## 2 Three-Dimensional Single Weight Zero Cross Correlation Code Construction
Table 2 can divide the cross-correlation section of the 3D-SWZCC code into two parts. The upper part contains \(R^{(1)}(g,h,l)\), \(R^{(2)}(g,h,l)\), \(R^{(3)}(g,h,l)\), and \(R^{(4)}(g,h,l)\), whereas the lower part contains \(R^{(5)}(g,h,l)\), \(R^{(6)}(g,h,l)\), \(R^{(7)}(g,h,l)\), and \(R^{(8)}(g,h,l)\). According to Table 2, it is clear that the two parts are identical and that each \(R^{(d)}\) has a value equal to one for one case and a value of zero for the other cases due to the zero cross-correlation property. \begin{table}
\begin{tabular}{c c c c c c c c c} \hline \hline  & & \multicolumn{6}{c}{\(Y_{1}=[0\] & \[0\] & \[1\] & \[Z_{1}=[0\] & \[0\] & \[1\] \\ \hline \(X_{0}= \begin{bmatrix}1\\ 0\end{bmatrix} \) & \(b_{1,2,2}=\begin{bmatrix}0\] & \[0\] & \[0\] & \[0\] & \[0\] & \[1\] & \[0\] & \[0\] & \[1\] \\ \hline \(X_{1}= \begin{bmatrix}0\\ 1\end{bmatrix} \) & \(b_{2,2,2}=\begin{bmatrix}0\] & \[0\] & \[0\] & \[0\] & \[0\] & \[0\] & \[0\] & \[0\] & \[1\] \\ \hline \(Y_{2}=[0\] & \[1\] & \[0]\] & \[0\]0 & \[0\]1 & \[0\]2 \\ \hline \(X_{0}= \begin{bmatrix}1\\ 0\end{bmatrix} \) & \(b_{1,3,2}=\begin{bmatrix}0\] & \[0\] & \[0\] & \[0\] & \[1\] & \[0\] & \[0\] & \[0\] & \[1\] \\ \hline \(X_{1}= \begin{bmatrix}0\\ 1\end{bmatrix} \) & \(b_{2,3,2}=\begin{bmatrix}0\] & \[0\] & \[0\] & \[0\] & \[0\] & \[0\] & \[0\] & \[0\] & \[0\] & \[1\] \\ \hline \(Y_{0}=[1\] & \[0\] & \[0\]0 & \[0\]1 & \[0\]2 & \[0\]3 & \[0\]4 & \[0\]5 & \[0\]6 & \[0\]7 & \[0\]8 \\ \hline \(X_{0}= \begin{bmatrix}1\\ 0\end{bmatrix} \) & \(b_{1,1,3}=\begin{bmatrix}1\] & \[0\] & \[0\] & \[0\] & \[0\] & \[0\] & \[0\] & \[0\] & \[0\] & \[0\] & \[0\] & \[0\]0 \\ \hline \(X_{1}= \begin{bmatrix}0\\ 1\end{bmatrix} \) & \(b_{2,1,3}=\begin{bmatrix}0\] & \[0\] & \[0\] & \[0\] & \[0\] & \[0\] & \[0\] & \[0\] & \[0\] & \[0\] & \[0\] \\ \hline \(Y_{1}= \begin{bmatrix}0\\ 0\end{bmatrix} \) & \(b_{1,2,3}=\begin{bmatrix}0\] & \[0\] & \[1\] & \[0\] & \[0\] & \[0\] & \[0\] & \[0\] & \[1\] & \[0\] \\ \hline \(X_{1}= \begin{bmatrix}0\\ 1\end{bmatrix} \) & \(b_{2,2,3}=\begin{bmatrix}0\] & \[0\] & \[0\] & \[0\] & \[0\] & \[0\] & \[0\] & \[0\] & \[1\] & \[0\] \\ \hline \(Y_{2}=[0\] & \[1\] & \[0\]0 & \[0\]1 & \[0\]2 & \[0\]3 & \[0\]4 & \[0\]5 & \[0\]6 & \[0\]7 \\ \hline \(X_{0}=\begin{bmatrix}  
\[A_{g,h,l}=\left[ \begin{array}{cccccccccccc}b_{1,1,1}&b_{1,2,1}&\ldots&b_{1,L_{2,1}}&b_{1,1,2}&b_{1,2,2}&\ldots&b_{1,L_{2,2}}&\ldots&b_{1,1,L_{3}}&b_{1,2,L_{3}} &\ldots&b_{1,L_{2,L_{3}}}\\ b_{2,1,1}&b_{2,2,1}&\ldots&b_{2,N,1}&b_{2,1,2}&b_{2,2,2}&\ldots&b_{2,L_{2,2}}& \ldots&b_{2,L_{1,L_{3}}}&b_{2,2,L_{3}}&\ldots&b_{2,L_{2,L_{3}}}\\ &&&&&&&&\vdots\vdots\vdots\vdots\\ b_{L_{1},1,1}&b_{L_{1},2,1}&\ldots&b_{L_{1},L_{2},1}&b_{L_{1},2}&b_{L_{1},2 }&\ldots&b_{L_{1},L_{2},2}&\ldots&b_{L_{1},1,L_{3}}&b_{L_{1},2,L_{3}}&\ldots&b _{L_{1},L_{2},L_{3}}\end{array} \right] \tag{10}\]  
Actually, the 'one' case comes from multiplying the code weights (\(w_{1}\cdot w_{2}\cdot w_{3}\)), and although the SWZCC code is characterized by the code weight, it is always fixed and equal to one, so the cross-correlation results are one. To construct the receiver, we can use any case. According to the upper and lower parts, we can write the two equations of cross-correlation as:  
\[\mathcal{R}^{(1)}(g,h,l)=\left\{ \begin{array}{ll}1&if\ g=0\cap h=0\cap l=0\\ 0&otherwise\end{array} \right. \tag{11}\]  
\[\mathcal{R}^{(1)}(g,h,l)=\left\{ \begin{array}{ll}1&if\ g=0\cap h=0\cap l=0\\ 0&otherwise\end{array} \right. \tag{12}\]