Study on Skills Enhancement Strategy Based on Sensing Technology in College Sports Football Teaching

The soccer movement process contains extremely complex human leg posture, before real-time detection of soccer leg posture, it is necessary to clarify the classification of soccer leg posture, scientific and reasonable leg posture classification helps to accurately detect the soccer leg posture. According to the limb state of the soccer player during the movement process, the soccer leg posture is mainly divided into the movement state and the static state. The overall structure of soccer leg posture classification is shown in Figure 1.

Figure 1.

Posture classification of football sportswear

Soccer is specifically divided into nine leg postures: stationary, crossing, jumping, shooting, passing, catching, grabbing, walking and running. The FSR [34] and gyroscope are mounted on the soccer player’s leg, respectively, and the FSR thick-film polymer sensor has a very small size, which is suitable for acquiring tactile signals. The sensors need to be mounted on the gastrocnemius muscle that matches the shape of the leg so that it is in full contact with the soccer playing leg.

The information collected by the gyroscope and FSR is used to initially determine whether the soccer player’s leg is in motion or not, and when the soccer player’s leg is in motion, the time-domain features and frequency-domain features of the data collected by the FSR sensor and the gyroscope sensor are extracted and sent to a support vector machine classifier, which realizes the accurate detection of the soccer player’s leg posture.

When the FSR is subjected to pressure, its own resistance value changes, according to the resistance material difference can be determined by the non-linear relationship between the applied pressure and the resistance value, according to the relationship between the voltage value on both sides of the FSR resistance and the resistance value can be obtained by the pressure value applied to the FSR.

When the FSR is subjected to force application, the display interface is 0~N value, and the pressure on both sides of the FSR is reflected by the displayed value, and the N setting value varies with the difference of processor type, and when the number of processor bits of the FSR sensor is n, the formula of N value can be obtained: (1)N=2n−1$$N = {2^n} - 1$$

The formula for calculating the voltage at the two ends of the FSR can be obtained: (2)V0V1=RR−Rr$$\frac{{{V_0}}}{{{V_1}}} = \frac{R}{{R - {R_r}}}$$

In the formula, V_o and R represent the initial voltage and resistance, R_r and V₁ represent the voltage divider in series with the force sensitive resistor and the maximum value of voltage detected by the processor, respectively, which can be seen in the equation of the relationship between voltage and resistance detected by FSR: (3)R=Vo2n−Vo$$R = \frac{{{V_o}}}{{{2^n} - {V_o}}}$$

Through the above formula to obtain the resistance value of the force sensitive resistor, based on the nonlinear relationship between the resistance R of the force sensitive resistor and the force f to obtain the pressure on the leg of the soccer player, and the collected pressure value extracted features to improve the accuracy of the leg posture detection using a support vector machine classifier.

Dispersion is a measure of the extent to which different observed variables differ at different values. The dispersion is the difference between the sample values of the gyro sensor signals, with ωnx$$\omega_n^x$$ and wn−1x$$w_{n - 1}^x$$ representing the x-axis angular velocity at times n and n − 1, respectively. Dispersion snz$$s_n^z$$ is the formula for the difference between the x-axis angular velocities of the gyro sensors at times n and n − 1: (4)snx=|ωnx−ωn−1x|$$s_n^x = |\omega_n^x - \omega_{n - 1}^x|$$

Considering the FSR sensor and gyroscope sensor [35] data characteristics to facilitate the accurate classification of soccer action leg posture, the dispersion of angular velocity and pressure of each axis of the sensor is represented by sna,sna,sna,sna,snb,snp,snp$$s_n^a,s_n^a,s_n^a,s_n^a,s_n^b,s_n^p,s_n^p$$, and Sna$$S_n^a$$ and Snp$$S_n^p$$ represent the dispersion of gyroscope sensor data at time n and FSR sensor data at time n, respectively, which can be obtained by Eq: (5)Sna=snxa+snya+snza$$S_n^a = s_{{n^x}}^a + s_{{n^y}}^a + s_{{n^z}}^a$$ (6)Snp=snp+snp+snp$$S_n^p = s_n^p + s_n^p + s_n^p$$

Angular velocity and pressure dispersion in the stationary state in the threshold α and threshold α_p, respectively, the sensor data in the soccer player in the soccer movement rapid changes in the degree of difference between the sensor data can be reflected by the degree of dispersion, based on the degree of dispersion to achieve the division of the leg posture of soccer movement.

