Use timing decorator for problems 71-80

Python/p071.py

@@ -12,12 +12,12 @@
 # of the fraction immediately to the left of 3/7.
 
 from math import gcd
-from timeit import default_timer
+
+from projecteuler import timing
 
 
-def main():
+@timing
-    start = default_timer()
+def p071():
-
     N = 1000000
 
     max_ = 0
@@ -43,13 +43,9 @@ def main():
 
             max_n = n
 
-    end = default_timer()
-
     print('Project Euler, Problem 71')
     print(f'Answer: {max_n}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p071()

Python/p072.py

@@ -10,14 +10,11 @@
 #
 # 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, timing
-
-from projecteuler import sieve, phi
 
 
-def main():
+@timing
-    start = default_timer()
+def p072():
-
     N = 1000001
 
     count = 0
@@ -30,13 +27,9 @@ def main():
     for i in range(2, N):
         count = count + phi(i, primes)
 
-    end = default_timer()
-
     print('Project Euler, Problem 72')
-    print(f'Answer: {count}')
+    print(f'Answer: {int(count)}')
-
-    print(f'Elapsed time: {end - start:.9f} seconds')
 
 
 if __name__ == '__main__':
-    main()
+    p072()

Python/p073.py

@@ -11,12 +11,12 @@
 # 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
+
+from projecteuler import timing
 
 
-def main():
+@timing
-    start = default_timer()
+def p073():
-
     count = 0
 
     # For each denominator q, we need to find the fractions p/q for which
@@ -30,13 +30,9 @@ def main():
             if gcd(j, i) == 1:
                 count = count + 1
 
-    end = default_timer()
-
     print('Project Euler, Problem 73')
     print(f'Answer: {count}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p073()

Python/p074.py

@@ -23,7 +23,8 @@
 # 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
+
+from projecteuler import timing
 
 
 N = 1000000
@@ -44,21 +45,21 @@ def len_chain(n):
 
         # Generate the next number of the chain by taking
         # the digits of the current value, calculating the
-#      factorials and adding them.*/
+        # factorials and adding them.
         while value != 0:
             tmp = tmp + factorial(value % 10)
             value = value // 10
 
         # If the chain length for the new value has already been
         # calculated before, use the saved value (only chains for
-#       values smaller than N are saved).*/
+        # values smaller than N are saved).
         if tmp < N and chains[tmp] != 0:
             return count + chains[tmp]
 
         value = tmp
 
         # If the current value is already present in the chain,
-#       the chain is finished.*/
+        # the chain is finished.
         for i in range(count):
             if chain[i] == value:
                 finished = 1
@@ -70,9 +71,8 @@ def len_chain(n):
     return count
 
 
-def main():
+@timing
-    start = default_timer()
+def p074():
-
     count = 0
 
     # Simple brute force approach, for every number calculate
@@ -81,13 +81,9 @@ def main():
         if len_chain(i) == 60:
             count = count + 1
 
-    end = default_timer()
-
     print('Project Euler, Problem 74')
     print(f'Answer: {count}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p074()

Python/p075.py

@@ -18,12 +18,12 @@
 # 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
+
+from projecteuler import timing
 
 
-def main():
+@timing
-    start = default_timer()
+def p075():
-
     N = 1500000
 
     l = [0] * (N+1)
@@ -68,13 +68,9 @@ def main():
         if l[i] == 1:
             count = count + 1
 
-    end = default_timer()
-
     print('Project Euler, Problem 75')
     print(f'Answer: {count}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p075()

Python/p076.py

@@ -11,14 +11,11 @@
 #
 # 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, timing
-
-from projecteuler import partition_fn
 
 
-def main():
+@timing
-    start = default_timer()
+def p076():
-
     partitions = [0] * 101
 
     # The number of ways a number can be written as a sum is given by the partition function
@@ -26,13 +23,9 @@ def main():
     # The function is implemented in projecteuler.py.
     n = partition_fn(100, partitions) - 1
 
