### 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:\[ T(n) = \Theta(f(n)) = \Theta(n^2) \]

Master's theorem: To apply master's theorem, the recurrence relation must be in this form: T(n) =…

Answered: Based on the master theorem, what is…

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

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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