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Write a recursive function power( base, exponent ) that, when invoked, returns base \(^{exponent}\) For example, power \((3,4)=3 \pm 3 * 3 * 3 .\) Assume that exponent is an integer greater than or equal to 1\. Hint: The recursion step would use the relationship base \(^{exponent}\) \(=\) base \(\cdot\) base \(^{exponent - 1}\) and the terminating condition occurs when exponent is equal to \(1,\) because base \(^{1}=\) base

Short Answer

Expert verified
The recursive function 'power(base, exponent)' returns 'base' raised to 'exponent' by recursively multiplying 'base' by 'power(base, exponent - 1)', until 'exponent' equals 1.

Step by step solution

01

Understanding the Problem

We are to write a recursive function named 'power' that takes two arguments: 'base' and 'exponent'. The function should return the 'base' raised to the power of 'exponent'. The recursive relationship we'll use is that base ^ exponent equals base times base ^ (exponent - 1). The base case occurs when exponent is equal to 1, at which point the function should return the base itself.
02

Defining the Base Case

We start by defining the base case of our recursive function. The base case occurs when the exponent is 1. In this situation, the function should simply return the base because any number raised to the power of 1 is itself.
03

Defining the Recursive Case

For the recursive case, we call the 'power' function with the same 'base' and 'exponent' decremented by 1. The function should return the result of multiplying the 'base' by the result of 'power(base, exponent - 1)'.
04

Writing the Recursive Function

We write the 'power' function by combining the base case and the recursive case in a function definition. If the exponent is 1, return base. Otherwise, recursively multiply the base by the power function of the base and exponent - 1.
05

Testing the Function

Finally, we can test our function by calling it with sample arguments to ensure it behaves as expected, for instance, calculating power(3,4) and checking if it equals 81.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Base Case in Recursion
The concept of a base case is central to understanding recursion in programming. Imagine a base case as the stopping point of the recursion, the condition under which the function will stop calling itself and begin to return values.

In our exercise, the base case occurs when the exponent is 1. This is a critical step in the design of a recursive algorithm, as it defines a simple, solvable instance of the problem for which the answer is known and can be returned directly without further recursion. For the function power(base, exponent), when exponent is 1, the function will simply return base, because mathematically, any number to the power of one is itself (\( base^1 = base \)). Properly identifying the base case is crucial, because without it, the recursion could lead to an infinite loop, causing a stack overflow error.
Recursive Relationship
The recursive relationship is the rule that breaks down the complex problem into simpler sub-problems by calling the function within itself with modified arguments. In recursion, this self-referential step is essential as it approaches the problem incrementally towards the base case.

In the context of our power function, the recursive relationship is established by the expression \( base^{exponent} = base \times base^{exponent - 1} \). With each recursive call, the exponent is decremented by one, which gradually moves the calculation closer to the base case. As each call waits for the next call to complete, they stack up creating a chain of unfinished operations, often visualized with a call stack. Once the base case is reached and the value is returned, the stack unwinds, completing each deferred operation sequentially until the initial call is resolved and the final result is returned.
Exponentiation Algorithm
The exponentiation algorithm we've discussed implements an efficient method to calculate the power of a number. Unlike iterative approaches that may use repeated multiplication, our recursive method efficiently reduces the problem size with each function call.

To understand the algorithm with our function power(base, exponent), let's consider an example: power(3, 4). This will lead to multiple calls: power(3, 3), power(3, 2), and finally power(3, 1). The last call returns 3 (the base case), and then the previous calls complete, eventually calculating 3 * 3 * 3 * 3, or 81.

This method highlights a divide-and-conquer strategy, where the larger problem is divided into smaller ones that are easier to manage and solve. By focusing on the recursive relationship and the base case, it's possible to implement an effective and clear solution to exponentiation without resorting to loops or iterative logic, which is particularly useful in languages like C++ that support recursion natively.

