Python:实现bailey borwein plouffe算法
def bailey_borwein_plouffe(digit_position: int, precision: int = 1000) -> str:
if (not isinstance(digit_position, int)) or (digit_position <= 0):
raise ValueError("Digit position must be a positive integer")
elif (not isinstance(precision, int)) or (precision < 0):
raise ValueError("Precision must be a nonnegative integer")
# compute an approximation of (16 ** (n - 1)) * pi whose fractional part is mostly
# accurate
sum_result = (
4 * _subsum(digit_position, 1, precision)
- 2 * _subsum(digit_position, 4, precision)
- _subsum(digit_position, 5, precision)
- _subsum(digit_position, 6, precision)
)
# return the first hex digit of the fractional part of the result
return hex(int((sum_result % 1) * 16))[2:]
def _subsum(
digit_pos_to_extract: int, denominator_addend: int, precision: int
) -> float:
# only care about first digit of fractional part; don't need decimal
sum = 0.0
for sum_index in range(digit_pos_to_extract + precision):
denominator = 8 * sum_index + denominator_addend
if sum_index < digit_pos_to_extract:
# if the exponential term is an integer and we mod it by the denominator
# before dividing, only the integer part of the sum will change;
# the fractional part will not
exponential_term = pow(
16, digit_pos_to_extract - 1 - sum_index, denominator
)
else:
exponential_term = pow(16, digit_pos_to_extract - 1 - sum_index)
sum += exponential_term / denominator
return sum
if __name__ == "__main__":
import doctest
doctest.testmod()
