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推导2
L
(
a
1
,
a
2
)
=
f
(
1
,
1
)
+
2
f
(
1
,
2
)
+
f
(
2
,
2
)
+
2
∑
j
=
3
N
f
(
1
,
j
)
+
2
∑
j
=
3
N
f
(
2
,
j
)
−
a
1
−
a
2
+
常数项
1
\begin{aligned} L(a_1, a_2) = f(1,1)+2f(1,2)+f(2,2) \\ +2\sum_{j=3}^Nf(1,j) +2\sum_{j=3}^Nf(2,j) \\ -a_1 - a_2 + \text{常数项}&& {1} \end{aligned}
L
(
a
1
,
a
2
)
=
f
(
1
,
1
)
+
2
f
(
1
,
2
)
+
f
(
2
,
2
)
+
2
j
=
3
∑
N
f
(
1
,
j
)
+
2
j
=
3
∑
N
f
(
2
,
j
)
−
a
1
−
a
2
+
常数项
1
其中:
f
(
i
,
j
)
=
a
i
a
j
y
i
y
j
K
(
x
i
,
x
j
)
f(i, j) = a_ia_jy_iy_jK(x_i,x_j)
f
(
i
,
j
)
=
a
i
a
j
y
i
y
j
K
(
x
i
,
x
j
)
对偶问题中的限制条件也可以分成变量和常量两部分:
∑
i
=
1
N
a
i
y
i
=
a
1
y
1
+
a
2
y
2
+
∑
i
=
3
N
a
i
y
i
=
0
2
\begin{aligned} \sum_{i=1}^Na_iy_i=a_1y_1+a_2y_2 + \sum_{i=3}^Na_iy_i = 0 && {2} \end{aligned}
i
=
1
∑
N
a
i
y
i
=
a
1
y
1
+
a
2
y
2
+
i
=
3
∑
N
a
i
y
i
=
0
2
令常数项为:
ξ
=
∑
i
=
3
N
a
i
y
i
3
\begin{aligned} \xi = \sum_{i=3}^Na_iy_i && {3} \end{aligned}
ξ
=
i
=
3
∑
N
a
i
y
i
3
根据公式(3),得到
a
1
a_1
a
1
和
a
2
a_2
a
2
的关系为:
a
1
=
−
ξ
−
a
2
y
2
y
1
a_1 = \frac {-\xi - a_2y_2}{y_1}
a
1
=
y
1
−
ξ
−
a
2
y
2
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Last updated
4 years ago