Generate a random point within a circle (uniformly

问题:

I need to generate a uniformly random point within a circle of radius R.

I realize that by just picking a uniformly random angle in the interval [0 ... 2π), and uniformly random radius in the interval (0 ... R) I would end up with more points towards the center, since for two given radii, the points in the smaller radius will be closer to each other than for the points in the larger radius.

I found a blog entry on this over here but I don\'t understand his reasoning. I suppose it is correct, but I would really like to understand from where he gets (2/R²)×r and how he derives the final solution.

---

Update: 7 years after posting this question I still hadn\'t received a satisfactory answer on the actual question regarding the math behind the square root algorithm. So I spent a day writing an answer myself. Link to my answer.

回答1:

Let\'s approach this like Archimedes would have.

How can we generate a point uniformly in a triangle ABC, where |AB|=|BC|? Let\'s make this easier by extending to a parallelogram ABCD. It\'s easy to generate points uniformly in ABCD. We uniformly pick a random point X on AB and Y on BC and choose Z such that XBYZ is a parallelogram. To get a uniformly chosen point in the original triangle we just fold any points that appear in ADC back down to ABC along AC.

Now consider a circle. In the limit we can think of it as infinitely many isoceles triangles ABC with B at the origin and A and C on the circumference vanishingly close to each other. We can pick one of these triangles simply by picking an angle theta. So we now need to generate a distance from the center by picking a point in the sliver ABC. Again, extend to ABCD, where D is now twice the radius from the circle center.

Picking a random point in ABCD is easy using the above method. Pick a random point on AB. Uniformly pick a random point on BC. Ie. pick a pair of random numbers x and y uniformly on [0,R] giving distances from the center. Our triangle is a thin sliver so AB and BC are essentially parallel. So the point Z is simply a distance x+y from the origin. If x+y>R we fold back down.

Here\'s the complete algorithm for R=1. I hope you agree it\'s pretty simple. It uses trig, but you can give a guarantee on how long it\'ll take, and how many random() calls it needs, unlike rejection sampling.

t = 2*pi*random()
u = random()+random()
r = if u>1 then 2-u else u
[r*cos(t), r*sin(t)]

Here it is in Mathematica.

f[] := Block[{u, t, r},
  u = Random[] + Random[];
  t = Random[] 2 Pi;
  r = If[u > 1, 2 - u, u];
  {r Cos[t], r Sin[t]}
]

ListPlot[Table[f[], {10000}], AspectRatio -> Automatic]

回答2:

How to generate a random point within a circle of radius R:

r = R * sqrt(random())
theta = random() * 2 * PI

(Assuming random() gives a value between 0 and 1 uniformly)

If you want to convert this to Cartesian coordinates, you can do

x = centerX + r * cos(theta)
y = centerY + r * sin(theta)

Why sqrt(random())?

Let\'s look at the math that leads up to sqrt(random()). Assume for simplicity that we\'re working with the unit circle, i.e. R = 1.

The average distance between points should be the same regardless of how far from the center we look. This means for example, that looking on the perimeter of a circle with circumference 2 we should find twice as many points as the number of points on the perimeter of a circle with circumference 1.

                

Since the circumference of a circle (2πr) grows linearly with r, it follows that the number of random points should grow linearly with r. In other words, the desired probability density function (PDF) grows linearly. Since a PDF should have an area equal to 1 and the maximum radius is 1, we have

                

So we know how the desired density of our random values should look like. Now: How do we generate such a random value when all we have is a uniform random value between 0 and 1?

We use a trick called inverse transform sampling

1. From the PDF, create the cumulative distribution function (CDF)
2. Mirror this along y = x
3. Apply the resulting function to a uniform value between 0 and 1.

Sounds complicated? Let me insert a yellow box with a little sidetrack that conveys the intuition:

Suppose we want to generate a random point with the following distribution:

                

That is

• 1/5 of the points uniformly between 1 and 2, and
• 4/5 of the points uniformly between 2 and 3.

The CDF is, as the name suggests, the cumulative version of the PDF. Intuitively: While PDF(x) describes the number of random values at x, CDF(x) describes the number of random values less than x.

In this case the CDF would look like:

                

To see how this is useful, imagine that we shoot bullets from left to right at uniformly distributed heights. As the bullets hit the line, they drop down to the ground:

                

See how the density of the bullets on the ground correspond to our desired distribution! We\'re almost there!

The problem is that for this function, the y axis is the output and the x axis is the input. We can only \"shoot bullets from the ground straight up\"! We need the inverse function!

