How to write PI calculation program in Java using multi-thread?

Question

I need to create a program that can calculate approximation to the constant PI, using Java multi-thread.

I'm intent to use Gregory-Leibniz Series to calculate the result for PI / 4, and then multiply by 4 to get the PI approximation.

But I have some concern about the program:

1. How can I seperate the calculation process so that I can implement a multi-thread processing for the program? Because the formula is for the total sum, I don't know how to split them into parts and then in the end I will collect them all.
2. I'm thinking about the fact that the program will execute the formula to infinite so user will need to provide some means of configuring the execution in order to determine when it should stop and return a result. Is it possible and how can I do that?

This is so far the most I can do by now.

public class PICalculate {

    public static void main(String[] args) {
        System.out.println(calculatePI(5000000) * 4);
    }

    static double calculatePI(int n) {
        double result = 0.0;
        if (n < 0) {
            return 0.0;
        }
        for (int i = 0; i <= n; i++) {
            result += Math.pow(-1, i) / ((2 * i) + 1);
        }
        return result;
    }
}

Answer 1

The most straightforward, but not the most optimal, approach is to distribute the sequence elements between threads you have. Ie, if you have 4 threads, thread one will work with n%4 == 0 elements, thread2 with n%4 == 1 elements and so on

public static void main(String ... args) throws InterruptedException {

    int threadCount = 4;
    int N = 100_000;
    PiThread[] threads = new PiThread[threadCount];
    for (int i = 0; i < threadCount; i++) {
        threads[i] = new PiThread(threadCount, i, N);
        threads[i].start();
    }
    for (int i = 0; i < threadCount; i++) {
        threads[i].join();
    }
    double pi = 0;
    for (int i = 0; i < threadCount; i++) {
        pi += threads[i].getSum();
    }
    System.out.print("PI/4 = " + pi);

}

static class PiThread extends Thread {

    private final int threadCount;
    private final int threadRemainder;
    private final int N;
    private double sum  = 0;

    public PiThread(int threadCount, int threadRemainder, int n) {
        this.threadCount = threadCount;
        this.threadRemainder = threadRemainder;
        N = n;
    }


    @Override
    public void run() {
        for (int i = 0; i <= N; i++) {
            if (i % threadCount == threadRemainder) {
                sum += Math.pow(-1, i) / (2 * i + 1);
            }
        }
    }

    public double getSum() {
        return sum;
    }
}

PiThread is more efficient, but arguably harder to read, if the loop is shorter:

public void run() {
    for (int i = threadRemainder; i <= N; i += threadCount) {
        sum += Math.pow(-1, i) / (2 * i + 1);
    }
}

In case you don't want to limit yourself with number of elements in sequence and just by time, you may follow an approach below. But note, that it is still limited with Long.MAX_VALUE and you'll have to use BigIntegers, BigDecimals or any other reasonable approach to improve it

public static volatile boolean running = true;

public static void main(String ... args) throws InterruptedException {
    int threadCount = 4;
    long timeoutMs = 5_000;
    final AtomicLong counter = new AtomicLong(0);
    PiThread[] threads = new PiThread[threadCount];
    for (int i = 0; i < threadCount; i++) {
        threads[i] = new PiThread(counter);
        threads[i].start();
    }

    Thread.sleep(timeoutMs);
    running = false;

    for (int i = 0; i < threadCount; i++) {
        threads[i].join();
    }

    double sum = 0;
    for (int i = 0; i < threadCount; i++) {
        sum += threads[i].getSum();
    }
    System.out.print("counter = " + counter.get());
    System.out.print("PI = " + 4*sum);

}

static class PiThread extends Thread {

    private AtomicLong counter;
    private double sum  = 0;

    public PiThread(AtomicLong counter) {
        this.counter = counter;
    }


    @Override
    public void run() {
        long i;
        while (running && isValidCounter(i = counter.getAndAdd(1))) {
            sum += Math.pow(-1, i) / (2 * i + 1);
        }
    }

    private boolean isValidCounter(long value) {
        return value >= 0 && value < Long.MAX_VALUE;
    }

    public double getSum() {
        return sum;
    }
}