2025-02-Algorithm/reviews/R3.md

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)|

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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))