Lecture 1

Big O Notation §

Given T(n) is the upper bound on a runtime of algorithm on the worst-case input of size n.

$$\begin{array}{rl}T(n)& =5(n-2)(n+1)+10\sqrt{n}\\ T(100)& =5\ast 98\ast 101+10^2=49,590\\ T(n)& =O(n^2)\end{array}$$

$f(n)$ is $O(g(n))$ if $\exists c>0,n_0>0$ such that $\forall n>n_0,f(n)\le c(g(n))$.

Example Problems §

$$\begin{array}{rl}f(n)& =50n^2-100n\\ f(n)& =O(n^2)\\ g(n)& =100n\log n+500\sqrt{n}\\ \therefore g(n)& =O(n\log n)\end{array}$$

$$\begin{array}{rl}f(n)& =(50\log n{)}^{100}\\ g(n)& =n^{\frac{1}{2}}\end{array}$$

$$\begin{gathered} (\log n{)}^{1000}<n^{.0001} \\ \therefore f(n)=O(g(n)) \end{gathered}$$

$$\begin{array}{rl}f(n)& =50{\log }_{10}(n^{100})\\ g(n)& =2{\log }_{2}n\end{array}$$

$$\begin{gathered} 3{\log }_{8}n=\log 8(n^3)={\log }_{2}n⟹lo{g}_{2}n=3{\log }_{8}n \\ \text{therefore }{\log }_{10}n=O({\log }_{2}n)\&{\log }_{2}n=O({\log }_{1}0n) \end{gathered}$$

$$\therefore f(g(n))\&g(f(n))$$

$$\begin{array}{rl}& \\ f(n)& =n^2\\ g(n)& =3^{{\log }_{2}n}\end{array}$$

$$\begin{gathered} g(n)=3^{{\log }_{2}n}=(2^{{\log }_{2}3}{)}^{{\log }_{2}n}=2^{{\log }_{2}3\times {\log }_{2}n}=(2^{{\log }_{2}n}{)}^{{\log }_{2}3}=n^{{\log }_{2}3} \\ \therefore n^{{\log }_{2}3}<n^2 \end{gathered}$$

Takeaway: To solve problems comparing $n$ and some number to the power of $\log n$ convert the base number to $n$ using log rules and compare.

$$\begin{array}{rl}& \\ f(n)& =2^{\sqrt{n}}\\ g(n)& =n^{{\log }_{2}n}\end{array}$$

$$\begin{gathered} g(n)=n^{{\log }_{2}n}=(2^{{\log }_{2}n}{)}^{{\log }_{2}n}=2^{({\log }_{2}n{)}^2} \\ \sqrt{n}>({\log }_2(n){)}^2 \\ \therefore g(n)=O(f(n)) \end{gathered}$$

Alg. Analysis §

Model of Computation:

• Basic arithmetic operations in O(1) time
• Infinite memory & O(1) time to access unit of memory
• All analysis in Big O
• Big O will be for a specific algorithm, NOT for the problem.

Max Substring Sum §

Input: Array $A=[a_1...a_n]$ of $n$ integers Output: max sum from a substring of A, i.e, max $a_i+a_{i+n}+...a_j$ where $i\le j$

Possible Solution 1, Brute Force Run time: $O(n)\times O(n)\times O(n)\times O(1)=O(n^3)$ times

max = 0
For i = 1 to n
	For j = i to n
		temp = 0
		for k = i to j
			temp = temp + a[k]
		if temp > max then max = temp
return max

Faster Solution, Sliding WIndow Run time: $O(n^2)$

max = 0
For i = 1 to n
	temp = 0
	for j = i to n
		temp = temp + a[j]
		if temp > max then max = temp
return max

There exists an $O(n)$ solution using DP