-    end = default_timer()
-
     print('Project Euler, Problem 76')
     print(f'Answer: {n}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p076()

Python/p077.py

@@ -10,16 +10,13 @@
 #
 # 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, timing
-
-from projecteuler import is_prime
 
 
 primes = [0] * 100
 
 
-# Function using a simple recursive brute force approach
+# Function using a simple recursive brute force approach to find all the partitions.
-# to find all the partitions.
 def count(value, n, i, target):
     for j in range(i, 100):
         value = value + primes[j]
@@ -36,9 +33,8 @@ def count(value, n, i, target):
     return n
 
 
-def main():
+@timing
-    start = default_timer()
+def p077():
-
     # Generate a list of the first 100 primes.
     i = 0
     j = 0
@@ -62,13 +58,9 @@ def main():
 
         i = i + 1
 
-    end = default_timer()
-
     print('Project Euler, Problem 77')
     print(f'Answer: {i}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p077()

Python/p078.py

@@ -13,32 +13,24 @@
 #
 # 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, timing
-
-from projecteuler import partition_fn
 
 
-def main():
+@timing
-    start = default_timer()
+def p078():
-
     N = 1000000
 
     partitions = [0] * N
 
     i = 0
 
-#   Using the partition function to calculate the number of partitions,
+    # 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(f'Answer: {i}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p078()

Python/p079.py

@@ -10,12 +10,12 @@
 
 import sys
 from itertools import permutations
-from timeit import default_timer
+
+from projecteuler import timing
 
 
 def check_passcode(passcode, len_, logins, n):
-#   For every login attempt, check if all the digits appear in the
+    # 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
@@ -32,9 +32,8 @@ def check_passcode(passcode, len_, logins, n):
     return 1
 
 
-def main():
+@timing
-    start = default_timer()
+def p079():
-
     try:
         with open('p079_keylog.txt', 'r', encoding='utf-8') as fp:
             logins = fp.readlines()
@@ -58,8 +57,7 @@ def main():
 
     j = 0
     for i in range(10):
-#       To generate the passcode, only use the digits present in the
+        # To generate the passcode, only use the digits present in the login attempts.
-#       login attempts.
         if digits[i] > 0:
             passcode_digits[j] = i
             j = j + 1
@@ -70,8 +68,7 @@ def main():
     while not found:
         passcode = [0] * len_
 
-#       For the current length, generate the first passcode with the
+        # For the current length, generate the first passcode with the digits in order.
-#       digits in order.
         for i in range(len_):
             passcode[i] = passcode_digits[i]
 
@@ -93,13 +90,9 @@ def main():
         # If the passcode has not yet been found, try a longer passcode.
         len_ = len_ + 1
 
-    end = default_timer()
-
     print('Project Euler, Problem 79')
     print(f'Answer: {res}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p079()

Python/p080.py

@@ -8,9 +8,10 @@
 # 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 projecteuler import timing
+
 
 def is_square(n):
     p = sqrt(n)
@@ -19,9 +20,8 @@ def is_square(n):
     return bool(p == m)
 
 
-def main():
+@timing
-    start = default_timer()
+def p080():
-
     # Set the precision to 100 digits
     mp.dps = 102
 
@@ -29,8 +29,7 @@ def main():
 
     for i in range(2, 100):
         if not is_square(i):
-#           Calculate the square root of the current number with the given precision
+            # Calculate the square root of the current number with the given precision and sum the digits to the total.
-#           and sum the digits to the total.
             root = str(sqrt(i))
 
             sum_ = sum_ + int(root[0])
@@ -38,13 +37,9 @@ def main():
             for j in range(2, 101):
                 sum_ = sum_ + int(root[j])
 
-    end = default_timer()
-
     print('Project Euler, Problem 80')
     print(f'Answer: {sum_}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p080()