Use timing decorator for problems 21-40

Python/p021.py

@@ -9,20 +9,18 @@
 # Evaluate the sum of all the amicable numbers under 10000.
 
 
-from timeit import default_timer
+from projecteuler import sum_of_divisors, timing
-
-from projecteuler import sum_of_divisors
 
 
-def main():
+@timing
-    start = default_timer()
+def p021():
-
     sum_ = 0
 
     for i in range(2, 10000):
         # Calculate the sum of proper divisors with the function
         # implemented in projecteuler.py.
         n = sum_of_divisors(i)
+
         # If i!=n and the sum of proper divisors of n=i,
         # sum the pair of numbers and add it to the total.
         if i != n and sum_of_divisors(n) == i:
@@ -30,13 +28,9 @@ def main():
 
     sum_ = sum_ // 2
 
-    end = default_timer()
-
     print('Project Euler, Problem 21')
     print(f'Answer: {sum_}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p021()

Python/p022.py

@@ -9,12 +9,12 @@
 # What is the total of all the name scores in the file?
 
 import sys
-from timeit import default_timer
+
+from projecteuler import timing
 
 
-def main():
+@timing
-    start = default_timer()
+def p022():
-
     try:
         with open('p022_names.txt', 'r', encoding='utf-8') as fp:
             names = list(fp.readline().replace('"', '').split(','))
@@ -37,13 +37,9 @@ def main():
         sum_ = sum_ + score
         i = i + 1
 
-    end = default_timer()
-
     print('Project Euler, Problem 22')
     print(f'Answer: {sum_}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p022()

Python/p023.py

@@ -12,18 +12,15 @@
 #
 # Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.
 
-from timeit import default_timer
+from projecteuler import sum_of_divisors, timing
-
-from projecteuler import sum_of_divisors
 
 
 def is_abundant(n):
     return sum_of_divisors(n) > n
 
 
-def main():
+@timing
-    start = default_timer()
+def p023():
-
     ab_nums = [False] * 28124
 
     # Find all abundant numbers smaller than 28123.
@@ -50,13 +47,9 @@ def main():
         if not sums[i]:
             sum_ = sum_ + i
 
-    end = default_timer()
-
     print('Project Euler, Problem 23')
     print(f'Answer: {sum_}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p023()

Python/p024.py

@@ -8,23 +8,19 @@
 # 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
+
+from projecteuler import timing
 
 
-def main():
+@timing
-    start = default_timer()
+def p024():
-
     # Generate all the permutations in lexicographic order and get the millionth one.
     perm = list(permutations('0123456789'))
     res = int(''.join(map(str, perm[999999])))
 
-    end = default_timer()
-
     print('Project Euler, Problem 24')
     print(f'Answer: {res}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p024()

Python/p025.py

@@ -21,12 +21,11 @@
 #
 # What is the index of the first term in the Fibonacci sequence to contain 1000 digits?
 
-from timeit import default_timer
+from projecteuler import timing
 
 
-def main():
+@timing
-    start = default_timer()
+def p025():
-
     fib1 = 1
     fib2 = 1
     fibn = fib1 + fib2
@@ -40,13 +39,9 @@ def main():
         fibn = fib1 + fib2
         i = i + 1
 
-    end = default_timer()
-
     print('Project Euler, Problem 25')
     print(f'Answer: {i}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p025()

Python/p026.py

@@ -16,27 +16,24 @@
 #
 # Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.
 
-from timeit import default_timer
+from projecteuler import timing
 
 
-def main():
+@timing
-    start = default_timer()
+def p026():
-
     max_ = 0
 
     for i in range(2, 1000):
         j = i
 
-#       The repeating cycle of 1/(2^a*5^b*p^c*...) is equal to
+        # The repeating cycle of 1/(2^a*5^b*p^c*...) is equal to that of 1/p^c*..., so factors 2 and 5 can be eliminated.
-#       that of 1/p^c*..., so factors 2 and 5 can be eliminated.
         while j % 2 == 0 and j > 1:
             j = j // 2
 
         while j % 5 == 0 and j > 1:
             j = j // 5
 
-#       If the denominator had only factors 2 and 5, there is no
+        # If the denominator had only factors 2 and 5, there is no repeating cycle.
-#       repeating cycle.
         if j == 1:
             n = 0
         else:
@@ -57,13 +54,9 @@ def main():
                 max_ = n
                 max_n = i
 
-    end = default_timer()
-
     print('Project Euler, Problem 26')
     print(f'Answer: {max_n}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p026()

Python/p027.py

@@ -19,14 +19,11 @@
 #
 # 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, timing
-
-from projecteuler import is_prime
 
 
-def main():
+@timing
-    start = default_timer()
+def p027():
-
     max_ = 0
 
     # Brute force approach, optimized by checking only values of b where b is prime.
@@ -51,13 +48,9 @@ def main():
                     save_a = a
                     save_b = b
 
-    end = default_timer()
-
     print('Project Euler, Problem 27')
     print(f'Answer: {save_a * save_b}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p027()

Python/p028.py

@@ -12,12 +12,11 @@
 #
 # What is the sum of the numbers on the diagonals in a 1001 by 1001 spiral formed in the same way?
 
