# An Algorithm for Fast Multiplication of Kaluza Numbers
## 2 Preliminary Remarks
\begin{bmatrix}\mathbf{I}_{N/2}& \mathbf{I}_{N/2}\\ \mathbf{I}_{N/2}&-\mathbf{I}_{N/2}\end{bmatrix}\begin{bmatrix}\frac{1}{2}( \mathbf{A}_{N/2}+\mathbf{B}_{N/2})&\mathbf{0}_{N/2}\\ \mathbf{0}_{N/2}&\frac{1}{2}(\mathbf{A}_{N/2}-\mathbf{B}_{N/2})\end{bmatrix} \begin{bmatrix}\mathbf{I}_{N/2}&\mathbf{I}_{N/2}\\ \mathbf{I}_{N/2}&-\mathbf{I}_{N/2}\end{bmatrix} , \tag{9}\]  
where \(\mathbf{A}_{N/2}\), \(\mathbf{B}_{N/2}\) are some matrices. Therefore, we can rewrite (6)-(8) as:  
\[\mathbf{\tilde{B}}_{16}^{(-)}= \begin{bmatrix}\mathbf{I}_{8}&\mathbf{I}_{8}\\ \mathbf{I}_{8}&-\mathbf{I}_{8}\end{bmatrix}\begin{bmatrix}\frac{1}{2}\mathbf{ B}_{8}^{(0+)}&\mathbf{0}_{8}\\ \mathbf{0}_{8}&\frac{1}{2}\mathbf{B}_{8}^{(0-)}\end{bmatrix}\begin{bmatrix} \mathbf{I}_{8}&\mathbf{I}_{8}\\ \mathbf{I}_{8}&-\mathbf{I}_{8}\end{bmatrix} ,\]  
\[\mathbf{\tilde{B}}_{16}^{(+)}= \begin{bmatrix}\mathbf{I}_{8}&\mathbf{I}_{8}\\ \mathbf{I}_{8}&-\mathbf{I}_{8}\end{bmatrix}\begin{bmatrix}\frac{1}{2}\mathbf{ B}_{8}^{(1+)}&\mathbf{0}_{8}\\ \mathbf{0}_{8}&\frac{1}{2}\mathbf{B}_{8}^{(1-)}\end{bmatrix}\begin{bmatrix} \mathbf{I}_{8}&\mathbf{I}_{8}\\ \mathbf{I}_{8}&-\mathbf{I}_{8}\end{bmatrix} ,\]  
where:  
\[\mathbf{B}_{8}^{(0+)}=\mathbf{B}_{8}^{(0)}+\mathbf{B}_{8}^{(1)}, \tag{10}\]  
\[\mathbf{B}_{8}^{(0-)}=\mathbf{B}_{8}^{(0)}-\mathbf{B}_{8}^{(1)}, \tag{11}\]  
\[\mathbf{B}_{8}^{(1+)}=\mathbf{B}_{8}^{(2)}+\mathbf{B}_{8}^{(3)}, \tag{12}\]  
\[\mathbf{B}_{8}^{(1-)}=\mathbf{B}_{8}^{(2)}-\mathbf{B}_{8}^{(3)}, \tag{13}\]  
\[\mathbf{B}_{8}^{(2+)}=\mathbf{B}_{8}^{(4)}+\mathbf{B}_{8}^{(5)}, \tag{14}\]  
\[\mathbf{B}_{8}^{(2-)}=\mathbf{B}_{8}^{(4)}-\mathbf{B}_{8}^{(5)}.\Combining partial decompositions in a single procedure we can rewrite procedure, (3) as following:  
\[\mathbf{Y}_{32\times 1}=\mathbf{M}_{32}^{(r)}\mathbf{T}_{32\times 48}\mathbf{S}_{ 48}^{(r)}\mathbf{W}_{48}^{(1)}\mathbf{\tilde{B}}_{48}\mathbf{W}_{48}^{(1)} \mathbf{S}_{48}^{(c)}\mathbf{T}_{48\times 32}\mathbf{M}_{32}^{(c)}\mathbf{X}_{32 \times 1},\]  
