## 3 Detection of Cracks on Asphalt Pavement of Expressway
\begin{table}
\begin{tabular}{c c c c c} \hline \hline
**Classification** & **Classification** & **Appearance Description** & **Grading Index** & **Unit of Measurement** \\ \hline \multirow{3}{*}{Transverse crack} & light & The seam wall does not fall off or falls off & \multirow{3}{*}{Seam width \(\leq\) 5 mm} & \multirow{3}{*}{M2(length \(\times\) 0.2 m)} \\  & heavy & The seam wall falls off more, and there & & \\  & & are many branch joints & & \\  & & The seam wall does not fall off or falls off & & \\ Longitudinal crack & light & less, and the buttless seam is less & & \\  & heavy & The seam wall falls off more, and there & & \\  & & are many branch joints & & \\  & & fine seam, no shedding or less shedding, & & \\ Reticular fissure & light & Large block size area & & \\  & & heavy & The seam is wide, shedding more, and & \\  & & the block is relatively broken & & \\  & & Initial cracking, fine seam, no falling off & & \\ Craze & light & or deformation & & \\  & in & Obvious cracks, wide seams, & & \\  & & slight deformation & & \\  & heavy & Broken blocks, wide seams, falling off & & \\  & & and serious deformation & & \textless{} 30\text{ cm}\) & \\ \hline \hline \end{tabular}
\end{table} 
Table 1: Classification and damage classification of pavement cracks. The proposed method selects the above three features as the input vectors of the support vector machine [21; 22] to detect the crack damage of expressway asphalt pavement. Assume that there is a linearly separable set of crack image samples in the feature space \(T=\{(x_{1},y_{1}),(x_{2},y_{2}),\cdots,(x_{n},y_{n})\}\), where \(x_{i}\in R^{d}\) is the first in the space \(i\) fracture samples of known categories, \(y_{i}\), \(i=1,2,\cdots,n\) the category properties of \(x_{i}\). According to the idea of using straight lines to divide categories in two-dimensional space, the straight line for classification in feature space is the classification hyperplane [23; 24], which can be expressed as \(w^{T}x+b=0\), where \(w\) is an \(n\) dimensional vector, \(b\) is the offset. For \(|w^{T}x+b|=1\), the sample points of the condition are closer to the ideal plane than other sample points, so the interval between the sample and the hyperplane is \(2/||w||\). The classification hyperplane is shown in Figure 2:  
In order to find the ideal plane for classification, the gap between the crack sample and the plane should be maximized, that is \(2/||w||\) maximum [25], which can be equivalently expressed as \(||w||_{2}/2\) minimum. Therefore, the solution for the optimal classification hyperplane \(\varphi(w)\) can be calculated using Formula (22) as follows:  
\[\left\{ \begin{array}{ll}\min&\varphi(w)=||w||_{2}/2\\ s.t.&y_{i}(w^{T}x_{i}+b)\geq 1,i=1,2,\cdots,n\end{array} \right. \tag{22}\]  
where \(w\) corresponds to the slope term in the straight line, which is the normal vector in high-dimensional space; \(b\) is the intercept term of the hyperplane in high-dimensional space. The above is the problem of calculating extreme values in convex optimization, which can be solved by the Lagrange multiplier method. Using the Lagrange multiplier method, the functional form of the above equation can be calculated using Formula (23):  
\[L(w,b,\alpha)=\frac{||w||^{2}}{2}-\sum_{i=1}^{n}\alpha_{i}[y_{i}(w^{T}x_{i}+b) -1] \tag{23}\]  
Figure 2: Classification hyperplane. Among them, \(\alpha_{i}>0\), as a Lagrange multiplier, to further solve the problem, first calculates the partial derivatives \(w\) and \(b\) and then makes the derivative result equal to 0. Further results can be obtained, as shown in Formula (24):  
\[\left\{ \begin{array}{l}w=\sum\limits_{i=1}^{n}\alpha_{i}y_{i}x_{i}\\ \sum\limits_{i=1}^{n}\alpha_{i}y_{i}=0\end{array} \right.