From: Caleb Clausen <vikkous@...>
Date: 2010-06-23T10:56:15+09:00
Subject: Re: [QUIZ] Random Points within a Circle (#234)

On 6/22/10, Colin Bartlett <colinb2r@googlemail.com> wrote:
> The random angle bit in your method is OK.
> For the square root: the area of an annulus radius r to r+e is
>     pi * ( (r+e)**2 - r**2 ) = pi * 2*r*e + pi * e**2
> So, if I treat e as an infinitesimal and neglect the e**2 (and hope that no
> real mathematicians are looking, or claim I'm using
>   Abraham Robinson's http://en.wikipedia.org/wiki/Abraham_robinson
>   Non-Standard Analysis http://en.wikipedia.org/wiki/Non-standard_analysis
> and hope that nobody asks me any awkward questions about how that really
> works)
> the area of the annulus is (approximately) pi * 2*r*e.
> (Think of it as a (very) "bent" rectangle of length pi * 2*r with a "width"
> of e.)
>
> If you use a uniform distribution for the radius to the random point then on
> average you'll be putting the same number of points in an annulus (a0) of
> width e very close to the centre of the circle as in an annulus (a1) of
> width e very close to the circumference of the circle. But the annulus (a1)
> has a much larger area than the annulus (a0), so the density of points will
> be greater in annulus (a0) than annulus (a1). Which is why Dave Howell made
> his comment.

That was beautiful! The best explanation yet. Thank you.

I keep wondering the use of sqrt will cause weird float rounding
errors; that is will there be valid [float, float] pairs which are in
the circle but can never be be found by an algorithm that includes
Math.sqrt(rand)? Maybe someone with a better grip on the math has an
answer to this.