Rename some input files

Python/p018.py

@@ -40,13 +40,13 @@ def main():
     start = default_timer()
 
     try:
-        with open('triang.txt', 'r', encoding='utf-8') as fp:
+        with open('p018_triangle.txt', 'r', encoding='utf-8') as fp:
             triang = []
 
             for line in fp:
                 triang.append(line.strip('\n').split())
     except FileNotFoundError:
-        print('Error while opening file trian.txt')
+        print('Error while opening file p018_triangle.txt')
         sys.exit(1)
 
     l = len(triang)

Python/p022.py

@@ -1,6 +1,6 @@
 #!/usr/bin/env python3
 
-# Using names.txt, a 46K text file containing over five-thousand first names, begin by sorting it into alphabetical order.
+# Using p022_names.txt, a 46K text file containing over five-thousand first names, begin by sorting it into alphabetical order.
 # Then working out the alphabetical value for each name, multiply this value by its alphabetical position in the list to obtain a name score.
 #
 # For example, when the list is sorted into alphabetical order, COLIN, which is worth 3 + 15 + 12 + 9 + 14 = 53, is the 938th name in the list.
@@ -16,10 +16,10 @@ def main():
     start = default_timer()
 
     try:
-        with open('names.txt', 'r', encoding='utf-8') as fp:
+        with open('p022_names.txt', 'r', encoding='utf-8') as fp:
             names = list(fp.readline().replace('"', '').split(','))
     except FileNotFoundError:
-        print('Error while opening file names.txt')
+        print('Error while opening file p022_names.txt')
         sys.exit(1)
 
     names.sort()

Python/p042.py

@@ -7,7 +7,7 @@
 # By converting each letter in a word to a number corresponding to its alphabetical position and adding these values we form a word value.
 # For example, the word value for SKY is 19 + 11 + 25 = 55 = t10. If the word value is a triangle number then we shall call the word a triangle word.
 #
-# Using words.txt (right click and 'Save Link/Target As...'), a 16K text file containing nearly two-thousand common English words,
+# Using p042_words.txt (right click and 'Save Link/Target As...'), a 16K text file containing nearly two-thousand common English words,
 # how many are triangle words?
 
 import sys
@@ -31,10 +31,10 @@ def main():
     start = default_timer()
 
     try:
-        with open('words.txt', 'r', encoding='utf-8') as fp:
+        with open('p042_words.txt', 'r', encoding='utf-8') as fp:
             words = list(fp.readline().replace('"', '').split(','))
     except FileNotFoundError:
-        print('Error while opening file words.txt')
+        print('Error while opening file p042_words.txt')
         sys.exit(1)
 
     count = 0

Python/p054.py

@@ -39,7 +39,7 @@
 #         Full House  	    Full House
 #      With Three Fours  With Three Threes
 #
-#      The file, poker.txt, contains one-thousand random hands dealt to two players. Each line of the file contains ten cards
+#      The file, p054_poker.txt, contains one-thousand random hands dealt to two players. Each line of the file contains ten cards
 #      (separated by a single space): the first five are Player 1's cards and the last five are Player 2's cards.
 #      You can assume that all hands are valid (no invalid characters or repeated cards), each player's hand is in no specific order,
 #      and in each hand there is a clear winner.
@@ -447,10 +447,10 @@ def main():
     start = default_timer()
 
     try:
-        with open('poker.txt', 'r', encoding='utf-8') as fp:
+        with open('p054_poker.txt', 'r', encoding='utf-8') as fp:
             games = fp.readlines()
     except FileNotFoundError:
-        print('Error while opening file poker.txt')
+        print('Error while opening file p054_poker.txt')
         sys.exit(1)
 
     count = 0

Python/p059.py

@@ -14,7 +14,7 @@
 # the message, which is likely, the key is repeated cyclically throughout the message. The balance for this method is using a sufficiently long
 # password key for security, but short enough to be memorable.
 #
-# Your task has been made easy, as the encryption key consists of three lower case characters. Using cipher.txt, a file containing the
+# Your task has been made easy, as the encryption key consists of three lower case characters. Using p059_cipher.txt, a file containing the
 # 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.
 
@@ -82,7 +82,7 @@ def main():
 
     enc_text = EncryptedText()
 
-    if enc_text.read_text('cipher.txt') == -1:
+    if enc_text.read_text('p059_cipher.txt') == -1:
         sys.exit(1)
 
     plain_text = enc_text.decrypt()

Python/p067.py

@@ -8,7 +8,7 @@
 # 8 5 9 3
 #
 # That is, 3 + 7 + 4 + 9 = 23.
-# Find the maximum total from top to bottom in triangle.txt (right click and 'Save Link/Target As...'), a 15K text file containing a triangle
+# Find the maximum total from top to bottom in p067_triangle.txt (right click and 'Save Link/Target As...'), a 15K text file containing a triangle
 # with one-hundred rows.
 #
 # NOTE: This is a much more difficult version of Problem 18. It is not possible to try every route to solve this problem, as there are 299 altogether!
@@ -27,11 +27,11 @@ def main():
     triang = []
 
     try:
-        with open('triangle.txt', 'r', encoding='utf-8') as fp:
+        with open('p067_triangle.txt', 'r', encoding='utf-8') as fp:
             for line in fp:
                 triang.append(line.strip('\n').split())
     except FileNotFoundError:
-        print('Error while opening file triangle.txt')
+        print('Error while opening file p067_triangle.txt')
         sys.exit(1)
 
     l = len(triang)

Python/p079.py

@@ -3,7 +3,7 @@
 # A common security method used for online banking is to ask the user for three random characters from a passcode.
 # For example, if the passcode was 531278, they may ask for the 2nd, 3rd, and 5th characters; the expected reply would be: 317.
 #
-# The text file, keylog.txt, contains fifty successful login attempts.
+# The text file, p079_keylog.txt, contains fifty successful login attempts.
 #
 # Given that the three characters are always asked for in order, analyse the file so as to determine the shortest possible
 # secret passcode of unknown length.
@@ -36,10 +36,10 @@ def main():
     start = default_timer()
 
     try:
-        with open('keylog.txt', 'r', encoding='utf-8') as fp:
+        with open('p079_keylog.txt', 'r', encoding='utf-8') as fp:
             logins = fp.readlines()
     except FileNotFoundError:
-        print('Error while opening file keylog.txt')
+        print('Error while opening file p079_keylog.txt')
         sys.exit(1)
 
     digits = [0] * 10

README.md

@@ -1,6 +1,6 @@
 # My Project Euler solutions
 These are my solutions in C and Python, not necessarily the best solutions. I've solved most of the first 100 problems, currently working on cleaning the code and uploading it. I will try to solve more problems in the future.
 
-# Issues
-- Solutions for problems 82 and 145 in Python run really slow.
+# Notes
+- Solutions for problems 82 and 145 in Python are really slow.
 - Solutions for problems 84, 85, 86, 87, 89, 92, 95, 96, 97. 99, 102, 112, 124 and 357 have been implemented in C but not in Python.