# On Orthogonal Double Covers and Decompositions of Complete Bipartite Graphs by Caterpillar Graphs
## 3 Main Results
\begin{array}{lll}0&\text{if}&i=0,2,4,\ldots,2x-2,\text{or}\\ 1&\text{if}&i=1,3,5,\ldots,2x-1,\text{or}\\ 2x&\text{if}&i=2x,2x+2,2x+4,\ldots,2 yx-2,\text{or}\\ 2x+1&\text{if}&i=2x+1,2x+3,2x+5,\ldots,2 yx-1.\end{array} \right. \tag{7}\]  
Hence,  
\[v_{-i}=\left\{ \begin{array}{lll}2x+1&\text{if}&i=1,3,5,\ldots,2 yx-2x-1,\text {or}\\ 2x&\text{if}&i=2,4,6,\ldots,2 yx-2x,\text{or}\\ 1&\text{if}&i=2 yx-2x+1,2 yx-2x+3,\ldots,2 yx-1,\text{or}\\ 0&\text{if}&i=2 yx-2x+2,2 yx-2x+4,\ldots,2 yx.\end{array} \right. \tag{8}\]  
Then,  
\[v_{i}-v_{-i}+i=\left\{ \begin{array}{lll}0&\text{if}&i=0,\text{or}\\ i-2x&\text{if}&i=1,2,3,\ldots,2x-1,\text{or}\\ i&\text{if}&i=2x,2x+1,2x+2,\ldots,2x(y-1),\text{or}\\ i+2x&\text{otherwise.}\end{array} \right. \tag{9}\]  
Note that \(\left\{v_{i}-v_{-i}+i:\,i\in\mathbb{Z}_{2 yx}\right\}=\mathbb{Z}_{2 yx}\). \(\square\)  
**Example 4**. : _Let \(y=3\) and \(x=3\), so the vector \(v(\mathbb{C}_{2}^{1,1}(K_{2,3}))=(0,1,0,1,0,1,6,7,6,7,6,7,6,7,\)\(6,7,6,7)\). Then, there exists an ODC of \(K_{18,18}\) by \(\mathbb{C}_{2}^{1,1}(K_{2,3})\); see Figure 9._  
**Theorem 4**. : _For all integers \(x\) and \(y\geq 3\), and \(gcd(x,3)=1\), then the vector \(v(x\mathbb{C}_{2}^{1,y-2}(K_{2,2}))=(0,1,2,\ldots,2x-1,0,1,2,\ldots,2x-1,4x,4x +1,4x+2,\ldots,6x-1,\ldots,4x,4x+1,4x+2,\ldots,\)\(6x-1)\in\mathbb{Z}_{4 xy}^{4 xy}\) is a symmetric starter vector of an ODC of \(K_{4 xy,4 xy}\) by \(x\mathbb{C}_{2}^{1,y-2}(K_{2,2})\)._  
In Theorem 5, we construct a symmetric starter of an \(ODC\) of \(K_{xy,xy}\) by \(\frac{x}{2}\mathbb{C}_{2}^{1,y-2}(P_{3})\) using a symmetric vector. For more illustration, see Figure 11.  
**Theorem 5**. : _Let \(y\geq 3\) and \(x\equiv 2,4\) mod \(6\) be integers, then, the vector \(v\Big{(}\frac{x}{2}\mathbb{C}_{2}^{1,y-2}(P_{3})\Big{)}=(0,1,2,\ldots,x-1,x,x+1, x+2,\ldots,2x-1,\ldots,x,x+1,x+2,\ldots,2x-1)\in\mathbb{Z}_{yx}^{yx}\) is a symmetric starter vector of an \(ODC\) of \(K_{xy,xy}\) by \(\frac{x}{2}\mathbb{C}_{2}^{1,y-2}(P_{3})\)._  
**Proof. ** From the vector \(v\Big{(}\frac{x}{2}\mathbb{C}_{2}^{1,y-2}(P_{3})\Big{)}\), we can define:  
\[v_{i}=\left\{ \begin{array}{ll}0&\text{if}\quad i=0,\text{or}\\ 1&\text{if}\quad i=1,\text{or}\\ \vdots&\vdots\quad\vdots\\ x-1&\text{if}\quad i=x-1,\text{or}\\ x&\text{if}\quad i=x,i=2x,\ldots,\quad i=(y-1)x,\text{or}\\ x+1&\text{if}\quad i=x+1,i=2x+1,\ldots,i=(y-1)x+1,\text{or}\\ \vdots&\vdots\quad\vdots\\ 2x-1&\text{if}\quad i=2x-1,i=3x-1,\ldots,\quad i=yx-1.\end{array} \right. \tag{13}\]  
Hence,  
\[v_{-i}=\left\{ \begin{array}{ll}2x-1&\text{if}\quad i=1,i=x+1,\ldots,\quad i =yx-2x+1,\text{or}\\ 2x-2&\text{if}\quad i=2,i=x+2,\ldots,\quad i=yx-2x+2,\text{or}\\ \vdots&\vdots\\ x&\text{if}\quad i=x,i=2x,\ldots,\quad i=yx-x,\text{or}\\ x-1&\text{if}\quad i=yx-x+1,\text{or}\\ x-2&\text{if}\quad i=yx-x+2,\text{or}\\ \vdots&\vdots\\ 0&\text{if}\quad i=yx.\end{array} \right. \tag{14}\]Then,  
\[v_{i}-v_{-i}+i=\left\{ \begin{array}{lll}0&\text{if}&i=0,\text{or}\\ 3-2x&\text{if}&i=1,\text{or}\\ 6-2x&\text{if}&i=2,\text{or}\\ \vdots&\vdots&\vdots\\ i-2&\text{if}&i=x-1,\text{or}\\ i&\text{if}&i=x,i=2x,\dots,\quad i=yx-x,\text{or}\\ i-x+2&\text{if}&i=x+1,i=2x+1,\dots,\quad i=yx-2x+1,\text{or}\\ i-x+4&\text{if}&i=x+2,i=2x+2,\dots,\quad i=yx-2x+2,\text{or}\\ \vdots&\vdots&\vdots\\ i+x-2&\text{if}&i=2x-1,i=3x-1,\dots,\quad i=yx-x-1,\text{or}\\ i+2&\text{if}&i=yx-x+1,\text{or}\\ i+4&\text{if}&i=yx-x+2,\text{or}\\ \vdots&\vdots&\vdots\\ i+2x-2&\text{if}&i=yx-1.\end{array} \right. \tag{15}\]  
Note that \(\{v_{i}-v_{-i}+i:\,i\in\mathbb{Z}_{yx}\}=\mathbb{Z}_{yx}\). \(\square\)  
Example 6 is a direct application of Theorem 5. **Example 6**. : _Let \(y=4\) and \(x=4\); so, the vector \(v(2\mathbb{C}_{2}^{1,2}(P_{3}))=(0,1,2,3,4,5,6,7,4,5,6,7,4,5,6,7)\). Then, there exists an \(ODC\) of \(K_{16,16}\) by \(2\mathbb{C}_{2}^{1,2}(P_{3})\); see Figures 12 and 13._  
**Theorem 6**.