javascript - 在 Bonferroni 不等式的 JS 实现中避免多重循环

Question

我正在尝试实现“Bonferroni 不等式”，它使用 Javascript UDF 为 GCP BigQuery 上的数据科学用例的许多独立事件的联合概率建模。但是我对 JS 很不熟悉，对好的做法一无所知。

要应用的公式如下:

P(U Ai) = SUM(P(Ai)) - SUM(P(Ai)*P(Aj)) + SUM(P(Ai)*P(Aj)*P(Ak) - ... i != j != k

我对此函数的输入是单个事件概率的数组:

[P(A1), P(A2), P(A3), ...] 

我本能地在行中创建了“for 循环”来获得结果，但是，看到如此丑陋的代码令人心痛，所以想知道你们中是否有人知道如何以更优雅和优化的方式实现它？

这是我为 4 级 Bonferroni 不等式编写的函数:

function unionBoundProbability(probList){

  var intersection2 = 0;
  var intersection3 = 0;
  var intersection4 = 0;
  var i = 0;
  var j = 0;
  var k = 0;
  var l = 0;
  var sum = 0;
  var product = 1;

  var sum = probList.reduce((a, b) => a + b, 0);

  for(i = 0; i < probList.length; i++){
    product *= probList[i];
    for(j = i+1; j < probList.length; j++){
      intersection2 += probList[i]*probList[j];
      for(k = j+1; k < probList.length; k++){
        intersection3 += probList[i]*probList[j]*probList[k];
        for(l = k+1; l < probList.length; l++){
          intersection4 += probList[i]*probList[j]*probList[k]*probList[l];
        }
      }
    }
  }

  switch (probList.length) {
    case 0:
      return 0;
      break;
    case 1:
      return probList[0];
      break;
    case 2:
      return sum - product;
      break;
    case 3:
      return sum - intersection2 + product;
      break
    case 4:
      return sum - intersection2 + intersection3 - product;
    case 5 :
      return sum - intersection2 + intersection3 - intersection4 + product;
    default:
      return Math.max((sum - intersection2 + intersection3  - intersection4), Math.max.apply(Math, probList));
  }
}

我想做的是计算作为输入传递的所有概率的并集概率的近似值。

如果我的概率小于 5，则 switch 语句应用确切的公式。否则，默认情况下应用 Bonferroni 近似，(因为我正在模拟信号被接收的机会，如果估计小于最好天线的概率，那么我保留最好的天线)。

谢谢你的帮助

Answer 1

此示例遵循来自 https://www.probabilitycourse.com/chapter6/6_2_1_union_bound_and_exten.php 的以下等式

P(⋃(i=1 => n)Ai)=∑(i=1 => n) P(Ai) − ∑(i<j) P(Ai ∩ Aj) + ∑(i<j<k) P(Ai ∩ Aj ∩ Ak) − ... +(−1)^n−1 P(⋂(i=1 => n) Ai)

我不知道您在给出的示例中包含阶乘的原因，但我没有包含阶乘，因为它们不在上面的等式中。

// Recursive function to update sums of each level
function updateSums(sums, probList, maxLevel, currentLevel = 1, currentProduct = 1, lastIndex = -1) {
  // Example case: maxLevel = 4, curentLevel = 3, path = { 1: 0, 2: 1 }, currentProduct = probList[0] * probList[1]
  // Loops through all entries except 0 and 1 and adds the products to sums[2], for each entry also calculates level 4 sums

  for (let i = lastIndex + 1; i < probList.length; i++) {
    const nextProduct = currentProduct * probList[i];
    sums[currentLevel - 1] += nextProduct;

    if (currentLevel < maxLevel) {
      // get the next level product sums for current product
      updateSums(sums, probList, maxLevel, currentLevel + 1, nextProduct, i);
    }
  }
}

// Main function
function inequality(probList) {
  probList = probList.sort((a, b) => b - a).slice(0, 4);

  // Calculate maxLevel
  const maxLevel = probList.length;
  if (!maxLevel) return 0;

  // create am array of sums, each entry represents 1 level
  const sums = (new Array(maxLevel)).fill(0);
  updateSums(sums, probList, maxLevel);

  return sums.reduce((a, c, i) => {
    return a + ((i % 2) ? -1 : 1) * c;
  }, 0);
}

console.log(inequality(probList));

PS: This is written in ES6