Algebraic Structure of Moore Cellular Automata - Finite Triangular Lattice

The interrelationship between elements of a system greatly influences its behavior. From the mid-20th century, various scientists started proposing multiple models for this approach. Cellular Automata (CA) is one of the magnificent models scientists welcome in this direction of study. Cellular Automata is a discrete model of computation that plays a significant role in the simulation of real-world systems. In 1996, von Neumann and Burks [1] initiated the concept of CA by modeling self-reproduction in biological systems. CA is modeled with a grid of cells that evolve over discrete generations depending upon the local transition rules. The major factors influencing the evolution of two-dimensional cellular automaton is the lattice shape, type of neighborhood, the set of states and the nature of the local transition rule. Cellular Automata rules can be categorized into two main classifications: linear rules and non - linear rules. The majority of research in Cellular Automata has focused on one-dimensional models where the lattice shape is not a critical factor. However, in two-dimensional Cellular Automata, the geometry of the underlying lattice significantly influences the system's dynamics and evolution. In this context, researchers have extended traditional square lattice models to explore more complex topologies such as hexagonal and triangular lattices. These alternative lattice structures introduce different neighborhood configurations, which directly affect rule formulation, propagation behavior, and pattern formation. Notable studies have examined the impact of such lattice variations on Cellular Automata behavior, particularly in modeling natural systems, pattern generation, and computational complexity.

The choice of lattice geometry such as square, triangular [2], hexagonal [3, 4], or pentagonal [5] significantly influences neighborhood configurations and consequently the system's evolution. Triangular grids produce certain shapes and patterns that are not possible to achieve with the typical square grid CA. Zawidzki [6] introduced the concept of Cellular Automata on a triangular tessellation. Saadat and Nagy [7, 8] and Morita [9] examined the approach of Cellular Automata on a triangular grid. One major factor of a Cellular Automata is the type of neighborhood considered. John von Neumann proposed a neighborhood model that depends on the current cell and its adjacent cells. In 1962, Moore [10] proposed an extended form of the neighborhood proposed by von Neumann. In this case, the neighborhood depends upon the center cell, its adjacent cells, and the cells connected with the nodes of the center cell. The Moore neighborhood of the triangular lattice consists 13 neighbors which is the highest of any other lattice for neighborhood of radius 1.

The geometric structure of Cellular Automata promotes curiosity about decoding the algebraic characterization behind the automata. A finite Cellular Automata comprises of a finite number of cells, each mapping to a state. A specific local transition rule computes each cell state at the next generation. A concise and clear representation of the rules using mathematical logic aids in understanding how the automata compute the next generation. Das et al. [11, 12], Ganguly et al. [13], and Pal Chaudhuri et al. [14] characterized the behavior of CA’s state transition using matrix algebra. The work by Kaspar et al. [15] on two-dimensional automata laid the motivation to explore the properties of two-dimensional Cellular Automata. Two-dimensional CAs, due to their variety of purposes, are investigated by Packard and Wolfram [16], Terrier [17], Durand [18], de Oliveira et al. [19], Kari [20], and many others. References [21-26] investigated the algebraic properties of two-dimensional CA using matrix algebra.

In infinite Cellular Automata, boundary conditions are inherently absent, and thus have no influence on the system's evolution. However, in finite Cellular Automata, boundary conditions become integral to the dynamics, as they directly affect the evolution of configurations over time. The interaction between the finite grid and its boundaries can lead to emergent behaviors that significantly impact the global evolution of the system. The two most prevalent boundary conditions are the null and periodic boundaries and their effects can be explored from the study of LuValle [27]. The work by Sahin et al. [28] explores the various boundary conditions of two-dimensional Cellular Automata using von Neumann neighborhood. In this work, we consider the finite Cellular Automata in a two-dimensional grid on a triangular lattice. In 2017, Uguz et al. [29] examined the reversibility and structure of the linear, triangular, two-dimensional von Neumann Cellular Automata with null boundary condition. However, in their formulation, the current (center) cell was not included as part of the neighborhood and was therefore excluded from the rule matrix construction. This omission deviates from the conventional CA framework, where the current cell is typically considered an essential part of the neighborhood. In the present work, we address this limitation by explicitly incorporating the center cell into the neighborhood definition and the corresponding rule matrix.

Triangular lattice Cellular Automata are known for their ability to generate aesthetically pleasing and highly symmetric patterns that are often difficult to replicate using square or hexagonal lattices. The inherent angular structure and rotational symmetry of the triangular grid allows for intricate interlocking motifs, radial formations, and fractal-like growth that are naturally aligned with equilateral geometry. These properties make triangular lattices particularly well-suited for visual complexity and symmetry, often resembling traditional tessellations and decorative art. The triangular lattice has gained significant attention in recent Cellular Automata research due to its geometric versatility and ability to generate complex, symmetric patterns which are not easily achievable with traditional square grids. Saadat and Nagy [8] further illustrated the expressive potential of triangular lattice Cellular Automata by generating intricate patterns like mandalas and trees through alternating rule applications. Ray et al. [30] utilized probabilistic Cellular Automata on two-dimensional structures to preserve quantum information, showing how lattice geometry influences memory stability under noise. Verhodubs [31] combined ontological frameworks with Cellular Automata to simulate cognitive processes, leveraging triangular lattices for spatial efficiency in representing thought dynamics.

The Moore neighborhood on a triangular lattice is more complicated than those on square and hexagonal lattices because it has more neighbors and different ways they interact. Despite its rich dynamics, the algebraic characterization of Triangular Lattice Moore Cellular Automata (TLMCA) remains largely unexplored in the current literature. This research aims to fill that gap by creating a way to represent the rules of how these automata change both locally and globally using a matrix. By using matrix algebra, we can break down the complicated local connections, making it easier to analyze the overall rule structure. This approach simplifies the complexity and opens a path to examine the reversibility.

In this work, we study the algebraic structure of finite linear TLMCA under null boundary condition. We consider the Moore neighborhood under null boundary condition. The structure of the paper is as follows: Section 2 provides some basic definitions. The rule matrix for finite linear TLMCA with null boundary condition is given in Section 3. In Section 4, we provided a method to find the $s^{th}$ configuration from the $s+1^{t h}$ configuration of the finite linear TLMCA with null boundary condition. Also presented an algorithm to generate patterns with finite linear TLMCA and the patterns generated using finite linear TLMCA in Section 4. The illustration of the pattern evolution and rule matrix evolution is presented in Section 5.

This section defines the Moore neighborhood and associated linear transition rule for a cell in a triangular lattice. Let $r\left(v_{i, j}\right)$ be the transition rule for each cell in the triangular lattice. The transition rule is defined in two distinct ways due to the presence of two categories of neighborhood states, which are determined by the position of the triangle (upward and downward). This finite linear TLMCA deals with finite number of states. We consider ternary state set throughout this work.

Definition 7. Moore neighborhood of a triangular lattice Cellular Automata is defined as a 13 − tuple.

If the cell $x_{i, j}$ lies inside the inverted triangle, then the neighborhood vector is defined as

$\begin{aligned} & \left(v_{i, j}, v_{i-1, j}, v_{i, j+1}, v_{i, j-1}, v_{i-1, j+1}, v_{i-1, j+2}, v_{i, j+2}\right.,\left.v_{i+1, j+1}, v_{i+1, j}, v_{i+1, j-1}, v_{i, j-2}, v_{i-1, j-2}, v_{i-1, j-1}\right)\end{aligned}$

If the cell $x_{i, j}$ lies inside the upright triangle, then the neighborhood vector is defined as

$\begin{aligned} & \left(v_{i, j}, v_{i, j+1}, v_{i+1, j}, v_{i, j-1}, v_{i, j+2}, v_{i+1, j+2}, v_{i+1, j+1},\right.\left.v_{i+1, j-1}, v_{i+1, j-2}, v_{i, j-2}, v_{i-1, j-1}, v_{i-1, j}, v_{i-1, j+1}\right)\end{aligned}$

where, each $v_{i, j}$ represents the state of the cell $x_{i, j}$.

Definition 8. Local transition rule of the finite linear TLMCA is defined as $R: C^t \rightarrow C^{t+1}$ where $R=\left\{r\left(v_{i, j}\right)\right\} . v_{i, j}$ represents the state of the cell $x_{i, j}$.

