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