Clean python code

Python/p001.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
 #
@@ -6,6 +6,7 @@
 
 from timeit import default_timer
 
+
 def main():
     start = default_timer()
 
@@ -20,9 +21,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 1')
-    print('Answer: {}'.format(sum_))
+    print(f'Answer: {sum}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p002.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
 #
@@ -8,6 +8,7 @@
 
 from timeit import default_timer
 
+
 def main():
 
     start = default_timer()
@@ -32,9 +33,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 2')
-    print('Answer: {}'.format(sum_))
+    print(f'Answer: {sum}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p003.py

@@ -1,14 +1,14 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # The prime factors of 13195 are 5, 7, 13 and 29.
 #
 # What is the largest prime factor of the number 600851475143?
 
-from math import floor, sqrt
-
 from timeit import default_timer
+
 from projecteuler import is_prime
 
+
 # Recursive approach: if num is prime, return num, otherwise
 # recursively look for the largest prime factor of num divided
 # by its prime factors until only the largest remains.
@@ -21,20 +21,21 @@ def max_prime_factor(num):
     if num % 2 == 0:
         return max_prime_factor(num // 2)
 
-    else:
-        i = 3
+    i = 3
 
-#       If num is divisible by i and i is prime, find largest
-#       prime factor of num/i.
-        while True:
-            if num % i == 0:
-                if is_prime(i):
-                    return max_prime_factor(num//i)
-            i = i + 2
+#   If num is divisible by i and i is prime, find largest
+#   prime factor of num/i.
+    while True:
+        if num % i == 0:
+            if is_prime(i):
+                return max_prime_factor(num//i)
+
+        i = i + 2
 
 #   Should never get here
     return -1
 
+
 def main():
     start = default_timer()
 
@@ -43,9 +44,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 3')
-    print('Answer: {}'.format(res))
+    print(f'Answer: {res}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p004.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.
 #
@@ -7,6 +7,7 @@
 from timeit import default_timer
 from projecteuler import is_palindrome
 
+
 def main():
     start = default_timer()
 
@@ -25,9 +26,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 4')
-    print('Answer: {}'.format(max_))
+    print(f'Answer: {max_}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p005.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # 2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
 #
@@ -7,6 +7,7 @@
 from timeit import default_timer
 from projecteuler import lcmm
 
+
 def main():
     start = default_timer()
 
@@ -19,9 +20,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 5')
-    print('Answer: {}'.format(res))
+    print(f'Answer: {res}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p006.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # The sum of the squares of the first ten natural numbers is,
 #
@@ -14,6 +14,7 @@
 
 from timeit import default_timer
 
+
 def main():
     start = default_timer()
 
@@ -30,9 +31,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 6')
-    print('Answer: {}'.format(square_sum - sum_squares))
+    print(f'Answer: {square_sum - sum_squares}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p007.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.
 #
@@ -7,6 +7,7 @@
 from timeit import default_timer
 from projecteuler import is_prime
 
+
 def main():
     start = default_timer()
 
@@ -25,9 +26,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 7')
-    print('Answer: {}'.format(n))
+    print(f'Answer: {n}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p008.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832.
 #
@@ -27,6 +27,7 @@
 
 from timeit import default_timer
 
+
 def main():
     start = default_timer()
 
@@ -69,9 +70,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 8')
-    print('Answer: {}'.format(max_))
+    print(f'Answer: {max_}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds'.format(end - start))
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p009.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
 #
@@ -11,9 +11,9 @@
 # Find the product abc.
 
 from math import gcd
-
 from timeit import default_timer
 
+
 def main():
     start = default_timer()
 
@@ -61,9 +61,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 9')
-    print('Answer: {}'.format(a * b * c))
+    print(f'Answer: {a * b * c}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p010.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
 #
@@ -7,12 +7,13 @@
 from timeit import default_timer
 from projecteuler import sieve
 
+
 def main():
     start = default_timer()
 
     N = 2000000
 
-#   Use the function in projecteuler.py implementing the 
+#   Use the function in projecteuler.py implementing the
 #   Sieve of Eratosthenes algorithm to generate primes.
     primes = sieve(N)
     sum_ = 0
@@ -25,9 +26,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 10')
-    print('Answer: {}'.format(sum_))
+    print(f'Answer: {sum_}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p011.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # In the 20×20 grid below, four numbers along a diagonal line have been marked in red.
 #
@@ -29,6 +29,7 @@
 
 from timeit import default_timer
 
+
 def main():
     start = default_timer()
 
@@ -51,7 +52,7 @@ def main():
             [4, 42, 16, 73, 38, 25, 39, 11, 24, 94, 72, 18, 8, 46, 29, 32, 40, 62, 76, 36],
             [20, 69, 36, 41, 72, 30, 23, 88, 34, 62, 99, 69, 82, 67, 59, 85, 74, 4, 36, 16],
             [20, 73, 35, 29, 78, 31, 90, 1, 74, 31, 49, 71, 48, 86, 81, 16, 23, 57, 5, 54],
-            [1, 70, 54, 71, 83, 51, 54, 69, 16, 92, 33, 48, 61, 43, 52, 1, 89, 19, 67, 48]];
+            [1, 70, 54, 71, 83, 51, 54, 69, 16, 92, 33, 48, 61, 43, 52, 1, 89, 19, 67, 48]]
 
     max_ = 0
 
@@ -112,9 +113,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 11')
-    print('Answer: {}'.format(max_))
+    print(f'Answer: {max_}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p012.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28.
 # The first ten terms would be:
@@ -22,6 +22,7 @@
 from timeit import default_timer
 from projecteuler import count_divisors
 
+
 def main():
     start = default_timer()
 
@@ -41,9 +42,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 12')
-    print('Answer: {}'.format(triang))
+    print(f'Answer: {triang}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p013.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # Work out the first ten digits of the sum of the following one-hundred 50-digit numbers.
 #
@@ -103,9 +103,9 @@
 # 20849603980134001723930671666823555245252804609722
 # 53503534226472524250874054075591789781264330331690
 
+from timeit import default_timer
 import numpy as np
 
-from timeit import default_timer
 
 def main():
     start = default_timer()
@@ -219,9 +219,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 13')
-    print('Answer: {}'.format(sum_[:10]))
+    print(f'Answer: {sum_[:10]}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p014.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3 
+#!/usr/bin/env python3 
 
 # The following iterative sequence is defined for the set of positive integers:
 #
@@ -16,13 +16,14 @@
 #
 # NOTE: Once the chain starts the terms are allowed to go above one million.
 
+from timeit import default_timer
 from numpy import zeros
 
-from timeit import default_timer
 
 N = 1000000
 collatz_found = zeros(N, dtype=int)
 
+
 # Recursive function to calculate the Collatz sequence for n.
 # If n is even, Collatz(n)=1+Collatz(n/2), if n is odd
 # Collatz(n)=1+Collatz(3*n+1).
@@ -38,8 +39,8 @@ def collatz_length(n):
 
     if n % 2 == 0:
         return 1 + collatz_length(n//2)
-    else:
-        return 1 + collatz_length(3*n+1)
+
+    return 1 + collatz_length(3*n+1)
 
 def main():
     start = default_timer()
@@ -60,9 +61,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 14')
-    print('Answer: {}'.format(max_))
+    print(f'Answer: {max_}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p015.py

@@ -1,12 +1,12 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # Starting in the top left corner of a 2×2 grid, and only being able to move to the right and down, there are exactly 6 routes to the bottom right corner
 # How many such routes are there through a 20×20 grid?
 
 from math import factorial
-
 from timeit import default_timer
 
+
 def main():
     start = default_timer()
 
@@ -22,9 +22,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 15')
-    print('Answer: {}'.format(count))
+    print(f'Answer: {count}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p016.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # 2^15 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.
 #
@@ -6,6 +6,7 @@
 
 from timeit import default_timer
 
+
 def main():
     start = default_timer()
 
@@ -21,9 +22,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 16')
-    print('Answer: {}'.format(sum_))
+    print(f'Answer: {sum_}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p017.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.
 #
@@ -9,6 +9,7 @@
 
 from timeit import default_timer
 
+
 def main():
     start = default_timer()
 
@@ -16,10 +17,10 @@ def main():
 #   the second letters for "twenty", "thirty", ..., "ninety",
 #   the third letters for "one hundred and", "two hundred and", ..., "nine hundre and",
 #   the last one-element one the number of letters of 1000
-    n_letters = [[3, 3, 5, 4, 4, 3, 5, 5, 4, 3, 6, 6, 8, 8, 7, 7, 9, 8, 8],\
-                [6, 6, 5, 5, 5, 7, 6, 6],\
-                [13, 13, 15, 14, 14, 13, 15, 15, 14],\
-                [11]]
+    n_letters = [[3, 3, 5, 4, 4, 3, 5, 5, 4, 3, 6, 6, 8, 8, 7, 7, 9, 8, 8],
+                 [6, 6, 5, 5, 5, 7, 6, 6],
+                 [13, 13, 15, 14, 14, 13, 15, 15, 14],
+                 [11]]
 
     sum_ = 0
 
@@ -59,9 +60,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 17')
-    print('Answer: {}'.format(sum_))
+    print(f'Answer: {sum_}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p018.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.
 #
@@ -30,24 +30,24 @@
 # NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge
 # with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o)
 
+import sys
 from timeit import default_timer
+
 from projecteuler import find_max_path
 
+
 def main():
     start = default_timer()
 
     try:
-        fp = open('triang.txt', 'r')
-    except:
-        print('Error while opening file {}'.format('triang.txt'))
-        exit(1)
+        with open('triang.txt', 'r', encoding='utf-8') as fp:
+            triang = []
 
-    triang = list()
-
-    for line in fp:
-        triang.append(line.strip('\n').split())
-
-    fp.close()
+            for line in fp:
+                triang.append(line.strip('\n').split())
+    except FileNotFoundError:
+        print('Error while opening file trian.txt')
+        sys.exit(1)
 
     l = len(triang)
 
@@ -60,9 +60,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 18')
-    print('Answer: {}'.format(max_))
+    print(f'Answer: {max_}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p019.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # You are given the following information, but you may prefer to do some research for yourself.
 #
@@ -14,9 +14,9 @@
 # How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)?
 
 import datetime
-
 from timeit import default_timer
 
+
 def main():
     start = default_timer()
 
@@ -31,9 +31,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 19')
-    print('Answer: {}'.format(count))
+    print(f'Answer: {count}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p020.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # n! means n × (n − 1) × ... × 3 × 2 × 1
 #
@@ -8,9 +8,9 @@
 # Find the sum of the digits in the number 100!
 
 from math import factorial
-
 from timeit import default_timer
 
+
 def main():
     start = default_timer()
 
@@ -25,9 +25,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 20')
-    print('Answer: {}'.format(sum_))
+    print(f'Answer: {sum_}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p021.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).
 # If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable pair and each of a and b are called amicable numbers.
@@ -8,11 +8,12 @@
 #
 # Evaluate the sum of all the amicable numbers under 10000.
 
