MIT algorithm 笔记（视频2）

1.
$f=O(g(n)) , O为上界\ \exists const\ c\gt0, n_0 \gt0,使得 0\le f(n) \le cg(n), for \ all \ n \ge n_0,$

$\qquad其中f(n)可看作是g(n)构成的函数集$

$\begin{align} 定义：O(g(n))=\text {$\{f(n)|\exists const \ c\gt 0,n_0\gt0,使得0\le f(n) \le c(g(n)), \quad for \ all\ n\ge n_0\}$} \end{align}$

$\quad Ex:2n^2=O((n)) \quad /\quad 2n^2\in O(n^3)$
2.
$\begin{align} \text{Macro convention:} \ \text{A set in a formula represents an anonymous function in that set.}\ \end{align}$

$\quad Ex1:\ f(n)=n^3+O(n^2)\ \text{means there is function}\ h(n)\in O(n^2)$
$\quad \text{such that}\ f(n)=n^3+h(n)$

$\quad Ex2:\ n^2+O(n)=O(n^2)\ \text{means for any}\ f(n)\in O(n)$
$\quad \text{there is an}\ h(n)\in O(n^2), \text{such that}\ n^2+f(n)=h(n)$

3.
$\Omega\ nonation(下界)$
$\Omega(g(n))=\{f(n)|\exists\ const\ c\gt0,n_0\gt0,such\ that\ 0\le cg(n)\le f(n)$
$\qquad \qquad \qquad \qquad for\ all\ n\ge n_0\}$

$\quad Ex1:\sqrt{n}=\Omega(logn)\qquad \Theta(g(n))=O(g(n))\cap\Omega(g(n))$

$o\ 和\ \omega\ 分别对应\ O\ 和\ \Omega,\ 只是没有等号。$

$\quad Ex2:2n^2=o(n^3)\ means\ 2n^2\lt cn^3,\frac 12n^2=\Theta(n^2)\neq o(n^2)$