Difficulty: ★ ★ ★ ☆

When I sought to write my own square root function back in 2020, I based it on a series of steps from the math.com website. I figured, “Hey! These guys are math nerds. They know their stuff.” Turns out, the ancient Babylonians beat them to the punch — and have a better method for calculating a square root.

When they weren’t conquering or being conquered, the Babylonians devised a method for calculating square roots that involves averages and division. I’ve seen it explained a few times, but the math nerds tend to make the procedure sound more complex than it is. Here’s my take:

1. Start with the positive integer to find its square root. For example, five.
2. Obtain the average of any two values greater-than and less-than what the square root might be. Continuing with the example, the average of 4 and 2 is 3.
3. Divide the original positive integer by the result of Step 2: 5/3 = 1.666667.
4. Repeat Steps 2 and 3 using the result of Step 3 as the less-than value in Step 2. So you continue with the average of 4 and 1.666667 = 2.833333, and then 5/2.833333, and so on.

Yes, even my version of the steps is confusing. So, Figure 1 attempts to illustrate the process, where Steps 1 and 2 (in the Figure) repeat a given number of times to obtain the square root of value root.

Figure 1. The Babylonian Method of finding the square root.

In Figure 1, I removed the part about guessing the greater-than and less-than values. Instead, and in my exercise solution, I start the process by using the positive integer itself (root) and 1.0. Then I repeat the steps from there, which may take longer but it codes more efficiently.

The more times you repeat Steps 1 and 2 (from Figure 1), the more precise the result. In my tests, seven times was adequate to obtain identical results as the C library’s sqrt() function, though some answers may suffer from the expected precision errors as floating-point calculations do.

Your task for this month’s Exercise is to write the function babylonian_sr(), which uses the Babylonian Method to calculate a square root. The function need not be recursive; I didn’t use recursion in my solution.

Here is a source code skeleton you can use for your solution:

2022_04_01-Lesson.c

#include <stdio.h>

double babylonian_sr(double root)
{
}

int main()
{
    double pv,sr;

    printf("Enter a positive value: ");
    scanf("%lf",&pv);
    if( pv <= 0 )
        return(1);
    sr = babylonian_sr(pv);
    printf("The square root of %.0f is %f\n",
            pv,
            sr
          );

    return(0);
}

This code prompts the user for a positive integer. This value is sent to the babylonian_sr() function, which calculates the square root to seven digits of precision. The value is returned from the function as a double, which is output in the main() function.

Here is a sample run of my solution:

Enter a positive value: 5
The square root of 5 is 2.236068

Please try this Exercise on your own before you peek at my solution.