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Most popular questions from this chapter

An integer is said to be a perfect number if the sum of its divisors, including 1 (but not the number itself), is equal to the number. For example, 6 is a perfect number, because 6=1 +2+ 3. Write a function is Perfect that determines whether parameter number is a perfect number. Use this function in a program that determines and prints all the perfect numbers between 1 and 1000. Print the divisors of each perfect number to confirm that the number is indeed perfect. Challenge the power of your computer by testing numbers much larger than 1000.

Show the value of x after each of the following statements is performed: a) x = fabs( 7.5 ) b) x = floor( 7.5 ) c) x = fabs( 0.0 ) d) x = ceil( 0.0 ) e) x = fabs( -6.4 ) f) x = ceil( -6.4 ) g) x = ceil( -fabs( -8 + floor( -5.5 ) ) )

Write a complete program that prompts the user for the radius ofasphere, and calculates and prints the volume of that sphere. Use an inline function sphereVolume that returns the result of the following expression: (4.0 / 3.0 * 3.14159 * pow(radius, 3)).

In this chapter, you studied functions that can be easily implemented both recursively and iteratively.. In this exercise, we present a problem whose recursive solution demonstrates the elegance of recursion, and whose iterative solution may not be as apparent. The Towers of Hanoi is one of the most famous classic problems every budding computer scientist must grapple with. Legend has it that in a temple in the Far East, priests are attempting to move a stack of golden disks from one diamond peg to another (Fig. 6.35). The initial stack has 64 disks threaded onto one peg and arranged from bottom to top by decreasing size. The priests are attempting to move the stack from one peg to another under the constraints that exactly one disk is moved at a time and at no time may a larger disk be placed above a smaller disk. Three pegs are provided, one being used for temporarily holding disks. Supposedly, the world will end when the priests complete their task, so there is little incentive for us to facilitate their efforts. Let’s assume that the priests are attempting to move the disks from peg 1 to peg 3. We wish to develop an algorithm that prints the precise sequence of peg-to-peg disk transfers. If we were to approach this problem with conventional methods, we would rapidly find ourselves hopelessly knotted up in managing the disks. Instead, attacking this problem with recursion in mind allows the steps to be simple. Moving n disks can be viewed in terms of moving only n – 1 disks (hence, the recursion), as follows: a) Move \(n-1\) disks from peg 1 to peg \(2,\) using peg 3 as a temporary holding area. b) Move the last disk (the largest) from peg 1 to peg 3. c) Move the \(n-1\) disks from peg 2 to peg \(3,\) using peg 1 as a temporary holding area. The process ends when the last task involves moving \(n=1\) disk (i.e., the base case). This task is accomplished by simply moving the disk, without the need for a temporary holding area. Write a program to solve the Towers of Hanoi problem. Use a recursive function with four parameters: a) The number of disks to be moved b) The peg on which these disks are initially threaded c) The peg to which this stack of disks is to be moved d) The peg to be used as a temporary holding area Display the precise instructions for moving the disks from the starting peg to the destination peg. To move a stack of three disks from peg 1 to peg 3, the program displays the following moves: \(1 \rightarrow 3\) (This means move one disk from peg 1 to peg \(3 .\) ) \\[ \begin{array}{l} 1 \rightarrow 2 \\ 3 \rightarrow 2 \\ 1 \rightarrow 3 \\ 2 \rightarrow 1 \\ 2 \rightarrow 3 \\ 1 \rightarrow 3 \end{array} \\]

Implement the following integer functions: a) Function Celsius returns the Celsius equivalent of a Fahrenheit temperature. b) Function Fahrenheit returns the Fahrenheit equivalent of a Celsius temperature. c) Use these functions to write a program that prints charts showing the Fahrenheit equivalents of all Celsius temperatures from 0 to 100 degrees, and the Celsius equivalents of all Fahrenheit temperatures from 32 to 212 degrees. Print the outputs in a neat tabular format that minimizes the number of lines of output while remaining readable.

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