When the time is n, the soccer player’s leg state is represented by, then β_n is 0 and 1, respectively, represents the stationary state and the movement state, which can be obtained by Eq: (7)βn={ 0,Sna<αa and Snp<αp 1,Sna≥αa or Snp≥αp$${\beta_n} = \left\{ {\begin{array}{*{20}{l}} {0,S_n^a < {\alpha_a}{\text{ and }}S_n^p < {\alpha_p}} \\ {1,S_n^a \geq {\alpha_a}{\text{ or }}S_n^p \geq {\alpha_p}} \end{array}} \right.$$

The dispersion of the FSR sensor and gyroscope sensor data is obtained through the above steps, and the soccer player’s leg movement is judged to be a stationary state and a sports state based on the obtained threshold value.

When the dispersion degree is used to determine that the soccer player’s leg is in motion, the time domain features and the frequency domain features of the soccer player’s leg posture are extracted, and a support vector machine classifier is used to realize the accurate detection of the soccer player’s leg posture.

The soccer leg attitude data collected by FSR and gyroscope mainly include pressure information and angular velocity information. anx,any,anz$$a_n^x,a_n^y,a_n^z$$ and pnx,pny,pnz$$p_n^x,p_n^y,p_n^z$$ are used to represent the angular velocity and pressure of the x, y, z-axis of the nrd sampling point, respectively, and the angular velocity vector sum and pressure vector sum equations of the nth sampling point can be obtained: (8)an=(anx)2+(any)2+(anz)2$${a_n} = \sqrt {{{(a_n^x)}^2} + {{(a_n^y)}^2} + {{(a_n^z)}^2}}$$ (9)pn=(pnx)2+(pny)2+(pnz)2$${p_n} = \sqrt {{{(p_n^x)}^2} + {{(p_n^y)}^2} + {{(p_n^z)}^2}}$$

The pressure vector and angular velocity vector obtained using the above equation and together with the pressure vector and angular velocity vector form a feature matrix of dimension 8. When the number of sampling points is N, one of the samples is the feature matrix of size N×8.

Obtain the real-time detection of soccer leg posture in real time domain feature sampling point mean value: (10)φa=1N∑n=1Nanpn$${\varphi_a} = \frac{1}{N}\sum\limits_{n = 1}^N {{a_n}} {p_n}$$

Real-time detection of soccer leg postures in real-time with time-domain feature sampling point variance formulas: (11)δ2=1N∑n=1N(anpn−φa)2$${\delta^2} = \frac{1}{N}\sum\limits_{n = 1}^N {{{({a_n}{p_n} - {\varphi_a})}^2}}$$

The extracted time-domain features [36] have 4 dimensions each of FSR sensor as well as gyroscope sensor x, y, z-axis vector and mean; 4 dimensions each of FSR sensor as well as gyroscope sensor x, y, z-axis vector and variance, and a total of 16 dimensions of the acquired soccer leg posture time-domain parameters.

The time-domain acquired data were converted to frequency-domain formulas using the Fourier transform principle: (12)Z(n)=∑n=0N−1ane−j2πNn$$Z(n) = \sum\limits_{n = 0}^{N - 1} {{a_n}} {e^{ - j\frac{{2\pi }}{N}n}}$$

Where, Z(n) represents the value of the nnd sampling point in the frequency domain, j represents the frequency domain dimension; e is the exponential function. The Fourier transform peak, i.e., the frequency domain feature of leg posture detection in soccer, is given by Eq: (13)f=HfsN$$f = \frac{{H{f_s}}}{N}$$

Where H and f, respectively, represent the number of sampling points in the frequency domain and the sampling frequency of the data acquisition sensor.

