Analysing the systemic implications of energy efficiency and circular economy strategies in the decarbonisation context

Panagiotis Fragkos; Panagiotis Fragkos

doi:10.3934/energy.2022011

AIMS Energy

2022, Volume 10, Issue 2: 191-218. doi: 10.3934/energy.2022011

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Research article Special Issues

Analysing the systemic implications of energy efficiency and circular economy strategies in the decarbonisation context

Panagiotis Fragkos ^,

E3-Modelling S.A., Panormou 70-72, Athens, Greece

Received: 22 November 2021 Revised: 30 March 2022 Accepted: 31 March 2022 Published: 07 April 2022

The Paris Agreement goals require a rapid and deep reduction in global greenhouse gas emissions. Recent studies have shown the large potential of circular economy to reduce global emissions by improving resource and material efficiency practices. However, most large-scale energy system and Integrated Assessment Models used for mitigation analysis typically ignore or do not adequately represent circular economy measures. This study aims to fill in this research gap by enhancing a leading global energy system model with a representation of energy efficiency and circular economy considerations. The scenario-based analysis offers an improved understanding of the potentials, costs and impacts of circular economy in the decarbonisation context. The study shows that enhanced energy efficiency and increased material circularity can reduce energy consumption in all sectors, but most importantly in the industrial sector. They can also reduce the required carbon price to achieve Paris goals and the dependence on expensive, immature, and risky technologies, like Carbon Capture and Storage. Circular economy measures should be properly integrated with broad climate policies to provide a holistic and self-consistent framework to deeply reduce carbon emissions.

Keywords:

Citation: Panagiotis Fragkos. Analysing the systemic implications of energy efficiency and circular economy strategies in the decarbonisation context[J]. AIMS Energy, 2022, 10(2): 191-218. doi: 10.3934/energy.2022011

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Abstract

1. Introduction

In graph theory, graph labeling is an assignment of labels or weights to the vertices and edges of a graph. Graph labeling plays an important role in many fields such as computer science, coding theory and physics ^[32]. Baca et al. ^[10] have introduced the definition of an edge irregular total $\ell$ -labeling of any graph as a labeling $\mathcal{L}:V \cup E\to \left\{\mathrm{1, 2}, 3, \dots, \ell\right\}$ in which every two distinct edges $fh$ and ${f}^{*}{h}^{*}$ of a graph $G$ have distinct weights, this means that ${w}_{\mathcal{L}}\left(fh\right)\ne {w}_{\mathcal{L}}\left({f}^{*}{h}^{*}\right)$ where ${w}_{\mathcal{L}}\left(fh\right) = \mathcal{L}\left(f\right)+\mathcal{L}\left(h\right)+\mathcal{L}\left(fh\right)$ . They have deduced inequality which gives a lower bound of $tes\left(G\right)$ for a graph $G$ ,

$tes\left(G\right)\ge max\left\{\left\lceil {\frac{\left|E\left(G\right)\right|+2}{3}} \right\rceil , \left\lceil {\frac{\Delta +1}{2}} \right\rceil \right\}$

(1)

Also, they have introduced the exact value of TEIS, $tes\left(G\right)$ for some families of graphs like fan graph ${F}_{n}$ and wheel graph ${W}_{n}$ ,

$tes\left({F}_{n}\right) = \left\lceil {\frac{3n+2}{3}} \right\rceil \text{, }$

$tes\left({W}_{n}\right) = \left\lceil {\frac{2n+2}{3}} \right\rceil \text{.}$

In ^[15] authors have proved that for any tree $T$

$tes\left(T\right) = max\left\{\left\lceil {\frac{k+1}{3}} \right\rceil , \left\lceil {\frac{\Delta +1}{2}} \right\rceil \right\},$

where $\Delta$ is maximum degree on $k$ vertices. In addition, Salama ^[26] investigated the exact value of TEIS for a polar grid graph,

$tes\left({P}_{m, n}\right) = \left\lceil {\frac{2mn+2}{3}} \right\rceil \text{.}$

Authors in ^[1] determined $\mathrm{T}\mathrm{E}\mathrm{I}\mathrm{S}$ for zigzag graphs. Also, the exact value of TEIS of the generalized web graph ${W}_{n, m}$ and some families has been determined, see ^[14]. Tilukay et al. ^[31] have investigated total irregularity strength for a wheel graph, a fan graph, a triangular Book graph and a friendship graph. On the other hand, in ^{[2,3,8,17,20,24,29]} the total edge irregularity strengths for hexagonal grid graphs, centralized uniform theta graphs, generalized helm graph, series parallel graphs, disjoint union of isomorphic copies of generalized Petersen graph, disjoint union of wheel graphs, subdivision of star ${S}_{n}$ and categorical product of two cycles have been investigated. For more details, see ^{[4,5,6,7,9,11,12,13,16,18,19,21,23,25,27,28,30]}.

A generalized theta graph $\theta ({t}_{1}, {t}_{2}, \dots, {t}_{n})$ is a pair of $n$ internal disjoint paths with lengths at least two joined by end vertices where the end vertices are named south pole $S$ and north pole $N$ and ${t}_{i}$ is the number of vertices in the nth path. Uniform theta graph $\theta (t, m)$ is a generalized theta graph in which all paths have the same numbers of internal vertices, for more details see ^[22].

In this paper, we have defined a new type of family of graph called uniform theta snake graph, ${\theta }_{n}(t, m)$ . Also, the exact value of TEIS for some special types of the new family has been determined.

2. Main results

In the following, we define a new type of graph which is called uniform theta snake graph.

Definition 1. If we replace each edge of a path ${P}_{n}$ by a uniform theta graph $\theta (t, m)$ , we have a uniform theta snake graph ${\theta }_{n}(t, m)$ . See Figure 1.

Figure 1. Uniform theta snake graph θ(t, m).

DownLoad: Full-Size Img PowerPoint

It is clear that for a uniform theta snake graph $\left|\mathrm{E}\left({\mathtt{θ }}_{\mathrm{n}}\right(\mathrm{t}, \mathrm{m}\left)\right)\right| = \mathrm{t}\left(\mathrm{m}+1\right)\mathrm{n}$ and $\left|\mathrm{V}\left({\mathtt{θ }}_{\mathrm{n}}\right(\mathrm{t}, \mathrm{m}\left)\right)\right| = \left(\mathrm{t}\mathrm{m}+1\right)\mathrm{n}+1$ . In this section, we determine the exact value of TEIS for uniform theta snake graph ${\mathtt{θ }}_{\mathrm{n}}\left(\mathrm{3, 3}\right)$ , ${\mathtt{θ }}_{\mathrm{n}}\left(3, \mathrm{m}\right)$ , ${\mathtt{θ }}_{\mathrm{n}}(\mathrm{t}, 3)$ , ${\mathtt{θ }}_{\mathrm{n}}\left(4, \mathrm{m}\right)$ , and ${\mathtt{θ }}_{\mathrm{n}}(\mathrm{t}, 4)$ .

Theorem 1. For a uniform theta snake graph ${\theta }_{n}\left(\mathrm{3, 3}\right)$ with 10 $n+1$ vertices and 12 $n$ edges, we have

$tes\left({\theta }_{n}\left(\mathrm{3, 3}\right)\right) = 4n+1\text{.}$

Proof. Since a uniform theta snake graph ${\theta }_{n}\left(\mathrm{3, 3}\right)$ has 12 $n$ edges and $\left({\theta }_{n}\left(\mathrm{3, 3}\right)\right) = 6$ , then from (1) we have:

$tes\left({\theta }_{n}\left(\mathrm{3, 3}\right)\right)\ge 4n+1\text{.}$

To prove the invers inequality, we show that $ħ-$ labeling is an edge irregular total for ${\theta }_{n}\left(\mathrm{3, 3}\right)$ , see , and $ħ = 4n+1$ . Let $ħ = 4n+1$ and a total $ħ-$ labeling $\alpha :V{(\theta }_{n}\left(\mathrm{3, 3}\right)) \cup E({\theta }_{n}\left(\mathrm{3, 3}\right))\to \left\{\mathrm{1, 2}, 3, \dots, \mathrm{ħ}\right\}$ is defined as:

Figure 2. Uniform theta snake graph θ(3, 3).

