推导4

根据对偶函数的限制条件可知：

0≤a1≤C0≤a2≤C1\begin{aligned} 0 \le a_1 \le C \\ 0 \le a_2 \le C && {1} \end{aligned}

把a2a_2用a1a_1表示得：

0≤−ξ−a2y2y1≤C2\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. y1=y2=1y_1 = y_2 = 1

    −C−ξ≤a2≤−ξa1+a2=−ξ\begin{aligned} -C - \xi \le a_2 \le -\xi \\ a_1+a_2=-\xi \end{aligned}

    得：

    −C+a1+a2≤a2new≤a1+a2-C+a_1+a_2 \le a_2^{new} \le a_1 + a_2

  2. y1=1,y2=−1y_1 = 1, y_2 = -1

    ξ≤a2≤C+ξa2−a1=ξ\begin{aligned} \xi \le a_2 \le C+\xi \\ a_2 - a_1 = \xi \end{aligned}

    得：

    a2−a1≤a2new≤C+a2−a1a_2 - a_1 \le a_2^{new} \le C + a_2 - a_1

  3. y1=−1,y2=1y_1 = -1, y_2 = 1

    −ξ≤a2≤C−ξa2−a1=−ξ\begin{aligned} -\xi \le a_2 \le C-\xi \\ a_2 - a_1 = -\xi \end{aligned}

    得：

    a2−a1≤a2new≤C+a2−a1a_2 - a_1 \le a_2^{new} \le C + a_2 - a_1

  4. y1=y2=−1y_1 = y_2 = -1

    ξ−C≤a2≤ξa1+a2=ξ\begin{aligned} \xi-C \le a_2 \le \xi \\ a_1 + a_2 = \xi \end{aligned}

    得：

    −C+a1+a2≤a2new≤a1+a2-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,a2−a1)≤a2new≤min⁡(C,C+a2−a1)\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+a1+a2)≤a2new≤min⁡(C,a1+a2)\max (0, -C+a_1+a_2) \le a_2^{new} \le \min(C, a_1 + a_2)
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