Use timing decorator for problems 41-60

Python/p041.py

@@ -5,34 +5,27 @@
 #
 # What is the largest n-digit pandigital prime that exists?
 
-from timeit import default_timer
+from projecteuler import is_pandigital, is_prime, timing
-
-from projecteuler import is_pandigital, is_prime
 
 
-def main():
+@timing
-    start = default_timer()
+def p041():
-
+    # 8- and 9-digit pandigital numbers can't be prime, because
-#   8- and 9-digit pandigital numbers can't be prime, because
+    # 1+2+...+8=36, which is divisible by 3, and 36+9=45 which is
-#   1+2+...+8=36, which is divisible by 3, and 36+9=45 which is
+    # also divisible by 3, and therefore the whole number is divisible
-#   also divisible by 3, and therefore the whole number is divisible
+    # by 3. So we can start from the largest 7-digit pandigital number,
-#   by 3. So we can start from the largest 7-digit pandigital number,
+    # until we find a prime.
-#   until we find a prime.
     i = 7654321
 
     while i > 0:
         if is_pandigital(i, len(str(i))) and is_prime(i):
             break
-#       Skipping the even numbers.
+        # Skipping the even numbers.
         i = i - 2
 
-    end = default_timer()
-
     print('Project Euler, Problem 41')
     print(f'Answer: {i}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p041()

Python/p042.py

@@ -11,7 +11,8 @@
 # how many are triangle words?
 
 import sys
-from timeit import default_timer
+
+from projecteuler import timing
 
 
 def is_triang(n):
@@ -27,9 +28,8 @@ def is_triang(n):
     return False
 
 
-def main():
+@timing
-    start = default_timer()
+def p042():
-
     try:
         with open('p042_words.txt', 'r', encoding='utf-8') as fp:
             words = list(fp.readline().replace('"', '').split(','))
@@ -39,7 +39,7 @@ def main():
 
     count = 0
 
-#   For each word, calculate its value and check if it's a triangle number.
+    # For each word, calculate its value and check if it's a triangle number.
     for word in words:
         value = 0
         l = len(word)
@@ -50,13 +50,9 @@ def main():
         if is_triang(value):
             count = count + 1
 
-    end = default_timer()
-
     print('Project Euler, Problem 42')
     print(f'Answer: {count}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p042()

Python/p043.py

@@ -16,7 +16,8 @@
 # Find the sum of all 0 to 9 pandigital numbers with this property.
 
 from itertools import permutations
-from timeit import default_timer
+
+from projecteuler import timing
 
 
 # Function to check if the value has the desired property.
@@ -59,26 +60,21 @@ def has_property(n):
     return True
 
 
-def main():
+@timing
-    start = default_timer()
+def p043():
-
+    # Find all the permutations
-#   Find all the permutations
     perm = list(permutations('0123456789'))
 
     sum_ = 0
 
-#   For each permutation, check if it has the required property
+    # For each permutation, check if it has the required property
     for i in perm:
         if has_property(i):
             sum_ = sum_ + int(''.join(map(str, i)))
 
-    end = default_timer()
-
     print('Project Euler, Problem 43')
     print(f'Answer: {sum_}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p043()

Python/p044.py

@@ -9,20 +9,17 @@
 # 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 timeit import default_timer
+from projecteuler import is_pentagonal, timing
-
-from projecteuler import is_pentagonal
 
 
-def main():
+@timing
-    start = default_timer()
+def p044():
-
     found = 0
     n = 2
 
-#   Check all couples of pentagonal numbers until the right one
+    # Check all couples of pentagonal numbers until the right one
-#   is found. Use the function implemented in projecteuler.py to
+    # is found. Use the function implemented in projecteuler.py to
-#   check if the sum and difference ot the two numbers is pentagonal.
+    # check if the sum and difference ot the two numbers is pentagonal.
     while not found:
         pn = n * (3 * n - 1) // 2
 
@@ -35,13 +32,9 @@ def main():
 
         n = n + 1
 
-    end = default_timer()
-
     print('Project Euler, Problem 44')
     print(f'Answer: {pn - pm}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p044()

Python/p045.py

@@ -10,34 +10,27 @@
 #
 # Find the next triangle number that is also pentagonal and hexagonal.
 
