Monte carlo area of a circle

Question

is it possible to use Monte carlo to compute the area of circle with a radius bigger than 1? i tried to make it this way but it only work for a circle of radius 1.

N = 10000
incircle = 0
count = 0
while (count<N)
 x = random()
 y = random()
 if sqrt((x-a)^2 +(y-b)^2) <= R then
    incircle = incircle+1
 endif
 count = count+1

so

I assume that: * a and b are the center of the circle (0.5) * the if with sqrt has the missing parenthesis added * cpt in the while loop is actually count * R is 0.5 in your case (or it is 1 but then it's not a radius, it's a diameter) Then you would see that a circle of different radius would just be a waste of time, since you're multiplying all the terms of the equation by a constant. — Bruno, Dec 06 '21 at 12:27
corrected, yes. So in short, for a circle of radius 2 i have to get a random number between 0 and 2 and not between 0 and 1? — Rasule, Dec 06 '21 at 12:42
Yes: basically you have to "fire" a random point in the square built around the circle. If the circle has a radius of 1, it means that the square has a side of 2. Every time you fire a point, then you check wether it landed inside the circle or not. Finally, by calculating the ratio between the points "fired" and those fell inside the circle, you can deduct the area of the circle and hence Pi. — Bruno, Dec 07 '21 at 13:06

Arty · Accepted Answer · 2021-12-06T13:14:59.977

Don't know what language you're using but I'll write following code in Python, hope you'll understand it.

I made computation for the case of area computation for circle of radius 2.

Here Monte Carlo algorithm is as follows. Lets create a square having X and Y in range [-R ; R]. Now lets randomly uniformly generate points inside this square.We will count total number of generated points and number of points inside a circle of radius R bounded by given square. A point falls into circle if X^2 + Y^2 <= R^2, otherwise its outside of circle. Finally area of square is (2 * R)^2, while circle area is (approximately) equal to num_points_in_circle / num_points_total-th part of area of a square.

In console output I also show error which shows absolute difference with expected value computed precisely through formula Pi * R^2.

In Python x ** 2 means x raised to the power of 2.

Try it online!

import random, math
R = 2
niters = 300_000
total, in_circle = 0, 0
for j in range(20):
    for i in range(niters):
        x = random.uniform(-R, R)
        y = random.uniform(-R, R)
        if x ** 2 + y ** 2 <= R ** 2:
            in_circle += 1
        total += 1
    area = (2 * R) ** 2 * in_circle / total
    expected = math.pi * R ** 2
    print(f'After {(j + 1) * niters:>8} iterations area is {area:.08f}, ' +
        f'error is {abs(area - expected):.08f}', flush = True)

Output:

After   300000 iterations area is 12.59130667, error is 0.02493605
After   600000 iterations area is 12.58341333, error is 0.01704272
After   900000 iterations area is 12.58072889, error is 0.01435827
After  1200000 iterations area is 12.57965333, error is 0.01328272
After  1500000 iterations area is 12.57741867, error is 0.01104805
After  1800000 iterations area is 12.57348444, error is 0.00711383
After  2100000 iterations area is 12.57067429, error is 0.00430367
After  2400000 iterations area is 12.57078000, error is 0.00440939
After  2700000 iterations area is 12.56853926, error is 0.00216864
After  3000000 iterations area is 12.56956800, error is 0.00319739
After  3300000 iterations area is 12.56826182, error is 0.00189120
After  3600000 iterations area is 12.56863111, error is 0.00226050
After  3900000 iterations area is 12.56763897, error is 0.00126836
After  4200000 iterations area is 12.56839238, error is 0.00202177
After  4500000 iterations area is 12.56855111, error is 0.00218050
After  4800000 iterations area is 12.56861667, error is 0.00224605
After  5100000 iterations area is 12.56857725, error is 0.00220664
After  5400000 iterations area is 12.56881185, error is 0.00244124
After  5700000 iterations area is 12.56925193, error is 0.00288132
After  6000000 iterations area is 12.56927733, error is 0.00290672

In this case, the variance is var = math.sqrt((in_circle^2-areap 2)/(total-1))? i tried it, if you have the time, let me know what do you think of it — Rasule, Dec 07 '21 at 18:38
[Try it online!](https://tio.run/##hVHBbsIwDL33K6xeSNsUQidNDI1J0/4A7TZNKCvJCGrTLg0IhPj2zklYy225xLGf33t22rPdNfqh71XdNsaC4Xrb1BRqbnfRGlZQRFpZYToM54xtGGORbSyvKCi9KZUpK4ElRoFFsjGwx7Qj@RakYMkyAjwur8Z84LvV3DkhQdCdHrRCdE3yNYV1MiDO/yKURJo0hQIyRPvgeQVrH41KHjnYznCkoeSHGlPcCI6ipIAUZQJhetc7Cw0ee@QGoW5j0@7HWEIGWJoWuWPCO5kR35HPk@BalBz3jXrTp8d0bEaymQfeUKdWlFZs/wRa5Qx5P77eGqUtkZNXiUuFC9nj/HP0C2HLy5fFFVzErWp0F8ZSHVxcsJyyhbxSmEA2rEFOhDHuvxzmqyO@IR9sJKFnQkFWh26Hrt7NQSR3VuK3Rku1FboUuC5UPvJqCTGF@COmji33g9OYhmd2e37GSRT1/S8 "Python 3 – Try It Online") — Rasule, Dec 07 '21 at 18:38
@Rasule I don't know exact formula for variance. Do you mean that computed error in my code is less than expected error variance? Is it good or bad? — Arty, Dec 07 '21 at 18:56
Oh all right. No your code is good. I was just checking the confidence interval. — Rasule, Dec 07 '21 at 20:33

Monte carlo area of a circle

1 Answers1