If $v_{i, j}$ lies inside the inverted triangle

$\begin{aligned} r\left(v_{i, j}\right)=\left(a_0 v_{i, j}+\right. & a v_{i-1, j}+b v_{i, j+1}+c v_{i, j-1}+a_1 v_{i-1, j+1}+a_2 v_{i-1, j+2}+a_3 v_{i, j+2}+b_1 v_{i+1, j+1} \\ & +b_2 v_{i+1, j}+b_3 v_{i+1, j-1}+c_1 v_{i, j-2}\left.+c_2 v_{i-1, j-2}+c_3 v_{i-1, j-1}\right) \bmod 3\end{aligned}$                     (1)

If $v_{i, j}$ lies inside the upright triangle

$\begin{gathered}r\left(v_{i, j}\right)=\left(d_0 v_{i, j}+d v_{i, j+1}+e v_{i+1, j}+f v_{i, j-1}+\right.d_1 v_{i, j+2}+d_2 v_{i+1, j+2}+d_3 v_{i+1, j+1}+e_1 v_{i+1, j-1}+ \\ e_2 v_{i+1, j-2}+e_3 v_{i, j-2}+f_1 v_{i-1, j-1}+f_2 v_{i-1, j}+\left.f_3 v_{i-1, j+1}\right) \bmod 3\end{gathered}$                      (2)

where,

$\begin{gathered}a_0, b, c, a_1, a_2, a_3, b_1, b_2, b_3, c_1, c_2, c_3, d_0, e, f, d_1, d_2, d_3, e_1, e_2, e_3, f_1, f_2, f_3 \in Z_k\end{gathered}$

Definition 9. A finite linear TLMCA is defined as a quadruple $T_M=(L, Q, N, R)$ where, $L$ is a two-dimensional triangular lattice $L \subseteq Z^2, Q$ is the finite set of states denoted by $v_{i, j}, N$ is the set of all 13-tuple neighborhood vectors $\left(n_1, n_2, \ldots, n_{13}\right)$ for all $m \cdot n$ cells of the triangular lattice Cellular Automata and $R$ is the local transition rule $R: C^{(s)} \rightarrow C^{(s+1)}$ and $R=\left\{r\left(v_{i, j}\right)\right\}$.

The rule matrices for obtaining the next step configurations of finite linear TLMCA are derived in the following theorems.

Theorem 1. Let $T_M$ be a finite linear TLMCA under null boundary condition over the field $Z_3$. Then the rule matrix of order $m \times m$ which updates the $s^{\text {th }}$ finite linear TLMCA configuration $C^{(s)}$ to $(s+1)^{\text {th }}$ configuration $C^{(s+1)}$ is given by

$\mathcal{T}_{\mathcal{R}}^{\mathcal{O}\mathcal{O}}=\left(\begin{array}{ccccccc}\alpha_1 & \beta_1 & \mathcal{O} & & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ \gamma_1 & \alpha_2 & \beta_2 & \cdots & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ \mathcal{O} & \gamma_2 & \alpha_1 & & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ & \vdots & & \ddots & & \vdots & \\ \mathcal{O} & \mathcal{O} & \mathcal{O} & & \alpha_1 & \beta_1 & \mathcal{O} \\ \mathcal{O} & \mathcal{O} & \mathcal{O} & \cdots & \gamma_1 & \alpha_2 & \beta_2 \\ \mathcal{O} & \mathcal{O} & \mathcal{O} & & \mathcal{O} & \gamma_2 & \alpha_1\end{array}\right)$

where, $\alpha_1, \alpha_2, \beta_1, \beta_2, \gamma_1, \gamma_2$ are $n \times n$ submatrices and $\mathcal{O}$ is an $n \times n$ null matrix. Here $m$ and $n$ are odd positive integers.

Proof. Let $T_M=(L, Q, N, R)$ be the finite linear TLMCA. $L$ is an $m \times n$ two-dimensional triangular lattice where $m$ and $n$ are odd positive integers. $Q$ is the finite set of states. Let us consider the cellular automaton has 3 states $\{0,1,2\} . N$ is the set neighborhood vectors. $R$ is the set of local transition rules. $R$ is defined as $R: C^{(s)} \rightarrow C^{(s+1)}$ where $R=\left\{r\left(v_{i, j}\right)\right\}$.

If $v_{i, j}$ lies inside the inverted triangle, from Eq. (1)

$\begin{gathered}r\left(v_{i, j}\right)=\left(a_0 v_{i, j}+a v_{i-1, j}+b v_{i, j+1}+c v_{i, j-1}+\right.a_1 v_{i-1, j+1}+a_2 v_{i-1, j+2}+a_3 v_{i, j+2} \\ +b_1 v_{i+1, j+1}+b_2 v_{i+1, j}+\left.b_3 v_{i+1, j-1}+c_1 v_{i, j-2}+c_2 v_{i-1, j-2}+c_3 v_{i-1, j-1}\right) \bmod 3\end{gathered}$

If $v_{i, j}$ lies inside the upright triangle, from Eq. (2)

$\begin{gathered}r\left(v_{i, j}\right)=\left(d_0 v_{i, j}+d v_{i, j+1}+e v_{i+1, j}+f v_{i, j-1}+d_1 v_{i, j+2}+\right.d_2 v_{i+1, j+2}+d_3 v_{i+1, j+1} \\+e_1 v_{i+1, j-1} +e_2 v_{i+1, j-2}+\left.e_3 v_{i, j-2}+f_1 v_{i-1, j-1}+f_2 v_{i-1, j}+f_3 v_{i-1, j+1}\right) \bmod 3\end{gathered}$

where,

$\begin{gathered}a_0, b, c, a_1, a_2, a_3, b_1, b_2, b_3, c_1, c_2, c_3, d_0, e, f, d_1, d_2, d_3, e_1, e_2, e_3, f_1, f_2, f_3 \in Z_k\end{gathered}$

Let $v_{i, j}=0 \quad$ when $i \notin\{1,2, \cdots m\}$ or $j \notin\{1,2, \cdots n\}$ because of employing null boundary conditions. By applying the local transition rule from Eqs. (1) and (2) to each cell in the automata, $m \cdot n$ linear equations are obtained.

$\begin{aligned} r\left(v_{1,1}\right)=\left(d_0 v_{1,1}\right. & +d v_{1,2}+e v_{2,1}+f v_{1,0}+d_1 v_{1,3}+d_2 v_{2,3}+d_3 v_{2,2} \\ &+e_1 v_{2,0} +e_2 v_{2,-1}+e_3 v_{1,-1}+f_1 v_{0,0}\left.+f_2 v_{0,1}+f_3 v_{0,2}\right) \bmod 3\end{aligned}$                      (3)

$\begin{aligned} r\left(v_{1,2}\right)=\left(a_0 v_{1,2}\right. & +a v_{0,2}+b v_{1,3}+c v_{1,1}+a_1 v_{0,3}+a_2 v_{0,4}+a_3 v_{1,4}+b_1 v_{2,3} \\ & +b_2 v_{2,2}+b_3 v_{2,1}+c_1 v_{1,0}+c_2 v_{0,0}\left.+c_3 v_{0,1}\right) \bmod 3\end{aligned}$                       (4)

$\begin{aligned} r\left(v_{1, n}\right)=\left(d_0 v_{1, n}\right. & +d v_{1, n+1}+e v_{2, n}+f v_{1, n-1}+d_1 v_{1, n+2}+d_2 v_{2, n+2}+d_3 v_{2, n+1} \\ & +e_1 v_{2, n-1}+e_2 v_{2, n-2}+e_3 v_{1, n-2}+f_1 v_{0, n-1}+f_2 v_{0, n}\left.+f_3 v_{0, n+1}\right) \bmod 3\end{aligned}$                      (5)

$\begin{aligned} r\left(v_{2,1}\right)=\left(a_0 v_{2,1}\right. & +a v_{1,1}+b v_{2,2}+c v_{2,0}+a_1 v_{1,2}+a_2 v_{1,3}+a_3 v_{2,3}+b_1 v_{3,2} \\ & +b_2 v_{3,1}+b_3 v_{3,0}+c_1 v_{2,-1}\left.+c_2 v_{1,-1}+c_3 v_{1,0}\right) \bmod 3\end{aligned}$                     (6)

$\begin{aligned} r\left(v_{2,2}\right)=\left(d_0 v_{2,2}\right. & +d v_{2,3}+e v_{3,2}+f v_{2,1}+d_1 v_{2,4}+d_2 v_{3,4}+d_3 v_{3,3} \\ & +e_1 v_{3,1}+e_2 v_{3,0}+e_3 v_{2,0}+f_1 v_{1,1}\left.+f_2 v_{1,2}+f_3 v_{1,3}\right) \bmod 3\end{aligned}$                     (7)

$\begin{aligned} r\left(v_{2, n}\right)=\left(a_0 v_{2, n}\right. & +a v_{1, n}+b v_{2, n+1}+c v_{2, n-1}+a_1 v_{1, n+1}+a_2 v_{1, n+2}+a_3 v_{2, n+2} \\ & +b_1 v_{3, n+1}+b_2 v_{3, n}+b_3 v_{3, n-1}+c_1 v_{2, n-2}+c_2 v_{1, n-2}\left.+c_3 v_{1, n-1}\right) \bmod 3\end{aligned}$                       (8)

$\begin{aligned} r\left(v_{m, 1}\right)=\left(d_0 v_{m, 1}\right. & +d v_{m, 2}+e v_{m+1,1}+f v_{m, 0}+d_1 v_{m, 3}+d_2 v_{m+1,3}+d_3 v_{m+1,2}+e_1 v_{m+1,0} \\ & +e_2 v_{m+1,-1}+e_3 v_{m,-1} +f_1 v_{m-1,0}+f_2 v_{m-1,1}\left.+f_3 v_{m-1,2}\right) \bmod 3\end{aligned}$                    (9)

$\begin{aligned} r\left(v_{m, 2}\right)=\left(a_0 v_{m, 2}\right. & +a v_{m-1,2}+b v_{m, 3}+c v_{m, 1}+a_1 v_{m-1,3}+a_2 v_{m-1,4}+a_3 v_{m, 4}+b_1 v_{m+1,3} \\ & +b_2 v_{m+1,2}+b_3 v_{m+1,1}+c_1 v_{m, 0}+c_2 v_{m-1,0}\left.+c_3 v_{m-1,1}\right) \bmod 3\end{aligned}$                   (10)

