Question

Suppose that, in a divide-and-conquer algorithm, we always divide an in stance of size \(n\) of a problem into 10 subinstances of size \(n / 3,\) and the dividing and combining steps take a time in \(\Theta\left(n^{2}\right)\). Write a recumence equation for the running time \(T(n),\) and solve the equation for \(T(n)\).

Short answer

The running time T(n) of the algorithm is \(\Theta(n^{2}log(n))\).

Step-by-step solution

Identify the Dividing and Combining Operation

The problem states that the algorithm is a divide-and-conquer variety, with a problem of size n being divided into 10 subproblems of size n/3. This means we can say the division operation happens with 10 cases each of size n/3. Also, it further asserts that the divide and combine steps take \(\Theta\left(n^{2}\right)\) time.

Define the Recurrence Relation

Combining the above information, we formulate the recurrence relation for the run-time T(n) of the algorithm as \(T(n) = 10T(n/3) + \Theta\left(n^{2}\right)\). This equation is made up of two parts, the first term 10T(n/3) represents the time taken to solve the 10 subproblems of size n/3 and the second term \(\Theta\left(n^{2}\right)\) represents time taken to divide and later combine the solutions.

Solve the Recurrence Relation

The solution to the recurrence relation is obtained using the Master Theorem. The Master Theorem applies to recurrences of the form \(T(n) = aT(n/b) + f(n)\). In this case, a=10, b=3, and f(n) = \(\Theta\left(n^{2}\right)\). Comparing this with \(n^{log_{b}{a}} = n^{log_{3}{10}}\), it is seen that \(f(n)\) grows faster. Therefore, the time complexity T(n) belongs to \(\Theta(n^{2}log(n))\).

Key concepts

Recurrence Equation

A Recurrence Equation is an essential concept in analyzing divide-and-conquer algorithms. It represents the algorithm's time complexity in terms of itself but with a reduced problem size. For our example, the recurrence equation is derived from the given scenario: the algorithm divides a problem of size \( n \) into 10 subproblems, each of size \( n/3 \), and the process of dividing and combining takes \( \Theta(n^{2}) \) time. This results in the recurrence equation:

• \( T(n) = 10T(n/3) + \Theta(n^{2}) \)

This reflects the total time for solving a problem of size \( n \), as it involves solving 10 smaller problems and then combining their solutions. An intuitive way to visualize this is by imagining the complexity of managing multiple smaller puzzles to solve the larger one.

Master Theorem

The Master Theorem is a powerful tool that provides a straightforward way to determine the asymptotic behavior of recurrence relations of the form:

• \( T(n) = aT(n/b) + f(n) \)

In our case, the equation \( T(n) = 10T(n/3) + \Theta(n^{2}) \) fits nicely with this form. Here, \( a = 10 \), \( b = 3 \), and \( f(n) = \Theta(n^{2}) \). Understanding the theorem involves comparing the function \( f(n) \) with \( n^{\log_b a} \).Since \( n^{\log_{3}10} \) (approximately \( n^{2.095} \)) grows faster than \( \Theta(n^{2}) \), we conclude based on the Master Theorem that the function \( f(n) \) dictates the overall time complexity. This comparison helps us deduce that the time complexity falls into \( \Theta(n^{2}\log n) \), offering a concise yet comprehensive way to solve recurrence equations like the one discussed.

Time Complexity Analysis

Time Complexity Analysis allows us to estimate the efficiency of an algorithm, especially its scalability with large inputs. For divide-and-conquer strategies, analyzing time complexity often involves solving recurrence relations through methods like the Master Theorem.For our specific example, the established time complexity \( \Theta(n^{2}\log n) \) indicates a performance that combines both polynomial and logarithmic elements. The polynomial part \((n^2)\) arises from the divide/combine steps, whereas the logarithmic factor \((\log n)\) is typical of divide-and-conquer approaches where the problem size is recurrently halved.This analysis becomes crucial when selecting algorithms for practical applications since it predicts how program efficiency will scale. Understanding this concept can aid you in choosing the right approach for handling large-scale problems efficiently.