From 8f8d39d19125b296ed5210497e8f0a73c35e24e0 Mon Sep 17 00:00:00 2001
From: Utkarsh Chaudhary <ucponia@gmail.com>
Date: Mon, 12 Oct 2020 19:46:15 +0530
Subject: [PATCH] Add a solution to Project Euler 72 (#2940)

* Added Problem 72

* Removed args from solution()

* Incorporated the suggested changes
---
 project_euler/problem_72/__init__.py |  0
 project_euler/problem_72/sol1.py     | 46 ++++++++++++++++++++++++++++
 2 files changed, 46 insertions(+)
 create mode 100644 project_euler/problem_72/__init__.py
 create mode 100644 project_euler/problem_72/sol1.py

diff --git a/project_euler/problem_72/__init__.py b/project_euler/problem_72/__init__.py
new file mode 100644
index 000000000..e69de29bb
diff --git a/project_euler/problem_72/sol1.py b/project_euler/problem_72/sol1.py
new file mode 100644
index 000000000..846396ab0
--- /dev/null
+++ b/project_euler/problem_72/sol1.py
@@ -0,0 +1,46 @@
+"""
+Problem 72 Counting fractions: https://projecteuler.net/problem=72
+
+Description:
+
+Consider the fraction, n/d, where n and d are positive integers. If n<d and HCF(n,d)=1,
+it is called a reduced proper fraction.
+If we list the set of reduced proper fractions for d ≤ 8 in ascending order of size, we
+get: 1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7,
+3/4, 4/5, 5/6, 6/7, 7/8
+It can be seen that there are 21 elements in this set.
+How many elements would be contained in the set of reduced proper fractions for
+d ≤ 1,000,000?
+
+Solution:
+
+Number of numbers between 1 and n that are coprime to n is given by the Euler's Totient
+function, phi(n). So, the answer is simply the sum of phi(n) for 2 <= n <= 1,000,000
+Sum of phi(d), for all d|n = n. This result can be used to find phi(n) using a sieve.
+
+Time: 3.5 sec
+"""
+
+
+def solution(limit: int = 1_000_000) -> int:
+    """
+    Returns an integer, the solution to the problem
+    >>> solution(10)
+    31
+    >>> solution(100)
+    3043
+    >>> solution(1_000)
+    304191
+    """
+
+    phi = [i - 1 for i in range(limit + 1)]
+
+    for i in range(2, limit + 1):
+        for j in range(2 * i, limit + 1, i):
+            phi[j] -= phi[i]
+
+    return sum(phi[2 : limit + 1])
+
+
+if __name__ == "__main__":
+    print(solution())