$\begin{aligned} r\left(v_{m, n}\right)=\left(d_0 v_{m, n}\right. & +d v_{m, n+1}+e v_{m+1, n}+f v_{m, n-1}+d_1 v_{m, n+2}+d_2 v_{m+1, n+2}+d_3 v_{m+1, n+1} +e_1 v_{m+1, n-1}\\ & +e_2 v_{m+1, n-2}+e_3 v_{m, n-2}+f_1 v_{m-1, n-1}+f_2 v_{m-1, n} \left.+f_3 v_{m-1, n+1}\right) \bmod 3\end{aligned}$     (11)

These equations can be represented in matrix form as:

$\mathcal{J}_{\mathcal{R}}^{\mathcal{O} \mathcal{O}}=\left(\begin{array}{ccccccc}\alpha_1 & \beta_1 & \mathcal{O} & & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ \gamma_1 & \alpha_2 & \beta_2 & \cdots & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ \mathcal{O} & \gamma_2 & \alpha_1 & & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ & \vdots & & \ddots & & \vdots & \\ \mathcal{O} & \mathcal{O} & \mathcal{O} & & \alpha_1 & \beta_1 & \mathcal{O} \\ \mathcal{O} & \mathcal{O} & \mathcal{O} & \cdots & \gamma_1 & \alpha_2 & \beta_2 \\ \mathcal{O} & \mathcal{O} & \mathcal{O} & & \mathcal{O} & \gamma_2 & \alpha_1\end{array}\right)$

where,

$\alpha_1=\left(\begin{array}{ccccccccc}\mathrm{d}_0 & \mathrm{~d} & \mathrm{~d}_1 & 0 & 0 & & 0 & 0 & 0 \\ \mathrm{c} & \mathrm{a}_0 & \mathrm{~b} & \mathrm{a}_3 & 0 & & 0 & 0 & 0 \\ \mathrm{e}_3 & \mathrm{f} & \mathrm{d}_0 & \mathrm{~d} & \mathrm{~d}_1 & \cdots & 0 & 0 & 0 \\ 0 & \mathrm{c}_1 & \mathrm{c} & 0 & \mathrm{~b} & & 0 & 0 & 0 \\ 0 & 0 & \mathrm{e}_3 & \mathrm{f} & \mathrm{d}_0 & & 0 & 0 & 0 \\ & & \vdots & & & \ddots & & \vdots & \\ 0 & 0 & 0 & 0 & 0 & & \mathrm{~d}_0 & \mathrm{~d} & \mathrm{~d}_1 \\ 0 & 0 & 0 & 0 & 0 & \cdots & \mathrm{c} & \mathrm{a}_0 & \mathrm{~b} \\ 0 & 0 & 0 & 0 & 0 & & \mathrm{e}_3 & \mathrm{f} & \mathrm{d}_0\end{array}\right)$

$\alpha_2=\left(\begin{array}{ccccccccc}\mathrm{a}_0 & \mathrm{~b} & \mathrm{a}_3 & 0 & 0 & & 0 & 0 & 0 \\ \mathrm{f} & \mathrm{d}_0 & \mathrm{~d} & \mathrm{~d}_1 & 0 & & 0 & 0 & 0 \\ \mathrm{c}_1 & \mathrm{c} & \mathrm{a}_0 & \mathrm{~b} & \mathrm{a}_3 & \cdots & 0 & 0 & 0 \\ 0 & \mathrm{e}_3 & \mathrm{f} & \mathrm{d}_0 & \mathrm{~d} & & 0 & 0 & 0 \\ 0 & 0 & \mathrm{c}_1 & \mathrm{c} & \mathrm{a}_0 & & 0 & 0 & 0 \\ & & \vdots & & & \ddots & & \vdots & \\ 0 & 0 & 0 & 0 & 0 & & \mathrm{a}_0 & \mathrm{~b} & \mathrm{a}_3 \\ 0 & 0 & 0 & 0 & 0 & \cdots & \mathrm{f} & \mathrm{d}_0 & \mathrm{~d} \\ 0 & 0 & 0 & 0 & 0 & & \mathrm{c}_1 & \mathrm{c} & \mathrm{a}_0\end{array}\right)$

$\beta_1=\left(\begin{array}{ccccccccc}\mathrm{e} & \mathrm{d}_3 & \mathrm{~d}_2 & 0 & 0 & & 0 & 0 & 0 \\ \mathrm{~b}_3 & \mathrm{~b}_2 & \mathrm{~b}_1 & 0 & 0 & & 0 & 0 & 0 \\ \mathrm{e}_2 & \mathrm{e}_1 & \mathrm{e} & \mathrm{d}_3 & \mathrm{~d}_2 & \cdots & 0 & 0 & 0 \\ 0 & 0 & \mathrm{~b}_3 & \mathrm{~b}_2 & \mathrm{~b}_1 & & 0 & 0 & 0 \\ 0 & 0 & \mathrm{e}_2 & \mathrm{e}_1 & \mathrm{e} & & 0 & 0 & 0 \\ & & \vdots & & & \ddots & & \vdots & \\ 0 & 0 & 0 & 0 & 0 & & \mathrm{e} & \mathrm{d}_3 & d_2 \\ 0 & 0 & 0 & 0 & 0 & \cdots & \mathrm{~b}_3 & \mathrm{~b}_2 & \mathrm{~b}_1 \\ 0 & 0 & 0 & 0 & 0 & & \mathrm{e}_2 & \mathrm{e}_1 & \mathrm{e}\end{array}\right)$

$\beta_2=\left(\begin{array}{ccccccccc}\mathrm{b}_2 & \mathrm{~b}_1 & 0 & 0 & 0 & & 0 & 0 & 0 \\ \mathrm{e}_1 & \mathrm{e} & \mathrm{d}_3 & \mathrm{~d}_2 & 0 & & 0 & 0 & 0 \\ 0 & \mathrm{~b}_3 & \mathrm{~b}_2 & \mathrm{~b}_1 & 0 & \cdots & 0 & 0 & 0 \\ 0 & \mathrm{e}_2 & \mathrm{e}_1 & \mathrm{e} & \mathrm{d}_3 & & 0 & 0 & 0 \\ 0 & 0 & 0 & \mathrm{~b}_3 & \mathrm{~b}_2 & & 0 & 0 & 0 \\ & & \vdots & & & \ddots & & \vdots & \\ 0 & 0 & 0 & 0 & 0 & & \mathrm{~b}_2 & \mathrm{~b}_1 & 0 \\ 0 & 0 & 0 & 0 & 0 & \cdots & \mathrm{e}_1 & \mathrm{e} & \mathrm{d}_3 \\ 0 & 0 & 0 & 0 & 0 & & 0 & \mathrm{~b}_3 & \mathrm{~b}_2\end{array}\right)$

$\gamma_1=\left(\begin{array}{ccccccccc}\mathrm{a} & \mathrm{a}_1 & \mathrm{a}_2 & 0 & 0 & & 0 & 0 & 0 \\ \mathrm{f}_1 & \mathrm{f}_2 & \mathrm{f}_3 & 0 & 0 & & 0 & 0 & 0 \\ \mathrm{c}_2 & \mathrm{c}_3 & \mathrm{a} & \mathrm{a}_1 & \mathrm{a}_2 & \cdots & 0 & 0 & 0 \\ 0 & 0 & \mathrm{f}_1 & \mathrm{f}_2 & \mathrm{f}_3 & & 0 & 0 & 0 \\ 0 & 0 & \mathrm{c}_2 & \mathrm{c}_3 & \mathrm{a} & & 0 & 0 & 0 \\ & & \vdots & & & \ddots & & \vdots & \\ 0 & 0 & 0 & 0 & 0 & & \mathrm{a} & \mathrm{a}_1 & a_3 \\ 0 & 0 & 0 & 0 & 0 & \cdots & \mathrm{f}_1 & \mathrm{f}_2 & \mathrm{f}_3 \\ 0 & 0 & 0 & 0 & 0 & & \mathrm{c}_2 & \mathrm{c}_3 & \mathrm{a}\end{array}\right)$

$\gamma_2=\left(\begin{array}{ccccccccc}\mathrm{f}_2 & \mathrm{f}_3 & 0 & 0 & 0 & & 0 & 0 & 0 \\ \mathrm{c}_3 & \mathrm{a} & \mathrm{a}_1 & \mathrm{a}_2 & 0 & & 0 & 0 & 0 \\ 0 & \mathrm{f}_1 & \mathrm{f}_2 & \mathrm{f}_3 & 0 & \cdots & 0 & 0 & 0 \\ 0 & \mathrm{c}_2 & \mathrm{c}_3 & \mathrm{a} & \mathrm{a}_1 & & 0 & 0 & 0 \\ 0 & 0 & 0 & \mathrm{f}_1 & \mathrm{f}_2 & & 0 & 0 & 0 \\ & & \vdots & & & \ddots & & \vdots & \\ 0 & 0 & 0 & 0 & 0 & & \mathrm{f}_2 & \mathrm{f}_3 & 0 \\ 0 & 0 & 0 & 0 & 0 & \cdots & \mathrm{c}_3 & \mathrm{a} & \mathrm{a}_1 \\ 0 & 0 & 0 & 0 & 0 & & 0 & \mathrm{f}_1 & \mathrm{f}_2\end{array}\right)$

Example 1. Consider a 3 × 3 finite linear TLMCA under null boundary with 3 states {0,1,2}. Find the next configuration of

$C^{(s)}=\left[\begin{array}{lll}1 & 2 & 1 \\ 1 & 1 & 2 \\ 2 & 1 & 2\end{array}\right]$

where,

Solution 1. Given m and n are odd. Therefore, the rule matrix for the CA is

$\mathcal{J}_{\mathcal{R}}^{\mathcal{O} \mathcal{O}}=\left[\begin{array}{lllllllll}0 & 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 \\ 1 & 0 & 1 & 1 & 1 & 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 1 & 1 & 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 & 1 & 1 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 & 0 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1 & 1 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 0\end{array}\right]$

We have

$C^{\prime(s+1)}=\mathcal{T}_{\mathcal{R}}^{m n} \cdot C^{\prime(s)}$                       (12)

$\bar{C}^{(s)}=\left[\begin{array}{lllllllll}1 & 2 & 1 & 1 & 1 & 2 & 2 & 1 & 2\end{array}\right]^T$

The superscript T denotes the transpose of the matrix.

