Total and Double Total Domination on Octagonal Grid

Abstract

A k-total dominating set is a set of vertices such that all vertices in the graph, including the vertices in the dominating set themselves, have at least k neighbors in the dominating set. The k-total domination number ${\gamma }_{kt}\left(G\right)$ is the cardinality of the smallest k-total dominating set. For $k=1,2$, the k-total dominating number is called the total and the double total dominating number, respectively. In this paper, we determine the exact values for the total domination number on a linear and on a double octagonal chain and an upper bound for the total domination number on a triple octagonal chain. Furthermore, we determine the exact values for the double total domination number on a linear and on a double octagonal chain and an upper bound for the double total domination number on a triple octagonal chain and on an octagonal grid $O_{m,n},m\ge 3,n\ge 3$. As each vertex in the octagonal system is either of degree two or of degree three, there is no k-total domination for $k\ge 3$.

1. Introduction

Graph domination can model problems such as resource allocation, vehicle routing, dominating queens, etc. Furthermore, graph dominations have significant applications in chemistry because chemical structures are often represented as graphs. For example, domination theory is utilized to encrypt binary strings into DNA sequences. Methods like the basic method and the insertion method employ graph domination to encrypt any chemical formula. Each chemical formula is converted into a binary string through graph domination and subsequently encrypted using DNA steganography [1].

The total and double total domination numbers have been widely studied on many chemical graphs such as the hexagonal grid, nanotubes, pyrene network, pyrene torus network, and hexabenzocoronene [2,3,4,5,6,7].

In this paper, we study total and double total domination on chains of octagons and octagonal grids. These systems are geometric objects obtained by arranging mutually congruent regular octagons in a plane. The octagonal chain and octagonal grid are special cases of octagonal graphs. An octagonal graph is a finite two-connected geometric graph where each interior face is bounded by a regular octagon or quadrangle with a side length of 1. Octagonal graphs have garnered significant interest from mathematicians due to their intriguing combinatorial properties. In [8], the number of isomers of tree-like octagonal graphs is determined by the generating functions. Further, in [9], it is demonstrated that every convex octagon has a side with a relative length of at most 1, a bound which is asymptotically tight. Combinatorial properties of fixed-boundary octagonal random tilings are explored in [10]. In [11], the connection between the number of perfect matchings in octagonal chain graphs and the Hosoya index of caterpillar trees is identified. More recently, in [12], the authors examined the Wiener indices in certain types of random octagonal chains, and in [13], the authors studied perfect matchings in random octagonal chain graphs.

We determine the exact values for the total domination number on a linear and on a double octagonal chain and an upper bound for the total domination number on a triple octagonal chain. Furthermore, we determine the exact values for the double total domination number on a linear and on a double octagonal chain and an upper bound for the double total domination number on a triple octagonal chain and on an octagonal grid $O_{m,n},m\ge 3,n\ge 3$.

Apart from this Introduction, the rest of this paper is organized as follows. Section 2 presents the necessary preliminaries regarding total and double total domination. This section also serves to provide the foundational concepts and definitions that are essential for understanding graph theory. Section 3 gives the exact values for the total domination number ${\gamma }_t$ on a linear and on a double octagonal chain and an upper bound for the total domination number on a triple octagonal chain. Section 4 deals with double total domination and gives the double total dominating number ${\gamma }_{2t}$ for a linear and a double octagonal chain and an upper bound for the double total domination number on a triple octagonal chain and for an arbitrary octagonal grid $O_{m,n},m,n\ge 3$.

2. Preliminaries

Total domination in graphs, initially introduced in [14], has garnered significant attention and has evolved into a thoroughly explored area within graph theory. This concept has become foundational in the study of various graph theoretical properties and applications. Some more recent work about total domination can be found in [15,16,17,18,19,20].

