主定理的证明
算法分析的那个定理.
Master Theorem
\[T(n) = \begin{cases} \Theta(1), & \text{if } n = 1,\\ aT(n/b) + f(n), & \text{if } n>1. \end{cases} \]
where \(a\ge1\), \(b>1\) are constants and \(f\) is nonnegative. Then
- If \(f(n) = O(n^{\log_b a-\varepsilon})\) for some constant \(\varepsilon >0\), then \(T(n) = \Theta(n^{\log_b a})\).
- If \(f(n) = \Theta(n^{\log_b a})\), then \(T(n) = \Theta(n^{\log_b a}\lg n)\), where \(\lg n\) stands for \(\log_2 n\).
- If \(f(n) = \Omega(n^{\log_b a+\varepsilon})\) for some constant \(\varepsilon >0\), and if \(af(n/b)\le cf(n)\) for some constant \(c<1\) and all sufficiently large \(n\), then \(T(n) = \Theta(f(n))\).
直观上看就是比较 \(f(n)\) 与 \(n^{\log_b a}\) 的阶数, 其中大的决定了 \(T(n)\) 的阶数; 如果阶数相同则乘 \(\lg n\).
证明: 先对 \(n\) 为 \(b^k\), \(k\in\mathbb N\) 的情形证明, 下面的渐进符号都是对 \(n\) 在 \(b^k\) 上的点而言的. 写出递归树, 高度为 \(\log_b n\), 故有 \(a^{\log_b n} = n^{\log_b a}\) 个叶, 从而
\[T(n) = \Theta(n^{\log_b a}) + g(n),
$$ 其中
\]
g(n) = \sum_{j=0}^{\log_b n-1}a^j f(n/b^j).
\[
Case 1. $f(n) = O(n^{\log_b a-\varepsilon})$, 易得 $g(n) = O(n^{\log_b a})$.
Case 2. $f(n) = \Theta(n^{\log_b a})$, 易得 $g(n) = \Theta(n^{\log_b a} \log_b n) = \Theta(n^{\log_b a} \lg n)$.
Case 3. 首先 $g(n) = \Omega(f(n))$. 又 $af(n/b)\le cf(n)$ for some constant $c<1$ and sufficiently large $n$. 即 $a^j f(n/b^j) \le c^j f(n)$. 得
\]
\begin{align}
g(n) &= \sum_{j=0}^{\log_b n-1}a^j f(n/b^j)\
&\le \sum_{j=0}^{\log_b n-1}c^j f(n) + O(1)\
&\le f(n)\sum_{j=0}^\infty c^j + O(1)\
&=O(f(n)).
\end{align}
\[故 $g(n)=\Theta(f(n))$.
对一般的 $n$ 证明略.\]
