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* Added solution to Project Euler 69 * Accept edits from code review Co-authored-by: Dhruv <dhruvmanila@gmail.com> * Added doctests * Renaming and exception handling * Apply suggestions from code review Co-authored-by: Dhruv <dhruvmanila@gmail.com> * Edited mistake. Co-authored-by: formal-acc <sarthak.agrawal@research.iiit.ac.in> Co-authored-by: Dhruv <dhruvmanila@gmail.com>
67 lines
1.7 KiB
Python
67 lines
1.7 KiB
Python
"""
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Totient maximum
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Problem 69: https://projecteuler.net/problem=69
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Euler's Totient function, φ(n) [sometimes called the phi function],
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is used to determine the number of numbers less than n which are relatively prime to n.
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For example, as 1, 2, 4, 5, 7, and 8,
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are all less than nine and relatively prime to nine, φ(9)=6.
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n Relatively Prime φ(n) n/φ(n)
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2 1 1 2
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3 1,2 2 1.5
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4 1,3 2 2
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5 1,2,3,4 4 1.25
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6 1,5 2 3
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7 1,2,3,4,5,6 6 1.1666...
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8 1,3,5,7 4 2
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9 1,2,4,5,7,8 6 1.5
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10 1,3,7,9 4 2.5
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It can be seen that n=6 produces a maximum n/φ(n) for n ≤ 10.
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Find the value of n ≤ 1,000,000 for which n/φ(n) is a maximum.
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"""
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def solution(n: int = 10 ** 6) -> int:
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"""
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Returns solution to problem.
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Algorithm:
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1. Precompute φ(k) for all natural k, k <= n using product formula (wikilink below)
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https://en.wikipedia.org/wiki/Euler%27s_totient_function#Euler's_product_formula
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2. Find k/φ(k) for all k ≤ n and return the k that attains maximum
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>>> solution(10)
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6
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>>> solution(100)
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30
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>>> solution(9973)
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2310
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"""
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if n <= 0:
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raise ValueError("Please enter an integer greater than 0")
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phi = list(range(n + 1))
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for number in range(2, n + 1):
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if phi[number] == number:
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phi[number] -= 1
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for multiple in range(number * 2, n + 1, number):
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phi[multiple] = (phi[multiple] // number) * (number - 1)
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answer = 1
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for number in range(1, n + 1):
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if (answer / phi[answer]) < (number / phi[number]):
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answer = number
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return answer
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if __name__ == "__main__":
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print(solution())
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