2025-02-Algorithm/reviews/R4a.md

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

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*}