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On total edge product cordial labeling of fullerenes
\hspace{2ex} $\arrowvert$\hspace{2ex}
Martin Ba\v{c}a et al.}


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\begin{document}

\begin{frontmatter}
% Write your paper title here
\papertitle{On total edge product cordial labeling of fu\-lle\-re\-nes}

%% use optional labels to link authors explicitly to addresses. The label may be more than one, use comma to separate

%\author[label1,label]{Edy Tri Baskoro${}^1$}

\author[label1]{Martin Ba\v{c}a}
\author[label2]{Muhammad Irfan}
\author[label2]{Aisha Javed}
\author[label1]{Andrea Semani\v{c}ov\'{a}-Fe\-\v{n}ov\-\v{c}\'{i}\-ko\-v\'a}


\address[label1]{\small Department of Applied Mathematics and Informatics, Technical University, Ko\v{s}ice, Slovak Republic}
\address[label2]{\small Abdus Salam School of Mathematical Sciences,   GC University, Lahore, Pakistan


\vspace*{2.5ex} 
 {\normalfont   martin.baca@tuke.sk, m.irfan.assms@gmail.com, aaishajaved@gmail.com, andrea.fenovcikova@tuke.sk}
 }

\begin{abstract}
For a simple graph $G=(V(G),E(G))$ this paper deals with the existence of an edge labeling  $\varphi :E(G) \to \{ 0,1,\dots,k - 1\} $, $2\leq k\leq |E(G)|$, which induces a vertex labeling ${\varphi ^*}:V(G) \to \{ 0,1,\dots,k - 1\} $ in such a way that for each vertex $v$, assigns the label $\varphi(e_1)\cdot\varphi(e_2)\cdot\ldots\cdot \varphi(e_n) \pmod k$, where $e_1, e_2,\dots, e_n$ are the edges incident to the vertex $v$. The labeling $\varphi$ is called a $k$-total edge product cordial labeling of $G$ if $|(e_{\varphi}(i)+v_{\varphi^*}(i))-(e_{\varphi}(j)+v_{\varphi^*}(j))|  \le 1$ for every $i,j$, $0 \le i < j \le k-1$, where $e_{\varphi}(i)$ and $v_{\varphi^*}(i)$ is the number of edges and vertices  with $\varphi(e)=i$ and $\varphi^*(v)=i$, respectively.

 The paper examines the existence of such labelings for toroidal fullerenes and for Klein-bottle fullerenes.

\let\thefootnote\relax\footnotetext{Received: xx xxxxx 20xx,\quad
  Accepted: xx xxxxx 20xx.\\[3ex]
  }
  
\end{abstract}

\begin{keyword}
% Separate keyword by \sep
cordial labeling \sep $k$-total edge product cordial labeling \sep toroidal fullerenes \sep  Klein-bottle fullerenes

% Write the classification number
Mathematics Subject Classification: 05C78 
\end{keyword}

\end{frontmatter}

%% Main text
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\section{Introduction}
Let $G=(V(G),E(G))$ be a finite graph without loops and multiple edges, where $V(G)$ and $E(G)$ are its vertex set and edge set, respectively. General reference for graph-theoretic notions is \cite{West}.

A labeling of a graph is any mapping that sends some set of graph elements to a set of numbers or colors. If we label only vertices (respectively edges), we call such a labeling a {\it vertex} (respectively {\it edge}) {\it labeling}.

A vertex labeling $\phi: V(G)\to \{0,1\}$ induces an edge labeling $\phi^*: E(G)\to \{0,1\}$ defined by $\phi^*(uv)= |\phi(u)-\phi(v)|$. For a vertex labeling $\phi$ and $i\in \{0,1\}$, a vertex $v$ is an $i$-vertex if $\phi(v)=i$ and an edge $e$ is an $i$-edge if $\phi^*(e)=i$. Denote the numbers of $0$-vertices, $1$-vertices, $0$-edges, and $1$-edges of $G$ under $\phi$ and $\phi^*$ by $v_{\phi}(0)$, $v_{\phi}(1)$, $e_{\phi^*}(0)$, and $e_{\phi^*}(1)$, respectively. A vertex labeling $\phi$ is called {\it cordial} if $|v_{\phi}(0)-v_{\phi}(1)|\le 1$ and $|e_{\phi^*}(0)-e_{\phi^*}(1)|\le 1$. 