This is why we mirror the whole thing; x becomes y and y becomes x:

                

We call this CDF^-1. To get values according to the desired distribution, we use CDF^-1(random()).

…so, back to generating random radius values where our PDF equals 2x.

Step 1: Create the CDF:

Since we\'re working with reals, the CDF is expressed as the integral of the PDF.

CDF(x) = ∫ 2x = x²

Step 2: Mirror the CDF along y = x:

Mathematically this boils down to swapping x and y and solving for y:

CDF:     y = x²
Swap:   x = y²
Solve:   y = √x
CDF^-1:  y = √x

Step 3: Apply the resulting function to a uniform value between 0 and 1

CDF^-1(random()) = √random()

Which is what we set out to derive :-)

回答3:

Here is a fast and simple solution.

Pick two random numbers in the range (0, 1), namely a and b. If b < a, swap them. Your point is (b*R*cos(2*pi*a/b), b*R*sin(2*pi*a/b)).

You can think about this solution as follows. If you took the circle, cut it, then straightened it out, you\'d get a right-angled triangle. Scale that triangle down, and you\'d have a triangle from (0, 0) to (1, 0) to (1, 1) and back again to (0, 0). All of these transformations change the density uniformly. What you\'ve done is uniformly picked a random point in the triangle and reversed the process to get a point in the circle.

回答4:

Note the point density in proportional to inverse square of the radius, hence instead of picking r from [0, r_max], pick from [0, r_max^2], then compute your coordinates as:

x = sqrt(r) * cos(angle)
y = sqrt(r) * sin(angle)

This will give you uniform point distribution on a disk.

http://mathworld.wolfram.com/DiskPointPicking.html

回答5:

Think about it this way. If you have a rectangle where one axis is radius and one is angle, and you take the points inside this rectangle that are near radius 0. These will all fall very close to the origin (that is close together on the circle.) However, the points near radius R, these will all fall near the edge of the circle (that is, far apart from each other.)

This might give you some idea of why you are getting this behavior.

The factor that\'s derived on that link tells you how much corresponding area in the rectangle needs to be adjusted to not depend on the radius once it\'s mapped to the circle.

Edit: So what he writes in the link you share is, \"That’s easy enough to do by calculating the inverse of the cumulative distribution, and we get for r:\".

The basic premise is here that you can create a variable with a desired distribution from a uniform by mapping the uniform by the inverse function of the cumulative distribution function of the desired probability density function. Why? Just take it for granted for now, but this is a fact.

Here\'s my somehwat intuitive explanation of the math. The density function f(r) with respect to r has to be proportional to r itself. Understanding this fact is part of any basic calculus books. See sections on polar area elements. Some other posters have mentioned this.

So we\'ll call it f(r) = C*r;

This turns out to be most of the work. Now, since f(r) should be a probability density, you can easily see that by integrating f(r) over the interval (0,R) you get that C = 2/R^2 (this is an exercise for the reader.)

Thus, f(r) = 2*r/R^2

OK, so that\'s how you get the formula in the link.

Then, the final part is going from the uniform random variable u in (0,1) you must map by the inverse function of the cumulative distribution function from this desired density f(r). To understand why this is the case you need to find an advanced probability text like Papoulis probably (or derive it yourself.)

Integrating f(r) you get F(r) = r^2/R^2

To find the inverse function of this you set u = r^2/R^2 and then solve for r, which gives you r = R * sqrt(u)

This totally makes sense intuitively too, u = 0 should map to r = 0. Also, u = 1 shoudl map to r = R. Also, it goes by the square root function, which makes sense and matches the link.

回答6:

It really depends on what you mean by \'uniformly random\'. This is a subtle point and you can read more about it on the wiki page here: http://en.wikipedia.org/wiki/Bertrand_paradox_%28probability%29, where the same problem, giving different interpretations to \'uniformly random\' gives different answers!

Depending on how you choose the points, the distribution could vary, even though they are uniformly random in some sense.

It seems like the blog entry is trying to make it uniformly random in the following sense: If you take a sub-circle of the circle, with the same center, then the probability that the point falls in that region is proportional to the area of the region. That, I believe, is attempting to follow the now standard interpretation of \'uniformly random\' for 2D regions with areas defined on them: probability of a point falling in any region (with area well defined) is proportional to the area of that region.

回答7:

The reason why the naive solution doesn\'t work is that it gives a higher probability density to the points closer to the circle center. In other words the circle that has radius r/2 has probability r/2 of getting a point selected in it, but it has area (number of points) pi*r^2/4.