-from timeit import default_timer
+from projecteuler import timing
 
 
-def main():
+@timing
-    start = default_timer()
+def p028():
-
     N = 1001
 
     limit = N * N
@@ -38,13 +37,9 @@ def main():
         sum_ = sum_ + j
         i = (i + 1) % 4
 
-    end = default_timer()
-
     print('Project Euler, Problem 28')
     print(f'Answer: {sum_}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p028()

Python/p029.py

@@ -13,14 +13,13 @@
 #
 # 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 projecteuler import timing
 
-def main():
-    start = default_timer()
 
+@timing
+def p029():
     powers = zeros(9801)
 
     # Generate all the powers
@@ -39,13 +38,9 @@ def main():
         if powers[i] != powers[i-1]:
             count = count + 1
 
-    end = default_timer()
-
     print('Project Euler, Problem 29')
     print(f'Answer: {count}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p029()

Python/p030.py

@@ -12,12 +12,11 @@
 #
 # Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.
 
-from timeit import default_timer
+from projecteuler import timing
 
 
-def main():
+@timing
-    start = default_timer()
+def p030():
-
     tot = 0
 
     # Straightforward brute force approach. The limit is chosen considering that
@@ -35,13 +34,9 @@ def main():
         if sum_ == i:
             tot = tot + i
 
-    end = default_timer()
-
     print('Project Euler, Problem 30')
     print(f'Answer: {tot}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p030()

Python/p031.py

@@ -10,7 +10,7 @@
 #
 # How many different ways can £2 be made using any number of coins?
 
-from timeit import default_timer
+from projecteuler import timing
 
 
 # Simple recursive function that tries every combination.
@@ -29,20 +29,15 @@ def count(coins, value, n, i):
     return n
 
 
-def main():
+@timing
-    start = default_timer()
+def p031():
-
     coins = [1, 2, 5, 10, 20, 50, 100, 200]
 
     n = count(coins, 0, 0, 0)
 
-    end = default_timer()
-
     print('Project Euler, Problem 31')
     print(f'Answer: {n}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p031()

Python/p032.py

@@ -9,17 +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 projecteuler import is_pandigital
+from projecteuler import is_pandigital, timing
 
 
-def main():
+@timing
-    start = default_timer()
+def p032():
-
     n = 0
     # Initially I used a bigger array, but printing the resulting products
     # shows that 10 values are sufficient.
@@ -79,13 +76,9 @@ def main():
         if products[i] != products[i-1]:
             sum_ = sum_ + products[i]
 
-    end = default_timer()
-
     print('Project Euler, Problem 32')
     print(f'Answer: {sum_}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p032()

Python/p033.py

@@ -11,12 +11,12 @@
 # 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
+
+from projecteuler import timing
 
 
-def main():
+@timing
-    start = default_timer()
+def p033():
-
     prod_n = 1
     prod_d = 1
 
@@ -39,13 +39,9 @@ def main():
     # Find the greater common divisor of the fraction found.
     div = gcd(prod_n, prod_d)
 
-    end = default_timer()
-
     print('Project Euler, Problem 33')
     print(f'Answer: {prod_d // div}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p033()

Python/p034.py

@@ -7,14 +7,14 @@
 # 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 projecteuler import timing
 
-def main():
-    start = default_timer()
 
+@timing
+def p034():
     a = 10
     sum_ = 0
     factorials = ones(10, int)
@@ -38,13 +38,9 @@ def main():
 
         a = a + 1
 
-    end = default_timer()
-
     print('Project Euler, Problem 34')
     print(f'Answer: {sum_}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p034()

Python/p035.py

@@ -6,9 +6,7 @@
 #
 # How many circular primes are there below one million?
 
-from timeit import default_timer
+from projecteuler import sieve, timing
-
-from projecteuler import sieve
 
 
 # Calculate all primes below one million, then check if they're circular.
@@ -47,9 +45,8 @@ def is_circular_prime(n):
     return True
 
 
-def main():
+@timing
-    start = default_timer()
+def p035():
-
     count = 13
 
     # Calculate all primes below one million, then check if they're circular.
@@ -57,13 +54,9 @@ def main():
         if is_circular_prime(i):
             count = count + 1
 
-    end = default_timer()
-
     print('Project Euler, Problem 35')
     print(f'Answer: {count}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p035()

Python/p036.py

@@ -6,14 +6,11 @@
 #
 # (Please note that the palindromic number, in either base, may not include leading zeros.)
 
