}}$, and generally $0\widetilde{.\overline{b_1}\overline{b_2}\cdots \overline{b_k}}\ne 0.\widetilde{b_1{b}_2\cdots b_k}$. $x=0.\widetilde{b_1{b}_2\cdots b_k}$ needs to be revised to make a $\mathit{k-}$periodic point of the N-type map. The revised way is to add $\overline{b_1}\overline{b_2}\cdots \overline{b_k}$ after $b_1{b}_2\cdots ,b_k$, which is a $2k$-digit cycle form:

$$x=0.b_1{b}_2\cdots b_k\overline{b_1}\overline{b_2}\cdots \overline{b_k}{b}_1{b}_2\cdots b_k\overline{b_1}\overline{b_2}\cdots \overline{b_k}\cdots$$

For instance, $\tilde{1}$ is a fixed point of $S$, and $\widetilde{001},\widetilde{010},\widetilde{100}$ are 3-periodic points of $S$. One can calculate the corresponding periodic point of the N-type map $T$ using Equation (5):

$$0.\widetilde{11}={\displaystyle \frac{1}{2}},0.\widetilde{001221}={\displaystyle \frac{1}{14}},0.\widetilde{012210}={\displaystyle \frac{3}{14}},0.\widetilde{122100}={\displaystyle \frac{9}{14}}$$

where ${\displaystyle \frac{1}{2}}$ is a fixed point of $T$, and ${\displaystyle \frac{1}{14}},{\displaystyle \frac{3}{14}},{\displaystyle \frac{9}{14}}$ are 3-periodic points of $T$. Similarly, one can calculate the periodic points whose periods can divide $k$:

$$x=0.b_1{b}_2\cdots b_k{\overline{b}}_1{\overline{b}}_2\cdots {\overline{b}}_k{b}_1{b}_2\cdots b_k{\overline{b}}_1{\overline{b}}_2\cdots {\overline{b}}_k\cdots ,$$

$$3^{k}x=b_1{b}_2\cdots b_k+0.{\overline{b}}_1{\overline{b}}_2\cdots {\overline{b}}_k{b}_1{b}_2\cdots b_k{\overline{b}}_1{\overline{b}}_2\cdots {\overline{b}}_k\cdots =b_1{b}_2\cdots b_k+1-x.$$

Assume the ternary number $b_1{b}_2\cdots b_k$ corresponds to decimal number $p$, then $x={\displaystyle \frac{p+1}{3^k+1}}$. For any ternary number $b_1{b}_2\cdots b_k$ such that $\#\{b_i=1,i=1,\cdots ,k\}$ is odd, one can obtain a periodic point whose period is the factor of $k$. For example, let $k=3$, the decimal number of ternary number $001$ is $p=1$, so $x={\displaystyle \frac{2}{3^3+1}}={\displaystyle \frac{1}{14}}$ is a 3-periodic point of $T$, and two other 3-periodic points are ${\displaystyle \frac{3}{14}},{\displaystyle \frac{9}{14}}$ corresponding to $012,122$, respectively.

According to the above-mentioned formula, one can calculate the periodic points whose period is the factor of $k$: ${\displaystyle \frac{p}{3^k-1}}$,${\displaystyle \frac{p+1}{3^k+1}}$. Then, the following periodic points with a period less than 4 are obtained.

fixed points: {0}, {1/2}, {1};

2-periodic points: {1/4, 3/4}, {1/5, 3/5}, {2/5, 4/5};

3-periodic points: {1/13, 3/13, 9/13}, {4/13, 12/13, 10/13}, {5/13, 11/13, 7/13},

{2/13, 6/13, 8/13}, {1/14, 3/14, 9/14}, {2/14, 6/14, 10/14}, {5/14, 13/14, 11/14},

{4/14, 12/14, 8/14}.

