虽然三种方法得到的弦都能“覆盖”整个圆，但弦长的分布不同，见上图。
from math import pi import numpy as np from PIL import Image as Img, ImageDraw as ID import matplotlib.pyplot as pltR = 64 # 圆的半径r为1，放大为R W = H = 2 * R + 16 # 图片比圆大一圈以免有些线画不全 N = 1_00_00 # 这么多次# 圆心为原点，左上(-1, -1)，右下(1, 1) # 输入为原始坐标, N * [x0, y0, x1, y1] # 画图时左上(0, 0)，右下(W, H) def line (draw, a) :a = R - a * R + 8 # a = W - (a + 1) * R - 8for i in range(N):draw.line( ((a[i,0],a[i,1]), (a[i,2],a[i,3])), fill=255 )def do_it (i) :img = Img.new('L', (W, H)) # 单色'1'比灰度'L'慢draw = ID.Draw(img)# AI说连续多次rand是均匀分布y = np.random.uniform(low=-1.0, high=1.0, size=N)x = np.sqrt(1 - y ** 2)if i == 0: a = np.column_stack((-x, y, x, y))else: # signs别用两次！x *= np.random.choice([-1, 1], size=N)if i == 1:# (x,y)到(0,-1)a = np.column_stack((np.full_like(x, 0), np.full_like(y, -1), x, y))else:y2 = np.random.uniform(low=-1.0, high=1.0, size=N)x2 = np.sqrt(1 - y2 ** 2) * np.random.choice([-1, 1], size=N)a = np.column_stack((x2, y2, x, y))line(draw, a); img.save(f'/t/{i}.png')# hypotenuse: 斜边a = np.hypot(a[:,2] - a[:,0], a[:,3] - a[:,1])prob = np.sum(a > 3 ** 0.5) / N # 概率# 弦覆盖圆吗？print(prob, (img.histogram()[255] / R ** 2) / pi)return a, str(prob)for i in range(3):a,p = do_it(i); plt.hist(a, bins=50, alpha=0.5, label=p)plt.xlabel('Chord Length', fontsize=12) plt.ylabel('Frequency', fontsize=12) plt.xticks(rotation=45, fontsize=9) plt.grid(axis='y', alpha=0.5) # 显示网格 plt.tight_layout() plt.legend() plt.show()
参考资料，中文搜“贝特朗悖论”即可，英文可Bing(PDF Bertrand paradox (probability) - Universitetet i Oslo).
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