where  
\[\mathbf{\tilde{B}}_{48}=\text{quasidiag}\bigg{(}\frac{1}{2}\mathbf{B}_{8}^{(0 +)},\frac{1}{2}\mathbf{B}_{8}^{(0-)},\frac{1}{2}\mathbf{B}_{8}^{(1+)},\frac{1} {2}\mathbf{B}_{8}^{(1-)},\frac{1}{2}\mathbf{B}_{8}^{(2+)},\frac{1}{2}\mathbf{B }_{8}^{(2-)}\bigg{)},\]  
\[\mathbf{S}_{48}^{(r)}=\mathbf{I}_{3}\otimes\mathbf{S}_{16}^{(r)},\]  
\[\mathbf{S}_{48}^{(c)}=\mathbf{I}_{3}\otimes\mathbf{S}_{16}^{(c)},\]  
\[\mathbf{W}_{48}^{(1)}=\mathbf{I}_{3}\otimes\mathbf{H}_{2}\otimes\mathbf{I}_{8},\]  
\(\mathbf{H}_{2}\) is the order 2 Hadamard matrix, i.e. :  
\[\mathbf{H}_{2}= \begin{bmatrix}1&1\\ 1&-1\end{bmatrix} .\]  
Introducing the following notation:  
\[ \begin{array}{llllll}d_{0}=c_{0}+c_{8},&d_{1}=c_{1}+c_{9},&d_{2}=c_{2}+c_{ 10},&d_{3}=c_{11}-c_{3},\\ d_{4}=c_{4}-c_{12},&d_{5}=c_{5}-c_{13},&d_{6}=c_{14}-c_{6},&d_{7}=c_{7}+c_{15},\\ d_{8}=c_{2}-c_{10},&d_{9}=c_{3}+c_{11},&d_{10}=c_{0}-c_{8},&d_{11}=c_{9}-c_{1},\\ d_{12}=c_{6}+c_{14},&d_{13}=c_{7}-c_{15},&d_{14}=c_{4}+c_{12},&d_{15}=c_{5}+c_{ 13},\\ d_{16}=c_{16}+c_{24},&d_{17}=c_{17}+c_{25},&d_{18}=c_{18}+c_{26},&d_{19}=c_{27}- c_{19},\\ d_{20}=c_{20}-c_{28},&d_{21}=c_{21}-c_{29},&d_{22}=c_{30}-c_{22},&d_{23}=c_{23} +c_{31},\\ d_{24}=c_{26}-c_{18},&d_{25}=c_{19}+c_{27},&d_{26}=c_{24}-c_{16},&d_{27}=c_{17}- c_{25},\\ d_{28}=c_{22}+c_{30},&d_{29}=c_{31}-c_{23},&d_{30}=c_{20}+c_{28},&d_{31}=c_{21}+c_{ 29}.\end{array} \]  
to (10)-(13), we obtain:  
\[\mathbf{B}_{8}^{(0+)}= \begin{bmatrix}d_{0}&d_{1}&d_{2}&d_{3}&d_{4}&d_{5}&d_{6 }&d_{7}\\ d_{1}&d_{0}&d_{3}&d_{2}&d_{5}&d_{4}&d_{7}&d_{6}\\ d_{8}&d_{9}&d_{10}&d_{11}&d_{12}&d_{13}&d_{14}&d_{15}\\ -d_{9}&d_{8}&d_{11}&d_{10}&d_{13}&d_{12}&d_{15}&d_{14}\\ d_{16}&d_{17}&d_{18}&d_{19}&d_{20}&d_{21}&d_{22}&d_{23}\\ d_{17}&d_{16}&d_{19}&d_{18}&d_{21}&d_{20}&d_{23}&d_{22}\\ d_{24}&d_{25}&d_{26}&d_{27}&d_{28}&d_{29}&d_{30}&d_{31}\\ d_{25}&d_{24}&d_{27}&d_{26}&d_{29}&d_{28}&d_{31}&d_{30}\end{bmatrix} ,\]  
\[\mathbf{B}_{8}^{(0-)}=