The points within each bracket form a cycle. There are three fixed points. The total number of 2-periodic points and fixed points is 9, and the total number of 3-periodic points and fixed points is 27. In general, if the number of $k$-periodic points is $kP(k)$, then the number of distinct $k$-periodic cycles is $P(k)$. Based on the fact that the number of $x$ such that $T^k(x)=x$ is $3^k$, one can use Equation (6) to calculate $kP(k)$.

$$kP(k)=3^k-{\displaystyle \sum _{\begin{array}{c}m|k\\ 1\le m<k\end{array}}mP(m)}$$

(6)

4. Chaos of the N-Type Map

In this section, the chaotic nature of the N-type map is proven. Theorem 1 states that the N-type map is topologically conjugate to $\{{\Sigma }_3;S\}$ with the homeomorphism $h$. This topological conjugacy is then used to demonstrate the chaotic nature of the N-type map.

Theorem 3.

The N-type map is chaotic.

Proof of Theorem 3.

The theorem is proven in three parts.

(a) There is one dense orbit of the N-type map. Firstly, a dense orbit for $\{{\Sigma }_3;S\}$ is constructed, and then this dense orbit is mapped by $h^{-1}$ to obtain one dense orbit of the N-type map. It can be easily demonstrated that the orbit of $\sigma$ defined by Equation (7) is dense within ${\Sigma }_3$.

$$\sigma =\underset{\mathrm{length} 1}{\underbrace{0 1 2}} \underset{\mathrm{length} 2}{\underbrace{00 01 02 10 11 12 20 21 22}} \underset{\mathrm{length} 3}{\underbrace{000 001 002 010 011 012 020 021 022 \cdots }} \cdots$$

(7)

where $\sigma$ is formed as follows. Let $A_n$ represent a collection of digit strings of length $n$ consisting of 0, 1, 2. The number of elements in $A_n$ is $\#\left\{A_n\right\}=3^n$. Denote $A_n$ by $A_n=\{r_1^n,r_2^n,\cdots ,r_{3^n}^n\}$, where $r_1^n,r_2^n,\cdots ,r_{3^n}^n$ are arranged in ascending order. One places all $A_n$ together $A_1{A}_2\cdots A_n\cdots$ to obtain an infinite string $\sigma$ given by Equation (7). Next, one maps $\sigma$ using $h^{-1}$ to obtain $x_0\in [0,1]$ given by Equation (8).

$$x_0=0.\underset{\mathrm{length} 1}{\underbrace{0 1 0}} \underset{\mathrm{length} 2}{\underbrace{22 21 02 12 11 12 20 21 00}} \underset{\mathrm{length} 3}{\underbrace{222 221 002 012 211 212 020 021 200 \cdots }} \cdots$$

(8)

One can demonstrate that the orbit of $x_0$ is dense in [0, 1] for the N-type map. Although there are duplicates $s_1\cdots s_n$ in the ternary representation of digit strings of length n (for example, there are two “12”s and two “21”s in the set of all digit strings of length 2), one can observe that the parity of 1s in all the preceding digit strings of the two duplicates is different. Suppose that the first digits of two duplicates are located at positions $K_1$ and $K_2$ ($K_1<K_2$) in the ternary expressions of $x_0$, respectively. Then, the forward orbit points $x_{K_1-1}$ and $x_{K_2-1}$ should be $x_{K_1-1}=0.s_1\cdots s_n$, $x_{K_2-1}=0.{\overline{s}}_1\cdots {\overline{s}}_n$. As a matter of fact, the total number of digits composed of all the digit strings of length $1~n-1$ is $T_1=1\times 3^1+2\times 3^2+\cdots +(n-1)\times 3^{n-1}=((2n-3)\times 3^n+3)/4$. Then, in the continuous orbit point set $\{x_{T_1+1},\cdots ,x_{T_1+n\times 3^n}\}$, there are $3^n$ points $x_{T_1+1+(i-1)\times n},i=1,\cdots ,3^n$, such that the first n-digit strings of their ternary expressions are different. This can be proven by contradiction. Assume that there are two points $X_1=x_{T_1+1+(i-1)\times n},X_2=x_{T_1+1+(j-1)\times n}$($1\le i<j\le 3^n$) and the first n digits of their ternary expressions are the same, i.e.,

$$X_1=0.s_1\cdots s_n{s}_{n+1}\cdots ,X_2=0.t_1\cdots t_n{t}_{n+1}\cdots ,t_i=s_i,i=1,\cdots ,n.$$

