Problem 22 - Discrete Mathematics with Applications

Question

Algorithm 5.10 computes the \(n\)th power of a positive real number \(x,\) where \(n \geq 0 .\) Use it to answer Exercises 18-24 . Algorithm exponentiation(x,n) (* This algorithm computes the nth power of \(x\) using recursion and returns the value in the variable answer.*) 0\. Begin (* algorithm *) 1\. if \(n=0\) then 2\. answer \(\leftarrow 1\) 3\. else 1 if \(n=1\) then 4\. answer \(\leftarrow x\) 5\. else 6\. begin (* else *) 7\. value \(\leftarrow\) exponentiation \((x,\lfloor n / 2\rfloor)\) 8\. answer \(\leftarrow\) value \(\cdot\) value 9\. If \(n\) is odd then 10\. answer \(\leftarrow\) answer \(\cdot\) \(x\) 11\. endelse 12\. End (* algorithm *) Find the recurrence relation satisfied by \(a_{n}.\)

Step-by-Step Solution

Answer

The recurrence relation satisfied by \(a_n\) is: \[a_n = \begin{cases} c & \text{for } n = 0 \ \text{or} \ n = 1 \\ a_{\lfloor n/2 \rfloor} + c & \text{for } n > 1 \end{cases}\]

Step 1: Understand the Algorithm

Let's first go through the algorithm step by step. Algorithm 5.10 computes the \(n\)th power of a positive real number \(x.\) If \(n=0\), the answer is 1, as any number raised to the power 0 is 1. If \(n=1\), the answer is \(x\), as any number raised to the power 1 is the number itself. Else, we calculate the value of exponentiation for \(x\) raised to the power \(\lfloor n/2 \rfloor\): value = exponentiation(x, \(\lfloor n/2 \rfloor\)) We store the result of value * value in the answer variable: answer = value * value If \(n\) is odd, we multiply the answer by \(x\): answer = answer * x

Step 2: Analyze the Algorithm for Recurrence

We can see that the algorithm computes the result using a divide-and-conquer approach by dividing the power in half and then squaring the result. Now, let's analyze each step of the algorithm to determine the recurrence relation satisfied by \(a_n\). 1. If \(n = 0\), there is a constant time operation, and we can represent it as \(a_0 = c\), where \(c\) is a constant. 2. If \(n = 1\), there is a constant time operation, and we can represent it as \(a_1 = c\), where \(c\) is a constant. 3. For \(n > 1\), the algorithm divides the problem into a smaller problem of size \(\lfloor n/2 \rfloor\). The solution of the smaller problem (value) is squared to get the answer, and if \(n\) is odd, the answer is multiplied by \(x\). This operation is performed in constant time, so the relation can be represented as \(a_n = a_{\lfloor n/2 \rfloor} + c\)

Step 3: Determine Recurrence Relation

Now that we have analyzed the algorithm, we can state the recurrence relation satisfied by \(a_n\) as: $a_n = \begin{cases} c & \text{for } n = 0 \ \text{or} \ n = 1 \\ a_{\lfloor n/2 \rfloor} + c & \text{for } n > 1 \end{cases}$

Key Concepts

Exponentiation Algorithm

Exponentiation algorithms are designed to compute powers of a number in an efficient way. In this particular algorithm, we aim to find the power of a base number, say, the n-th power of a number \( x \).

While it might be tempting to simply multiply \( x \) by itself \( n \) times, this direct approach is quite inefficient, especially for large values of \( n \). That's where this recursive algorithm comes in handy.

The key idea is to reduce the problem into smaller, manageable parts by harnessing the power of recursion. For example, you can square the result of calculating the half power: if you know \( x^{\lfloor n/2 \rfloor} \), then \( x^n \) can be computed by squaring this result and adjusting for odd powers. By dividing the task, the algorithm performs the calculations much faster, avoiding repetitive multiplications.

Recurrence Relations

Recurrence relations are fundamental in understanding algorithms that rely on repeating steps. They express the overall process in terms of smaller computations.

In the context of our exponentiation algorithm, the recurrence relation is used to explain how the algorithm breaks down the power computation problem.

The essence of this method is encoded in the relation \( a_n = a_{\lfloor n/2 \rfloor} + c \) for \( n > 1 \). Simply put, this tells us that to compute \( x^n \), we first calculate the power for half of \( n \) (or its floor division by 2), and then perform a constant number of additional operations.

Base cases such as \( a_0 = c \) and \( a_1 = c \) introduce constants as they can be solved directly without any further breakdown.

Divide-and-Conquer Approach

The divide-and-conquer approach is a strategic method used in computer science to solve complex problems by breaking them down into simpler ones. It promotes efficiency by ensuring that each subtask is easier to manage and solve.

In the case of our exponentiation algorithm, this approach is crucial.

• **Divide:** Split the power \( n \) into halves. This step ensures that we are working with a smaller problem size, reducing complexity.
• **Conquer:** Solve each part, meaning compute the power of the base number for these reduced powers. We then square the result to complete most of the power operation.
• **Combine:** Merge the results back together. For instance, if \( n \) is odd, multiply the already computed square by \( x \) to adjust the final result.

This efficient breakdown greatly enhances performance by minimizing redundant calculations.

Power Computation

Power computation involves determining a number raised to an exponent, which is an essential operation in many mathematical and algorithmic exercises. The basic concept is simple: multiplying a base number by itself a certain number of times defined by the exponent.

However, achieving this efficiently, especially for large exponents, presents challenges. This is where algorithms like our recursive exponentiation come into play,
leveraging principles of mathematics and programming to streamline the process.

By breaking down the power into smaller parts through a strategic plan, the algorithm enhances computational speed and accuracy. Understanding these fundamentals of power computation equips us with a powerful tool to tackle numerous problems across various fields of study.