$\mathcal{J}_{\mathcal{R}}^{O O} \times C^{(s)}=\left[\begin{array}{lllllllll}1 & 0 & 1 & 1 & 2 & 2 & 1 & 1 & 0\end{array}\right]^T$

The configuration at the next time step is

$C^{(s+1)}=\left[\begin{array}{lll}1 & 0 & 1 \\ 1 & 2 & 2 \\ 1 & 1 & 0\end{array}\right]$

Theorem 2. Let $T_M$ be a finite linear TLMCA under null boundary condition over the field $Z_3$. Then the rule matrix of order $m \times m$ which updates the $s^{\text {th }}$ finite linear TLMCA configuration $C^{(s)}$ to $(s+1)^{\text {th }}$ configuration $C^{(s+1)}$ is given by

$\mathcal{T}_{\mathcal{R}}^{\mathcal{O} \mathcal{E}}=\left(\begin{array}{ccccccc}\lambda_1 & \mu_1 & \mathcal{O} & & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ \omega_1 & \lambda_2 & \mu_2 & \cdots & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ \mathcal{O} & \omega_2 & \lambda_1 & & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ & \vdots & & \ddots & & \vdots & \\ \mathcal{O} & \mathcal{O} & \mathcal{O} & & \lambda_1 & \mu_1 & \mathcal{O} \\ \mathcal{O} & \mathcal{O} & \mathcal{O} & \cdots & \omega_1 & \lambda_2 & \mu_2 \\ \mathcal{O} & \mathcal{O} & \mathcal{O} & & \mathcal{O} & \omega_2 & \lambda_1\end{array}\right)$

where, $\lambda_1, \lambda_2, \mu_1, \mu_2, \omega_1, \omega_2$ are $n \times n$ submatrices and $\mathcal{O}$ is an $n \times n$ null matrix. Here m is an odd positive integer and is an even positive integer.

Proof. Let $T_M=(L, Q, N, R)$ be the finite linear TLMCA. $L$ is an $m \times n$ two-dimensional triangular lattice where $m$ and $n$ are odd and even positive integers. $Q$ is the finite set of states. Let us consider the cellular automaton has $k$ states. $N$ is the set neighborhood vectors. $R$ is the set of local transition rules. $R$ is defined as $R: C^{(s)} \rightarrow C^{(s+1)}$ where $R=\left\{r\left(v_{i, j}\right)\right\}$. Let $v_{i, j}=0$ when $i \notin\{1,2, \cdots m\}$ or $j \notin \{1,2, \cdots n\}$ because of employing null boundary conditions. By applying the local transition rule from Eqs. (1) and (2) to each cell in the automata, $m \cdot n$ linear equations are obtained.

$\begin{aligned} r\left(v_{1,1}\right)=\left(d_0 v_{1,1}\right. & +d v_{1,2}+e v_{2,1}+f v_{1,0}+d_1 v_{1,3}+d_2 v_{2,3}+d_3 v_{2,2}+e_1 v_{2,0} \\ & +e_2 v_{2,-1}+e_3 v_{1,-1} \left.+f_1 v_{0,0}+f_2 v_{0,1}+f_3 v_{0,2}\right) \bmod 3\end{aligned}$                      (13)

$\begin{aligned} r\left(v_{1,2}\right)=\left(a_0 v_{1,2}\right. & +a v_{0,2}+b v_{1,3}+c v_{1,1}+a_1 v_{0,3}+a_2 v_{0,4}+a_3 v_{1,4}+b_1 v_{2,3} \\ & +b_2 v_{2,2}+b_3 v_{2,1}+c_1 v_{1,0}+c_2 v_{0,0}\left.+c_3 v_{0,1}\right) \bmod 3\end{aligned}$                     (14)

$\begin{aligned} r\left(v_{1, n}\right)=\left(a_0 v_{1, n}\right. & +a v_{0, n}+b v_{1, n+1}+c v_{1, n-1}+a_1 v_{0, n+1}+a_0 v_{1, n}+a v_{0, n}+b v_{1, n+1} \\ & +c v_{1, n-1}+a_1 v_{0, n+1}+b_3 v_{2, n-1}+c_1 v_{1, n-2}+c_2 v_{0, n-2}\left.+c_3 v_{0, n-1}\right) \bmod 3\end{aligned}$                (15)

$\begin{aligned} r\left(v_{2,1}\right)=\left(a_0 v_{2,1}\right. & +a v_{1,1}+b v_{2,2}+c v_{2,0}+a_1 v_{1,2}+a_2 v_{1,3}+a_3 v_{2,3}+b_1 v_{3,2} \\ & +b_2 v_{3,1}+b_3 v_{3,0}+c_1 v_{2,-1}\left.+c_2 v_{1,-1}+c_3 v_{1,0}\right) \bmod 3\end{aligned}$                     (16)

$\begin{aligned} r\left(v_{2,2}\right)=\left(d_0 v_{2,2}\right. & +d v_{2,3}+e v_{3,2}+f v_{2,1}+d_1 v_{2,4}+d_2 v_{3,4}+d_3 v_{3,3} \\ & +e_1 v_{3,1}+e_2 v_{3,0}+e_3 v_{2,0}+f_1 v_{1,1}\left.+f_2 v_{1,2}+f_3 v_{1,3}\right) \bmod 3\end{aligned}$                    (17)

$\begin{aligned} r\left(v_{2, n}\right)=\left(d_0 v_{2, n}\right. & +d v_{2, n+1}+e v_{3, n}+f v_{2, n-1}+d_1 v_{2, n+2}+d_2 v_{3, n+2}+d_3 v_{3, n+1} \\ & +e_1 v_{3, n-1}+e_2 v_{3, n-2}+e_3 v_{2, n-2}+f_1 v_{1, n-1}+f_2 v_{1, n}\left.+f_3 v_{1, n+1}\right) \bmod 3\end{aligned}$                  (18)

$\begin{aligned} r\left(v_{m, 1}\right)=\left(d_0 v_{m, 1}\right. & +d v_{m, 2}+e v_{m+1,1}+f v_{m, 0}+d_1 v_{m, 3}+d_2 v_{m+1,3}+d_3 v_{m+1,2}+e_1 v_{m+1,0} \\ & +e_2 v_{m+1,-1}+e_3 v_{m,-1}+f_1 v_{m-1,0}+f_2 v_{m-1,1}\left.+f_3 v_{m-1,2}\right) \bmod 3\end{aligned}$                 (19)

$\begin{aligned} r\left(v_{m, 2}\right)=\left(a_0 v_{m, 2}\right. & +a v_{m-1,2}+b v_{m, 3}+c v_{m, 1}+a_1 v_{m-1,3}+a_2 v_{m-1,4}+a_3 v_{m, 4}+b_1 v_{m+1,3} \\ & +b_2 v_{m+1,2}+b_3 v_{m+1,1}+c_1 v_{m, 0}+c_2 v_{m-1,0}\left.+c_3 v_{m-1,1}\right) \bmod 3\end{aligned}$               (20)

$\begin{aligned} r\left(v_{m, n}\right)=\left(a_0 v_{m, n}\right. & +a v_{m-1, n}+b v_{m, n+1}+c v_{m, n-1}+a_1 v_{m-1, n+1}+a_2 v_{m-1, n+2}+a_3 v_{m, n+2} \\ &+b_1 v_{m+1, n+1} +b_2 v_{m+1, n}+b_3 v_{m+1, n-1}+c_1 v_{m, n-2}+c_2 v_{m-1, n-2}\left.+c_3 v_{m-1, n-1}\right) \bmod 3\end{aligned}$                   (21)

These equations can be represented in matrix form as:

