Monte Carlo py solves the area problem! (save pupils Series)

Recently, I saw such a topic in my circle of friends , Once again refresh the difficulty of primary school students' question bank .

Excellent young bloggers are in front of excellent young bloggers , The curve problem of this ghost animal can naturally be easily solved by integral , But the title has already limited the level of grade 6 , How can you kill a chicken with an ox knife , Even smart as a blogger ,6 Don't talk about points when you're old ,cos Not at all ,（ A female pupil said she could use tan,cos Angle solution ）, Therefore, this is obviously too reluctant for the majority of primary school students to understand .

that , From the perspective of primary school students, is there any way to deal with this malicious problem ？

The answer is yes

As for the method , For those who are good at smashing cards Pupils throwing darts , Monte Carlo is obviously perfect for , About introducing and connecting a blog , Please read by yourself .

https://blog.csdn.net/binlin199012/article/details/80604080

First, we draw the outline of the figure , Just pick the part we need .

Then start throwing needles wildly

# Take the radius as 1 The circle of , Take the area calculation in the first quadrant as an example 
import random
import math
import matplotlib.pyplot as plt
import numpy as  np
from matplotlib.patches import Ellipse, Circle
C=5000# cycles 
s=0# Counter 
# The field of view of the drawing 
plt.xlim(xmax=4,xmin=0)
plt.ylim(ymax=4,ymin=0)

x1=np.arange(0, 4, 0.1)
y1=0.5*x1
plt.plot(x1, y1)

x2 = y2 = np.arange(0, 4, 0.1)
x2, y2 = np.meshgrid(x2,y2)
plt.contour(x2, y2, (x2-4)**2 + (y2-4)**2, [16])     #x**2 + y**2 = 9  Round 
plt.show()
for i in range(C):
	# Generate random coordinates (x,y)
    x=4*random.random()# Generate 0-1 Random number between 
    #print(x)
    y=4*random.random()# Generate 0-1 Random number between 
    #print(y)
    y_predict=-math.sqrt(16-(x-4)**2) + 4
    y_predict2=0.5*x
    #print(y_predict)
    if y_predict>=y and y_predict2>=y and x<=4:# Coordinates y Less than y_predict, It's a round point , That is to meet the conditions , Counter plus 1
        s=s+1
        plt.plot(x,y,'r+')
    else:
        plt.plot(x, y, 'b+')
plt.show()  
predict_value=16*s/C
print(predict_value)

The injection result is shown in the figure

This is a vote 5000 Time , The calculation result is 1.23 about , Just an approximate solution .

But pupils can say to their teachers “ I can cast it indefinitely to approach the real solution ！”

Of course, teachers don't like this “ Because what the computer produces is only pseudo-random , Can't get a real natural pointer ”

But smart pupils have long thought of a solution

“ The air resistance in Brownian motion is purely random , So I just stand high enough , You can get the real frequency by hand casting the pipeline ！ ”