Use timing decorator for problems 11-20

Python/p011.py

@@ -27,12 +27,11 @@
 #
 # What is the greatest product of four adjacent numbers in the same direction (up, down, left, right, or diagonally) in the 20×20 grid?
 
-from timeit import default_timer
+from projecteuler import timing
 
 
-def main():
+@timing
-    start = default_timer()
+def p011():
-
     grid = [[8, 2, 22, 97, 38, 15, 0, 40, 0, 75, 4, 5, 7, 78, 52, 12, 50, 77, 91, 8],
             [49, 49, 99, 40, 17, 81, 18, 57, 60, 87, 17, 40, 98, 43, 69, 48, 4, 56, 62, 0],
             [81, 49, 31, 73, 55, 79, 14, 29, 93, 71, 40, 67, 53, 88, 30, 3, 49, 13, 36, 65],
@@ -56,12 +55,12 @@ def main():
 
     max_ = 0
 
-#   Brute-force approach: for each number in the grid, try products with its three
+    # Brute-force approach: for each number in the grid, try products with its three
-#   adjacent numbers in every direction (horizontal, vertical and the two diagonals).
+    # adjacent numbers in every direction (horizontal, vertical and the two diagonals).
-#   If the product is larger than the current  maximum, save it.
+    # If the product is larger than the current  maximum, save it.
     for i in range(17):
         for j in range(17):
-#           Horizontal direction.
+            # Horizontal direction.
             prod = 1
             k = j
             while k < j + 4:
@@ -71,7 +70,7 @@ def main():
             if prod > max_:
                 max_ = prod
 
-#           Vertical direction.
+            # Vertical direction.
             prod = 1
             k = i
             while k < i + 4:
@@ -81,7 +80,7 @@ def main():
             if prod > max_:
                 max_ = prod
 
-#           Diagonal direction, from top left to bottom right.
+            # Diagonal direction, from top left to bottom right.
             prod = 1
             k = i
             w = j
@@ -94,10 +93,10 @@ def main():
             if prod > max_:
                 max_ = prod
 
-#   The last diagonal is handled separately
+    # The last diagonal is handled separately
     for i in range(17):
         for j in range(3, 20):
-#           Diagonal direction, from top right to bottom left.
+            # Diagonal direction, from top right to bottom left.
             prod = 1
             k = i
             w = j
@@ -110,13 +109,9 @@ def main():
             if prod > max_:
                 max_ = prod
 
-    end = default_timer()
-
     print('Project Euler, Problem 11')
     print(f'Answer: {max_}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p011()

Python/p012.py

@@ -19,33 +19,27 @@
 #
 # What is the value of the first triangle number to have over five hundred divisors?
 
-from timeit import default_timer
+from projecteuler import count_divisors, timing
-from projecteuler import count_divisors
 
 
-def main():
+@timing
-    start = default_timer()
+def p012():
-
     i = 0
     triang = 0
     finished = 0
 
-#   Generate all triangle numbers until the first one with more than 500 divisors is found.
+    # Generate all triangle numbers until the first one with more than 500 divisors is found.
     while not finished:
         i = i + 1
         triang = triang + i
 
-#       Use the function implemented in projecteuler.py to count divisors of a number.
+        # Use the function implemented in projecteuler.py to count divisors of a number.
         if count_divisors(triang) > 500:
             finished = 1
 
-    end = default_timer()
-
     print('Project Euler, Problem 12')
     print(f'Answer: {triang}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p012()

Python/p013.py

@@ -103,13 +103,13 @@
 # 20849603980134001723930671666823555245252804609722
 # 53503534226472524250874054075591789781264330331690
 
-from timeit import default_timer
 import numpy as np
 
+from projecteuler import timing
 
-def main():
-    start = default_timer()
 
+@timing
+def p013():
     numbers = [37107287533902102798797998220837590246510135740250,
                46376937677490009712648124896970078050417018260538,
                74324986199524741059474233309513058123726617309629,
@@ -211,18 +211,14 @@ def main():
                20849603980134001723930671666823555245252804609722,
                53503534226472524250874054075591789781264330331690]
 
-#   Convert the list of numbers in a numpy array and calculate the sum
+    # Convert the list of numbers in a numpy array and calculate the sum
     numbers = np.array(numbers)
 
     sum_ = str(numbers.sum())
 
-    end = default_timer()
-
     print('Project Euler, Problem 13')
     print(f'Answer: {sum_[:10]}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p013()

Python/p014.py

@@ -16,9 +16,10 @@
 #
 # NOTE: Once the chain starts the terms are allowed to go above one million.
 
