硬间隔
首先描述硬间隔问题,我们记为式(1):
$$\min_{w,b} \frac{1}{2}||w||^2 \tag{1}$$
$$st. \quad y^{(i)}(w^Tx^{(i)}+b) \ge 1$$
构造拉格朗日乘子:
$$L(w,b,\alpha) = \frac{1}{2}||w||^2 + \sum_{i=1}^m \alpha_i [1-y^{(i)}(w^Tx^{(i)} + b)] \tag{2}$$
依据拉格朗日乘子法,我们有问题的转化关系:
$$primal \quad problem \rightarrow \min_{w,b} \max_{\alpha} L(w,b,\alpha) \rightarrow \max_{\alpha} \min_{w,b}L(w,b,\alpha)$$
先求解内层的极小值部分:\(\min_{w,b} L(w,b,\alpha)\),即\(\frac{\partial}{\partial w} L(w,b,\alpha) =0\),\(\frac{\partial}{\partial b} L(w,b,\alpha) =0 \),将这两个式子解出来得:
$$\begin{align} &\frac{\partial}{\partial w} L(w,b,\alpha) = w + \sum_{i=1}^m \alpha_i y^{(i)}x^{(i)} = 0 \Longrightarrow w=\sum_{i=1}^m \alpha_iy^{(i)}x^{(i)} \tag{3} \\ &\frac{\partial}{\partial b} L(w,b,\alpha) = \sum_{i=1}^m \alpha_iy^{(i)} = 0 \Longrightarrow \sum_{i=1}^m \alpha_i y^{(i)} = 0 \tag{4}\end{align}$$
将关于\(w, b\)的解代入拉格朗日乘子中,并进行化简:
$$ \begin{align