9. 二叉搜索树

Tree insert: searchs for that element A[i]
Tree walk:

Theorem:

E[height of a randomly built binary search tree] = O(lg n)

Proof outline

Prove Jensen's inequality:
$f(E[X]) \leq E[f(X)]$ for convex function f
Instead of analyzing $X_n$ = the random variable of the height of a randomly constructed BST on n nodes, we analize $Y_n = 2^{X_n}$
Prove that $E[Y_n] = O(n^3)$
Conclude that
$2^{E[X_n]} \leq E[2^{X_n}] = E[Y_n] = O(n^3)$
$⇒ E[X_n] \leq lg(O(n^3)) = 3lg(n) + O(1)$

$E[X_n] \approx 2.9882lg\,n$ (by Luke Devroy, 1986)

Proof 1.Jensen's inequality

$f: \mathbb{R} \rightarrow \mathbb{R}$ is convex if for all $x,y \in \mathbb{R}$ and all $\alpha, \beta \geq 0$ with $\alpha + \beta = 1$ , $f(\alpha x + \beta y) \leq \alpha f(x) + \beta f(y)$

vector and convex function

Lemma

if $\mathbb{R} \rightarrow \mathbb{R}$ is convex & $x_1, x_2, ... , x_n \in \mathbb{R}$ & $\alpha_1, \alpha_2, ... , \alpha_n \geq 0$ & $\sum^n_{k=1}{x_k} = 1$ ,
then $f \left( \sum_{k=1}^n{\alpha_kx_k} \right) \leq \sum_{k=1}^n{\alpha_kf(x_k)}$

灰色区域某一点意味着不等号右边

Induction
Base Case: When n = 1,

\alpha_1 = 1

, 不等式

f \left( \sum_{k=1}^n{\alpha_kx_k} \right) \leq \sum_{k=1}^n{\alpha_kf(x_k)}

成立。
Induction Step:

f \left( \sum_{k=1}^n{\alpha_kx_k} \right)

= f \left( \alpha_nx_n +(1 - \alpha_n) \sum^{n-1}_{k=1}{\frac{\alpha_k}{1-\alpha_n}x_k} \right)

\leq \alpha_nf(x_n) + (1 - \alpha_n)f\left( \sum^{n-1}_{k=1}{\frac{\alpha_k}{1-\alpha_n}x_k} \right)

= \alpha_nf(x_n) + f \left( \sum^{n-1}_{k=1}{\alpha_kx_k} \right) = \sum_{k=1}^n{\alpha_kf(x_k)}

Jensen's inequality

$f(E[X]) \leq E[f(X)]$ (f is a convex and X is a integer random variable)
proof:
$f(E[X]) = f \left(\sum^{∞}_{-∞}{x * Pr\{X = x\}} \right)$
$\leq \sum^{∞}_{-∞}{Pro\{X = x\} * f(x)} = E[f(X)]$

Proof 2. Analysis Y_n

X_n = the random variable of the height of a randomly constructed BST on n nodes. Y_n = 2^X_n

If root r has rank k, then Xn = 1 + max{X_k-1, X_n-k}, Y_n = 2 * max{Y_k-1, Y_n-k}

define indicator r.v.'s
$Z_{nk}=\begin{cases} 1, rank(root) = k\\ 0, otherwise \end{cases}$
Pr{Z_nk = 1} = E[Z_nk] = 1/n
$Y_n = \sum^n_{k = 1}{Z_{nk}(2max\{Y_{k-1}, Y_{n-k}\})}$

Proof 3. Analysis E[Y_n]

$E[Y_n] = E \left[ \sum^n_{k = 1}{Z_{nk}(2max\{Y_{k-1}, Y_{n-k}\})} \right]$
$= \sum^n_{k = 1}{E[Z_{nk}(2max\{Y_{k-1}, Y_{n-k}\})]}$
$= 2\sum^n_{k=1}{E[Z_{nk}]E[max\{Y_{k-1}, Y_{n-k}\}]}$
$= \frac{2}{n} \sum^n_{k=1}{E[max\{Y_{k-1}, Y_{n-k}\}]}$
$\leq \frac{2}{n} \sum^n_{k=1}{E[Y_{k-1} + Y_{n-k}]}$
$= \frac{4}{n} \sum^{n-1}_{k=0}{E[Y_k]}$
Then, we are going to use substitution.
Claim: $E[X_n] \leq cn^3$
Proof:
The base case must be true when c is sufficiently large.
Induction Step: $E[Y_n] \leq \frac{4}{n} \sum^{n-1}_{k=0}{E[Y_k]} \leq \frac{4}{n} \sum^{n-1}_{k=0}{ck^3} \leq \frac{4c}{n} \int_0^n {x^3}\,{\rm d}x = cn^3$

Proof 4.

$2^{E[X_n]} \leq E[2^{X_n}] = E[Y_n] = O(n^3)$ （简单地使用了Jensen's inequality）
$⇒ E[X_n] \leq lg(O(n^3)) = 3lg(n) + O(1)$ ( lg(O(n³)) = lg(cn³) = 3lg(n) + lg(c) )

最后编辑于：2018.08.19 10:01:23

©著作权归作者所有,转载或内容合作请联系作者
【社区内容提示】社区部分内容疑似由AI辅助生成，浏览时请结合常识与多方信息审慎甄别。
平台声明：文章内容（如有图片或视频亦包括在内）由作者上传并发布，文章内容仅代表作者本人观点，简书系信息发布平台，仅提供信息存储服务。

9. 二叉搜索树