# 推导4

根据对偶函数的限制条件可知：

$$
\begin{aligned}
0 \le a\_1 \le C  \\
0 \le a\_2 \le C  && {1}
\end{aligned}
$$

把$$a\_2$$用$$a\_1$$表示得：

$$
\begin{aligned}
0 \le \frac {-\xi - a\_2y\_2}{y\_1} \le C  &&{2}
\end{aligned}
$$

由于$y\_1$和$y\_2$的取值只能是1或者-1，而1或-1会对不等号的化简影响不同，所以把公式（2）分成4种情况：

1. $$y\_1 = y\_2 = 1$$

   $$
   \begin{aligned}
   -C - \xi \le a\_2 \le -\xi \\
   a\_1+a\_2=-\xi
   \end{aligned}
   $$

   得：

   $$
   -C+a\_1+a\_2 \le a\_2^{new} \le a\_1 + a\_2
   $$
2. $$y\_1 = 1, y\_2 = -1$$

   $$
   \begin{aligned}
   \xi \le a\_2 \le C+\xi  \\
   a\_2 - a\_1 = \xi
   \end{aligned}
   $$

   得：

   $$
   a\_2 - a\_1 \le a\_2^{new} \le C + a\_2 - a\_1
   $$
3. $$y\_1 = -1, y\_2 = 1$$

   $$
   \begin{aligned}
   -\xi \le a\_2 \le C-\xi \\
   a\_2 - a\_1 = -\xi
   \end{aligned}
   $$

   得：

   $$
   a\_2 - a\_1 \le a\_2^{new} \le C + a\_2 - a\_1
   $$
4. $$y\_1 = y\_2 = -1$$

   $$
   \begin{aligned}
   \xi-C \le a\_2 \le \xi \\
   a\_1 + a\_2 = \xi
   \end{aligned}
   $$

   得：

   $$
   -C+a\_1+a\_2 \le a\_2^{new} \le a\_1 + a\_2
   $$

综合以上结果得：\
当y1y2<0时，

$$
\max (0, a\_2 - a\_1) \le a\_2^{new} \le \min(C, C + a\_2 - a\_1)
$$

当y1y2>0时，

$$
\max (0, -C+a\_1+a\_2) \le a\_2^{new} \le \min(C, a\_1 + a\_2)
$$


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