## 4 CSH-SVM Classification Algorithm
### CSH-SVM Improvement Algorithm  
Based on the method proposed in the literature [15], when a sample satisfies multiple hyperplanes at the same time, it can be sorted by the distance between the sample and the hyperplane, as shown in Figure 8. This process can find which distance between the sample point and its actual category under the hyperplane is the maximum to be considered part of that category. This method is targeted to special samples without changing the SVM model, which is difficult to realize in practical production and application. In this paper, this was improved by finding a new hyperplane to divide the transition area by the original hyperplane. In the high-dimensional space, any two hyperplanes must intersect and produce an angle. It is because these angles exist that some samples meet the hyperplane equations on both sides at the same time. By finding the central partition in the two hyperplanes, we can accurately determine the specific category to which the sample belongs, as shown in Figure 8.  
Take the three-dimensional data as an example: the intersection of two hyperplanes must produce an angle, and there is a certain connection between the angle of the two planes and the normal vectors of the two planes, as shown in Figure 9. The two hyperplane equations are as follows:  
\[\mathrm{OH}_{1}:\omega_{1}{}^{\mathrm{T}}\mathrm{x}+\mathrm{b}_{1}=0, \tag{15}\]  
\[\mathrm{OH}_{2}:\omega_{2}{}^{\mathrm{T}}\mathrm{x}+\mathrm{b}_{2}=0, \tag{16}\]  
Figure 8: (**a**) Relationship between samples in the transition area and the corresponding category hyperplane; (**b**) Relationship between the improved hyperplane and the corresponding category samples. By the definition of a hyperplane, \(\omega_{1}\) and \(\omega_{2}\) are the normal vectors of the two original hyperplanes \(OH_{1}\) and \(OH_{2}\). The angle formed by the intersection of \(\omega_{1}\) and \(\omega_{2}\) is \(\theta\), which is the same as the angle formed by the two hyperplanes \(OH_{1}\) and \(OH_{2}\). When \(0>\theta\geq\frac{\pi}{2}\), the intersection of \(\omega_{1}\) and \(\omega_{2}\) gives \(\frac{\omega_{1}}{||\omega_{1}||}\) and \(\frac{\omega_{2}}{||\omega_{2}||}\) when the two vectors have the same length and the same direction as \(\omega_{1}\) and \(\omega_{2}\). \(\frac{\omega_{1}}{||\omega_{1}||}\) and \(\frac{\omega_{2}}{||\omega_{2}||}\) The angle parallels of the two vectors in the plane where they are located are perpendicular to the improved hyperplane \(OH_{12}\), thus \(\frac{\omega_{1}}{||\omega_{1}||}\) and \(\frac{\omega_{2}}{||\omega_{2}||}\) are added to the vector \(\omega_{12}\), which is the normal vector of the improved hyperplane \(OH_{12}\). \[\omega_{12}=\frac{\omega_{1}}{||\omega_{1}||}+\frac{\omega_{2}}{||\omega_{2} ||}, \tag{17}\]  
When \(-\frac{\pi}{2}>\theta\geq 0\), \(\frac{\omega_{1}}{||\omega_{1}||}\) and \(\frac{\omega_{2}}{||\omega_{2}||}\) in the plane of the two vectors' angle bisection are parallel to the improved hyperplane, the size of the angle formed by the two planes should be complementary to the two vectors' angle. Thus, the improved hyperplane \(OH_{12}\) normal vector \(\omega_{12}\) and \(\omega_{1}\) and \(\omega_{2}\) can be obtained by calculating the relationship as follows:  
\[\omega_{12}=\frac{\omega_{1}}{||\omega_{1}||}-\frac{\omega_{2}}{||\omega_{2} ||}, \tag{18}\]  
The improved hyperplane \(OH_{12}\) has a common solution \(X_{0}\) with the hyperplanes \(OH_{1}\) and \(OH_{2}\), and \(X_{0}\) is a nonzero vector with the same characteristic dimension. The solution of \(X_{0}\) by the original hyperplane equations \(OH_{1}\) and \(OH_{2}\) is as follows:  
\[X_{0}=\frac{b_{2}-b_{1}}{(\omega_{1}-\omega_{2})^{T}} \tag{19}\]  
An improved construction of the hyperplane equation of \(OH_{12}\) is as follows:  
\[OH_{12}:\omega_{12}{}^{T}x+b_{12}=0 \tag{20}\]  
\(X_{0}\) is a special solution of the improved hyperplane equation \(OH_{12}\), which can be obtained by substituting the following into Equation (19):  
Figure 8.0