From d8f5b31fab57cc009e87a8d62c8d03075f66e9bd Mon Sep 17 00:00:00 2001
From: fpringle <freddyjepringle@gmail.com>
Date: Fri, 16 Oct 2020 11:42:51 +0200
Subject: [PATCH] Add solution for Project Euler problem 72 (#3122)

* Added solution for Project Euler problem 72.

* Update type annotations and 0-padding of the directory name. Reference: #3256

* Rename sol1.py to sol2.py

* Added newline at the end of sol2.py

* Revert sol1.py
---
 project_euler/problem_072/sol2.py | 45 +++++++++++++++++++++++++++++++
 1 file changed, 45 insertions(+)
 create mode 100644 project_euler/problem_072/sol2.py

diff --git a/project_euler/problem_072/sol2.py b/project_euler/problem_072/sol2.py
new file mode 100644
index 0000000..08e92c1
--- /dev/null
+++ b/project_euler/problem_072/sol2.py
@@ -0,0 +1,45 @@
+"""
+Project Euler Problem 72: https://projecteuler.net/problem=72
+
+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?
+"""
+
+
+def solution(limit: int = 1000000) -> int:
+    """
+    Return the number of reduced proper fractions with denominator less than limit.
+    >>> solution(8)
+    21
+    >>> solution(1000)
+    304191
+    """
+    primes = set(range(3, limit, 2))
+    primes.add(2)
+    for p in range(3, limit, 2):
+        if p not in primes:
+            continue
+        primes.difference_update(set(range(p * p, limit, p)))
+
+    phi = [float(n) for n in range(limit + 1)]
+
+    for p in primes:
+        for n in range(p, limit + 1, p):
+            phi[n] *= 1 - 1 / p
+
+    return int(sum(phi[2:]))
+
+
+if __name__ == "__main__":
+    print(f"{solution() = }")
-- 
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