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Problem14.m

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+%ProjectEuler/Octave/Problem14.m
+%Matthew Ellison
+% Created: 
+%Modified: 03-28-19
+%This is a script to answer Problem 14 for Project Euler
+%{
+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?
+%}
+%{
+	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/>.
+%}
+
+
+%Setup your variables
+currentLength = 0;
+maxLength = 0;
+currentNum = 1;
+maxNum = 0;
+
+%Start the timer
+startTime = clock();
+
+%Step through every number less than 1000000
+while(currentNum < 1000000)
+	currentLength = 0;
+	workingNum = currentNum;	%So you can do math on the current number
+	while(workingNum > 1)
+		if(mod(workingNum, 2) == 0)
+			workingNum = workingNum / 2;
+		else
+			workingNum = (3 * workingNum) + 1;
+		end
+		++currentLength;
+	end
+	%If the current number has a longer chain than the current max it becomes the max
+	if(currentLength > maxLength)
+		maxLength = currentLength;
+		maxNum = currentNum;
+	end
+	++currentNum;
+end
+
+%Stop the timer
+endTime = clock();
+
+%Print the results
+printf("The longest chain with a starting number < 1000000 is %d with a length of %d\n", maxNum, maxLength)
+printf("It took %f seconds to run this algorithm\n", etime(endTime, startTime))
+
+%Cleanup the variables
+clear maxLength;
+clear maxNum;
+clear endTime;
+clear startTime;
+clear currentNum;
+clear currentLength;
+clear workingNum;
+clear ans;
+
+%{
+Results:
+The longest chain with a starting number < 1000000 is 837799 with a length of 524
+It took 1136.180946 seconds to run this algorithm
+%}