Pi computation algorithm

Question

I need to write an algorithm to find pi, using the formula

pi/4 = 1 - 1/3 + 1/5 - 1/7 +- ...

They also want us to make it stop after it has calculated six digits.
And does anyone know how to make it round off?

My code currently is:

public class Pi 
{
    public static void main(String[] args) 
    {
        double sum = 1; 
        int sign = -1; 
        double value = 3; 
        while (value < 10_000_000_000) 
        {
            sum = sum + sign / value; 
            value = value + 2; 
            sign = sign * -1; 
        }
        System.out.println("The value of pi is: " + (4 * sum)); 
    }
}

Right now it runs like this

The value of pi is: 3.1415926335902506

I need it to calculate and output 3.141593.

Can you please state whats the problem with your current code? What does it currently output and is it wrong? Or do you only want to know how to implement your two questions but the pi computation itself is already working? — Zabuzard
– Zabuzard, Commented Aug 22, 2017 at 20:09
@BerkleyLamb I think the questions aims more for the two highlighted sub-questions and not for the algorithm itself. In this case it is no duplicate to this question. But possibly to two other questions like How to round a number to n decimal places in Java and Number of decimal digits in a double. — Zabuzard
– Zabuzard, Commented Aug 22, 2017 at 20:13
You can stop when "it finds 6 digit" because you won't now these 6 digits ^^ you have to do more than 1000 iterations at least to have the 6 correct digits — azro
– azro, Commented Aug 22, 2017 at 20:21
The tricky part is determining when it has calculated six digits. There's not a perfectly generic way to determine such a thing. For all values of a (a > 0), there exists a value b such that incrementing b by a will alter all digits of b. So, you're going to need a solution that that wouldn't work for all numbers, but should work for the initial digits of pi. Stopping after a fixed number of iterations is a good approach. I'm guessing 10^7 iterations would be plenty. — mrog
– mrog, Commented Aug 22, 2017 at 20:59

Joris Schellekens · Accepted Answer · 2017-08-22 22:25:02Z

2

Try this code:

/** This method rounds a double to specified number of digits
 * The idea is to multiply the incoming number with the appropriate
 * power of 10, round it to the nearest integer, and then divide it.
 */
public static double round(double x, int n)
{
    double exp = Math.pow(10, n);
    return Math.round(x * exp) / exp;
}

public static void main (String[] args)
{
    double sum = 1; 
    int sign = -1; 
    double value = 3; 

    /* The convergence rate for this series is known,
     * As suggested, 10^7 iterations are enough.
     */
    int nofIterations = Math.pow(10, 7);
    for(int i=0;i<nofIterations;i++) 
    {
        sum = sum + sign / value; 
        value = value + 2; 
        sign = sign * -1; 
    }
    double pseudoPi = 4 * sum;

    /** Now we print both the calculated value and the rounded value
     */
    System.out.println("The calculated value of pi is : " + pseudoPi); 
    System.out.println("The rounded value of pi is    : " + round(pseudoPi, 6));
}

That prints:

The calculated value of pi is : 3.141592663589326
The rounded value of pi is : 3.141593

edited Aug 22, 2017 at 22:25

answered Aug 22, 2017 at 21:06

Joris Schellekens

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5 Comments

mrog Over a year ago

I believe calling Math.pow in the for loop's condition would impact performance quite a bit. Perhaps this computation should be done on a different line so it isn't repeated with every iteration.

Joris Schellekens Over a year ago

Normally the compiler ought to realize that this is a constant, and make the suggested optimization.

Zabuzard Over a year ago

Can you explain your code and state what you have changed? Will be more helpful than a code only answer.

mrog Over a year ago

@JorisSchellekens The compiler can't make that optimization because there's no guarantee that a given method will always return the same value for a given input. I timed for (int i = 0; i < 10000000; i++) {} and for (int i = 0; i < Math.pow(10, 7); i++) {}. The latter version was an order of magnitude slower.

Joris Schellekens Over a year ago

@mrog Thanks for your input. I made the change to my code-sample. Please upvote.

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