-from timeit import default_timer
 from numpy import zeros
 
+from projecteuler import timing
+
 
 N = 1000000
 collatz_found = zeros(N, dtype=int)
@@ -32,8 +33,8 @@ def collatz_length(n):
     if n == 1:
         return 1
 
-#   If Collatz(n) has been previously calculated,
+    # If Collatz(n) has been previously calculated,
-#   just return the value.
+    # just return the value.
     if n < N and collatz_found[n]:
         return collatz_found[n]
 
@@ -42,14 +43,14 @@ def collatz_length(n):
 
     return 1 + collatz_length(3*n+1)
 
-def main():
-    start = default_timer()
 
+@timing
+def p014():
     max_l = 0
     max_ = 0
 
-#   For each number from 1 to 1000000, find the length of the sequence
+    # For each number from 1 to 1000000, find the length of the sequence
-#   and save its value, so that it can be used for the next numbers.
+    # and save its value, so that it can be used for the next numbers.
     for i in range(1, N):
         count = collatz_length(i)
         collatz_found[i] = count
@@ -58,13 +59,9 @@ def main():
             max_l = count
             max_ = i
 
-    end = default_timer()
-
     print('Project Euler, Problem 14')
     print(f'Answer: {max_}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p014()

Python/p015.py

@@ -4,28 +4,24 @@
 # How many such routes are there through a 20×20 grid?
 
 from math import factorial
-from timeit import default_timer
+
+from projecteuler import timing
 
 
-def main():
+@timing
-    start = default_timer()
+def p015():
-
+    # Using a combinatorial solution: in a 20x20 grid there will always be
-#   Using a combinatorial solution: in a 20x20 grid there will always be
+    # 20 movements to the right and 20 movements down, that can be represented
-#   20 movements to the right and 20 movements down, that can be represented
+    # as a string of Rs and Ds. The number of routes is the number of combinations.
-#   as a string of Rs and Ds. The number of routes is the number of combinations.
+    # This is obtained calculating n!/(k!*(n-k)!), where n=40 and k=20.
-#   This is obtained calculating n!/(k!*(n-k)!), where n=40 and k=20.
     count = factorial(40)
     tmp = factorial(20)
     tmp = tmp * tmp
     count = count // tmp
 
-    end = default_timer()
-
     print('Project Euler, Problem 15')
     print(f'Answer: {count}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p015()

Python/p016.py

@@ -4,14 +4,13 @@
 #
 # What is the sum of the digits of the number 2^1000?
 
-from timeit import default_timer
+from projecteuler import timing
 
 
-def main():
+@timing
-    start = default_timer()
+def p016():
-
+    # Simply calculate 2^1000, convert the result to string and calculate
-#   Simply calculate 2^1000, convert the result to string and calculate
+    # the sum of the digits
-#   the sum of the digits
     res = str(2 ** 1000)
 
     sum_ = 0
@@ -19,13 +18,9 @@ def main():
     for i in res:
         sum_ = sum_ + int(i)
 
-    end = default_timer()
-
     print('Project Euler, Problem 16')
     print(f'Answer: {sum_}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p016()

Python/p017.py

@@ -7,16 +7,15 @@
 # NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contains 23 letters and 115 (one hundred and fifteen)
 # contains 20 letters. The use of "and" when writing out numbers is in compliance with British usage.
 
-from timeit import default_timer
+from projecteuler import timing
 
 
-def main():
+@timing
-    start = default_timer()
+def p017():
-
+    # First list contains number of letters for numbers from 1 to 19,
-#   First list contains number of letters for numbers from 1 to 19,
+    # the second letters for "twenty", "thirty", ..., "ninety",
-#   the second letters for "twenty", "thirty", ..., "ninety",
+    # the third letters for "one hundred and", "two hundred and", ..., "nine hundre and",
-#   the third letters for "one hundred and", "two hundred and", ..., "nine hundre and",
+    # the last one-element one the number of letters of 1000
-#   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],
@@ -24,46 +23,43 @@ def main():
 
     sum_ = 0
 
-#   Sum the letters of the first 19 numbers.
+    # Sum the letters of the first 19 numbers.
     for i in range(19):
         sum_ = sum_ + n_letters[0][i]
 
