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def calculate_pi(limit: int) -> str:

"""

https://en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80

Leibniz Formula for Pi

The Leibniz formula is the special case arctan(1) = pi / 4.

Leibniz's formula converges extremely slowly: it exhibits sublinear convergence.

Convergence (https://en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80#Convergence)

We cannot try to prove against an interrupted, uncompleted generation.

https://en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80#Unusual_behaviour

The errors can in fact be predicted, but those calculations also approach infinity

for accuracy.

Our output will be a string so that we can definitely store all digits.

>>> import math

>>> float(calculate_pi(15)) == math.pi

True

Since we cannot predict errors or interrupt any infinite alternating series

generation since they approach infinity, or interrupt any alternating series, we'll

need math.isclose()

>>> math.isclose(float(calculate_pi(50)), math.pi)

True

>>> math.isclose(float(calculate_pi(100)), math.pi)

True

Since math.pi contains only 16 digits, here are some tests with known values:

>>> calculate_pi(50)

'3.14159265358979323846264338327950288419716939937510'

>>> calculate_pi(80)

'3.14159265358979323846264338327950288419716939937510582097494459230781640628620899'

"""

# Variables used for the iteration process

q = 1

r = 0

t = 1

k = 1

n = 3

l = 3

decimal = limit

counter = 0

result = ""

# We can't compare against anything if we make a generator,

# so we'll stick with plain return logic

while counter != decimal + 1:

if 4 * q + r - t < n * t:

result += str(n)

if counter == 0:

result += "."

if decimal == counter:

break

counter += 1

nr = 10 * (r - n * t)

n = ((10 * (3 * q + r)) // t) - 10 * n

q *= 10

r = nr

else:

nr = (2 * q + r) * l

nn = (q * (7 * k) + 2 + (r * l)) // (t * l)

q *= k

t *= l

l += 2

k += 1

n = nn

r = nr

return result

def main() -> None:

print(f"{calculate_pi(50) = }")

import doctest

doctest.testmod()

if __name__ == "__main__":

main()

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