a more precise distance algorithm

[Gary Herron]

On 05/25/2015 12:21 PM, ravas wrote:
> I read an interesting comment:
> """
> The coolest thing I've ever discovered about Pythagorean's Theorem is an alternate way to calculate it. If you write a program that uses the distance form c = sqrt(a^2 + b^2) you will suffer from the lose of half of your available precision because the square root operation is last. A more accurate calculation is c = a * sqrt(1 + b^2 / a^2). If a is less than b, you should swap them and of course handle the special case of a = 0.
> """
>
> Is this valid?

> Does it apply to python?

This is a statement about floating point numeric calculations on a 
computer,.  As such, it does apply to Python which uses the underlying 
hardware for floating point calculations.

Validity is another matter.  Where did you find the quote?

Gary Herron


> Any other thoughts? :D
>
> My imagining:
>
> def distance(A, B):
>      """
>      A & B are objects with x and y attributes
>      :return: the distance between A and B
>      """
>      dx = B.x - A.x
>      dy = B.y - A.y
>      a = min(dx, dy)
>      b = max(dx, dy)
>      if a == 0:
>          return b
>      elif b == 0:
>          return a
>      else:
>          return a * sqrt(1 + (b / a)**2)