Lec06 顺序统计、中值

MIT算法导论课程：Lec06 顺序统计、中值，对应书上的章节：Chapter 9

1. Order statistics

选出n个元素中第i小的数，i=1:minimum，i = n: maximum，i = ⎣(n+1)/2⎦ or ⎡(n+1)/2⎤: median

很容易想到的一个方法是先排序在索引第i个元素。复杂度：\(\Theta(n\lg n)\)

2. Randomized divide-and conquer algorithm

伪代码

RAND-SELECT(A,p,q,i) //ith smallest of A[p...q]
	if p=q then return A[p]
	r=RAND-PARTITION(A,p,q)
	k=r-p+1 //k=rank(A[r])
	if i=k then return A[r]
	if i<k
		then return RAND-SELECT(A,p,r-1,i)
		else return RAND-SELECT(A,r+1,q,i-k)

Analysis of expected time

划分指示器

\[ X_{k}=\left\{\begin{array}{ll} 1 & \text { if PARTITION generates a } k: n-k-1 \text { split, } \\ 0 & \text { otherwise. } \end{array}\right. \]

\[ \begin{aligned} T(n)=&\left\{\begin{array}{cl} T(\max \{0, n-1\})+\Theta(n) & \text { if } 0: n-1 \text { split, } \\ T(\max \{1, n-2\})+\Theta(n) & \text { if } 1: n-2 \text { split, } \\ \vdots \\ T(\max \{n-1,0\})+\Theta(n) & \text { if } n-1: 0 \text { split } \end{array}\right.\\ &=\sum_{k=0}^{n-1} X_{k}(T(\max \{k, n-k-1\})+\Theta(n)) \end{aligned} \]

再用替换法证明，\(E[T(n)]\leq cn\)for constant c>0。 \[ \begin{aligned} E[T(n)] & \leq \frac{2}{n} \sum_{k=\lfloor n / 2\rfloor}^{n-1} c k+\Theta(n) \\ & \leq \frac{2 c\left(\frac{3}{8} n^{2}\right)+\Theta(n)}{n} \\ &=c n-\left(\frac{c n}{4}-\Theta(n)\right) \\ & \leq c n \end{aligned} \] if c is chosen large enough so that cn/4 dominates the Θ(n)

Summary of randomized order-statistic selection

线性期望时间
worst case：\(\Theta(n^2)\)

3. Worst-case linear-time order statistics

[Blum, Floyd, Pratt, Rivest, Tarjan]

IDEA: Generate a good pivot recursively.

SELECT(i, n)
1. Divide the n elements into groups of 5. Find the median of each 5-element group by rote.
2. Recursively SELECT the median x of the ⎣n/5⎦ group medians to be the pivot.
3. Partition around the pivot x. Let k = rank(x).
4. if i = k then return x
	elseif i < k
		then recursively SELECT the ith smallest element in the lower part
	else recursively SELECT the (i–k)th smallest element in the upper part
	
3 4步和RAND-SELECT一样

Analysis

实际上，此算法运行缓慢，因为n前面的常数很大
随机算法更加实用。

练习：为什么不分成3组？