-from math import floor, sqrt
 
 from timeit import default_timer
+
 from projecteuler import sum_of_divisors
 
+
 def main():
     start = default_timer()
 
@@ -32,9 +33,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 21')
-    print('Answer: {}'.format(sum_))
+    print(f'Answer: {sum_}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p022.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # Using names.txt, a 46K text file containing over five-thousand first names, begin by sorting it into alphabetical order.
 # Then working out the alphabetical value for each name, multiply this value by its alphabetical position in the list to obtain a name score.
@@ -8,18 +8,20 @@
 #
 # What is the total of all the name scores in the file?
 
+import sys
 from timeit import default_timer
 
+
 def main():
     start = default_timer()
 
     try:
-        fp = open('names.txt', 'r')
-    except:
-        print('Error while opening file {}'.format('names.txt'))
-        exit(1)
+        with open('names.txt', 'r', encoding='utf-8') as fp:
+            names = list(fp.readline().replace('"', '').split(','))
+    except FileNotFoundError:
+        print('Error while opening file names.txt')
+        sys.exit(1)
 
-    names = list(fp.readline().replace('"', '').split(','))
     names.sort()
 
     sum_ = 0
@@ -38,9 +40,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 22')
-    print('Answer: {}'.format(sum_))
+    print(f'Answer: {sum_}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p023.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # A perfect number is a number for which the sum of its proper divisors is exactly equal to the number.
 # For example, the sum of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number.
@@ -12,14 +12,15 @@
 #
 # Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.
 
-from math import floor, sqrt
-
 from timeit import default_timer
+
 from projecteuler import sum_of_divisors
 
+
 def is_abundant(n):
     return sum_of_divisors(n) > n
 
+
 def main():
     start = default_timer()
 
@@ -52,9 +53,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 23')
-    print('Answer: {}'.format(sum_))
+    print(f'Answer: {sum_}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p024.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # A permutation is an ordered arrangement of objects. For example, 3124 is one possible permutation of the digits 1, 2, 3 and 4.
 # If all of the permutations are listed numerically or alphabetically, we call it lexicographic order. The lexicographic permutations of 0, 1 and 2 are:
@@ -8,9 +8,9 @@
 # What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?
 
 from itertools import permutations
-
 from timeit import default_timer
 
+
 def main():
     start = default_timer()
 
@@ -21,9 +21,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 24')
-    print('Answer: {}'.format(res))
+    print(f'Answer: {res}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p025.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # The Fibonacci sequence is defined by the recurrence relation:
 #
@@ -23,6 +23,7 @@
 
 from timeit import default_timer
 
+
 def main():
     start = default_timer()
 
@@ -42,9 +43,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 25')
-    print('Answer: {}'.format(i))
+    print(f'Answer: {i}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p026.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:
 #
@@ -18,6 +18,7 @@
 
 from timeit import default_timer
 
+
 def main():
     start = default_timer()
 
@@ -59,9 +60,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 26')
-    print('Answer: {}'.format(max_n))
+    print(f'Answer: {max_n}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p027.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # Euler discovered the remarkable quadratic formula:
 #
@@ -20,8 +20,10 @@
 # Find the product of the coefficients, a and b, for the quadratic expression that produces the maximum number of primes for consecutive values of n, starting with n=0.
 
 from timeit import default_timer
+
 from projecteuler import is_prime
 
+
 def main():
     start = default_timer()
 
@@ -52,9 +54,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 27')
-    print('Answer: {}'.format(save_a * save_b))
+    print(f'Answer: {save_a * save_b}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p028.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows:
 #
@@ -14,6 +14,7 @@
 
 from timeit import default_timer
 
+
 def main():
     start = default_timer()
 
@@ -40,9 +41,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 28')
-    print('Answer: {}'.format(sum_))
+    print(f'Answer: {sum_}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p029.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # Consider all integer combinations of ab for 2 ≤ a ≤ 5 and 2 ≤ b ≤ 5:
 #
@@ -13,9 +13,10 @@
 #
 # How many distinct terms are in the sequence generated by ab for 2 ≤ a ≤ 100 and 2 ≤ b ≤ 100?
 
+from timeit import default_timer
+
 from numpy import zeros
 
-from timeit import default_timer
 
 def main():
     start = default_timer()
@@ -31,7 +32,7 @@ def main():
 #   Sort the values and count the different values.
     powers = list(powers)
     powers.sort()
-    
+
     count = 1
 
     for i in range(1, 9801):
@@ -41,9 +42,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 29')
-    print('Answer: {}'.format(count))
+    print(f'Answer: {count}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p030.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # Surprisingly there are only three numbers that can be written as the sum of fourth powers of their digits:
 #
@@ -14,6 +14,7 @@
 
 from timeit import default_timer
 
+
 def main():
     start = default_timer()
 
@@ -37,9 +38,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 30')
-    print('Answer: {}'.format(tot))
+    print(f'Answer: {tot}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p031.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # In England the currency is made up of pound, £, and pence, p, and there are eight coins in general circulation:
 #
@@ -12,20 +12,23 @@
 
 from timeit import default_timer
 
+
 # Simple recursive function that tries every combination.
 def count(coins, value, n, i):
     for j in range(i, 8):
         value = value + coins[j]
         if value == 200:
             return n + 1
-        elif value > 200:
+
+        if value > 200:
             return n
-        else:
-            n = count(coins, value, n, j)
-            value = value - coins[j]
+
+        n = count(coins, value, n, j)
+        value = value - coins[j]
 
     return n
 
+
 def main():
     start = default_timer()
 
@@ -36,9 +39,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 31')
-    print('Answer: {}'.format(n))
+    print(f'Answer: {n}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p032.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once; for example, the 5-digit number, 15234,
 # is 1 through 5 pandigital.
@@ -9,12 +9,14 @@
 #
 # HINT: Some products can be obtained in more than one way so be sure to only include it once in your sum.
 
+from timeit import default_timer
+
 import numpy as np
 from numpy import zeros
 
-from timeit import default_timer
 from projecteuler import is_pandigital
 
+
 def main():
     start = default_timer()
 
@@ -26,14 +28,14 @@ def main():
 #   To get a 1 to 9 pandigital concatenation of the two factors and product,
 #   we need to multiply a 1 digit number times a 4 digit numbers (the biggest
 #   one digit number 9 times the biggest 3 digit number 999 multiplied give
-#   8991 and the total digit count is 8, which is not enough), or a 2 digit 
+#   8991 and the total digit count is 8, which is not enough), or a 2 digit
 #   number times a 3 digit number (the smallest two different 3 digits number,
 #   100 and 101, multiplied give 10100, and the total digit count is 11, which
 #   is too many). The outer loop starts at 2 because 1 times any number gives
-#   the same number, so its digit will be repeated and the result can't be 
-#   pandigital. The nested loop starts from 1234 because it's the smallest 
+#   the same number, so its digit will be repeated and the result can't be
+#   pandigital. The nested loop starts from 1234 because it's the smallest
 #   4-digit number with no repeated digits, and it ends at 4987 because it's
-#   the biggest number without repeated digits that multiplied by 2 gives a 
+#   the biggest number without repeated digits that multiplied by 2 gives a
 #   4 digit number.
     for i in range(2, 9):
         for j in range(1234, 4987):
@@ -80,9 +82,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 32')
-    print('Answer: {}'.format(sum_))
+    print(f'Answer: {sum_}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p033.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # The fraction 49/98 is a curious fraction, as an inexperienced mathematician in attempting to simplify it may incorrectly believe that 49/98 = 4/8,
 # which is correct, is obtained by cancelling the 9s.
@@ -11,9 +11,9 @@
 # If the product of these four fractions is given in its lowest common terms, find the value of the denominator.
 
 from math import gcd
-
 from timeit import default_timer
 
+
 def main():
     start = default_timer()
 
@@ -25,7 +25,7 @@ def main():
 #           If the example is non-trivial, check if cancelling the digit that's equal
 #           in numerator and denominator gives the same fraction.
             if i % 10 != 0 and j % 10 != 0 and\
-            i != j and i % 10 == j // 10:
+                    i != j and i % 10 == j // 10:
                 n = i // 10
                 d = j % 10
 
@@ -42,9 +42,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 33')
-    print('Answer: {}'.format(prod_d // div))
+    print(f'Answer: {prod_d // div}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p034.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # 145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.
 #
@@ -7,10 +7,10 @@
 # Note: as 1! = 1 and 2! = 2 are not sums they are not included.
 
 from math import factorial
+from timeit import default_timer
 
 from numpy import ones
 
-from timeit import default_timer
 
 def main():
     start = default_timer()
@@ -41,9 +41,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 34')
-    print('Answer: {}'.format(sum_))
+    print(f'Answer: {sum_}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p035.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # The number, 197, is called a circular prime because all rotations of the digits: 197, 971, and 719, are themselves prime.
 #
@@ -7,18 +7,23 @@
 # How many circular primes are there below one million?
 