Let G be a graph with the vertex set $V\left(G\right)$ and edge set $E\left(G\right)$. The degree of a vertex $v\in V\left(G\right)$, denoted by $deg\left(v\right)$, is the number of edges that are incident to the vertex v. The maximum degree of a graph G, denoted by $\Delta \left(G\right)$, is the highest degree among the vertices of G. Similarly, the minimum degree of a graph G, denoted by $\delta \left(G\right)$, is the lowest degree among the vertices of G.

The open neighborhood of v is $N\left(v\right)=\{u\in V(G\left)\right|uv\in E\left(G\right)\}$, and the closed neighborhood of v is $N\left[v\right]=v\cup N\left(v\right)$. For a set $S\subseteq V\left(G\right)$, its open neighborhood is the set $N\left(S\right)={\cup }_{v\in S}N\left(v\right)$, and its closed neighborhood is the set $N\left[S\right]=N\left(S\right)\cup S$. A set $S\subseteq G\left(V\right)$ is an open packing of G if the open neighborhoods of the vertices of S are pairwise disjoint in G. The open packing number of G, denoted by ${\rho }^o\left(G\right)$, is the maximum cardinality among all open packings of G.

A set $D\subset V\left(G\right)$ is a dominating set of a graph G if every vertex y in $V\left(G\right)\setminus D$ is adjacent to some vertex in D. The domination number $\gamma \left(G\right)$ is the cardinality of the smallest dominating set. Open packings are the natural dual object of total dominating sets [21]. Moreover, for every graph G, it holds that ${\rho }^o\left(G\right)\le \gamma \left(G\right)$, and if $S\subseteq V\left(G\right)$ is both a dominating set and an open packing, then ${\rho }^o\left(G\right)=\gamma \left(G\right)=\left|S\right|$ [22].

Total domination is the stronger version of domination. A set $D\subset V\left(G\right)$ is a total dominating set of a graph G with no isolated vertex if every vertex y in $V\left(G\right)$ is adjacent to some vertex in D. The total domination number ${\gamma }_t\left(G\right)$ is the cardinality of the smallest total dominating set. A set $D\subseteq V\left(G\right)$ is a k-total dominating set of a graph G with no isolated vertex if each vertex in $V\left(G\right)$ has at least k neighbors in D. The k-total domination number ${\gamma }_{kt}\left(G\right)$ is the cardinality of the smallest k-total dominating set. For $k=2$, the 2-total dominating set is called the double total dominating set.

Each vertex in the octagonal system is either of degree two or of degree three. It follows that on the octagonal system, there is no k-total domination for $k\ge 3$.

The domination polynomial of G is the polynomial $D(G,x)={\sum }_{i=1}^{\left|V\right(G\left)\right|}d(G,i)x^i$, where $d(G,i)$ is the number of dominating sets of size i. A root of $D(G,x)$ is called a domination root of G. We denote the set of distinct domination roots by $Z\left(D\right(G,x\left)\right)$. The characterization of graphs using domination polynomials is studied in [23].

For more foundational concepts and definitions that are crucial for understanding graph theory, see [24].

3. Total Domination Number of Octagonal Grid ${\mathit{O}}_{\mathit{m},\mathit{n}}$, $\mathbf{1}\le \mathit{n}\le \mathbf{3}$

We define an octagonal grid $O_{m,n}$ as a grid with m octagons in a row and n octagons in a column. The octagonal row is also called an octagonal chain because it is constructed by consecutively connected octagons. There are n octagonal chains on $O_{m,n}$. When $n=1$, it forms a linear octagonal chain; for $n=2$, it becomes a double octagonal chain; and for $n=3$, it is a triple octagonal chain. To label vertices in $O_{m,n}$, we use $3n+1$ subrows and $3m+1$ subcolumns. The vertex set $V\left(O_{m,n}\right)$ is defined as

$$\begin{array}{cc}\hfill V\left(O_{m,n}\right)=& \{v_{i,j}:i\equiv 1mod3,j\equiv 2,0mod3\}\cup \{v_{i,j}:i\equiv 2,0mod3,j\equiv 1mod3\},\hfill \end{array}$$

where $1\le i\le 3n+1$, $1\le j\le 3m+1$. See Figure 1.

