使用 Stack 机使用 Brouncker 公式逼近 PI

Question

我正在研究如何将此公式转换为堆栈计算机以获得 PI 的近似值（基于 Brouncker's）。

我正在使用的公式是 4/(1 + 1^2/(2 + 3^2/(2 + 5^2/(2 + 7^2/(2 + 9^2/2)))))，大约是 2.97。我如何将其翻译成堆栈机器代码？

这是我目前所知道的，但它是错误的：

DIV
PUSH 4
DIV
ADD
PUSH 1
POW
PUSH 1
PUSH 2
DIV
ADD
POW 
PUSH 3
PUSH 2
PUSH 2
DIV
ADD
POW
PUSH 5
PUSH 2
PUSH 2
DIV
ADD
POW
PUSH 7
PUSH 2
PUSH 2
ADD
DIV
POW
PUSH 9
PUSH 2
PUSH 2
PUSH 2

（说明是从下往上读的）

4 除以 4.0909 结束，但 4 除以 1.344 应该结束。那里有一些数学语法是错误的，但我试过了但还没有弄清楚我错在哪里...

如果需要，这是我为阅读说明而构建的class，它是用 C# 编写的，但应该被广泛理解：

using System;
using System.Collections.Generic;
using System.Globalization;
using System.Linq;

namespace Máquinas
{
    /**
    * @author toyomitsu
    *
    * @date - 2020.05.17 
    */
    public class StackMachine
    {
        private Stack<string> _Stack;
        private string[] _Instructions;

        public StackMachine(string[] Instructions)
        {
            Array.Reverse(Instructions);

            _Stack = new Stack<string>();
            _Instructions = Instructions;
        }

        public void PerformInstructions()
        {
            Console.WriteLine();
            Console.WriteLine("*** Stack machine ***");

            foreach (string Instruction in _Instructions)
            {
                string[] InstData = Instruction.Split(' ');

                switch (InstData[0])
                {
                    case "PUSH":
                        Push(InstData[1]);
                        break;

                    case "POP":
                        Pop();
                        break;

                    case "ADD":
                        Add();
                        break;

                    case "SUB":
                        Sub();
                        break;

                    case "MOD":
                        Mod();
                        break;

                    case "MUL":
                        Mul();
                        break;

                    case "DIV":
                        Div();
                        break;

                    case "POW":
                        Pow();
                        break;

                    case "SQR":
                        Sqr();
                        break;

                    case "EXP":
                        Exp();
                        break;

                    default:
                        continue;
                }

                List<string> Registry = _Stack.ToList<string>();

                Console.WriteLine();
                Console.WriteLine(string.Format("** Instruction: {0} **", Instruction));
                for (int i = 0; i < Registry.Count; i++)
                {
                    Console.WriteLine(string.Format("Registry {0}: {1}", i + 1, Registry[i]));
                }
                Console.WriteLine("**********************");
            }
        }

        #region Functions
        private void Push(string Data)
        {
            _Stack.Push(Data);
        }

        private string Pop()
        {
            return _Stack.Pop();
        }

        private void Add()
        {
            string Data1 = Pop();
            string Data2 = Pop();

            double Num1 = double.Parse(Data1, CultureInfo.InvariantCulture);
            double Num2 = double.Parse(Data2, CultureInfo.InvariantCulture);

            double Result = Num1 + Num2;
            Push(Result.ToString(CultureInfo.InvariantCulture));
        }

        private void Sub()
        {
            string Data1 = Pop();
            string Data2 = Pop();

            double Num1 = double.Parse(Data1, CultureInfo.InvariantCulture);
            double Num2 = double.Parse(Data2, CultureInfo.InvariantCulture);

            double Result = Num1 - Num2;
            Push(Result.ToString(CultureInfo.InvariantCulture));
        }

        private void Mul()
        {
            string Data1 = Pop();
            string Data2 = Pop();

            double Num1 = double.Parse(Data1, CultureInfo.InvariantCulture);
            double Num2 = double.Parse(Data2, CultureInfo.InvariantCulture);

            double Result = Num1 * Num2;
            Push(Result.ToString(CultureInfo.InvariantCulture));
        }

        private void Mod()
        {
            string Data1 = Pop();
            string Data2 = Pop();

            double Num1 = double.Parse(Data1, CultureInfo.InvariantCulture);
            double Num2 = double.Parse(Data2, CultureInfo.InvariantCulture);

            double Result = Num1 % Num2;
            Push(Result.ToString(CultureInfo.InvariantCulture));
        }

        private void Div()
        {
            string Data1 = Pop();
            string Data2 = Pop();

            double Num1 = double.Parse(Data1, CultureInfo.InvariantCulture);
            double Num2 = double.Parse(Data2, CultureInfo.InvariantCulture);

            double Result = Num1 / Num2;
            Push(Result.ToString(CultureInfo.InvariantCulture));
        }

        private void Pow()
        {
            string Data1 = Pop();
            string Data2 = Pop();

            double Num1 = double.Parse(Data1, CultureInfo.InvariantCulture);
            double Num2 = double.Parse(Data2, CultureInfo.InvariantCulture);

            double Result = Math.Pow(Num1, Num2);
            Push(Result.ToString(CultureInfo.InvariantCulture));
        }

        private void Sqr()
        {
            string Data1 = Pop();

            double Num1 = double.Parse(Data1, CultureInfo.InvariantCulture);

            double Result = Math.Sqrt(Num1);
            Push(Result.ToString(CultureInfo.InvariantCulture));
        }

        private void Exp()
        {
            string Data1 = Pop();

            double Num1 = double.Parse(Data1, CultureInfo.InvariantCulture);

            double Result = Math.Exp(Num1);
            Push(Result.ToString(CultureInfo.InvariantCulture));

        }
        #endregion
    }
}

编辑：我解决了这个。我的数学语法一团糟。正确的叠机是：

DIV
PUSH 4
ADD
PUSH 1
DIV
POW
PUSH 1
PUSH 2
ADD
PUSH 2
DIV
POW
PUSH 3
PUSH 2
ADD
PUSH 2
DIV
POW
PUSH 5
PUSH 2
ADD
PUSH 2
DIV
POW
PUSH 7
PUSH 2
ADD
PUSH 2
DIV
POW
PUSH 9
PUSH 2
PUSH 2

Answer 1

您已将 2 + 7^2/(..x..) 翻译成 x, 2, 2, 7, ^, +, /（其中 x 是临时堆栈结果）？

这不会导致 (2 + 7^2)/x 吗？你不想x, 2, 7, ^, /, 2, +吗？

使用 Stack 机使用 Brouncker 公式逼近 PI

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