 from timeit import default_timer
+
 from projecteuler import sieve
 
-def is_circular_prime(n):
-    global primes
 
+# Calculate all primes below one million, then check if they're circular.
+N = 1000000
+primes = sieve(N)
+
+
+def is_circular_prime(n):
 #   If n is not prime, it's obviously not a circular prime.
     if primes[n] == 0:
         return False
 
 #   The primes below 10 are circular primes.
     if primes[n] == 1 and n < 10:
-            return True
+        return True
 
     tmp = n
     count = 0
@@ -32,7 +37,7 @@ def is_circular_prime(n):
         count = count + 1
         tmp = tmp // 10
 
-    for i in range(1, count):
+    for _ in range(1, count):
 #       Generate rotations and check if they're prime.
         n = n % (10 ** (count - 1)) * 10 + n // (10 ** (count - 1))
 
@@ -41,15 +46,10 @@ def is_circular_prime(n):
 
     return True
 
+
 def main():
     start = default_timer()
 
-    global primes
-
-    N = 1000000
-
-#   Calculate all primes below one million, then check if they're circular.
-    primes = sieve(N)
     count = 13
 
 #   Calculate all primes below one million, then check if they're circular.
@@ -60,9 +60,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 35')
-    print('Answer: {}'.format(count))
+    print(f'Answer: {count}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p036.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # The decimal number, 585 = 1001001001_2 (binary), is palindromic in both bases.
 #
@@ -7,8 +7,10 @@
 # (Please note that the palindromic number, in either base, may not include leading zeros.)
 
 from timeit import default_timer
+
 from projecteuler import is_palindrome
 
+
 def main():
     start = default_timer()
 
@@ -26,9 +28,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 36')
-    print('Answer: {}'.format(sum_))
+    print(f'Answer: {sum_}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p037.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # The number 3797 has an interesting property. Being prime itself, it is possible to continuously remove digits from left to right,
 # and remain prime at each stage: 3797, 797, 97, and 7. Similarly we can work from right to left: 3797, 379, 37, and 3.
@@ -8,8 +8,10 @@
 # NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.
 
 from timeit import default_timer
+
 from projecteuler import is_prime
 
+
 def is_tr_prime(n):
 #   One-digit numbers and non-prime numbers are
 #   not truncatable primes.
@@ -40,6 +42,7 @@ def is_tr_prime(n):
 #   If it gets here, the number is truncatable prime.
     return True
 
+
 def main():
     start = default_timer()
 
@@ -57,9 +60,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 37')
-    print('Answer: {}'.format(sum_))
+    print(f'Answer: {sum_}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p038.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # Take the number 192 and multiply it by each of 1, 2, and 3:
 #
@@ -13,11 +13,11 @@
 #
 # What is the largest 1 to 9 pandigital 9-digit number that can be formed as the concatenated product of an integer with (1,2, ... , n) where n > 1?
 
-from numpy import zeros
-
 from timeit import default_timer
+
 from projecteuler import is_pandigital
 
+
 def main():
     start = default_timer()
 
@@ -57,9 +57,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 38')
-    print('Answer: {}'.format(max_))
+    print(f'Answer: {max_}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p039.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # If p is the perimeter of a right angle triangle with integral length sides, {a,b,c}, there are exactly three solutions for p = 120.
 #
@@ -6,9 +6,10 @@
 #
 # For which value of p ≤ 1000, is the number of solutions maximised?
 
+from timeit import default_timer
+
 from numpy import zeros
 
-from timeit import default_timer
 
 def main():
     start = default_timer()
@@ -68,9 +69,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 39')
-    print('Answer: {}'.format(res))
+    print(f'Answer: {res}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p040.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # An irrational decimal fraction is created by concatenating the positive integers:
 #
@@ -10,9 +10,10 @@
 #
 # d_1 × d_10 × d_100 × d_1000 × d_10000 × d_100000 × d_1000000
 
+from timeit import default_timer
+
 from numpy import zeros
 
-from timeit import default_timer
 
 def main():
     start = default_timer()
@@ -66,9 +67,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 40')
-    print('Answer: {}'.format(n))
+    print(f'Answer: {n}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p041.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital
 # and is also prime.
@@ -6,8 +6,10 @@
 # What is the largest n-digit pandigital prime that exists?
 
 from timeit import default_timer
+
 from projecteuler import is_pandigital, is_prime
 
+
 def main():
     start = default_timer()
 
@@ -18,18 +20,19 @@ def main():
 #   until we find a prime.
     i = 7654321
 
-    while(i > 0):
+    while i > 0:
         if is_pandigital(i, len(str(i))) and is_prime(i):
             break
 #       Skipping the even numbers.
         i = i - 2
-    
+
     end = default_timer()
 
     print('Project Euler, Problem 41')
-    print('Answer: {}'.format(i))
-    
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
+    print(f'Answer: {i}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
+
 
 if __name__ == '__main__':
     main()

Python/p042.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # The nth term of the sequence of triangle numbers is given by, tn = ½n(n+1); so the first ten triangle numbers are:
 #
@@ -10,8 +10,10 @@
 # Using words.txt (right click and 'Save Link/Target As...'), a 16K text file containing nearly two-thousand common English words,
 # how many are triangle words?
 
+import sys
 from timeit import default_timer
 
+
 def is_triang(n):
     i = 1
     j = 1
@@ -24,18 +26,16 @@ def is_triang(n):
 
     return False
 
+
 def main():
     start = default_timer()
 
     try:
-        fp = open('words.txt', 'r')
-    except:
-        print('Error while opening file {}'.format('words.txt'))
-        exit(1)
-
-    words = list(fp.readline().replace('"', '').split(','))
-
-    fp.close()
+        with open('words.txt', 'r', encoding='utf-8') as fp:
+            words = list(fp.readline().replace('"', '').split(','))
+    except FileNotFoundError:
+        print('Error while opening file words.txt')
+        sys.exit(1)
 
     count = 0
 
@@ -53,9 +53,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 42')
-    print('Answer: {}'.format(count))
+    print(f'Answer: {count}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p043.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # The number, 1406357289, is a 0 to 9 pandigital number because it is made up of each of the digits 0 to 9 in some order, but it also has a
 # rather interesting sub-string divisibility property.
@@ -16,9 +16,9 @@
 # Find the sum of all 0 to 9 pandigital numbers with this property.
 
 from itertools import permutations
-
 from timeit import default_timer
 
+
 # Function to check if the value has the desired property.
 def has_property(n):
     value = int(n[1]) * 100 + int(n[2]) * 10 + int(n[3])
@@ -50,7 +50,7 @@ def has_property(n):
 
     if value % 13 != 0:
         return False
-    
+
     value = int(n[7]) * 100 + int(n[8]) * 10 + int(n[9])
 
     if value % 17 != 0:
@@ -58,6 +58,7 @@ def has_property(n):
 
     return True
 
+
 def main():
     start = default_timer()
 
@@ -74,9 +75,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 43')
-    print('Answer: {}'.format(sum_))
+    print(f'Answer: {sum_}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p044.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # Pentagonal numbers are generated by the formula, Pn=n(3n−1)/2. The first ten pentagonal numbers are:
 #
@@ -9,11 +9,11 @@
 # Find the pair of pentagonal numbers, Pj and Pk, for which their sum and difference are pentagonal and D = |Pk − Pj| is minimised;
 # what is the value of D?
 
-from math import sqrt
-
 from timeit import default_timer
+
 from projecteuler import is_pentagonal
 
+
 def main():
     start = default_timer()
 
@@ -38,9 +38,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 44')
-    print('Answer: {}'.format(pn-pm))
+    print(f'Answer: {pn - pm}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p045.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:
 #
@@ -10,11 +10,11 @@
 #
 # Find the next triangle number that is also pentagonal and hexagonal.
 
-from math import sqrt
-
 from timeit import default_timer
+
 from projecteuler import is_pentagonal
 
+
 def main():
     start = default_timer()
 
@@ -34,9 +34,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 45')
-    print('Answer: {}'.format(n))
+    print(f'Answer: {n}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p046.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # It was proposed by Christian Goldbach that every odd composite number can be written as the sum of a prime and twice a square.
 #
@@ -14,11 +14,15 @@
 # What is the smallest odd composite that cannot be written as the sum of a prime and twice a square?
 
 from timeit import default_timer
+
 from projecteuler import sieve
 
-def goldbach(n):
-    global primes
 
+N = 10000
+primes = sieve(N)
+
+
+def goldbach(n):
 #   Check every prime smaller than n.
     for i in range(3, n, 2):
         if primes[i] == 1:
@@ -43,12 +47,6 @@ def goldbach(n):
 def main():
     start = default_timer()
 
-    global primes
-
-    N = 10000
-
-    primes = sieve(N)
-
     found = 0
     i = 3
 
@@ -64,9 +62,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 46')
-    print('Answer: {}'.format(i-2))
+    print(f'Answer: {i - 2}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p047.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # The first two consecutive numbers to have two distinct prime factors are:
 #
@@ -15,6 +15,7 @@
 
 from timeit import default_timer
 
+
 # Function using a modified sieve of Eratosthenes to count
 # the distinct prime factors of each number.
 def count_factors(n):
@@ -29,6 +30,7 @@ def count_factors(n):
 
     return factors
 
+
 def main():
     start = default_timer()
 
@@ -53,9 +55,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 47')
-    print('Answer: {}'.format(res))
+    print(f'Answer: {res}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p048.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # The series, 1^1 + 2^2 + 3^3 + ... + 10^10 = 10405071317.
 #
@@ -6,6 +6,7 @@
 
 from timeit import default_timer
 
+
 def main():
     start = default_timer()
 
@@ -19,9 +20,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 48')
-    print('Answer: {}'.format(str(sum_)[-10:]))
+    print(f'Answer: {str(sum_)[-10:]}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p049.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # The arithmetic sequence, 1487, 4817, 8147, in which each of the terms increases by 3330, is unusual in two ways: (i) each of the three terms are prime,
 # and, (ii) each of the 4-digit numbers are permutations of one another.
@@ -8,11 +8,11 @@
 #
 # What 12-digit number do you form by concatenating the three terms in this sequence?
 