Figure 1. $O_{3,2}$ with 7 subrows and 10 subcolumns. Labeled vertices are presented in gray color, while others are presented in black.

Specifically, the border vertices on $O_{m,n}$ are vertices which are located on the top and on the bottom, on the left and on the right border of the graph.

Theorem 1.

For a linear octagonal chain with m octagons, it holds that

$${\gamma }_t\left(O_{m,1}\right)=2m+2.$$

Proof of Theorem 1.

We define $D=\{v_{2,1+3j},j=0,\cdots ,m\}\cup \{v_{3,1+3j},j=0,\cdots ,m\}$, where $\left|D\right|=2m+2.$

D is the total dominating set on $O_{m,1}$, with each vertex being total dominated by exactly one vertex. See Figure 2. Any total dominating set where each vertex in the graph is total dominated by exactly one vertex is minimal. If you remove just one arbitrary vertex from the total dominating set, at least one vertex on the graph will not be total dominated. It follows that ${\gamma }_t\left(O_{m,1}\right)=2m+2.$ □

Figure 2. Black vertices are in the total dominating set on $O_{5,1}$, while other vertices are omitted for simplicity, i.e., there are no black circles for omitted vertices.

Theorem 2.

For a double octagonal chain with m octagons, it holds that

$${\gamma }_t\left(O_{m,2}\right)=4m+4.$$

Proof of Theorem 2.

We define $D=\{v_{2,1+3j},j=0,\cdots ,m\}$∪$\{v_{3,1+3j},j=0,\cdots ,m\}$∪ $\{v_{6,1+3j},j=0,\cdots ,m\}$∪$\{v_{7,2+3j},j=0,\cdots ,m-1\}$∪$\left\{v_{7,3m}\right\}$, where

$$\left|D\right|=3(m+1)+m+1=4m+4.$$

It is easy to see that all vertices are total dominated by set D, so ${\gamma }_t\left(O_{m,2}\right)\le 4m+4.$ See Figure 3. Vertices $\{v_{2,1+3j},j=0,\cdots ,m\}$∪$\{v_{3,1+3j},j=0,\cdots ,m\}$ total dominate all vertices on the first four subrows on $O_{m,2}$. This structure is optimal as follows. Firstly, each vertex is total dominated by exactly one vertex. Secondly, because of the grid structure, vertices from some subrow can only total dominate vertices from the subrow below or subrow above.

Figure 3. Black vertices are in the total dominating set on $O_{5,2}$, while other vertices are omitted for simplicity.

Further, vertices from the fifth, sixth and seventh subrows must be total dominated. Vertices from the fifth row must not be in the total dominating set because vertices from the fourth row would then be total dominated by two vertices. To total dominate vertices from the fifth subrow, we need at least all vertices from the sixth subrow. This ensures that every vertex in the first five subrows is also total dominated by exactly one vertex.

Finally, to total dominate vertices from the sixth subrow, we need at least $m+1$ vertices from the seventh subrow which are adjacent to the vertices from the sixth subrow. In such a case, all vertices from the sixth subrow are total dominated by exactly one vertex. Furthermore, all vertices from the seventh subrow are total dominated by exactly one vertex except for vertices $v_{7,3m-1}$ and $v_{7,3m}$, which are total dominated by two vertices. However, both vertices $v_{7,3m-1}$ and $v_{7,3m}$ must be in the total dominating set to total dominate $v_{6,3m-2}$ and $v_{6,3m+1}$, respectively. See Figure 3. It follows that

$${\gamma }_t\left(O_{m,2}\right)\ge 2m+2+2(m+1)=4m+4.$$

□

Theorem 3.