DownLoad: Full-Size Img PowerPoint

$\mathtt{α }\left({\mathrm{c}}_{0}\right) = 1,$

$\mathtt{α }\left({\mathrm{c}}_{\mathrm{s}}\right) = 4\mathrm{s}\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\;\mathrm{f}\mathrm{o}\mathrm{r}\;\mathrm{ }1\le \mathrm{s}\le \mathrm{n}-1$

$\mathtt{α }\left({\mathrm{c}}_{\mathrm{n}}\right) = \mathrm{ħ},$

$\alpha \left({x}_{i, j}\right) = \left\{\begin{array}{cc}j& for\;1\le j\le 3\\ j+1& for\;4\le j\le 6\\ .& .\\ .& .\\ j+n-1& for\;3n-2\le j\le 3n-1\end{array}, i = 1, \mathrm{ }2, \mathrm{ }3\right.\text{, }$

$\alpha \left({x}_{i, 3n}\right) = ħ-1 \; for\; i = 1, 2, 3$

$\alpha \left({c}_{0}{x}_{i, 1}\right) = i \; for\; i = i1, 2, 3$

$\alpha \left({c}_{S}{x}_{i, 3S}\right) = 4S+i \; for\; 1\le S\le n-1, i = 1, 2, 3$

$\alpha \left({c}_{S}{x}_{i, 3S+1}\right) = 4S+i+1 \; for\; 1\le S\le n-1, i = 1, 2, 3$

$\alpha \left({c}_{n}{x}_{i, 3n}\right) = \left\{\begin{array}{cc}ħ-2& for\; i = 1\\ ħ-1& for\; i = 2\\ ħ& for\; i = 3\end{array}\right.\text{, }$

$\alpha \left({{x}_{i, j}x}_{i, j+1i}\right) = \left\{\begin{array}{cc}j+i+1& for\; 1\le j\le 2\\ j+i+2& for\; 4\le j\le 5\\ .& .\\ .& .\\ j+i+n-\mathrm{I}1& for\; 3n-5\le j\le 3n-4\\ ħ+i-3& for\; 3n-2\le j\le 3n-1\end{array}, i = 1, 2, 3\right..$

It is clear that $ħ$ is the greatest used label. The weights of edges of ${\theta }_{n}\left(\mathrm{3, 3}\right)$ are given by:

${w}_{\alpha }\left({c}_{0}{x}_{i, 1}\right) = i+2 \; for\; i = 1, 2, 3\text{, }$

${w}_{\alpha }\left({c}_{S}{x}_{i, 3S}\right) = 12S+i-1 \; for\; 1\le S\le n-1 , i = 1, 2, 3$

${w}_{\alpha }\left({c}_{S}{x}_{i, 3S+1}\right) = 12S+i+2\; for\; 1\le S\le n-1, i = 1, 2, 3,$

${w}_{\alpha }\left({c}_{n}{x}_{i, 3n}\right) = \left\{\begin{array}{cc}3\left(ħ-1\right)& for\; i = 1\\ 3ħ-2& for\; i = 2\\ 3ħ-1& for\; i = 3\end{array}\right.\text{, }$

${w}_{\alpha }\left({{x}_{i, j}x}_{i, j+1}\right) = \left\{\begin{array}{cc}3j+i+2& for\; 1\le j\le 2\\ 3j+i+5& for\; 4\le j\le 5\\ .& .\\ .& .\\ 3j+i+3n-4& for\; 3n-5\le j\le 3n-4\\ 3ħ+i-10& for\; j = 3n-2\\ 3ħ+i-7& for\; j = 3n-1\end{array}, i = 1, 2, 3\right.$

Obviously, the weights of edges are distinct. So $\alpha$ is an edge irregular total $\mathrm{ħ}-$ labeling. Hence

$tes\left({\theta }_{n}\left(\mathrm{3, 3}\right)\right) = 4n+1\text{.}$

Theorem 2. For ${\theta }_{n}\left(3, m\right), m>3$ be a uniform theta snake graph. Then

$tes\left({\theta }_{n}(3, m)\right) = (m+1)n+1\text{.}$

Proof. Since $\left|E\left({\theta }_{n}\right(3, m\left)\right)\right| = 3\left(m+1\right)n$ and $\Delta \left({\theta }_{n}\right(3, m\left)\right) = 6$ . Substituting in (1), we find

$tes\left({\theta }_{n}(3, m)\right)\ge (m+1)n+1\text{.}$

The existence of an edge irregular total $ƛ-$ labeling for ${\theta }_{n}(3, m)$ , See , $m>3$ will be shown, with $ƛ = (m+1)n+1$ . Define a total $ƛ-$ labeling $\beta :V\left({\theta }_{n}(3, m)\right)\cup E\left({\theta }_{n}(3, m)\right)\to \left\{\mathrm{1, 2}, 3, \dots, ƛ\right\}$ for ${\theta }_{n}(3, m)$ as:

Figure 3. Uniform theta snake graph θ(3, m).

DownLoad: Full-Size Img PowerPoint

$\beta \left({c}_{0}\right) = 1,$

$\beta \left({c}_{s}\right) = (m+1)s \; for\; 1\le s\le n-1\text{, }$

$\beta \left({c}_{n}\right) = ƛ$

$\beta \left({x}_{i, j}\right) = \left\{\begin{array}{cc}j& for\; 1\le j\le m\\ j+1& for\; m+1\le j\le 2m\\ .& .\\ .& .\\ j+n-1& for\; m\left(n-1\right)+1\le j\le mn-1\end{array}\right.\text{, }$

$\beta \left({x}_{i, mn}\right) = ƛ-1 \; for\; i = 1, 2, 3$

$\beta \left({c}_{0}{x}_{i, 1}\right) = 1 \; for\; i = 1, 2, 3$

$\beta \left({c}_{S}{x}_{i, mS}\right) = (m+1)S+i \; for\; 1\le S\le n-1, i = 1, 2, 3$

$\beta \left({c}_{S}{x}_{i, mS+1}\right) = (m+1)S+i+1 \; for\; 1\le S\le n-1, i = 1, 2, 3$

$\beta \left({c}_{n}{x}_{i, mn}\right) = \left\{\begin{array}{cc}ƛ-2& for\; i = 1\\ ƛ-1& for\; i = 2\\ ƛ& for\; i = 3\end{array}\right.\text{, }$

$\beta \left({{x}_{i, j}x}_{i, j+1}\right) = \left\{\begin{array}{cc}j+i+1& for\; 1\le j\le m-1\\ j+i+2& for\; m+1\le j\le 2m-1\\ .& .\\ .& .\\ j+i+n& for\; m\left(n-1\right)+1\le j\le mn-2\\ j+i+n-1& for\; j = mn-1\end{array}.\right.$

Clearly, $ƛ$ is the most label of edges and vertices. The edges weights are given as follows:

${w}_{\beta }\left({c}_{0}{x}_{i, 1}\right) = i+2\; for\; i = 1, 2, 3\text{, }$

${w}_{\beta }\left({c}_{S}{x}_{i, mS}\right) = 3\left(m+1\right)S+i-1 \; for\; 1\le S\le n-1, i = 1, 2, 3$

${w}_{\beta }\left({c}_{S}{x}_{i, mS+1}\right) = 3\left(m+1\right)S+i+2 \; for\; 1\le S\le n-1, i = 1, 2, 3\text{, }$

${w}_{\beta }\left({c}_{n}{x}_{i, mn}\right) = \left\{\begin{array}{cc}3ƛ-3& for\; i = 1\\ 3ƛ-2& for\; i = 2\\ 3ƛ-1& for\; i = 3\end{array}\right.\text{, }$

${w}_{\beta }\left({{x}_{i, j}x}_{i, j+1}\right) = \left\{\begin{array}{cc}3j+i+2& for\; 1\le j\le m-1\\ 3j+i+5& for\; m+1\le j\le 2m-1\\ .& .\\ .& .\\ 3j\mathrm{I}+i+3n-1& for\; m\left(n-1\right)+1\le j\le mn-2\\ 3ƛ+i-7& for\; j = mn-1\end{array}, \right.$

It is obvious that the weights of edges are different, thus $\beta$ is an edge irregular total $ƛ-$ labeling of ${\theta }_{n}(3, m)$ . Hence

$tes\left({\theta }_{n}(3, m)\right) = (m+1)n+1\text{.}$

Theorem 3. Let ${\theta }_{n}\left(t, 3\right)$ be a theta snake graph for $t>3$ . Then

$tes\left({\theta }_{n}(t, 3)\right) = \left\lceil {\frac{4tn+2}{3}} \right\rceil \text{.}$

Proof. A size of the graph ${\theta }_{n}(t, 3)$ equals $4tn$ and $\Delta \left({\theta }_{n}\left(t, 3\right)\right) = 2t$ , then from (1) we have

$tes\left({\theta }_{n}(t, 3)\right)\ge \left\lceil {\frac{4tn+2}{3}} \right\rceil \text{.}$

We define an edge irregular total $ħ-$ labeling for ${\theta }_{n}(t, 3)$ to get upper bound. So, let $ħ = \left\lceil {\frac{4tn+2}{3}} \right\rceil$ and a total $ħ-$ labeling $\gamma :V\left({\theta }_{n}(t, 3)\right)\cup E\left({\theta }_{n}(t, 3)\right)\to \left\{\mathrm{1, 2}, 3, \dots, ħ\right\}$ is defined in the following three cases:

Case 1. $4tn+2\equiv 0\left(mod \; 3\right)$

$\gamma$ is defined as:

$\gamma \left({c}_{0}\right) = 1,$

$\gamma \left({c}_{S}\right) = \left(t+1\right)S\; for\; 1\le S\le n-1\text{, }$

$\gamma \left({c}_{n}\right) = ħ$

$\gamma \left({x}_{i, j}\right)\mathrm{I} = \left\{\begin{array}{cc}i& for\; 1\le j\le 3, i = 1, 2, \dots , t\\ i+t+1& for\; 4\le j\le 6, i = 1, 2, \dots , t\\ i+2(t+1)& for\; 7\le j\le 9, i = 1, 2, \dots , t\\ .& .\\ .& .\\ .& .\\ i+(n-1)(t+1)& for\; 3n-5\le j\le 3n-3, i = 1, 2, \dots , t\\ ħ-1& for\; 3n-2\le j\le 3n, i = 1\\ ħ& for\; 3n-2\le j\le 3n, i = 2, 3, \dots , t\end{array}\right.\text{, }$

$\gamma \left({c}_{0}{x}_{i, 1}\right) = 1 \; for\; i = 1, 2, \dots , t$

$\gamma \left({c}_{S}{x}_{i, 3S}\right) = 2St-2S+3 \; for\; 1\le S\le n-1, i = 1, 2, \dots , t$

$\gamma \left({c}_{n}{x}_{i, 3n}\right) = \left\{\begin{array}{cc}ħ-t+2& for\; i = 1\\ ħ-t+i& for\; i = 2, 3, \dots , t\end{array}\right.\text{, }$

$\gamma \left({c}_{S}{x}_{i, 3S+1}\right) = 2St-2S+2 \; for\; 1\le S\le n-1, i = 1, 2, \dots , t$

$\gamma \left({{c}_{n-1}x}_{i, 3n-2}\right) = \left\{\begin{array}{c}\left\{\begin{array}{cc}\left(t+2\right)n-t-5& for\; i = 1\\ \left(t+2\right)n-t+i-7& for\; i = 2, 3, \dots , t\end{array}, n = 2, 3\right.\\ \begin{array}{c}\left\{\begin{array}{cc}\left(t+1\right)n-t-1& for\; i = 1\\ \left(t+1\right)n-t+i-3& for\; i = 2, 3, \dots , t\end{array}, n\ne 2, 3\right.\end{array}\end{array}\right.$

$\gamma \left({{x}_{i, j}x}_{i, j+1}\right) = \left\{\begin{array}{c}\left\{\begin{array}{cc}t+j& for\; 1\le j\le 2\\ 3t+j-5& for\; 4\le j\le 5\\ 5t+j-10& for\; 7\le j\le 8\\ .& .\\ .& .\\ .& .\\ \left(2n-3\right)t+j-5\left(n-2\right)& for\; 3n-5\le j\le 3n-4\end{array}, i = 1, 2, \dots , t\right.\\ \begin{array}{cc}ħ-3\left(t+n\right)+j+5& for\; 3n-2\le j\le 3n-1, i = 1\\ ħ-3\left(t+n\right)+j+5+2(i-2)& for\; 3n-2\le j\le 3n-1, i = 2, 3, \dots , t\end{array}\end{array}\right.$

Obviously, $ħ$ is the greatest label. The edges weights of ${\theta }_{n}(t, 3)$ can be expressed as:

${w}_{\gamma }\left({c}_{0}{x}_{i, 1}\right) = i+2 \; for\; i = 1, 2, \dots , t$

${w}_{\gamma }\left({c}_{S}{x}_{i, 3S}\right) = t\left(4S-1\right)+i+2\; for\; 1\le S\le n-1, i = 1, 2, \dots , t$

${w}_{\gamma }\left({c}_{S}{x}_{i, 3S+1}\right) = 4St+i+2 \; for\; 1\le S\le n-1, i = 1, 2, \dots , t\text{, }$

${w}_{\gamma }\left({c}_{n-1}{x}_{i, 3n-2}\right) = \left\{\begin{array}{cc}2nt+3n-2t+ħ+i-8& for\; n = 2, 3\\ 2nt+2n-2t+ħ+i-4& for\; n\ne 2, 3\end{array}, i = 1, 2, \dots , t\right.$

${w}_{\gamma }\left({{x}_{i, j}x}_{i, j+1}\right) = \left\{\begin{array}{c}\left\{\begin{array}{cc}t+j+2i& for\; 1\le j\le 2\\ 5t+j+2i-4& for\; 4\le j\le 5\\ 9t+j+2i-6& for\; 7\le j\le 8\\ .& .\\ .& .\\ .& .\\ \left(4n-5\right)t+j+2i-3n+8& for\; 3ni-5\le j\le 3n-4\end{array}, i = 1, 2, \dots , t\right.\\ \begin{array}{cc}3ħ-3\left(t+in\right)+j+3& for\; 3n-2\le j\le 3n-1, i = 1\\ 3ħ-3\left(t+in\right)+j+2i+3& for\; 3n-2\le j\le 3n-1, i = 2, 3, \dots , t\end{array}\end{array}\right.$

It implies that the edges weights have distinct values. So $\gamma$ is the desired edge irregular total $ħ-$ labeling, $ħ = \left\lceil {\frac{4tn+2}{3}} \right\rceil$ . Hence

$tes\left({\theta }_{n}(t, 3)\right) = \left\lceil {\frac{4tn+2}{3}} \right\rceil \text{.}$

Case 2. $4tn+2\equiv 1\left(mod \; 3\right)$

$\mathrm{D}\mathrm{e}\mathrm{f}\mathrm{i}\mathrm{n}\mathrm{e}\;\gamma$ as:

$\gamma \left({c}_{0}\right) = 1,$

$\gamma \left({c}_{S}\right) = \left(t+1\right)S \; for\; 1\le S\le n-1\text{, }$

$\gamma \left({c}_{n}\right) = ħ$

$\gamma \left({x}_{i, j}\right) = \left\{\begin{array}{cc}i& for\; 1\le j\le 3, i = 1, 2, \dots , t\\ i+t+1& for\; 4\le j\le 6, i = 1, 2, \dots , t\\ i+2(t+1)& for\; 7\le j\le 9, i = 1, 2, \dots , t\\ .& .\\ .& .\\ .& .\\ i+(n+1)(t+1)& for\; 3n-5\le j\le 3n-3, i = 1, 2, \dots , t\\ ħ-1& for\; 3n-2\le j\le 3n, i = 1\\ ħ& for\; 3n-2\le j\le 3n, i = 2, 3, \dots , t\end{array}\right.\text{, }$

$\gamma \left({c}_{0}{x}_{i, 1}\right) = 1\; for\; i = 1, 2, \dots , t$

$\gamma \left({c}_{S}{x}_{i, 3S}\right) = 2St-2S+3 \; for\; 1\le S\le n-1, i = 1, 2, \dots,\rm{t}$

$\gamma \left({c}_{n}{x}_{i, 3n}\right) = \left\{\begin{array}{cc}ħ-t& for\; i = 1\\ ħ-t+i-2& for\; i = 2, 3, \dots , t\end{array}\right.\text{, }$

$\gamma \left({c}_{S}{x}_{i, 3S+1}\right) = 2St-2S+2 \; for\; 1\le \mathrm{I}S\le n-1, i = 1, 2, \dots t$

$\gamma \left({{c}_{n-1}x}_{i3n-2}\right) = \left\{\begin{array}{c}\left\{\begin{array}{cc}\left(t+2\right)n-t-5& for\; i = 1\\ \left(t+2\right)n-t+i-7& for\; i = 2, 3, \dots , t\end{array}, n = 2, 3\right.\\ \begin{array}{c}\left\{\begin{array}{cc}\left(t+1\right)n-t-1& for\; i = 1\\ \left(\mathrm{I}t+\mathrm{I}1\right)n-t+i-3& for\; i = 2, 3, \dots , t\end{array}, n\ne 2, 3\right.\end{array}\end{array}\right.$

$\gamma \left({{x}_{i, j}x}_{i, j+1}\right) = \left\{\begin{array}{c}\left\{\begin{array}{cc}t+j& for\; 1\le j\le 2\\ 3t+j-5& for\; 4\le j\le 5\\ 5t+j-10& for\; 7\le j\le 8\\ .& .\\ .& .\\ .& .\\ \left(2n-3\right)t+j-5(n-2)& for\; 3n-5\le j\le 3n-4\end{array}, i = 1, 2, \dots , t\right.\\ \begin{array}{cc}ħ-3\left(t+n\right)+j+3& for\; 3n-2\le j\le 3n-1, i = 1\\ ħ-3\left(t+n\right)+j+2(i-2)& for\; 3n-2\le j\le 3n-1, i = 2, 3, \dots , t\end{array}\end{array}\right.$

It is clear that the greatest label is $ħ$ . We define the weights of edges of ${\theta }_{n}(t, 3)$ as:

${w}_{\gamma }\left({c}_{0}{x}_{i, 1}\right) = i+2\; for\; i = 1, 2, \dots , t$

${w}_{\gamma }\left({c}_{S}{x}_{i, 3S}\right) = t\left(4S-1\right)+i+2 \; for\; 1\le S\le n-1, i = 1, 2, \dots , t{w}_{\gamma }\left({c}_{n}{x}_{i, 3n}\right) = 3ħ-t+i-2 \; for\; 1\le S\le n-1, i = 1, 2, \dots , t$

