From d02f6bbfbd8ff1c99e1e1e1a85572477412c6927 Mon Sep 17 00:00:00 2001 From: sarthaka1310 <56290744+sarthaka1310@users.noreply.github.com> Date: Sun, 11 Oct 2020 23:38:30 +0530 Subject: [PATCH] Added solution to Project Euler 69 (#2934) * Added solution to Project Euler 69 * Accept edits from code review Co-authored-by: Dhruv * Added doctests * Renaming and exception handling * Apply suggestions from code review Co-authored-by: Dhruv * Edited mistake. Co-authored-by: formal-acc Co-authored-by: Dhruv --- project_euler/problem_69/__init__.py | 0 project_euler/problem_69/sol1.py | 66 ++++++++++++++++++++++++++++ 2 files changed, 66 insertions(+) create mode 100644 project_euler/problem_69/__init__.py create mode 100644 project_euler/problem_69/sol1.py diff --git a/project_euler/problem_69/__init__.py b/project_euler/problem_69/__init__.py new file mode 100644 index 000000000..e69de29bb diff --git a/project_euler/problem_69/sol1.py b/project_euler/problem_69/sol1.py new file mode 100644 index 000000000..d148dd79a --- /dev/null +++ b/project_euler/problem_69/sol1.py @@ -0,0 +1,66 @@ +""" +Totient maximum +Problem 69: https://projecteuler.net/problem=69 + +Euler's Totient function, φ(n) [sometimes called the phi function], +is used to determine the number of numbers less than n which are relatively prime to n. +For example, as 1, 2, 4, 5, 7, and 8, +are all less than nine and relatively prime to nine, φ(9)=6. + +n Relatively Prime φ(n) n/φ(n) +2 1 1 2 +3 1,2 2 1.5 +4 1,3 2 2 +5 1,2,3,4 4 1.25 +6 1,5 2 3 +7 1,2,3,4,5,6 6 1.1666... +8 1,3,5,7 4 2 +9 1,2,4,5,7,8 6 1.5 +10 1,3,7,9 4 2.5 + +It can be seen that n=6 produces a maximum n/φ(n) for n ≤ 10. + +Find the value of n ≤ 1,000,000 for which n/φ(n) is a maximum. +""" + + +def solution(n: int = 10 ** 6) -> int: + """ + Returns solution to problem. + Algorithm: + 1. Precompute φ(k) for all natural k, k <= n using product formula (wikilink below) + https://en.wikipedia.org/wiki/Euler%27s_totient_function#Euler's_product_formula + + 2. Find k/φ(k) for all k ≤ n and return the k that attains maximum + + >>> solution(10) + 6 + + >>> solution(100) + 30 + + >>> solution(9973) + 2310 + + """ + + if n <= 0: + raise ValueError("Please enter an integer greater than 0") + + phi = list(range(n + 1)) + for number in range(2, n + 1): + if phi[number] == number: + phi[number] -= 1 + for multiple in range(number * 2, n + 1, number): + phi[multiple] = (phi[multiple] // number) * (number - 1) + + answer = 1 + for number in range(1, n + 1): + if (answer / phi[answer]) < (number / phi[number]): + answer = number + + return answer + + +if __name__ == "__main__": + print(solution())