-from numpy import zeros
-
 from timeit import default_timer
+
 from projecteuler import sieve
 
+
 def main():
     start = default_timer()
 
@@ -35,18 +35,20 @@ def main():
                 if i + 2 * j < N and primes[i+j] == 1 and primes[i+2*j] == 1 and\
                         ''.join(sorted(str(i))) == ''.join(sorted(str(i+j))) and\
                         ''.join(sorted(str(i))) == ''.join(sorted(str(i+2*j))):
-                            found = 1
-                            break
-        if(found):
+                    found = 1
+                    break
+        if found:
             break
+
         i = i + 2
 
     end = default_timer()
 
     print('Project Euler, Problem 49')
-    print('Answer: {}'.format(str(i)+str(i+j)+str(i+2*j)))
+    print(f'Answer: {str(i)+str(i+j)+str(i+2*j)}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p050.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # The prime 41, can be written as the sum of six consecutive primes:
 #
@@ -11,8 +11,10 @@
 # Which prime, below one-million, can be written as the sum of the most consecutive primes?
 
 from timeit import default_timer
+
 from projecteuler import sieve
 
+
 def main():
     start = default_timer()
 
@@ -64,9 +66,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 50')
-    print('Answer: {}'.format(max_p))
+    print(f'Answer: {max_p}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p051.py

@@ -10,8 +10,16 @@
 # value family.
 
 from timeit import default_timer
+
 from projecteuler import sieve
 
+
+N = 1000000
+
+# N set to 1000000 as a reasonable limit, which turns out to be enough.
+primes = sieve(N)
+
+
 def count_digit(n, d):
     count = 0
 
@@ -22,9 +30,8 @@ def count_digit(n, d):
 
     return count
 
-def replace(n):
-    global primes
 
+def replace(n):
     n_string = list(str(n))
     l = len(n_string)
     max_ = 0
@@ -60,35 +67,29 @@ def replace(n):
 def main():
     start = default_timer()
 
-    global primes
-
-    N = 1000000
-
-#   N set to 1000000 as a reasonable limit, which turns out to be enough.
-    primes = sieve(N)
-
 #   Checking only odd numbers with at least 4 digits.
     i = 1001
 
     while i < N:
-#       The family of numbers needs to have at least one of 0, 1 or 2 as 
-#       repeated digits, otherwise we can't get a 8 number family (there 
+#       The family of numbers needs to have at least one of 0, 1 or 2 as
+#       repeated digits, otherwise we can't get a 8 number family (there
 #       are only 7 other digits). Also, te number of repeated digits must
 #       be 3, otherwise at least 3 resulting numbers will be divisible by 3.
 #       So the smallest number of this family must have three 0s, three 1s or
 #       three 2s.
         if count_digit(i, 0) >= 3 or count_digit(i, 1) >= 3 or\
                 count_digit(i, 2) >= 3:
-                    if primes[i] == 1 and replace(i) == 8:
-                        break
+            if primes[i] == 1 and replace(i) == 8:
+                break
         i = i + 2
-    
+
     end = default_timer()
 
     print('Project Euler, Problem 51')
-    print('Answer: {}'.format(i))
+    print(f'Answer: {i}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p052.py

@@ -4,10 +4,9 @@
 #
 # Find the smallest positive integer, x, such that 2x, 3x, 4x, 5x, and 6x, contain the same digits.
 
-from numpy import zeros
-
 from timeit import default_timer
 
+
 def main():
     start = default_timer()
 
@@ -20,15 +19,16 @@ def main():
                 ''.join(sorted(str(i))) == ''.join(sorted(str(4*i))) and\
                 ''.join(sorted(str(i))) == ''.join(sorted(str(5*i))) and\
                 ''.join(sorted(str(i))) == ''.join(sorted(str(6*i))):
-                    break
+            break
         i = i + 1
-    
+
     end = default_timer()
 
     print('Project Euler, Problem 52')
-    print('Answer: {}'.format(i))
+    print(f'Answer: {i}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p053.py

@@ -12,9 +12,10 @@
 #
 # How many, not necessarily distinct, values of (n r) for 1≤n≤100, are greater than one-million?
 
+from timeit import default_timer
+
 from scipy.special import comb
 
-from timeit import default_timer
 
 def main():
     start = default_timer()
@@ -32,9 +33,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 53')
-    print('Answer: {}'.format(count))
+    print(f'Answer: {count}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p054.py

@@ -46,10 +46,11 @@
 #
 #      How many hands does Player 1 win?
 
+import sys
 from enum import IntEnum
-
 from timeit import default_timer
 
+
 class Value(IntEnum):
     Two = 1
     Three = 2
@@ -65,10 +66,11 @@ class Value(IntEnum):
     King = 12
     Ace = 13
 
+
 class Card():
 
     def __init__(self, *args, **kwargs):
-        super(Card, self).__init__(*args, **kwargs)
+        super().__init__(*args, **kwargs)
 
         self.value = None
         self.suit = None
@@ -103,20 +105,22 @@ class Card():
 
         self.suit = card[1]
 
+
 class Hand():
 
     def __init__(self, *args, **kwargs):
-        super(Hand, self).__init__(*args, **kwargs)
+        super().__init__(*args, **kwargs)
 
         self.cards = list()
 
     def sort(self):
         self.cards.sort(key=lambda x: x.value)
 
+
 class Game():
 
     def __init__(self, *args, **kwargs):
-        super(Game, self).__init__(*args, **kwargs)
+        super().__init__(*args, **kwargs)
 
         self.hand1 = None
         self.hand2 = None
@@ -130,7 +134,8 @@ class Game():
                 self.hand1.cards[0].suit == self.hand1.cards[2].suit and\
                 self.hand1.cards[0].suit == self.hand1.cards[3].suit and\
                 self.hand1.cards[0].suit == self.hand1.cards[4].suit:
-                    return 1
+            return 1
+
 #       If player 2 has a Royal Flush, player 2 wins.
         if self.hand2.cards[4].value == Value.Ace and self.hand2.cards[3].value == Value.King and\
                 self.hand2.cards[2].value == Value.Queen and self.hand2.cards[1].value == Value.Jack and\
@@ -138,7 +143,7 @@ class Game():
                 self.hand2.cards[0].suit == self.hand2.cards[2].suit and\
                 self.hand2.cards[0].suit == self.hand2.cards[3].suit and\
                 self.hand2.cards[0].suit == self.hand2.cards[4].suit:
-                    return -1
+            return -1
 
         straightflush1 = 0
         straightflush2 = 0
@@ -146,35 +151,35 @@ class Game():
 #       Check if player 1 has a straight flush.
         if self.hand1.cards[0].suit == self.hand1.cards[1].suit and self.hand1.cards[0].suit == self.hand1.cards[2].suit and\
                 self.hand1.cards[0].suit == self.hand1.cards[3].suit and self.hand1.cards[0].suit == self.hand1.cards[4].suit:
-                    value = self.hand1.cards[0].value
+            value = self.hand1.cards[0].value
 
-                    straightflush1 = 1
+            straightflush1 = 1
 
-                    for i in range(1, 5):
-                        value = value + 1
+            for i in range(1, 5):
+                value = value + 1
 
-                        if self.hand1.cards[i].value != value:
-                            straightflush1 = 0
-                            break
+                if self.hand1.cards[i].value != value:
+                    straightflush1 = 0
+                    break
 
 #       Check if player 2 has a straight flush.
         if self.hand2.cards[0].suit == self.hand2.cards[1].suit and self.hand2.cards[0].suit == self.hand2.cards[2].suit and\
                 self.hand2.cards[0].suit == self.hand2.cards[3].suit and self.hand2.cards[0].suit == self.hand2.cards[4].suit:
-                    value = self.hand2.cards[0].value
+            value = self.hand2.cards[0].value
 
-                    straightflush2 = 1
+            straightflush2 = 1
 
-                    for i in range(1, 5):
-                        value = value + 1
+            for i in range(1, 5):
+                value = value + 1
 
-                        if self.hand2.cards[i].value != value:
-                            straightflush2 = 0
-                            break
+                if self.hand2.cards[i].value != value:
+                    straightflush2 = 0
+                    break
 
 #       If player 1 has a straight flush and player 2 doesn't, player 1 wins
         if straightflush1 and not straightflush2:
             return 1
-        
+
 #       If player 2 has a straight flush and player 1 doesn't, player 2 wins
         if not straightflush1 and straightflush2:
             return -1
@@ -185,8 +190,8 @@ class Game():
         if straightflush1 and straightflush2:
             if self.hand1.cards[0].value > self.hand2.cards[0].value:
                 return 1
-            else:
-                return -1
+
+            return -1
 
         four1 = 0
         four2 = 0
@@ -221,12 +226,12 @@ class Game():
 #       Check if player 1 has a full house.
         if self.hand1.cards[0].value == self.hand1.cards[1].value and self.hand1.cards[3].value == self.hand1.cards[4].value and\
                 (self.hand1.cards[1].value == self.hand1.cards[2].value or self.hand1.cards[2].value == self.hand1.cards[3].value):
-                    full1 = 1
+            full1 = 1
 
 #       Check if player 2 has a full house.
         if self.hand2.cards[0].value == self.hand2.cards[1].value and self.hand2.cards[3].value == self.hand2.cards[4].value and\
                 (self.hand2.cards[1].value == self.hand2.cards[2].value or self.hand2.cards[2].value == self.hand2.cards[3].value):
-                    full2 = 1
+            full2 = 1
 
 #       If player 1 has a full house and player 2 doesn't, player 1 wins.
         if full1 and not full2:
@@ -242,7 +247,8 @@ class Game():
         if full1 and full2:
             if self.hand1.cards[2].value > self.hand2.cards[2].value:
                 return 1
-            elif self.hand1.cards[2].value < self.hand2.cards[2].value:
+
+            if self.hand1.cards[2].value < self.hand2.cards[2].value:
                 return -1
 
         flush1 = 0
@@ -251,12 +257,12 @@ class Game():
 #       Check if player 1 has a flush.
         if self.hand1.cards[0].suit == self.hand1.cards[1].suit and self.hand1.cards[0].suit == self.hand1.cards[2].suit and\
                 self.hand1.cards[0].suit == self.hand1.cards[3].suit and self.hand1.cards[0].suit == self.hand1.cards[4].suit:
-                    flush1 = 1
+            flush1 = 1
 
