JavaScript：PI（？）计算器

Question

有没有办法在 JavaScript 中计算 pi？我知道你可以用它Math.PI来找到这样的馅饼：

var pie = Math.PI;
alert(pie); // output "3.141592653589793"

但这是不准确的。我想要的是能够计算它，拥有尽可能多的数字，而不是像pie = 3.141592.... 但是，我想要的不仅仅是更多的数字，而是尽可能多的数字（比如有一千位数字，但我需要更多）。

Answer 1

下面是 Jeremy Gibbons 在Unbounded Spigot Algorithms for the Digits of Pi (2004), Chaper 6中描述的流算法的实现：

function * generateDigitsOfPi() {
    let q = 1n;
    let r = 180n;
    let t = 60n;
    let i = 2n;
    while (true) {
        let digit = ((i * 27n - 12n) * q + r * 5n) / (t * 5n);
        yield Number(digit);
        let u = i * 3n;
        u = (u + 1n) * 3n * (u + 2n);
        r = u * 10n * (q * (i * 5n - 2n) + r - t * digit);
        q *= 10n * i * (i++ * 2n - 1n);
        t *= u;
    }
}

// Demo
let iter = generateDigitsOfPi();

let output = document.querySelector("div");
(function displayTenNextDigits() {
    let digits = "";
    for (let i = 0; i < 10; i++) digits += iter.next().value;
    output.insertAdjacentHTML("beforeend", digits);
    scrollTo(0, document.body.scrollHeight);
    requestAnimationFrame(displayTenNextDigits);
})();

div { word-wrap:break-word; font-family: monospace }

<div></div>

Answer 2

我在这个网站上找到了这段代码：

mess = "";
Base = Math.pow(10, 11);
cellSize = Math.floor(Math.log(Base) / Math.LN10);
a = Number.MAX_VALUE;
MaxDiv = Math.floor(Math.sqrt(a));

function makeArray(n, aX, Integer) {
  var i = 0;
  for (i = 1; i < n; i++) aX[i] = null;
  aX[0] = Integer
}

function isEmpty(aX) {
  var empty = true
  for (i = 0; i < aX.length; i++)
    if (aX[i]) {
      empty = false;
      break
    }
  return empty
}

function Add(n, aX, aY) {
  carry = 0
  for (i = n - 1; i >= 0; i--) {
    aX[i] += Number(aY[i]) + Number(carry);
    if (aX[i] < Base) carry = 0;
    else {
      carry = 1;
      aX[i] = Number(aX[i]) - Number(Base)
    }
  }
}

function Sub(n, aX, aY) {
  for (i = n - 1; i >= 0; i--) {
    aX[i] -= aY[i];
    if (aX[i] < 0) {
      if (i > 0) {
        aX[i] += Base;
        aX[i - 1]--
      }
    }
  }
}

function Mul(n, aX, iMult) {
  carry = 0;
  for (i = n - 1; i >= 0; i--) {
    prod = (aX[i]) * iMult;
    prod += carry;
    if (prod >= Base) {
      carry = Math.floor(prod / Base);
      prod -= (carry * Base)
    } else carry = 0;
    aX[i] = prod
  }
}

function Div(n, aX, iDiv, aY) {
  carry = 0;
  for (i = 0; i < n; i++) {
    currVal = Number(aX[i]) + Number(carry * Base);
    theDiv = Math.floor(currVal / iDiv);
    carry = currVal - theDiv * iDiv;
    aY[i] = theDiv
  }
}

function arctan(iAng, n, aX) {
  iAng_squared = iAng * iAng;
  k = 3;
  sign = 0;
  makeArray(n, aX, 0);
  makeArray(n, aAngle, 1);
  Div(n, aAngle, iAng, aAngle);
  Add(n, aX, aAngle);
  while (!isEmpty(aAngle)) {
    Div(n, aAngle, iAng_squared, aAngle);
    Div(n, aAngle, k, aDivK);
    if (sign) Add(n, aX, aDivK);
    else Sub(n, aX, aDivK);
    k += 2;
    sign = 1 - sign
  }
  mess += "aArctan=" + aArctan + "<br>"
}