$\mathcal{T}_{\mathcal{R}}^{\mathcal{O} \mathcal{E}}=\left(\begin{array}{ccccccc}\lambda_1 & \mu_1 & \mathcal{O} & & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ \omega_1 & \lambda_2 & \mu_2 & \cdots & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ \mathcal{O} & \omega_2 & \lambda_1 & & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ & \vdots & & \ddots & & \vdots & \\ \mathcal{O} & \mathcal{O} & \mathcal{O} & & \lambda_1 & \mu_1 & \mathcal{O} \\ \mathcal{O} & \mathcal{O} & \mathcal{O} & \cdots & \omega_1 & \lambda_2 & \mu_2 \\ \mathcal{O} & \mathcal{O} & \mathcal{O} & & \mathcal{O} & \omega_2 & \lambda_1\end{array}\right)$

where,

$\lambda_1=\left(\begin{array}{ccccccccc}\mathrm{d}_0 & \mathrm{~d} & \mathrm{~d}_1 & 0 & 0 & & 0 & 0 & 0 \\ c & \mathrm{a}_0 & \mathrm{~b} & \mathrm{a}_3 & 0 & & 0 & 0 & 0 \\ \mathrm{e}_3 & \mathrm{f} & \mathrm{d}_0 & \mathrm{~d} & \mathrm{~d}_1 & \cdots & 0 & 0 & 0 \\ 0 & \mathrm{c}_1 & \mathrm{c} & \mathrm{a}_0 & \mathrm{~b} & & 0 & 0 & 0 \\ 0 & 0 & \mathrm{e}_3 & \mathrm{f} & \mathrm{d}_0 & & 0 & 0 & 0 \\ & & \vdots & & & \ddots & & \vdots & \\ 0 & 0 & 0 & 0 & 0 & & \mathrm{a}_0 & \mathrm{~b} & a_3 \\ 0 & 0 & 0 & 0 & 0 & \cdots & \mathrm{f} & \mathrm{d}_0 & \mathrm{~d} \\ 0 & 0 & 0 & 0 & 0 & & \mathrm{c}_1 & \mathrm{c} & \mathrm{a}_0\end{array}\right)$

$\lambda_2=\left(\begin{array}{ccccccccc}\mathrm{a}_0 & \mathrm{~b} & \mathrm{a}_3 & 0 & 0 & & 0 & 0 & 0 \\ \mathrm{f} & \mathrm{d}_0 & \mathrm{~d} & \mathrm{~d}_1 & 0 & & 0 & 0 & 0 \\ \mathrm{c}_1 & \mathrm{c} & \mathrm{a}_0 & \mathrm{~b} & \mathrm{a}_3 & \cdots & 0 & 0 & 0 \\ 0 & \mathrm{e}_3 & \mathrm{f} & \mathrm{d}_0 & \mathrm{~d} & & 0 & 0 & 0 \\ 0 & 0 & \mathrm{c}_1 & \mathrm{c} & \mathrm{a}_0 & & 0 & 0 & 0 \\ & & \vdots & & & \ddots & & \vdots & \\ 0 & 0 & 0 & 0 & 0 & & \mathrm{~d}_0 & \mathrm{~d} & d_1 \\ 0 & 0 & 0 & 0 & 0 & \cdots & \mathrm{c} & \mathrm{a}_0 & \mathrm{~b} \\ 0 & 0 & 0 & 0 & 0 & & \mathrm{e}_3 & \mathrm{f} & \mathrm{d}_0\end{array}\right)$

$\mu_1=\left(\begin{array}{ccccccccc}\mathrm{e} & \mathrm{d}_3 & \mathrm{~d}_2 & 0 & 0 & & 0 & 0 & 0 \\ \mathrm{~b}_3 & \mathrm{~b}_2 & \mathrm{~b}_1 & 0 & 0 & & 0 & 0 & 0 \\ \mathrm{e}_3 & \mathrm{f} & 0 & \mathrm{~d} & \mathrm{~d}_1 & \cdots & 0 & 0 & 0 \\ 0 & \mathrm{c}_1 & \mathrm{c} & 0 & \mathrm{~d} & & 0 & 0 & 0 \\ 0 & 0 & \mathrm{e}_3 & \mathrm{f} & 0 & & 0 & 0 & 0 \\ & & \vdots & & & \ddots & & \vdots & \\ 0 & 0 & 0 & 0 & 0 & & \mathrm{~b}_2 & \mathrm{~b}_1 & 0 \\ 0 & 0 & 0 & 0 & 0 & \cdots & \mathrm{e}_1 & \mathrm{e} & \mathrm{d}_3 \\ 0 & 0 & 0 & 0 & 0 & & 0 & \mathrm{~b}_3 & \mathrm{~b}_2\end{array}\right)$

$\mu_2=\left(\begin{array}{ccccccccc}\mathrm{b}_2 & \mathrm{~b}_1 & 0 & 0 & 0 & & 0 & 0 & 0 \\ \mathrm{e}_1 & \mathrm{e} & \mathrm{d}_3 & \mathrm{~d}_2 & 0 & & 0 & 0 & 0 \\ 0 & \mathrm{~b}_3 & \mathrm{~b}_2 & \mathrm{~b}_1 & 0 & \cdots & 0 & 0 & 0 \\ 0 & \mathrm{e}_2 & \mathrm{e}_1 & \mathrm{e} & \mathrm{d}_3 & & 0 & 0 & 0 \\ 0 & 0 & 0 & \mathrm{~b}_3 & \mathrm{~b}_2 & & 0 & 0 & 0 \\ & & \vdots & & & \ddots & & \vdots & \\ 0 & 0 & 0 & 0 & 0 & & \mathrm{e} & \mathrm{d}_3 & d_2 \\ 0 & 0 & 0 & 0 & 0 & \cdots & \mathrm{~b}_3 & \mathrm{~b}_2 & \mathrm{~b}_1 \\ 0 & 0 & 0 & 0 & 0 & & \mathrm{e}_2 & \mathrm{e}_1 & \mathrm{e}\end{array}\right)$

$\omega_1=\left(\begin{array}{ccccccccc}\mathrm{a} & \mathrm{a}_1 & \mathrm{a}_2 & 0 & 0 & & 0 & 0 & 0 \\ \mathrm{f}_1 & \mathrm{f}_2 & \mathrm{f}_3 & 0 & 0 & & 0 & 0 & 0 \\ \mathrm{c}_2 & \mathrm{c}_3 & \mathrm{a} & \mathrm{a}_1 & \mathrm{a}_2 & \cdots & 0 & 0 & 0 \\ 0 & 0 & \mathrm{f}_1 & \mathrm{f}_2 & \mathrm{f}_3 & & 0 & 0 & 0 \\ 0 & 0 & \mathrm{c}_2 & \mathrm{c}_3 & \mathrm{a} & & 0 & 0 & 0 \\ & & \vdots & & & \ddots & & \vdots & \\ 0 & 0 & 0 & 0 & 0 & & \mathrm{f}_2 & \mathrm{f}_3 & 0 \\ 0 & 0 & 0 & 0 & 0 & \cdots & \mathrm{c}_3 & \mathrm{a} & \mathrm{a}_1 \\ 0 & 0 & 0 & 0 & 0 & & 0 & \mathrm{f}_1 & \mathrm{f}_2\end{array}\right)$

$\omega_2=\left(\begin{array}{ccccccccc}\mathrm{f}_2 & \mathrm{f}_3 & 0 & 0 & 0 & & 0 & 0 & 0 \\ \mathrm{c}_3 & \mathrm{a} & \mathrm{a}_1 & \mathrm{a}_2 & 0 & & 0 & 0 & 0 \\ 0 & \mathrm{f}_1 & \mathrm{f}_2 & \mathrm{f}_3 & 0 & \cdots & 0 & 0 & 0 \\ 0 & \mathrm{c}_2 & \mathrm{c}_3 & \mathrm{a} & \mathrm{a}_1 & & 0 & 0 & 0 \\ 0 & 0 & 0 & \mathrm{f}_1 & \mathrm{f}_2 & & 0 & 0 & 0 \\ & & \vdots & & & \ddots & & \vdots & \\ 0 & 0 & 0 & 0 & 0 & & \mathrm{a} & \mathrm{a}_1 & a_2 \\ 0 & 0 & 0 & 0 & 0 & \cdots & \mathrm{f}_1 & \mathrm{f}_2 & \mathrm{f}_3 \\ 0 & 0 & 0 & 0 & 0 & & \mathrm{c}_2 & \mathrm{c}_3 & \mathrm{a}\end{array}\right)$

Example 2. Consider a 3 × 4 finite linear TLMCA under null boundary with 3 states {0,1,2}. Find the next configuration of

$C^{(s)}=\left[\begin{array}{llll}1 & 2 & 0 & 1 \\ 1 & 1 & 2 & 1 \\ 2 & 1 & 1 & 0\end{array}\right]$

where,

Solution. Here m is an odd positive integer and n is an even positive integer.

$\mathcal{T}_{\mathcal{R}}^{\mathcal{O} \mathcal{E}}=\left[\begin{array}{llllllllllll}0 & 1 & 2 & 0 & 1 & 2 & 2 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 2 & 2 & 2 & 2 & 0 & 0 & 0 & 0 & 0 \\ 2 & 1 & 0 & 1 & 2 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 2 & 1 & 0 & 0 & 2 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 2 & 2 & 0 & 0 & 1 & 2 & 0 & 2 & 2 & 0 & 0 \\ 2 & 2 & 2 & 0 & 1 & 0 & 1 & 2 & 2 & 1 & 2 & 2 \\ 2 & 2 & 1 & 2 & 2 & 1 & 0 & 1 & 0 & 2 & 2 & 2 \\ 0 & 0 & 2 & 2 & 0 & 2 & 1 & 0 & 0 & 2 & 2 & 1 \\ 0 & 0 & 0 & 0 & 2 & 2 & 0 & 0 & 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 0 & 2 & 1 & 2 & 2 & 1 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 & 0 & 2 & 2 & 2 & 2 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 2 & 2 & 1 & 0 & 2 & 1 & 0\end{array}\right]$

$\bar{C}^{(s)}=\left[\begin{array}{llllllllllll}1 & 2 & 0 & 1 & 1 & 1 & 2 & 1 & 2 & 1 & 1 & 0\end{array}\right]^T$

The superscript T denotes the transpose of the matrix.

