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49 lines
1.5 KiB
Python
49 lines
1.5 KiB
Python
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"""
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Project Euler Problem 57: https://projecteuler.net/problem=57
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It is possible to show that the square root of two can be expressed as an infinite
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continued fraction.
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sqrt(2) = 1 + 1 / (2 + 1 / (2 + 1 / (2 + ...)))
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By expanding this for the first four iterations, we get:
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1 + 1 / 2 = 3 / 2 = 1.5
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1 + 1 / (2 + 1 / 2} = 7 / 5 = 1.4
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1 + 1 / (2 + 1 / (2 + 1 / 2)) = 17 / 12 = 1.41666...
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1 + 1 / (2 + 1 / (2 + 1 / (2 + 1 / 2))) = 41/ 29 = 1.41379...
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The next three expansions are 99/70, 239/169, and 577/408, but the eighth expansion,
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1393/985, is the first example where the number of digits in the numerator exceeds
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the number of digits in the denominator.
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In the first one-thousand expansions, how many fractions contain a numerator with
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more digits than the denominator?
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"""
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def solution(n: int = 1000) -> int:
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"""
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returns number of fractions containing a numerator with more digits than
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the denominator in the first n expansions.
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>>> solution(14)
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2
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>>> solution(100)
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15
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>>> solution(10000)
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1508
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"""
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prev_numerator, prev_denominator = 1, 1
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result = []
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for i in range(1, n + 1):
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numerator = prev_numerator + 2 * prev_denominator
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denominator = prev_numerator + prev_denominator
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if len(str(numerator)) > len(str(denominator)):
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result.append(i)
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prev_numerator = numerator
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prev_denominator = denominator
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return len(result)
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if __name__ == "__main__":
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print(f"{solution() = }")
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