The notion of the cordial labeling was first introduced by Cahit \cite{Cahit} as a weaker version of graceful labeling. He proved in \cite{Cahit-37-90} that every tree is cordial, the complete bipartite graph $K_{m,n}$ is cordial for all $m$ and $n$, and the complete graph $K_n$ is cordial if and only if $n\le 3$. Cordial labelings of various families of graphs were studied in \cite{Ho-Lee-90,Kirchherr-91,Kuoa-169-97}. For related results see \cite{Ho-Shee-93,Lee-Liu-91} and for generalizations see  \cite{Collins-90,Hovey-91}. Cairnie and Edwards \cite{Cairnie-216} have determined the computational complexity of cordial labelings. They proved the conjecture of Kirchherr \cite{Kirchherr-93} that deciding whether a graph admitting a cordial labeling is NP-complete.

A binary vertex labeling $\phi: V(G)\to \{0,1\}$ with induced edge labeling $\phi^*: E(G)\to \{0,1\}$ defined by $\phi^*(uv)= \phi(u)\phi(v)$ is called a {\it product cordial labeling} if $|v_{\phi}(0)-v_{\phi}(1)|\le 1$ and $|e_{\phi^*}(0)-e_{\phi^*}(1)|\le 1$. The concept of the product cordial labeling was introduced by Sundaram et al. \cite{SPS}. Some labelings with variations in cordial theme, namely an edge product cordial labeling and  a total edge product cordial  labeling have been introduced by Vaidya and Barasara in \cite{VB,VB1}.

Let $k$ be an integer, $2 \le k \le |E(G)|$. An edge labeling $\varphi: E(G)\to \{0,1,\dots,k-1\}$ with induced vertex labeling $\varphi^*: V(G)\to \{0,1, \dots,k-1\}$ defined by $\varphi^*(v)= \varphi(e_1)\cdot\varphi(e_2)\cdot\ldots\cdot \varphi(e_n) \pmod k$, where $e_1, e_2,\dots, e_n$ are the edges incident to the vertex $v$, is called a {\it $k$-total edge product cordial labeling} of $G$ if $|(e_{\varphi}(i) + v_{\varphi^*}(i))- (e_{\varphi}(j) + v_{\varphi^*}(j))|\le 1$ for every $i,j$, $0 \le i < j \le k-1$. 

The concept of the $k$-total edge product cordial labeling was introduced by Azaizeh et al. in \cite{AHAL}. A~graph $G$ with a $k$-total edge product cordial labeling is called a {\it $k$-total edge product cordial graph}.

In the paper, we investigate the existence of $3$-total edge product cordial labeling for toroidal fullerenes and for Klein-bottle fullerenes. 

The discovery of the fullerene molecules and related forms of carbon such as nanotubes has ge\-ne\-ra\-ted an~explosion of activity in chemistry, physics, and materials science. Classical fullerene is an~all-carbon molecule in which the atoms are arranged on a~pseudospherical framework made up entirely of pentagons and hexagons. Its molecular graph is a~finite trivalent graph embedded on the surface of a~sphere with only hexagonal and (exactly $12$) pentagonal faces. Deza et al. \cite{Deza-40-00} considered fullerene's extension to other closed surfaces and showed that only four surfaces are possible, namely sphere, torus, Klein bottle and projective plane. Unlike spherical fullerenes, toroidal and Klein bottle's fullerenes have been regarded as tessellations of entire hexagons on their surfaces since they must contain no pentagons, see \cite{Deza-40-00, Klein-34-94}.

Let $L$ be a~regular hexagonal lattice and let $P_m^n$ be an~$m\times n$ quadrilateral section (with $m\ge 2$ hexagons on the top and bottom sides and $n\ge 2$ hexagons on the lateral sides, $n$ is even) cut from the regular hexagonal lattice $L$, (see Figure~\ref{obr1}). 

\begin{figure}[ht!]
	\centering
		\includegraphics[scale=.7]{figure2/figure101.pdf}
	\caption{Quadrilateral section $P_m^n$ cut from the regular  hexagonal lattice $L$.}
	\label{obr1}
\end{figure}

If we identify two lateral sides of $P_m^n$ then we form a~cylinder. If we identify the top and bottom sides of the cylinder such that we identify the vertices $u_i^0$ and $u_i^{n}$, and the vertices $v_i^0$ and $v_i^{n}$, for $i=1,2,\dots, m$, we are able to obtain the {\it toroidal fullerene} (toroidal polyhex) $\mathbb{H}_m^n$ with $mn$ hexagons. We can see that the  toroidal fullerene is a~cubic bipartite graph embedded on the torus such that each face is a~hexagon. If we identify the top and bottom sides of the cylinder in such a~way that we identify the vertices $u_1^0$ and $u_1^{n}$, the vertices $u_i^0$ and $u_{m+2-i}^{n}$, for $i=2,3,\dots, m$, and the vertices $v_i^0$ and $v_{m+1-i}^{n}$, for $i=1,2,\dots, m$, we  obtain the {\it Klein-bottle fullerene} (Klein-bottle polyhex) $\mathbb{KB}_m^n$ with $mn$ hexagons. In this case $\mathbb{KB}_m^n$ is a~cubic bipartite graph of order $2mn$ and size $3mn$ embedded on the Klein-bottle and contains only hexagons.


