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102 lines
2.6 KiB
Python
102 lines
2.6 KiB
Python
"""
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Project Euler Problem 70: https://projecteuler.net/problem=70
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Euler's Totient function, φ(n) [sometimes called the phi function], is used to
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determine the number of positive numbers less than or equal to n which are
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relatively prime to n. For example, as 1, 2, 4, 5, 7, and 8, are all less than
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nine and relatively prime to nine, φ(9)=6.
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The number 1 is considered to be relatively prime to every positive number, so
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φ(1)=1.
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Interestingly, φ(87109)=79180, and it can be seen that 87109 is a permutation
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of 79180.
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Find the value of n, 1 < n < 10^7, for which φ(n) is a permutation of n and
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the ratio n/φ(n) produces a minimum.
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-----
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This is essentially brute force. Calculate all totients up to 10^7 and
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find the minimum ratio of n/φ(n) that way. To minimize the ratio, we want
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to minimize n and maximize φ(n) as much as possible, so we can store the
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minimum fraction's numerator and denominator and calculate new fractions
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with each totient to compare against. To avoid dividing by zero, I opt to
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use cross multiplication.
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References:
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Finding totients
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https://en.wikipedia.org/wiki/Euler's_totient_function#Euler's_product_formula
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"""
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from __future__ import annotations
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def get_totients(max_one: int) -> list[int]:
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"""
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Calculates a list of totients from 0 to max_one exclusive, using the
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definition of Euler's product formula.
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>>> get_totients(5)
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[0, 1, 1, 2, 2]
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>>> get_totients(10)
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[0, 1, 1, 2, 2, 4, 2, 6, 4, 6]
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"""
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totients = [0] * max_one
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for i in range(0, max_one):
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totients[i] = i
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for i in range(2, max_one):
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if totients[i] == i:
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for j in range(i, max_one, i):
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totients[j] -= totients[j] // i
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return totients
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def has_same_digits(num1: int, num2: int) -> bool:
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"""
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Return True if num1 and num2 have the same frequency of every digit, False
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otherwise.
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>>> has_same_digits(123456789, 987654321)
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True
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>>> has_same_digits(123, 23)
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False
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>>> has_same_digits(1234566, 123456)
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False
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"""
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return sorted(str(num1)) == sorted(str(num2))
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def solution(max: int = 10000000) -> int:
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"""
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Finds the value of n from 1 to max such that n/φ(n) produces a minimum.
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>>> solution(100)
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21
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>>> solution(10000)
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4435
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"""
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min_numerator = 1 # i
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min_denominator = 0 # φ(i)
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totients = get_totients(max + 1)
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for i in range(2, max + 1):
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t = totients[i]
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if i * min_denominator < min_numerator * t and has_same_digits(i, t):
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min_numerator = i
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min_denominator = t
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return min_numerator
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if __name__ == "__main__":
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print(f"{solution() = }")
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