Question

Recursion trees work, even if the problems do not break up geometrically or if the work per level is not \(n^{c}\) units. Draw recursion trees and find the best big \(O\) bounds you can for solutions to the following recurrences. For each, assume that \(T(1)=1\). a. \(T(n)=T(n-1)+n\) b. \(T(n)=2 T(n-1)+n\) c. \(T(n)=T(\lfloor\sqrt{n}\rfloor)+1\) (Assume \(n\) has the form \(n=2^{2^{i}}\).) d. \(T(n)=2 T(n / 2)+n \log n\) (Assume \(n\) is a power of 2 .)

Short answer

a. \(O(n^2)\); b. \(O(2^n)\); c. \(O(\log \log n)\); d. \(O(n \log^2 n)\).

Step-by-step solution

Understanding Recurrence a

The recurrence relation is given by \( T(n) = T(n-1) + n \). This implies that each recursive call reduces the problem size by 1 and adds \( n \) amount of work. The base case is \( T(1) = 1 \). We will construct a recursion tree to help find the big \( O \) bound.

Drawing the Recursion Tree for a

At the top (root), we have \( T(n) \), contributing \( n \) units of work. At the next level, we have the contribution \( T(n-1) \), which accounts for \( n-1 \) units of work. This pattern continues down to the base case where \( T(1) = 1 \), with each level contributing decreasing amounts: \( n, n-1, n-2, \ldots, 2, 1 \).

Calculating Total Work for a

The work can be seen as the sum of the first \( n \) natural numbers: \( T(n) = n + (n-1) + (n-2) + \ldots + 1 \). The sum of the first \( n \) natural numbers is known to be \( \frac{n(n+1)}{2} \). Thus, the total work is \( O(n^2) \).

Understanding Recurrence b

The recurrence relation is \( T(n) = 2T(n-1) + n \). This implies each level of the tree contributes \( 2T(n-1) \) plus linear work \( n \). The base case is \( T(1) = 1 \).

Drawing the Recursion Tree for b

For level 0 (root), the work is \( n \). At level 1, there are 2 subproblems each contributing \( T(n-1) \) with linear overhead operations totaling 2n. For level \( k \), there are \( 2^k \) calls to \( T(n-k) \), each contributing \( n-k \).

Calculating Total Work for b

The total work contributes \( n + 2n + 4(n-1) + 8(n-2) + \ldots \) until the base case. This process is more complex but leads to an exponential formula. Approximately, when summing, it results in \( O(2^n) \).

Understanding Recurrence c

The recurrence \( T(n) = T(\lfloor \sqrt{n} \rfloor) + 1 \) reduces the problem size significantly each time by taking the square root, with minor constant work at each step. Assume \( n = 2^{2^i} \) and \( T(1) = 1 \).

Drawing the Recursion Tree for c

The recursion tree depth increases by reducing by powers of 2. Each level carries constant work 1. At each reduction, the problem reduces by squares: \( n \to \sqrt{n} \to \sqrt{\sqrt{n}} \).

Calculating Total Work for c

In this form, the number of steps is \( O(\log \log n) \), as each reduction is exponentially less. Hence, total work is \( O(\log \log n) \).

Understanding and Solving Recurrence d

The relation is \( T(n) = 2 T(n / 2) + n \log n \). Using the Master Theorem, where \( a = 2, b = 2, f(n) = n\log n \), the solution falls into the third case of the Master's Theorem.

Application of Master Theorem for d

Since \( f(n) = n \log n \) is more than \( n^{\log_b a} = n \), according to the third case of the Master Theorem, the dominating term for complexity is \( f(n) = n\log n \). Thus, the solution is \( O(n \log^2 n) \).

Key concepts

Recurrence Relations

Recurrence relations are mathematical expressions that define sequences and other structures using their own preceding terms. They are widely used in *algorithm analysis* to establish how an algorithm runs on different inputs. Consider them like a formula that repeats over itself with a changing variable. This way, we can understand how the complexity evolves as the input grows.

A simple example is the Fibonacci sequence, where each term is the sum of the two preceding ones. Here, a recurrence relation allows us to capture the essence of recursive problems by describing their progression step by step.

In recursion trees, these relations help visualize how recursive calls evolve. Each node in a tree represents a subproblem, and the descendants show how it splits into smaller subproblems, dictated by the recurrence relation itself.

Big O Notation

Big O Notation is a mathematical concept used to describe the upper bound of an algorithm's runtime or space requirements. It focuses on the behavior of an algorithm as the input size approaches infinity, which helps us compare its efficiency regardless of hardware or other external factors.

Understanding Big O is crucial for algorithm analysis, allowing us to prioritize efficiency in coding practices. It defines how functions grow and lets us quickly identify whether an algorithm is practical for large inputs or not.

In the context of recursion trees, determining the Big O complexity involves analyzing the nature and depth of the tree. For instance, summing the nodes' contributions helps establish how quickly the recursive calls accumulate work, leading to a final complexity expression.

Master Theorem

The Master Theorem provides a straightforward method to determine the time complexity of recurrence relations that fit a specific form. It is mostly used with divide-and-conquer algorithms, which solve problems by breaking them into subproblems of similar type and recombining their solutions.

It applies to recurrences of the form: \[ T(n) = a \cdot T\left(\frac{n}{b}\right) + f(n) \] where:

• \( a \) is the number of subproblems in the recursion,
• \( b \) is the factor by which the subproblem size reduces in each call,
• \( f(n) \) represents the cost of work done outside the recursive calls.

By evaluating \( f(n) \) relative to \( n^{\log_b a} \), we classify the recurrence into three cases:

• If \( f(n) = \Theta(n^{\log_b a}) \), the solution is \( \Theta(n^{\log_b a} \cdot \log n) \).
• If \( f(n) = o(n^{\log_b a}) \), the solution is \( \Theta(n^{\log_b a}) \).
• If \( f(n) = \Omega(n^{\log_b a}) \), the solution is \( \Theta(f(n)) \).

By using the Master Theorem, we're able to deduce outcomes efficiently, making it easier to analyze complex recursive algorithms.

Algorithm Analysis

Algorithm analysis is the process of determining the amount of resources necessary to execute an algorithm. The primary aim is to ascertain both time and space complexity, allowing an understanding of an algorithm's efficiency.

When analyzing algorithms, we often rely on tools such as Big O Notation and the Master Theorem to help determine their efficiency in worst-case scenarios or average cases. This gives us insights into whether an algorithm is suitable for particular applications or data sizes.

Key considerations in algorithm analysis:

• Understand the problem and the algorithm: Get familiar with how the algorithm systematically breaks down and resolves the problem, especially in terms of recursion.
• Identify recurrence relations: These relations outline how the input problem splits into smaller subproblems.
• Evaluate the solution using Big O and Master Theorem: Determine how the recursive nature affects time and space complexity.

Algorithm analysis allows computer scientists and programmers to design algorithms that deliver solutions efficiently and perform reliably across a variety of use cases.