Therefore we want a radius probability density to have the following property:

The probability of choosing a radius smaller or equal to a given r has to be proportional to the area of the circle with radius r. (because we want to have a uniform distribution on the points and larger areas mean more points)

In other words we want the probability of choosing a radius between [0,r] to be equal to its share of the overall area of the circle. The total circle area is pi*R^2, and the area of the circle with radius r is pi*r^2. Thus we would like the probability of choosing a radius between [0,r] to be (pi*r^2)/(pi*R^2) = r^2/R^2.

Now comes the math:

The probability of choosing a radius between [0,r] is the integral of p(r) dr from 0 to r (that\'s just because we add all the probabilities of the smaller radii). Thus we want integral(p(r)dr) = r^2/R^2. We can clearly see that R^2 is a constant, so all we need to do is figure out which p(r), when integrated would give us something like r^2. The answer is clearly r * constant. integral(r * constant dr) = r^2/2 * constant. This has to be equal to r^2/R^2, therefore constant = 2/R^2. Thus you have the probability distribution p(r) = r * 2/R^2

Note: Another more intuitive way to think about the problem is to imagine that you are trying to give each circle of radius r a probability density equal to the proportion of the number of points it has on its circumference. Thus a circle which has radius r will have 2 * pi * r \"points\" on its circumference. The total number of points is pi * R^2. Thus you should give the circle r a probability equal to (2 * pi * r) / (pi * R^2) = 2 * r/R^2. This is much easier to understand and more intuitive, but it\'s not quite as mathematically sound.

回答8:

Let ρ (radius) and φ (azimuth) be two random variables corresponding to polar coordinates of an arbitrary point inside the circle. If the points are uniformly distributed then what is the disribution function of ρ and φ?

For any r: 0 < r < R the probability of radius coordinate ρ to be less then r is

P[ρ < r] = P[point is within a circle of radius r] = S1 / S0 =(r/R)²

Where S1 and S0 are the areas of circle of radius r and R respectively. So the CDF can be given as:

          0          if r<=0
  CDF =   (r/R)**2   if 0 < r <= R
          1          if r > R

And PDF:

PDF = d/dr(CDF) = 2 * (r/R**2) (0 < r <= R).

Note that for R=1 random variable sqrt(X) where X is uniform on [0, 1) has this exact CDF (because P[sqrt(X) < y] = P[x < y**2] = y**2 for 0 < y <= 1).

The distribution of φ is obviously uniform from 0 to 2*π. Now you can create random polar coordinates and convert them to Cartesian using trigonometric equations:

x = ρ * cos(φ)
y = ρ * sin(φ)

Can\'t resist to post python code for R=1.

from matplotlib import pyplot as plt
import numpy as np

rho = np.sqrt(np.random.uniform(0, 1, 5000))
phi = np.random.uniform(0, 2*np.pi, 5000)

x = rho * np.cos(phi)
y = rho * np.sin(phi)

plt.scatter(x, y, s = 4)

You will get

回答9:

Here is my Python code to generate num random points from a circle of radius rad:

import matplotlib.pyplot as plt
import numpy as np
rad = 10
num = 1000

t = np.random.uniform(0.0, 2.0*np.pi, num)
r = rad * np.sqrt(np.random.uniform(0.0, 1.0, num))
x = r * np.cos(t)
y = r * np.sin(t)

plt.plot(x, y, \"ro\", ms=1)
plt.axis([-15, 15, -15, 15])
plt.show()

回答10:

Solution in Java and the distribution example (2000 points)

public void getRandomPointInCircle() {
    double t = 2 * Math.PI * Math.random();
    double r = Math.sqrt(Math.random());
    double x = r * Math.cos(t);
    double y = r * Math.sin(t);
    System.out.println(x);
    System.out.println(y);
}

based on previus solution https://stackoverflow.com/a/5838055/5224246 from @sigfpe

回答11:

I think that in this case using polar coordinates is a way of complicate the problem, it would be much easier if you pick random points into a square with sides of length 2R and then select the points (x,y) such that x^2+y^2<=R^2.

回答12:

First we generate a cdf[x] which is

The probability that a point is less than distance x from the centre of the circle. Assume the circle has a radius of R.

obviously if x is zero then cdf[0] = 0

obviously if x is R then the cdf[R] = 1

obviously if x = r then the cdf[r] = (Pi r^2)/(Pi R^2)

This is because each \"small area\" on the circle has the same probability of being picked, So the probability is proportionally to the area in question. And the area given a distance x from the centre of the circle is Pi r^2

so cdf[x] = x^2/R^2 because the Pi cancel each other out

we have cdf[x]=x^2/R^2 where x goes from 0 to R

So we solve for x

R^2 cdf[x] = x^2

x = R Sqrt[ cdf[x] ]

We can now replace cdf with a random number from 0 to 1

x = R Sqrt[ RandomReal[{0,1}] ]

Finally

r = R Sqrt[  RandomReal[{0,1}] ];
theta = 360 deg * RandomReal[{0,1}];
{r,theta}

we get the polar coordinates {0.601168 R, 311.915 deg}

回答13:

There is a linear relationship between the radius and the number of points \"near\" that radius, so he needs to use a radius distribution that is also makes the number of data points near a radius r proportional to r.