$\mathcal{J}_{\mathcal{R}}^{\mathcal{O E}} \times \bar{C}^{(s)} = [9\,\,11\,\,9\,\,8\,\, 16\,\,18\,\,16\,\,10\,\,7\,\,12\,\,13\,\,10]^T$

$\bar{C}^{(s+1)}=\left[\begin{array}{llllllllllll}0 & 2 & 0 & 2 & 1 & 0 & 1 & 1 & 1 & 0 & 1 & 1\end{array}\right]^T$

The configuration at next time step is

$C^{(s+1)}=\left[\begin{array}{llll}0 & 2 & 0 & 2 \\ 1 & 0 & 1 & 1 \\ 1 & 0 & 1 & 1\end{array}\right]$

Theorem 3. Let $T_M$ be a finite linear TLMCA under null boundary condition over the field $Z_3$. Then the rule matrix of order $m \times m$ which updates the $s^{\text {th }}$ finite linear TLMCA configuration $C^{(s)}$ to $(s+1)^{\text {th }}$ configuration $C^{(s+1)}$ is given by

$\mathcal{T}_{\mathcal{R}}^{\mathcal{E} \mathcal{O}}=\left(\begin{array}{ccccccc}\alpha_1 & \beta_1 & \mathcal{O} & & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ \gamma_1 & \alpha_2 & \beta_2 & \cdots & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ \mathcal{O} & \gamma_2 & \alpha_1 & & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ & \vdots & & \ddots & & \vdots & \\ \mathcal{O} & \mathcal{O} & \mathcal{O} & & \alpha_2 & \beta_2 & \mathcal{O} \\ \mathcal{O} & \mathcal{O} & \mathcal{O} & \cdots & \gamma_2 & \alpha_1 & \beta_1 \\ \mathcal{O} & \mathcal{O} & \mathcal{O} & & \mathcal{O} & \gamma_1 & \alpha_2\end{array}\right)$

where, $\alpha_1, \alpha_2, \beta_1, \beta_2, \gamma_1, \gamma_2$ are $n \times n$ submatrices and $\mathcal{O}$ is an $n \times n$ null matrix. Here m is an even positive integer and n is an odd positive integer.

Proof. Let $T_M=(L, Q, N, R)$ be the finite linear TLMCA. $L$ is an $m \times n$ two-dimensional triangular lattice where $m$ and $n$ are even and odd positive integers. $Q$ is the finite set of states. Let us consider the cellular automaton has 3 states $\{0,1,2\} . N$ is the set neighborhood vectors. $R$ is the set of local transition rules. $R$ is defined as $R: C^{(s)} \rightarrow C^{(s+1)}$ where $R=\left\{r\left(v_{i, j}\right)\right\}$. Let $v_{i, j}=0$ when $i \notin\{1,2, \cdots m\}$ or $j \notin \{1,2, \cdots n\}$ because of employing null boundary conditions. By applying the local transition rule from Eqs. (1) and (2) to each cell in the automata, $m \cdot n$ linear equations are obtained.

$\begin{aligned} r\left(v_{1,1}\right)=\left(d_0 v_{1,1}\right. & +d v_{1,2}+e v_{2,1}+f v_{1,0}+d_1 v_{1,3}+d_2 v_{2,3}+d_3 v_{2,2}+e_1 v_{2,0} \\ & +e_2 v_{2,-1}+e_3 v_{1,-1}\left.+f_1 v_{0,0}+f_2 v_{0,1}+f_3 v_{0,2}\right) \bmod 3\end{aligned}$                     (22)

$\begin{aligned} r\left(v_{1,2}\right)=\left(a_0 v_{1,2}\right. & +a v_{0,2}+b v_{1,3}+c v_{1,1}+a_1 v_{0,3}+a_2 v_{0,4}+a_3 v_{1,4} \\ & +b_1 v_{2,3}+b_2 v_{2,2}+b_3 v_{2,1}+c_1 v_{1,0}+c_2 v_{0,0}\left.+c_3 v_{0,1}\right) \bmod 3\end{aligned}$                  (23)

$\begin{aligned} r\left(v_{1, n}\right)=\left(d_0 v_{1, n}\right. & +d v_{1, n+1}+e v_{2, n}+f v_{1, n-1}+d_1 v_{1, n+2}+d_2 v_{2, n+2}+d_3 v_{2, n+1}+e_1 v_{2, n-1} \\ & +e_2 v_{2, n-2}+e_3 v_{1, n-2}+f_1 v_{0, n-1}+f_2 v_{0, n}\left.+f_3 v_{0, n+1} \right) \bmod 3\end{aligned}$                 (24)

$\begin{aligned} r\left(v_{2,1}\right)=\left(a_0 v_{2,1}\right. & +a v_{1,1}+b v_{2,2}+c v_{2,0}+a_1 v_{1,2}+a_2 v_{1,3}+a_3 v_{2,3}+b_1 v_{3,2} \\ & +b_2 v_{3,1}+b_3 v_{3,0}+c_1 v_{2,-1}\left.+c_2 v_{1,-1}+c_3 v_{1,0}\right) \bmod 3\end{aligned}$                        (25)

$\begin{gathered}r\left(v_{2,2}\right)=\left(d_0 v_{2,2}+d v_{2,3}+e v_{3,2}+f v_{2,1}+d_1 v_{2,4}\right. +d_2 v_{3,4}+d_3 v_{3,3} \\ +e_1 v_{3,1}+e_2 v_{3,0}+e_3 v_{2,0}+f_1 v_{1,1}\left.+f_2 v_{1,2}+f_3 v_{1,3}\right) \bmod 3\end{gathered}$               (26)

$\begin{gathered}r\left(v_{2, n}\right)=\left(a_0 v_{2, n}+a v_{1, n}+b v_{2, n+1}+c v_{2, n-1}\right.+a_1 v_{1, n+1}+a_2 v_{1, n+2}+a_3 v_{2, n+2} \\ +b_1 v_{3, n+1}+b_2 v_{3, n}+b_3 v_{3, n-1}+c_1 v_{2, n-2}+c_2 v_{1, n-2}+\left.c_3 v_{1, n-1}\right) \bmod 3\end{gathered}$                  (27)

$\begin{aligned} r\left(v_{m, 1}\right)=\left(a_0 v_{m, 1}\right. & +a v_{m-1,1}+b v_{m, 2}+c v_{m, 0}+a_1 v_{m-1,2}+a_2 v_{m-1,3}+a_3 v_{m, 3}+b_1 v_{m+1,2} \\ & +b_2 v_{m+1,1}+b_3 v_{m+1,0}+c_1 v_{m,-1}+c_2 v_{m-1,-1}\left.+c_3 v_{m-1,0}\right) \bmod 3\end{aligned}$                 (28)

$\begin{gathered}r\left(v_{m, 2}\right)=\left(d_0 v_{m, 2}+d v_{m, 3}+e v_{m+1,2}+f v_{m, 1}\right.+d_1 v_{m, 4}+d_2 v_{m+1,4}+d_3 v_{m+1,3} \\ +e_1 v_{m+1,1}+e_2 v_{m+1,0}+e_3 v_{m, 0}+f_1 v_{m-1,1}+f_2 v_{m-1,2}\left.+f_3 v_{m-1,3}\right) \bmod 3\end{gathered}$               (29)

$\begin{gathered}r\left(v_{m, n}\right)=\left(a_0 v_{m, n}+a v_{m-1, n}+b v_{m, n+1}+c v_{m, n-1}\right.+a_1 v_{m-1, n+1}+a_2 v_{m-1, n+2}+a_3 v_{m, n+2} \\ +b_1 v_{m+1, n+1}+b_2 v_{m+1, n}+b_3 v_{m+1, n-1}+c_1 v_{m, n-2}+c_2 v_{m-1, n-2}\left.+c_3 v_{m-1, n-1}\right) \bmod 3\end{gathered}$                   (30)

These equations can be represented in matrix form as:

$\mathcal{J}_{\mathcal{R}}^{\mathcal{E} \mathcal{O}}=\left(\begin{array}{ccccccc}\alpha_1 & \beta_1 & \mathcal{O} & & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ \gamma_1 & \alpha_2 & \beta_2 & \cdots & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ \mathcal{O} & \gamma_2 & \alpha_1 & & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ & \vdots & & \ddots & & \vdots & \\ \mathcal{O} & \mathcal{O} & \mathcal{O} & & \alpha_2 & \beta_2 & \mathcal{O} \\ \mathcal{O} & \mathcal{O} & \mathcal{O} & \cdots & \gamma_2 & \alpha_1 & \beta_1 \\ \mathcal{O} & \mathcal{O} & \mathcal{O} & & \mathcal{O} & \gamma_1 & \alpha_2\end{array}\right)$

where, $\alpha_1, \alpha_2, \beta_1, \beta_2, \gamma_1, \gamma_2$ are $n \times n$ submatrices which can be referred from Theorem 1 and $\mathcal{O}$ is an $n \times n$ null matrix.