-from timeit import default_timer
+from projecteuler import is_pentagonal, timing
-
-from projecteuler import is_pentagonal
 
 
-def main():
+@timing
-    start = default_timer()
+def p045():
-
     found = 0
     i = 143
 
     while not found:
         i = i + 1
-#       Hexagonal numbers are also triangle numbers, so it's sufficient
+        # Hexagonal numbers are also triangle numbers, so it's sufficient
-#       to generate hexagonal numbers (starting from H_144) and check if
+        # to generate hexagonal numbers (starting from H_144) and check if
-#       they're also pentagonal.
+        # they're also pentagonal.
         n = i * (2 * i - 1)
 
         if is_pentagonal(n):
             found = 1
 
-    end = default_timer()
-
     print('Project Euler, Problem 45')
     print(f'Answer: {n}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p045()

Python/p046.py

@@ -13,9 +13,7 @@
 #
 # 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, timing
-
-from projecteuler import sieve
 
 
 N = 10000
@@ -23,13 +21,13 @@ primes = sieve(N)
 
 
 def goldbach(n):
-#   Check every prime smaller than n.
+    # Check every prime smaller than n.
     for i in range(3, n, 2):
         if primes[i] == 1:
             j = 1
 
-#           Check if summing twice a square to the prime number
+            # Check if summing twice a square to the prime number
-#           gives n. Return 1 when succeeding.
+            # gives n. Return 1 when succeeding.
             while True:
                 tmp = i + 2 * j * j
 
@@ -41,31 +39,27 @@ def goldbach(n):
                 if tmp >= n:
                     break
 
-#   Return 0 if no solution is found.
+    # Return 0 if no solution is found.
     return False
 
-def main():
-    start = default_timer()
 
+@timing
+def p046():
     found = 0
     i = 3
 
-#   For every odd number, check if it's prime, if it is check
+    # For every odd number, check if it's prime, if it is check
-#   if it satisfies the Goldbach property. Continue until the
+    # if it satisfies the Goldbach property. Continue until the
-#   first number that doesn't is found.
+    # first number that doesn't is found.
     while not found and i < N:
         if primes[i] == 0:
             if not goldbach(i):
                 found = 1
         i = i + 2
 
-    end = default_timer()
-
     print('Project Euler, Problem 46')
     print(f'Answer: {i - 2}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p046()

Python/p047.py

@@ -13,7 +13,7 @@
 #
 # Find the first four consecutive integers to have four distinct prime factors each. What is the first of these numbers?
 
-from timeit import default_timer
+from projecteuler import timing
 
 
 # Function using a modified sieve of Eratosthenes to count
@@ -31,17 +31,16 @@ def count_factors(n):
     return factors
 
 
-def main():
+@timing
-    start = default_timer()
+def p047():
-
     N = 150000
 
     factors = count_factors(N)
 
     count = 0
 
-#   Find the first instance of four consecutive numbers
+    # Find the first instance of four consecutive numbers
-#   having four distinct prime factors.
+    # having four distinct prime factors.
     for i in range(N):
         if factors[i] == 4:
             count = count + 1
@@ -52,13 +51,9 @@ def main():
             res = i - 3
             break
 
-    end = default_timer()
-
     print('Project Euler, Problem 47')
     print(f'Answer: {res}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p047()

Python/p048.py

@@ -4,26 +4,21 @@
 #
 # Find the last ten digits of the series, 1^1 + 2^2 + 3^3 + ... + 1000^1000.
 
-from timeit import default_timer
+from projecteuler import timing
 
 
-def main():
+@timing
-    start = default_timer()
+def p048():
-
     sum_ = 0
 
-#   Simply calculate the sum of the powers
+    # Simply calculate the sum of the powers
     for i in range(1, 1001):
         power = i ** i
         sum_ = sum_ + power
 
-    end = default_timer()
-
     print('Project Euler, Problem 48')
     print(f'Answer: {str(sum_)[-10:]}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p048()

Python/p049.py

@@ -8,14 +8,11 @@
 #
 # What 12-digit number do you form by concatenating the three terms in this sequence?
 