According to the iteration rule, the corresponding strings of length $n$ in the itinerary $\sigma$ of $x_0$ should be the same. Since the itinerary map is a homeomorphism, one can derive the itinerary of $x_0$ to obtain $\sigma$ given by Equation (7), where all the strings of length $n$ are different, and a conflict has arisen. Therefore, there are $3^n$ points $x_{T_1+1+(i-1)\times n},i=1,\cdots ,3^n$ in the orbit point set $\{x_{T_1+1},\cdots ,x_{T_1+n\times 3^n}\}$ such that the first $n$ digit strings of their ternary expressions are different. As a result, for any given $x$ = $0.b_1{b}_2\cdots b_n\cdots$, its first $n$ digit string $b_1{b}_2\cdots b_n$ must appear in the first $n$ digits of the ternary representation of a number in $x_{T_1+1},\cdots ,x_{T_1+n\times 3^n}$. For an arbitrary $\epsilon >0$, choose a sufficiently large $n$ such that ${\displaystyle \frac{1}{3^n}}<\epsilon$. Then, iterate the constructed $x_0$ given by Equation (8) for $K$ times. Suppose the first $n$ digits of $x_K=T^K(x_0)$ are $b_1{b}_2\cdots b_n$, then $|T^K(x_0)-x|\le {\displaystyle \frac{1}{3^n}}<\epsilon$. For example, if the first two digits $b_1{b}_2$ of $x$ are ‘00’, then $K=3$, $x_3=0.00\cdots$; if the first 3 digits $b_1{b}_2{b}_3$ of $x$ are ‘000’, then $K=21$, $x_{21}=0.000\cdots$.

(b) The periodic points of the N-type map are dense in [0, 1]. Similar to the proof in part (a), for any $\epsilon >0$, choose a sufficiently large $n$ such that ${\displaystyle \frac{1}{3^n}}<\epsilon$. Suppose the ternary representation of $x$ is $0.b_1{b}_2\cdots b_n\cdots$. Then, there exists either a $\mathit{n-}$periodic point $q=0.\widetilde{b_1{b}_2\cdots b_n}$ or $q=0.\widetilde{b_1{b}_2\cdots b_n{\overline{b}}_1{\overline{b}}_2\cdots {\overline{b}}_n}$. In the case of the former, $\#\{b_i=1,i=1,\cdots ,k\}$ is even, and in the case of the latter, it is odd. Obviously, $|q-x|\le {\displaystyle \sum _{i=n+1}^{\infty }\frac{2}{3^i}=}{\displaystyle \frac{1}{3^n}}<\epsilon$. Therefore, for any $x\in [0,1]$ and any ball $B(x,\epsilon )$ with radius $\epsilon >0$, there is a periodic point $q\in B(x,\epsilon )$ such that $|x-q|<\epsilon$.

(c) The N-type map is sensitive to initial conditions. For any given $\epsilon >0$, choose $n$ such that ${\displaystyle \frac{2}{3^n}}<\epsilon$. It is easy to see that $d_c(\sigma ,\tau )\le {\displaystyle \frac{1}{3^n}}$ if and only if ${\sigma }_i={\tau }_i,i=1,\cdots ,n$, $\sigma ={\sigma }_1{\sigma }_2\cdots ,\tau ={\tau }_1{\tau }_2\cdots$. If ${\sigma }_i={\tau }_i,i=1,\cdots ,n$, ${\sigma }_{n+1}\ne {\tau }_{n+1}$, then $d_c(S^n(\sigma ),S^n(\tau ))\ge {\displaystyle \frac{1}{3}}$. For any given $x=0.s_1{s}_2\cdots s_n\cdots$, $\sigma =h(x)={\sigma }_1{\sigma }_2\cdots$, one can construct $\tau \in {\Sigma }_3$ and apply the map $h^{-1}$ to it to obtain $y=h^{-1}(\tau )\in [0,1]$, such that $|x-y|<\epsilon$, but $|T^n(x)-T^n(y)|\ge {\displaystyle \frac{1}{3}}$.