The extracted frequency domain features are pressure and angular velocity sensor x, y, z-axis frequency domain peaks and their corresponding frequency values of 6 dimensions, the pressure vector, angular velocity vector, and the frequency domain peaks and the corresponding peaks corresponding to the frequency of 2 dimensions, containing a total of 16 dimensions of frequency domain features.

The raw signals of 3D acceleration {ω_x, ω_y, ω_z} and 3D angular velocity {a_x, a_y, a_z} are too noisy and repetitive for accurate soccer action recognition and evaluation; for foot-based sport action recognition, in addition to the two important features of angular velocity and acceleration, the rotation of the ankle also plays a crucial role in soccer action recognition and evaluation, and mining the information of the stance angle can obtain features containing more information in the action. In this section, a stance angle model is used to extract more useful features.

The attitude angle consists of three Euler angles: yaw, pitch and roll.

Typically, three-dimensional rotation problems are solved by rotation matrices. Quaternions are used as the quotient of two oriented lines in 3D space to solve the angle problem in the angular trajectory model. A quaternion is represented as p = p_ω + p_xi + p_yj + p_zk.

There is a real part p_ω and three imaginary part parameters p_x, p_y and p_z. To compute the initial attitude angle: (14){ θ=arcsin(ax0) φ=arctanmagy0magx0 ψ=arctan(−ay0az0)$$\left\{ {\begin{array}{*{20}{l}} {\theta = \arcsin ({a_{x0}})} \\ {\varphi = \arctan \frac{{mag{y_0}}}{{mag{x_0}}}} \\ {\psi = \arctan ( - \frac{{{a_{y0}}}}{{{a_{z0}}}})} \end{array}} \right.$$

where a_x0, a_y0 and a_z0 represent the initial acceleration, and θ, ψ and ϕ represent the initial yaw, roll and pitch, respectively. magx₀ and magy₀ are calculated from a_x0, a_y0 and a_z0, as specified in equation (15): (15){ magx0=ax0ay0+ax0az01−2(ay02+az02) magy0=ax0ay0+ay0az01−2(ax02+az02)$$\left\{ {\begin{array}{*{20}{l}} {mag{x_0} = \frac{{{a_{x0}}{a_{y0}} + {a_{x0}}{a_{z0}}}}{{1 - 2(a_{y0}^2 + a_{z0}^2)}}} \\ {mag{y_0} = \frac{{{a_{x0}}{a_{y0}} + {a_{y0}}{a_{z0}}}}{{1 - 2(a_{x0}^2 + a_{z0}^2)}}} \end{array}} \right.$$

The initial four elements are calculated from the initial angles to obtain equation (16): (16)[ p0 p1 p2 p3]=[ sinψ2sinθ2sinφ2+cosψ2cosθ2cosφ2 sinψ2cosθ2cosφ2−cosψ2sinθ2sinφ2 cosψ2sinθ2cosφ2+sinψ2cosθ2sinφ2 cosψ2cosθ2sinφ2−sinψ2sinθ2cosφ2]$$\left[ {\begin{array}{*{20}{c}} {{p_0}} \\ {{p_1}} \\ {{p_2}} \\ {{p_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\sin \frac{\psi }{2}\sin \frac{\theta }{2}\sin \frac{\varphi }{2} + \cos \frac{\psi }{2}\cos \frac{\theta }{2}\cos \frac{\varphi }{2}} \\ {\sin \frac{\psi }{2}\cos \frac{\theta }{2}\cos \frac{\varphi }{2} - \cos \frac{\psi }{2}\sin \frac{\theta }{2}\sin \frac{\varphi }{2}} \\ {\cos \frac{\psi }{2}\sin \frac{\theta }{2}\cos \frac{\varphi }{2} + \sin \frac{\psi }{2}\cos \frac{\theta }{2}\sin \frac{\varphi }{2}} \\ {\cos \frac{\psi }{2}\cos \frac{\theta }{2}\sin \frac{\varphi }{2} - \sin \frac{\psi }{2}\sin \frac{\theta }{2}\cos \frac{\varphi }{2}} \end{array}} \right]$$