-from timeit import default_timer
+from projecteuler import sieve, timing
-
-from projecteuler import sieve
 
 
-def main():
+@timing
-    start = default_timer()
+def p049():
-
     N = 10000
 
     primes = sieve(N)
@@ -23,15 +20,14 @@ def main():
     found = 0
     i = 1489
 
-#   Starting from i=1489 (bigger than the first number in the sequence given in the problem),
+    # Starting from i=1489 (bigger than the first number in the sequence given in the problem),
-#   check odd numbers. If they're prime, loop on even numbers j (odd+even=odd, odd+odd=even and
+    # check odd numbers. If they're prime, loop on even numbers j (odd+even=odd, odd+odd=even and
-#   we need odd numbers because we're looking for primes) up to 4254 (1489+2*4256=10001 which has
+    # we need odd numbers because we're looking for primes) up to 4254 (1489+2*4256=10001 which has
-#   5 digits.
+    # 5 digits.
     while i < N:
         if primes[i] == 1:
             for j in range(1, 4255):
-#               If i, i+j and i+2*j are all primes and they have
+                # If i, i+j and i+2*j are all primes and they have all the same digits, the result has been found.
-#               all the same digits, the result has been found.
                 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))):
@@ -42,13 +38,9 @@ def main():
 
         i = i + 2
 
-    end = default_timer()
-
     print('Project Euler, Problem 49')
     print(f'Answer: {str(i)+str(i+j)+str(i+2*j)}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p049()

Python/p050.py

@@ -10,14 +10,11 @@
 #
 # 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, timing
-
-from projecteuler import sieve
 
 
-def main():
+@timing
-    start = default_timer()
+def p050():
-
     N = 1000000
 
     primes = sieve(N)
@@ -25,12 +22,12 @@ def main():
     max_ = 0
     max_p = 0
 
-#   Starting from a prime i, add consecutive primes until the
+    # Starting from a prime i, add consecutive primes until the
-#   sum exceeds the limit, every time the sum is also a prime
+    # sum exceeds the limit, every time the sum is also a prime
-#   save the value and the count if the count is larger than the
+    # save the value and the count if the count is larger than the
-#   current maximum. Repeat for all primes below N.
+    # current maximum. Repeat for all primes below N.
-#   A separate loop is used for i=2, so later only odd numbers are
+    # A separate loop is used for i=2, so later only odd numbers are
-#   checked for primality.
+    # checked for primality.
     i = 2
     j = i + 1
     count = 1
@@ -63,13 +60,9 @@ def main():
 
                 j = j + 2
 
-    end = default_timer()
-
     print('Project Euler, Problem 50')
     print(f'Answer: {max_p}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p050()

Python/p051.py

@@ -9,13 +9,10 @@
 # Find the smallest prime which, by replacing part of the number (not necessarily adjacent digits) with the same digit, is part of an eight prime
 # value family.
 
-from timeit import default_timer
+from projecteuler import sieve, timing
-
-from projecteuler import sieve
 
 
 N = 1000000
-
 # N set to 1000000 as a reasonable limit, which turns out to be enough.
 primes = sieve(N)
 
@@ -38,8 +35,8 @@ def replace(n):
 
     for i in range(l-3):
         for j in range(i+1, l-2):
-#             Replacing the last digit can't give 8 primes, because at least
+            # Replacing the last digit can't give 8 primes, because at least
-#             six of the numbers obtained will be divisible by 2 and/or 5.
+            # six of the numbers obtained will be divisible by 2 and/or 5.
             for k in range(j+1, l-1):
                 count = 0
 
@@ -64,32 +61,28 @@ def replace(n):
 
     return max_
 
-def main():
-    start = default_timer()
 
-#   Checking only odd numbers with at least 4 digits.
+@timing
+def p051():
+    # 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
+        # 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
+        # repeated digits, otherwise we can't get a 8 number family (there
-#       are only 7 other digits). Also, te number of repeated digits must
+        # 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.
+        # 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
+        # So the smallest number of this family must have three 0s, three 1s or
-#       three 2s.
+        # 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
         i = i + 2
 
-    end = default_timer()
-
     print('Project Euler, Problem 51')
     print(f'Answer: {i}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p051()

Python/p052.py

@@ -4,15 +4,14 @@
 #
 # Find the smallest positive integer, x, such that 2x, 3x, 4x, 5x, and 6x, contain the same digits.
 