 #       Check if player 2 has a flush.
         if self.hand2.cards[0].suit == self.hand2.cards[1].suit and self.hand2.cards[0].suit == self.hand2.cards[2].suit and\
                 self.hand2.cards[0].suit == self.hand2.cards[3].suit and self.hand2.cards[0].suit == self.hand2.cards[4].suit:
-                    flush2 = 1
+            flush2 = 1
 
 #       If player 1 has a flush and player 2 doesn't, player 1 wins.
         if flush1 and not flush2:
@@ -304,13 +310,13 @@ class Game():
         if (self.hand1.cards[0].value == self.hand1.cards[1].value and self.hand1.cards[0].value == self.hand1.cards[2].value) or\
                 (self.hand1.cards[1].value == self.hand1.cards[2].value and self.hand1.cards[1].value == self.hand1.cards[3].value) or\
                 (self.hand1.cards[2].value == self.hand1.cards[3].value and self.hand1.cards[2].value == self.hand1.cards[4].value):
-                    three1 = 1
+            three1 = 1
 
 #       Check if player 2 has three of a kind.
         if (self.hand2.cards[0].value == self.hand2.cards[1].value and self.hand2.cards[0].value == self.hand2.cards[2].value) or\
                 (self.hand2.cards[1].value == self.hand2.cards[2].value and self.hand2.cards[1].value == self.hand2.cards[3].value) or\
                 (self.hand2.cards[2].value == self.hand2.cards[3].value and self.hand2.cards[2].value == self.hand2.cards[4].value):
-                    three2 = 1
+            three2 = 1
 
 #       If player 1 has three of a kind and player 2 doesn't, player 1 wins.
         if three1 and not three2:
@@ -323,7 +329,8 @@ class Game():
         if three1 and three2:
             if self.hand1.cards[2].value > self.hand2.cards[2].value:
                 return 1
-            elif self.hand1.cards[2].value < self.hand2.cards[2].value:
+
+            if self.hand1.cards[2].value < self.hand2.cards[2].value:
                 return -1
 
         twopairs1 = 0
@@ -333,13 +340,13 @@ class Game():
         if (self.hand1.cards[0].value == self.hand1.cards[1].value and self.hand1.cards[2].value == self.hand1.cards[3].value) or\
                 (self.hand1.cards[0].value == self.hand1.cards[1].value and self.hand1.cards[3].value == self.hand1.cards[4].value) or\
                 (self.hand1.cards[1].value == self.hand1.cards[2].value and self.hand1.cards[3].value == self.hand1.cards[4].value):
-                    twopairs1 = 1
+            twopairs1 = 1
 
 #       Check if player 2 has two pairs.
         if (self.hand2.cards[0].value == self.hand2.cards[1].value and self.hand2.cards[2].value == self.hand2.cards[3].value) or\
                 (self.hand2.cards[0].value == self.hand2.cards[1].value and self.hand2.cards[3].value == self.hand2.cards[4].value) or\
                 (self.hand2.cards[1].value == self.hand2.cards[2].value and self.hand2.cards[3].value == self.hand2.cards[4].value):
-                    twopairs2 = 1
+            twopairs2 = 1
 
 #       If player 1 has two pairs and player 2 doesn't, player 1 wins.
         if twopairs1 and not twopairs2:
@@ -354,17 +361,21 @@ class Game():
         if twopairs1 and twopairs2:
             if self.hand1.cards[3].value > self.hand2.cards[3].value:
                 return 1
-            elif self.hand1.cards[3].value < self.hand2.cards[3].value:
+
+            if self.hand1.cards[3].value < self.hand2.cards[3].value:
                 return -1
-            elif self.hand1.cards[1].value > self.hand2.cards[1].value:
+
+            if self.hand1.cards[1].value > self.hand2.cards[1].value:
                 return 1
-            elif self.hand1.cards[1].value < self.hand2.cards[1].value:
+
+            if self.hand1.cards[1].value < self.hand2.cards[1].value:
                 return -1
 
             for i in range(4, -1, -1):
                 if self.hand1.cards[i].value > self.hand2.cards[i].value:
                     return 1
-                elif self.hand1.cards[i].value < self.hand2.cards[i].value:
+
+                if self.hand1.cards[i].value < self.hand2.cards[i].value:
                     return -1
 
         pair1 = 0
@@ -373,12 +384,12 @@ class Game():
 #       Check if player 1 has a pair of cards.
         if self.hand1.cards[0].value == self.hand1.cards[1].value or self.hand1.cards[1].value == self.hand1.cards[2].value or\
                 self.hand1.cards[2].value == self.hand1.cards[3].value or self.hand1.cards[3].value == self.hand1.cards[4].value:
-                    pair1 = 1
+            pair1 = 1
 
 #       Check if player 2 has a pair of cards.
         if self.hand2.cards[0].value == self.hand2.cards[1].value or self.hand2.cards[1].value == self.hand2.cards[2].value or\
                 self.hand2.cards[2].value == self.hand2.cards[3].value or self.hand2.cards[3].value == self.hand2.cards[4].value:
-                    pair2 = 1
+            pair2 = 1
 
 #       If player 1 has a pair of cards and player 2 doesn't, player 1 wins.
         if pair1 and not pair2:
@@ -427,22 +438,20 @@ class Game():
                 return 1
             if self.hand1.cards[i].value < self.hand2.cards[i].value:
                 return -1
-        
+
 #       If everything is equal, return 0
         return 0
 
+
 def main():
     start = default_timer()
 
     try:
-        fp = open('poker.txt', 'r')
-    except:
-        print('Error while opening file {}'.format('poker.txt'))
-        exit(1)
-
-    games = fp.readlines()
-
-    fp.close()
+        with open('poker.txt', 'r', encoding='utf-8') as fp:
+            games = fp.readlines()
+    except FileNotFoundError:
+        print('Error while opening file poker.txt')
+        sys.exit(1)
 
     count = 0
 
@@ -478,9 +487,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 54')
-    print('Answer: {}'.format(count))
+    print(f'Answer: {count}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p055.py

@@ -24,13 +24,15 @@
 # NOTE: Wording was modified slightly on 24 April 2007 to emphasise the theoretical nature of Lychrel numbers.
 
 from timeit import default_timer
+
 from projecteuler import is_palindrome
 
+
 def is_lychrel(n):
     tmp = n
 
 #   Run for 50 iterations
-    for i in range(50):
+    for _ in range(50):
         reverse = 0
 
 #       Find the reverse of the given number
@@ -50,6 +52,7 @@ def is_lychrel(n):
 
     return True
 
+
 def main():
     start = default_timer()
 
@@ -64,9 +67,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 55')
-    print('Answer: {}'.format(count))
+    print(f'Answer: {count}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p056.py

@@ -7,6 +7,7 @@
 
 from timeit import default_timer
 
+
 def main():
     start = default_timer()
 
@@ -28,9 +29,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 56')
-    print('Answer: {}'.format(max_))
+    print(f'Answer: {max_}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p057.py

@@ -18,6 +18,7 @@
 
 from timeit import default_timer
 
+
 def main():
     start = default_timer()
 
@@ -27,10 +28,10 @@ def main():
 
 #   If n/d is the current term of the expansion, the next term can be calculated as
 #   (n+2d)/(n+d).
-    for i in range(1, 1000):
+    for _ in range(1, 1000):
         d2 = 2 * d
         d = n + d
-        n = n +  d2
+        n = n + d2
 
         if len(str(n)) > len(str(d)):
             count = count + 1
@@ -38,9 +39,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 57')
-    print('Answer: {}'.format(count))
+    print(f'Answer: {count}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p058.py

@@ -17,14 +17,16 @@
 # what is the side length of the square spiral for which the ratio of primes along both diagonals first falls below 10%?
 
 from timeit import default_timer
+
 from projecteuler import is_prime
 
+
 def main():
     start = default_timer()
 
 #   Starting with 1, the next four numbers in the diagonal are 3 (1+2), 5 (1+2+2), 7 (1+2+2+2)
-#   and 9 (1+2+2+2+2). Check which are prime, increment the counter every time a new prime is 
-#   found, and divide by the number of elements of the diagonal, which are increase by 4 at 
+#   and 9 (1+2+2+2+2). Check which are prime, increment the counter every time a new prime is
+#   found, and divide by the number of elements of the diagonal, which are increase by 4 at
 #   every cycle. The next four number added to the diagonal are 13 (9+4), 17 (9+4+4), 21 and 25.
 #   Then 25+6 etc., at every cycle the step is increased by 2. Continue until the ratio goes below 0.1.
     i = 1
@@ -55,16 +57,17 @@ def main():
         step = step + 2
         diag = diag + 4
         l = l + 2
-        
+
         if ratio < 0.1:
             break
 
     end = default_timer()
 
     print('Project Euler, Problem 58')
-    print('Answer: {}'.format(l))
+    print(f'Answer: {l}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p059.py

@@ -20,66 +20,70 @@
 
 from timeit import default_timer
 
+
 class EncryptedText():
 
     def __init__(self, *args, **kwargs):
-        super(EncryptedText, self).__init__(*args, **kwargs)
+        super().__init__(*args, **kwargs)
 
         self.text = None
         self.len = 0
 
     def read_text(self, filename):
         try:
-            fp = open(filename, 'r')
-        except:
-            print('Error while opening file {}'.format(filename))
+            with open(filename, 'r', encoding='utf-8') as fp:
+                self.text = list(map(int, list(fp.readline().split(','))))
+        except FileNotFoundError:
+            print(f'Error while opening file {filename}')
+
             return -1
 
-        self.text = list(map(int, list(fp.readline().split(','))))
         self.len = len(self.text)
 
