algorithm - 最大归一化子数组和的快速算法？

Question

最大子数组和问题有一个非常简单的线性时间解 https://en.m.wikipedia.org/wiki/Maximum_subarray_problem .

如果相反，我们想要最大化 sum(subarray)/sqrt(subarray length)，是否存在次二次时间解？

输入数组的元素将是 -infinity 到 +infinity 范围内的浮点值。

Answer 1

更新

我在下面的测试中添加了一个版本的 estabroo 基于 Kadane 的代码。在我的测试中似乎显示出高达 10% 的差异(运行代码段进行随机测试)。

(结束更新)

就近似值而言，我能想到的最好办法是在搜索过程中使用窗口大小的随机样本对目标进行二分搜索 O(log m * n * num_samples_constant)，其中m 是范围。在测试中，我看到暴力破解(限于 5000 个元素阵列，范围为 ±1000000000)和后者之间的变化在 0% 到 30% 之间变化，样本大小为 200 个窗口长度。 (也许另一个例程可以进一步完善？)

下面的 JavaScript 代码运行 10 次测试并报告最小和最大的差异，然后仅在更长的数组上进行二进制搜索。

其他想法包括使用 FFT 生成总和，但我不知道是否有一种有效的方法可以将每个总和与生成它的子数组长度相关联；以及尝试找到问题的另一种表示形式:

f = sqrt(i - j) * (si - sj), for j < i
f^2 = sqrt(i - j) * (si - sj) * sqrt(i - j) * (si - sj)
    = (i - j) * (si^2 - 2si*sj + sj^2)
    = i*si^2 - 2i*si*sj + i*sj^2
      -j*si^2 + 2j*si*sj - j*sj^2

    = i*si^2 + 
      (-2sj, sj^2, -j, 2j*sj, -j*sj^2) // known before i
        dot (i*si, 1, si^2, si, 1)

(因此，如果我们在对数时间内解决了一个 5 维凸包更新，即 5 维极值点问题，并弄清楚我们的候选对象是正数还是负数，我们就可以开始了:)

function prefix_sums(A){
  let ps = new Array(A.length + 1).fill(0)
  for (let i=0; i<A.length; i++)
    ps[i + 1] = A[i] + ps[i]
  return ps
}

function brute_force(ps){
  let best = -Infinity
  let best_idxs = [-1, -1]
  for (let i=1; i<ps.length; i++){
    for (let j=0; j<i; j++){
      let s = (ps[i] - ps[j]) / Math.sqrt(i - j)
      if (s > best){
        best = s
        best_idxs = [j, i - 1]
      }
    }
  }
  return [best, best_idxs]
}

function get_high(A){
  return A.reduce((acc, x) => x > 0 ? acc + x : acc, 0)
}

function get_low(A){
  return Math.min.apply(null, A)
}


function f(A){
  let n = A.length
  let ps = prefix_sums(A)
  let high = get_high(A)
  let low = get_low(A)
  let best = [-1, -1]
  let T = low + (high - low) / 2
  let found = false

  while (low + EPSILON < high){
    T = low + (high - low) / 2
    // Search for T
    found = false

    for (let l=0; l<NUM_SAMPLES; l++){
      let w = Math.max(1, ~~(Math.random() * (n + 1)))

      for (let i=w; i<ps.length; i++){
        let s = (ps[i] - ps[i - w]) / Math.sqrt(w)
        if (s >= T){
          found = true
          best = [i - w, i - 1]
          break
        }
      }
      if (found)
        break
    }
    // Binary search
    if (found)
      low = T
    else
      high = T - EPSILON 
  }

  return [low, best]
}

function max_subarray(A){
    var max_so_far = max_ending_here = A[0]
    var startOld = start = end = 0
    var divb = divbo = 1
    //for i, x in enumerate(A[1:], 1):
    for (let i=1; i<A.length; i++){
        var x = A[i]
        divb = i - start + 1
        divbo = divb - 1
        if (divb <= 1){
            divb = 1
            divbo = 1
        }
        undo = max_ending_here * Math.sqrt(divbo)
        max_ending_here = Math.max(x, (undo + x)/Math.sqrt(divb))
        if (max_ending_here == x)
            start = i
        max_so_far = Math.max(max_so_far, max_ending_here)
        if (max_ending_here < 0)
            start = i + 1
        else if (max_ending_here == max_so_far){
            startOld = start
            end = i
        }
    }
    if (end == A.length-1){
        start = startOld + 1
        var new_max = max_so_far
        divbo = end - startOld + 1
        divb = divbo - 1
        while (start < end){
            new_max = (new_max * Math.sqrt(divbo) - A[start-1])/Math.sqrt(divb)
            if (new_max > max_so_far){
                max_so_far = new_max
                startOld = start
            }
            start += 1
        }
    }
    return [max_so_far , startOld, end]
}

const EPSILON = 1
const NUM_SAMPLES = 200

let m = 1000000000
let n = 5000
let A

let max_diff = 0
let min_diff = Infinity
let max_diff2 = 0
let min_diff2 = Infinity
let num_tests = 10

for (let i=0; i<num_tests; i++){
  A = []
  for (let i=0; i<n; i++)
    A.push([-1, 1][~~(2 * Math.random())] * Math.random() * m + Math.random())

  let f_A = f(A)
  let g_A = brute_force(prefix_sums(A))
  let m_A = max_subarray(A)
  let diff = (g_A[0] - f_A[0]) / g_A[0]
  max_diff = Math.max(max_diff, diff)
  min_diff = Math.min(min_diff, diff)
  let diff2 = (g_A[0] - m_A[0]) / g_A[0]
  max_diff2 = Math.max(max_diff2, diff2)
  min_diff2 = Math.min(min_diff2, diff2)
}

console.log(`${ n } element array`)
console.log(`${ num_tests } tests`)
console.log(`min_diff: ${ min_diff * 100 }%`)
console.log(`max_diff: ${ max_diff * 100 }%`)
console.log(`min_diff (Kadane): ${ min_diff2 * 100 }%`)
console.log(`max_diff (Kadane): ${ max_diff2 * 100 }%`)

n = 100000
A = []
for (let i=0; i<n; i++)
  A.push([-1, 1][~~(2 * Math.random())] * Math.random() * m + Math.random())

var start = +new Date()
console.log(`\n${ n } element array`)
console.log(JSON.stringify(f(A)))
console.log(`${ (new Date() - start) / 1000 } seconds`)