Based on the master theorem, what is the solution to T (n) = 3T (2/2) + n² (n²) oe (n² logn) (n³) e (2¹)

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
icon
Related questions
Question
### Master Theorem Application

**Question:**
Based on the master theorem, what is the solution to \( T(n) = 3T\left(\frac{n}{2}\right) + n^2 \)?

**Options:**
- \( \Theta(n^2) \)
- \( \Theta(n^2 \log n) \)
- \( \Theta(n^3) \)
- \( \Theta(2^n) \)
  
_Selected Answer:_
- \( \Theta(2^n) \)

**Explanation:**
The given recurrence relation is \( T(n) = 3T\left(\frac{n}{2}\right) + n^2 \).

To solve this using the master theorem for divide-and-conquer recurrences of the form \( T(n) = aT\left(\frac{n}{b}\right) + f(n) \):

- \( a = 3 \)
- \( b = 2 \)
- \( f(n) = n^2 \)

Calculate \( \log_b{a} \):
\[ \log_2{3} \approx 1.585 \]

Compare \( f(n) \) to \( n^{\log_b{a}} \):
- \( f(n) = n^2 \)
- \( n^{\log_2{3}} \approx n^{1.585} \)

Since \( f(n) = n^2 \) grows faster than \( n^{\log_2{3}} \), we fall into Case 3 of the master theorem, where \( f(n) = \Omega(n^{c}) \) for \( c > \log_b{a} \).

Thus, the solution is:
\[ T(n) = \Theta(f(n)) = \Theta(n^2) \]
Transcribed Image Text:### Master Theorem Application **Question:** Based on the master theorem, what is the solution to \( T(n) = 3T\left(\frac{n}{2}\right) + n^2 \)? **Options:** - \( \Theta(n^2) \) - \( \Theta(n^2 \log n) \) - \( \Theta(n^3) \) - \( \Theta(2^n) \) _Selected Answer:_ - \( \Theta(2^n) \) **Explanation:** The given recurrence relation is \( T(n) = 3T\left(\frac{n}{2}\right) + n^2 \). To solve this using the master theorem for divide-and-conquer recurrences of the form \( T(n) = aT\left(\frac{n}{b}\right) + f(n) \): - \( a = 3 \) - \( b = 2 \) - \( f(n) = n^2 \) Calculate \( \log_b{a} \): \[ \log_2{3} \approx 1.585 \] Compare \( f(n) \) to \( n^{\log_b{a}} \): - \( f(n) = n^2 \) - \( n^{\log_2{3}} \approx n^{1.585} \) Since \( f(n) = n^2 \) grows faster than \( n^{\log_2{3}} \), we fall into Case 3 of the master theorem, where \( f(n) = \Omega(n^{c}) \) for \( c > \log_b{a} \). Thus, the solution is: \[ T(n) = \Theta(f(n)) = \Theta(n^2) \]
Expert Solution
steps

Step by step

Solved in 2 steps

Blurred answer
Knowledge Booster
Time complexity
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, computer-science and related others by exploring similar questions and additional content below.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Database System Concepts
Database System Concepts
Computer Science
ISBN:
9780078022159
Author:
Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:
McGraw-Hill Education
Starting Out with Python (4th Edition)
Starting Out with Python (4th Edition)
Computer Science
ISBN:
9780134444321
Author:
Tony Gaddis
Publisher:
PEARSON
Digital Fundamentals (11th Edition)
Digital Fundamentals (11th Edition)
Computer Science
ISBN:
9780132737968
Author:
Thomas L. Floyd
Publisher:
PEARSON
C How to Program (8th Edition)
C How to Program (8th Edition)
Computer Science
ISBN:
9780133976892
Author:
Paul J. Deitel, Harvey Deitel
Publisher:
PEARSON
Database Systems: Design, Implementation, & Manag…
Database Systems: Design, Implementation, & Manag…
Computer Science
ISBN:
9781337627900
Author:
Carlos Coronel, Steven Morris
Publisher:
Cengage Learning
Programmable Logic Controllers
Programmable Logic Controllers
Computer Science
ISBN:
9780073373843
Author:
Frank D. Petruzella
Publisher:
McGraw-Hill Education