回答14:

I used once this method: This may be totally unoptimized (ie it uses an array of point so its unusable for big circles) but gives random distribution enough. You could skip the creation of the matrix and draw directly if you wish to. The method is to randomize all points in a rectangle that fall inside the circle.

bool[,] getMatrix(System.Drawing.Rectangle r) {
    bool[,] matrix = new bool[r.Width, r.Height];
    return matrix;
}

void fillMatrix(ref bool[,] matrix, Vector center) {
    double radius = center.X;
    Random r = new Random();
    for (int y = 0; y < matrix.GetLength(0); y++) {
        for (int x = 0; x < matrix.GetLength(1); x++)
        {
            double distance = (center - new Vector(x, y)).Length;
            if (distance < radius) {
                matrix[x, y] = r.NextDouble() > 0.5;
            }
        }
    }

}

private void drawMatrix(Vector centerPoint, double radius, bool[,] matrix) {
    var g = this.CreateGraphics();

    Bitmap pixel = new Bitmap(1,1);
    pixel.SetPixel(0, 0, Color.Black);

    for (int y = 0; y < matrix.GetLength(0); y++)
    {
        for (int x = 0; x < matrix.GetLength(1); x++)
        {
            if (matrix[x, y]) {
                g.DrawImage(pixel, new PointF((float)(centerPoint.X - radius + x), (float)(centerPoint.Y - radius + y)));
            }
        }
    }

    g.Dispose();
}

private void button1_Click(object sender, EventArgs e)
{
    System.Drawing.Rectangle r = new System.Drawing.Rectangle(100,100,200,200);
    double radius = r.Width / 2;
    Vector center = new Vector(r.Left + radius, r.Top + radius);
    Vector normalizedCenter = new Vector(radius, radius);
    bool[,] matrix = getMatrix(r);
    fillMatrix(ref matrix, normalizedCenter);
    drawMatrix(center, radius, matrix);
}

回答15:

The area element in a circle is dA=rdr*dphi. That extra factor r destroyed your idea to randomly choose a r and phi. While phi is distributed flat, r is not, but flat in 1/r (i.e. you are more likely to hit the boundary than \"the bull\'s eye\").

So to generate points evenly distributed over the circle pick phi from a flat distribution and r from a 1/r distribution.

Alternatively use the Monte Carlo method proposed by Mehrdad.

EDIT

To pick a random r flat in 1/r you could pick a random x from the interval [1/R, infinity] and calculate r=1/x. r is then distributed flat in 1/r.

To calculate a random phi pick a random x from the interval [0, 1] and calculate phi=2*pi*x.

回答16:

I don\'t know if this question is still open for a new solution with all the answer already given, but I happened to have faced exactly the same question myself. I tried to \"reason\" with myself for a solution, and I found one. It might be the same thing as some have already suggested here, but anyway here it is:

in order for two elements of the circle\'s surface to be equal, assuming equal dr\'s, we must have dtheta1/dtheta2 = r2/r1. Writing expression of the probability for that element as P(r, theta) = P{ r1< r< r1 + dr, theta1< theta< theta + dtheta1} = f(r,theta)*dr*dtheta1, and setting the two probabilities (for r1 and r2) equal, we arrive to (assuming r and theta are independent) f(r1)/r1 = f(r2)/r2 = constant, which gives f(r) = c*r. And the rest, determining the constant c follows from the condition on f(r) being a PDF.

