4

I have thoroughly searched for this topic all over the internet, and the threads are either dead, or use a different method than what is described in my book.

For example, http://www.geeksforgeeks.org/square-root-of-a-perfect-square/ . This doesn't work for me because my algorithm needs to loop until it reaches 1% of the last "guess".

Here is the question from the text.

The Babylonian algorithm to compute the square root of a number n is as follows:

  1. Make a guess at the number (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 an integer for n, iterates through the Babylonian algorithm until guess is within 1% of the previous guess, and outputs the answer as a double.

I have written the following code:

#include <iostream>

using std::cout;
using std::cin;
using std::endl;

int main()
{
int  n;
double r, guess(4), lastGuess;

cout << "Enter a number to find the square root of: ";
cin >> n;

do
{

    r = n / guess;
    lastGuess = guess;
    guess = ( guess + r ) / 2;

//  cout <<"Guess: " << guess << endl;
//  cout <<"Last Guess: " << lastGuess << endl;

    cout << "Guess : " << guess  << endl;
    cout << "Last Guess 1% = " << lastGuess + ( lastGuess * 0.01 ) << endl;
    cout << "r = " << r << endl;

} while( guess >= lastGuess * 0.01 );
cout << r;

return 0;
}

The program computes the right answer for r, but the loop doesn't terminate despite guess being greater than 1% added to lastGuess.

This program produces the following output when inputting 144 as n.

....
r = 12
Guess : 12
Last Guess 1% = 12.12
r = 12
Guess : 12
Last Guess 1% = 12.12
r = 12
Guess : 12
Last Guess 1% = 12.12
r = 12
Guess : 12
Last Guess 1% = 12.12
....

The root (r) is correct (12). The guess is LESS than lastGuess (12 < 12.12), which should return a false to the condition, correct?. Why is the loop not ending?

Community
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Anthony Santiago
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4 Answers4

4

If you want to add 1% you need to multiply by 1.01, not 0.01.

while( guess >= lastGuess * 1.01 );

By the way, this iterates while guess is growing by more than 1%. You should also allow for the opposite, that it may have shrunk by more than 1%. The approximation could approach the answer from either direction. (It will approach positive roots from the right and negative roots from the left.)

John Kugelman
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  • Right! This is correct. My original expression was lastGuess +( lastGuess * 0.01 ). I took it out for a reason I cannot recall. It makes sense now. That should be the same thing as (lastGuess * 1.01). I also see what you mean now when n is less than 4. Thanks a lot! – Anthony Santiago Sep 22 '14 at 03:39
  • "You should also allow for the opposite, that it may have shrunk by more than 1%" Fortunately the iterates are alternately too high and too low. – David Eisenstat Sep 22 '14 at 03:46
  • @DavidEisenstat Newton's method can alternate. The Babylonian method tends to converge from the right. – John Kugelman Sep 22 '14 at 03:57
  • Actually, it always converges from the right, so my comment about the other check being redundant stands. – David Eisenstat Sep 22 '14 at 04:35
  • @DavidEisenstat Technically, when you take negative roots into account, it will converge from the *outside*. If you guess -20 for the square root of 100, it will converge to -10 from the left. – John Kugelman Sep 22 '14 at 04:45
  • I just tried it myself and I had to take both sides into consideration. I knew I had to apply both checks but I didn't know why until you explained it. I didn't include the code here because the solution would reflect on what I needed to do for the second check. It was left out in the post to make the code more clear so you guys would be able to better assist me and focus on one clear topic. – Anthony Santiago Sep 22 '14 at 12:49
3

While printing your lastGuess you are using 

 lastGuess + ( lastGuess * 0.01 )

But while checking loop condition you are using

lastGuess*0.01

So in loop condition use the same equation which you are using for printing lastGuess value.

Hemant Gangwar
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0

To properly exit the loop, use something similar to this.

void f(int N)
{
    double x = N / 4;
    double prev = 0.0f;

    while(1)
    {
        x = 0.5 * (x + N / x);
        if (prev == x)
            break;

        prev = x;
        printf("val: %f\n", x);
    }

    printf("SQRT(%d) = %f\n", N, x);
}
Ashot Madatyan
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0

I had the same problem in my book. This is what I wrote and it works perfect.

#include <iostream>
using namespace std;
int main()
{
    double guess, root, previousGuess;
    int number;
    cout << "Enter a number to find the Babylonian square root.\n";
    cin >> number;
    cout << "You want to find the square root of " << number << ".\n";
    cout << "Enter a guess for the square root.\n";
    cin >> guess;

    do
    {
        root = number / guess;
        previousGuess = guess;
        guess = (guess + root ) / 2;

    } while(guess < 0.99*previousGuess || guess > 1.01*previousGuess);

    cout << "The answer is " << guess << ".\n";
    return 0;
}
FrankGray
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