${w}_{\gamma }\left({c}_{S}{x}_{i, 3S+1}\right) = 4St+i+2\; for\; 1\le S\le n-1, i = 1, 2, \dots , t\text{, }$

${w}_{\gamma }\left({c}_{n-1}{x}_{i, 3n-2}\right) = \left\{\begin{array}{cc}2nt+3n-2t+ħ+i-8& for\; n = 2, 3\\ 2nt+2n-2t+ħ+i-4& for\; n\ne 2, 3\end{array}, i = 1, 2, \dots , t\right.$

${w}_{\gamma }\left({{x}_{i, j}x}_{i, j+1}\right) = \left\{\begin{array}{c}\left\{\begin{array}{cc}t+j+2i& for\; 1\le j\le 2\\ 5t+j+2i-4& for\; 4\le j\le 5\\ 9t+j+2i-6& for\; 7\le j\le 8\\ .& .\\ .& .\\ .& .\\ \left(4n-5\right)t+j+2i-3n+8& for\; 3n-5\le j\le 3n-4\end{array}, i = 1, 2, \dots , t\right.\\ \begin{array}{cc}3ħ-3\left(t+n\right)+j+1& for\; 3n-2\le j\le 3n-1, i = 1\\ 3ħ-3\left(t+n\right)+j+2(i-2)& for\; 3n-2\le j\le 3n-1, i = 2, 3, \dots , t\end{array}\end{array}\right.$

It is obvious that the edges weights are different. Then

$tes\left({\theta }_{n}(t, 3)\right) = \left\lceil {\frac{4tn+2}{3}} \right\rceil \text{.}$

Case 3. $4tn+2\equiv 2\left(mod \; 3\right)$

$\gamma$ is defined as follows:

$\gamma \left({c}_{0}\right) = 1,$

$\gamma \left({c}_{S}\right) = \left(t+1\right)S \; for\; 1\le S\le n-1\text{, }$

$\gamma \left({c}_{n}\right) = ħ$

$\gamma \left({x}_{i, j}\right) = \left\{\begin{array}{cc}i& for\; 1\le j\le 3, i = 1, 2, \dots , t\\ i+t+1& for\; 4\le j\le 6, i = 1, 2, \dots , t\\ i+2(t+1)& for\; 7\le j\le 9, i = 1, 2, \dots , t\\ .& .\\ .& .\\ .& .\\ i+\left(n-1\right)\left(t+1\right)& for\; 3n-5\le j\le 3n-3, i = 1, 2, \dots , t\\ ħ-1& for\; 3n-2\le j\le 3n, i = 1\\ ħ& for\; 3n-2\le j\le 3n, i = 2, 3, \dots , t\end{array}, \right.$

$\gamma \left({c}_{0}{x}_{i, 1}\right) = 1\; for\; i = 1, 2, \dots , t$

$\gamma \left({c}_{S}{x}_{i, 3S}\right) = 2St-2S+3 \; for\; 1\le S\le n-1, i = 1, 2, \dots , t$

$\gamma \left({c}_{n}{x}_{i, 3n}\right) = \left\{\begin{array}{cc}ħ-t+1& for\; i = 1\\ ħ-t+i-1& for\; i = 2, 3, \dots , t\end{array}\right.\text{, }$

$\gamma \left({c}_{S}{x}_{i, 3S+1}\right) = 2St-2S+2 \; for\; 1\le S\le n-2, i = 1, 2, \dots , t$

$\gamma \left({{x}_{i, j}x}_{i, j+1}\right) = \left\{\begin{array}{c}\left\{\begin{array}{cc}t+j& for\; 1\le j\le 2\\ 3t+j-5& for\; 4\le j\le 5\\ 5t+j-10& for\; 7\le j\le 8\\ .& .\\ .& .\\ .& .\\ \left(2n-3\right)t+j-5(n-2)& for\; 3n-5\le j\le 3n-4\end{array}, i = 1, 2, \dots t\right.\\ \begin{array}{cc}ħ-3\left(t+i\right)+j+4& for\; 3n-2\le j\le 3ni-1, i = 1\\ ħ-3\left(t+n\right)+j+2i& for\; 3n-2\le j\le 3n-1, i = 2, 3, \dots , t\end{array}\end{array}\right.$

We can see that $ħ$ is the greatest label. For edges weights of ${\theta }_{n}(t, 3)$ , we have

${w}_{\gamma }\left({c}_{0}{x}_{i, 1}\right) = i+2 \; for\; i = 1, 2, \dots , t$

${w}_{\gamma }\left({c}_{0}{x}_{i, 3S}\right) = t\left(4S-1\right)+i+2 \; for\; 1\le S\le n-1, i = 1, 2, \dots , t{w}_{\gamma }\left({c}_{n}{x}_{i, 3n}\right) = 3ħ-t+i-1 \; for\; 1\le S\le n-1, i = 1, 2, \dots , t$

${w}_{\gamma }\left({c}_{S}{x}_{i, 3S+1}\right) = 4St+i+2 \; for\; 1\le S\le n-1, i = 1, 2, \dots , t\text{, }$

${w}_{\gamma }\left({c}_{n}{x}_{i, 3n-2}\right) = \left\{\begin{array}{cc}2nt-3n-2t+ħ+i-8& for\; n = 2, 3\\ 2nt+2n-2t+ħ+i-4& for\; n\ne 2, 3\end{array}, i = 1, 2, \dots , t\right.$

${w}_{\gamma }\left({{x}_{i, j}x}_{i, j+1}\right) = \left\{\begin{array}{c}\left\{\begin{array}{cc}t+j+2i& for\; 1\le j\le 2\\ 5t+j+2i-4& for\; 4\le j\le 5\\ 9t+j+2i-6& for\; 7\le j\le 8\\ .& .\\ .& .\\ .& .\\ \left(4n-5\right)t+j+2i-3n+8& for\; 3n-5\le j\le 3n-4\end{array}, i = 1, 2, \dots , t\right.\\ \begin{array}{cc}3ħ-3\left(t+n\right)+j+2& for\; 3n-2\le j\le 3n-1, i = 1\\ 3ħ-3\left(t+n\right)+j+2i& for\; 3n-2\le j\le 3n-1, i = 2, 3, \dots , t\end{array}\end{array}\right.$

It clears that the edges weights are i distinct. So $\gamma$ is the desired edge irregular total $ħ-$ labeling, $ħ = \left\lceil {\frac{4tn+2}{3}} \right\rceil$ . Hence

$tes\left({\theta }_{n}(t, 3)\right) = \left\lceil {\frac{4tn+2}{3}} \right\rceil \text{.}$

Theorem 4. For ${\theta }_{n}\left(4, m\right)$ be a theta snake graph for $t>3$ . Then

$tes\left({\theta }_{n}(4, m)\right) = \left\lceil {\frac{4(m+1)n+2}{3}} \right\rceil \text{.}$

Proof. Since $\left|E\left({\theta }_{n}\left(4, m\right)\right)\right| = 4(m+1)n$ and $\Delta \left({\theta }_{n}\left(4, m\right)\right) = 8$ , then from (1) we have

$tes\left({\theta }_{n}(4, m)\right)\ge \left\lceil {\frac{4(m+1)n+2}{3}} \right\rceil \text{.}$

The existence of an edge irregular total $ƛ-$ labeling for ${\theta }_{n}(4, m)$ , $m>3$ will be shown, with $ƛ = \left\lceil {\frac{4(m+1)n+2}{3}} \right\rceil$ . Define a total $ƛ-$ labeling $\beta :V\left({\theta }_{n}(4, m)\right)\cup E\left({\theta }_{n}(4, m)\right)\to \left\{\mathrm{1, 2}, 3, \dots, ƛ\right\}$ for ${\theta }_{n}(4, m)$ in the following three cases as:

Case 1. $4\left(m+1\right)n+2\equiv 0\left(mod \; 3\right)$ , $i = 1, 2, \mathrm{3, 4}$

$\beta$ is defined as:

$\beta \left({c}_{s}\right) = \left\{\begin{array}{c}\begin{array}{c}1 \; for\; s = 0\\ \left(m+1\right)s \; for\; 1\le s\le \left\lceil {\frac{n}{2}} \right\rceil \end{array}\\ ƛ+s-n \; for\;\left\lceil {\frac{n}{2}} \right\rceil \le s\le n\end{array}\right.\text{, }$

$\beta \left({x}_{i, j}\right) = \left\{\begin{array}{cc}j& for\; 1\le j\le m\\ j+1& for\; m+1\le j\le 2m\\ .& .\\ .& .\\ \begin{array}{c}j+\left\lceil {\frac{n}{2}} \right\rceil -1\\ ƛ-j+22\\ ƛ\end{array}& \begin{array}{c}for\; m\left(\left\lceil {\frac{n}{2}} \right\rceil -1\right)+1\le j\le m\left\lceil {\frac{n}{2}} \right\rceil +1\\ for\; m\left\lceil {\frac{n}{2}} \right\rceil +2\le j\le m\left(n-1\right)\\ for\; m\left(n-1\right)+1\le j\le mn-1\end{array}\end{array}, \right.$