(c.1) If ${\sigma }_{n+1}=1$ and there are an even number of 1s in ${\sigma }_1{\sigma }_2\cdots {\sigma }_n$ or equivalently in $s_1{s}_2\cdots s_n$, let $\tau ={\sigma }_1{\sigma }_2\cdots {\sigma }_n\tilde{0}$. According to Theorem 2 and Theorem 1, one can obtain $x=0.s_1\cdots s_{n}1s_{n+2}\cdots$ and $y=h^{-1}(\tau )=0.s_1{s}_2\cdots s_n\tilde{0}$, respectively. Then,

$$|x-y|\le {\displaystyle \frac{1}{3^n}}<\epsilon , T^n(x)=0.1s_{n+2}\cdots , T^n(y)=0.\tilde{0}, |T^n(x)-T^n(y)|\ge {\displaystyle \frac{1}{3}}$$

(c.2) If ${\sigma }_{n+1}=1$ and there are an odd number of 1s existing in ${\sigma }_1{\sigma }_2\cdots {\sigma }_n$, let $\tau ={\sigma }_1{\sigma }_2\cdots {\sigma }_n\tilde{2}$. One can obtain $x=0.s_1{s}_2\cdots s_{n}1s_{n+2}\cdots$, $y=h^{-1}(\tau )=0.s_1{s}_2\cdots s_n\tilde{0}$. Then, $|x-y|\le {\displaystyle \frac{1}{3^n}}<\epsilon$, but $T^n(x)=0.1{\overline{s}}_{n+2}\cdots$, $T^n(y)=0.\tilde{2}$, $|T^n(x)-T^n(y)|=|{\displaystyle \frac{1}{3}}+{\displaystyle \sum _{i=2}^{\infty }\frac{(2-{\overline{s}}_{n+i})}{3^i}|\ge }{\displaystyle \frac{1}{3}}$.

(c.3) If ${\sigma }_{n+1}=0$, let $\tau ={\sigma }_1{\sigma }_2\cdots {\sigma }_n\tilde{2}$. One can obtain

$$(x,y)=(x,h^{-1}(\tau ))=\left\{\begin{array}{c}(0.s_1{s}_2\cdots s_{n}0s_{n+2}\cdots ,0.s_1{s}_2\cdots s_n\tilde{2}), \mathrm{if} \#\{{\sigma }_i=1,i=1,\dots ,n\} \mathrm{is} \mathrm{even};\\ (0.s_1{s}_2\cdots s_{n}2s_{n+2}\cdots ,0.s_1{s}_2\cdots s_n\tilde{0}), \mathrm{if} \#\{{\sigma }_i=1,i=1,\dots ,n\} \mathrm{is} \mathrm{odd}.\end{array}\right.$$

Then, $|x-y|\le {\displaystyle \frac{1}{3^n}}<\epsilon$ holds for both cases. For the former case, $T^n(x)=0.0s_{n+2}\cdots$, $T^n(y)=0.\tilde{2}$, then $|T^n(x)-T^n(y)|\ge {\displaystyle \frac{1}{3}}$. For the latter case, $T^n(x)=0.0{\overline{s}}_{n+2}\cdots$, $T^n(y)=0.\tilde{2}$, then $|T^n(x)-T^n(y)|\ge {\displaystyle \frac{1}{3}}$.

(c.4) If ${\sigma }_{n+1}=2$, let $\tau ={\sigma }_1{\sigma }_2\cdots {\sigma }_n\tilde{0}$. One can similarly obtain

$$(x,y)=(x,h^{-1}(\tau ))=\left\{\begin{array}{c}(0.s_1{s}_2\cdots s_{n}2s_{n+2}\cdots ,0.s_1{s}_2\cdots s_n\tilde{0}), \mathrm{if} \#\{{\sigma }_i=1,i=1,\dots ,n\} \mathrm{is} \mathrm{even};\\ (0.s_1{s}_2\cdots s_{n}0s_{n+2}\cdots ,0.s_1{s}_2\cdots s_n\tilde{2}), \mathrm{if} \#\{{\sigma }_i=1,i=1,\dots ,n\} \mathrm{is} \mathrm{odd}.\end{array}\right.$$

Consequently, $|x-y|\le {\displaystyle \frac{1}{3^n}}<\epsilon$, but $|T^n(x)-T^n(y)|\ge {\displaystyle \frac{1}{3}}$. □

