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Problem14.py

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+#ProjectEuler/Python/Problem14.py
+#Matthew Ellison
+# Created: 01-31-19
+#Modified: 03-28-19
+"""
+The following iterative sequence is defined for the set of positive integers:
+n → n/2 (n is even)
+n → 3n + 1 (n is odd)
+Which starting number, under one million, produces the longest chain?
+"""
+#Unless otherwise listed, all of my non-standard imports can be gotten from my pyClasses repository at https://bitbucket.org/Mattrixwv/pyClasses
+"""
+	Copyright (C) 2019  Matthew Ellison
+
+	This program is free software: you can redistribute it and/or modify
+	it under the terms of the GNU Lesser General Public License as published by
+	the Free Software Foundation, either version 3 of the License, or
+	(at your option) any later version.
+
+	This program is distributed in the hope that it will be useful,
+	but WITHOUT ANY WARRANTY; without even the implied warranty of
+	MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
+	GNU Lesser General Public License for more details.
+
+	You should have received a copy of the GNU Lesser General Public License
+	along with this program.  If not, see <https://www.gnu.org/licenses/>.
+"""
+
+
+from Stopwatch import Stopwatch
+
+__topNum = 1000000	#The largest number that you will check against the chain
+
+
+#This function returns a list of numbers created by the chain
+def getChain(startNum: int) -> list:
+	#Put the starting number in the list
+	chain = [startNum]
+
+	#Starting with the current number perform the correct opperations on the numbers until that number reaches 1
+	while(startNum > 1):
+		#Determine if the number is odd or even and perform the correct operation and add the new number to the list
+		if((startNum % 2) == 0):
+			startNum /= 2
+		else:
+			startNum = (3 * startNum) + 1
+		#Add the new number to the chain
+		chain.append(startNum)
+
+	#Return the list
+	return chain
+
+def Problem14():
+	#Setup your variables
+	largestChain = []	#Holds the longest chain of numbers
+
+	#Start at 1 and run up to 1000000, checking how long the chain is when started with each number
+	for startingNumber in range(1, __topNum):
+		currentChain = getChain(startingNumber)	#Get the chain
+		#If the new chain is longer than the current longest chain replace it
+		if(len(currentChain) > len(largestChain)):
+			largestChain = currentChain.copy()
+
+	#Print the results
+	print("The longest chain with a starting number < " + str(__topNum) + " is " + str(largestChain[0]) + " with a length of " + str(len(largestChain)))
+
+
+#If you are running this file, automatically start the correct function
+if __name__ == "__main__":
+	timer = Stopwatch()	#Determines the algorithm's run time
+	timer.start()	#Start the timer
+	Problem14()		#Call the function that answers the question
+	timer.stop()	#Stop the timer
+	#Print the results of the timer
+	print("It took " + timer.getString() + " to run this algorithm")
+
+"""Results:
+The longest chain with a starting number < 1000000 is 837799 with a length of 525
+It took 28.893 seconds to run this algorithm
+"""