Question
(This is a version of Programming Project 2.1 from Chapter 2.) The Babylonian algorithm to compute the square root of a positive number n is as follows:
1. Make a guess at the answer (you can pick n/2 as your initial guess).
2. Compute r = n / guess.
3. Set guess = (guess +r) / 2.
4. Go back to step 2 until the last two guess values are within 1% of each
other.
Write a program that inputs a double for n , iterates through the Babylonian algorithm until the guess is within 1% of the previous guess, and outputs the answer as a double to two decimal places. Your answer should be accurate even for large values of n.
Answer
Program:
// BabylonianAlgorithm.java
import java.util.Scanner;
public class BabylonianAlgorithm
{
public static void main(String[] args)
{
Scanner keyboard = new Scanner(System.in);
double n;
double guess;
double r;
double previousGuess;
double diff;
System.out.print("Enter a value for n: ");
n = keyboard.nextDouble();
guess = n / 2;
do
{
r = n / guess;
previousGuess = guess;
guess = (guess + r) / 2;
diff = (previousGuess - guess) / (0.01 * previousGuess);
}while(diff > 0);
System.out.printf(
"The square root of %.2f is approximately: %.2f\n", n, guess);
keyboard.close();
}
}
Output:
Enter a value for n: 225
The square root of 225.00 is approximately: 15.00
Discussions
sinaaa
import java.util.Scanner;
public class Review1 {
public static void main(String[] args) {
Scanner keyboard = new Scanner(System.in);
System.out.print("Enter a value for n: ");
double n = keyboard.nextDouble();
double guess = n/2;
double r, pGuess, percentage;
for(int i=0; i>=0; i++)
{
r = n/guess;
pGuess = guess;
guess = (guess+r) /2;
percentage = ((pGuess-guess)/((pGuess+guess)/2)*100);
System.out.printf("%.2f\n", percentage);
if (percentage<0.1)
break;
}
System.out.printf("The square root of %.1f is approximately: %.5f\n", n, guess);
keyboard.close();
}
}
sinaaa
import java.util.Scanner;
public class Review1 {
public static void main(String[] args) {
Scanner keyboard = new Scanner(System.in);
System.out.print("Enter a value for n: ");
double n = keyboard.nextDouble();
double guess = n/2;
double r, pGuess, percentage;
for(int i=0; i>=0; i++)
{
r = n/guess;
pGuess = guess;
guess = (guess+r) /2;
percentage = ((pGuess-guess)/((pGuess+guess)/2)*100);
System.out.printf("%.2f\n", percentage);
if (percentage<0.1)
break;
}
System.out.printf("The square root of %.1f is approximately: %.5f\n", n, guess);
keyboard.close();
}
}
Post the discussion to improve the above solution.