Am 25.05.15 um 21:21 schrieb ravas:
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?
Yes. Valid for floating point math, which can overflow and lose precision.
Does it apply to python?
Yes. Python uses floating point math by default
Any other thoughts? :D
My imagining:
def distance(A, B):
Wrong. Just use the built-in function Math.hypot() - it should handle
these cases and also overflow, infinity etc. in the best possible way.
Apfelkiste:~ chris$ python
Python 2.7.2 (default, Oct 11 2012, 20:14:37)
[GCC 4.2.1 Compatible Apple Clang 4.0 (tags/Apple/clang-418.0.60)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> import math
>>> math.hypot(3,4)
5.0
Christian
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