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\section{Product cordial labeling of toroidal polyhex $\mathbb{H}_m^n$}

In the following segments  by an {\it open edge} we mean the edge with only one end vertex. We will use two operations for gluing the segments, where the symbol $\oplus_v$ denotes the operation of gluing two segments  in the vertical direction and analogously  the symbol $\oplus_h$ denotes the operation of gluing two segments in the horizontal direction. 
 
 The next theorem shows that the toroidal polyhex $\mathbb{H}_3^n$,   $n$ even, admits a $3$-total edge product cordial labeling.

\begin{theorem}    \label{theorem for H-3-n}
 For $n$ even, $n\ge 4$, the toroidal polyhex $\mathbb{H}_3^n$ is $3$-total edge product cordial.
\end{theorem}
\begin{proof}
For obtaining the toroidal polyhex $\mathbb{H}_3^n$ we will use the labeled segments $A_3^2$, $B_3^2$ and $C_3^2$ illustrated in Figure \ref{segment A-3-2, B-3-2, C-3-2}. We can see that each labeled segment has the same number of 0, 1 and 2, namely in the segment $A_3^2$ every number of zeros, ones and twos is used 14 times, in the segment $B_3^2$ every number is used 10 times and in the segment $C_3^2$ every number is used 6 times as a label. 

   \begin{figure}[ht!]
 \centering \includegraphics{figure2/figure1}  \hspace{1.08cm} \includegraphics{figure2/figure2}
 
 \bigskip
 \centering \includegraphics{figure2/figure3}
  \caption{The labeled segments $A_3^2$, $B_3^2$ and $C_3^2$.}
 \label{segment A-3-2, B-3-2, C-3-2}
  \end{figure}

%   \begin{figure}[ht!]
% \centering \includegraphics{figure-A-3-2}
%  \caption{The labeled segments $A_3^2$, $B_3^2$ and $C_3^2$.}
% \label{segment A-3-2, B-3-2, C-3-2}
%  \end{figure}

First we glue $(\frac{n}{2}-2)$  segments $B_3^2$ together in the vertical direction. Since the open edges in the segment $B_3^2$ are labeled with number 1 it follows that by gluing these segments we do not change the vertex labels in the segment   $B_3^2 \oplus_v B_3^2 \oplus_v \dots \oplus_v B_3^2 = (\frac{n}{2}-2)B_3^2$. Then we glue in the vertical direction the segment $A_3^2$ to the segment $(\frac{n}{2}-2)B_3^2$ to obtain  $A_3^2 \oplus_v (\frac{n}{2}-2)B_3^2$. Finally we glue the segment $A_3^2 \oplus_v (\frac{n}{2}-2)B_3^2$ in the vertical direction to the segment $C_3^2$. All open edges used in the gluing  operations are labeled with the number 1 therefore these operations do not have any impact to the vertex labels in the resulting segment $H_3^n$, where
$$H_3^n =    \begin{bmatrix} 
		                     {A}_3^2\\ 
		                         \oplus_v\\ 
		                         (\frac{n}{2}-2)B_3^2 \\
		                         \oplus_v\\ 
		                               C_3^2
		                      \end{bmatrix}.$$

Now we identify two lateral sides of the segment $H_3^n$ to form a~cylinder or nanotube.  Then we identify the top and bottom sides of the cylinder such that we join the open edges in the bottom side of the cylinder labeled by $0$ to the corresponding vertices in the top side of the cylinder (labeled by $0$) and we obtain the toroidal polyhex $\mathbb{H}_3^n$ with $3n$ hexagons. Table \ref{table1} shows multiplicity of numbers $0$, $1$ and $2$ used in the segments $A_3^2$, $B_3^2$, $C_3^2$ and $H_3^n$. 