The Runge-Kutta method eliminates the complicated process of solving differential equations mainly when the information about the derivatives and initial values of the equations is known, and the method is used to solve the four elements with Eq. (17): (17)[ p0 p1 p2 p3]t+Δt=[ p0 p1 p2 p3]t+Δt2[ −ωx⋅p1−ωy⋅p2−ωz⋅p3 +ωx⋅p0−ωz⋅p2−ωy⋅p3 +ωy⋅p0−ωz⋅p1+ωx⋅p3 −ωz⋅p0+ωy⋅p1−ωx⋅p2]$${\left[ {\begin{array}{*{20}{c}} {{p_0}} \\ {{p_1}} \\ {{p_2}} \\ {{p_3}} \end{array}} \right]_{t + \Delta t}} = {\left[ {\begin{array}{*{20}{c}} {{p_0}} \\ {{p_1}} \\ {{p_2}} \\ {{p_3}} \end{array}} \right]_t} + \frac{{\Delta t}}{2}\left[ {\begin{array}{*{20}{c}} { - {\omega_x} \cdot {p_1} - {\omega_y} \cdot {p_2} - {\omega_z} \cdot {p_3}} \\ { + {\omega_x} \cdot {p_0} - {\omega_z} \cdot {p_2} - {\omega_y} \cdot {p_3}} \\ { + {\omega_y} \cdot {p_0} - {\omega_z} \cdot {p_1} + {\omega_x} \cdot {p_3}} \\ { - {\omega_z} \cdot {p_0} + {\omega_y} \cdot {p_1} - {\omega_x} \cdot {p_2}} \end{array}} \right]$$

where t denotes the current time value and Δt denotes the interval to the next data. The following quaternion is used to solve the problem of attitude angle. Coordinate Transformation Matrix The coordinate transformation matrix D from the object’s coordinate system to the Earth’s coordinate system is known to be: (18)D=[ 100 010 001]+2cosθ2[ 0 −nsinθ2 msinθ2 nsinθ2 0 −lsinθ2 −msinθ2 lsinθ2 0] +2[ −(m2+n2)sin2θ2 lmsin2θ2 nlsin2θ2 lmsin2θ2 −(l2+n2)sin2θ2 mnsin2θ2 nlsin2θ2 mnsin2θ2 −(m2+l2)sin2]$$\begin{array}{l} D = \left[ {\begin{array}{*{20}{c}} {100} \\ {010} \\ {001} \end{array}} \right] + 2\cos \frac{\theta }{2}\left[ {\begin{array}{*{20}{c}} 0&{ - n\sin \frac{\theta }{2}}&{ m\sin \frac{\theta }{2}} \\ {n\sin \frac{\theta }{2}}&0&{ - l\sin \frac{\theta }{2}} \\ { - m\sin \frac{\theta }{2}}&{ l\sin \frac{\theta }{2}}&0 \end{array}} \right] \\ + 2\left[ {\begin{array}{*{20}{c}} { - ({m^2} + {n^2}){{\sin }^2}\frac{\theta }{2}}&{ lm{{\sin }^2}\frac{\theta }{2}}&{ nl{{\sin }^2}\frac{\theta }{2}} \\ {lm{{\sin }^2}\frac{\theta }{2}}&{ - ({l^2} + {n^2}){{\sin }^2}\frac{\theta }{2}}&{ mn{{\sin }^2}\frac{\theta }{2}} \\ {nl{{\sin }^2}\frac{\theta }{2}}&{ mn{{\sin }^2}\frac{\theta }{2}}&{ - ({m^2} + {l^2}){{\sin }^2}} \end{array}} \right] \\ \end{array}$$

