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(Recursive Greatest Common Divisor) The greatest common divisor of integers \(x\) and \(y\) is the largest integer that evenly divides both \(x\) and \(y .\) Write a recursive function gcd that returns the greatest common divisor of \(x\) and \(y,\) defined recursively as follows: If y is equal to \(0,\) then \(\operatorname{gcd}(x, y)\) is \(x ;\) otherwise, gcd \((x, y)\) is gcd \((y, x \% y),\) where g is the modulus operator. [ Note: For this algorithm, x must be larger than y.]

Short Answer

Expert verified
The GCD is found using a recursive function: return gcd(y, x % y) when y ≠ 0, and return x when y == 0.

Step by step solution

01

Understanding GCD

The greatest common divisor (GCD) of two integers is the largest positive integer that divides both numbers without leaving a remainder. For example, the GCD of 8 and 12 is 4.
02

Defining the Base Case

The base case of the recursive function is when the second number, y, is 0. In this case, the GCD is x because any number is divisible by itself, and the remainder when dividing x by x is 0.
03

Defining the Recursive Case

If y is not equal to 0, the GCD of x and y can be found by finding the GCD of y and x mod y. The modulus operator (%) is used to find the remainder of x divided by y. This continues until y becomes 0.
04

Creating the Recursive Function

To implement the recursive GCD function in code: 1. Define a function "gcd(x, y)". 2. Check if y == 0: If true, return x. 3. Otherwise, return "gcd(y, x % y)".

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Key Concepts

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

Greatest Common Divisor
The greatest common divisor (GCD) is a fundamental concept in number theory and computing. It represents the largest positive integer that can perfectly divide two given numbers without leaving a remainder. Take, for example, the numbers 20 and 30. The divisors of 20 are 1, 2, 4, 5, 10, 20 and the divisors of 30 are 1, 2, 3, 5, 6, 10, 15, 30. The common divisors are 1, 2, 5, and 10, with 10 being the largest. Therefore, 10 is the GCD of 20 and 30.

The concept of GCD isn't just academic; it is used extensively in simplifying fractions, optimizing algorithms in computer science, cryptography, and more. The GCD plays an important role in computational efficiency, especially when dealing with large numbers.
Modulus Operator
The modulus operator, represented by the symbol % in many programming languages, is used to find the remainder of a division operation. For instance, when you divide 11 by 4, it goes two times into 11, leaving a remainder of 3. Therefore, 11 % 4 = 3.

This operator is particularly important in the recursive algorithm for finding the GCD. In the recursive step, we replace one of the numbers with the result of the modulus operation between the two numbers -
  • This helps in breaking down the problem progressively.
  • It reduces the size of the numbers involved, making the recursive operation work towards a base case efficiently.
  • The operation continues reshaping the problem until a certain stopping condition, like the base case, is met.
Base Case
A base case is a crucial part of a recursive function that provides a condition under which the function stops calling itself. In the GCD algorithm, the base case is straightforward: when the second integer, denoted as \( y \), becomes 0, the function will return the first integer \( x \) as the GCD.

Why does this work? When \( y \) is 0, it means that \( x \) is the largest number by itself that divides exactly, as any number divided by itself yields a remainder of 0. Hence, reaching the base case confirms that the solution is found. Without a base case, recursive functions risk running indefinitely, leading to stack overflow errors or unintended behavior.
Algorithm
An algorithm is a step-by-step procedure used to solve a particular problem, and the recursive GCD computation is a classic example of an efficient algorithm. This algorithm:
  • Begins with two numbers where the first must be larger than the second.
  • Checks if the second number is 0 in its base case.
  • Uses the modulus operator to find the remainder when the first number is divided by the second.
  • Recursively calls itself with the second number and the result from the modulus operation until reaching the base case.
The recursive nature of the algorithm simplifies the calculations progressively while ensuring that each step moves toward a solution. As a result, this avoids unnecessary calculations and ensures a quick resolution to finding the GCD. Recursive algorithms, in general, are powerful tools for solving problems that can be broken down into simpler, repeated steps.

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Most popular questions from this chapter

Find the error in each of the following program segments and explain how to correct it: a. float cube( float ); // function prototype double cube( float number ) // function definition { return number * number * number; } b. register auto int x = 7; c. int randomNumber = srand(); d. float y = 123.45678; int x; x = y; cout << static_cast < float > ( x ) << endl; e. double square( double number ) { double number; return number * number; } f.int sum( int n ) { if ( n == 0 ) return 0; else return n + sum( n ); }

For example, power \((3,4)=3 * 3 * 3 * 3 .\) 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

Write statements that assign random integers to the variable \(n\) in the following ranges: a. \(1 \leq n \leq 2\) b. \(1 \leq n \leq 100\) c. \(0 \leq n \leq 9\) d. \(1000 \leq n \leq 1112\) e. \(1 \leq n \leq 1\) f. \(3 \leq n \leq 11\)

Write function distance that calculates the distance between two points \((x 1, y 1)\) and \((x 2,\) \(y 2\) ). All numbers and return values should be of type double.

Any program that can be implemented recursively can be implemented iteratively. although sometimes with more difficulty and less clarity. Try writing an iterative version of the Towers of Hanoi. If you succeed, compare your iterative version with the recursive version developed in Exercise 6.42 . Investigate issues of performance, clarity and your ability to demonstrate the correctness of the programs.

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