%Table 1

\begin{table}[ht!]
	\renewcommand{\arraystretch}{1.5}
\setlength{\tabcolsep}{2.5pt}
\centering
		\begin{tabular}{|l||l|l|l|}
		\hline
		segment  & $e_{\varphi}(0)+v_{\varphi^*}(0)$ &  $e_{\varphi}(1)+v_{\varphi^*}(1)$& $e_{\varphi}(2)+v_{\varphi^*}(2)$\\
		\hline\hline
		$A_3^2$& $14$&$14$&$14$\\
		\hline
		$B_3^2$& $10$&$10$&$10$\\
		\hline
		$C_3^2$& $6$&$6$&$6$\\
		\hline
		$(\frac{n}{2}-2)B_3^2$& $5n-20$&$5n-20$&$5n-20$\\
		\hline
		$H_3^n = A_3^2 \oplus_v (\frac{n}{2}-2)B_3^2 \oplus_v C_3^2$& $5n$ & $5n$& $5n$\\
		\hline
			\end{tabular}
	\caption{Multiplicity of     0s,   of 1s and   of 2s in the segments 	$A_3^2$, 	$B_3^2$, 	$C_3^2$, 	$(\frac{n}{2}-2)B_3^2$ and $H_3^n$.}
	\label{table1}
\end{table}		

It is only routine checking that the toroidal polyhex $\mathbb{H}_3^n$ contains every number of 0, 1 and 2 exactly $5n$ times.
\end{proof}

Next we extend the construction of the segment $H_3^n$ for arbitrary $m$ to obtain the segment $H_m^n$ and then to obtain the  graph of the toroidal polyhex $\mathbb{H}_m^n$.


\begin{theorem}    \label{theorem for H-m-n}
 For $n$ even, $n\ge 4$ and $m\ge 3$ the, toroidal polyhex $\mathbb{H}_m^n$ is $3$-total edge product cordial.
\end{theorem}
\begin{proof}
For obtaining a $3$-total edge product cordial labeling of $\mathbb{H}_m^4$ we need the next labeled segments $D_1^4$ and $D_2^4$ depicted in Figure \ref{segment D-1-4, D-2-4}.

   \begin{figure}[ht!]
 \centering \includegraphics{figure2/figure5}  \hspace{1.08cm} \includegraphics{figure2/figure12}
  \caption{The labeled segments $D_1^4$ and $D_2^4$.}
 \label{segment D-1-4, D-2-4}
  \end{figure}

For $m=3t$, $t\ge 1$, we glue the segment $A_3^2$ in the vertical direction to the segment $C_3^2$ and the resulting segment $A_3^2 \oplus_v C_3^2$ we glue in the horizontal direction $t$ times. Hence 
$$H_m^4 =   {\underbrace{ \begin{bmatrix} 
		                     {A}_3^2\\ 
		                         \oplus_v\\ 
		                               C_3^2
		                      \end{bmatrix} 
		                      \oplus_h
		                       \begin{bmatrix} 
		                     {A}_3^2\\ 
		                         \oplus_v\\ 
		                               C_3^2
		                      \end{bmatrix} 
		                      \oplus_h
		                      \cdots
		                      \oplus_h
		                       \begin{bmatrix} 
		                     {A}_3^2\\ 
		                         \oplus_v\\ 
		                               C_3^2
		                      \end{bmatrix}}_{t} }.
		                      $$
For $m=3t+1$, $t\ge 1$, we glue the segment $  \left[ A_3^2 \oplus_v C_3^2 \right]$ horizontally $t$ times and moreover, we glue horizontally the segment $D_1^4$ and we obtain  $$H_m^4 = {\underbrace{ \begin{bmatrix} 
		                     {A}_3^2\\ 
		                         \oplus_v\\ 
		                               C_3^2
		                      \end{bmatrix} 
		                      \oplus_h
		                       \begin{bmatrix} 
		                     {A}_3^2\\ 
		                         \oplus_v\\ 
		                               C_3^2
		                      \end{bmatrix} 
		                      \oplus_h
		                      \cdots
		                      \oplus_h
		                       \begin{bmatrix} 
		                     {A}_3^2\\ 
		                         \oplus_v\\ 
		                               C_3^2
		                      \end{bmatrix}}_{t} }\oplus_h D_1^4.$$
For $m=3t+2$, $t\ge 1$, we glue the segment $  \left[ A_3^2 \oplus_v C_3^2 \right]$  horizontally $t$ times and moreover, we glue horizontally  the segment $D_2^4$ and we create the segment  $$H_m^4 = {\underbrace{ \begin{bmatrix} 
		                     {A}_3^2\\ 
		                         \oplus_v\\ 
		                               C_3^2
		                      \end{bmatrix} 
		                      \oplus_h
		                       \begin{bmatrix} 
		                     {A}_3^2\\ 
		                         \oplus_v\\ 
		                               C_3^2
		                      \end{bmatrix} 
		                      \oplus_h
		                      \cdots
		                      \oplus_h
		                       \begin{bmatrix} 
		                     {A}_3^2\\ 
		                         \oplus_v\\ 
		                               C_3^2
		                      \end{bmatrix}}_{t} }\oplus_h D_2^4.$$

Table \ref{table2} shows how many times the numbers $0$, $1$ and $2$ are used as edge and vertex labels in the segments $D_1^4$ and $D_2^4$ and in the resulting segment  $H_m^4$.