m, l, and n refer to the direction vectors projected onto geographic coordinates along the direction of the rotation vector. The rotation of the object corresponds to the rotation of the stem about the axis. The diagonal function of the quaternion is equation (19): (19)cosθ2+(li→+mj→+nk→)sinθ2=cosθ2+u→sinθ2$$\cos \frac{\theta }{2} + (l\vec i + m\vec j + n\vec k)\sin \frac{\theta }{2} = \cos \frac{\theta }{2} + \vec u\sin \frac{\theta }{2}$$

where i→,j→,k→$$\vec i,\vec j,\vec k$$ denotes the unit direction vector in the geographic lattice system has Eq. (20): (20){ p0=cosθ2 p1=lsinθ2 p2=msinθ2 p3=nsinθ2$$\left\{ {\begin{array}{*{20}{l}} {{p_0} = \cos \frac{\theta }{2}} \\ {{p_1} = l\sin \frac{\theta }{2}} \\ {{p_2} = m\sin \frac{\theta }{2}} \\ {{p_3} = n\sin \frac{\theta }{2}} \end{array}} \right.$$

The constructed quaternion describes the problem of fixed-point rotation of matter. Accordingly, the system of objects is formed by a one-time equivalent rotation of the Earth system. Equation (20) is replaced by D₁. (21)D1=[ p02+p12−p22−p32 2(p1p2−p0p3) 2(p0p2+p1p3) 2(p0p3+p1p2) p02−p12+p22−p32 2(p2p3−p0p1) 2(p1p3−p0p2) 2(p0p1+p2p3) p02−p12−p22+p]$${D_1} = \left[ {\begin{array}{*{20}{c}} {p_0^2 + p_1^2 - p_2^2 - p_3^2}&{ 2({p_1}{p_2} - {p_0}{p_3})}&{ 2({p_0}{p_2} + {p_1}{p_3})} \\ {2({p_0}{p_3} + {p_1}{p_2})}&{ p_0^2 - p_1^2 + p_2^2 - p_3^2}&{ 2({p_2}{p_3} - {p_0}{p_1})} \\ {2({p_1}{p_3} - {p_0}{p_2})}&{ 2({p_0}{p_1} + {p_2}{p_3})}&{ p_0^2 - p_1^2 - p_2^2 + p} \end{array}} \right]$$

Let the unit vector in the coordinate system X be (ex₁, ey₁, ez₁)^T and correspond to (ex₂, ey₂, ez₂) in Y. The projection direction of (ex₂, ey₂, ez₂)^T is Cbⁿ. Assuming that there is a vector R with magnitude (x, y, z)^T on X and (x₂, y₂, z₂)^T on Y, the coordinate transformation is: (x, y, z)T = Cbⁿ(x₂, y₂, z₂)^T. The attitude angle transformation matrix is also converted into the form of the coordinate transformation matrix from the object coordinate system to the geographic coordinate system, with Eq. (22) Cbⁿ yielding the following result: (22)Cbn=Rot(z,ψ)Rot(y,θ)Rot(x,φ) =[ cosψ −sinψ 0 cosψ cosψ 0 0 0 1][ 1 0 0 0 cosφ −sinφ 0 sinφ cosφ][ cosθ 0 sinθ 0 1 0 −sinθ 0 cosθ] =[ cosψcosθ −sinψcosφ+cosψsinθsinφ sinψsinφ+cos sinψcosθ cosψcosφ+sinψsinθsinφ −cosψsinφ+sin −sinθ cosθsinφ cosθcos]$$\begin{array}{l} C_b^n = Rot(z,\psi )Rot(y,\theta )Rot(x,\varphi ) \\ = \left[ {\begin{array}{*{20}{c}} {\cos \psi }&{ - \sin \psi }&0 \\ {\cos \psi }&{ \cos \psi }&0 \\ 0&0&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 0&{ \cos \varphi }&{ - \sin \varphi } \\ 0&{ \sin \varphi }&{ \cos \varphi } \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\cos \theta }&0&{ \sin \theta } \\ 0&1&0 \\ { - \sin \theta }&0&{ \cos \theta } \end{array}} \right] \\ = \left[ {\begin{array}{*{20}{c}} {\cos \psi \cos \theta }&{ - \sin \psi \cos \varphi + \cos \psi \sin \theta \sin \varphi }&{ \sin \psi \sin \varphi + \cos } \\ {\sin \psi \cos \theta }&{ \cos \psi \cos \varphi + \sin \psi \sin \theta \sin \varphi }&{ - \cos \psi \sin \varphi + \sin } \\ { - \sin \theta }&{ \cos \theta \sin \varphi }&{ \cos \theta \cos } \end{array}} \right] \\ \end{array}$$