$\beta \left({c}_{0}{x}_{i, 1}\right) = 1 \; for\; i = 1, 2, \mathrm{3, 4}$

$\beta \left({c}_{S}{x}_{i, mS}\right) = \left\{\begin{array}{c}{2c}_{S}+i-1\; for\; 1\le S\le \left\lceil {\frac{n}{2}} \right\rceil -1\\ \begin{array}{c}{c}_{S}+i-4\left(m+1\right)for\;\left\lceil {\frac{n}{2}} \right\rceil \le s\le n-1\\ ƛ-4+i \; for\; s = n\end{array}\end{array}\right., i = 1, 2, \mathrm{3, 4}$

$\beta \left({c}_{S}{x}_{i, mS+1}\right) = \left\{\begin{array}{c}{2c}_{S}+i+1 \; for\; 1\le S\le \left\lceil {\frac{n}{2}} \right\rceil , i = 1, 2, \mathrm{3, 4}\\ {c}_{S}+i-4\left(m+1\right)+2 \; for\;\left\lceil {\frac{n}{2}} \right\rceil +1\le s\le n-1\end{array}\right.$

$\beta \left({c}_{n}{x}_{i, mn}\right) = \left\{\begin{array}{cc}ƛ-3& for\; i = 1\\ ƛ-2& for\; i = 2\\ \begin{array}{c}ƛ-1\\ ƛ\end{array}& \begin{array}{c}for\; i = 3\\ for\; i = 4\end{array}\end{array}\right.\text{, }$

$\beta \left({{x}_{i, j}x}_{i, j+1}\right) = \left\{\begin{array}{cc}j+i+1& for\; 1\le j\le m-1\\ j+i+2& for\; m+1\le j\le 2m-1\\ .& .\\ .& .\\ j+i+\left\lceil {\frac{n}{2}} \right\rceil & for\; j = m\left(\left\lceil {\frac{n}{2}} \right\rceil -1\right)+1\\ 2j+i-2[nm\left(\left\lceil {\frac{n}{2}} \right\rceil -1\right)+1]& for\; m\left(\left\lceil {\frac{n}{2}} \right\rceil -1\right)+2\le j\le mn-1\end{array}.\right.$

It is clear that $ƛ$ is the greatest used label. The weights of edges of ${\theta }_{n}(4, m)$ are given by:

${w}_{\beta }\left({c}_{0}{x}_{i, 1}\right) = i+2 \; for\; i = 1, 2, \mathrm{3, 4}\text{, }$

${w}_{\beta }\left({c}_{S}{x}_{i, mS}\right) = \left\{\begin{array}{c}{2ms+s+2c}_{S}+i-1 \; for\; 1\le S\le \left\lceil {\frac{n}{2}} \right\rceil -1, \\ \begin{array}{c}{c}_{S}+i+ƛ+\left(s-4\right)\left(m+1\right)-n+\left\lceil {\frac{n}{2}} \right\rceil -1\; for\;\left\lceil {\frac{n}{2}} \right\rceil \le s\le n-1\\ 3ƛ-4+i+s-n\; for\; s = n\end{array}\end{array}\right., i = 1, \mathrm{2, 3}, 4$

${w}_{\beta }\left({c}_{S}{x}_{i, mS+1}\right) = \left\{\begin{array}{c}\left(2m+1\right)s+{2c}_{S}+i+1 \; for\; 1\le S\le \left\lceil {\frac{n}{2}} \right\rceil, \\ 2ƛ+s-n+{c}_{S}+i-4\left(m+1\right)+2 \;for\; \left\lceil {\frac{n}{2}} \right\rceil \le s\le n-1\end{array}\right.i = 1, 2, \mathrm{3, 4},$

${w}_{\beta }\left({c}_{n}{x}_{i, mn}\right) = \left\{\begin{array}{cc}3ƛ+s-n-3& for\; i = 1\\ 3ƛ+s-n-2& for\; i = 2\\ \begin{array}{c}3ƛ+s-n-1\\ 3ƛ+s-n\end{array}& \begin{array}{c}for\; i = 3\\ for\; i = 4\end{array}\end{array}\right.\text{, }$

${w}_{\beta }\left({{x}_{i, j}x}_{i, j+1}\right) = \left\{\begin{array}{cc}3j+i+2& for\; 1\le j\le m-1\\ 3j+i+4& for\; m+1\le j\le 2m-1\\ .& .\\ .& .\\ 3j+i+3\left\lceil {\frac{n}{2}} \right\rceil -1& for\; j = m\left(\left\lceil {\frac{n}{2}} \right\rceil -1\right)+1\\ \begin{array}{c}4j+2ƛ+45+i-2[nm\left(\left\lceil {\frac{n}{2}} \right\rceil -1\right)+1]\\ 2j+2ƛ+i-2[nm\left(\left\lceil {\frac{n}{2}} \right\rceil -1\right)+1]\end{array}& \begin{array}{c}for\; m\left\lceil {\frac{n}{2}} \right\rceil +2\le j\le m\left(n-1\right)\\ for\; m\left(n-1\right)+1\le j\le mn-1\end{array}\end{array}, \right.$

It is obvious that the weights of edges are different, thus $\beta$ is an edge irregular total $ƛ-$ labeling of ${\theta }_{n}(4, m)$ . Hence

$tes\left({\theta }_{n}(4, m)\right) = \left\lceil {\frac{4(m+1)n+2}{3}} \right\rceil \text{.}$

Case 2. $4\left(m+1\right)n+2\equiv 1\left(mod \; 3\right)$ , $i = 1, 2, \mathrm{3, 4}$

$\beta$ is defined as:

$\beta \left({c}_{s}\right) = \left\{\begin{array}{c}\begin{array}{c}1\; for\; s = 0\\ \left(m+1\right)s \; for\; 1\le s\le \left\lceil {\frac{n}{2}} \right\rceil \end{array}\\ ƛ+s-n \; for\;\left\lceil {\frac{n}{2}} \right\rceil \le s\le n\end{array}\right.\text{, }$

$\beta \left({x}_{i, j}\right) = \left\{\begin{array}{cc}j& for\; 1\le j\le m\\ j+1& for\; m+1\le j\le 2m\\ .& .\\ .& .\\ \begin{array}{c}j+\left\lceil {\frac{n}{2}} \right\rceil -1\\ ƛ-j+22\\ ƛ\end{array}& \begin{array}{c}for\; m\left(\left\lceil {\frac{n}{2}} \right\rceil -1\right)+1\le j\le m\left\lceil {\frac{n}{2}} \right\rceil \\ for\; m\left\lceil {\frac{n}{2}} \right\rceil +1\le j\le m\left(n-1\right)\\ for\; m\left(n-1\right)+1\le j\le mn-1\end{array}\end{array}, \right.$

$\beta \left({c}_{0}{x}_{i, 1}\right) = 1 \; for\; i = 1, 2, \mathrm{3, 4}$

$\beta \left({c}_{S}{x}_{i, mS}\right) = \left\{\begin{array}{c}{2c}_{S}+i-1 \; for\; 1\le S\le \left\lceil {\frac{n}{2}} \right\rceil -1, \\ \begin{array}{c}\begin{array}{c}ƛ-7+i \; for\; s = \left\lceil {\frac{n}{2}} \right\rceil \\ {c}_{S}+i-4m-2 \; for\;\left\lceil {\frac{n}{2}} \right\rceil +1\le s\le n-1\end{array}\\ ƛ-6+i \; for\; s = n\end{array}\end{array}\right., i = 1, 2, \mathrm{3, 4}$

$\beta \left({c}_{S}{x}_{i, mS+1}\right) = \left\{\begin{array}{c}{2c}_{S}+i+1 \; for\; 1\le S\le \left\lceil {\frac{n}{2}} \right\rceil \\ {c}_{S}+i-4m \; for\;\left\lceil {\frac{n}{2}} \right\rceil \le s\le n-1\end{array}, i = 1, 2, \mathrm{3, 4}\right.$

$\beta \left({c}_{n}{x}_{i, mn}\right) = \left\{\begin{array}{cc}ƛ-5& for\; i = 1\\ ƛ-4& for\; i = 2\\ \begin{array}{c}ƛ-3\\ ƛ-2\end{array}& \begin{array}{c}for\; i = 3\\ for\; i = 4\end{array}\end{array}\right.\text{, }$

It is clear that $ƛ$ is the greatest used label. The weights of edges of ${\theta }_{n}(4, m)$ are given by:

${w}_{\beta }\left({c}_{0}{x}_{i, 1}\right) = i+2\; for\; i = 1, 2, \mathrm{3, 4}\text{, }$

