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From: Ian Kelly <ian.g.kelly@gmail.com>
Date: Mon, 25 May 2015 22:42:51 -0600
Subject: Re: a more precise distance algorithm
To: Python <python-list@python.org>
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On Mon, May 25, 2015 at 1:21 PM, ravas <ravas@outlook.com> 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 distanc=
e form c =3D 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 accur=
ate calculation is c =3D 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 =3D 0.
> """
>
> Is this valid? Does it apply to python?
> 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 =3D B.x - A.x
>     dy =3D B.y - A.y
>     a =3D min(dx, dy)
>     b =3D max(dx, dy)
>     if a =3D=3D 0:
>         return b
>     elif b =3D=3D 0:
>         return a

This branch is incorrect because a could be negative.

You don't need this anyway; the a =3D=3D 0 branch is only there because of
the division by a in the else branch.

>     else:
>         return a * sqrt(1 + (b / a)**2)

Same issue; if a is negative then the result will have the wrong sign.