From Ref. [25], for $n\ge 3$, it holds that

$${\gamma }_t\left(P_n\right)={\gamma }_t\left(C_n\right)=\left\{\begin{array}{cc}{\displaystyle \frac{n}{2}}+1\hfill & \mathit{if}\phantom{a}n\equiv 2mod4,\hfill \\ \lceil {\displaystyle \frac{n}{2}}\rceil \hfill & \mathit{otherwise}.\hfill \end{array}\right.$$

The statement proved in Theorem 3 will be used in our following Theorem 4.

Theorem 4.

For a triple octagonal chain with m octagons, it holds that

$${\gamma }_t\left(O_{m,3}\right)\le \left\{\begin{array}{cc}4m+4+{\displaystyle \frac{3m+1}{2}}+1\hfill & \mathit{if}\phantom{a}m\equiv 3mod4,\hfill \\ 4m+4+⌈{\displaystyle \frac{3m+1}{2}}⌉\hfill & \mathit{otherwise}.\hfill \end{array}\right.$$

Proof of Theorem 4.

We define $D_1=\{v_{2,1+3j},j=0,\cdots ,m\}$∪$\{v_{3,1+3j},j=0,\cdots ,m\}$∪ $\{v_{6,1+3j},j=0,\cdots ,m\}$∪$\{v_{7,2+3j},j=0,\cdots ,m-1\}$∪$\left\{v_{7,3m}\right\}$. According to Theorem 2, this is a minimal total dominating set on $O_{m,2}$ and $|D_1|=4m+4$.

It is easy to see that if we have $D_1$ on $O_{m,3}$, all vertices from the first eight subrows on $O_{m,3}$ are total dominated. This is optimal because a maximum of eight subrows can be total dominated by vertices from two octagonal rows. In more detail, the last subrow on the second octagonal row is the seventh subrow. Therefore, only vertices from the third octagonal row which can be total dominated from vertices belonging to the second octagonal row are vertices on the eighth subrow.

Further, we notice that vertices on the 9th and 10th subrows make a path $P_{3m+1}$. See Figure 4.

Figure 4. $D_1$ and $P_{3m+1}$ on $O_{5,3}$. Black vertices are in $D_1$, while other vertices are omitted for simplicity. Vertices making a path $P_{3m+1}$ are circled with a dashed line.

From Theorem 3, it follows that

$${\gamma }_t\left(P_{3m+1}\right)=\left\{\begin{array}{cc}{\displaystyle \frac{3m+1}{2}}+1\hfill & \mathit{if}\phantom{a}3m+1\equiv 2mod4,\hfill \\ ⌈{\displaystyle \frac{3m+1}{2}}⌉\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

Furthermore, if $3m+1\equiv 2mod4$, then $m\equiv 3mod4$. Then, $D_1$ together with the minimal total dominating set on $P_{3m+1}$ is the total dominating set on $O_{m,3}$, and

$${\gamma }_t\left(O_{m,3}\right)\le \left|D_1\right|+{\gamma }_t\left(P_{3m+1}\right)=\left\{\begin{array}{cc}4m+4+{\displaystyle \frac{3m+1}{2}}+1\hfill & \mathrm{if}\phantom{a}m\equiv 3mod4,\hfill \\ 4m+4+⌈{\displaystyle \frac{3m+1}{2}}⌉\hfill & otherwise.\hfill \end{array}\right.$$

□

4. Double Total Domination on Octagonal Grid

Theorem 5.

For a linear octagonal chain with m octagons, it holds that

$${\gamma }_{2t}\left(O_{m,1}\right)=6m+2.$$

Proof of Theorem 5.

In the linear octagonal chain, every vertex is adjacent to at least one vertex of degree two. To achieve double total domination, all vertices on the graph must be in the double total domination set. There are $6m+2$ vertices on the linear octagonal chain; hence,

$${\gamma }_{2t}\left(O_{m,1}\right)=6m+2.$$

□

Lemma 1.