-from timeit import default_timer
+from projecteuler import timing
 
 
-def main():
+@timing
-    start = default_timer()
+def p052():
-
     i = 1
 
-#   Brute force approach, try every integer number until the desired one is found.
+    # Brute force approach, try every integer number until the desired one is found.
     while True:
         if ''.join(sorted(str(i))) == ''.join(sorted(str(2*i))) and\
                 ''.join(sorted(str(i))) == ''.join(sorted(str(3*i))) and\
@@ -22,13 +21,9 @@ def main():
             break
         i = i + 1
 
-    end = default_timer()
-
     print('Project Euler, Problem 52')
     print(f'Answer: {i}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p052()

Python/p053.py

@@ -12,31 +12,26 @@
 #
 # 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 projecteuler import timing
 
-def main():
-    start = default_timer()
 
+@timing
+def p053():
     LIMIT = 1000000
 
     count = 0
 
-#   Use the scipy comb function to calculate the binomial values
+    # Use the scipy comb function to calculate the binomial values
     for i in range(23, 101):
         for j in range(1, i+1):
             if comb(i, j) > LIMIT:
                 count = count + 1
 
-    end = default_timer()
-
     print('Project Euler, Problem 53')
     print(f'Answer: {count}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p053()

Python/p054.py

@@ -48,7 +48,8 @@
 
 import sys
 from enum import IntEnum
-from timeit import default_timer
+
+from projecteuler import timing
 
 
 class Value(IntEnum):
@@ -111,7 +112,7 @@ class Hand():
     def __init__(self, *args, **kwargs):
         super().__init__(*args, **kwargs)
 
-        self.cards = list()
+        self.cards = []
 
     def sort(self):
         self.cards.sort(key=lambda x: x.value)
@@ -126,8 +127,7 @@ class Game():
         self.hand2 = None
 
     def play(self):
-
+        # If player 1 has a Royal Flush, player 1 wins.
-#       If player 1 has a Royal Flush, player 1 wins.
         if self.hand1.cards[4].value == Value.Ace and self.hand1.cards[3].value == Value.King and\
                 self.hand1.cards[2].value == Value.Queen and self.hand1.cards[1].value == Value.Jack and\
                 self.hand1.cards[0] == Value.Ten and self.hand1.cards[0].suit == self.hand1.cards[1].suit and\
@@ -136,7 +136,7 @@ class Game():
                 self.hand1.cards[0].suit == self.hand1.cards[4].suit:
             return 1
 
-#       If player 2 has a Royal Flush, player 2 wins.
+        # 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\
                 self.hand2.cards[0] == Value.Ten and self.hand2.cards[0].suit == self.hand2.cards[1].suit and\
@@ -148,7 +148,7 @@ class Game():
         straightflush1 = 0
         straightflush2 = 0
 
-#       Check if player 1 has a straight flush.
+        # 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
@@ -162,7 +162,7 @@ class Game():
                     straightflush1 = 0
                     break
 
-#       Check if player 2 has a straight flush.
+        # 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
@@ -176,17 +176,17 @@ class Game():
                     straightflush2 = 0
                     break
 
-#       If player 1 has a straight flush and player 2 doesn't, player 1 wins
+        # 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 player 2 has a straight flush and player 1 doesn't, player 2 wins
         if not straightflush1 and straightflush2:
             return -1
 
-#       If both players have a straight flush, the one with the highest value wins.
+        # If both players have a straight flush, the one with the highest value wins.
-#       Any card can be compared, because in the straight flush the values are consecutive
+        # Any card can be compared, because in the straight flush the values are consecutive
-#       by definition.
+        # by definition.
         if straightflush1 and straightflush2:
             if self.hand1.cards[0].value > self.hand2.cards[0].value:
                 return 1
@@ -196,24 +196,24 @@ class Game():
         four1 = 0
         four2 = 0
 
