## 3 Edge Decomposition of Bipartite Circulant Graphs \(C_{2 nm,nm}\) Based on Cartesian Products
The edge set of the second graph \(K_{1,2}\cup\ K_{1,2(m-1)}\) can be represented by \((c_{0},d_{0}+2m)\), \((c_{1},d_{1}+2m),\ldots,(c_{2m-1},d_{2m-1}+2m)\), where  
\[c_{i}=\left\{ \begin{matrix}0&if&i=0,m;\\ m&if&1\leq i\leq m-1,m+1\leq i\leq 2m-1\end{matrix} \right.\] \[\left.d_{i}=c_{i}+i.\right. \tag{26}\]  
Hence, \(\{d_{j}-c_{j}:j\in\mathbb{Z}_{2m}\}=\mathbb{Z}_{2m}\). Suppose that the first graph \(K_{1,n}\) is represented by the vector \(v(K_{1,n})=(a_{0},a_{1},a_{2},\ldots,\)\(a_{n-1})\), and the second graph \(K_{1,2}\cup\ K_{1,2(m-1)}\) is represented by the \(\left(K_{1,2}\cup\ K_{1,2(m-1)}\right)=(c_{0},c_{1},c_{2},\ldots,c_{2m-1})\), then \(v(K_{1,n})\times w(K_{1,2}\cup\ K_{1,2(m-1)})=(a_{0}c_{0},a_{0}c_{1},\ldots,a_{ p}c_{q},\ldots,\)\(a_{n-1}c_{2m-1})\), where \(p\in\mathbb{Z}_{n}\) and \(q\in\mathbb{Z}_{2m}\). Then, construct the ordered pairs \((a_{0}c_{0},b_{0}d_{0})\), \((a_{0}c_{1},b_{0}d_{1}),\ldots,\)\((a_{p}c_{q},b_{p}d_{q}),\ldots,(a_{n-1}c_{2m-1},b_{n-1}d_{2m-1})\), where \(p\in\mathbb{Z}_{n}\) and \(q\in\mathbb{Z}_{2m}\). Hence, from (25) and (26), we can conclude that  
\[\left\{(b_{i}-a_{i})\big{(}d_{j}-c_{j}\big{)}:i\in\mathbb{Z}_{n},j\in\mathbb{Z }_{2m}\right\}=\mathbb{Z}_{n}\times\mathbb{Z}_{2m}. \tag{27}\]  
Then, let \(\varphi:\mathbb{Z}_{n}\times\mathbb{Z}_{2m}\rightarrow\mathbb{Z}_{2 nm}\) be a one-one mapping defined by  
\[\varphi\big{(}a_{i}c_{j}\big{)}=e_{2 mi+j}=2 ma_{i}+c_{j},i\in\mathbb{Z}_{n},j \in\mathbb{Z}_{2m}, \tag{28}\]  
and \(\psi:\mathbb{Z}_{n}\times\mathbb{Z}_{2m}\rightarrow\mathbb{Z}_{2 nm}\) be a one-one mapping defined by  
\[\psi\big{(}b_{i}d_{j}\big{)}=f_{2 mi+j}=2 mb_{i}+d_{j},i\in\mathbb{Z}_{n},j\in \mathbb{Z}_{2m}. \tag{29}\]  
Hence, we can obtain the following chain \((e_{0},f_{0}),(e_{1},f_{1}),\ldots,(e_{2 nm-1},f_{2 nm-1})\). The translation for the previous chain is \((e_{0}+\sigma,f_{0}+\sigma),(e_{1}+\sigma,f_{1}+\sigma),\ldots,\)\((e_{2 nm-1}+\sigma,f_{2 nm-1}+\sigma),\sigma\in\mathbb{Z}_{2 mn}\), where the additions are calculated modulo \(nm\). Let \(\pi:\mathbb{Z}_{2 nm}\rightarrow\mathbb{Z}_{4 nm}\) be a one-one mapping defined by  
\[\pi(f_{a,\sigma})=2 nm+f_{a,\sigma},f_{a,\sigma}=f_{a}+\sigma,\alpha,\sigma\in \mathbb{Z}_{2 nm}. \tag{30}\]  
We now find novel chains of ordered pairs that are defined by \((e_{0}+\sigma,\pi(f_{0,\sigma}))\), \((e_{1}+\sigma,\pi(f_{1,\sigma})),\ldots,(e_{2 nm-1}+\sigma,\pi(f_{2 nm-1},\sigma))\). These novel chains represent the edge decomposition of \(C_{4 nm,2 nm}\) i.e., \(E(\mathbb{C}_{\sigma})=\{(e_{0}+\sigma,\pi(f_{0,\sigma})),(e_{1}+\sigma,\pi(f_ {1,\sigma})),\ldots,\)\((e_{2 nm-1}+\sigma,\pi(f_{2 nm-1},\sigma))\},\sigma\in\mathbb{Z}_{2 nm}\) and \(\cup_{\sigma\in\mathbb{Z}_{2 nm}}E(\mathbb{C}_{\sigma})=E(C_{4 nm,2 nm})\) that can be proved as follows. Let \(\lambda_{i}=f_{i}-e_{i}\), which is unique for every pair \((e_{i},f_{i});i\in\mathbb{Z}_{2 nm}\). For \(u\
eq v\in\mathbb{Z}_{2 nm}\), let \((e_{i}+u  
\(2 nm\) to the second component of the ordered pairs \((e_{0}+\sigma,f_{0}+\sigma),(e_{1}+\sigma,f_{1}+\sigma),\dots,\)\((e_{2 nm-1}+\sigma,f_{2 nm-1}+\sigma),\sigma\in\mathbb{Z}_{2 nm}\) constructs the edge decomposition of \(C_{4 nm,2 nm}\). Hence, \(\cup_{\sigma\in\mathbb{Z}_{2 nm}}E(\mathbb{C}_{\sigma})=E(C_{4 nm,2 nm})\). \(\square\)  
**Theorem 3**. : _There is an edge decomposition of \(C_{4n,2n}\) by the graph \(K_{1,2}\cup\ K_{1,2(n-1)}\) and an edge decomposition of \(C_{4m,2m}\) by the graph \(K_{1,2}\cup\ K_{1,2(m-1)}\), then there is an edge decomposition of \(C_{8 nm,Ann}\) by the graph \(K_{1,4}\cup\ K_{1,4(n-1)}\cup\ K_{1,4(n-1)}\cup\ K_{1,4(n-1)(m-1)};m\geq 1,n\geq 2\)._  
Proof. : The edge set of \(K_{1,2}\cup\ K_{1,2(n-1)}\) can be represented by \((a_{0},b_{0}+2n),(a_{1},b_{1}+2n),\dots,\)\((a_{2n-1},b_{2n-1}+2n)\), where  
\[a_{i}= \begin{cases}0&if\qquad\qquad\qquad i=0,n;\\ n&if\quad 1\leq i\leq n-1,n+1\leq i\leq 2n-1\end{cases} \]  
\[b_{i}=a_{i}+i. \tag{31}\]  
Hence, \(\{b_{i}-a_{i}:i\in\mathbb{Z}_{2n}\}=\mathbb{Z}_{2n}\).