-#   Add the letters of the numbers from 20 to 99.
+    # Add the letters of the numbers from 20 to 99.
     for i in range(8):
-#       "Twenty", "thirty", ... "ninety" are used ten times each
+        # "Twenty", "thirty", ... "ninety" are used ten times each
-#       (e.g. "twenty", "twenty one", "twenty two", ..., "twenty nine").
+        # (e.g. "twenty", "twenty one", "twenty two", ..., "twenty nine").
         n_letters[1][i] = n_letters[1][i] * 10
 
-#       Add "one", "two", ..., "nine".
+        # Add "one", "two", ..., "nine".
         for j in range(9):
             n_letters[1][i] = n_letters[1][i] + n_letters[0][j]
+
         sum_ = sum_ + n_letters[1][i]
 
-#   Add the letters of the numbers from 100 to 999.
+    # Add the letters of the numbers from 100 to 999.
     for i in range(9):
-#       "One hundred and", "two hundred and",... are used 100 times each.
+        # "One hundred and", "two hundred and",... are used 100 times each.
         n_letters[2][i] = n_letters[2][i] * 100
 
-#       Add "one" to "nineteen".
+        # Add "one" to "nineteen".
         for j in range(19):
             n_letters[2][i] = n_letters[2][i] + n_letters[0][j]
-#       Add "twenty" to "ninety nine", previously calculated.
+        # Add "twenty" to "ninety nine", previously calculated.
         for j in range(8):
             n_letters[2][i] = n_letters[2][i] + n_letters[1][j]
-#       "One hundred", "two hundred", ... don't have the "and", so remove
+        # "One hundred", "two hundred", ... don't have the "and", so remove
-#       three letters for each of them.
+        # three letters for each of them.
         sum_ = sum_ + n_letters[2][i] - 3
 
-#   Add "one thousand".
+    # Add "one thousand".
     sum_ = sum_ + n_letters[3][0]
 
-    end = default_timer()
-
     print('Project Euler, Problem 17')
     print(f'Answer: {sum_}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p017()

Python/p018.py

@@ -31,14 +31,12 @@
 # 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
+from projecteuler import find_max_path, timing
 
 
-def main():
+@timing
-    start = default_timer()
+def p018():
-
     try:
         with open('p018_triangle.txt', 'r', encoding='utf-8') as fp:
             triang = []
@@ -54,16 +52,12 @@ def main():
     for i in range(l):
         triang[i] = list(map(int, triang[i]))
 
-#   Use the function implemented in projecteuler.c to find the maximum path.
+    # Use the function implemented in projecteuler.c to find the maximum path.
     max_ = find_max_path(triang, 15)
 
-    end = default_timer()
-
     print('Project Euler, Problem 18')
     print(f'Answer: {max_}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p018()

Python/p019.py

@@ -14,27 +14,23 @@
 # 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
+
+from projecteuler import timing
 
 
-def main():
+@timing
-    start = default_timer()
+def p019():
-
     count = 0
 
-#   Use the datetime library to find out which first day of the month is a Sunday
+    # Use the datetime library to find out which first day of the month is a Sunday
     for year in range(1901, 2001):
         for month in range(1, 13):
             if datetime.datetime(year, month, 1).weekday() == 6:
                 count = count + 1
 
-    end = default_timer()
-
     print('Project Euler, Problem 19')
     print(f'Answer: {count}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p019()

Python/p020.py

@@ -8,13 +8,13 @@
 # Find the sum of the digits in the number 100!
 
 from math import factorial
-from timeit import default_timer
+
+from projecteuler import timing
 
 
-def main():
+@timing
-    start = default_timer()
+def p020():
-
+    # Calculate the factorial, convert the result to string and sum the digits.
-#   Calculate the factorial, convert the result to string and sum the digits.
     n = str(factorial(100))
 
     sum_ = 0
@@ -22,13 +22,9 @@ def main():
     for i in n:
         sum_ = sum_ + int(i)
 
-    end = default_timer()
-
     print('Project Euler, Problem 20')
     print(f'Answer: {sum_}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p020()