J

```
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 = 1/4 Pi .
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 always be a string since we can defintely store all digits in there.
For simplicity' sake, let's just compare against known values and since our outpit
is a string, we need to convert to float.
>>> 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 are going to need math.isclose()
>>> math.isclose(float(calculate_pi(50)), math.pi)
True
>>> math.isclose(float(calculate_pi(100)), math.pi)
True
Since math.pi-constant contains only 16 digits, here some test with preknown values:
>>> calculate_pi(50)
'3.14159265358979323846264338327950288419716939937510'
>>> calculate_pi(80)
'3.14159265358979323846264338327950288419716939937510582097494459230781640628620899'
To apply the Leibniz formula for calculating pi,
the variables q, r, t, k, n, and l are used for the iteration process.
"""
q = 1
r = 0
t = 1
k = 1
n = 3
l = 3
decimal = limit
counter = 0
result = ""
"""
We will avoid using yield since we otherwise get a Generator-Object,
which we can't just compare against anything. We would have to make a list out of it
after the generation, so we will just stick to 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()
```