-#       Check if player 1 has four of a kind. Since cards are ordered, it is sufficient
+        # Check if player 1 has four of a kind. Since cards are ordered, it is sufficient
-#       to check if the first card is equal to the fourth or if the second is equal to the fifth.
+        # to check if the first card is equal to the fourth or if the second is equal to the fifth.
         if self.hand1.cards[0].value == self.hand1.cards[3].value or self.hand1.cards[1].value == self.hand1.cards[4].value:
             four1 = 1
 
-#       Check if player 2 has four of a kind
+        # Check if player 2 has four of a kind
         if self.hand2.cards[0].value == self.hand2.cards[3].value or self.hand2.cards[1].value == self.hand2.cards[4].value:
             four2 = 1
 
-#       If player 1 has four of a kind and player 2 doesn't, player 1 wins.
+        # If player 1 has four of a kind and player 2 doesn't, player 1 wins.
         if four1 and not four2:
             return 1
 
-#       If player 2 has four of a kind and player 1 doesn't, player 2 wins.
+        # If player 2 has four of a kind and player 1 doesn't, player 2 wins.
         if not four1 and four2:
             return -1
 
-#       If both players have four of a kind, check who has the highest value for those four cards.
+        # If both players have four of a kind, check who has the highest value for those four cards.
         if four1 and four2:
             if self.hand1.cards[1].value > self.hand2.cards[1].value:
                 return 1
@@ -223,27 +223,26 @@ class Game():
         full1 = 0
         full2 = 0
 
-#       Check if player 1 has a full house.
+        # 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
 
-#       Check if player 2 has a full house.
+        # 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
 
-#       If player 1 has a full house and player 2 doesn't, player 1 wins.
+        # If player 1 has a full house and player 2 doesn't, player 1 wins.
         if full1 and not full2:
             return 1
 
-#       If player 2 has a full house and player 1 doesn't, player 2 wins.
+        # If player 2 has a full house and player 1 doesn't, player 2 wins.
         if not full1 and full2:
             return -1
 
-#       If both players have a full house, check who has the highest value
+        # If both players have a full house, check who has the highest value for the three equal cards (the third card in the array will be part
-#       for the three equal cards (the third card in the array will be part
+        # of the set of three).
-#       of the set of three).
         if full1 and full2:
             if self.hand1.cards[2].value > self.hand2.cards[2].value:
                 return 1
@@ -254,28 +253,28 @@ class Game():
         flush1 = 0
         flush2 = 0
 
-#       Check if player 1 has a flush.
+        # 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
 
-#       Check if player 2 has a flush.
+        # 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
 
-#       If player 1 has a flush and player 2 doesn't, player 1 wins.
+        # If player 1 has a flush and player 2 doesn't, player 1 wins.
         if flush1 and not flush2:
             return 1
 
-#       If player 2 has a flush and player 1 doesn't, player 2 wins.
+        # If player 2 has a flush and player 1 doesn't, player 2 wins.
         if not flush1 and flush2:
             return -1
 
         straight1 = 1
         straight2 = 1
 
-#       Check if player 1 has a straight.
+        # Check if player 1 has a straight.
         value = self.hand1.cards[0].value
 
         for i in range(1, 5):
@@ -285,7 +284,7 @@ class Game():
                 straight1 = 0
                 break
 
-#       Check if player 2 has a straight.
+        # Check if player 2 has a straight.
         value = self.hand2.cards[0].value
 
         for i in range(1, 5):
@@ -295,34 +294,34 @@ class Game():
                 straight2 = 0
                 break
 
-#       If player 1 has a straight and player 2 doesn't, player 1 wins.
+        # If player 1 has a straight and player 2 doesn't, player 1 wins.
         if straight1 and not straight2:
             return 1
 
-#       If player 2 has a straight and player 1 doesn't, player 2 wins.
+        # If player 2 has a straight and player 1 doesn't, player 2 wins.
         if not straight1 and straight2:
             return -1
 
         three1 = 0
         three2 = 0
 
-#       Check if player 1 has three of a kind.
+        # Check if player 1 has three of a kind.
         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
 
