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88 lines
2.3 KiB
Python
88 lines
2.3 KiB
Python
def calculate_pi(limit: int) -> str:
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"""
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https://en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80
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Leibniz Formula for Pi
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The Leibniz formula is the special case arctan(1) = pi / 4.
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Leibniz's formula converges extremely slowly: it exhibits sublinear convergence.
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Convergence (https://en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80#Convergence)
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We cannot try to prove against an interrupted, uncompleted generation.
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https://en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80#Unusual_behaviour
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The errors can in fact be predicted, but those calculations also approach infinity
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for accuracy.
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Our output will be a string so that we can definitely store all digits.
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>>> import math
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>>> float(calculate_pi(15)) == math.pi
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True
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Since we cannot predict errors or interrupt any infinite alternating series
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generation since they approach infinity, or interrupt any alternating series, we'll
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need math.isclose()
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>>> math.isclose(float(calculate_pi(50)), math.pi)
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True
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>>> math.isclose(float(calculate_pi(100)), math.pi)
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True
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Since math.pi contains only 16 digits, here are some tests with known values:
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>>> calculate_pi(50)
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'3.14159265358979323846264338327950288419716939937510'
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>>> calculate_pi(80)
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'3.14159265358979323846264338327950288419716939937510582097494459230781640628620899'
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"""
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# Variables used for the iteration process
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q = 1
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r = 0
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t = 1
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k = 1
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n = 3
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m = 3
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decimal = limit
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counter = 0
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result = ""
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# We can't compare against anything if we make a generator,
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# so we'll stick with plain return logic
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while counter != decimal + 1:
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if 4 * q + r - t < n * t:
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result += str(n)
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if counter == 0:
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result += "."
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if decimal == counter:
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break
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counter += 1
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nr = 10 * (r - n * t)
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n = ((10 * (3 * q + r)) // t) - 10 * n
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q *= 10
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r = nr
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else:
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nr = (2 * q + r) * m
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nn = (q * (7 * k) + 2 + (r * m)) // (t * m)
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q *= k
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t *= m
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m += 2
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k += 1
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n = nn
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r = nr
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return result
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def main() -> None:
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print(f"{calculate_pi(50) = }")
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import doctest
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doctest.testmod()
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if __name__ == "__main__":
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main()
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