Question

Suppose that, in a divide-and-conquer algorithm, we always divide an instance 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 recurrence equation for the running time \(T(n),\) and solve the equation for \(T(n)\).

Short answer

Using the Master's theorem, the solution for the recurrence equation \(T(n) = 10T(n / 3) + n^{2}\) is \(T(n) \in \Theta(n^2)\).

Step-by-step solution

Define the Recurrence Relation

For the given problem scenario, the divide-and-conquer recurrence equation can be defined as \(T(n) = 10T(n / 3) + \Theta(n^{2})\). This is because the problem of size \(n\) is divided into 10 sub-problems of size \(n / 3\) (hence, 10T(n / 3)) and the time to divide and combine the problems is \(\Theta(n^{2})\).

Determine the form of Recurrence

The problem follows a multiplicative decrease-increase model, i.e., \(aT(n/b) + f(n)\), where \(a > 1, b > 1\), and \(f(n)\) is asymptotically positive. In this case, \(a = 10, b = 3\) and \(f(n) = n^{2}\). Set \(k = \log_b{a} = \log_3{10}\). Depending on \(k\) and the degree of \(f(n)\), different cases of the master theorem can be applied to solve the recurrence.

Solving the Recurrence

By Master's Theorem, for \(aT(n/b) + f(n)\) if \(f(n) \in \Theta (n^c)\), where \(c > \log_b{a}\) then \(T(n) \in \Theta (n^c)\). Here \(c = 2 > \log_3{10}\), criteria for case-3 of Master's theorem is satisfied. Hence, \(T(n) \in \Theta (n^2)\).

Key concepts

Recurrence Relation

To grasp the divide-and-conquer algorithm in the given exercise, understanding recurrence relations is crucial. A recurrence relation is a mathematical expression of how a problem's solution relates to solutions of smaller instances of the same problem. It is a way to describe the time complexity of recursive algorithms through equations that represent these recursive steps.

In the context of our problem, we have the recurrence relation:

• \( T(n) = 10T(n/3) + \Theta(n^2) \)

This indicates that the algorithm breaks down a problem of size \(n\) into 10 smaller problems, each of size \(n/3\). The term \(10T(n/3)\) quantifies how the problem scales, while \(\Theta(n^2)\) represents the time taken for the combining phase of the algorithm.

By defining the recurrence relation, we set the groundwork to analyze and solve the problem using methods such as the Master’s Theorem, allowing us to find a closed-form solution for the running time \(T(n)\).

Master's Theorem

Master's Theorem provides a straightforward way to determine the time complexity of recurrence relations that arise in divide-and-conquer algorithms. It helps us solve recurrence relations of the form:

• \( T(n) = aT(n/b) + f(n) \)

Here, \(a\) is the number of subproblems, \(b\) is the factor by which the subproblem size decreases, and \(f(n)\) represents the cost of combining the solutions.

With our problem:

• \(a = 10\), \(b = 3\), and \(f(n) = n^2\)
• The key comparison is \(c\), the exponent in \(f(n)\), against \(\log_b{a}\)

Since \(c = 2\) and \(\log_3{10}\) is less than 2, we apply case 3 of the Master's Theorem. This case shows us that when \(f(n)\) grows faster than the critical threshold, the time complexity is driven by \(f(n)\).

Thus, Master's Theorem tells us that the solution to the recurrence is \(T(n) \in \Theta(n^2)\), simplifying the complex recurrence to a straightforward result.

Algorithm Complexity

Algorithm complexity is a measure of the time and space required for an algorithm to solve a problem. In divide-and-conquer algorithms, it's important to understand how to express these complexities when the algorithm involves recursive processes. The problem in question defines the complexity of dividing, conquering, and combining steps through recurrence relations.

In our case, after applying the Master's Theorem, we derive that the running complexity for the algorithm is:

• \( T(n) \in \Theta(n^2) \)

This complexity indicates that the time required grows quadratically with the size of the input \(n\). Simple polynomial complexities like \(\Theta(n^2)\) are easily interpretable, providing clear insights into how scaling up the problem size impacts performance, making it practical to anticipate how efficient an algorithm will be for large data.

Understanding algorithm complexity not only helps in theoretical analysis but also informs decisions in practical applications, optimizing algorithms for better performance in real-world scenarios.