Question

Give a big \(\Theta\) bound on the solution to the recurrence $$ T(n)= \begin{cases}3 T(\lceil n / 2\rceil)+\sqrt{n+3} & \text { if } n>1 \\ d & \text { if } n=1\end{cases} $$

Short answer

The solution to the recurrence is \( T(n) = \Theta(n^{1.585}) \).

Step-by-step solution

Identify the Recurrence Relation

The given recurrence is \( T(n) = 3T(\lceil n / 2 \rceil) + \sqrt{n+3} \) for \( n > 1 \) and \( T(n) = d \) for \( n = 1 \), where \( d \) is a constant. This recurrence is similar to the Divide and Conquer type.

Apply the Master Theorem

To apply the Master Theorem, we need to identify parameters in the form \( T(n) = aT(n/b) + f(n) \). Here, \( a = 3 \), \( b = 2 \), and \( f(n) = \sqrt{n+3} \). According to the theorem, we compare \( f(n) = \sqrt{n+3} \) with \( n^{\log_b a} = n^{\log_2 3} \).

Compare \(f(n)\) with \(n^{\log_b a}\)

Calculate \( n^{\log_2 3} \). The value of \( \log_2 3 \) is approximately 1.585. Therefore, \( n^{\log_2 3} \approx n^{1.585} \). Now compare this with \( \sqrt{n+3} \), which simplifies to approximately \( \sqrt{n} \).

Determine the Dominant Term

Since \( n^{1/2} \) grows slower than \( n^{1.585} \), \( f(n) \) is asymptotically smaller than \( n^{\log_2 3} \). This places \( f(n) \) into the Master Theorem's Case 1, where \( f(n) = O(n^c) \) with \( c < \log_b a \).

Use Case 1 of the Master Theorem

Since the function \( f(n) \) is smaller than \( n^{\log_b a} \), by Case 1 of the Master Theorem, the solution to the recurrence \( T(n) = \Theta(n^{\log_b a}) = \Theta(n^{1.585}) \).

Key concepts

Recurrence Relation

In the world of algorithms, a recurrence relation is a mathematical way to define a sequence recursively, connecting each term to its predecessors. Specifically, when evaluating problems through a divide and conquer approach, recurrence relations become invaluable. For the given exercise, the relation is defined as follows: - For large values of \( n \), the problem is reduced into three smaller subproblems of size approximately half of \( n \), mathematically represented by \( T(n) = 3T(\lceil n / 2 \rceil) + \sqrt{n+3} \). - For the smallest size, \( n = 1 \), the solution is given as a constant \( d \).Recurrence relations help us break down and understand how the complexity of a problem grows with the size of the input \( n \). They guide us in expressing complex problems succinctly and analyzing their computational complexity.

Master Theorem

The Master Theorem is a handy tool for solving recurrence relations of the type \( T(n) = aT(n/b) + f(n) \). It gives us a way to determine the asymptotic behavior or the growth rate of these relations, especially prominent in divide and conquer algorithms. In our case:- We identify \( a = 3 \), \( b = 2 \), and \( f(n) = \sqrt{n+3} \).- We then compute \( n^{\log_b a} = n^{\log_2 3} \). Here, \( \log_2 3 \approx 1.585 \).The Master Theorem compares the function \( f(n) \) with \( n^{\log_b a} \):- If \( f(n) \) grows slower, equal, or faster than \( n^{\log_b a} \), we apply the respective case of the theorem to determine \( T(n) \).In this exercise, \( f(n) = \sqrt{n+3} \), which grows slower than \( n^{1.585} \).Thus, applying **Case 1** of the theorem, we conclude that \( T(n) = \Theta(n^{1.585}) \).

Asymptotic Analysis

Asymptotic analysis is crucial for understanding the efficiency of algorithms as input sizes become very large. It helps provide a high-level understanding by concentrating on the dominant factors and ignoring lower order terms and constants.In the context of our recurrence relation:- We examined how \( T(n) = \Theta(n^{1.585}) \) was derived, focusing on the relationship between \( f(n) = \sqrt{n+3} \) and \( n^{1.585} \).Asymptotically, we express the solution using "big Theta" notation, which defines an average-case running time. This allows us to present a tight bound on growth:- The expression \( \Theta(n^{1.585}) \) suggests that as \( n \) grows larger, the time complexity will be proportional to \( n^{1.585} \).By focusing only on the highest growth rates, asymptotic analysis simplifies comparing algorithms, especially when typical constant factors are not feasible to compute or compare directly.