Since the N-type map is composed of three line-segments with slopes 3, −3, and 3, respectively, its Lyapunov exponent is ln3 > 0. This fact verifies the chaotic nature of the system from the perspective of the Lyapunov exponent. Although it is not possible to conduct a bifurcation analysis, we can numerically verify the chaotic nature of this map through aspects such as its sensitivity to initial conditions, the autocorrelation and cross-correlation of the trajectories, and the distribution of the trajectories. These results help prove the theoretical findings numerically. The sensitivity of the N-type map to initial conditions is clearly illustrated in Figure 2. As can be seen from Figure 2, the truncated orbit $\{T^n(x_0):n=1,\cdots ,60\}$ of $x_0=0.371$ and the truncated orbit $\{T^n(x_0):n=1,\cdots ,60\}$ of $x_0=0.371+{10}^{-12}$ will exhibit a significant difference after $n=21$. This indicates that the N-type map displays extremely high sensitivity to initial conditions. The N-type map also exhibits other favorable chaotic characteristics. For example, the sequences it generates traverse the interval [0, 1] and have dense orbits. Moreover, the sequences possess desirable autocorrelation and cross-correlation properties. All these excellent chaotic properties enable the N-type map to generate outstanding pseudo-random sequences, making it highly suitable for designing cryptosystems.

Figure 2. The sensitivity of the N-type map to initial conditions. Two orbits are derived from $x_0=0.371$ and its slightly different version $x_0=0.371+{10}^{-12}$.

The definitions of the autocorrelation coefficient and the cross-correlation coefficient are presented in Equations (9) and (10), respectively. The autocorrelation coefficient quantifies the correlation degree between a sequence and itself at a time delay. The cross-correlation coefficient quantifies the correlation degree between one sequence and another sequence at a time delay.

$$autocorr(k)={\displaystyle \frac{{\displaystyle \sum _{i=k+1}^n(x_i-\overline{x})(x_{i-k}-\overline{x})}}{{\displaystyle \sum _{i=1}^n{(x_i-\overline{x})}^2}}},$$

(9)

$$crosscorr(k)={\displaystyle \frac{{\displaystyle \sum _{i=k+1}^n(x_i-\overline{x})(y_{i-k}-\overline{y})}}{\sqrt{{\displaystyle \sum _{i=1}^n{(x_i-\overline{x})}^2}}\sqrt{{\displaystyle \sum _{i=1}^n{(y_i-\overline{y})}^2}}}},$$

(10)

where $n$ is the length of the sequence; $k$ is the delay order; $\overline{x}$ and $\overline{y}$ are the mean values of sequence $x$ and sequence $y$, respectively. Let the delay order $k=0,\cdots ,200$, and then calculate the autocorrelation and cross-correlation coefficients of the chaotic sequences. The results are presented in Figure 3 and Figure 4, respectively. A good autocorrelation coefficient graph should be very close to the Dirac $\delta$ function. As can be seen from Figure 3, the autocorrelation coefficients of sequence $x=\{T^n(x_0):n=1,\cdots ,100000\}$ are extremely close to the Dirac $\delta$ function, indicating that the chaotic sequence generated by the N-type map has excellent autocorrelation. An excellent cross-correlation coefficient graph should be very close to the zero function. Figure 4 shows the cross-correlation coefficient graph between sequence $x$ and sequence $y=\{T^n(y_0):n=0,1,\cdots ,100000\}$ with $y_0=0.371+{10}^{-12}$. One can observe that the sequences generated by the N-type map have excellent cross-correlation.

Figure 3. The autocorrelation of the orbit sequence $\{x_n=T^n(x_0):n=0,1,\cdots ,100000\}$ with $x_0=0.371$.

Figure 4. The cross-correlation of $\{x_n=T^n(x_0):n=0,1,\cdots ,100000\}$ with $x_0=0.371$ and $\{y_n=T^n(y_0):n=0,1,\cdots ,100000\}$ with $y_0=0.371+{10}^{-12}$.

Figure 5 presents the histogram of the sequence generated by the N-type map. It vividly demonstrates that the trajectory points are densely distributed in the interval [0, 1], and their distribution is remarkably close to a uniform distribution.

Figure 5. The histogram of the orbit sequence $\{x_n=T^n(x_0):n=0,1,\cdots ,100000\}$ with $x_0=0.371$.