-#       Check if player 2 has three of a kind.
+        # 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
 
-#       If player 1 has three of a kind and player 2 doesn't, player 1 wins.
+        # If player 1 has three of a kind and player 2 doesn't, player 1 wins.
         if three1 and not three2:
             return 1
 
-#       If player 2 has three of a kind and player 1 doesn't, player 2 wins.
+        # If player 2 has three of a kind and player 1 doesn't, player 2 wins.
         if not three1 and three2:
             return -1
 
@@ -336,28 +335,28 @@ class Game():
         twopairs1 = 0
         twopairs2 = 0
 
-#       Check if player 1 has two pairs.
+        # Check if player 1 has two pairs.
         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
 
-#       Check if player 2 has two pairs.
+        # 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
 
-#       If player 1 has two pairs and player 2 doesn't, player 1 wins.
+        # If player 1 has two pairs and player 2 doesn't, player 1 wins.
         if twopairs1 and not twopairs2:
             return 1
 
-#       If player 2 has two pairs and player 1 doesn't, player 2 wins.
+        # If player 2 has two pairs and player 1 doesn't, player 2 wins.
         if not twopairs1 and twopairs2:
             return -1
 
-#       If both players have two pairs, check who has the highest pair. If it's equal,
+        # If both players have two pairs, check who has the highest pair. If it's equal,
-#       check the other pair. If it's still equal, check the remaining card.
+        # check the other pair. If it's still equal, check the remaining card.
         if twopairs1 and twopairs2:
             if self.hand1.cards[3].value > self.hand2.cards[3].value:
                 return 1
@@ -381,28 +380,28 @@ class Game():
         pair1 = 0
         pair2 = 0
 
-#       Check if player 1 has a pair of cards.
+        # 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
 
-#       Check if player 2 has a pair of cards.
+        # 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
 
-#       If player 1 has a pair of cards and player 2 doesn't, player 1 wins.
+        # If player 1 has a pair of cards and player 2 doesn't, player 1 wins.
         if pair1 and not pair2:
             return 1
 
-#       If player 2 has a pair of cards and player 1 doesn't, player 2 wins.
+        # If player 2 has a pair of cards and player 1 doesn't, player 2 wins.
         if not pair1 and pair2:
             return -1
 
-#       If both players have a pair of cards, check who has the highest pair. Since cards are
+        # If both players have a pair of cards, check who has the highest pair. Since cards are
-#       ordered by value, either card[1] will be part of the pair (card[0]=card[1] or
+        # ordered by value, either card[1] will be part of the pair (card[0]=card[1] or
-#       card[1]=card[2]) or card[3] will be part of the pair (card[2]=card[3] or
+        # card[1]=card[2]) or card[3] will be part of the pair (card[2]=card[3] or
-#       card[3]=card[4]).
+        # card[3]=card[4]).
         if pair1 and pair2:
             if self.hand1.cards[0].value == self.hand1.cards[1].value or self.hand1.cards[1].value == self.hand1.cards[2].value:
                 value = self.hand1.cards[1].value
@@ -432,20 +431,19 @@ class Game():
                     if value < self.hand2.cards[3].value:
                         return -1
 
-#       If all other things are equal, check who has the highest card, if it's equal check the second highest and so on.
+        # If all other things are equal, check who has the highest card, if it's equal check the second highest and so on.
         for i in range(4, -1, -1):
             if self.hand1.cards[i].value > self.hand2.cards[i].value:
                 return 1
             if self.hand1.cards[i].value < self.hand2.cards[i].value:
                 return -1
 
-#       If everything is equal, return 0
+        # If everything is equal, return 0
         return 0
 
 
-def main():
+@timing
-    start = default_timer()
+def p054():
-
     try:
         with open('p054_poker.txt', 'r', encoding='utf-8') as fp:
             games = fp.readlines()
@@ -476,7 +474,7 @@ def main():
         hand1.sort()
         hand2.sort()
 
-        for k in hand1.cards:
+        for _ in hand1.cards:
             game = Game()
             game.hand1 = hand1
             game.hand2 = hand2
@@ -484,13 +482,9 @@ def main():
         if game.play() == 1:
             count = count + 1
 
