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reviews/R4a.md

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+# Review 4-1
+
+* Hajin Ju, 2024062806
+
+## Problem 1
+
+Show that the solution of $T(n) = 2T(\lfloor n / 2 \rfloor) + n $ is $O(n \lg n)$ by the substitution method. (Show the inductive step only.)
+
+### Solution 1
+
+* inductive step
+
+$$T(n) \leq c n \lg n, \quad (\text{for some}\; c > 0,\, n > n_0)$$
+
+We can assume the hypothesis($T(n) = O(n\lg n)$) holds for all positive int smaller than $n$.
+then, 
+
+$$T(\lfloor n / 2 \rfloor) \leq c \lfloor n / 2 \rfloor \lg {\lfloor n / 2 \rfloor} \leq c (n / 2) \lg ( n / 2)\\= c(n/2)(\lg n - \lg 2)$$
+
+$$\begin{align*}
+T(n) &= 2T(\lfloor n / 2 \rfloor) + n \leq cn(\lg n - \lg 2) + n\\
+&=cn\lg n - cn \lg 2 + n\\
+&=cn \lg - cn + n\\
+&\leq cn\lg n
+\end{align*}$$
+
+therefore, $T(n) = O(n \lg n)$
+
+## Problem 2
+
+Use a recursion tree to determine a good asymptotic upper bound on the recurrence $T(n) = 3T(\lfloor n / 4 \rfloor) + \theta(n^2)$.
+
+### Solution 2
+```mermaid
+flowchart TD
+    A["$$T(n):\;cn^2$$"] --> B1["$$T(n/4):\;cn^2/16$$"];
+    A --> B2["$$T(n/4):\;cn^2/16$$"];
+    A --> B3["$$T(n/4):\;cn^2/16$$"];
+
+    subgraph L0[ ]
+        A
+    end
+
+    subgraph L1[ ]
+        B1
+        B2
+        B3
+    end
+```
+
+* level 0
+  * $cn^2$
+* level 1
+  * $\frac{3}{16}cn^2$
+* level $k$
+  * $(\frac{3}{16})^k cn^2$
+* level $\log_4 n$
+  * $n^{\log_4 3}$
+
+
+therefore, total cost is 
+
+$$\begin{align*}
+T(n) &= \sum^{\log_4 n - 1}_{i = 0}(\frac{3}{16})^i cn^2 + \Theta(n^{\log_4 3})\\
+&< \sum^{\infty}_{i = 0}(\frac{3}{16})^i cn^2 + \Theta(n^{\log_4 3})\\
+&=\frac{16}{13}cn^2 +\Theta(n^{\log_4 3}) = O(n^2)
+\end{align*}
+$$