5. Application to Image Scrambling Encryption

To utilize the itineraries of the N-type map for achieving image pixel position scrambling, the intervals corresponding to the string at the start of the N-type map’s itineraries are defined as follows:

$$I_{{\sigma }_1{\sigma }_2\cdots {\sigma }_n}=\{x:T^j(x)\in I_{{\sigma }_{j+1}},0\le j\le n-1\}={\displaystyle \underset{j=0}{\overset{n-1}{\cap }}{T}^{-j}(I_{{\sigma }_{j+1}})},{\sigma }_i\in \{0,1,2\},i=1,\cdots ,n.$$

If $x\in I_{{\sigma }_1{\sigma }_2\cdots {\sigma }_n}$, then $T^{j-1}(T(x))\in I_{{\sigma }_{j+1}},1\le j\le n-1$, $T(x)\in I_{{\sigma }_2\cdots {\sigma }_n}$. Further, [0, 1] can be evenly partitioned into distinct intervals $I_{{\sigma }_1{\sigma }_2\cdots {\sigma }_n}$ with length $2^{-n}$. There are $2^n$ intervals $I_{{\sigma }_1{\sigma }_2\cdots {\sigma }_n}$ such that ${\displaystyle \underset{{\sigma }_1,\cdots {\sigma }_n}{\cup }{I}_{{\sigma }_1{\sigma }_2\cdots {\sigma }_n}}=[0,1]$. The points that lie in $I_{{\sigma }_1{\sigma }_2\cdots {\sigma }_n}$ have the itineraries with the same first $n$ symbols ${\sigma }_1{\sigma }_2\cdots {\sigma }_n,{\sigma }_i\in \{0,1,2\}$. For $n=1,2$, the interval divisions are shown in Figure 6.

$$\begin{array}{l}[0,1]=[0,1/3]\cup [1/3,2/3]\cup [2/3,1]=I_0\cup I_1\cup I_2;\\ [0,1]=[0,1/9]\cup \cdots \cup [8/9,1]=I_{00}\cup I_{01}\cup I_{02}\cup I_{12}\cup I_{11}\cup I_{10}\cup I_{20}\cup I_{21}\cup I_{22}.\end{array}$$

Figure 6. The N-type map and its truncated itinerary for (a) $n=1,2$. $I_0$, $I_1$, $I_2$ are the three distinct intervals with length 1/3; (b) $I_{00,}$, $I_{01}$, $I_{02}$, $I_{12,}$, $I_{11}$, $I_{10}$, $I_{20,}$, $I_{21}$, $I_{22}$ are the nine distinct intervals with length 1/9. The subscript sequence numbers of the intervals form an itinerary.

Obviously, $I_{{\sigma }_1{\sigma }_2\cdots {\sigma }_n}$ are closed interval sets such that $I_{{\sigma }_1}\supset I_{{\sigma }_1{\sigma }_2}\supset \cdots \supset I_{{\sigma }_1{\sigma }_2\cdots {\sigma }_n}$.

According to the definition of itinerary, to generate a truncated itinerary of length $n$, one needs to uniformly partition the interval [0, 1] into $3^n$ subintervals of length $1/3^n$. By selecting any point within each interval (e.g., the midpoint) and iterating the N-type map for $n$ times, one can record the ordinal indices of intervals where each iterated trajectory point falls, thereby forming the truncated itinerary for any point in the interval. However, this definition-based computation method is inefficient and requires improvement. The improvement approach leverages the inherent structural properties of the N-type map itself, revealing a recursive relationship between adjacent levels of itineraries. Figure 6 illustrates the recursive relationship from first-level to second-level itineraries. The first-level truncated itineraries correspond to three interval indices 0, 1, 2. The second-level itineraries can be generated by prepending 0, 1, or 2 to the first-level itineraries, respectively. Specifically, prepending 0 yields 00, 01, 02; prepending 1 first generates 10, 11, 12, then reverses their order to obtain 12, 11, 10; prepending 2 generates 20, 21, 22. Thus, the length-2 truncated itineraries are 00, 01, 02, 12, 11, 10, 20, 21, 22. This recursive construction significantly improves computational efficiency compared to the original definition-based method. The recursive itinerary generation algorithm is outlined in Algorithm 1.