-        fp.close()
-
     def decrypt(self):
         found = 0
 
         for c1 in range(ord('a'), ord('z')+1):
-           if found:
+            if found:
                break
-           for c2 in range(ord('a'), ord('z')+1):
-               if found:
-                   break
-               for c3 in range(ord('a'), ord('z')+1):
-                   if found:
-                       break
-                   plain_text = [''] * self.len
-                   for i in range(0, self.len-2, 3):
-                       plain_text[i] = str(chr(self.text[i]^c1))
-                       plain_text[i+1] = str(chr(self.text[i+1]^c2))
-                       plain_text[i+2] = str(chr(self.text[i+2]^c3))
+            for c2 in range(ord('a'), ord('z')+1):
+                if found:
+                    break
+               
+                for c3 in range(ord('a'), ord('z')+1):
+                    if found:
+                        break
 
-                   if i == self.len - 2:
-                       plain_text[i] = str(chr(self.text[i]^c1))
-                       plain_text[i+1] = str(chr(self.text[i+1]^c2))
+                    plain_text = [''] * self.len
 
-                   if i == self.len - 1:
-                       plain_text[i] = str(chr(self.text[i]^c1))
+                    for i in range(0, self.len-2, 3):
+                        plain_text[i] = str(chr(self.text[i]^c1))
+                        plain_text[i+1] = str(chr(self.text[i+1]^c2))
+                        plain_text[i+2] = str(chr(self.text[i+2]^c3))
 
-                   plain_text = ''.join(plain_text)
+                    if i == self.len - 2:
+                        plain_text[i] = str(chr(self.text[i]^c1))
+                        plain_text[i+1] = str(chr(self.text[i+1]^c2))
+
+                    if i == self.len - 1:
+                        plain_text[i] = str(chr(self.text[i]^c1))
+
+                    plain_text = ''.join(plain_text)
+
+                    if 'the' in plain_text and 'be' in plain_text and 'to' in plain_text and 'of' in plain_text and\
+                            'and' in plain_text and 'in' in plain_text and 'that' in plain_text and 'have' in plain_text:
+                        found = 1
 
-                   if 'the' in plain_text and 'be' in plain_text and 'to' in plain_text and 'of' in plain_text and\
-                           'and' in plain_text and 'in' in plain_text and 'that' in plain_text and 'have' in plain_text:
-                               found = 1
-        
         return plain_text
 
+
 def main():
     start = default_timer()
 
     enc_text = EncryptedText()
 
     if enc_text.read_text('cipher.txt') == -1:
-        exit(1)
+        sys.exit(1)
 
     plain_text = enc_text.decrypt()
 
@@ -91,9 +95,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 59')
-    print('Answer: {}'.format(sum_))
+    print(f'Answer: {sum_}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p060.py

@@ -7,8 +7,10 @@
 # Find the lowest sum for a set of five primes for which any two primes concatenate to produce another prime.
 
 from timeit import default_timer
+
 from projecteuler import sieve, is_prime
 
+
 def main():
     start = default_timer()
 
@@ -29,7 +31,7 @@ def main():
         p2 = p1 + 2
 
         while p2 < N and not found:
-#           If p2 is not prime, or at least one of the possible concatenations of 
+#           If p2 is not prime, or at least one of the possible concatenations of
 #           p1 and p2 is not prime, go to the next number.
             if primes[p2] == 0 or not is_prime(int(str(p1)+str(p2))) or not is_prime(int(str(p2)+str(p1))):
                 p2 = p2 + 2
@@ -42,8 +44,9 @@ def main():
 #               p1, p2 and p3 is not prime, got to the next number.
                 if primes[p3] == 0 or not is_prime(int(str(p1)+str(p3))) or not is_prime(int(str(p3)+str(p1))) or\
                         not is_prime(int(str(p2)+str(p3))) or not is_prime(int(str(p3)+str(p2))):
-                            p3 = p3 + 2
-                            continue
+                    p3 = p3 + 2
+
+                    continue
 
                 p4 = p3 + 2
 
@@ -53,8 +56,9 @@ def main():
                     if primes[p4] == 0 or not is_prime(int(str(p1)+str(p4))) or not is_prime(int(str(p4)+str(p1))) or\
                             not is_prime(int(str(p2)+str(p4))) or not is_prime(int(str(p4)+str(p2))) or\
                             not is_prime(int(str(p3)+str(p4))) or not is_prime(int(str(p4)+str(p3))):
-                                p4 = p4 + 2
-                                continue
+                        p4 = p4 + 2
+
+                        continue
 
                     p5 = p4 + 2
 
@@ -65,8 +69,9 @@ def main():
                                 not is_prime(int(str(p2)+str(p5))) or not is_prime(int(str(p5)+str(p2))) or\
                                 not is_prime(int(str(p3)+str(p5))) or not is_prime(int(str(p5)+str(p3))) or\
                                 not is_prime(int(str(p4)+str(p5))) or not is_prime(int(str(p5)+str(p4))):
-                                    p5 = p5 + 2
-                                    continue
+                            p5 = p5 + 2
+
+                            continue
 
 #                       If it gets here, the five values have been found.
                         n = p1 + p2 + p3 + p4 + p5
@@ -83,9 +88,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 60')
-    print('Answer: {}'.format(n))
+    print(f'Answer: {n}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p061.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # Triangle, square, pentagonal, hexagonal, heptagonal, and octagonal numbers are all figurate (polygonal) numbers and are generated
 # by the following formulae:
@@ -19,18 +19,22 @@
 # Find the sum of the only ordered set of six cyclic 4-digit numbers for which each polygonal type: triangle, square, pentagonal,
 # hexagonal, heptagonal, and octagonal, is represented by a different number in the set.
 
+from timeit import default_timer
+
 from numpy import zeros
 
-from timeit import default_timer
+
+polygonal = zeros((6, 10000), int)
+chain = [0] * 6
+flags = [0] * 6
+sum_ = 0
+
 
 # Recursive function to find the required set. It finds a polygonal number,
 # check if it can be part of the chain, then use recursion to find the next
 # number. If a solution can't be found with the current numbers, it uses
 # backtracking and tries the next polygonal number.
 def find_set(step):
-    global polygonal
-    global chain
-    global flags
     global sum_
 
 #   Use one polygonal number per type, starting from triangular.
@@ -49,7 +53,7 @@ def find_set(step):
 #                   If it's the first number, just add it as first step in the chain.
                     if step == 0:
                         chain[step] = j
-                        sum_ = sum_ + j
+                        sum_ += j
 
 #                       Recursively try to add other numbers to the chain. If a solution
 #                       is found, return 1.
@@ -58,42 +62,34 @@ def find_set(step):
 
 #                       If a solution was not found, backtrack, subtracting the value of
 #                       the number from the total.
-                        sum_ = sum_ - j
+                        sum_ -= j
 #                   If this is the last step and the current number can be added to the chain,
 #                   add it, update the sum and return 1. A solution has been found.
                     elif step == 5 and j % 100 == chain[0] // 100 and j // 100 == chain[step-1] % 100:
                         chain[step] = j
-                        sum_ = sum_ + j
+                        sum_ += j
                         return 1
 #                   For every other step, add the number to the chain if possible, then recursively
 #                   try to add other numbers.
                     elif step < 5 and j // 100 == chain[step-1] % 100:
                         chain[step] = j
-                        sum_ = sum_ + j
+                        sum_ += + j
 
                         if find_set(step+1):
                             return 1
 
 #                       If a solution was not found, backtrack.
-                        sum_ = sum_ - j
+                        sum_ -= j
 
 #           Remove the flag for the current polygonal type.
             flags[i] = 0
 
     return 0
 
+
 def main():
     start = default_timer()
 
-    global polygonal
-    global chain
-    global flags
-    global sum_
-
-    polygonal = zeros((6, 10000), int)
-    chain = [0] * 6
-    flags = [0] * 6
-    sum_ = 0
     i = 1
     n = 1
 
@@ -172,9 +168,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 61')
-    print('Answer: {}'.format(sum_))
+    print(f'Answer: {sum_}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p062.py

@@ -5,9 +5,10 @@
 #
 # Find the smallest cube for which exactly five permutations of its digits are cube.
 
+from timeit import default_timer
+
 from numpy import zeros
 
-from timeit import default_timer
 
 def main():
     start = default_timer()
@@ -45,9 +46,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 62')
-    print('Answer: {}'.format(cubes[i]))
+    print(f'Answer: {cubes[i]}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p063.py

@@ -6,6 +6,7 @@
 
 from timeit import default_timer
 
+
 def main():
     start = default_timer()
 
@@ -30,9 +31,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 63')
-    print('Answer: {}'.format(count))
+    print(f'Answer: {count}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p064.py

@@ -42,19 +42,18 @@
 #
 # How many continued fractions for N≤10000 have an odd period?
 
