Solution to problem 35

src/Python/Problem035.py

@@ -0,0 +1,45 @@
+#!/usr/bin/env python3
+"""
+Created on 02 Apr 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for problem 35 of Project Euler
+https://projecteuler.net/problem=35
+"""
+
+from utils import timeit, is_prime
+
+def circular_number(n):
+    n = str(n)
+    result = []
+    for i in range(len(n)):
+        result.append(int(n[i:] + n[:i]))
+    return result
+
+@timeit("Problem 35")
+def compute():
+    """
+    The number, 197, is called a circular prime because all rotations of the
+    digits: 197, 971, and 719, are themselves prime.
+
+    There are thirteen such primes below 100:
+    2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97.
+
+    How many circular primes are there below one million?
+    """
+    circular_primes = []
+    for i in range(2, 1_000_000):
+        for j in circular_number(i):
+            if not is_prime(j):
+                break
+        else:
+            circular_primes.append(i)
+    
+    return len(circular_primes)
+
+
+if __name__ == "__main__":
+
+    print(f"Result for Problem 35: {compute()}")