[Note: parts of this message were removed to make it a legal post.]

(Apologies for this oldish "re-post": it seems that changing from googlemail

to gmail makes emails not recognised by ruby-talk.)

I guess this is why Benoit had that sqrt in there. I don't quite get

why it's necessary.

I kinda like Yaser's solution.

Yaser's solution is a "Monte Carlo" simulation? My initial reaction was that

Yaser's method might not end up with the correct distribution, but on

further thought it (I think):

1. Generates points uniformly in a square of side r.

2. Redistributes the points uniformly in a square of side 2*r. (I wondered

what the "Flip coin to reflect" bits were doing until I realised that.)

3. Ignores any points outside the circle. Which should leave points

uniformly distributed (by area) inside the circle.

So Yaser's solution does have the correct distribution of points. And shows

that sometimes a way to generate a correct distribution is to use an

incorrect distribution and then ignore some values generated from the

incorrect distribution.

(Strictly speaking, is the test that "v.length > 0" necessary, and should

the other test be "v.length <= radius"? It won't make much difference

though.)

** The rest of this has now been covered by other posts, but the links might

be interesting.

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.

So if we want to use a method which selects a random angle and a random

distance from the centre of the circle, for the random distance from the

centre of the circle we need a 0 to 1 distribution which isn't uniform but

which has more weight for higher than lower values in 0 to 1. When I saw

this quiz I thought of using the random angle and distance method, and after

a bit of work guessed that the correct distribution for the distance was

probably to take the square root of random numbers generated from a uniform

0 to 1 distribution. BUT: I couldn't immediately see a way to prove that! So

I didn't post anything! I suspect Benoit can *prove* that taking the square

root gives the correct distribution for the radius!

def uniform_random_point_in_

circle( r, x, y )

# circle radius r, centre at x, y

r_to_point = Math.sqrt( Kernel.rand ) * r

radians_to_point = 2 * Math:

I * Kernel.rand

return x + r_to_point * Math.cos( radians_to_point ),

y + r_to_point * Math.sin( radians_to_point )

end