Use timing decorator for last solved problems

Python/p081.py

@@ -13,7 +13,8 @@
 # by only moving right and down.
 
 import sys
-from timeit import default_timer
+
+from projecteuler import timing
 
 
 def sum_paths(matrix, m, n):
@@ -31,9 +32,8 @@ def sum_paths(matrix, m, n):
     return matrix[0][0]
 
 
-def main():
+@timing
-    start = default_timer()
+def p081():
-
     try:
         with open('matrix.txt', 'r', encoding='utf-8') as fp:
             matrix = fp.readlines()
@@ -49,13 +49,9 @@ def main():
 
     dist = sum_paths(matrix, 80, 80)
 
-    end = default_timer()
-
     print('Project Euler, Problem 81')
     print(f'Answer: {dist}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p081()

Python/p082.py

@@ -16,16 +16,14 @@
 # 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 projecteuler import dijkstra
+from projecteuler import dijkstra, timing
 
 
-def main():
+@timing
-    start = default_timer()
+def p082():
-
     try:
         with open('matrix.txt', 'r', encoding='utf-8') as fp:
             matrix = fp.readlines()
@@ -43,24 +41,19 @@ def main():
 
     min_path = 999999999
 
-#   Use Dijkstra's algorithm starting from all possible nodes
+    # Use Dijkstra's algorithm starting from all possible nodes in the first column.
-#   in the first column.
     for i in range(80):
         dijkstra(matrix, distances, 80, 80, 1, 0, i)
 
-#       For the current starting node, check if there is an ending node
+        # For the current starting node, check if there is an ending node
-#       with a smaller path cost than the current minimum.
+        # with a smaller path cost than the current minimum.
         for j in range(80):
             if distances[j][79] < min_path:
                 min_path = distances[j][79]
 
-    end = default_timer()
-
     print('Project Euler, Problem 82')
     print(f'Answer: {min_path}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p082()

Python/p083.py

@@ -15,16 +15,14 @@
 # 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 projecteuler import dijkstra
+from projecteuler import dijkstra, timing
 
 
-def main():
+@timing
-    start = default_timer()
+def p083():
-
     try:
         with open('matrix.txt', 'r', encoding='utf-8') as fp:
             matrix = fp.readlines()
@@ -43,13 +41,9 @@ def main():
 
     dist = distances[79][79]
 
-    end = default_timer()
-
     print('Project Euler, Problem 83')
     print(f'Answer: {dist}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p083()

Python/p088.py

@@ -18,7 +18,7 @@
 #
 # What is the sum of all the minimal product-sum numbers for 2<=k<=12000?
 
-from math import floor, sqrt, prod
+from math import prod
 
 from projecteuler import sieve, timing
 

Python/p104.py

@@ -10,35 +10,33 @@
 # 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.
 
 import sys
-from timeit import default_timer
 
 from mpmath import matrix
 
-from projecteuler import is_pandigital
+from projecteuler import is_pandigital, timing
 
 
 sys.set_int_max_str_digits(70000)
 
 
-def main():
+@timing
-    start = default_timer()
+def p104():
-
     found = 0
     fib1 = 1
     fib2 = 1
     i = 2
 
     while not found:
-#       Calculate the next Fibonacci number modulo 10^9 and check if the result is 1-9 pandigital.
+        # Calculate the next Fibonacci number modulo 10^9 and check if the result is 1-9 pandigital.
         fibn = (fib1 + fib2) % 1000000000
         fib1 = fib2
         fib2 = fibn
         i = i + 1
 
-#       If the last 9 digits of fib_n are pandigital, calculate the ith Fibonacci number using
+        # If the last 9 digits of fib_n are pandigital, calculate the ith Fibonacci number using
-#       the matrix representation, with sufficient precision so that the at least the first
+        # the matrix representation, with sufficient precision so that the at least the first
-#       9 digits are correct (we don't need the whole number. If the first 9 digits are also
+        # 9 digits are correct (we don't need the whole number. If the first 9 digits are also
-#       pandigital, we found the solution.
+        # pandigital, we found the solution.
         if is_pandigital(fibn, 9):
             fib_matrix = matrix(2)
             fib_matrix[0, 0] = 1
@@ -52,13 +50,9 @@ def main():
             if is_pandigital(int(str(fib)[:9]), 9):
                 found = 1
 
-    end = default_timer()
-
     print('Project Euler, Problem 104')
     print(f'Answer: {i}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p104()

Python/p145.py

@@ -8,18 +8,17 @@
 #
 # How many reversible numbers are there below one-billion (109)?
 
-from timeit import default_timer
+from projecteuler import timing
 
 
-def main():
+@timing
-    start = default_timer()
+def p145():
-
     N = 1000000000
 
     count = 0
 
-#   Brute force approach, sum each number and their reverse and
+    # Brute force approach, sum each number and their reverse and
-#   check if there are only odd digits.
+    # check if there are only odd digits.
     for i in range(11, N):
         if i % 10 != 0:
             s = str(i + int(''.join(reversed(str(i)))))
@@ -28,13 +27,9 @@ def main():
                     '6' not in s and '8' not in s:
                 count = count + 1
 
-    end = default_timer()
-
     print('Project Euler, Problem 145')
     print(f'Answer: {count}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p145()

Python/p206.py

@@ -3,26 +3,25 @@
 # 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
+from projecteuler import timing
 
 
-def main():
+@timing
-    start = default_timer()
+def p206():
-
+    # Since the square of n has 19 digits, n must be at least 10^9.
-#   Since the square of n has 19 digits, n must be at least 10^9.
     n = 1000000000
     found = 0
 
     while not found:
-#       If the square on n ends with 10, n must be divisible by 10.
+        # If the square on n ends with 10, n must be divisible by 10.
         n = n + 10
         p = n * n
 
-#       A square divisible by 10 is also divisible by 100.
+        # A square divisible by 10 is also divisible by 100.
         if p % 100 != 0:
             continue
 
-#       Check if the digits of the square correspond to the given pattern.
+        # Check if the digits of the square correspond to the given pattern.
         i = 9
         p = p // 100
 
@@ -35,13 +34,9 @@ def main():
         if p == 0:
             found = 1
 
-    end = default_timer()
-
     print('Project Euler, Problem 206')
     print(f'Answer: {n}')
 
-    print(f'Elapsed time: {end - start:.9f} seconds')
-
 
 if __name__ == '__main__':
-    main()
+    p206()