Algorithm 1. (Recursive itinerary generation)

Initialization: define the itinerary set of length 1 as the base ternary unit $V_1$= (0, 1, 2). The corresponding ternary encodings are 0, 1, 2 (stored as decimal values 0, 1, 2).

Recursive construction: given the decimal representation of $V_n$, generate $V_{n+1}$ in four steps.

(a) Left branch: directly inherit $V_n$, equivalent to prepending 0 to each itinerary (decimal values remain unchanged): $VL=V_n$.

(b) Middle branch: add $3^n$ to each element of $V_n$, then reverse the entire sequence: $VM=\mathrm{fliplr}(3^n+V_n)$.

(c) Right branch: add $2\times 3^n$ to each element of $V_n$, preserving the original order: $VM=2\times 3^n+V_n$.

(d) Merge: combine all branches: $V_{n+1}=(VL,VM,VR)$.

One can convert the final decimal values generated by Algorithm 1 to $n$-digit ternary strings to obtain the truncated itinerary ${\sigma }_1{\sigma }_2\cdots {\sigma }_n$ of length $n$. If the truncated itinerary ${\sigma }_1{\sigma }_2\cdots {\sigma }_n$ is considered as a ternary presentation number, then $3^n$ distinct integers belonging to $[0,3^n-1]$ will be produced. When these integers are arranged in order, a permutation of the $3^n$ numbers from 0 to $3^n-1$ is obtained. Thus, this permutation can be used to alter the original positions of the plain image pixels to achieve the purpose of image scrambling.

Regarding the application of the itineraries of the N-type map, two scrambling algorithms are designed. The proposed image scrambling algorithms can be applied to any image with arbitrary height $H$ and width $W$. In Algorithm 2, a plain image is first converted to a one-dimensional vector. Then, the N-type map is applied to generate 1D itinerary set $\{{\sigma }_1{\sigma }_2\cdots {\sigma }_n:{\sigma }_i\in \{0,1,2\},i=1,\cdots ,n\}$, which is used to permute the 1D pixel positions in the vector. In Algorithm 3, two N-type maps are utilized to generate 2D itinerary set to permute the 2D pixel positions in the image matrix.

Algorithm 2. (1D itinerary-based image scrambling)

Step 1. Read a plain image $P$ with height $H$ and width $W$. Rearrange $P$ into a vector $V(i),i=0,\cdots ,L-1$ of length $L=H\times W$ in column-priority order.

Step 2. Let $n$ be the least integer $n$ such that $3^n\ge L$ is met. Apply Algorithm 1 to obtain the truncated itinerary set $S=\{{\sigma }_1{\sigma }_2\cdots {\sigma }_n:{\sigma }_i\in \{0,1,2\left\}\right\}$, which contains $3^n$ elements. For each truncated itinerary, reverse its order to obtain ${\sigma }_n{\sigma }_{n-1}\cdots {\sigma }_1$. Convert it into a decimal number ${\sigma }_n\times 3^{n-1}+{\sigma }_{n-1}\times 3^{n-2}+\cdots +{\sigma }_2\times 3+{\sigma }_1$. Arrange these decimal numbers orderly to form a permutation of the set $\{0,1,\cdots ,3^n-1\}$. Then, reject the numbers larger than $L-1$ to obtain a permutation $ind(i),i=0,\cdots ,L-1$ of $0,1,\cdots ,H\times W-1$.

Step 3. Use $ind$ to rearrange the elements of $V$ and get $V1$:

$V1(i)=V(ind(i)),i=0,1,\cdots ,L-1.$

Step 4. Rearrange $V1$ into 2D matrix in column-priority order and the final scrambled image with height $H$ and width $W$ is obtained.

Algorithm 3. (2D itinerary-based image scrambling)

Step 1. Input a plain image $P(i,j),i=0,1,\cdots ,H-1,j=0,1,\cdots ,W-1$.