${w}_{\beta }\left({c}_{S}{x}_{i, mS}\right) = \left\{\begin{array}{c}\begin{array}{c}{2ms+s+2c}_{S}+i-1 \; for\; 1\le S\le \left\lceil {\frac{n}{2}} \right\rceil -1, \\ 2ƛ-m\left\lceil {\frac{n}{2}} \right\rceil +\left(m+1\right)s+i+15 \; for\; s = \left\lceil {\frac{n}{2}} \right\rceil \end{array}\\ \begin{array}{c}{c}_{S}+i+ƛ+(s-4)\left(m+1\right)-n+\left\lceil {\frac{n}{2}} \right\rceil -1 \; for\;\left\lceil {\frac{n}{2}} \right\rceil \le s\le n-1\\ 3ƛ-4+i+s-n \; for\; s = n\end{array}\end{array}\right.$

${w}_{\beta }\left({c}_{S}{x}_{i, mS+1}\right) = \left\{\begin{array}{c}\left(2m+1\right)s+{2c}_{S}+i+1\; for\; 1\le S\le \left\lceil {\frac{n}{2}} \right\rceil , i = 1, 2, \mathrm{3, 4}\\ 2ƛ+s-n+{c}_{S}+i-4m\; for\;\left\lceil {\frac{n}{2}} \right\rceil \le s\le n-1\end{array}\right.i = 1, 2, \mathrm{3, 4},$

${w}_{\beta }\left({c}_{n}{x}_{i, mn}\right) = \left\{\begin{array}{cc}3ƛ+s-n-5& for\; i = 1\\ 3ƛ+s-n-4& for\; i = 2\\ \begin{array}{c}3ƛ+s-n-3\\ 3ƛ+s-n-2\end{array}& \begin{array}{c}for\; i = 3\\ for\; i = 4\end{array}\end{array}\right.\text{, }$

It is obvious that the weights of edges are different, thus $\beta$ is an edge irregular total $ƛ-$ labeling of ${\theta }_{n}(4, m)$ . Hence

$tes\left({\theta }_{n}(4, m)\right) = \left\lceil {\frac{4(m+1)n+2}{3}} \right\rceil \text{.}$

Case 3. $4\left(m+1\right)n+2\equiv 2\left(mod \; 3\right)$ , $i = 1, 2, \mathrm{3, 4}$

$\beta$ is defined as:

$\beta \left({x}_{i, j}\right) = \left\{\begin{array}{cc}j& for\; 1\le j\le m\\ j+1& for\; m+1\le j\le 2m\\ .& .\\ .& .\\ \begin{array}{c}j+\left\lceil {\frac{n}{2}} \right\rceil -1\\ ƛ-j+22\\ ƛ\end{array}& \begin{array}{c}for\; m\left(\left\lceil {\frac{n}{2}} \right\rceil -2\right)+1\le j\le m\left(\left\lceil {\frac{n}{2}} \right\rceil -1\right)\\ for\; m\left(\left\lceil {\frac{n}{2}} \right\rceil -1\right)+1\le j\le m\left(n-1\right)\\ for\; m\left(n-1\right)+1\le j\le mn-1\end{array}\end{array}, \right.$

$\beta \left({c}_{0}{x}_{i, 1}\right) = 1 \; for\; i = 1, 2, \mathrm{3, 4}$

$\beta \left({c}_{S}{x}_{i, mS}\right) = \left\{\begin{array}{c}{2c}_{S}+i-1 \; for\; 1\le S\le \left\lceil {\frac{n}{2}} \right\rceil -1, i = 1, 2, \mathrm{3, 4}\\ \begin{array}{c}\begin{array}{c}ƛ-7+i \; for\; s = \left\lceil {\frac{n}{2}} \right\rceil \\ {c}_{S}+i-4m-2 \; for\;\left\lceil {\frac{n}{2}} \right\rceil +1\le s\le n-1\end{array}\\ ƛ-5+i \; for\; s = n\end{array}\end{array}\right.$

$\beta \left({c}_{S}{x}_{i, mS+1}\right) = \left\{\begin{array}{c}{2c}_{S}+i+1 \; for\; 1\le S\le \left\lceil {\frac{n}{2}} \right\rceil -1, i = 1, 2, \mathrm{3, 4}\\ \begin{array}{c}{c}_{S}+1+i \; for\; s = \left\lceil {\frac{n}{2}} \right\rceil \\ {c}_{S}+i-4m+1 \; for\;\left\lceil {\frac{n}{2}} \right\rceil +1\le s\le n-1\end{array}\end{array}\right.$

$\beta \left({c}_{n}{x}_{i, mn}\right) = \left\{\begin{array}{cc}ƛ-4& for\; i = 1\\ ƛ-3& for\; i = 2\\ \begin{array}{c}ƛ-2\\ ƛ-1\end{array}& \begin{array}{c}for\; i = 3\\ for\; i = 4\end{array}\end{array}\right.\text{, }$

$\beta \left({{x}_{i, j}x}_{i, j+1}\right) = \left\{\begin{array}{cc}j+i+1& for\; 1\le j\le m-1\\ j+i+2& for\; m+1\le j\le 2m-1\\ .& .\\ .& .\\ j+i+\left\lceil {\frac{n}{2}} \right\rceil & for\; j = m\left(\left\lceil {\frac{n}{2}} \right\rceil -1\right)+1\\ 2j+i-2\left[nm\left(\left\lceil {\frac{n}{2}} \right\rceil -1\right)+1\right]+1& for\; m\left(\left\lceil {\frac{n}{2}} \right\rceil -1\right)+2\le j\le mn-1\end{array}.\right.$

It is clear that $ƛ$ is the greatest used label. The weights of edges of ${\theta }_{n}(4, m)$ are given by:

${w}_{\beta }\left({c}_{0}{x}_{i, 1}\right) = i+2 \; for\; i = 1, 2, \mathrm{3, 4}\text{, }$

${w}_{\beta }\left({c}_{S}{x}_{i, mS}\right) = \left\{\begin{array}{c}\begin{array}{c}{2ms+s+2c}_{S}+i-1 \; for\; 1\le S\le \left\lceil {\frac{n}{2}} \right\rceil -1, \\ 2ƛ-m\left\lceil {\frac{n}{2}} \right\rceil +\left(m+1\right)s+i+15 \; for\; s = \left\lceil {\frac{n}{2}} \right\rceil \end{array}\\ \begin{array}{c}{c}_{S}+i+ƛ+(s-4)\left(m+1\right)-n+\left\lceil {\frac{n}{2}} \right\rceil -1 \; for\;\left\lceil {\frac{n}{2}} \right\rceil \le s\le n-1\\ 3ƛ-3+i+s-n \; for\; s = n\end{array}\end{array}\right.$

${w}_{\beta }\left({c}_{S}{x}_{i, mS+1}\right) = \left\{\begin{array}{c}\left(2m+1\right)s+{2c}_{S}+i+1 \; for\; 1\le S\le \left\lceil {\frac{n}{2}} \right\rceil, i = 1, 2, \mathrm{3, 4}\\ \begin{array}{c}\\ 2ƛ+s-n+{c}_{S}+i-4m+1\; for\; \left\lceil {\frac{n}{2}} \right\rceil \le s\le n-1\end{array}\end{array}\right.,$

It is obvious that the weights of edges are different, thus $\beta$ is an edge irregular total $ƛ-$ labeling of ${\theta }_{n}(4, m)$ . Hence

$tes\left({\theta }_{n}(4, m)\right) = \left\lceil {\frac{4(m+1)n+2}{3}} \right\rceil$

Theorem 5. If ${\theta }_{n}\left(t, 4\right)$ is theta snake graph for $t>3$ . Then

$tes\left({\theta }_{n}(t, 4)\right) = \left\lceil {\frac{5tn+2}{3}} \right\rceil \text{.}$

Proof. Since $\left|E\left({\theta }_{n}\right(t, 4\left)\right)\right| = 5tn$ and $\Delta \left({\theta }_{n}\right(t, 4\left)\right) = 2t$ . Substituting in (1), we have

$tes\left({\theta }_{n}(t, 4)\right)\ge \left\lceil {\frac{5tn+2}{3}} \right\rceil \text{.}$

We define an edge irregular total $ħ-$ labeling for ${\theta }_{n}(t, 4)$ to get upper bound. Let $ħ = \left\lceil {\frac{5tn+2}{3}} \right\rceil$ and a total $ħ-$ labeling $\gamma :V\left({\theta }_{n}(t, 4)\right)\cup E\left({\theta }_{n}(t, 4)\right)\to \left\{\mathrm{1, 2}, 3, \dots, ħ\right\}$ is defined in the following three cases:

Case 1. $5tn+2\equiv 0\left(mod \; 3\right)$

$\mathrm{D}\mathrm{e}\mathrm{f}\mathrm{i}\mathrm{n}\mathrm{e}\; \gamma$ as:

$\gamma \left({c}_{0}\right) = 1,$

$\gamma \left({c}_{S}\right) = \left(t+1\right)S \; for\; 1\le S\le n-1\text{, }$

$\gamma \left({c}_{n}\right) = ħ$

$\gamma \left({x}_{i, j}\right) = \left\{\begin{array}{cc}i& for\; 1\le j\le 4, i = 1, 2, \dots , t\\ i+t+1& for\; 5\le j\le 8, i = 1, 2, \dots , t\\ i+2(t+1)& for\; 9\le j\le 12, i = 1, 2, \dots , t\\ .& .\\ .& .\\ .& .\\ i+(n-1)(t+1)& for\; 4n-7\le j\le 4n-4, i = 1, 2, \dots , t\\ ħ-1& for\; 4n-3\le j\le 4n, i = 1\\ ħ& for\; 4n-3\le j\le 4n, i = 2, 3, \dots , t\end{array}\right.\text{, }$

$\gamma \left({c}_{0}{x}_{i, 1}\right) = 1 \; for\; i = 1, 2, \dots , t$

$\gamma \left({c}_{S}{x}_{i, 4S}\right) = 3St-2S+3 \; for\; 1\le S\le n-1, i = 1, 2, \dots , t$

$\gamma \left({c}_{n}{x}_{i, 4n}\right) = \left\{\begin{array}{cc}ħ-t+2& for\; i = 1\\ ħ-t+i& for\; i = 2, 3, \dots , t\end{array}\right.\text{, }$

$\gamma \left({c}_{S}{x}_{i, 4S+1}\right) = 3St-2S+2 \; for\; 1\le S\le n-1, i = 1, 2, \dots , t$

$\gamma \left({{c}_{n-1}x}_{i, 4n-3}\right) = \left\{\begin{array}{c}\left\{\begin{array}{cc}\left(t+2\right)n-t-5& for\; i = 1\\ \left(t+2\right)n-t+i-7& for\; i = 2, 3, \dots , t\end{array}, n = 2, 3\right.\\ \left(t+1\right)n-t+i-3 \; for\; i = 2, 3, \dots , t , n\ne 2, 3\end{array}\right.$

$\gamma \left({{x}_{i, j}x}_{i, j+1}\right) = \left\{\begin{array}{c}\left\{\begin{array}{cc}t+j& for\; 1\le j\le 2\\ 3t+j-5& for\; 4\le j\le 5\\ 5t+j-10& for\; 7\le j\le 8\\ .& .\\ .& .\\ .& .\\ \left(2n-3\right)t+j-5\left(n-2\right)& for\; 4n-5\le j\le 4n-4\end{array}, i = 1, 2, \dots , t\right.\\ \begin{array}{cc}ħ-3\left(t+n\right)+j+5& for\; 4n-2\le j\le 4n, i = 1\\ ħ-3\left(t+n\right)+j+5+2(i-2)& for\; 4n-2\le j\le 4n, i = 2, 3, \dots , t\end{array}\end{array}\right.$

It is clear that, $ħ$ is the greatest label. The edges weights of ${\theta }_{n}(t, 4)$ can be expressed as:

${w}_{\gamma }\left({c}_{0}{x}_{i, 1}\right) = i+2 \; for\; i = 1, 2, \dots , t$

${w}_{\gamma }\left({c}_{S}{x}_{i, 4S}\right) = t\left(5S-1\right)+i+2 \; for\; 1\le S\le n-1, i = 1, 2, \dots , t{w}_{\gamma }\left({c}_{n}{x}_{i, 4n}\right) = 3ħ-t+i \; for\; i = 1, 2, \dots , t$

${w}_{\gamma }\left({c}_{S}{x}_{i, 4S+1}\right) = 5St+i+2 \; for\; 1\le S\le n-1, i = 1, 2, \dots , t\text{, }$

${w}_{\gamma }\left({c}_{n-1}{x}_{i, 4n-2}\right) = \left\{\begin{array}{cc}2nt+3n-2t+ħ+i-8& for\; n = 2, 3\\ 2nt+2n-2t+ħ+i-6& for\; n\ne 2, 3\end{array}, i = 1, 2, \dots , t\right.$

${w}_{\gamma }\left({{x}_{i, j}x}_{i, j+1}\right) = \left\{\begin{array}{c}\left\{\begin{array}{cc}t+j+2i& for\; 1\le j\le 2\\ 5t+j+2i-4& for\; 4\le j\le 5\\ 9t+j+2i-6& for\; 7\le j\le 8\\ .& .\\ .& .\\ .& .\\ \left(4n-5\right)t+j+2i-3n+8& for\; 4n-5\le j\le 4n-4\end{array}, i = 1, 2, \dots , t\right.\\ \begin{array}{cc}3ħ-3\left(t+n\right)+j+3& for\; 4n-2\le j\le 4n-1, i = 1\\ 3ħ-3\left(t+n\right)+j+2i+3& for\; 4n-2\le j\le 4n-1, i = 2, 3, \dots , t\end{array}\end{array}\right.$

It implies that the edges weights have distinct values. So $\gamma$ is the desired edge irregular total $ħ-$ labeling, $ħ = \left\lceil {\frac{5tn+2}{3}} \right\rceil$ . Hence

$tes\left({\theta }_{n}(t, 4)\right) = \left\lceil {\frac{5tn+2}{3}} \right\rceil \text{.}$

Case 2. $5tn+2\equiv 1\left(mod \; 3\right)$

$\mathrm{D}\mathrm{e}\mathrm{f}\mathrm{i}\mathrm{n}\mathrm{e}\; \gamma$ as:

$\gamma \left({c}_{0}\right) = 1,$

$\gamma \left({c}_{S}\right) = \left(t+1\right)S \; for\; 1\le S\le n-1\text{, }$

$\gamma \left({c}_{n}\right) = ħ$

$\gamma \left({x}_{i, j}\right) = \left\{\begin{array}{cc}i& for\; 1\le j\le 4, i = 1, 2, \dots , t\\ i+t+1& for\; 5\le j\le 8, i = 1, 2, \dots , t\\ i+2(t+1)& for\; 9\le j\le 12, i = 1, 2, \dots , t\\ .& .\\ .& .\\ .& .\\ i+(n+1)(t+1)& for\; 4n-7\le j\le 4n-4, i = 1, 2, \dots , t\\ ħ-1& for\; 4n-3\le j\le 4n, i = 1\\ ħ& for\; 4n-3\le j\le 4n, i = 2, 3, \dots , t\end{array}\right.\text{, }$

$\gamma \left({c}_{0}{x}_{i, 1}\right) = 1 \; for\; i = 1, 2, \dots , t$

$\gamma \left({c}_{S}{x}_{i, 4S}\right) = 3St-2S+3 \; for\; 1\le S\le n-1, i = 1, 2, \dots, \rm{t}$

$\gamma \left({c}_{n}{x}_{i, 4n}\right) = \left\{\begin{array}{cc}ħ-t& for\; i = 1\\ ħ-t+i-2& for\; i = 2, 3, \dots , t\end{array}\right.\text{, }$

$\gamma \left({c}_{S}{x}_{i, 4S+1}\right) = 3St-2S+2$

$for\; 1\le S\le n-1, i = 1, 2, \dots t$

$\gamma \left({{c}_{n-1}x}_{i, 4n-3}\right) = \left\{\begin{array}{c}\left\{\begin{array}{cc}\left(t+2\right)n-t-5& for\; i = 1\\ \left(t+2\right)n-t+i-7& for\; i = 2, 3, \dots , t\end{array}, n = 2, 3\right.\\ \begin{array}{c}\left\{\begin{array}{cc}\left(t+1\right)n-t-1& for\; i = 1\\ \left(t+1\right)n-t+i-3& for\; i = 2, 3, \dots , t\end{array}, n\ne 2, 3\right.\end{array}\end{array}\right.$

$\gamma \left({{x}_{i, j}x}_{i, j+1}\right) = \left\{\begin{array}{c}\left\{\begin{array}{cc}t+j& for\; 1\le j\le 3\\ 3t+j-5& for\; 5\le j\le 7\\ 5t+j-10& for\; 9\le j\le 11\\ .& .\\ .& .\\ .& .\\ \left(2n-3\right)t+j-5(n-2)& for\; 4n-7\le j\le 4n-5\end{array}, i = 1, 2, \dots , t\right.\\ \begin{array}{cc}ħ-4\left(t+n\right)+j+3& for\; 4n-3\le j\le 4n-1, i = 1\\ ħ-4\left(t+n\right)+j+2(i-2)& for\; 4n-3\le j\le 4n-1, i = 2, 3, \dots , t\end{array}\end{array}\right.$

It is clear that the i greatest label is $ħ$ . We define the weights of edges of ${\theta }_{n}(t, 4)$ as:

${w}_{\gamma }\left({c}_{0}{x}_{i, 1}\right) = i+2\; for\; i = 1, 2, \dots , t$

${w}_{\gamma }\left({c}_{S}{x}_{i, 4S}\right) = t\left(5S-1\right)+i+2\; for\; 1\le S\le n-1, i = 1, 2, \dots , t{w}_{\gamma }\left({c}_{n}{x}_{i, 4n}\right) = 3ħ-t+i-2 \; for\; 1\le S\le n-1, i = 1, 2, \dots , t$

${w}_{\gamma }\left({c}_{S}{x}_{i, 4S+1}\right) = 5St+i+2 \; for\; 1\le S\le n-1, i = 1, 2, \dots , t\text{, }$

