随同hypot() [英] Companion to hypot()

问题描述

hypot 函数在该语言的1999年修订版中引入C中，以给定另一侧作为参数来计算直角三角形的斜边，但是要小心避免上溢/下溢，天真的实现会导致

The hypot function, introduced into C in the 1999 revision of the language, calculates the hypotenuse of a right triangle given the other sides as arguments, but with care taken to avoid the over/underflow which would result from the naive implementation as

double hypot(double a, double b)
{
  return sqrt(a*a + b*b);
}

我发现自己需要伴随功能:给定一个边和一个三角形的斜边，找到第三个边(避免出现下溢/溢出).我可以想到几种方法，但是想知道是否存在现有的最佳实践"?

I find myself with the need for companion functionality: given a side and the hypotenuse of a triangle, find the third side (avoiding under/overflow). I can think of a few ways to do this, but wondered if there was an existing "best practice"?

我的目标是Python，但实际上我正在寻找算法指针.

My target is Python, but really I'm looking for algorithm pointers.

感谢您的答复.如果有人对结果感兴趣，可以在此处和Python版本此处，是<一个href ="https://hypothesis.works/" rel ="nofollow noreferrer">假设项目.

Thanks for the replies. In case anyone is interested in the result, my C99 implementation can be found here and a Python version here, part of the Hypothesis project.

推荐答案

要做的第一件事是分解:

The first thing to do is factorize:

b = sqrt(h*h - a*a) = sqrt((h-a)*(h+a))

我们不仅避免了一些溢出，而且还获得了准确性.

如果有任何因素接近 1E + 154 = sqrt(1E + 308)(IEEE 754 64位浮点数最多)，那么我们还必须避免溢出:

We have not only avoided some overflow, but also gained accuracy.

If any factor is close to 1E+154 = sqrt(1E+308) (max with IEEE 754 64 bits float) then we must also avoid overflow:

sqrt((h-a)*(h+a)) = sqrt(h-a) * sqrt(h+a)

这种情况不太可能发生，因此两个 sqrt 都是合理的，即使它比 sqrt 慢.

This case is very unlikely, so the two sqrt's are justified, even if its slower than just a sqrt.

请注意，如果 h〜5E + 7 * a ，则 h〜b ，这意味着没有足够的数字表示 b 与 h 不同.

Notice that if h ~ 5E+7 * a then h ~ b which means that there are not enough digits to represent b as different from h.

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