%Table 2

\begin{table}[ht!]
	\renewcommand{\arraystretch}{1.5}
\setlength{\tabcolsep}{2.5pt}
\centering
		\begin{tabular}{|l||l|l|l|}
		\hline
		segment  & $e_{\varphi}(0)+v_{\varphi^*}(0)$ &  $e_{\varphi}(1)+v_{\varphi^*}(1)$& $e_{\varphi}(2)+v_{\varphi^*}(2)$\\
		\hline\hline
		$D_1^4$& $6$&$7$&$7$\\
		\hline
		$D_2^4$& $14$&$13$&$13$\\
		\hline
		$H_m^4$, $m=3t$, $t\ge 1$& $20t$&$20t$&$20t$\\
		\hline
			$H_m^4$, $m=3t+1$, $t\ge 1$& $20t+6$&$20t+7$&$20t+7$\\
		\hline
			$H_m^4$, $m=3t+2$, $t\ge 1$& $20t+14$&$20t+13$&$20t+13$\\
		\hline
			\end{tabular}
	\caption{Multiplicity of     0s,   of 1s and   of 2s in the segments 	$D_1^4$, 	$D_2^4$ and $H_m^4$.}
	\label{table2}
\end{table}		

One can see that the resulting segments $H_m^4$ in each previous case satisfies the property of having a  $3$-total edge product cordial labeling. We identify two lateral sides of the segment $H_m^4$ to form a~cylinder and then we identify the top and bottom sides of the cylinder such that we join the open edges in the bottom side of the cylinder to the corresponding vertices in the top side of the cylinder labeled by $0$. Since the open edges in the corresponding segments are  labeled with number 1 (except for open edges incident to vertices in the top side) it follows that gluing these segments does not have any impact on the vertex labels in the resulting toroidal polyhex $\mathbb{H}_m^4$.  

The graph $\mathbb{H}_m^n$ with a $3$-total edge product cordial labeling for $n$ even, $n\ge 6$, we obtain by using the segment $H_3^n = A_3^2 \oplus_v (\frac{n}{2}-2)B_3^2 \oplus_v C_3^2$ from Theorem \ref{theorem for H-3-n} such that the certain multiple of the segment $H_3^n$ we glue in horizontal direction with a special segment   which depends on   $n$ and $m$.

For next operations we need new segments $A_1^4$, $C_2^4$,  $C_1^2$ and $C_1^4$ depicted in Figure \ref{segment c-1-2, C-1-4} and the segments $B_1^6$, $C_1^6$ and $C_2^6$ shown in Figure \ref{segment B-1-6, C-1-6, C-2-6}.


   \begin{figure}[ht!]
 \centering \includegraphics{figure2/figure4}  \hspace{1.08cm} \includegraphics{figure2/figure11}
 % \caption{The labeled segments $A_1^4$ and $C_2^4$.}
% \label{segment A-1-4, C-2-4}
%  \end{figure}
%
%   \begin{figure}[ht!]
 \hspace{1.08cm} \includegraphics{figure2/figure6}  \hspace{1.08cm} \includegraphics{figure2/figure7}
  \caption{The labeled segments  $A_1^4$, $C_2^4$,   $C_1^2$ and $C_1^4$.}
 \label{segment c-1-2, C-1-4}
  \end{figure}

   \begin{figure}[ht!]
 \centering \includegraphics{figure2/figure8}  \hspace{1.08cm} \includegraphics{figure2/figure9}     \hspace{1.08cm} 
 \includegraphics{figure2/figure10}
  \caption{The labeled segments  $B_1^6$, $C_1^6$ and $C_2^6$.}
 \label{segment B-1-6, C-1-6, C-2-6}
  \end{figure}

Table \ref{table3} gives multiplicity of numbers 0, 1 and 2 used in the segments $A_1^4$, $C_2^4$, $C_1^2$, $C_1^4$, $B_1^6$, $C_1^6$ and $C_2^6$.

