Solution

The problem of this contest was to test two methods to approximate pi and measure how accurate they get compared to how many digits it gets right.
I wrote a simple code which uses the following two methods:
1. Count how many grid dots are in a circle(can be done in linear time when using the circle function).
2. Use the following sequence to apprixmate it:[IMAGE: http://www.people.vcu.edu/%7Epdattalo/sum_infinite_series.jpg]

I decided to go with a maximum accuracy of twenty binary digits. Here is my code

import javax.swing.JFrame;
import java.awt.Graphics;
import java.awt.Color;

public class abcd extends JFrame {
    int[] firstMethod;
    int[] secondMethod;
    public abcd() {
        setSize(800, 400);
        setVisible(true);

        doFirstMethod();
        doSecondMethod();
        repaint();
    }
    // return accuracy in bits:
    public int getAccuracy(double approx) {
        if((int)approx != 3)
            return 0;
        long approxL = Double.doubleToLongBits(approx);
        long piL = Double.doubleToLongBits(Math.PI);
        long mantissaStart = 0x0008000000000000L;
        int prec = 0;   
        for(long i = mantissaStart; i >= 0; i >>= 1) {
            if((approxL & i) != (piL & i))
                return prec;
            prec++;
        }
        return prec;
    }
    public void doFirstMethod() {
        firstMethod = new int[20];
        int index = 0;
        for(int i = 0; index < 20; i += 1+(i>>3)) {
            double π = countDots(i);
            int numDigits = getAccuracy(π);
            if(numDigits >= index+1) {
                // Repeat the calculation a couple of times to increase time accuracy
                long t1 = System.nanoTime();
                for(int j = 0; j < 1000000; j++)
                    getAccuracy(π);
                long t2 = System.nanoTime();
                firstMethod[index] = (int)((t2-t1)/1000000.0);
                index++;
            }
        }
    }
    public void doSecondMethod() {
        secondMethod = new int[20];
        int index = 0;
        for(int i = 0; index < 20; i += 1+(i>>3)) {
            double π = sumUp(i);
            int numDigits = getAccuracy(π);
            if(numDigits >= index+1) {
                // Repeat the calculation a couple of times to increase time accuracy
                long t1 = System.nanoTime();
                for(int j = 0; j < 1000000; j++)
                    getAccuracy(π);
                long t2 = System.nanoTime();
                secondMethod[index] = (int)((t2-t1)/1000000.0);
                index++;
            }
        }
    }

    // Use the partial sums of the infinite sequence sum 1/n² = π²/6:
    public double sumUp(int n) {
        double sum = 0;
        for(int i = 1; i <= n; i++) {
            sum += 1.0/i/i;
        }
        return Math.sqrt(sum*6);
    }

    // Counts how many grid points lie in-/outside a quarter circle. Kind of similar to integrating the circle function.
    public double countDots(int r) {
        long num = 0;
        long rSqr = r*r;
        for(int i = 0; i < r; i++) {
            num += (long)(Math.sqrt(rSqr-i*i)); // Use the floor function to find the number of grid points inside the circle at this position.
        }
        return 4*num/(double)rSqr;
    }

    public void paint(Graphics g) {
        g.setColor(Color.BLACK);
        for(int i = 1; i < 20; i++) {
            g.drawLine(i*10, 200-firstMethod[i-1], i*10+10, 200-firstMethod[i]);
            g.drawLine(200+i*10, 200-secondMethod[i-1], 200+i*10+10, 200-secondMethod[i]);
        }
    }

    public static void main(String[] args) {
        new abcd();
    }
}

This code outputs the following two graphs:
[IMAGE: https://cdn.steemitimages.com/DQmYSwjWAxGnkxfca2Qyp87HuPV5n8VGMPWdLMCNyqXBucA/Screenshot%20from%202019-12-14%2019-52-06.png]
Not much to see except from the fact that both graphs are growing kind of linear.
↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓

List of participants with their entries:

↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓

Winners:

Congratulations @lallo, you won 2 SBI!

↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓

The next contest starts tomorrow. Don't miss it!