One of my favorite calculations of pi is to pick random coordinates in a unit square and count how many of them are in a circle. it's so stupid and so clever at the same time.
This was recreated from memory. I think it is close but I may have a bounding bug.
import random
def pi(count):
inside = 0
for i in range(count):
test_x = random.random()
test_y = random.random()
if test_x ** 2 + test_y ** 2 < 1:
inside += 1
return inside / count * 4 #above is a quarter circle
print(pi(2 ** 30) )
With the metatopic of this thread being obscure languages, I had some fun squeezing this into some list comprehensions (maybe someone's got an idea of how to keep track of the state within the list):
```
$ cat << EOF > pi.py
state = [0, 0, 2*8, 2*12]; _ = [print(f'\rRun {state.__setitem__(0, state[0] + 1) or state[0]}/{state[3]} | Last \u03c0: {current_pi:.6f} | *Average \u03c0: {(state.__setitem__(1, state[1] + current_pi) or state[1]) / state[0]:.6f}*', end='', flush=True) for current_pi in [(4 * sum([1 for _ in range(state[2]) if __import__("random").random()*2 + __import__("random").random()*2 < 1]) / state[2]) for _ in range(state[3])]]; print()
EOF
$ time python3 pi.py
Run 4096/4096 | Last π: 3.140625 | *Average π: 3.143051*
python3 pi.py 0.41s user 0.01s system 99% cpu 0.429 total
```
Play around with the `2*8` and `2*10` values in the state, they control the amount of rounds and the range in which the random values get generated respectively.
https://esolangs.org/wiki/StupidStackLanguage