1. 凸二次规划问题求解

要求解的带约束最优化问题为:
minw,b12∣∣w∣∣2s.t.yi(w⋅xi+b)−1≥0,i=1,2,⋯,N1 \begin{aligned} min_{w,b} \quad \frac {1}{2}||w||^2 \\ s.t. \quad y_i(w \cdot x_i + b)-1 \ge 0, i = 1,2,\cdots,N && {1} \end{aligned}

将原始问题转换为对偶最优化问题(推导过程):
mina12∑i=1N∑j=1Naiajyiyj(xi⋅xj)−∑i=1nais.t.∑i=1Naiyi=0ai≥0,i=1,2,⋯,N2 \begin{aligned} \min_a \quad \frac{1}{2}\sum_{i=1}^N\sum_{j=1}^Na_ia_jy_iy_j(x_i\cdot x_j) - \sum_{i=1}^na_i \\ s.t. \quad \sum_{i=1}^Na_iy_i = 0 \\ \quad a_i \ge 0, i = 1,2,\cdots, N && {2} \end{aligned}

通过公式(2)求解a∗a*(推导过程)

通过a∗a*求出w∗w*和b∗b*:

w∗=∑i=1Nai∗yixib∗=yj−∑i=1Nai∗yi(xi⋅xj)3 \begin{aligned} w* = \sum_{i=1}^Na_i^*y_ix_i \\ b* = y_j - \sum_{i=1}^Na_i^*y_i(x_i \cdot x_j) && {3} \end{aligned}

通过公式(3)得到:
分离超平面:
∑i=1Nai∗yi(x⋅xi)+b∗=0 \sum_{i=1}^Na_i^*y_i(x\cdot x_i) + b^* = 0 分类决策函数(推导过程):
f(x)=sign(∑i=1Nai∗yi(x⋅xi)+b∗) f(x) = sign(\sum_{i=1}^Na_i^*y_i(x\cdot x_i) + b^*)

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