# Review 3

* Hajin Ju, 2024062806

## Problem 1

Write `O` if an entry is true or `X` otherwise.

### Solution 1

|               | $$O(n \lg n)$$ | $$\Omega(n \lg n)$$ | $$\Theta(n\lg n)$$ |
| :-----------: | :------------: | :-----------------: | :----------------: |
|   $$\lg n$$   |       O        |          X          |         X          |
|     $$n$$     |       O        |          X          |         X          |
|  $$n \lg n$$  |       O        |          O          |         O          |
| $$n \lg^2 n$$ |       X        |          O          |         X          |
|    $$n^2$$    |       X        |          O          |         X          |


## Problem 2

Show $3n + 1 = O(n^2)$ by the definition of $O$.


### Solution 2

A function $f(n) = O(g(n))$ if there exist constants $c\geq 0$ and $n_0\geq 0$, s.t.

$$n \geq n_0 \Rightarrow \leq |f(n)| \leq c |g(n)|$$

---

let $g(n) = n^2$
let $f(n) = 3n+1$

suppose $c=4,\, n_0 = 1$

and then, for all $n \geq 1 \to |3n+1| \leq 4n^2$

therefore, $3n+1 = O(n^2)$ by the above definition.


## Problem 3

Write asymptotic notations that satisfy each relation and explain why.

1. Transitivity
2. Reflexivity
3. Symmetry

### Solution 3

1. Transitivity

* $O$ is transitive
because $f(n) = O(g(n))$ and $g(n) = O(h(n))$ implies $f(n) = O(h(n))$
there must exists $n_0\geq 0$, s.t.  $n \geq n_0 \Rightarrow f(n) \leq c_0 g(n) \leq c_1 c_0 h(n)$

* $\Omega$ is transitive
because $f(n) = \Omega(g(n))$ and $g(n) = \Omega(h(n))$ implies $f(n) = \Omega(h(n))$
there must exists $n_0\geq 0$, s.t.  $n \geq n_0 \Rightarrow f(n) \geq c_0g(n) \geq c_1 c_0 h(n)$

* $\Theta$ is transitive
because $f(n) = \Theta(g(n))$ and $g(n) = \Theta(h(n))$ implies $f(n) = \Theta(h(n))$
$$f(n) = O(h(n)) \land f(n) = \Omega(h(n))$$

2. Reflexivity

* $O$ is reflexive
because $f(n) = O(f(n))$ where $c=1$
* $\Omega$ is reflexive
because $f(n) = \Omega(f(n))$ where $c=1$
* $\Theta$ is reflexive
* because $f(n) = \Theta(f(n))$

3. Symmetry

* $O$ is **not** symmetric
because $f(n) = O(g(n))$ does not imply $g(n) = O(g(n))$
for example, $n = O(n^2)$ cannot imply $n^2 = O(n)$
* $\Omega$ is **not** symmetric
because $f(n) = \Omega(g(n))$ does not imply $g(n) = \Omega(g(n))$
for example, $n^2 = \Omega(n)$ cannot imply $n = \Omega(n^2)$
* $\Theta$ is symmetric
because $f(n) = \Theta(g(n))$ implies $g(n) = \Theta(g(n))$