如何做BBP算法的基数16以获取PI [英] How to do base 16 for BBP algorithm to get PI

问题描述

我正在使用以下代码来计算Pi，但我得到了错误的答案，因为该公式某些使用基数16的方式.Wikipedia说

I am using the following code to calculate Pi but I get the wrong answer because the formula some how uses base 16. Wikipedia says here that formula has to do with hexadecimal but I don't know which part of formula process. What part of the formula do you use base 16, is it the input of the formula or just the output? For eg Do I convert k to base 16 before entering k into the formula or do I input normal base 10 number into formula and convert output of formula to base 10 from base 16. The wrong answer gives a few decimal places wrong for this code:

import java.io.Console;

public class Main {

public static void main(String[] args) {
        System.out.println("Enter how many decimal places you want for Pi?");
        String line = System.console().readLine();
        double x = Double.parseDouble(line);
        double y = 0.0;

        for(double i = 0; i <= x; i++)
        {
            y = y + (1.0/Math.pow(16.0,i))*((4.0/(8.0*i + 1.0))-(2.0/(8.0*i + 4.0)) - 
            (1.0/(8.0*i + 5.0)) - (1.0/(8.0*i + 6.0)));
        }
        System.out.println(y);
    }
}

推荐答案

您的代码正确实现了Bailey–Borwein–Plouffe公式.它计算的π估计随着您使用的迭代次数的增加而逐渐增加:

Your code implements the Bailey–Borwein–Plouffe formula correctly. It calculates an estimation of π that increases in accurary the more iterations you use:

• 输入1，您将得到:3.1414224664224664
• 输入2，您将得到:3.1415873903465816
• 输入3，您将得到:3.1415924575674357
• 输入4，您将得到:3.1415926454603365
• 输入5，您将得到:3.141592653228088
• 输入6，您将得到:3.141592653572881
• 输入7，您将得到:3.141592653588973
• 输入8，您将得到:3.1415926535897523
• 输入9，您将得到:3.1415926535897913
• 输入10，您将得到:3.141592653589793
• 输入11，您将得到:3.141592653589793

• Enter 1, you get: 3.1414224664224664
• Enter 2, you get: 3.1415873903465816
• Enter 3, you get: 3.1415924575674357
• Enter 4, you get: 3.1415926454603365
• Enter 5, you get: 3.141592653228088
• Enter 6, you get: 3.141592653572881
• Enter 7, you get: 3.141592653588973
• Enter 8, you get: 3.1415926535897523
• Enter 9, you get: 3.1415926535897913
• Enter 10, you get: 3.141592653589793
• Enter 11, you get: 3.141592653589793

就是这样.任何更高的输入都是无用的，因为使用double时无法获得更精确的输入.

And that's it. Any higher input is useless, because you can't get more precise when you use double.

如果要获得更高的精度，则应使用比double更高精度的类型，例如BigDecimal:

If you want to get more precision, you should use a type with more precision than double, e.g. BigDecimal:

BigDecimal pi = BigDecimal.ZERO;
for (int i = 0; i <= x; i++) {
    BigDecimal a = BigDecimal.valueOf(1).divide(BigDecimal.valueOf(16).pow(i), 30, RoundingMode.HALF_UP);
    BigDecimal b1 = BigDecimal.valueOf(4).divide(BigDecimal.valueOf(8).multiply(BigDecimal.valueOf(i)).add(BigDecimal.valueOf(1)), 30, RoundingMode.HALF_UP); 
    BigDecimal b2 = BigDecimal.valueOf(2).divide(BigDecimal.valueOf(8).multiply(BigDecimal.valueOf(i)).add(BigDecimal.valueOf(4)), 30, RoundingMode.HALF_UP); 
    BigDecimal b3 = BigDecimal.valueOf(1).divide(BigDecimal.valueOf(8).multiply(BigDecimal.valueOf(i)).add(BigDecimal.valueOf(5)), 30, RoundingMode.HALF_UP); 
    BigDecimal b4 = BigDecimal.valueOf(1).divide(BigDecimal.valueOf(8).multiply(BigDecimal.valueOf(i)).add(BigDecimal.valueOf(6)), 30, RoundingMode.HALF_UP); 
    BigDecimal b = b1.subtract(b2).subtract(b3).subtract(b4);
    pi = pi.add(a.multiply(b));
}

• 输入10，您将得到:3.141592653589793129614170564040940187329961242598487417146011
• 输入100，您将得到:3.141592653589793238462643383279097710627677399356070777774427
• Enter 10, you get: 3.141592653589793129614170564040940187329961242598487417146011
• Enter 100, you get: 3.141592653589793238462643383279097710627677399356070777774427

您使用的精度越高(在此由BigDecimal的divide方法中的参数30设置)，并且运行的迭代次数越多，π的近似值就越好.

The more precision you use (here set by the parameter 30 in the divide method of BigDecimal) and the more iterations you run, the better the approximation of π will be.

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