${w}_{\gamma }\left({c}_{n-1}{x}_{i, 4n-3}\right) = \left\{\begin{array}{cc}3nt+3n-2t+ħ+i-8& for\; n = 2, 3\\ 3nt+2n-2t+ħ+i-6& for\; n\ne 2, 3\end{array}, i = 1, 2, \dots , t\right.$

${w}_{\gamma }\left({{x}_{i, j}x}_{i, j+1}\right) = \left\{\begin{array}{c}\left\{\begin{array}{cc}t+j+2i& for\; 1\le j\le 3\\ 5t+j+2i-4& for\;i 5\le j\le 7\\ 9t+j+2i-6& for\; 9\le j\le 11\\ .& .\\ .& .\\ .& .\\ \left(4n-5\right)t+j+2i-3n+8& for\; 4n-7\le j\le 4n-5\end{array}, i = 1, 2, \dots , t\right.\\ \begin{array}{cc}3ħ-4\left(t+n\right)+j+1& for\; 4n-3\le j\le 4n-1, i = 1\\ 3ħ-4\left(t+n\right)+j+2(i-2)& for\; 4n-3\le j\le 4n-1, i = 2, 3, \dots , t\end{array}\end{array}\right.$

It is obvious that the edges weights are different. Then

$tes\left({\theta }_{n}(t, 4)\right) = \left\lceil {\frac{5tn+2}{3}} \right\rceil \text{.}$

Case 3. $5tn+2\equiv 2\left(mod \; 3\right)$

$\mathrm{D}\mathrm{e}\mathrm{f}\mathrm{i}\mathrm{n}\mathrm{e} \; \gamma$ as:

$\gamma \left({c}_{0}\right) = 1,$

$\gamma \left({c}_{S}\right) = \left(t+1\right)S \; for\; 1\le S\le n-1\text{, }$

$\gamma \left({c}_{n}\right) = ħ$

$\gamma \left({x}_{i, j}\right) = \left\{\begin{array}{cc}i& for\; 1\le j\le 4, i = 1, 2, \dots , t\\ i+t+1& for\; 5\le j\le 8, i = 1, 2, \dots , t\\ i+2(t+1)& for\; 9\le j\le 12, i = 1, 2, \dots , t\\ .& .\\ .& .\\ .& .\\ i+(in-1)(t+1)& for\; 4n-7\le j\le 4n-4, i = 1, 2, \dots , t\\ ħ-1& for\; 4n-3\le j\le 4n, i = 1\\ ħ& for\; 4n-3\le j\le 4n, i = 2, 3, \dots , t\end{array}, \right.$

$\gamma \left({c}_{0}{x}_{i, 1}\right) = 1 \; for\; i = 1, 2, \dots , t$

$\gamma \left({c}_{S}{x}_{i, 4S}\right) = 3St-2S+3 \; for\; 1\le S\le n-1, i = 1, 2, \dots , t$

$\gamma \left({c}_{n}{x}_{i, 4n}\right) = \left\{\begin{array}{cc}ħ-t+1& for\; i = 1\\ ħ-t+i-1& for\; i = 2, 3, \dots , t\end{array}\right.\text{, }$

$\gamma \left({c}_{S}{x}_{i, 4S+1}\right) = 3St-2S+2 \; for\; 1\le S\le n-2, i = 1, 2, \dots , t$

$\gamma \left({{c}_{n-1}x}_{i, 4n-3}\right) = \left\{\begin{array}{c}\left\{\begin{array}{cc}\left(t+2\right)n-t-5& for\; i = 1\\ \left(t+2\right)n-t+i-7& for\; i = 2, 3, \dots , t\end{array}, n = 2, 3\right.\\ \begin{array}{c}\left\{\begin{array}{cc}\left(t+1\right)n-t-1& for\; i = 1\\ \left(t+1\right)n-t+i-3& for\; i = 2, 3, \dots , t\end{array}, n\ne 2, 3\right.\end{array}\end{array}\right.$

$\gamma \left({{x}_{i, j}x}_{i, j+1}\right) = \left\{\begin{array}{c}\left\{\begin{array}{cc}t+j& for\; 1\le j\le 3\\ 3t+j-5& for\; 5\le j\le 7\\ 5t+j-10& for\; 9\le j\le 11\\ .& .\\ .& .\\ .& .\\ \left(2n-3\right)t+j-5(n-2)& for\; 4n-7\le j\le 4n-5\end{array}, i = 1, 2, \dots t\right.\\ \begin{array}{cc}ħ-4\left(t+n\right)+j+4& for\; 4n-3\le j\le 4n-1, i = 1\\ ħ-4\left(t+n\right)+j+2i& for\; 4n-3\le j\le 4n-1, i = 2, 3, \dots , t\end{array}\end{array}\right.$

We can see that $ħ$ is the greatest label. For edges weights of ${\theta }_{n}(t, 4)$ , we have:

${w}_{\gamma }\left({c}_{0}{x}_{i, 1}\right) = i+2\; for\; i = 1, 2, \dots , t$

${w}_{\gamma }\left({c}_{0}{x}_{i, 4S}\right) = t\left(5S-1\right)+i+2 \; for\; 1\le S\le n-1, i = 1, 2, \dots , t{w}_{\gamma }\left({c}_{n}{x}_{i, 4n}\right) = 3ħ-t+i-1 \; for\; 1\le S\le in-1, i = 1, 2, \dots , t$

${w}_{\gamma }\left({c}_{S}{x}_{i, 4S+1}\right) = 5St+i+2 \; for\; 1\le S\le n-1, i = 1, 2, \dots , t\text{, }$

${w}_{\gamma }\left({c}_{n}{x}_{i, 4n-3}\right) = \left\{\begin{array}{cc}2nt-3n-2t+ħ+i-8& for\; n = 2, 3\\ 2nt+2n-2t+ħ+i-6& for\; n\ne 2, 3\end{array}, i = 1, 2, \dots , t\right.$

${w}_{\gamma }\left({{x}_{i, j}x}_{i, j+1}\right) = \left\{\begin{array}{c}\left\{\begin{array}{cc}t+j+2i& for\; 1\le j\le 3\\ 5t+j+2i-4& for\; 5\le j\le 7\\ 9t+j+2i-6& for\; 9\le j\le 11\\ .& .\\ .& .\\ .& .\\ \left(4n-5\right)t+j+2i-3n+8& for\; 4n-7\le j\le 4n-5\end{array}, i = 1, 2, \dots , t\right.\\ \begin{array}{cc}3ħ-4\left(t+n\right)+j+2& for\; 4n-3\le j\le 3n-1, i = 1\\ 3ħ-4\left(t+n\right)+j+2i& for\; 4n-3\le j\le 4n-1, i = 2, 3, \dots , t\end{array}\end{array}\right.$

It is obvious that the edges weights are distinct. So $\gamma$ is the desired edge irregular total $ħ-$ labeling, $ħ = \; \left\lceil {\frac{5tn+2}{3}} \right\rceil$ . Hence

$tes\left({\theta }_{n}(t, 4)\right) = \left\lceil {\frac{5tn+2}{3}} \right\rceil \text{.}$

The previous results lead us to introduce the following conjecture for a general case of a uniform theta snake graph ${\theta }_{n}(t, m)$ .

Conjecture. For uniform theta snake graph ${\theta }_{n}(t, m)$ , $n\ge 2, t\ge 3, and m\ge 3$ we have

$tes\left({\theta }_{n}(t, m)\right) = \left\lceil {\frac{(m+1)tn+2}{3}} \right\rceil \text{.}$

3. Conclusions

In the current paper, we have defined a new type of a family of graph called uniform theta snake graph, ${\theta }_{n}(t, m)$ . Also, the exact i value of TEISs for ${\mathtt{θ }}_{\mathrm{n}}\left(\mathrm{3, 3}\right)$ , ${\mathtt{θ }}_{\mathrm{n}}(3, \mathrm{m})$ and ${\mathtt{θ }}_{\mathrm{n}}(\mathrm{t}, 3)$ has been determined. Finally, we have generalized for t, m and found TEIS of a uniform theta snake graph ${\mathtt{θ }}_{\mathrm{n}}(\mathrm{t}, \mathrm{m})$ for $\mathrm{m}\ge 3$ , $\mathrm{t}\ge 3$ .

$tes\left({\theta }_{n}\left(\mathrm{3, 3}\right)\right) = 4n+1\text{.}$

$tes\left({\theta }_{n}(3, im)\right) = (im+1)in+1\text{.}$

$tes\left({\theta }_{n}(t, 3)\right) = \left\lceil {\frac{4tn+2}{3}} \right\rceil$

$tes\left({\theta }_{n}(t, m)\right) = \left\lceil {\frac{(m+1)tn+2}{3}} \right\rceil \text{.}$

Conflict of interest

All authors declare no conflict of interest in this paper.

Acknowledgment

We are so grateful to the reviewer for his many valuable suggestions and comments that significantly improved the paper.

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