-from math import floor, sqrt
-
+from math import sqrt
 from timeit import default_timer
+
 from projecteuler import build_sqrt_cont_fraction
 
+
 def is_square(n):
     p = sqrt(n)
     m = int(p)
 
-    if p == m:
-        return True
-    else:
-        return False
+    return bool(p == m)
+
 
 def main():
     start = default_timer()
@@ -74,9 +73,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 64')
-    print('Answer: {}'.format(count))
+    print(f'Answer: {count}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p065.py

@@ -31,6 +31,7 @@
 
 from timeit import default_timer
 
+
 def main():
     start = default_timer()
 
@@ -41,10 +42,10 @@ def main():
     n1 = 8
     n2 = 11
 
-#   For a continued fractions [a_0; a_1, a_2, ...], the numerator of the 
+#   For a continued fractions [a_0; a_1, a_2, ...], the numerator of the
 #   next convergent N_n=a_n*N_(n-1)+N_(n-2). The first three values for e are
 #   3, 8 and 11, the next ones are easily calculated, considering that a_n
-#   follows a simple pattern: 
+#   follows a simple pattern:
 #   a_1=1, a_2=2, a_3=1
 #   a_4=1, a_5=4, a_6=1
 #   a_7=1, a_8=6, a_9=1
@@ -70,9 +71,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 65')
-    print('Answer: {}'.format(sum_))
+    print(f'Answer: {sum_}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p066.py

@@ -21,18 +21,17 @@
 # Find the value of D ≤ 1000 in minimal solutions of x for which the largest value of x is obtained.
 
 from math import sqrt
-
 from timeit import default_timer
+
 from projecteuler import pell_eq
 
+
 def is_square(n):
     p = sqrt(n)
     m = int(p)
 
-    if p == m:
-        return True
-    else:
-        return False
+    return bool(p == m)
+
 
 def main():
     start = default_timer()
@@ -52,9 +51,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 66')
-    print('Answer: {}'.format(max_d))
+    print(f'Answer: {max_d}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p067.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.
 #
@@ -15,24 +15,24 @@
 # If you could check one trillion (1012) routes every second it would take over twenty billion years to check them all.
 # There is an efficient algorithm to solve it. ;o)
 
+import sys
 from timeit import default_timer
+
 from projecteuler import find_max_path
 
+
 def main():
     start = default_timer()
 
+    triang = []
+
     try:
-        fp = open('triangle.txt', 'r')
-    except:
-        print('Error while opening file {}'.format('triangle.txt'))
-        exit(1)
-
-    triang = list()
-
-    for line in fp:
-        triang.append(line.strip('\n').split())
-
-    fp.close()
+        with open('triangle.txt', 'r', encoding='utf-8') as fp:
+            for line in fp:
+                triang.append(line.strip('\n').split())
+    except FileNotFoundError:
+        print('Error while opening file triangle.txt')
+        sys.exit(1)
 
     l = len(triang)
 
@@ -45,9 +45,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 67')
-    print('Answer: {}'.format(max_))
+    print(f'Answer: {max_}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p068.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # Consider the following "magic" 3-gon ring, filled with the numbers 1 to 6, and each line adding to nine.
 #
@@ -46,9 +46,9 @@
 #
 
 from itertools import permutations
-
 from timeit import default_timer
 
+
 # Function to evaluate the ring. The ring is represented as a vector of 2*n elements,
 # the first n elements represent the external nodes of the ring, the next n elements
 # represent the internal ring.
@@ -91,8 +91,9 @@ def eval_ring(ring, n):
 
     return res
 
+
 def list_to_int(l):
-    if l == None:
+    if l is None:
         return 0
 
     res = 0
@@ -108,6 +109,7 @@ def list_to_int(l):
 
     return res
 
+
 def main():
     start = default_timer()
 
@@ -118,8 +120,8 @@ def main():
     n = None
 
     for ring in rings:
-        eval_ = eval_ring(ring, 5);
-   
+        eval_ = eval_ring(ring, 5)
+
 #       Convert the list into an integer number.
         n = list_to_int(eval_)
 
@@ -129,9 +131,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 68')
-    print('Answer: {}'.format(max_))
+    print(f'Answer: {max_}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p069.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # 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.
@@ -19,8 +19,10 @@
 # Find the value of n ≤ 1,000,000 for which n/φ(n) is a maximum.
 
 from timeit import default_timer
+
 from projecteuler import is_prime
 
+
 def main():
     start = default_timer()
 
@@ -47,9 +49,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 69')
-    print('Answer: {}'.format(res))
+    print(f'Answer: {res}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p070.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # Euler's Totient function, φ(n) [sometimes called the phi function], is used to determine the number of positive numbers
 # less than or equal to n which are relatively prime to n. For example, as 1, 2, 4, 5, 7, and 8, are all less than nine and
@@ -9,11 +9,11 @@
 #
 # Find the value of n, 1 < n < 10^7, for which φ(n) is a permutation of n and the ratio n/φ(n) produces a minimum.
 
-from numpy import zeros
-
 from timeit import default_timer
+
 from projecteuler import sieve, is_semiprime, phi_semiprime
 
+
 def main():
     start = default_timer()
 
@@ -26,14 +26,14 @@ def main():
 
     for i in range(2, N):
 #       When n is prime, phi(n)=(n-1), so to minimize n/phi(n) we should
-#       use n prime. But n-1 can't be a permutation of n. The second best 
+#       use n prime. But n-1 can't be a permutation of n. The second best
 #       bet is to use semiprimes. For a semiprime n=p*q, phi(n)=(p-1)(q-1).
 #       So we check if a number is semiprime, if yes calculate phi, finally
 #       check if phi(n) is a permutation of n and update the minimum if it's
 #       smaller.
         semi_p, a, b = is_semiprime(i, primes)
-        
-        if semi_p == True:
+
+        if semi_p is True:
             p = phi_semiprime(i, a, b)
 
             if ''.join(sorted(str(p))) == ''.join(sorted(str(i))) and i / p < min_:
@@ -43,9 +43,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 70')
-    print('Answer: {}'.format(n))
+    print(f'Answer: {n}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p071.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # 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.
 #
@@ -12,9 +12,9 @@
 # of the fraction immediately to the left of 3/7.
 
 from math import gcd
-
 from timeit import default_timer
 
+
 def main():
     start = default_timer()
 
@@ -46,9 +46,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 71')
-    print('Answer: {}'.format(max_n))
+    print(f'Answer: {max_n}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p072.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # 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.
 #
@@ -11,8 +11,10 @@
 # How many elements would be contained in the set of reduced proper fractions for d ≤ 1,000,000?
 
 from timeit import default_timer
+
 from projecteuler import sieve, phi
 
+
 def main():
     start = default_timer()
 
@@ -21,7 +23,7 @@ def main():
     count = 0
     primes = sieve(N)
 
-#   For any denominator d, the number of reduced proper fractions is 
+#   For any denominator d, the number of reduced proper fractions is
 #   the number of fractions n/d where gcd(n, d)=1, which is the definition
 #   of Euler's Totient Function phi. It's sufficient to calculate phi for each
 #   denominator and sum the value.
@@ -31,9 +33,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 72')
-    print('Answer: {}'.format(count))
+    print(f'Answer: {count}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p073.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # 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.
 #
@@ -11,9 +11,9 @@
 # How many fractions lie between 1/3 and 1/2 in the sorted set of reduced proper fractions for d ≤ 12,000?
 
 from math import gcd
-
 from timeit import default_timer
 
+
 def main():
     start = default_timer()
 
@@ -33,9 +33,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 73')
-    print('Answer: {}'.format(count))
+    print(f'Answer: {count}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p074.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # The number 145 is well known for the property that the sum of the factorial of its digits is equal to 145:
 #
@@ -23,13 +23,14 @@
 # How many chains, with a starting number below one million, contain exactly sixty non-repeating terms?
 
 from math import factorial
-
 from timeit import default_timer
 
-def len_chain(n):
-    global N
-    global chains
 
+N = 1000000
+chains = [0] * N
+
+
+def len_chain(n):
     chain = [0] * 60
     count = 0
     finished = 0
@@ -68,16 +69,10 @@ def len_chain(n):
 
     return count
 
+
 def main():
     start = default_timer()
 
-    global N
-    global chains
-
-    N = 1000000
-
-    chains = [0] * N
-
     count = 0
 
 #   Simple brute force approach, for every number calculate
@@ -89,9 +84,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 74')
-    print('Answer: {}'.format(count))
+    print(f'Answer: {count}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p075.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # It turns out that 12 cm is the smallest length of wire that can be bent to form an integer sided right angle triangle in exactly one way,
 # but there are many more examples.
@@ -18,9 +18,9 @@
 # Given that L is the length of the wire, for how many values of L ≤ 1,500,000 can exactly one integer sided right angle triangle be formed?
 
 from math import gcd
-
 from timeit import default_timer
 
+
 def main():
     start = default_timer()
 
@@ -46,7 +46,7 @@ def main():
                 b = 2 * m * n
                 c = m * m + n * n
 
-                if(a + b + c <= N):
+                if a + b + c <= N:
                     l[a+b+c] = l[a+b+c] + 1
 
                 i = 2
@@ -54,7 +54,7 @@ def main():
                 tmpb = i * b
                 tmpc = i * c
 
-                while(tmpa + tmpb + tmpc <= N):
+                while tmpa + tmpb + tmpc <= N:
                     l[tmpa+tmpb+tmpc] = l[tmpa+tmpb+tmpc] + 1
 
                     i = i + 1
@@ -71,9 +71,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 75')
-    print('Answer: {}'.format(count))
+    print(f'Answer: {count}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p076.py

@@ -1,8 +1,21 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
+
+# It is possible to write five as a sum in exactly six different ways:
+#
+# 4 + 1
+# 3 + 2
+# 3 + 1 + 1
+# 2 + 2 + 1
+# 2 + 1 + 1 + 1
+# 1 + 1 + 1 + 1 + 1
+#
+# How many different ways can one hundred be written as a sum of at least two positive integers?*/
 
 from timeit import default_timer
+
 from projecteuler import partition_fn
 
+
 def main():
     start = default_timer()
 
@@ -16,9 +29,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 76')
-    print('Answer: {}'.format(n))
+    print(f'Answer: {n}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p077.py

@@ -11,33 +11,34 @@
 # What is the first value which can be written as the sum of primes in over five thousand different ways?
 
 from timeit import default_timer
+
 from projecteuler import is_prime
 
+
+primes = [0] * 100
+
+
 # Function using a simple recursive brute force approach
 # to find all the partitions.
 def count(value, n, i, target):
-    global primes
-
     for j in range(i, 100):
         value = value + primes[j]
 
         if value == target:
             return n + 1
-        elif value > target:
+
+        if value > target:
             return n
-        else:
-            n = count(value, n, j, target)
-            value = value - primes[j]
+
+        n = count(value, n, j, target)
+        value = value - primes[j]
 
     return n
 
+
 def main():
     start = default_timer()
 
-    global primes
-
-    primes = [0] * 100
-
 #   Generate a list of the first 100 primes.
     i = 0
     j = 0
@@ -64,9 +65,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 77')
-    print('Answer: {}'.format(i))
+    print(f'Answer: {i}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p078.py

@@ -3,19 +3,21 @@
 # Let p(n) represent the number of different ways in which n coins can be separated into piles.
 # For example, five coins can be separated into piles in exactly seven different ways, so p(5)=7.
 #
-#       OOOOO
-#      OOOO  O
-#      OOO  OO
-#     OOO  O  O
-#     OO  OO  O
-#    OO  O  O  O
+#   OOOOO
+#   OOOO  O
+#   OOO  OO
+#   OOO  O  O
+#   OO  OO  O
+#   OO  O  O  O
 #   O  O  O  O  O
 #
 # Find the least value of n for which p(n) is divisible by one million.
 
 from timeit import default_timer
+
 from projecteuler import partition_fn
 
+
 def main():
     start = default_timer()
 
@@ -26,16 +28,17 @@ def main():
     i = 0
 
 #   Using the partition function to calculate the number of partitions,
-#   giving the result modulo N.*/   
+#   giving the result modulo N.*/
     while partition_fn(i, partitions, N) != 0:
         i = i + 1
 
     end = default_timer()
 
     print('Project Euler, Problem 78')
-    print('Answer: {}'.format(i))
+    print(f'Answer: {i}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p079.py

@@ -8,13 +8,14 @@
 # Given that the three characters are always asked for in order, analyse the file so as to determine the shortest possible
 # secret passcode of unknown length.
 