Example 3. Consider a $4 \times 3$ finite linear TLMCA under null boundary with 4 states $\{0,1,2,3\}$. Find the next configuration of

$C^{(s)}=\left[\begin{array}{lll}0 & 1 & 2 \\ 3 & 0 & 1 \\ 2 & 3 & 0 \\ 1 & 2 & 3\end{array}\right]$

with

Solution. Here m is an even positive integer and n is an even positive integer.

$\mathcal{T}_{\mathcal{R}}^{\mathcal{E} \mathcal{O}}=\left[\begin{array}{llllllllllll}0 & 2 & 1 & 2 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & 0 & 2 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 2 & 0 & 1 & 1 & 2 & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & 1 & 1 & 0 & 2 & 1 & 1 & 1 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 2 & 0 & 2 & 1 & 2 & 1 & 0 & 0 & 0 \\ 1 & 1 & 2 & 1 & 2 & 0 & 0 & 1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 & 0 & 0 & 2 & 1 & 2 & 1 & 1 \\ 0 & 0 & 0 & 1 & 2 & 1 & 2 & 0 & 2 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 & 1 & 2 & 0 & 1 & 1 & 2 \\ 0 & 0 & 0 & 0 & 0 & 0 & 2 & 1 & 1 & 0 & 2 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 2 & 0 & 2 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 2 & 1 & 2 & 0\end{array}\right]$

$\bar{C}^{(s)}=\left[\begin{array}{llllllllllll}0 & 1 & 2 & 3 & 0 & 1 & 2 & 3 & 0 & 1 & 2 & 3\end{array}\right]^T$

$\mathcal{J}_{\mathcal{R}}^{\mathcal{E} \mathcal{O}} \times \bar{C}^{(s)} = \left[\begin{array}{llllllllllll}11 & 8 & 7 & 9 & 19 & 11 & 16 & 14 & 18 & 14 & 13 & 10\end{array}\right]^T$

$\bar{C}^{(s+1)}=\left[\begin{array}{llllllllllll}2 & 2 & 1 & 0 & 1 & 2 & 1 & 2 & 0 & 2 & 1 & 1\end{array}\right]^T$

The configuration at next time step is

$C^{(s+1)}=\left[\begin{array}{lll}2 & 2 & 1 \\ 0 & 1 & 2 \\ 1 & 2 & 0 \\ 2 & 1 & 1\end{array}\right]$

Theorem 4. Let $T_M$ be a finite linear TLMCA under null boundary condition over the field $Z_3$. Then the rule matrix of order $m \times m$ which updates the $s^{\text {th }}$ finite linear TLMCA configuration $C^{(s)}$ to $(s+1)^{\text {th }}$ configuration $C^{(s+1)}$ is given by

$\mathcal{T}_{\mathcal{R}}^{\mathcal{E} \mathcal{E}}=\left(\begin{array}{ccccccc}\lambda_1 & \mu_1 & \mathcal{O} & & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ \omega_1 & \lambda_2 & \mu_2 & \cdots & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ \mathcal{O} & \omega_2 & \lambda_1 & & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ & \vdots & & \ddots & & \vdots & \\ \mathcal{O} & \mathcal{O} & \mathcal{O} & & \lambda_2 & \mu_2 & \mathcal{O} \\ \mathcal{O} & \mathcal{O} & \mathcal{O} & \cdots & \omega_2 & \lambda_1 & \mu_1 \\ \mathcal{O} & \mathcal{O} & \mathcal{O} & & \mathcal{O} & \omega_1 & \lambda_2\end{array}\right)$

where, $\lambda_1, \lambda_2, \mu_1, \mu_2, \omega_1, \omega_2$ are $\mathrm{n} \times \mathrm{n}$ submatrices and $\mathcal{O}$ is an $\mathrm{n} \times \mathrm{n}$ null matrix. Here m and n are even positive integers.

Proof. Let $T_M=(L, Q, N, R)$ be the finite linear TLMCA. $L$ is an $m \times n$ two-dimensional triangular lattice where $m$ and $n$ are even positive integers. $Q$ is the finite set of states. Let us consider the cellular automaton has $k$ states. $N$ is the set neighborhood vectors. $R$ is the set of local transition rules. $R$ is defined as $R: C^{(s)} \rightarrow C^{(s+1)}$ where $R=\left\{r\left(v_{i, j}\right)\right\}$. Let $v_{i, j}=0$ when $i \notin\{1,2, \cdots m\}$ or $j \notin\{1,2, \cdots n\}$ because of employing null boundary conditions. By applying the local transition rule from Eqs. (1) and (2) to each cell in the automata, $m \cdot n$ linear equations are obtained.

$\begin{aligned} & r\left(v_{1,1}\right)=\left(\begin{array}{llll}d_0 &v_{1,1} & +d v_{1,2}\end{array}+e v_{2,1}\right.+f v_{1,0}+d_1 \quad v_{1,3}+d_2 \quad v_{2,3}+d_3 \quad v_{2,2} \\ &+e_1 \quad v_{2,0}+e_2 \quad v_{2,-1}+e_3 \quad v_{1,-1}+f_1 \quad v_{0,0}+f_2 \quad v_{0,1}+f_3 \quad v_{0,2})\bmod 3\end{aligned}$                  (31)

$\begin{aligned} r\left(v_{1,2}\right)=\left(a_0 v_{1,2}\right. & +a v_{0,2}+b v_{1,3}+c v_{1,1}+a_1 v_{0,3}+a_2 v_{0,4}+a_3 v_{1,4} \\ & +b_1 v_{2,3}+b_2 v_{2,2}+b_3 v_{2,1}+c_1 v_{1,0}+c_2 v_{0,0}\left.+c_3 v_{0,1}\right) \bmod 3\end{aligned}$                           (32)

$\begin{aligned} r\left(v_{1, n}\right)=\left(a_0 v_{1, n}\right. & +a v_{0, n}+b v_{1, n+1}+c v_{1, n-1} +a_1 v_{0, n+1}a_0 v_{1, n}+a v_{0, n}+b v_{1, n+1} \\ & +c v_{1, n-1}+a_1 v_{0, n+1}+b_3 v_{2, n-1}+c_1 v_{1, n-2}+c_2 v_{0, n-2}\left.+c_3 v_{0, n-1}\right) \bmod 3\end{aligned}$                       (33)

$\begin{aligned} r\left(v_{2,1}\right)=\left(a_0 v_{2,1}\right. & +a v_{1,1}+b v_{2,2}+c v_{2,0}+a_1 v_{1,2}+a_2 v_{1,3}+a_3 v_{2,3}+b_1 v_{3,2} \\ & +b_2 v_{3,1}+b_3 v_{3,0}+c_1 v_{2,-1}\left.+c_2 v_{1,-1}+c_3 v_{1,0}\right) \bmod 3\end{aligned}$                    (34)

$\begin{aligned} r\left(v_{2,2}\right)=\left(d_0 v_{2,2}\right. & +d v_{2,3}+e v_{3,2}+f v_{2,1}+d_1 v_{2,4}+d_2 v_{3,4}+d_3 v_{3,3} \\ & +e_1 v_{3,1}+e_2 v_{3,0}+e_3 v_{2,0}+f_1 v_{1,1}+f_2 v_{1,2}\left.+f_3 v_{1,3}\right) \bmod 3\end{aligned}$                       (35)

$\begin{aligned} r\left(v_{2, n}\right)=\left(d_0 v_{2, n}\right. & +d v_{2, n+1}+e v_{3, n}+f v_{2, n-1}+d_1 v_{2, n+2}+d_2 v_{3, n+2}+d_3 v_{3, n+1} \\ & +e_1 v_{3, n-1}+e_2 v_{3, n-2}+e_3 v_{2, n-2}+f_1 v_{1, n-1}+f_2 v_{1, n}\left.+f_3 v_{1, n+1}\right) \bmod 3\end{aligned}$                             (36)

$\begin{aligned} r\left(v_{m, 1}\right)=\left(a_0 v_{m, 1}\right. & +a v_{m-1,1}+b v_{m, 2}+c v_{m, 0}+a_1 v_{m-1,2}+a_2 v_{m-1,3}+a_3 v_{m, 3}+b_1 v_{m+1,2} \\ & +b_2 v_{m+1,1}+b_3 v_{m+1,0}+c_1 v_{m,-1}+c_2 v_{m-1,-1}\left.+c_3 v_{m-1,0}\right) \bmod 3\end{aligned}$                   (37)

$\begin{aligned} r\left(v_{m, 2}\right)=\left(d_0 v_{m, 2}\right. & +d v_{m, 3}+e v_{m+1,2}+f v_{m, 1}+d_1 v_{m, 4}+d_2 v_{m+1,4}+d_3 v_{m+1,3}+e_1 v_{m+1,1} \\ & +e_2 v_{m+1,0}+e_3 v_{m, 0}+f_1 v_{m-1,1}+f_2 v_{m-1,2}\left.+f_3 v_{m-1,3}\right) \bmod 3\end{aligned}$                 (38)

