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* ci(pre-commit): Add pep8-naming to `pre-commit` hooks (#7038) * refactor: Fix naming conventions (#7038) * Update arithmetic_analysis/lu_decomposition.py Co-authored-by: Christian Clauss <cclauss@me.com> * [pre-commit.ci] auto fixes from pre-commit.com hooks for more information, see https://pre-commit.ci * refactor(lu_decomposition): Replace `NDArray` with `ArrayLike` (#7038) * chore: Fix naming conventions in doctests (#7038) * fix: Temporarily disable project euler problem 104 (#7069) * chore: Fix naming conventions in doctests (#7038) Co-authored-by: Christian Clauss <cclauss@me.com> Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com>
78 lines
2.0 KiB
Python
78 lines
2.0 KiB
Python
"""
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Project Euler Problem 64: https://projecteuler.net/problem=64
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All square roots are periodic when written as continued fractions.
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For example, let us consider sqrt(23).
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It can be seen that the sequence is repeating.
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For conciseness, we use the notation sqrt(23)=[4;(1,3,1,8)],
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to indicate that the block (1,3,1,8) repeats indefinitely.
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Exactly four continued fractions, for N<=13, have an odd period.
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How many continued fractions for N<=10000 have an odd period?
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References:
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- https://en.wikipedia.org/wiki/Continued_fraction
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"""
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from math import floor, sqrt
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def continuous_fraction_period(n: int) -> int:
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"""
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Returns the continued fraction period of a number n.
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>>> continuous_fraction_period(2)
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1
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>>> continuous_fraction_period(5)
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1
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>>> continuous_fraction_period(7)
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4
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>>> continuous_fraction_period(11)
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2
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>>> continuous_fraction_period(13)
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5
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"""
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numerator = 0.0
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denominator = 1.0
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ROOT = int(sqrt(n)) # noqa: N806
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integer_part = ROOT
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period = 0
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while integer_part != 2 * ROOT:
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numerator = denominator * integer_part - numerator
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denominator = (n - numerator**2) / denominator
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integer_part = int((ROOT + numerator) / denominator)
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period += 1
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return period
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def solution(n: int = 10000) -> int:
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"""
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Returns the count of numbers <= 10000 with odd periods.
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This function calls continuous_fraction_period for numbers which are
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not perfect squares.
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This is checked in if sr - floor(sr) != 0 statement.
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If an odd period is returned by continuous_fraction_period,
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count_odd_periods is increased by 1.
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>>> solution(2)
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1
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>>> solution(5)
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2
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>>> solution(7)
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2
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>>> solution(11)
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3
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>>> solution(13)
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4
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"""
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count_odd_periods = 0
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for i in range(2, n + 1):
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sr = sqrt(i)
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if sr - floor(sr) != 0:
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if continuous_fraction_period(i) % 2 == 1:
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count_odd_periods += 1
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return count_odd_periods
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if __name__ == "__main__":
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print(f"{solution(int(input().strip()))}")
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