%Table 3

\begin{table}[ht!]
	\renewcommand{\arraystretch}{1.5}
\setlength{\tabcolsep}{2.5pt}
\centering
		\begin{tabular}{|l||l|l|l|}
		\hline
		segment  & $e_{\varphi}(0)+v_{\varphi^*}(0)$ &  $e_{\varphi}(1)+v_{\varphi^*}(1)$& $e_{\varphi}(2)+v_{\varphi^*}(2)$\\
		\hline\hline
		$A_1^4$& $8$&$8$&$8$\\
		\hline
		$C_2^4$& $10$&$11$&$11$\\
		\hline
		$C_1^2$& $2$&$2$&$2$\\
		\hline
		$C_1^4$& $5$&$6$&$5$\\
		\hline
		$B_1^6$& $10$&$10$&$10$\\
		\hline
		$C_1^6$& $9$&$9$&$8$\\
		\hline
		$C_2^6$& $17$&$18$&$17$\\
		\hline
			\end{tabular}
	\caption{Multiplicity of     0s,   of 1s and   of 2s in the segments 	$A_1^4$, $C_2^4$, $C_1^2$, $C_1^4$, $B_1^6$, $C_1^6$ and $C_2^6$.}
	\label{table3}
\end{table}		

Let us consider the following 7 cases.

{\it Case 1.} For $m=3t$, $t\ge 1$, and every $n$ even, $n\ge 6$, we get
$$ H_m^n =  \underbrace{H_3^n \oplus_h H_3^n \oplus_h \cdots \oplus_h H_3^n}_{t}     .$$

{\it Case 2.} For $m=3t+1$, $t\ge 1$, and $n=6s$, $s\ge 1$, the special segment   we obtain by gluing vertically the segment $A_1^4$ with vertically $(s-1)$ times of the segment $B_1^6$ and the resulting segment $A_1^4 \oplus_v (s-1)B_1^6$ we glue in the vertical  direction with the segment $C_1^2$. Then we have
$$ H_m^n =  \left[ \underbrace{H_3^n \oplus_h H_3^n \oplus_h \cdots \oplus_h H_3^n}_{t}\right] \oplus_h \begin{bmatrix} 
		                     A_1^4\\ 
		                         \oplus_v\\ 
		                              (s-1)B_1^6\\ 
		                         \oplus_v\\  
		                         C_1^2
		                      \end{bmatrix} .$$

{\it Case 3.} For $m=3t+2$, $t\ge 1$, and $n=6s$, $s\ge 1$, the segment $H_m^n$ we obtain as follows 
$$ H_m^n =  \left[  \underbrace{H_3^n \oplus_h H_3^n \oplus_h \cdots \oplus_h H_3^n}_{t} \right] \oplus_h   
\begin{bmatrix} 
		                     A_1^4\\ 
		                         \oplus_v\\ 
		                              (s-1)B_1^6\\ 
		                         \oplus_v\\  
		                         C_1^2
		                      \end{bmatrix}\oplus_h   
\begin{bmatrix} 
		                     A_1^4\\ 
		                         \oplus_v\\ 
		                              (s-1)B_1^6\\ 
		                         \oplus_v\\  
		                         C_1^2
		                      \end{bmatrix} .$$

{\it Case 4.} For $m=3t+1$, $t\ge 1$, and $n=6s+2$, $s\ge 1$, the special segment   we create by gluing the segment $A_1^4$ in the vertical  direction with vertically  $(s-1)$ times of the segment $B_1^6$ and the resulting segment $A_1^4 \oplus_v (s-1)B_1^6$ we glue in the  vertical  direction with the segment $C_1^4$. Then we get
$$ H_m^n =  \left[  \underbrace{H_3^n \oplus_h H_3^n \oplus_h \cdots \oplus_h H_3^n}_{t}\right] \oplus_h \begin{bmatrix} 
		                     A_1^4\\ 
		                         \oplus_v\\ 
		                              (s-1)B_1^6\\ 
		                         \oplus_v\\  
		                         C_1^4
		                      \end{bmatrix} .$$

{\it Case 5.} For $m=3t+2$, $t\ge 1$, and $n=6s+2$, $s\ge 1$, the special segment  we obtain by gluing two segments $A_1^4$ in the vertical  direction with vertically $2(s-1)$ times of the segment $B_1^6$ and the resulting segment $2A_1^4 \oplus_v 2(s-1)B_1^6$ we glue in the  vertical  direction with the segment $C_2^4$. Then we have
$$ H_m^n =\left[  \underbrace{H_3^n \oplus_h H_3^n \oplus_h \cdots \oplus_h H_3^n}_{t}\right] \oplus_h \begin{bmatrix} 
		                    2 A_1^4\\ 
		                         \oplus_v\\ 
		                             2 (s-1)B_1^6\\ 
		                         \oplus_v\\  
		                         C_2^4
		                      \end{bmatrix} .$$