There are exactly $6m+6n-4$ border vertices on $O_{m,n}$, and each of them must be in any double total dominating set.

Proof of Lemma 1.

First, we see that there are exactly $6m+6n-4$ vertices on the border of $O_{m,n}$. There are $3m-1$ on the top and on the bottom borders, respectively. Furthermore, there are $3n-1$ on each side border. Hence, there are $2(3m-1)+2(3n-1)=6m+6n-4$ border vertices. Further, all of them are adjacent to at least one vertex of degree two. So, they all must be in any double total dominating set. □

Lemma 2.

$${\gamma }_{2t}\left(O_{2,2}\right)=22.$$

Proof of Lemma 2.

From Lemma 1, it follows that the border on $O_{2,2}$ has 20 vertices, and each of them must be in any double total dominating set. The remaining four internal vertices which form $C_4$ are only total dominated once. To double total dominate $C_4$, we need at least two more vertices. Therefore, ${\gamma }_{2t}\left(O_{2,2}\right)=22.$ □

Theorem 6.

For a double octagonal chain with $m\ge 3$ octagons, it holds that

$${\gamma }_{2t}\left(O_{m,2}\right)=9m+3.$$

Proof of Theorem 6.

From Lemma 1, it follows that the border on $O_{m,2}$ has $6m+8$ vertices, and each of them must be in any double total dominating set. We note the border vertices with B. Next, we define $D=B$∪$\{v_{3,4+3j},j=0,\cdots ,m-2\}$∪$\{v_{4,5+3j},j=0,\cdots ,m-3\}$∪$\{v_{4,6+3j},j=0,\cdots ,m-3\}$, $m\ge 3$. See Figure 5.

Figure 5. Black vertices are in the double total dominating set on $O_{5,2}$, while other vertices are omitted for simplicity. Vertices which are double total dominated by 3 vertices are highlighted with dashed rectangles.

D is the double total domination set on $O_{m,2}$, and $\left|D\right|=6m+8+(m-1)+2(m-2)=9m+3.$ Therefore,

$${\gamma }_{2t}\left(O_{m,2}\right)\le 9m+3,m\ge 3.$$

Now we will prove that D is the minimal double total dominating set. First, according to Lemma 1, all border vertices must be in D. Second, most of internal vertices on $O_{m,2}$ are double total dominated by exactly two vertices. Only vertices $\{v_{3,7+3j},j=0,\cdots ,m-4\}$ are double total dominated by three vertices. See Figure 5. They are highlighted with dashed rectangles. If we were to remove vertices $\{v_{3,7+3j},j=0,\cdots ,m-4\}$ from D, then vertices $\{v_{4,6+3j},j=0,\cdots ,m-4\}$∪$\{v_{4,8+3j},j=0,\cdots ,m-4\}$ would be just total, not double total dominated. Furthermore, vertices $\{v_{3,7+3j},j=0,\cdots ,m-4\}$ are the best choice to double total dominate vertices which are only total dominated. This is because each vertex total dominates two different vertices, which is the best possible in this scenario. Hence, D is minimal. □

Lemma 3.

$${\gamma }_{2t}\left(O_{1,3}\right)={\gamma }_{2t}\left(O_{3,1}\right)\mathit{and}{\gamma }_{2t}\left(O_{2,3}\right)={\gamma }_{2t}\left(O_{3,2}\right)$$

Proof of Lemma 3.

If we rotate $O_{3,1}$ and $O_{3,2}$, we obtain $O_{1,3}$ and $O_{2,3}$, respectively. Hence, the equalities hold. □

Theorem 7.

