Based on the master theorem, what is the solution to T (n) = 16 (7) + n ○e (n³) e (n² log n) ○e (¹) 00 (n2)
Based on the master theorem, what is the solution to T (n) = 16 (7) + n ○e (n³) e (n² log n) ○e (¹) 00 (n2)
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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![### Master Theorem: Solving Recurrence Relations
#### Question:
Based on the master theorem, what is the solution to \( T(n) = 16 \left( \frac{n}{4} \right) + n \)?
#### Options:
1. \( \Theta(n^3) \)
2. \( \Theta(n^2 \log n) \) *(selected option)*
3. \( \Theta(n^4) \)
4. \( \Theta(n^2) \)
#### Explanation:
To determine the solution to the recurrence relation \( T(n) = 16 \left(\frac{n}{4}\right) + n \), we apply the Master Theorem for divide-and-conquer recurrences of the form:
\[ T(n) = aT\left(\frac{n}{b}\right) + f(n) \]
Here, \( a = 16 \), \( b = 4 \), and \( f(n) = n \).
We compare \( f(n) \) with \( n^{\log_b a} \):
- Calculate \( \log_b a \):
\[
\log_b a = \log_4 16 = 2
\]
So, \( f(n) = n \) is compared with \( n^2 \).
The Master Theorem has three cases:
1. If \( f(n) = O\left(n^{c-\epsilon}\right) \) for some constant \( \epsilon > 0 \), then \( T(n) = \Theta\left(n^c\right) \).
2. If \( f(n) = \Theta\left(n^c \log^k n \right) \), then \( T(n) = \Theta\left(n^c \log^{k+1} n \right) \).
3. If \( f(n) = \Omega\left(n^{c+\epsilon}\right) \), and if \( af\left(\frac{n}{b}\right) \leq kf(n) \) for some \( k < 1 \) and large n, then \( T(n) = \Theta(f(n)) \).
In this case, \( f(n) = n \) and \( n^2 = n^{\log_4 16} \), so we are in the first case as \( f(n) \) grows slower than \(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb4fac401-f688-4a8e-b637-a2500ce6ac46%2F97d353af-067f-42ca-92c4-67b9c35b2652%2Fstw97kr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Master Theorem: Solving Recurrence Relations
#### Question:
Based on the master theorem, what is the solution to \( T(n) = 16 \left( \frac{n}{4} \right) + n \)?
#### Options:
1. \( \Theta(n^3) \)
2. \( \Theta(n^2 \log n) \) *(selected option)*
3. \( \Theta(n^4) \)
4. \( \Theta(n^2) \)
#### Explanation:
To determine the solution to the recurrence relation \( T(n) = 16 \left(\frac{n}{4}\right) + n \), we apply the Master Theorem for divide-and-conquer recurrences of the form:
\[ T(n) = aT\left(\frac{n}{b}\right) + f(n) \]
Here, \( a = 16 \), \( b = 4 \), and \( f(n) = n \).
We compare \( f(n) \) with \( n^{\log_b a} \):
- Calculate \( \log_b a \):
\[
\log_b a = \log_4 16 = 2
\]
So, \( f(n) = n \) is compared with \( n^2 \).
The Master Theorem has three cases:
1. If \( f(n) = O\left(n^{c-\epsilon}\right) \) for some constant \( \epsilon > 0 \), then \( T(n) = \Theta\left(n^c\right) \).
2. If \( f(n) = \Theta\left(n^c \log^k n \right) \), then \( T(n) = \Theta\left(n^c \log^{k+1} n \right) \).
3. If \( f(n) = \Omega\left(n^{c+\epsilon}\right) \), and if \( af\left(\frac{n}{b}\right) \leq kf(n) \) for some \( k < 1 \) and large n, then \( T(n) = \Theta(f(n)) \).
In this case, \( f(n) = n \) and \( n^2 = n^{\log_4 16} \), so we are in the first case as \( f(n) \) grows slower than \(
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