{\it Case 6.} For $m=3t+1$, $t\ge 1$, and $n=6s+4$, $s\ge 1$, the special segment   we generate by gluing the segment $A_1^4$ in the vertical  direction with vertically $(s-1)$ times of the segment $B_1^6$ and the resulting segment $A_1^4 \oplus_v (s-1)B_1^6$ we glue in the  vertical  direction with the segment $C_1^6$. The expected resulting segment will be as follows 
$$ H_m^n = \left[  \underbrace{H_3^n \oplus_h H_3^n \oplus_h \cdots \oplus_h H_3^n}_{t}\right] \oplus_h \begin{bmatrix} 
		                     A_1^4\\ 
		                         \oplus_v\\ 
		                              (s-1)B_1^6\\ 
		                         \oplus_v\\  
		                         C_1^6
		                      \end{bmatrix} .$$

{\it Case 7.} For $m=3t+2$, $t\ge 1$, and $n=6s+4$, $s\ge 1$, the special segment   we create by gluing two segments $A_1^4$ in the vertical  direction with vertically $2(s-1)$ times of the segment $B_1^6$ and the resulting segment $2A_1^4 \oplus_v 2(s-1)B_1^6$ we glue in the  vertical  direction with the segment $C_2^6$. The required segment has the following form 
$$ H_m^n = \left[  \underbrace{H_3^n \oplus_h H_3^n \oplus_h \cdots \oplus_h H_3^n}_{t}\right] \oplus_h \begin{bmatrix} 
		                   2  A_1^4\\ 
		                         \oplus_v\\ 
		                            2  (s-1)B_1^6\\ 
		                         \oplus_v\\  
		                         C_2^6
		                      \end{bmatrix} .$$

%Table 4


\begin{table}[ht!]
	\renewcommand{\arraystretch}{1.5}
\setlength{\tabcolsep}{2.5pt}
\centering
		\begin{tabular}{|l|l||l|l|l|}
		\hline
	Case&	segment $H_m^n$ & $e_{\varphi}(0)+v_{\varphi^*}(0)$ &  $e_{\varphi}(1)+v_{\varphi^*}(1)$& $e_{\varphi}(2)+v_{\varphi^*}(2)$\\
		\hline\hline
	1.&	$m=3t$, $t\ge 1$& $5nt$&$5nt$&$5nt$\\
	&$n\ge 6$, $n$ even &&&\\
		\hline		
	2.&	$m=3t+1$, $t\ge 1$& $5nt+10s$&$5nt+10s$&$5nt+10s$\\
	&$n=6s$, $s\ge 1$ &&&\\
		\hline
	3.&	$m=3t+2$, $t\ge 1$& $5nt+20s$&$5nt+20s$&$5nt+20s$\\
	&$n=6s$, $s\ge 1$ &&&\\
		\hline
	4.&	$m=3t+1$, $t\ge 1$& $5nt+10s+3$&$5nt+10s+4$&$5nt+10s+3$\\
	&$n=6s+2$, $s\ge 1$ &&&\\
		\hline
	5.&	$m=3t+2$, $t\ge 1$& $5nt+20s+6$&$5nt+20s+7$&$5nt+20s+7$\\
	&$n=6s+2$, $s\ge 1$ &&&\\
		\hline
	6.&	$m=3t+1$, $t\ge 1$& $5nt+10s+7$&$5nt+10s+7$&$5nt+10s+6$\\
	&$n=6s+4$, $s\ge 1$ &&&\\
		\hline
	7.&	$m=3t+2$, $t\ge 1$& $5nt+20s+13$&$5nt+20s+14$&$5nt+20s+13$\\
	&$n=6s+4$, $s\ge 1$ &&&\\
		\hline
			\end{tabular}
	\caption{Multiplicity of     0s,   of 1s and   of 2s in the segment  	$H_m^n$ for $n$ even, $n\ge 6$ and $m\ge 3$.}
	\label{table4}
\end{table}		


All possible cases for obtaining the segment $H_m^n$ for $n$ even, $n\ge 6$ and $m\ge 3$, are described in Table \ref{table4}, where it is shown how many times the numbers 0, 1  and 2 are used as edge and vertex labels. 

Identifying two lateral sides of the segment $H_m^n$ we form a~cylinder and then identifying the top and bottom sides of the cylinder such that the open edges in the bottom side of the cylinder are joined to the corresponding vertices in the top side of the cylinder, we obtain the toroidal polyhex $\mathbb{H}_m^n$. We can see that the labels of the open edges used to glue the corresponding segments has no effect on the vertex labels in the resulting segments/graphs. From Table 4 it follows that for every case the resulting toroidal polyhex $\mathbb{H}_m^n$ is $3$-total edge product cordial.

