## 2 Three-Dimensional Ray-Tracing Model
### Path Correction of the Multipath Transmission Model
The support vector machine algorithm constructs the optimal classification plane based on linear separability and limited sample information, separating samples of different classifications while considering model complexity and accuracy [20; 21; 27; 28]. As shown in Figure 3, the sample data to be segmented are divided into two parts, L1 and L2, with the distance between them called the classification margin. The optimal classification line can be obtained when the classification margin is maximized. When extended to high-dimensional space, it becomes the optimal classification hyperplane. For a linearly separable problem, let the training samples be represented as \(D=\left\{(\vec{x_{1}},y_{1}),(\vec{x_{2}},y_{2})\cdots(\vec{x_{i}},y_{i}),i=1,2,3\cdots m\right\}\), \(y\in\left\{+1,-1\right\}\). Here, \(\vec{x_{i}}\) represents the \(i\)-th sample feature variable, and \(\vec{y_{i}}\) is the corresponding desired output. Correspondingly, the support vector classification interval C=\(2/\left\|\vec{\omega}\right\|\) is required to accommodate two different categories, \(\vec{\omega}\) is hyperplanar normal, and the objective hyperplane requires minimizing \(\left\|\vec{\omega}\right\|^{2}/2\) by maximizing the support vector interval. This leads to a quadratic programming problem, namely:  
\[\left\{ \begin{array}{l}\min\frac{1}{2}\left\|\vec{\omega}\right\|^{2}\\ s.t.y_{i}(\vec{\omega}\vec{x}\vec{x}+b)\geq 1,i=1,2,3\cdots m\end{array} \right. \tag{16}\]  
Figure 3: SVM hyperplane partition. (Orange dots and green squares are different samples. L, L1 and L2 are classification lines. C is classification margin). where \(b\) is the intercept of the hyperplane. By introducing the Lagrange function transformation, we can obtain:  
\[L(\overrightarrow{\omega},b,\overrightarrow{a})=\frac{1}{2}\left\|\overrightarrow {\omega}\right\|^{2}-\sum\limits_{i=1}^{m}\alpha_{i}\left[y_{i}( \overrightarrow{\omega}^{T}\overrightarrow{x_{i}}+b)-1\right] \tag{17}\]  
In Equation (19), \(\overrightarrow{\alpha}=(\alpha_{1},\alpha_{2},\cdots\alpha_{m})^{T}\) represents the Lagrange multiplier, and \(L(\overrightarrow{\omega},b,\overrightarrow{\alpha})\) is a convex quadratic function associated with \(\overrightarrow{\omega}\). We differentiate with respect to \(\overrightarrow{\omega}\) and b, and substitute them into the Lagrangian expression, then simplify it. \[L(\overrightarrow{\omega},b,\overrightarrow{\alpha})=\sum\limits_{i=1}^{m} \alpha_{i}-\frac{1}{2}\sum\limits_{i=1}^{m}\sum\limits_{j=1}^{m}y_{i}y_{j} \alpha_{i}\alpha_{j}\overrightarrow{x_{i}}^{T}\overrightarrow{x_{j}} \tag{18}\]  
The corresponding duality problem is obtained through the change of Lagrange function:  
\[\left\{ \begin{array}{l}\max\!L(\overrightarrow{\omega},b,\overrightarrow{ \alpha})=\sum\limits_{i=1}^{m}\alpha_{i}-\frac{1}{2}\sum\limits_{i=1}^{m} \sum\limits_{j=1}^{m}y_{i}y_{j}\alpha_{i}\alpha_{j}\overrightarrow{x_{i}}^{ T} \overrightarrow{x_{j}}\\ s.t.\sum\limits_{i=1}^{m}\alpha_{i}y_{i}=0,\alpha\geq 0,i=1,2,3,\cdots,m\end{array} \right. \tag{19}\]  
Utilizing quadratic programming to solve this problem, assuming \(\alpha_{i}^{\overrightarrow{*}}=(\alpha_{1}^{*},\alpha_{2}^{*},\cdots\alpha_ {m}^{*})^{T}\) as the global optimal solution, we can obtain:  
\[\left\{ \begin{array}{l}\overrightarrow{\omega^{*}}=\sum\limits_{i=1}^{m} \alpha_{i}^{\overrightarrow{*}}y_{i}\overrightarrow{x_{i}}\\ b^{*}=y_{i}-\sum\limits_{i=1}^{m}\alpha_{i}^{\overrightarrow{*}}y_{i}( \overrightarrow{x_{i}},\overrightarrow{x_{j}})\\ b^{*}=-(\omega^{*\sum\limits_{i=1}^{m}\alpha_{i}^{\overrightarrow{*}}\overrightarrow {x_{i}})/(2\sum\limits_{i=1}^{m}\alpha_{i}^{*})}\end{array} \right.