$\begin{aligned} r\left(v_{m, n}\right)=\left(d_0 v_{m, n}\right. & +d v_{m, n+1}+e v_{m+1, n}+f v_{m, n-1}+d_1 v_{m, n+2}+d_2 v_{m+1, n+2}+d_3 v_{m+1, n+1}+e_1 v_{m+1, n-1} \\ & +e_2 v_{m+1, n-2}+e_3 v_{m, n-2}+f_1 v_{m-1, n-1}+f_2 v_{m-1, n} \left.+f_3 v_{m-1, n+1}\right) \bmod 3\end{aligned}$                        (39)

These equations can be represented in matrix form as:

$\mathcal{J}_{\mathcal{R}}^{\mathcal{E} \mathcal{E}}=\left(\begin{array}{ccccccc}\lambda_1 & \mu_1 & \mathcal{O} & & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ \omega_1 & \lambda_2 & \mu_2 & \cdots & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ \mathcal{O} & \omega_2 & \lambda_1 & & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ & \vdots & & \ddots & & \vdots & \\ \mathcal{O} & \mathcal{O} & \mathcal{O} & & \lambda_2 & \mu_2 & \mathcal{O} \\ \mathcal{O} & \mathcal{O} & \mathcal{O} & \cdots & \omega_2 & \lambda_1 & \mu_1 \\ \mathcal{O} & \mathcal{O} & \mathcal{O} & & \mathcal{O} & \omega_1 & \lambda_2\end{array}\right)$

where, $\lambda_1, \lambda_2, \mu_1, \mu_2, \omega_1, \omega_2$ are $n \times n$ submatrices which can be referred from Theorem 2 and $\mathcal{O}$ is an $\mathrm{n} \times \mathrm{n}$ null matrix.

Example 4 . Consider a $4 \times 4$ finite linear TLMCA under null boundary with 5 states $\{0,1,2,3,4\}$. Find the next configuration of

$C^{(s)}=\left[\begin{array}{llll}0 & 1 & 2 & 3 \\ 4 & 0 & 1 & 2 \\ 3 & 4 & 0 & 1 \\ 2 & 3 & 4 & 0\end{array}\right]$

where,

Solution. Given m and n are even. Therefore, the rule matrix for the cellular automaton is

$\mathcal{T}_{\mathcal{R}}^{\mathcal{O} \varepsilon}=\left[\begin{array}{llllllllllllllll}0 & 0 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 2 & 2 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & 0 & 0 & 0 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 & 0 & 3 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 & 0 & 1 & 0 & 2 & 2 & 0 & 0 & 0 & 0 & 0 & 0 \\ 3 & 3 & 3 & 0 & 0 & 0 & 0 & 1 & 2 & 0 & 1 & 1 & 0 & 0 & 0 & 0 \\ 3 & 3 & 0 & 1 & 3 & 0 & 0 & 0 & 0 & 2 & 2 & 2 & 0 & 0 & 0 & 0 \\ 0 & 0 & 3 & 3 & 0 & 2 & 0 & 0 & 0 & 2 & 2 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 3 & 3 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 3 & 0 & 1 & 1 & 0 & 0 & 0 & 1 & 2 & 2 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 3 & 3 & 3 & 2 & 0 & 0 & 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 3 & 3 & 0 & 0 & 3 & 0 & 0 & 0 & 3 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 3 & 3 & 3 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 3 & 3 & 0 & 1 & 3 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 3 & 3 & 0 & 2 & 0 & 0\end{array}\right]$

$\bar{C}^{(s)}=\left[\begin{array}{llllllllllllllll}0 & 1 & 2 & 3 & 4 & 0 & 1 & 2 & 3 & 4 & 0 & 1 & 2 & 3 & 4 & 0\end{array}\right]^T$

The superscript T denotes the transpose of the matrix.

$\mathcal{J}_{\mathcal{R}}^{\varepsilon \varepsilon} \times \bar{C}^{(s)}=\left[\begin{array}{llllllllllllllll}3 & 11 & 8 & 3 & 18 & 18 & 28 & 23 & 19 & 34 & 19 & 24 & 8 & 21 & 28 & 9\end{array}\right]^T$

$\bar{C}^{(s+1)}=\left[\begin{array}{llllllllllllllll}0 & 2 & 2 & 0 & 0 & 0 & 1 & 2 & 1 & 1 & 1 & 0 & 2 & 0 & 1 & 0\end{array}\right]^T$

The configuration at next time step is

$C^{(s+1)}=\left[\begin{array}{llll}0 & 2 & 2 & 0 \\ 0 & 0 & 1 & 2 \\ 1 & 1 & 1 & 0 \\ 2 & 0 & 1 & 0\end{array}\right]$

Transition rules are employed in computing the next configuration of the CA. Using a matrix allows for efficient application of transition rules across all cells simultaneously. The patterns evolved by applying the transition rule individually to each cell and the configurations obtained by applying the rule matrix method produce the same result for any finite $m \times n$ cellular automaton under any linear rule of the finite linear TLMCA.

In particular, the Triangular Lattice Moore neighborhood CA has to deal with 13 neighborhood including itself. For instance, the configuration of a $5 \times 5$ TLMCA is taken here. The next configuration of current configuration is computed using both transition rule and transition rule matrices. Figure 13 depicts the lattice representing of states of the TLMCA. Black colour represents state 2, grey colour represents state 1 and white colour represents state 0.

13.png

Figure 13. Configuration at time ‘s’

5.1 Transition rule

Form the Figure 13, we have $\begin{aligned}v_{1,1}=v_{1,5}=v_{2,2}= v_{3,3}=v_{4,1}=v_{5,4}=2, v_{1,2}=v_{1,4}=v_{2,1}=v_{2,3}=v_{2,4}=v_{3,2}=v_{3,4}=v_{3,5}=v_{4,3}=v_{4,4}=v_{4,5}=1\text { and } v_{1,3}=v_{2,5}=v_{3,1}=v_{4,2}=v_{5,5}=0 .

\end{aligned}$

The values $\begin{aligned}a_0=d_0=0, a=b=c=a_1=a_2=a_3=b_1=b_2=b_3=c_1=c_2=c_3=d=e=f=d_1=d_2=d_3=e_1=e_2=e_3=f_1=f_2=f_3=1

\end{aligned}$

By applying the transition rule to each cell individually, we get

$r\left(v_{1,1}\right)=r\left(v_{1,3}\right)=r\left(v_{1,4}\right)=r\left(v_{2,3}\right)=r\left(v_{2,4}\right)$

$=r\left(\nu_{3,2}\right)=r\left(\nu_{3,4}\right)=r\left(\nu_{4,5}\right)=r\left(\nu_{5,2}\right)$

$=r\left(v_{5,4}\right)=r\left(v_{5,5}\right)=2, r\left(v_{1,2}\right)=r\left(v_{2,1}\right)$

$=r\left(v_{2,2}\right)=r\left(v_{2,5}\right)=r\left(v_{3,5}\right)=r\left(v_{4,3}\right)=r\left(v_{4,4}\right)$

$=r\left(v_{5,1}\right)=1$ and $r\left(v_{1,5}\right)=r\left(v_{3,1}\right)=r\left(v_{3,3}\right)$

$=r\left(v_{4,1}\right)=r\left(v_{4,2}\right)=r\left(v_{5,3}\right)=0$.

Therefore, the resulting next step configuration of the considered triangular lattice is depicted in Figure 14.

14.png

Figure 14. Configuration at time ‘s+1’

5.2 Rule matrix

The order of the considered CA is 5 × 5. Here m and n are odd positive integers. By proceeding with the methodology involved in theorem 1, the rule matrix is as follows of order 25 × 25.

$\mathcal{T}_{\mathcal{R}}^{55}=\left[\begin{array}{ccccc}\alpha_1 & \beta_1 & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ \gamma_1 & \alpha_2 & \beta_2 & \mathcal{O} & \mathcal{O} \\ \mathcal{O} & \gamma_2 & \alpha_1 & \beta_1 & \mathcal{O} \\ \mathcal{O} & \mathcal{O} & \gamma_1 & \alpha_2 & \beta_2 \\ \mathcal{O} & \mathcal{O} & \mathcal{O} & \gamma_2 & \alpha_1\end{array}\right]$

where,

$\begin{array}{ll}\alpha_1=\left[\begin{array}{lllll}0 & 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 1 & 0 \\ 1 & 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 & 0\end{array}\right] & \alpha_2=\left[\begin{array}{lllll}0 & 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 1 & 0 \\ 1 & 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 & 0\end{array}\right] \\ \beta_1=\left[\begin{array}{lllll}1 & 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1\end{array}\right] & \beta_2=\left[\begin{array}{lllll}1 & 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1\end{array}\right]\end{array}$

$\gamma_1=\left[\begin{array}{lllll}1 & 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1\end{array}\right] \quad \gamma_2=\left[\begin{array}{lllll}1 & 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1\end{array}\right]$

$\left[T_R^{m n}\right] \cdot C^{\prime(s)}=C^{\prime(s+1)}$

We obtain a matrix which is the next step configuration of the given CA by applying rule matrix concept and is denoted by $C^{(s+1)}$

$C^{(s+1)}=\left[\begin{array}{lllll}2 & 1 & 2 & 2 & 0 \\ 1 & 1 & 2 & 2 & 1 \\ 0 & 2 & 0 & 2 & 1 \\ 0 & 0 & 1 & 1 & 2 \\ 1 & 2 & 0 & 2 & 2\end{array}\right]$

The configuration matrix $C^{(s+1)}$ exactly resembles the configuration obtained by applying the transition rule to all cells individually.