### 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 \(

Answer is: Θ(n2) The running time of divide and conquer algorithms may be calculated using the…

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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)

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Chapter1: Introduction

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