Scala尾递归方法具有除法和余数误差

本文介绍了Scala尾递归方法具有除法和余数误差的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我目前正在通过在Scala中编写尾递归来计算两个自然数的二项式系数.但是我的代码在除数方面存在问题，像我那样将整数除以k会给您一个非零的余数，从而导致舍入误差.那么谁能帮我弄清楚，如何解决?

I'm currently computing the binomial coefficient of two natural numbers by write a tail recursion in Scala. But my code has something wrong with the dividing numbers, integer division by k like I did as that will give you a non-zero remainder and hence introduce rounding errors. So could anyone help me figure it out, how to fix it ?

def binom(n: Int, k: Int): Int = { require(0 <= k && k <= n) def binomtail(n: Int, k: Int, ac: Int): Int = { if (n == k || k == 0) ac else binomtail(n - 1, k - 1, (n*ac)/k) } binomtail(n,k,1) }

推荐答案

通常，它保持:

binom(n, k) = if (k == 0 || k == n) 1 else binom(n - 1, k - 1) * n / k

如果要在线性时间内进行计算，则必须确保每个中间结果都是整数.现在，

If you want to compute it in linear time, then you have to make sure that each intermediate result is an integer. Now,

binom(n - k + 1, 1)

肯定是整数(只是 n-k + 1 ).从这个数字开始，然后将两个参数都增加一个，您可以通过以下中间步骤到达 binom(n，k):

is certainly an integer (it's just n - k + 1). Starting with this number, and incrementing both arguments by one, you can arrive at binom(n, k) with the following intermediate steps:

binom(n - k + 1, 1) binom(n - k + 2, 2) ... binom(n - 2, k - 2) binom(n - 1, k - 1) binom(n, k)

这意味着您必须以正确的顺序累积"，从 1 到 k ，而不是从 k 到 1 -那么可以保证所有中间结果都对应于实际的二项式系数，因此是整数(而不是分数).这是尾递归函数的样子:

It means that you have to "accumulate" in the right order, from 1 up to k, not from k down to 1 - then it is guaranteed that all intermediate results correspond to actual binomial coefficients, and are therefore integers (not fractions). Here is what it looks like as tail-recursive function:

def binom(n: Int, k: Int): Int = { require(0 <= k && k <= n) @annotation.tailrec def binomtail(nIter: Int, kIter: Int, ac: Int): Int = { if (kIter > k) ac else binomtail(nIter + 1, kIter + 1, (nIter * ac) / kIter) } if (k == 0 || k == n) 1 else binomtail(n - k + 1, 1, 1) }

小小的视觉测试:

val n = 12 for (i <- 0 to n) { print(" " * ((n - i) * 2)) for (j <- 0 to i) { printf(" %3d", binom(i, j)) } println() }

打印:

1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1 1 10 45 120 210 252 210 120 45 10 1 1 11 55 165 330 462 462 330 165 55 11 1 1 12 66 220 495 792 924 792 495 220 66 12 1

好吧，如果需要，可以将其与此进行比较

Looks ok, compare it with this, if you want.