回答17:

A programmer solution:

• Create a bit map (a matrix of boolean values). It can be as large as you want.
• Draw a circle in that bit map.
• Create a lookup table of the circle\'s points.
• Choose a random index in this lookup table.

const int RADIUS = 64;
const int MATRIX_SIZE = RADIUS * 2;

bool matrix[MATRIX_SIZE][MATRIX_SIZE] = {0};

struct Point { int x; int y; };

Point lookupTable[MATRIX_SIZE * MATRIX_SIZE];

void init()
{
  int numberOfOnBits = 0;

  for (int x = 0 ; x < MATRIX_SIZE ; ++x)
  {
    for (int y = 0 ; y < MATRIX_SIZE ; ++y)
    {
      if (x * x + y * y < RADIUS * RADIUS) 
      {
        matrix[x][y] = true;

        loopUpTable[numberOfOnBits].x = x;
        loopUpTable[numberOfOnBits].y = y;

        ++numberOfOnBits;

      } // if
    } // for
  } // for
} // ()

Point choose()
{
  int randomIndex = randomInt(numberOfBits);

  return loopUpTable[randomIndex];
} // ()

The bitmap is only necessary for the explanation of the logic. This is the code without the bitmap:

const int RADIUS = 64;
const int MATRIX_SIZE = RADIUS * 2;

struct Point { int x; int y; };

Point lookupTable[MATRIX_SIZE * MATRIX_SIZE];

void init()
{
  int numberOfOnBits = 0;

  for (int x = 0 ; x < MATRIX_SIZE ; ++x)
  {
    for (int y = 0 ; y < MATRIX_SIZE ; ++y)
    {
      if (x * x + y * y < RADIUS * RADIUS) 
      {
        loopUpTable[numberOfOnBits].x = x;
        loopUpTable[numberOfOnBits].y = y;

        ++numberOfOnBits;
      } // if
    } // for
  } // for
} // ()

Point choose()
{
  int randomIndex = randomInt(numberOfBits);

  return loopUpTable[randomIndex];
} // ()

回答18:

I am still not sure about the exact \'(2/R2)×r\' but what is apparent is the number of points required to be distributed in given unit \'dr\' i.e. increase in r will be proportional to r2 and not r.

check this way...number of points at some angle theta and between r (0.1r to 0.2r) i.e. fraction of the r and number of points between r (0.6r to 0.7r) would be equal if you use standard generation, since the difference is only 0.1r between two intervals. but since area covered between points (0.6r to 0.7r) will be much larger than area covered between 0.1r to 0.2r, the equal number of points will be sparsely spaced in larger area, this I assume you already know, So the function to generate the random points must not be linear but quadratic, (since number of points required to be distributed in given unit \'dr\' i.e. increase in r will be proportional to r2 and not r), so in this case it will be inverse of quadratic, since the delta we have (0.1r) in both intervals must be square of some function so it can act as seed value for linear generation of points (since afterwords, this seed is used linearly in sin and cos function), so we know, dr must be quadratic value and to make this seed quadratic, we need to originate this values from square root of r not r itself, I hope this makes it little more clear.

回答19:

Such a fun problem.
The rationale of the probability of a point being chosen lowering as distance from the axis origin increases is explained multiple times above. We account for that by taking the root of U[0,1]. Here\'s a general solution for a positive r in Python 3.

import numpy
import math
import matplotlib.pyplot as plt

def sq_point_in_circle(r):
    \"\"\"
    Generate a random point in an r radius circle 
    centered around the start of the axis
    \"\"\"

    t = 2*math.pi*numpy.random.uniform()
    R = (numpy.random.uniform(0,1) ** 0.5) * r

    return(R*math.cos(t), R*math.sin(t))

R = 200 # Radius
N = 1000 # Samples

points = numpy.array([sq_point_in_circle(R) for i in range(N)])
plt.scatter(points[:, 0], points[:,1])

回答20:

You can also use your intuition.

The area of a circle is pi*r^2

For r=1

This give us an area of pi. Let us assume that we have some kind of function fthat would uniformly distrubute N=10 points inside a circle. The ratio here is 10 / pi

Now we double the area and the number of points

For r=2 and N=20

This gives an area of 4pi and the ratio is now 20/4pi or 10/2pi. The ratio will get smaller and smaller the bigger the radius is, because its growth is quadratic and the N scales linearly.

To fix this we can just say

x = r^2
sqrt(x) = r

If you would generate a vector in polar coordinates like this

length = random_0_1();
angle = random_0_2pi();

More points would land around the center.

length = sqrt(random_0_1());
angle = random_0_2pi();

length is not uniformly distributed anymore, but the vector will now be uniformly distributed.

回答21:

1) Choose a random X between -1 and 1.

var X:Number = Math.random() * 2 - 1;

2) Using the circle formula, calculate the maximum and minimum values of Y given that X and a radius of 1:

var YMin:Number = -Math.sqrt(1 - X * X);
var YMax:Number = Math.sqrt(1 - X * X);

3) Choose a random Y between those extremes:

var Y:Number = Math.random() * (YMax - YMin) + YMin;

4) Incorporate your location and radius values in the final value:

var finalX:Number = X * radius + pos.x;
var finalY:Number = Y * radois + pos.y;