For a triple octagonal chain with $m\ge 3$ octagons, it holds that

$${\gamma }_{2t}\left(O_{m,3}\right)\le \left\{\begin{array}{cc}6m+16{\displaystyle \frac{m}{3}}+6\hfill & \mathit{if}\phantom{a}m\equiv 0mod3,\hfill \\ 6m+16⌊{\displaystyle \frac{m}{3}}⌋+12\hfill & \mathit{if}\phantom{a}m\equiv 1mod3,\hfill \\ 6m+16⌊{\displaystyle \frac{m}{3}}⌋+18\hfill & \mathit{if}\phantom{a}m\equiv 0mod3.\hfill \end{array}\right.$$

Proof of Theorem 7.

Let $O_{m,3}$ be an octagonal chain with $m\ge 3$ and B its border vertices. Because of Lemma 1, $\left|B\right|=6m+14$.

• $m\equiv 0mod3$.
  Let D = B∪$\{v_{4,5+9j},v_{4,6+9j},v_{5,4+9j},v_{5,7+9j},v_{6,4+9j},v_{6,7+9j},v_{7,5+9j},v_{7,6+9j},$$j=0,\cdots ,{\displaystyle \frac{m}{3}}-1\}$∪$\{v_{3,10+9j},v_{4,9+9j},v_{4,11+9j},v_{5,10+9j},v_{6,10+9j},v_{7,9+9j},v_{7,11+9j},$ $v_{8,10+9j},j=0,\cdots ,{\displaystyle \frac{m}{3}}-2\}$, $m>3$. See Figure 6.
  

  Figure 6. Black vertices are in the double total dominating set on $O_{6,3}$, while other vertices are omitted for simplicity.
  D is the double total dominating set, and

  $$\left|D\right|=6m+14+8{\displaystyle \frac{m}{3}}+8\left({\displaystyle \frac{m}{3}}-1\right)=6m+16{\displaystyle \frac{m}{3}}+6.$$
  For $m=3$, we construct the double total dominating set with all border vertices and vertices $\{v_{4,5},v_{4,6},v_{5,4},v_{5,7},v_{6,4},v_{6,6},v_{7,5},v_{7,6}\}.$ Then

  $${\gamma }_{2t}\left(O_{3,3}\right)\le 32+8=40=6·3++16{\displaystyle \frac{3}{3}}+6=6m+16{\displaystyle \frac{m}{3}}+6.$$
• $m\equiv 1mod3$.
  Let $D=$B∪$\{v_{4,5+9j},v_{4,6+9j},v_{5,4+9j},v_{5,7+9j},v_{6,4+9j},v_{6,7+9j},v_{7,5+9j},v_{7,6+9j},$$j=0,\cdots ,\lfloor {\displaystyle \frac{m}{3}}\rfloor -1\}$∪$\{v_{3,10+9j},v_{4,9+9j},v_{4,11+9j},v_{5,10+9j},$$v_{6,10+9j},v_{7,9+9j},v_{7,11+9j}$ $,v_{8,10+9j},$$j=0,\cdots ,\lfloor {\displaystyle \frac{m}{3}}\rfloor -2\}\cup \{v_{4,3m-4},v_{4,3m-3},$$v_{5,3m-2},v_{6,3m-2},v_{7,3m-4},$ $v_{7,3m-3}\},$$m>4$. See Figure 7.
  

  Figure 7. Black vertices are in the double total dominating set on $O_{7,3}$, while other vertices are omitted for simplicity.
  D is the double total dominating set, and

  $$\left|D\right|=6m+14+8⌊{\displaystyle \frac{m}{3}}⌋+8\left(⌊{\displaystyle \frac{m}{3}}⌋-1\right)+6=6m+16⌊{\displaystyle \frac{m}{3}}⌋+12.$$
  For $m=4$, we construct the double total dominating set with all border vertices and vertices $\{v_{4,5},v_{4,6},v_{5,4},v_{5,7},v_{6,4},v_{6,7},v_{7,5},v_{7,6},v_{4,8},v_{4,9},v_{5,10},v_{6,10},v_{7,8},v_{7,9}\}.$ Then

