Lecture4

Building a heap¶

对应的算法：

A.heap-size = A.length

for i = A.length / 2 downto 1
    MAX-HEAPIFY(A, i)

这实际上是一个线性的算法，我们下面来说明它是线性的。

考虑一个高为 $\log n$ 的堆，考虑它的第 $h$ 层

Property

An n-element heap has height $\lfloor \log n \rfloor$ and at most $\lceil n / 2^{h+1} \rceil$ nodes of any height $h$.

There is a upper bound

\[ \sum_{h = 0}^{\lfloor \log n \rfloor} \frac{n}{2^{h+1}}O(h) = O \left( n\sum_{h = 0}^{\lfloor \log n \rfloor} \frac{h}{2^h} \right) = O\left( n\sum_{h = 0}^{\infty} \frac{h}{2^h} \right) = O(n)\]

Quick sort¶

Expected time analysis¶

Let $T(n)$ be the random variable for the running time of RANDOMIZED-QUICKSORT

Define the indicator random variable

\[ X_k = 1_{it generate spilt $k$ and $n - k - 1$} \]

We have

\[ E(X_k) = P{X_k = 1} = \frac{1}{n} \]

\[ E(T(n)) = E(\sum_{k = 0}^{n - 1} X_k(T_k) + T(n -k - 1) + \Theta(n)) \]

\[ E[T(n)] = \frac{2}{n} \sum_{k = 0}^{n - 1} E[T(k)] + \Theta(n) \]

Assume that $E[T(n)] = O(n \log n)$

\[ E[T(n)] \leq \frac{2}{n} \sum_{k=2}^{n-1} ak \log k + \Theta(n) \leq \frac{2a}{n} (\frac{1}{2}n^2\log n - \frac{1}{8}n^2) + \Theta = O(n\log n)\]

Remark

\[ \sum_{k=2}^{n-1} k \log k \leq \frac{1}{2}n^2 \log n - \frac{1}{8} n^2\]

maybe the following inequality also works

\[ \sum_{k=2}^{n-1} k \log k \leq \frac{n(n-1)}{2} \log n \]

Lemma

Let $X$ be # comparisons in Partition over entire execution of Quciksort on an $n$-element array. Then the running time of QuickSort is $O(n + X)$

We define the indicator

\[ X_{ij} = I_A\]

$A$ means that $z_i$ is compared to z_j

We have the partition of random variable

\[ X = \sum_{i = 1}^{n - 1} \sum_{j = i +1}^{n} X_{ij}\]

We have the following Expection

\[E[X_{ij}] = \frac{2}{j - i + 1}\]

Then by the linearity of Expection

\[ E[X] =\sum_{i = 1}^{n - 1} \sum_{j = i +1}^{n} \frac{2}{j - i + 1} \]

By the knowledge of harmonic series

\[E[x] = O(n\log n)\]

Why the quicksort quick¶

it compare with a continue array in memory

It hit the crash more often !