【算法】均勻的生成圓內的隨機點

 

算法 1

設半徑為$R$。

$x = r \ast cos(\theta)$

$y = r \ast sin(\theta)$

其中 $0\leqslant r \leqslant R$，$t$為0-1均勻分布產生的隨機數，$r = sqrt(t) \ast R$，$\theta = 2\pi \ast t, t \sim U(0, 1)$ 證明：url

import numpy as np
import random
import math
import matplotlib.pyplot as plt
import matplotlib as mpl
mpl.rcParams['agg.path.chunksize'] = 10000 # the default is 0
N = int(10000)
x1 = 1.5
x2 = 0.5
radius = 2
a = 2 * math.pi * np.array([random.random() for _ in range(N)])
r = np.array([random.random() for _ in range(N)])
x = radius*np.sqrt(r)*np.cos(a) + x1;
y = radius*np.sqrt(r)*np.sin(a) + x2;
plt.scatter(x, y, s=1)
plt.show()

 

下面的算法是錯誤的

原因在於對$R$求開方，導致$r = sqrt(R)*t$平方后不在滿足均勻分布。

import numpy as np
import random
import math
import matplotlib.pyplot as plt
import matplotlib as mpl
mpl.rcParams['agg.path.chunksize'] = 10000 # the default is 0
N = int(10000)
x1 = 1.5
x2 = 0.5
radius = 2
a = 2 * math.pi * np.array([random.random() for _ in range(N)])
r = np.array([random.random() for _ in range(N)])
#x = radius*np.sqrt(r)*np.cos(a) + x1;
#y = radius*np.sqrt(r)*np.sin(a) + x2;
x = np.sqrt(radius)*r*np.cos(a) + x1;
y = np.sqrt(radius)*r*np.sin(a) + x2;
plt.scatter(x, y, s=1)
plt.show()

 參考鏈接：JustDoIT