# Answered Essay: NOTE: DO the *Challenging* Version!

NOTE: DO the *Challenging* Version!

Program Info:

The Babylonian algorithm to compute the square root of a 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 for as many iterations as necessary. The more that steps 2 and 3 are repeated, the closer guess will become to the square root of n.

Write a program that inputs a double for n and iterates through the Babylonian algorithm 100 times. For a more challenging version, iterate until guess is within 1% of the previous guess, and outputs the answer as a double.

More Important Info:

Babylonian Algorithm for square root:

Do the Challenging version!!! (I would do the simple version first, then add the challenging loop condition)

This is a math problem:

The challenge is to find the condition that will stop the loop when guess is within 1% of the previous guess. This is not so hard to do on paper. It’s a simple inequality:   | guess – previous_guess | < 1% * previous_guess

You will NOT use absolute value function (if you know what that is)

You will have to convert this expression to a simple boolean expression, removing the absolute value.

#include <iostream>

using namespace std;

int main() {

double n;

cout << “Please enter the number whose square root u want to know :”;

cin >> n;

/* declare necessary variable */

double guess;

double previous_guess;

double r;

guess = n / 2;

do {

previous_guess = guess;

r = n / guess;

guess = (guess + r) / 2;

when ever u want to visualise the working of code , then simply print their data

and observe it

you will definitely see the pattern

below printing line will print the various data which will be helfull for visualisation

delete the below line after your satisfaction

you will also able to judge then what should be used for termination i.e

previous_guess – guess or

guess – previous_guess

you can print other data 2 for more visualisation

and

important thing –

if infinite loop condition is there then

U GUESS IT RIGHT

check the condition variable of loop

*/

cout << previous_guess << ” ” << ” ” << guess << ” ” << previous_guess – guess << ” ” << guess – previous_guess << ” ” << previous_guess * .01 << endl;

} while ((previous_guess – guess) > previous_guess * .01);

cout << endl << endl;

cout << “Square root is : ” << guess << endl << endl;

return 0;

}

//Sample output

Pages (550 words)
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