function calcPI(numDec) {
  var ans = "";
  t1 = new Date();
  numDec = Number(numDec) + 5;
  iAng = new Array(10);
  coeff = new Array(10);
  arrayLength = Math.ceil(1 + numDec / cellSize);
  aPI = new Array(arrayLength);
  aArctan = new Array(arrayLength);
  aAngle = new Array(arrayLength);
  aDivK = new Array(arrayLength);
  coeff[0] = 4;
  coeff[1] = -1;
  coeff[2] = 0;
  iAng[0] = 5;
  iAng[1] = 239;
  iAng[2] = 0;
  makeArray(arrayLength, aPI, 0);
  makeArray(arrayLength, aAngle, 0);
  makeArray(arrayLength, aDivK, 0);
  for (var i = 0; coeff[i] != 0; i++) {
    arctan(iAng[i], arrayLength, aArctan);
    Mul(arrayLength, aArctan, Math.abs(coeff[i]));
    if (coeff[i] > 0) Add(arrayLength, aPI, aArctan);
    else Sub(arrayLength, aPI, aArctan)
  }
  Mul(arrayLength, aPI, 4);
  sPI = "";
  tempPI = "";
  for (i = 0; i < aPI.length; i++) {
    aPI[i] = String(aPI[i]);
    if (aPI[i].length < cellSize && i != 0) {
      while (aPI[i].length < cellSize) aPI[i] = "0" + aPI[i]
    }
    tempPI += aPI[i]
  }
  for (i = 0; i <= numDec; i++) {
    if (i == 0) sPI += tempPI.charAt(i) + ".<br>";
    else {
      if (document.getElementById("cbCount").checked) addcount = " (" + (i) + ")";
      else addcount = "";
      if (document.getElementById("cbSpace").checked) thespace = " ";
      else thespace = "";
      if ((i) % 50 == 0 && i != 0) sPI += tempPI.charAt(i) + addcount + "<br>";
      else if (i % 5 == 0) sPI += tempPI.charAt(i) + thespace;
      else sPI += tempPI.charAt(i)
    }
  }
  ans += ("PI (" + numDec + ")=" + sPI + "<br>");
  ans += ("Win PI=<br>3.1415926535897932384626433832795<br>");
  t2 = new Date();
  timeTaken = (t2.getTime() - t1.getTime()) / 1000;
  ans += "It took: " + timeTaken + " seconds";
  var myDiv = document.getElementById("d1");
  myDiv.innerHTML = ans
}

<form name="" onsubmit="calcPI(this.t1.value);return false;">
  Number of Digits:<br>
  <input type="text" name="t1" id="t1" value="100" size="25" maxlength="25">
  <br>Add Count:
  <input type="checkbox" name="cbCount" id="cbCount" value="" checked="checked">
  <br>Add Spaces:
  <input type="checkbox" name="cbSpace" id="cbSpace" value="" checked="checked">
  <br>
  <input type="button" value="Calculate Pi" onclick="calcPI(this.form.t1.value)">
</form>
<div id="d1"></div>

Answer 3

您可以通过使用蒙特卡罗模拟来近似 π 的值。生成 [-1,1] 范围内的随机 X 和 Y 则 (X, Y) 在以原点为中心的单位圆内的似然度为 π/4。更多样本可以更好地估计其价值。然后，您可以通过比较单位圆内的样本比率与样本总数并乘以 4 来估计 π。

this.pi = function(count) {
    var inside = 0;

    for (var i = 0; i < count; i++) {
        var x = random()*2-1;
        var y = random()*2-1;
        if ((x*x + y*y) < 1) {
            inside++
        }
    }

    return 4.0 * inside / count;
}