# 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 

$$\begin{align*}
T(n) &= \sum^{\log_4 n - 1}_{i = 0}(\frac{3}{16})^i cn^2 + \Theta(n^{\log_4 3})\\
&< \sum^{\infty}_{i = 0}(\frac{3}{16})^i cn^2 + \Theta(n^{\log_4 3})\\
&=\frac{16}{13}cn^2 +\Theta(n^{\log_4 3}) = O(n^2)
\end{align*}
$$