+import sys
 from itertools import permutations
-
 from timeit import default_timer
 
+
 def check_passcode(passcode, len_, logins, n):
 #   For every login attempt, check if all the digits appear in the
-#   passcode in the correct order. Return 0 if a login attempt 
+#   passcode in the correct order. Return 0 if a login attempt
 #   incompatible with the current passcode is found.
     for i in range(n):
         k = 0
@@ -30,28 +31,26 @@ def check_passcode(passcode, len_, logins, n):
 
     return 1
 
+
 def main():
     start = default_timer()
 
     try:
-        fp = open('keylog.txt', 'r')
-    except:
-        print('Error while opening file {}'.format('keylog.txt'))
-        exit(1)
-
-    logins = fp.readlines()
-
-    fp.close()
+        with open('keylog.txt', 'r', encoding='utf-8') as fp:
+            logins = fp.readlines()
+    except FileNotFoundError:
+        print('Error while opening file keylog.txt')
+        sys.exit(1)
 
     digits = [0] * 10
     passcode_digits = [0] * 10
 
     for i in logins:
         keylog = int(i)
-       
+
 #       Check which digits are present in the login attempts.
         while True:
-            digits[keylog%10] = digits[keylog%10] + 1
+            digits[keylog % 10] = digits[keylog % 10] + 1
             keylog = keylog // 10
 
             if keylog == 0:
@@ -97,9 +96,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 79')
-    print('Answer: {}'.format(res))
+    print(f'Answer: {res}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p080.py

@@ -8,18 +8,16 @@
 # For the first one hundred natural numbers, find the total of the digital sums of the first one hundred decimal digits
 # for all the irrational square roots
 
+from timeit import default_timer
 from mpmath import sqrt, mp
 
-from timeit import default_timer
 
 def is_square(n):
     p = sqrt(n)
     m = int(p)
 
-    if p == m:
-        return True
-    else:
-        return False
+    return bool(p == m)
+
 
 def main():
     start = default_timer()
@@ -43,9 +41,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 80')
-    print('Answer: {}'.format(sum_))
+    print(f'Answer: {sum_}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p081.py

@@ -12,8 +12,10 @@
 # Find the minimal path sum, in matrix.txt, a 31K text file containing a 80 by 80 matrix, from the top left to the bottom right
 # by only moving right and down.
 
+import sys
 from timeit import default_timer
 
+
 def sum_paths(matrix, m, n):
     for i in range(m-2, -1, -1):
         matrix[m-1][i] = matrix[m-1][i] + matrix[m-1][i+1]
@@ -27,21 +29,20 @@ def sum_paths(matrix, m, n):
                 matrix[i][j] = matrix[i][j] + matrix[i][j+1]
 
     return matrix[0][0]
-    
+
+
 def main():
     start = default_timer()
 
     try:
-        fp = open('matrix.txt', 'r')
-    except:
-        print('Error while opening file {}'.format('matrix.txt'))
-        exit(1)
-
-    matrix = fp.readlines()
-
-    fp.close()
+        with open('matrix.txt', 'r', encoding='utf-8') as fp:
+            matrix = fp.readlines()
+    except FileNotFoundError:
+        print('Error while opening file matrix.txt')
+        sys.exit(1)
 
     j = 0
+
     for i in matrix:
         matrix[j] = list(map(int, i.strip().split(',')))
         j = j + 1
@@ -51,9 +52,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 81')
-    print('Answer: {}'.format(dist))
+    print(f'Answer: {dist}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p082.py

@@ -15,24 +15,28 @@
 #
 # Find the minimal path sum, in matrix.txt, a 31K text file containing a 80 by 80 matrix, from the left column to the right column.
 
+import sys
+from timeit import default_timer
+
 from numpy import zeros
 
-from timeit import default_timer
 from projecteuler import dijkstra
 
+
 def main():
     start = default_timer()
 
     try:
-        fp = open('matrix.txt', 'r')
-    except:
-        print('Error while opening file {}'.format('matrix.txt'))
-        exit(1)
+        with open('matrix.txt', 'r', encoding='utf-8') as fp:
+            matrix = fp.readlines()
+    except FileNotFoundError:
+        print('Error while opening file matrix.txt')
+        sys.exit(1)
 
-    matrix = fp.readlines()
     distances = zeros((80, 80), int)
 
     j = 0
+
     for i in matrix:
         matrix[j] = list(map(int, i.strip().split(',')))
         j = j + 1
@@ -53,9 +57,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 82')
-    print('Answer: {}'.format(min_path))
+    print(f'Answer: {min_path}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p083.py

@@ -14,21 +14,24 @@
 # Find the minimal path sum, in matrix.txt, a 31K text file containing a 80 by 80 matrix,
 # from the top left to the bottom right by moving left, right, up, and down.
 
+import sys
+from timeit import default_timer
+
 from numpy import zeros
 
-from timeit import default_timer
 from projecteuler import dijkstra
 
+
 def main():
     start = default_timer()
 
     try:
-        fp = open('matrix.txt', 'r')
-    except:
-        print('Error while opening file {}'.format('matrix.txt'))
-        exit(1)
+        with open('matrix.txt', 'r', encoding='utf-8') as fp:
+            matrix = fp.readlines()
+    except FileNotFoundError:
+        print('Error while opening file matrix.txt')
+        sys.exit(1)
 
-    matrix = fp.readlines()
     distances = zeros((80, 80), int)
 
     j = 0
@@ -36,16 +39,17 @@ def main():
         matrix[j] = list(map(int, i.strip().split(',')))
         j = j + 1
 
-    dijkstra(matrix, distances, 80, 80, 1, 1) 
+    dijkstra(matrix, distances, 80, 80, 1, 1)
 
     dist = distances[79][79]
 
     end = default_timer()
 
     print('Project Euler, Problem 83')
-    print('Answer: {}'.format(dist))
+    print(f'Answer: {dist}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p104.py

@@ -1,3 +1,5 @@
+#! /usr/bin/env python3
+
 # The Fibonacci sequence is defined by the recurrence relation:
 #
 # F_n = F_n−1 + F_n−2, where F_1 = 1 and F_2 = 1.
@@ -7,13 +9,17 @@
 #
 # Given that F_k is the first Fibonacci number for which the first nine digits AND the last nine digits are 1-9 pandigital, find k.
 
-#!/usr/bin/python
-
-from mpmath import matrix, mp
-
+import sys
 from timeit import default_timer
+
+from mpmath import matrix
+
 from projecteuler import is_pandigital
 
+
+sys.set_int_max_str_digits(70000)
+
+
 def main():
     start = default_timer()
 
@@ -45,13 +51,14 @@ def main():
 
             if is_pandigital(int(str(fib)[:9]), 9):
                 found = 1
- 
+
     end = default_timer()
 
     print('Project Euler, Problem 104')
-    print('Answer: {}'.format(i))
+    print(f'Answer: {i}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p145.py

@@ -10,11 +10,12 @@
 
 from timeit import default_timer
 
+
 def main():
     start = default_timer()
 
-    N = 10000000
-    
+    N = 1000000000
+
     count = 0
 
 #   Brute force approach, sum each number and their reverse and
@@ -23,16 +24,17 @@ def main():
         if i % 10 != 0:
             s = str(i + int(''.join(reversed(str(i)))))
 
-            if not '0' in s and not '2' in s and not '4' in s and\
-                    not '6' in s and not '8' in s:
+            if '0' not in s and '2' not in s and '4' not in s and\
+                    '6' not in s and '8' not in s:
                 count = count + 1
 
     end = default_timer()
 
     print('Project Euler, Problem 145')
-    print('Answer: {}'.format(count))
+    print(f'Answer: {count}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()

Python/p206.py

@@ -1,10 +1,11 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 # Find the unique positive integer whose square has the form 1_2_3_4_5_6_7_8_9_0,
 # where each “_” is a single digit.
 
 from timeit import default_timer
 
+
 def main():
     start = default_timer()
 
@@ -37,10 +38,10 @@ def main():
     end = default_timer()
 
     print('Project Euler, Problem 206')
-    print('Answer: {}'.format(n))
+    print(f'Answer: {n}')
+
+    print(f'Elapsed time: {end - start:.9f} seconds')
 
-    print('Elapsed time: {:.9f} seconds'.format(end - start))
 
 if __name__ == '__main__':
     main()
-

Python/projecteuler.py

@@ -1,4 +1,4 @@
-#!/usr/bin/python3
+#!/usr/bin/env python3
 
 from math import sqrt, floor, ceil, gcd
 

README.md

@@ -1,3 +1,6 @@
 # My Project Euler solutions
 These are my solutions in C and Python, not necessarily the best solutions. I've solved most of the first 100 problems, currently working on cleaning the code and uploading it. I will try to solve more problems in the future.
 
+# Issues
+- Solutions for problems 82 and 145 in Python run really slow.
+- Solutions for problems 84, 85, 86, 87, 89, 92, 95, 96, 97. 99, 102, 112, 124 and 357 have been implemented in C but not in Python.