分治法主定理
主定理的证明
假设有递归式：
T
(
n
)
=
a
T
(
n
b
)
+
f
(
n
)
T(n) = aT(\frac{n}{b}) + f(n)
T(n)=aT(bn)+f(n)
证明：
T
(
n
)
=
a
T
(
n
/
b
)
+
f
(
n
)
T(n) = aT(n/b) + f(n)
T(n)=aT(n/b)+f(n)
=
a
[
a
T
(
n
/
b
2
)
+
f
(
n
/
b
)
]
+
f
(
n
)
= a\big[aT(n/b^2)+f(n/b)\big] + f(n)
=a[aT(n/b2)+f(n/b)]+f(n)
=
a
2
T
(
n
/
b
2
)
+
f
(
n
)
+
a
f
(
n
/
b
)
= a^2T(n/b^2) + f(n) + af(n/b) ~\,
=a2T(n/b2)+f(n)+af(n/b)
=
a
k
T
(
n
/
b
k
)
+
∑
j
=
0
k
−
1
a
j
f
(
n
/
b
j
)
= a^{k}T(n/b^k) + \sum_{j=0}^{k-1} a^j f(n/b^j) ~~~~
=akT(n/bk)+j=0∑k−1ajf(n/bj)
=
⋯
(
不
妨
设
n
是
b
的
幂
)
= \cdots ~(不妨设n是b的幂)~~~~~~~~~~~~
=⋯ (不妨设n是b的幂)
=
a
l
o
g
b
n
T
(
1
)
+
∑
j
=
0
l
o
g
b
n
−
1
a
j
f
(
n
/
b
j
)
= a^{log_b^n}T(1) + \sum_{j=0}^{log_b^n-1} a^j f(n/b^j) ~~
=alogbnT(1)+j=0∑logbn−1ajf(n/bj)
由对数的换底公式：
a
l
o
g
b
n
=
b
l
o
g
b
(
a
l
o
g
b
n
)
=
b
l
o
g
b
a
l
o
g
b
n
=
b
l
o
g
b
n
l
o
g
b
a
=
b
l
o
g
b
(
n
l
o
g
b
a
)
=
n
l
o
g
b
a
a^{log_b^n} =b^{log_b^{(a^{log_b^n})}} =b^{log_b^a log_b^n} =b^{log_b^n log_b^a}= b^{log_b^{(n^{log_b^a})}}=n^{log_b^a}
alogbn=blogb(alogbn)=blogbalogbn=blogbnlogba=blogb(nlogba)=nlogba
所以原递归式的渐进复杂度为 O ( n l o g b a ) + ∑ j = 0 l o g b n − 1 a j f ( n / b j ) O(n^{log_b^a}) + \sum_{j=0}^{log_b^n-1} a^j f(n/b^j) O(nlogba)+j=0∑logbn−1ajf(n/bj)
简单起见，下面只考虑
f
(
n
)
=
n
k
f(n) = n^k
f(n)=nk 的情形：
∑
j
=
0
l
o
g
b
n
−
1
a
j
f
(
n
/
b
j
)
\sum_{j=0}^{log_b^n-1} a^j f(n/b^j)
j=0∑logbn−1ajf(n/bj)
=
∑
j
=
0
l
o
g
b
n
−
1
a
j
(
n
/
b
j
)
k
= \sum_{j=0}^{log_b^n-1} a^j (n/b^j)^k ~~~~~~~
=j=0∑logbn−1aj(n/bj)k
=
n
k
∑
j
=
0
l
o
g
b
n
−
1
(
a
/
b
k
)
j
= n^k\sum_{j=0}^{log_b^n-1} (a/b^k)^j ~~~~~~~
=nkj=0∑logbn−1(a/bk)j
=
{
n
k
(
1
−
(
a
/
b
k
)
l
o
g
b
n
1
−
a
/
b
k
)
,
如果
a
≠
b
k
n
k
(
l
o
g
b
n
−
1
)
,
如果
a
=
b
k
= \left\{ \begin{array}{lr} n^k \left(\frac{1-(a/b^k)^{log_b^n}}{1-a/b^k}\right), & \text{如果$a\neq b^k$}\\ &\\ n^k(log_b^n-1), & \text{如果$a = b^k$} \end{array} \right.
=⎩⎪⎨⎪⎧nk(1−a/bk1−(a/bk)logbn),nk(logbn−1),如果a=bk如果a=bk
=
{
O
(
n
k
)
,
如果
a
<
b
k
(
k
>
l
o
g
b
a
)
O
(
n
l
o
g
b
a
)
,
如果
a
>
b
k
O
(
n
k
l
o
g
b
n
)
=
O
(
n
l
o
g
b
a
l
o
g
b
n
)
,
如果
a
=
b
k
= \left\{ \begin{array}{lr} O(n^k) , & \text{如果$a < b^k(k>log_b^a)$}\\ &\\ O(n^{log_b^a}), & \text{如果$a> b^k$}\\ &\\ O(n^klog_b^n)=O(n^{log_b^a}log_b^n), & \text{如果$a = b^k$} \end{array} \right.
=⎩⎪⎪⎪⎪⎨⎪⎪⎪⎪⎧O(nk),O(nlogba),O(nklogbn)=O(nlogbalogbn),如果a<bk(k>logba)如果a>bk如果a=bk
综上所述，对
T
(
n
)
=
a
T
(
n
b
)
+
f
(
n
)
T(n) = aT(\frac{n}{b}) + f(n)
T(n)=aT(bn)+f(n) 的情形：
T
(
n
)
=
{
O
(
n
k
)
,
如果
a
<
b
k
(
k
>
l
o
g
b
a
)
O
(
n
l
o
g
b
a
)
,
如果
a
>
b
k
O
(
n
k
l
o
g
b
n
)
=
O
(
n
l
o
g
b
a
l
o
g
b
n
)
,
如果
a
=
b
k
T(n)= \left\{ \begin{array}{lr} O(n^k) , & \text{如果$a< b^k(k>log_b^a)$}\\ &\\ O(n^{log_b^a}), & \text{如果$a> b^k$}\\ &\\ O(n^klog_b^n)=O(n^{log_b^a}log_b^n), & \text{如果$a = b^k$} \end{array} \right.
T(n)=⎩⎪⎪⎪⎪⎨⎪⎪⎪⎪⎧O(nk),O(nlogba),O(nklogbn)=O(nlogbalogbn),如果a<bk(k>logba)如果a>bk如果a=bk
减治法主定理
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