Generate a random point within a rectangle (uniformly)

Question

Generate a random point within a rectangle (uniformly)

This suppose to be a simple problem.

However, in RANDOM_DATA homepage I found the following note:

However, we will not achieve uniform distribution in the simple case of a rectangle of nonequal sides [0,A] x [0,B], if we naively scale the random values (u1,u2) to (A*u1,B*u2). In that case, the expected point density of a wide, short region will differ from that of a narrow tall region. The absence of uniformity is most obvious if the points are plotted.

I found it quite of strange... I can't figure out why such scaling will affect the uniformity.

What am I missing?

Edit:

Thank you Patrick87 and missingno. I was searching for a theoretical reason for the statement. I now understand that the reason is not theoretical, but practical - the granularity of floating-point values.

If I'll generate two uniform floating-points between 0 and 1 (which is an issue by itself due to the nature of floating-point value representation. Look here for an algorithm) - the granularity will be limited.

Suppose that there are X different values between 0 and 1. By scaling (u1,u2) to (u1,2*u2) we'll have X different values in the range [0,u1] and X different values in the range [0,2*u2]. For area uniformity we should have twice as many different values in [0,2*u2] than in [0,u1].

Given that, Allow me to change my question:

How should I generate a random point within a rectangle (with uniform distribution by area)?

Answer 1

That statement is incorrect, direct product of two independent uniform measures is a uniform measure. This can be shown as follows:

A probability for a random point to hit a rectangle with sides a and b is equal to probability for the first coordinate to hit the segment with the length a and the second coordinate to hit the segment with the length b. (We are talking about projections of a rectangle to axes).

First probability is a / A, the second one is b / B. As these variables are independent, the probabilities multiply, so the resulting probability is ab / AB, so we have a uniform 2D distribution as the probability is proportional to the area of the rectangle. This formula is symmetric with respect to a and b, so the observation in the quote is wrong about narrow and wide rectangles.