-from timeit import default_timer
+from projecteuler import is_palindrome, timing
-
-from projecteuler import is_palindrome
 
 
-def main():
+@timing
-    start = default_timer()
+def p036():
-
     N = 1000000
 
     sum_ = 0
@@ -25,13 +22,9 @@ def main():
         if is_palindrome(i, 10) and is_palindrome(i, 2):
             sum_ = sum_ + i
 
-    end = default_timer()
-
     print('Project Euler, Problem 36')
     print(f'Answer: {sum_}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p036()

Python/p037.py

@@ -7,9 +7,7 @@
 #
 # NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.
 
-from timeit import default_timer
+from projecteuler import is_prime, timing
-
-from projecteuler import is_prime
 
 
 def is_tr_prime(n):
@@ -27,8 +25,7 @@ def is_tr_prime(n):
             return False
         tmp = tmp // 10
 
-#   Starting from the last digit, check if it's prime, then
+    # Starting from the last digit, check if it's prime, then add back one digit at a time on the left and check if it
-#   add back one digit at a time on the left and check if it
     # is prime. Return 0 when it isn't.
     i = 10
     tmp = n % i
@@ -43,9 +40,8 @@ def is_tr_prime(n):
     return True
 
 
-def main():
+@timing
-    start = default_timer()
+def p037():
-
     i = 0
     n = 1
     sum_ = 0
@@ -57,13 +53,9 @@ def main():
             i = i + 1
         n = n + 1
 
-    end = default_timer()
-
     print('Project Euler, Problem 37')
     print(f'Answer: {sum_}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p037()

Python/p038.py

@@ -13,14 +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 timeit import default_timer
+from projecteuler import is_pandigital, timing
-
-from projecteuler import is_pandigital
 
 
-def main():
+@timing
-    start = default_timer()
+def p038():
-
     max_ = 0
 
     # A brute force approach is used, starting with 1 and multiplying
@@ -54,13 +51,9 @@ def main():
             if n > 987654321:
                 break
 
-    end = default_timer()
-
     print('Project Euler, Problem 38')
     print(f'Answer: {max_}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p038()

Python/p039.py

@@ -6,14 +6,13 @@
 #
 # For which value of p ≤ 1000, is the number of solutions maximised?
 
-from timeit import default_timer
-
 from numpy import zeros
 
+from projecteuler import timing
 
-def main():
-    start = default_timer()
 
+@timing
+def p039():
     max_ = 0
     savedc = zeros(1000, int)
 
@@ -32,8 +31,7 @@ def main():
                 b = 2 * m * n
                 c = m * m + n * n
 
-#               Increase counter if a+b+c=p and the triplet is new,
+                # Increase counter if a+b+c=p and the triplet is new, then save the value of c to avoid counting the same
-#               then save the value of c to avoid counting the same
                 # triplet more than once.
                 if a + b + c == p and savedc[c] == 0:
                     savedc[c] = 1
@@ -44,8 +42,7 @@ def main():
                 tmpb = b
                 tmpc = c
 
-#               Check all the triplets obtained multiplying a, b and c
+                # Check all the triplets obtained multiplying a, b and c for integer numbers, until the perimeters exceeds p.
-#               for integer numbers, until the perimeters exceeds p.
                 while tmpa + tmpb + tmpc < p:
                     tmpa = a * i
                     tmpb = b * i
@@ -66,13 +63,9 @@ def main():
             max_ = count
             res = p
 
-    end = default_timer()
-
     print('Project Euler, Problem 39')
     print(f'Answer: {res}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p039()

Python/p040.py

@@ -10,14 +10,13 @@
 #
 # d_1 × d_10 × d_100 × d_1000 × d_10000 × d_100000 × d_1000000
 
-from timeit import default_timer
-
 from numpy import zeros
 
+from projecteuler import timing
 
-def main():
-    start = default_timer()
 
+@timing
+def p040():
     digits = zeros(1000005, int)
     i = 1
     value = 1
@@ -64,13 +63,9 @@ def main():
     # Calculate the product.
     n = digits[0] * digits[9] * digits[99] * digits[999] * digits[9999] * digits[99999] * digits[999999]
 
-    end = default_timer()
-
     print('Project Euler, Problem 40')
     print(f'Answer: {n}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p040()