  $${\gamma }_{2t}\left(O_{4,3}\right)\le 38+14=52=6m+16⌊{\displaystyle \frac{m}{3}}⌋+12.$$
• $m\equiv 2mod3$.
  Let $D=$B$\cup \{v_{4,5+9j},v_{4,6+9j},v_{5,4+9j},v_{5,7+9j},v_{6,4+9j},$$v_{6,7+9j},v_{7,5+9j},v_{7,6+9j},$
  $j=0,\cdots ,\lfloor {\displaystyle \frac{m}{3}}\rfloor -1\}$$\cup \{v_{3,10+9j},v_{4,9+9j},v_{4,11+9j},$$v_{5,10+9j},$$v_{6,10+9j},v_{7,9+9j},$$v_{7,11+9j},$$v_{8,10+9j},$$j=0,\cdots ,\lfloor {\displaystyle \frac{m}{3}}\rfloor -1\}\cup \{v_{4,3m-1},v_{5,3m-2},v_{6,3m-2},v_{7,3m-1}\},$$m>5$. See Figure 8.
  

  Figure 8. Black vertices are in the double total dominating set on $O_{8,3}$, while other vertices are omitted for simplicity.
  D is the double total dominating set, and

  $$\left|D\right|=6m+14+8⌊{\displaystyle \frac{m}{3}}⌋+8⌊{\displaystyle \frac{m}{3}}⌋+4=6m+16⌊{\displaystyle \frac{m}{3}}⌋+18.$$
  For $m=5$, we construct the double total dominating set with all border vertices and vertices $\{v_{4,5},v_{4,6},v_{5,4},v_{5,7},v_{6,4},v_{6,7},v_{7,5},v_{7,6}\}$$\cup \{v_{3,10},v_{4,9},v_{4,11},v_{5,10},$
  $v_{6,10},v_{7,9},v_{7,11},v_{8,10}\}\cup \{v_{4,14},v_{5,13},v_{6,13},v_{7,14}\}.$ Then

  $${\gamma }_{2t}\left(O_{5,3}\right)\le 44+20=64=6m+16⌊{\displaystyle \frac{m}{3}}⌋+18.$$

□

Theorem 8.

$${\gamma }_{2t}\left(O_{m,n}\right)\le 6m+6n-4+{\gamma }_{2t}\left(O_{m-2,n-2}\right),m\ge 4,n\ge 4.$$

Proof of Theorem 8.

From Lemma 1, it follows that the border on $O_{m,n}$ has $6m+6n-4$ vertices, and each of them must be in any double total dominating set. The border vertices double total dominate border vertices and total dominate only a few internal vertices, but most of the internal vertices are not dominated at all. See Figure 9. The border vertices are colored black. White internal vertices are total dominated, and gray internal vertices are not dominated at all. The gray vertices make an octagonal grid $O_{m-2,n-2}$. Again, any double total dominating set on $O_{m-2,n-2}$ will contain its border vertices. Therefore,

$${\gamma }_{2t}\left(O_{m,n}\right)\le 6m+6n-4+{\gamma }_{2t}\left(O_{m-2,n-2}\right).$$

Figure 9. $O_{2,2}$ inside of $O_{4,4}$.

□

5. Conclusions

In this paper, we conducted a study to determine the exact values of the total domination number for linear and double octagonal chains. Additionally, we established an upper bound for the total domination number on a triple octagonal chain. Furthermore, our investigation extended to the exact values of the double total domination number for both linear and double octagonal chains. We also derived an upper bound for the double total domination number in the context of a triple octagonal chain and an octagonal grid, $O_{m,n}$, where $m\ge 3,n\ge 3$.

Our findings suggest that some of these established bounds may indeed be minimal. This opens up intriguing possibilities for further research to verify these minimality claims and potentially identify tighter bounds for various configurations.

In future work, we plan to expand our exploration to other chemical graph structures such as more general octagon graphs and tiling graphs.