Python/project_euler/problem_065/sol1.py
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Co-authored-by: Christian Clauss <cclauss@me.com>
Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com>
Co-authored-by: Dhruv Manilawala <dhruvmanila@gmail.com>
2022-10-13 19:53:59 +05:30

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Python

"""
Project Euler Problem 65: https://projecteuler.net/problem=65
The square root of 2 can be written as an infinite continued fraction.
sqrt(2) = 1 + 1 / (2 + 1 / (2 + 1 / (2 + 1 / (2 + ...))))
The infinite continued fraction can be written, sqrt(2) = [1;(2)], (2)
indicates that 2 repeats ad infinitum. In a similar way, sqrt(23) =
[4;(1,3,1,8)].
It turns out that the sequence of partial values of continued
fractions for square roots provide the best rational approximations.
Let us consider the convergents for sqrt(2).
1 + 1 / 2 = 3/2
1 + 1 / (2 + 1 / 2) = 7/5
1 + 1 / (2 + 1 / (2 + 1 / 2)) = 17/12
1 + 1 / (2 + 1 / (2 + 1 / (2 + 1 / 2))) = 41/29
Hence the sequence of the first ten convergents for sqrt(2) are:
1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, ...
What is most surprising is that the important mathematical constant,
e = [2;1,2,1,1,4,1,1,6,1,...,1,2k,1,...].
The first ten terms in the sequence of convergents for e are:
2, 3, 8/3, 11/4, 19/7, 87/32, 106/39, 193/71, 1264/465, 1457/536, ...
The sum of digits in the numerator of the 10th convergent is
1 + 4 + 5 + 7 = 17.
Find the sum of the digits in the numerator of the 100th convergent
of the continued fraction for e.
-----
The solution mostly comes down to finding an equation that will generate
the numerator of the continued fraction. For the i-th numerator, the
pattern is:
n_i = m_i * n_(i-1) + n_(i-2)
for m_i = the i-th index of the continued fraction representation of e,
n_0 = 1, and n_1 = 2 as the first 2 numbers of the representation.
For example:
n_9 = 6 * 193 + 106 = 1264
1 + 2 + 6 + 4 = 13
n_10 = 1 * 193 + 1264 = 1457
1 + 4 + 5 + 7 = 17
"""
def sum_digits(num: int) -> int:
"""
Returns the sum of every digit in num.
>>> sum_digits(1)
1
>>> sum_digits(12345)
15
>>> sum_digits(999001)
28
"""
digit_sum = 0
while num > 0:
digit_sum += num % 10
num //= 10
return digit_sum
def solution(max_n: int = 100) -> int:
"""
Returns the sum of the digits in the numerator of the max-th convergent of
the continued fraction for e.
>>> solution(9)
13
>>> solution(10)
17
>>> solution(50)
91
"""
pre_numerator = 1
cur_numerator = 2
for i in range(2, max_n + 1):
temp = pre_numerator
e_cont = 2 * i // 3 if i % 3 == 0 else 1
pre_numerator = cur_numerator
cur_numerator = e_cont * pre_numerator + temp
return sum_digits(cur_numerator)
if __name__ == "__main__":
print(f"{solution() = }")