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reviews/R3.md

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+# 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 $g(n) = \Theta(g(n))$