# On Orthogonal Double Covers and Decompositions of Complete Bipartite Graphs by Caterpillar Graphs
## 3 Main Results
: _Let \(y,x\geq 3\) be integers, then the vector \(v(\mathbb{C}_{2}^{1,y-2}(C_{2}(1,x-2)))=(0,1,1,\dots,1,x,x+1,x+1,\dots,x+1, \dots,x+1,x+1,\dots,x+1,x+1,\dots,x+1,x+1,\dots,x+1)\in\mathbb{Z}_{xy}\) is a symmetric starter vector of an \(ODC\) of \(K_{xy,xy}\) by \(\mathbb{C}_{2}^{1,y-2}(C_{2}(1,x-2))\)._  
**Proof. ** From the vector \(v\big{(}\mathbb{C}_{2}^{1,y-2}(C_{2}(1,x-2))\big{)}\), we can define:  
\[v_{i}=\left\{ \begin{array}{ll}0&\text{if}\quad i=0,\text{or}\\ 1&\text{if}\quad 1\leq i<x,\text{or}\\ x&\text{if}\quad i=x,i=2x,i=3x,\ldots,i=yx-x,\quad\text{or}\\ x+1&\text{otherwise}.\end{array} \right. \tag{16}\]  
Hence,  
\[v_{-i}=\left\{ \begin{array}{ll}0&\text{if}\quad i=0,\text{or}\\ x&\text{if}\quad i=x,i=2x,i=3x,\ldots,i=yx-x,\quad\text{or}\\ 1&\text{if}\quad yx-x<i\leq yx,\text{or}\\ x+1&\text{otherwise}.\end{array} \right. \tag{17}\]  
Then,  
\[v_{i}-v_{-i}+i=\left\{ \begin{array}{ll}0&\text{if}\quad i=0,\text{or}\\ i-x&\text{if}\quad 1\leq i<x,\text{or}\\ i&\text{if}\quad x\leq i\leq yx-x,\text{or}\\ i+x&\text{if}\quad yx-x<i<yx.\end{array} \right. \tag{18}\]  
Note that \(\{v_{i}-v_{-i}+i\colon i\in\mathbb{Z}_{yx}\}=\mathbb{Z}_{yx}\). Example 7 is a direct application of Theorem 6. \(\square\)  
**Example 7. **_Let \(y=3\) and \(x=5\); so, the vector \(v(\mathbb{C}_{2}^{1,1}(C_{2}(1,3)))=(0,1,1,1,1,5,6,6,6,6,5,6,\)\(6,6,6)\). Then, there exists an \(ODC\) of \(K_{15,15}\) by \(\mathbb{C}_{2}^{1,1}(C_{2}(1,3))\); see Figure 14._  
In Theorem 7, we construct a symmetric starter of an \(ODC\) of \(K_{2 xy,2 xy}\) by \(K_{y,2x}\) using symmetric vectors. For more information, see Figure 15.  
Figure 14: Symmetric starter of an \(ODC\) of \(K_{15,15}\) by \(\mathbb{C}_{2}^{1,1}(C_{2}(1,3))\). Figure 15: General symmetric starter of an \(ODC\) of \(K_{2 xy,2 xy}\) by \(K_{y,2x}\). **Theorem 7**. : _Let \(y,x\geq 2\) be integers, then the vector \(v(K_{y,2x})=((x-1),(x-2),(x-3),\)\(\ldots,0,(2x-1),(2x-2),(2x-3),\ldots,x,\ldots,(x-1),(x-2),(x-3),\ldots,0,(2x-1),(2x -2),\)\((2x-3),\ldots,x)\in\mathbb{Z}_{2 yx}^{2 yx}\), is a symmetric starter vector of an ODC of \(K_{2 xy,2 xy}\) by \(K_{y,2x}\)._  
**Proof. ** From the vector \(v(K_{y,2x})\), we can define:  
\[v_{i}=\left\{ \begin{array}{ll}x-1&\text{if}\quad i=0,i=2x,\ldots,\quad i=2 yx -2x,\text{or}\\ x-2&\text{if}\quad i=1,i=2x+1,\ldots,\quad i=2 yx-2x+1,\text{or}\\ \vdots&\vdots&\vdots\\ 0&\text{if}\quad i=x-1,i=3x-1,\ldots,\quad i=2 yx-x-1,\text{or}\\ 2x-1&\text{if}\quad i=x,i=3x,\ldots,\quad i=2 yx-x,\text{or}\\ 2x-2&\text{if}\quad i=x+1,i=3x+1,\ldots,i=2 yx-x+1,\text{or}\\ \vdots&\vdots&\vdots\\ x&\text{if}\quad i=2x-1,i=4x-1,\ldots,\quad i=2 yx-1.\end{array} \right. \tag{19}\]  
Hence,  
\[v_{-i}=\left\{ \begin{array}{ll}x&\text{if}\quad i=1,i=1+2x,\ldots,\quad i=2 yx-2x+1,\text{or}\\ x+1&\text{if}\quad i=2,i=2+2x,\ldots,\quad i=2 yx-2x+2,\text{or}\\ \vdots&\vdots&\vdots\\ 2x-1&\text{if}\quad i=x,i=3x,\ldots,\quad i=2 yx-x,\text{or}\\ 0&\text{if}\quad i=x+1,i=3x+1,\ldots,\quad i=2 yx-x+1,\text{or}\\ 1&\text{if}\quad i=x+1,i=3x+1,\ldots,i=2 yx-x+2,\text{or}\\ \vdots&\vdots&\vdots\\ x-1&\text{if}\quad i=2x,i=4x,\ldots,\quad i=2 yx.\end{array} \right. \tag{20}\]  
Then,  
\[v_{i}-v_{-i}+i=\left\{ \begin{array}{ll}0&\text{if}\quad i=0,\text{or}\\ i-2&\text{if}\quad i=1,i=1+2x,\ldots,\quad i=2 yx-2x+1,\text{or}\\ i-4&\text{if}\quad i=2,i=2+2x,\ldots,\quad i=2 yx-2x+2,\text{or}\\ \vdots&\vdots&\vdots\\ i-2x+2&\text{if}\quad i=x-1,i=3x-1,\ldots,\quad i=2 yx-x-1,\text{or}\\ i&\text{if}\quad i=x,i=2x,\ldots,\quad i=2 yx-x,\text{or}\\ i+2x-2&\text{if}\quad i=x+1,i=3x+1,\ldots,\quad i=2 yx-x+1,\text{or}\\ i+2x-4&\text{if}\quad i=x+2,i=3x+2,\ldots,\quad i=2 yx-x+2,\text{or}\\ \vdots&\vdots&\vdots\\ i+2&\text{if}\quad i=2x-1,i=4x-1,\ldots,\quad i=2 yx-1.\end{array} \right. \tag{21}\]  
Note that \(\left\{v_{i}-v_{-i}+i:i\in\mathbb{Z}_{2 yx}\right\}=\mathbb{Z}_{2 yx}\). \(\square\)  
Example 8 is a direct application of Theorem 7. **Example 8**. : _Let \(y=2\) and \(x=4\); so, the vector \(v(K_{2,8})=(3,2,1,0,7,6,5,4,3,2,1,0,7,6,5,4)\).