-    end = default_timer()
-
     print('Project Euler, Problem 54')
     print(f'Answer: {count}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p054()

Python/p055.py

@@ -23,28 +23,26 @@
 #
 # 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, timing
-
-from projecteuler import is_palindrome
 
 
 def is_lychrel(n):
     tmp = n
 
-#   Run for 50 iterations
+    # Run for 50 iterations
     for _ in range(50):
         reverse = 0
 
-#       Find the reverse of the given number
+        # Find the reverse of the given number
         while tmp > 0:
             reverse = reverse * 10
             reverse = reverse + tmp % 10
             tmp = tmp // 10
 
-#       Add the reverse to the original number
+        # Add the reverse to the original number
         tmp = n + reverse
 
-#       If the sum is palindrome, the number is not a Lychrel number.
+        # If the sum is palindrome, the number is not a Lychrel number.
         if is_palindrome(tmp, 10):
             return False
 
@@ -53,24 +51,19 @@ def is_lychrel(n):
     return True
 
 
-def main():
+@timing
-    start = default_timer()
+def p055():
-
     count = 0
 
-#   For each number, use the is_lychrel function to check if the number
+    # For each number, use the is_lychrel function to check if the number
-#   is a Lychrel number.
+    # is a Lychrel number.
     for i in range(1, 10000):
         if is_lychrel(i):
             count = count + 1
 
-    end = default_timer()
-
     print('Project Euler, Problem 55')
     print(f'Answer: {count}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p055()

Python/p056.py

@@ -5,15 +5,14 @@
 #
 # Considering natural numbers of the form, a^b, where a, b < 100, what is the maximum digital sum?
 
-from timeit import default_timer
+from projecteuler import timing
 
 
-def main():
+@timing
-    start = default_timer()
+def p056():
-
     max_ = 0
 
-#   Straightforward brute force approach
+    # Straightforward brute force approach
     for a in range(1, 100):
         for b in range(1, 100):
             p = a ** b
@@ -26,13 +25,9 @@ def main():
             if sum_ > max_:
                 max_ = sum_
 
-    end = default_timer()
-
     print('Project Euler, Problem 56')
     print(f'Answer: {max_}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p056()

Python/p057.py

@@ -16,18 +16,17 @@
 #
 # In the first one-thousand expansions, how many fractions contain a numerator with more digits than the denominator?
 
-from timeit import default_timer
+from projecteuler import timing
 
 
-def main():
+@timing
-    start = default_timer()
+def p057():
-
     n = 1
     d = 1
     count = 0
 
-#   If n/d is the current term of the expansion, the next term can be calculated as
+    # If n/d is the current term of the expansion, the next term can be calculated as
-#   (n+2d)/(n+d).
+    # (n+2d)/(n+d).
     for _ in range(1, 1000):
         d2 = 2 * d
         d = n + d
@@ -36,13 +35,9 @@ def main():
         if len(str(n)) > len(str(d)):
             count = count + 1
 
-    end = default_timer()
-
     print('Project Euler, Problem 57')
     print(f'Answer: {count}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p057()

Python/p058.py

@@ -16,19 +16,16 @@
 # If one complete new layer is wrapped around the spiral above, a square spiral with side length 9 will be formed. If this process is continued,
 # 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, timing
-
-from projecteuler import is_prime
 
 
-def main():
+@timing
-    start = default_timer()
+def p058():
-
+    # Starting with 1, the next four numbers in the diagonal are 3 (1+2), 5 (1+2+2), 7 (1+2+2+2)
-#   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
-#   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
-#   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.
-#   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.
-#   Then 25+6 etc., at every cycle the step is increased by 2. Continue until the ratio goes below 0.1.
     i = 1
     l = 1
     step = 2
@@ -61,13 +58,9 @@ def main():
         if ratio < 0.1:
             break
 
-    end = default_timer()
-
     print('Project Euler, Problem 58')
     print(f'Answer: {l}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p058()

Python/p059.py

@@ -18,7 +18,9 @@
 # encrypted ASCII codes, and the knowledge that the plain text must contain common English words, decrypt the message and find the sum of the
 # ASCII values in the original text.
 
