86 lines
2.3 KiB
Markdown
86 lines
2.3 KiB
Markdown
# Review 3
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* Hajin Ju, 2024062806
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## Problem 1
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Write `O` if an entry is true or `X` otherwise.
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### Solution 1
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| | $$O(n \lg n)$$ | $$\Omega(n \lg n)$$ | $$\Theta(n\lg n)$$ |
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| :-----------: | :------------: | :-----------------: | :----------------: |
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| $$\lg n$$ | O | X | X |
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| $$n$$ | O | X | X |
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| $$n \lg n$$ | O | O | O |
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| $$n \lg^2 n$$ | X | O | X |
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| $$n^2$$ | X | O | X |
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## Problem 2
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Show $3n + 1 = O(n^2)$ by the definition of $O$.
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### Solution 2
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A function $f(n) = O(g(n))$ if there exist constants $c\geq 0$ and $n_0\geq 0$, s.t.
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$$n \geq n_0 \Rightarrow \leq |f(n)| \leq c |g(n)|$$
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---
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let $g(n) = n^2$
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let $f(n) = 3n+1$
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suppose $c=4,\, n_0 = 1$
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and then, for all $n \geq 1 \to |3n+1| \leq 4n^2$
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therefore, $3n+1 = O(n^2)$ by the above definition.
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## Problem 3
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Write asymptotic notations that satisfy each relation and explain why.
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1. Transitivity
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2. Reflexivity
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3. Symmetry
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### Solution 3
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1. Transitivity
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* $O$ is transitive
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because $f(n) = O(g(n))$ and $g(n) = O(h(n))$ implies $f(n) = O(h(n))$
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there must exists $n_0\geq 0$, s.t. $n \geq n_0 \Rightarrow f(n) \leq c_0 g(n) \leq c_1 c_0 h(n)$
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* $\Omega$ is transitive
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because $f(n) = \Omega(g(n))$ and $g(n) = \Omega(h(n))$ implies $f(n) = \Omega(h(n))$
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there must exists $n_0\geq 0$, s.t. $n \geq n_0 \Rightarrow f(n) \geq c_0g(n) \geq c_1 c_0 h(n)$
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* $\Theta$ is transitive
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because $f(n) = \Theta(g(n))$ and $g(n) = \Theta(h(n))$ implies $f(n) = \Theta(h(n))$
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$$f(n) = O(h(n)) \land f(n) = \Omega(h(n))$$
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2. Reflexivity
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* $O$ is reflexive
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because $f(n) = O(f(n))$ where $c=1$
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* $\Omega$ is reflexive
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because $f(n) = \Omega(f(n))$ where $c=1$
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* $\Theta$ is reflexive
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* because $f(n) = \Theta(f(n))$
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3. Symmetry
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* $O$ is **not** symmetric
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because $f(n) = O(g(n))$ does not imply $g(n) = O(g(n))$
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for example, $n = O(n^2)$ cannot imply $n^2 = O(n)$
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* $\Omega$ is **not** symmetric
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because $f(n) = \Omega(g(n))$ does not imply $g(n) = \Omega(g(n))$
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for example, $n^2 = \Omega(n)$ cannot imply $n = \Omega(n^2)$
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* $\Theta$ is symmetric
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because $f(n) = \Theta(g(n))$ implies $g(n) = \Theta(g(n))$
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