D₁ and Cbⁿ are both matrices that transform the pose of an object’s coordinate system to a geographic coordinate system, i.e., Eq. (23): (23){ 2(p1p3−p0p2)=−sinθ=g1 2(p0p1+p2p3)=sinφcosθ=g2 p02−p12−p22+p32=cosθcosφ=g3 2(p0p3+p1p2)=sinψcosθ=g4 p02+p12−p22−p32=cosψcosθ=g5$$\left\{ {\begin{array}{*{20}{l}} {2({p_1}{p_3} - {p_0}{p_2}) = - \sin \theta = {g_1}} \\ {2({p_0}{p_1} + {p_2}{p_3}) = \sin \varphi \cos \theta = {g_2}} \\ {p_0^2 - p_1^2 - p_2^2 + p_3^2 = \cos \theta \cos \varphi = {g_3}} \\ {2({p_0}{p_3} + {p_1}{p_2}) = \sin \psi \cos \theta = {g_4}} \\ {p_0^2 + p_1^2 - p_2^2 - p_3^2 = \cos \psi \cos \theta = {g_5}} \end{array}} \right.$$

PCA is performed on the data that has been extracted to characterize the variables. The model using PCA can achieve higher accuracy than other nonlinear dimensionality reduction methods. PCA is essentially a basis transformation that maximizes the variance of the transformed data, i.e., it minimizes the variance between the axes (principal axes) and the points by rotating the axes and translating the origin of the coordinates. Assume that matrix F has dimension n × m, i.e., a total of n samples in m dimensions. F can be decomposed into UΣV^T, where U is of the same size as F. The orthogonal matrix V is m × m, and Σ is a diagonal matrix of the same size as V with YrUΣ_r = F(ΣV^T)⁽⁻¹⁾Σ_r.

A total of 34 dimensions are extracted from a vector feature, where F = [F₁, F₂, …, F₃₄]. After PCA, the new feature obtained can be denoted as Y_r = [Y_r1, Y_r2, ⋯, Y_rm] and m represents the computational dimension. (24)Yrm=F1aij+F2aij+⋯+Fmaij$${Y_{rm}} = {F_1}{a_{ij}} + {F_2}{a_{ij}} + \cdots + {F_m}{a_{ij}}$$

where a_ij is the eigenvalue of the covariance matrix. The equation can be simplified to the following equation: (25)Yrm=F1a1+F2a2+⋯+Fmam$${Y_{rm}} = {F_1}{a_1} + {F_2}{a_2} + \cdots + {F_m}{a_m}$$

Classification basis, the problem in this study belongs to a small-sample linear inseparable problem, so the SVM algorithm model [37] is chosen to solve it.

Let the training dataset be T={(Xk,yk)|X∈Rm,yk∈{0,1}k=1n}$$T = \{ ({X_k},{y_k})|X \in {R^m},{y_k} \in \{ 0,1\}_{k = 1}^n\}$$, which contains n samples, X represents a m-dimensional matrix, the classification label y_k has a value of 0 or 1, and the current sample is X_k. Therefore, the optimization objective function is represented as: (26)minω,b12||ω||2 s.t.yk(ω⋅xk+b)−1≥0,k=1,2,…,N$$\begin{array}{c} {\min_{\omega ,b}}\frac{1}{2}||\omega |{|^2} \\ s.t.\quad {y_k}(\omega \cdot {x_k} + b) - 1 \geq 0,k = 1,2, \ldots ,N \\ \end{array}$$

