478. Generate Random Point in a Circle

Given the radius and x-y positions of the center of a circle, write a function randPoint which generates a uniform random point in the circle.

Note:

1. input and output values are in floating-point.
2. radius and x-y position of the center of the circle is passed into the class constructor.
3. a point on the circumference of the circle is considered to be in the circle.
4. randPoint returns a size 2 array containing x-position and y-position of the random point, in that order.

Example 1:

Input: 
["Solution","randPoint","randPoint","randPoint"]
[[1,0,0],[],[],[]]
Output: [null,[-0.72939,-0.65505],[-0.78502,-0.28626],[-0.83119,-0.19803]]

Example 2:

Input: 
["Solution","randPoint","randPoint","randPoint"]
[[10,5,-7.5],[],[],[]]
Output: [null,[11.52438,-8.33273],[2.46992,-16.21705],[11.13430,-12.42337]]

Explanation of Input Syntax:

The input is two lists: the subroutines called and their arguments. Solution's constructor has three arguments, the radius, x-position of the center, and y-position of the center of the circle. randPoint has no arguments. Arguments are always wrapped with a list, even if there aren't any.

Approach #1: Math. [Java]

class Solution {
    double radius, x_center, y_center;
    public Solution(double radius, double x_center, double y_center) {
        this.radius = radius;
        this.x_center = x_center;
        this.y_center = y_center;
    }
    
    public double[] randPoint() {
        double r = Math.sqrt(Math.random()) * radius;
        double deg = Math.random() * 2 * Math.PI;
        double x = x_center + r * Math.cos(deg);
        double y = y_center + r * Math.sin(deg);
        return new double[]{x, y};
    }
}

/**
 * Your Solution object will be instantiated and called as such:
 * Solution obj = new Solution(radius, x_center, y_center);
 * double[] param_1 = obj.randPoint();
 */

　　

Analysis:

The task is to generate uniformly distributed numbers within a circle of radius P in the (x, y) plane. At first polar coordinates seems like a great idea, and the naive solution is to pick a radius r uniformly distributed in [0, R], and then an angle theta uniformly distributed in [0, 2*PI]. But, you end up with an exess of points near the origin (0, 0)! This is wrong because if we look at a certain angle interval, say [theta, theta+dtheda], there needs to be more points generated further out (at large r), than close to zero. The radius must not be picked from a uniform distribution, but one that foes as pdf_r = (2 / R^2) * r

That's easy enough to do by calculating the inverse of the cumulative distribution, and we get for r:

r = R * sqrt(rand())

where rand() is a uniform random number in [0, 1].

Analysis:

https://leetcode.com/problems/generate-random-point-in-a-circle/