Distance from the discriminant to the closest instances on either side
Distance of x to the hyperplane: $|w^Tx^t+w_0|/||w||$
→ $r^t(w^x+w_0)/||w|| \geq \rho, \forall t$ ($\rho$ is margin, 2-classes case)
→ infinity solutions
For a unique solution, fix $\rho ||w||=1$ and to maximize margin, we minimize $||w||$
$\text{find } ||w|| \text{ to } \min \frac{1}{2} ||w||^2 \text{ subject to } r^t(w^Tx^t+w_0) \geq +1, \forall t$
Circled instances are the support vectors
Use Lagrange function:
$L_p = \frac{1}{2}||w||^2 - \sum^N_{t=1} \alpha^t[r^t(w^Tx^t+w_0)-1] \text{ with respect to }\alpha$
$L_p = -\frac{1}{2}\sum_t\sum_s \alpha^t \alpha^s r^t r^s (x^t)^Tx^s + \sum_t \alpha^t \text{ subject to } \sum_t \alpha^t r^t =0 \text{ and }\alpha^t \geq 0, \forall t$
Not linearly separable
$r^t(w^Tx^t+w_0) \geq 1 - \xi^t$ for slack variable $\xi$
Soft error : #{$\xi>0$}
$\sum_t \xi^t$
New primal is
$L_p = \frac{1}{2}||w||^2 + C\sum_t\xi^t - \sum_t\alpha^t[r^t(w^Tx^t+w_0)-1+\xi^t]-\sum_t\mu^t\xi^t$
(a) $r^tg(x^t)>1, \xi^t=0, \alpha^t = 0$
(b) it is on the right side and on the margin
$\xi^t = 0, 0 < \alpha^t < C$
(c) it is on the right side but is in the margin
$0 \leq \xi^t < 1$, $\xi^t = 1-g(x^t), \alpha^t = C$
(d) it is on the wrong side: misclassification
$\xi^t= 1+g(x^t) \geq 1, \alpha^t = C$