Here, ω is a vector on the hyperplane and the offset b of the hyperplane is along ω from the origin. According to the theory of convex optimization, Eq. (26) is transformed into an unconstrained problem. The optimization function can be expressed as: (27)L(ω,b,α)=12||ω||2−∑k=1Nαkyk(ω⋅xk+b)+∑k=1Nαk$$L(\omega ,b,\alpha ) = \frac{1}{2}||\omega |{|^2} - \sum\limits_{k = 1}^N {{\alpha_k}} {y_k}(\omega \cdot {x_k} + b) + \sum\limits_{k = 1}^N {{\alpha_k}}$$

where α_k is the Lagrange multiplier [38] and α_k≥0(k = 1, 2, 3, …, n). The original problem can be expressed as: (28)maxαminωbL(ω,b,α)$${\max_\alpha }{\min_{\omega b}}L(\omega ,b,\alpha )$$

Solving ω and b as a minimal problem gives the values of ω and b, i.e., we have Eq. (29): (29){ ω=∑k=1Nαkykxk ∑k=1Nαkyk=0$$\left\{ {\begin{array}{*{20}{l}} {\omega = \sum\limits_{k = 1}^N {{\alpha_k}} {y_k}{x_k}} \\ {\sum\limits_{k = 1}^N {{\alpha_k}} {y_k} = 0} \end{array}} \right.$$

The obtained solution is substituted into the Lagrangian function of the minimization problem and the optimization function is obtained after the substitution: (30)minα12∑k=1N∑j=1Nαiαjykyj(xk⋅xj)−∑k=1Nαk s.t.∑k=1Nαkyk=0 αk≥0,k=1,2,…,N$$\begin{array}{c} {\min_\alpha }\frac{1}{2}\sum\limits_{k = 1}^N {\sum\limits_{j = 1}^N {{\alpha_i}} } {\alpha_j}{y_k}{y_j}({x_k} \cdot {x_j}) - \sum\limits_{k = 1}^N {{\alpha_k}} \\ s.t.\quad \sum\limits_{k = 1}^N {{\alpha_k}} {y_k} = 0 \\ {\alpha_k} \geq 0,k = 1,2, \ldots ,N \\ \end{array}$$

Then some anomalous sample points with a slack variable ζ_k of (x_k, y_k) are introduced to make the training set linearly inseparable, with a penalty parameter C ≥ 0. The original problem is described as equation (31): (31)minω,b,ζ12||ω||2+C∑k=1Nζk s.t.yk(ω⋅xk+b)≥1−ζk,k=1,2,…,N ζk≥0,k=1,2,…,N$$\begin{array}{c} {\min_{\omega ,b,\zeta }}\frac{1}{2}||\omega |{|^2} + C\sum\limits_{k = 1}^N {{\zeta_k}} \\ s.t.\quad {y_k}(\omega \cdot {x_k} + b) \geq 1 - {\zeta_k},k = 1,2, \ldots ,N \\ {\zeta_k} \geq 0,k = 1,2, \ldots ,N \\ \end{array}$$

The final function is obtained after conversion: (32)minα12∑k=1N∑j=1Nαkαjykyj(xk⋅xj)−∑k=1Nαk s.t.∑k=1Nαkyk=0 0≤αk≤C,k=1,2,…,N$$\begin{array}{c} {\min_\alpha }\frac{1}{2}\sum\limits_{k = 1}^N {\sum\limits_{j = 1}^N {{\alpha_k}} } {\alpha_j}{y_k}{y_j}({x_k} \cdot {x_j}) - \sum\limits_{k = 1}^N {{\alpha_k}} \\ s.t.\quad \sum\limits_{k = 1}^N {{\alpha_k}} {y_k} = 0 \\ 0 \leq {\alpha_k} \leq C,k = 1,2, \ldots ,N \\ \end{array}$$