\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

A {\it helical torus} decorated with graphite can be formed by joining the opposite ends of a chiral or helical nanotube, such that the cycle of one end of nanotube is rotated relative to the cycle of the second end of nanotube, see \cite{Terrones-5-03}. More precisely, we suppose that a nanotube is our cylinder created identifying two lateral sides of $P_m^n$ in Figure \ref{obr1}. To create the helical torus we identify the top and bottom sides of the cylinder/nanotube such that we identify the vertices $u_i^0$ and $u_{i+j}^n$ and the vertices $v_i^0$ and $v_{i+j}^n$, for $i=1,2,\dots, m$ and $j=1,2, \dots, m-1$, where $i+j$ is taken modulo $m$. 

Since in the proof of Theorem \ref{theorem for H-m-n} the segment $H_m^n$, for $n$ even, $n\ge 4$ and $m\ge 3$, contains in the top side only the vertices labeled by $0$, it follows that any gluing of the open edges in the bottom side of $H_m^n$ to the vertices in the top side is possible and does not have any impact on the vertex labels in the resulting graph of the helical torus $\mathbb{H}_m^n$. It means that the gluing operation satisfies the property of having a $3$-total edge product cordial labeling. Hence we have 

\begin{theorem}    \label{theorem helical torus}
 For $n$ even, $n\ge 4$ and $m\ge 3$, the helical torus $\mathbb{H}_m^n$ admits a $3$-total edge product cordial labeling.
\end{theorem}

By joining the opposite ends of a nanotube we can obtain the {\it Klein-bottle polyhex}. More precisely, to create the Klein-bottle polyhex $\mathbb{KB}_m^n$ we identify the top and bottom sides of the cylinder/nanotube (see Figure \ref{obr1}) such that we identify the vertices  $u_1^0$ and $u_1^{n}$, the vertices $u_i^0$ and $u_{m+2-i}^{n}$, for $i=2,3,\dots, m$, and the vertices $v_i^0$ and $v_{m+1-i}^{n}$, for $i=1,2,\dots, m$. 

With respect to the fact that the gluing operation does not have any impact on the vertex labels in the resulting graph of the Klein-bottle polyhex $\mathbb{KB}_m^n$, therefore the property of admitting a $3$-total edge product cordial labeling also holds. Consequently we get 

\begin{theorem}    \label{theorem Klein-bottle}
 For $n$ even, $n\ge 4$ and $m\ge 3$, the Klein-bottle polyhex $\mathbb{KB}_m^n$ admits a $3$-total edge product cordial labeling.
\end{theorem}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Conclusion}

In this paper we proved the existence of the $3$-total edge product cordial labeling for the toroidal polyhex, respectively helical torus, $\mathbb{H}_m^n$, and for the Klein-bottle polyhex $\mathbb{KB}_m^n$, for $n$ even, $n\ge 4$ and $m\ge 3$. In this case the Klein-bottle polyhex $\mathbb{KB}_m^n$ is a~cubic bipartite graph. The next construction describes the existence of the  Klein-bottle polyhex $\mathbb{KB}_{m+1/2}^n$ as a~cubic non-bipartite graph.

Let $L$ be a~regular hexagonal lattice and  let $P_{m+1/2}^n$ be a~quadrilateral section (with $m+1/2$ hexagons on the top and bottom sides, $m\ge 1$, and $n\ge 2$ hexagons on the lateral sides, $n$ is even) cut from the regular hexagonal lattice $L$, (see Figure~\ref{obr2}). 

\begin{figure}[ht!]
	\centering
		\includegraphics[scale=.7]{figure2/figure102.pdf}
	\caption{Quadrilateral section $P_{m+1/2}^n$ cut  from the regular  hexagonal lattice $L$.}
	\label{obr2}
\end{figure}

We identify the top and bottom sides of $P_{m+1/2}^n$ to form a~cylinder. Then we identify the lateral sides of the cylinder such that we identify the vertices $u_1^0$ and $v_{m+1}^{0}$, and the vertices $u_1^j$ and $v_{m+1}^{n-j}$, for $j=1,2,\dots, n-1$, to obtain the  Klein-bottle polyhex   $\mathbb{KB}_{m+1/2}^n$. We can see that $\mathbb{KB}_{m+1/2}^n$ is a~cubic non-bipartite graph of order $2n(m+1/2)$ and size $3n(m+1/2)$ embedded on the Klein-bottle and contains $n(m+1/2)$ hexagons.

Let us suggest the following open problem.

\begin{problem} Find a   $3$-total edge product cordial labeling for the Klein-bottle polyhex  $\mathbb{KB}_{m+1/2}^n$, for $n$ even, $n\ge 2$ and $m\ge 3$.
\end{problem}



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%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%
\section*{Acknowledgement}
This work was supported by the Slovak Science and Technology Assistance Agency under the contract No. APVV-15-0116.


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%%%%%%%%%%%%%%%%
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