Math.Sqrt() 的时间复杂度？

Question

How can I find the complexity of this function?

private double EuclideanDistance(MFCC.MFCCFrame vec1, MFCC.MFCCFrame vec2)
{
  double Distance = 0.0;
  for (int K = 0; K < 13; K++)
     Distance += (vec1.Features[K] - vec2.Features[K]) * (vec1.Features[K] - vec2.Features[K]);
  return Math.Sqrt(Distance);
}

I know that the below section is O(1):

double Distance = 0.0;
for (int K = 0; K < 13; K++)
   Distance += (vec1.Features[K]-vec2.Features[K])*(vec1.Features[K]-vec2.Features[K]);

But I can't figure out what the complexity of Math.Sqrt() is.

Answer 1

You can consider it O(1):

In other words, Math.Sqrt() translates to a single floating point machine code instruction

source: c# Math.Sqrt Implementation

Answer 2

As mentioned by BlackBear , the Math.Sqrt implementation translates to a to a single floating point machine code instruction (fsqrt). The number of cycles of this instruction is bounded ( here are some examples). And that means its complexity is O(1).

That is only true, because we use a limited number of floating-point values. The "actual" complexity of that operation depends on the number of bits of the input. Here you can find a list of the complexity of basic arithmetic functions. According to that list the squareroot function has the complexity of the multiplication function (O(n log n) for two n-digit numbers).

You said, that you assume addition and multiplication function have complexity O(1). That means, you can assume that the squareroot function, allthough much slower, has complexity O(1), too.