mirror of
https://github.com/TheAlgorithms/Python.git
synced 2024-11-24 05:21:09 +00:00
d5a9f649b8
Ignore `A003` 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>
100 lines
2.5 KiB
Python
100 lines
2.5 KiB
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() = }")
|