Question

Find the best big \(O\) upper bound you can to any solution to the recurrence defined on nonnegative integers by $$ T(n) \leq 2 T\left(\left[\frac{n}{2}\right]+1\right)+c n . $$ (There is nothing wrong with informed guesswork.) Prove by induction that your answer is correct.

Short answer

The best big \(O\) upper bound is \(O(n \log n)\).

Step-by-step solution

Identify the Recurrence Relation

The given recurrence relation is \( T(n) \leq 2T\left(\left\lfloor \frac{n}{2} \right\rfloor + 1\right) + cn \), where \( cn \) is the cost of splitting the problem, and \( 2 \) indicates the number of subproblems. The floor function adjusts the input size for recursive calls.

Establish the Guess

We make an informed guess that \( T(n) \) has an upper bound of \( O(n \log n) \). This is based on its similarity to the Merge Sort recurrence \( T(n) = 2T\left(\frac{n}{2}\right) + cn \), which has a solution of \( O(n \log n) \). Due to the increment inside \( T \), we suspect a similar but slightly adjusted bound might still hold.

Induction Hypothesis

Assume for some \( k \geq 0 \) that \( T(k) \leq ck \log k \) for our induction hypothesis. We want to demonstrate it holds for \( n \).

Base Case

Consider \( n = 0 \). \( T(0) \) is typically some constant, say \( T(0) = C \leq c0 \log 1 = 0 \), which satisfies our hypothesis as \( \log 1 = 0 \).

Induction Step

For the induction step, assume \( T(k) \leq ck \log k \) is true for all \( k < n \). We assert it for \( n \):\[ T(n) \leq 2T\left(\left\lfloor \frac{n}{2} \right\rfloor + 1\right) + cn \]By the induction hypothesis:\[ T\left(\left\lfloor \frac{n}{2} \right\rfloor + 1\right) \leq c\left(\left\lfloor \frac{n}{2} \right\rfloor + 1\right) \log\left(\left\lfloor \frac{n}{2} \right\rfloor + 1\right) \]Thus:\[ T(n) \leq 2c\left(\left\lfloor \frac{n}{2} \right\rfloor + 1\right) \log\left(\left\lfloor \frac{n}{2} \right\rfloor + 1\right) + cn \]We show this satisfies \( O(n \log n) \) by substituting back:

Simplify and Conclude

Using properties of logarithms and that \( \log\left(\left\lfloor \frac{n}{2} \right\rfloor + 1\right) \leq \log n \), we have:\[ T(n) \leq 2c\left(\frac{n}{2} + 1\right) \log n + cn \]Simplify:\[ \leq cn \log n + 2c \log n + cn \]This can be expressed as:\[ = cn \log n + O(n) \]Therefore, \( T(n) \leq O(n \log n) \). The hypothesis is proven by induction.

Key concepts

Big O Notation

Big O Notation is a mathematical concept used to describe the upper bound of an algorithm's runtime. It is especially helpful when analyzing how an algorithm's performance scales with input size. In this context, we use big O notation to express how the execution time or space requirement grows relative to the size of the input.

Here are some key points about Big O Notation:

• It provides a high-level understanding of an algorithm's efficiency.
• The notation focuses on the most significant terms and ignores constant factors.
• Big O notation helps us compare the efficiency of different algorithms.
• Common examples include \( O(1) \) for constant time operations, \( O(n) \) for linear time, and \( O(n^2) \) for quadratic time algorithms.

In the provided exercise, we aim to find the best Big O upper bound for a recurrence relation. We suspect the upper bound to be in terms of \( O(n \log n) \), similar to other divide and conquer algorithms like Merge Sort. This is because such a notation succinctly captures the balance of splitting a problem into smaller subproblems and the cost associated with this division.

Induction Proof

An induction proof is a powerful method to prove statements that are true for all natural numbers. It works very much like climbing a ladder. The base case acts as the first step, demonstrating the truth for an initial value, typically \( n = 0 \) or \( n = 1 \). Then, we assume the hypothesis holds for some arbitrary number \( k \), and we use this to prove it for \( k + 1 \).

In this exercise, we're using induction to prove that \( T(n) \leq O(n \log n) \). Here’s how it unfolds:

• Base Case: Show the hypothesis is true for the smallest value of \( n \), which is usually 0. In our case, \( T(0) \leq 0 \), confirming the base case.
• Induction Hypothesis: Assume the hypothesis is true for some number \( k \), that is, \( T(k) \leq ck \log k \).
• Induction Step: Use the hypothesis to prove the statement for \( n \). We use the induction hypothesis to establish the boundary for a more complex expression like \( T(n) \).

The combination of these steps confirms that our original guess of \( T(n) = O(n \log n) \) is valid and accurately describes the behavior of the recurrence relation.

Divide and Conquer

Divide and conquer is an essential algorithm design paradigm used to solve complex problems by breaking them down into subproblems of a similar kind. Each subproblem is solved independently, and their solutions are combined to solve the original problem.

Key traits of the divide and conquer strategy include:

• Division: Splitting the problem into one or more smaller subproblems.
• Conquer: Solving each subproblem optimally, typically through recursion.
• Combine: Merging the solutions of the subproblems to form a solution for the original problem.

In continuity with our exercise, this strategy is seen in the recurrence relation \( T(n) \leq 2T(\left\lfloor \frac{n}{2} \right\rfloor + 1) + cn \). The recurrence can be understood as dividing the problem into two roughly equal parts and incurring a linear cost \( cn \) for splitting. This mirrors the recursive approach found in algorithms like Merge Sort, where achieving \( O(n \log n) \) complexity is typical because dividing and solving cost are mated properly.