# Answered Essay: The Babylonian algorithm to compute the square root of a number n is

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.

Dear Student,

here i have written the C++ program as per the requirement.I have also inclueded the comment for better understanding.

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Note: Please note that the below program has been tested on ubuntu 16.04 system and compiled under g++ compiler. This code will also work on other IDE’s and C++ microsoft visual studio express.

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Program:

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#include <iostream>

#include<stdlib.h>

//Namespace declration

using namespace std;

int main()
{

//variable data type declration

double n, count = 0;

//input a double number

cout << “Please enter a double number: “;

cin >> n;

//input a guess

cout << “Please enter a ‘guess’ number to divide by double number: “;

cin >> guess;

//assign the guess to a new variable previous_guess
double previous_guess = guess;

double new_guess;

int new_count;

//loop until the guess is less than the 1% of the previous guess

while(count<100)
{

r = n/previous_guess;

new_guess = (previous_guess + r)/2;

//when guess is less than the 1% of the previous guess
//break the loop
if((new_guess – previous_guess) < (previous_guess * 0.01))

{

new_count = count;

break;

}

//else execute this condition

else

{

r = n/new_guess;

previous_guess = (new_guess + r)/2;

}

//increment count

count++;

}

//display the count

cout<<“guess is within 1% of the previous guess. in “<< new_count<<” counts”<<endl;

//display the squre root

cout << “The sqaure root of “<< n << ” is ” << answer;

cout << endl;

return 0;

}//end of the main function

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here i have attached the output of the program as a screen shot…

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Output:

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Pages (550 words)
Approximate price: -

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