-from timeit import default_timer
+import sys
+
+from projecteuler import timing
 
 
 class EncryptedText():
@@ -40,6 +42,8 @@ class EncryptedText():
 
         self.len = len(self.text)
 
+        return 0
+
     def decrypt(self):
         found = 0
 
@@ -57,16 +61,16 @@ class EncryptedText():
                     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] = str(chr(self.text[i] ^ c1))
-                        plain_text[i+1] = str(chr(self.text[i+1]^c2))
+                        plain_text[i+1] = str(chr(self.text[i+1] ^ c2))
-                        plain_text[i+2] = str(chr(self.text[i+2]^c3))
+                        plain_text[i+2] = str(chr(self.text[i+2] ^ c3))
 
                     if i == self.len - 2:
-                        plain_text[i] = str(chr(self.text[i]^c1))
+                        plain_text[i] = str(chr(self.text[i] ^ c1))
-                        plain_text[i+1] = str(chr(self.text[i+1]^c2))
+                        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[i] = str(chr(self.text[i] ^ c1))
 
                     plain_text = ''.join(plain_text)
 
@@ -77,9 +81,8 @@ class EncryptedText():
         return plain_text
 
 
-def main():
+@timing
-    start = default_timer()
+def p059():
-
     enc_text = EncryptedText()
 
     if enc_text.read_text('p059_cipher.txt') == -1:
@@ -92,13 +95,9 @@ def main():
     for i in list(plain_text):
         sum_ = sum_ + ord(i)
 
-    end = default_timer()
-
     print('Project Euler, Problem 59')
     print(f'Answer: {sum_}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p059()

Python/p060.py

@@ -6,14 +6,11 @@
 #
 # 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 is_prime, sieve, timing
-
-from projecteuler import sieve, is_prime
 
 
-def main():
+@timing
-    start = default_timer()
+def p060():
-
     N = 10000
 
     primes = sieve(N)
@@ -21,9 +18,9 @@ def main():
     found = 0
     p1 = 3
 
-#   Straightforward brute force approach
+    # Straightforward brute force approach
     while p1 < N and not found:
-#       If p1 is not prime, go to the next number.
+        # If p1 is not prime, go to the next number.
         if primes[p1] == 0:
             p1 = p1 + 2
             continue
@@ -31,8 +28,8 @@ 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.
+            # 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
                 continue
@@ -40,8 +37,8 @@ def main():
             p3 = p2 + 2
 
             while p3 < N and not found:
-#               If p3 is not prime, or at least one of the possible concatenations of
+                # If p3 is not prime, or at least one of the possible concatenations of
-#               p1, p2 and p3 is not prime, got to the next number.
+                # 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
@@ -51,8 +48,8 @@ def main():
                 p4 = p3 + 2
 
                 while p4 < N and not found:
-#                   If p4 is not prime, or at least one of the possible concatenations of
+                    # If p4 is not prime, or at least one of the possible concatenations of
-#                   p1, p2, p3 and p4 is not prime, go to the next number.
+                    # p1, p2, p3 and p4 is not prime, go to the next number.
                     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))):
@@ -63,8 +60,8 @@ def main():
                     p5 = p4 + 2
 
                     while p5 < N and not found:
-#                       If p5 is not prime, or at least one of the possible concatenations of
+                        # If p5 is not prime, or at least one of the possible concatenations of
-#                       p1, p2, p3, p4 and p5 is not prime, go to the next number
+                        # p1, p2, p3, p4 and p5 is not prime, go to the next number
                         if primes[p5] == 0 or not is_prime(int(str(p1)+str(p5))) or not is_prime(int(str(p5)+str(p1))) or\
                                 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\
@@ -73,7 +70,7 @@ def main():
 
                             continue
 
-#                       If it gets here, the five values have been found.
+                        # If it gets here, the five values have been found.
                         n = p1 + p2 + p3 + p4 + p5
                         found = 1
 
@@ -85,13 +82,9 @@ def main():
 
         p1 = p1 + 2
 
-    end = default_timer()
-
     print('Project Euler, Problem 60')
     print(f'Answer: {n}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p060()