Question

我用Java编写了一个简单的程序来通过蒙特卡罗方法计算Pi。因为你需要大量的下降才能让Pi'精确'而且它变得越来越慢我决定实现多线程。现在我的问题是：有没有办法加快计算速度？并且每次只为每个物理线程计算一次迭代，或者我的多线程概念是否完全错误？

以下是代码：

public class PiMC{
    public static void main(String[] args) {
        ExecutorService exec=Executors.newCachedThreadPool();
        //Future object is used to get result from a thread
        Future<Double> result0=exec.submit(new Thread1());
        Future<Double> result1=exec.submit(new Thread1());
        Future<Double> result2=exec.submit(new Thread1());
        Future<Double> result3=exec.submit(new Thread1());
        Future<Double> result4=exec.submit(new Thread1());
        Future<Double> result5=exec.submit(new Thread1());
        Future<Double> result6=exec.submit(new Thread1());
        Future<Double> result7=exec.submit(new Thread1());

        try
        {
            System.out.println(((result0.get() + result1.get() + result2.get() + result3.get() + result4.get()+ result5.get() + result6.get() + result7.get()) / Long.MAX_VALUE) * 4);
        }
        catch(InterruptedException e){System.out.println(e);}
        catch(ExecutionException e){System.out.println(e);}
    }
}

class Thread1 implements Callable {
    @Override
    public Double call() throws Exception {
        long drops = Long.MAX_VALUE / 8;
        //long drops = 500;
        double in = 0;
        for (long i = 0; i <= drops; i++) {
            double x = Math.random();
            double y = Math.random();
            if (x * x + y * y <= 1) {
                in++;
            }
        }
        return in;
    }
}

Answer 1

你意识到Long.MAX_VALUE / 8有多大？它是（2 ^ 63 - 1）/ 8，大约是1e18 ......相当大的数字（即使是今天最好的计算机填满整个建筑物至少需要1000s，请参阅评论）。

更好的方法是将先前计算的值与pi的当前值进行比较并进行比较。如果差异为0（由于精度受限而发生 - > eps是最小数字＆gt; 0，其中1 + eps！= 1）取消执行并返回值：

int sum = 0, drops = 0;
double pi = 0, oldPi;
do {
     oldPi = pi;
     double x = Math.random(), y = Math.random();
     if (x * x + y * y <= 1)
         sum++;
     drops++;
     pi = 4.0 * sum / drops;
} while (pi != oldPi || pi < 3); // pi < 3 to avoid problems when the first
// drops are outside of the circle, pi == 0 would also work, BUT setting
// pi to a value different from 0 at the beginning can still fail with only pi != oldPi

如果你想使用多个线程很难，因为pi的值的更新必须同步，我猜你不会获得太多。但是，有几个线程可以独立计算pi，你可以比较（或平均）结果。

蒙特卡罗在多个线程上计算Pi

1 个答案: