用 (x,y) 生成圆盘均匀点

取

$$\theta = 2\pi x,\quad r=\sqrt{y},$$

再输出点

$$(r\cos\theta,\ r\sin\theta).$$

原因：面积元素为 $r\,dr\,d\theta$，要让半径分布与面积匹配，需要 $r^2$ 均匀，因此 $r=\sqrt{y}$。

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Original Explanation

$\theta = 2 \pi x$ and $r = \sqrt{y}$, then take the point as $(r \cos(\theta),r \sin(\theta))$

Solution

We could generate a random point in the square and reject it if it is not in the circle. However, the question asks to transform the point, not filter it. We need to somehow manipulate the given random point $(x, y)$

Initial Misstep

The coordinates of the point in the circle are then given by $(r \cos(\theta), r \sin{(\theta)})$ where $r$ is the radius and $\theta$ is the angle. Thus, we can use polar coordinates, $x$ can be converted to the angle and $y$ can be used as the radius.

1. Let $\theta = 2 \pi x$, its value ranged between $0$ to $2\pi$

2. Let $r = y$, the radius in the polar coordinates, its value is between 0 and 1.

3. The Polar Coordinate $(r, \theta)$ can be converted to the cartesian coordinate. Thus, $(y \cos(2 \pi x), y \sin(2\pi x))$ will generate a random point within the disk of radius 1.

4. However, to ensure uniform distribution, we can't just take $r=y$. This is because points near the center of the circle are much more likely to be chosen if we just use a uniform distribution for the radius. Consider a small ring in the circle with radius $r$ and thickness $dr$. The area of this ring (which is a very thin circular strip) is $2\pi \cdot r \cdot dr$. A larger ring has more area and should contain more points. If we take $r=y$, we end up with the same number of points in each ring. This means that larger rings have a lower density of points. This is not uniform.

Correct Solution

We wish to use $(r \cos(\theta), r \sin(\theta))$ as the random point, but we need a way to make it uniform.

Uniform distribution means that the probability of finding a point in any part is proportional to the size of that part.

In the case of 1D range $[0, 1]$, given $y$ is uniform, i.e. $P(y_1 \le y \le y_2) = y_2 - y_1$ (where $0 \le y_1 < y_2 \le 1$)

We want the same property for the disk.

Consider an annulus (ring) between radius $r_1$ and $r_2$ (where $0 \leq r_1 < r_2 \leq 1$)

$\text{Area of annulus} = \pi r_2^2 - \pi r_1^2$

We want the probability to be in proportion to this area, i.e. we want $P(r_1 \le r \le r_2 ) \propto (r_2^2 - r_1^2)$

Taking $r = y$ gives $P(r_1 \le y \le r_2 ) = r_2 - r_1$, which is not proportional to the area.

But, with a leap of faith, if we take $r = \sqrt{y}$, we get

$P(r_1 \le \sqrt{y} \le r_2 ) =P(r_1^2 \le y \le r_1^2)$

$= r_2^2 - r_1^2$ (because $y$ is uniform in the range $[0, 1]$)

Thus, we get the desired property.

Therefore, we can take $r = \sqrt{y}$ and $\theta = 2 \pi x$ to get a uniform distribution of points in the disk.