Initial commit with existing files

Problem24.m

@@ -0,0 +1,87 @@
+function [] = Problem24()
+%ProjectEuler/Octave/Problem24.m
+%Matthew Ellison
+% Created: 03-25-19
+%Modified: 03-28-19
+%What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?
+%{
+	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/>.
+%}
+
+
+NEEDED_PERM = 1000000;	%The number of the permutation that you need
+
+
+%Setup the variables
+nums = "0123456789";
+
+%Start the time
+startTime = clock();
+
+%Get all permutations of the string
+permutations = getPermutations(nums);
+
+%Stop the timer
+endTime = clock();
+
+%Print the results
+printf("The 1 millionth permutation is %s\n", permutations{NEEDED_PERM})
+printf("It took %f second to run this algorithm\n", etime(endTime, startTime))
+
+
+end	%Problem24()
+
+function [perms] = getPermutations(master, num = 1)
+	perms = {};
+	%Check if the number is out of bounds
+	if((num > size(master)(2)) || (num <= 0))
+		%Do nothing and return and empty list
+		perms;
+	%If this is the last possible recurse just return the current string
+	elseif(num == size(master)(2))
+		perms(end + 1) = master;
+	%If there are more possible recurses, recurse with the current permutations
+	else
+		temp = getPermutations(master, num + 1);
+		perms = {perms{:}, temp{:}};
+		%You need to swap the current letter with every possible letter after it
+		%The ones needed to swap before will happen automatically when the function recurses
+		for cnt = 1 : (size(master)(2) - num)
+			%Swap two elements
+			temp = master(num);
+			master(num) = master(num + cnt);
+			master(num + cnt) = temp;
+			%Get the permutations after swapping two elements
+			temp = getPermutations(master, num + 1);
+			perms = {perms{:}, temp{:}};
+			%Swap the elements back
+			temp = master(num);
+			master(num) = master(num + cnt);
+			master(num + cnt) = temp;
+		end
+	end
+
+	%The array is not necessarily in alpha-numeric order. So if this is the full array sort it before returning
+	if(num == 1)
+		perms = sort(perms);
+	end
+end	%getPermutations()
+
+%{
+Results:
+The 1 millionth permutation is 2783915460
+It took 433.922920 second to run this algorithm
+%}