Step 2. Select the least integers $m,n$ such that $2^m\ge H,2^n\ge W$. Then, use Algorithm 1 to separately generate the truncated itinerary sets $\{{\sigma }_1{\sigma }_2\cdots {\sigma }_m\}$ and $\{{\tau }_1{\tau }_2\cdots {\tau }_n\}$. Reverse the order of each itinerary to obtain $\{{\sigma }_m{\sigma }_{m-1}\cdots {\sigma }_1\}$, $\{{\tau }_n{\tau }_{n-1}\cdots {\tau }_1\}$. Convert the elements in $\{{\sigma }_m{\sigma }_{m-1}\cdots {\sigma }_1\}$ and $\{{\tau }_n{\tau }_{n-1}\cdots {\tau }_1\}$ into decimal numbers and arrange these decimal numbers to form permutations of $0,1,\cdots ,2^m-1$ and $0,1,\cdots ,2^n-1$, respectively. Next, filter out the numbers in $\{{\sigma }_m{\sigma }_{m-1}\cdots {\sigma }_1\}$ that are larger than $H-1$ and those in $\{{\tau }_n{\tau }_{n-1}\cdots {\tau }_1\}$ that are larger than $W-1$. After filtering, we obtain the permutations of $0,1,\cdots ,H-1$ and $0,1,\cdots ,W-1$, respectively. The permutations are denoted as $indx(i),i=0,\cdots ,H-1$ and $indy(i),i=0,\cdots ,W-1$, respectively.

Step 3. $indx,indy$ are used to rearrange the pixel positions of plain image $P$ and obtain the scrambled image $C$:

$C(i,j)=P(indx(i),indy(j)),i=0,1,\cdots ,H-1,j=0,1,\cdots ,W-1.$

The proposed algorithms are capable of scrambling plain images of arbitrary sizes. The corresponding decryption algorithms can restore the plain images without any distortion. The experimental results are presented in Figure 7 and Figure 8. In the simulation, the sizes of the plain images of Lena and rice are 225 × 217 and 121 × 236, respectively. In Figure 7, Algorithm 2 can achieve a favorable encryption effect after one round of scrambling, and no texture trace can be detected in the encrypted image. Figure 8 illustrates the scrambling effect of the plain images when using Algorithm 3. As can be observed from Figure 7 and Figure 8, for the same number of encryption rounds, the scrambling effect of Algorithm 2 is generally superior to that of Algorithm 3. However, Algorithm 2 requires converting 2D images to 1D vectors and vice versa. Additionally, it needs to generate longer truncated itineraries. Consequently, for the same number of encryption rounds, it takes more time to encrypt the plain images.

Figure 7. Algorithm 2: (a–c) are plain images of rice, cipher-image of one round, cipher-image of two rounds, respectively; (d–f) are plain images Lena, cipher-image of one round, cipher-image of two rounds, respectively.

Figure 8. Algorithm 3: (a–c) are plain images of rice, cipher-image of one round, cipher-image of two rounds, respectively; (d–f) are plain images of Lena, cipher-image of one round, cipher-image of two rounds, respectively.

6. Conclusions

In this paper, the topological conjugation between the N-type map $T$ defined in [0, 1] and the shift map $S$ in three-symbol space ${\Sigma }_3$ is proven. We precisely constructed the homeomorphism $h:[0,1]\to {\Sigma }_3$. As a result, it becomes straightforward to calculate the periodic points of the N-type map and prove its chaotic nature. This is achieved by using the homeomorphism $h$ to map the periodic points and orbit points of the shift map $S$ to those of the N-type map. As an application, two image scrambling encryption algorithms are proposed based on the truncated itineraries of the N-type map. These algorithms can showcase the application value of mathematical theories and methods. The itinerary method of dynamical systems not only has theoretical significance but also can be applied to encode the positions of image pixels. It is a new field worthy of in-depth theoretical research and application development. The itinerary calculation method for the N-type map proposed in this paper serves as an example for the itinerary calculation of other dynamical systems. For a general map, whether there exists a universal itinerary computational encoding method remains an open question that requires further exploration. In particular, the problem of calculating the itineraries for two-dimensional piecewise linear maps is a research topic that has not yet been explored. The research results are of great significance both for characterizing the chaotic behavior of two-dimensional maps and for encoding applications. This is a topic that can be further explored in the future. It is hoped that more valuable results regarding the itineraries of dynamical systems will be obtained.