Transciber’s notes:

• $\iff$ is “if and only if”
• $\implies$ is “implies”
• $\forall$ is “for all”
• $\exists$ is “exists”

Big-Oh–Part I

CMPT 225

Slide 1

• Recall: if f,g are functions f:N->N,g:N->N
• f is O(g) means there are constants $n_{0},c > 0$ s.t. for every $n>n_{0},f(n) \leq \text{c.g}(n)$
• that is for all but finitely many “small” values of n: $f(n) \leq cg(n)$
• or f grows no faster than g (asymtotically)
• we typically write $f(n) = O(g(n))$

Consider f(n),g(n):

Claim: $f(n) = O(g(n))$

Why:

• When $f(n) > g(n)$ for all n?
• But $f(n) < 2g(n)$ for almost all values of n.
• choose $n_{0}$ to “exclude” the real values of n.
• Now: $f(n) < 2g(n)$ for all $n>n_{0}$
• So: $f(n) = O(g(n))$

Also:

(A graph a cannot transcribe because the notation doesn’t make any sense. All future graphs have complexity on the y axis and the value of n [input to the function] as the x asxis.)

Claim: f(n) is not O(g(n))

However large we choose $c,n_{0}$, there will be some k (larger than $n_0$) s.d.: $n > k \implies f(n) > c g(n)$

Consider O(1) (i.e. $g(n)=1$)

$f(n) = O(1)$ if $\exists n_{0},c > 0$ s.t. $\forall n>n_{0} \space \space f(n) \leq c\times 1$

(Another graph that makes no Goddamn sense whatsoever.)

• for every $n>n_{0}, f(n)<c$ <– >–>
• so, f(n) grwos no faster than a constant
• so f(n) is asymptotically bounded by a constant

The constant does not matter

Fact: $f(n) = O(1) \iff f(n)=O(\text{10}^{27}) \iff f(n) = O(\frac{1}{\text{10}^{27}})$

Suppose: $f(n) = O(\text{10}^{27})$ *

Claim: $f(n)$ is also $O(\frac{1}{\text{10}^{27}})$

* means $\exists n_{0},c > 0$ s.t. $n > n_{0} \implies f(n) \leq c\times\text{10}^{27}$

Want to show: $\exists n_{0},c^{1}>0$ s.t. $n>n_{0} \implies f(n) \leq c^{1}\times \frac{1}{\text{10}^{27}}$

Choose $c^1$ big enough that $c^{1} \frac{1}{\text{10}^{27}} \leq c \times \text{10}^{27}$

e.g: $c^{1} = c \times \text{10}^{54}$

Then:

$\forall n>n_{0}, f(n) \leq c \times \text{10}^{27} \\ \leq c \times \text{10}^{-27} - \text{10}^{54}\\ \leq c^1 - \text{10}^{-27}$

(54-27=27)

So: $f(n)=O(\text{10}^{-27})$

Asymtotic Notation (e.g. Big-Oh)

• Is not “about” algorithm
• Is a tool for describing (growth of) functions
• It is useful for describing functions related to algorithms + data structures, e.g.:
  • minimum or maximum time taken
  • minimum or maximum space needed
  • etc.
• We use it so often for worst-case time for an algorithm that we often leave implciit a statement like “let T(n) be the max time taken by algorithm A as an input of size as most n.” This statement is essential.

Ex. Complexity of Palindrome Checking

• using a stack & queue
• Algorithm:
  1. Insert all tokens into a stack & a queue
  2. Repeat pop one token; dequeue one token. if different report ‘no’.
• size of input = number of symbols or token
• each token is (all together O(1)):
  • pushed on the stack, O(1)
  • enqueued on the queue, O(1)
  • popped off the stack, O(1)
  • dequeued from the queue, O(1)
  • compared to one other token, O(1)
• $\text{n tokens} \implies n \times O(1)$ time in total
• So: $T(n) = n \times O(1) = O(n)$

What does $n\times O(1) = O(n)$ mean?

It means: $f(n) = O(1) \iff n\times f(n) = O(n)$

To see if it is true:

$f(n) = O(1) \iff \exists c>0 \space \text{s.t.} \space f(n) < c, \text{for any} \space n \in \Z\\ \iff \exists c>0 \text{ s.t. } n\times f(n) < c\times n, \text{for any } n \in \N\\ \iff n\times f(n) = O(n)$

End