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