Question

Consider algorithm solve given below. This algorithm solves problem \(P\) by finding the output (solution) \(O\) corresponding to any input \(l\). void solve (input I, output& O) { if (size (I) == 1) find solution O directly; else{ partition I into 5 inputs I1, I2, I3, I4, I5, where size (Ij) = size (I)/3 for j = 1, ..., 5; for (j = 1; j < = 5; j++) solve (Ij, Oj); combine O1, O2, O3, O4, O5 to get O for P with input I; } } Assume \(g(n)\) basic operations for partitioning and combining and no basic operations for an instance of size 1 a. Write a recurrence equation \(T(n)\) for the number of basic operations needed to solve \(P\) when the input size is \(n\) b. What is the solution to this recurrence equation if \(g(n) \in \Theta(n) ?\) (Proof is not required.) c. Assuming that \(g(n)=n^{2}\), solve the recurrence equation exactly for \(n=27\) d. Find the general solution for \(n\) a power of 3

Short answer

a. \(T(n) = 5T(n/3) + g(n)\) \nb. \(T(n) = \Theta(n^{\log_{3}5})\) \nc. \(T(27) = 2025\) \nd. \(T(n) = \sum_{i=0}^{\log_3 n} 5^{i} * (\frac{n}{3^i})^{2}\)

Step-by-step solution

Formulate the recurrence relation

First, based on the algorithm, we have that for a given input \(I\) of size \(n\), the algorithm will partition the input into 5 sub-inputs each of size \(n/3\). It will then solve each of these independently. So the recurrence relation is in the form: \(T(n) = 5T(n/3) + g(n)\)

Solve when \(g(n) \in \Theta(n)\)

Next, when \(g(n) \in \Theta(n)\), we apply the Master Theorem. The Master Theorem states that a recurrence relation of the form \(T(n) = aT(n/b) + \Theta(n^{c})\), \(a > 0, b > 1, c \geq 0\), has a solution in \n1. \(\Theta(n^{c}\log_{b}n)\), if \(a = b^{c}\) \n2. \(\Theta(n^{c})\), if \(a < b^{c}\) \n3. \(\Theta(n^{\log_{b}a})\), if \(a > b^{c}\) \nHere, \(a = 5, b = 3, c = 1\). So, \(a > b^{c}\), and therefore, \(T(n) = \Theta(n^{\log_{3}5})\)

Solve the recurrence for \(g(n)=n^{2}\) and \(n=27\)

Now, let's assume \(g(n)=n^{2}\). Then \(T(n) = 5T(n/3) + n^{2}\) For \(n=27\), we calculate the basic operations as follows: \(T(27) = 5T(9) + 27^{2}\), calculate \(T(9) = 5T(3)+9^{2}\), and finally \( T(3)=5T(1)+3^{2}\). Since for \(T(1)\) there are no basic operations, we get \(T(27) = 5^3 * T(1) + 5^2 * 9^{2} + 5 * 27^{2} + 27^{2} = 5^2 * 9^{2} + 5 * 27^{2} + 27^{2} = 2025.\)

Find the general solution for \(n\) a power of 3

Lastly, assume that \(n\) is a power of 3, i.e., \(n = 3^k\), then the total operations of the algorithm can be solved by summing up from level 0 to log_{3}n of the operations and must be multiplied by the corresponding numbers of nodes at each level. You'll get \(T(n) = \sum_{i=0}^{\log_3 n} 5^{i} * (\frac{n}{3^i})^{2}\). Note that \(n/3^i\) equals \(3^{\log_3 n - i}\), which in turn equals \(3^{k - i}\), where \(k = \log_3 n\).

Key concepts

Algorithm Analysis

Algorithm analysis is a critical field in computer science that aims to determine the computational complexity of algorithms in terms of time and space used. It involves understanding how the execution time or memory requirements of an algorithm grow relative to the size of the input.

Consider the algorithm in the task, which employs a divide and conquer strategy. Here, a problem of size n is divided into five smaller problems, each of size n/3. To analyze this algorithm, we use a recurrence relation that represents how the time complexity T(n) develops based on the size of the input.

The recurrence relation serves as a mathematical model that incorporates the time to split the problem and combine the solutions of subproblems. Specifically, for our algorithm, the relation is given by T(n) = 5T(n/3) + g(n), where g(n) encapsulates the basic operations for partitioning and combining. Using this model, we can predict and evaluate the cost of the algorithm over inputs of different sizes, which is particularly helpful when choosing the most efficient algorithm for a given task.

Master Theorem

The Master Theorem provides a method to get a quick solution for recurrence relations of divide and conquer algorithms, such as the one given in our task. These recurrences typically take the form T(n) = aT(n/b) + f(n), where n is the size of the problem, a is the number of subproblems, and b is the factor by which the problem size is reduced in every split.

In the context of the given exercise, we use the Master Theorem to deduce the time complexity when the combining function g(n) is in Theta(n). The theorem outlines three cases to consider, depending on the relationship between a, b, and c. These cases ultimately help us categorize the time complexity into distinct classes like Theta(n^{log_b a}), giving us a powerful tool to quickly determine an algorithm's efficiency without the need for extensive derivation.

Divide and Conquer

Divide and conquer is an algorithm design paradigm based on multi-branched recursion. An algorithm that uses the divide and conquer approach divides the problem into several subproblems that are similar to the original problem but smaller in size, solves these subproblems recursively, and then combines their solutions to create a solution to the original problem.

In the solve function outlined in our task, the problem is divided into 5 subproblems, each one-third the size of the original. This division is the 'divide' part of the paradigm. Each of these subproblems is then themselves solved using the same solve function, which represents the 'conquer' step. Finally, the solutions to the subproblems are combined, which demonstrates the 'combine' aspect of divide and conquer.

This paradigm is especially effective for problems where the solution can be constructed from solutions to smaller instances, such as sorting algorithms (like mergesort), and for optimization problems, where the best solution is chosen from among the solutions to the subproblems. By breaking down complex problems into simpler, manageable pieces, divide and conquer dramatically improves readability, maintainability, and often efficiency of algorithmic solutions.