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Problem21.lua

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+--ProjectEuler/lua/Problem21.lua
+--Matthew Ellison
+-- Created: 03-19-19
+--Modified: 03-28-19
+--Evaluate the sum of all the amicable numbers under 10000
+--All of my requires, unless otherwise listed, can be found at https://bitbucket.org/Mattrixwv/luaClasses
+--[[
+	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/>.
+]]
+
+
+require "Stopwatch"
+require "Algorithms"
+
+LIMIT = 10000;	--The top number that will be evaluated
+
+
+--Setup the timer
+timer = Stopwatch:create();
+
+--Setup the variables
+divisorSum = {};	--Holds the sum of the factors of the subscript number
+
+--Start the timer
+timer:start();
+
+--Generate the factors of all the numbers < 10000, get their sum, and add it to the list
+for cnt = 1, LIMIT do
+	divisors = getDivisors(cnt);	--Get all the divisors of a number
+	if(#divisors > 1) then
+		table.remove(divisors);	--Remove the last entry because it will be the number itself
+	end
+	divisorSum[#divisorSum + 1] = getSum(divisors);	--Add the sum of the divisors to the vector
+end
+--Check every sum of divisors in the list for a matching sum
+amicable = {};
+for cnt = 1, #divisorSum do
+	sum = divisorSum[cnt];
+	--If the sum is greater than the number of divisors then it is impossible to be amicable. Skip the number and continue
+	if(sum >= #divisorSum) then
+	--We know that divisorSum[cnt] == sum, do if divisorSum[sum] == cnt we have found an amicable number
+	elseif(divisorSum[sum] == cnt) then
+		--A number can't be amicable with itself, so skip those numbers
+		if(sum == cnt) then
+		--Add the number to the amicable vector
+		else
+			amicable[#amicable + 1] = cnt;
+		end
+	end
+end
+
+--Stop the timer
+timer:stop();
+
+--Sort the vector for neatness
+table.sort(amicable);
+
+--Print the results
+io.write("All amicable numbers less than " .. LIMIT .. " are \n");
+--Print the list of amicable numbers
+for location = 1, #amicable do
+	io.write(amicable[location] .. '\n');
+end
+io.write("The sum of all of these amicable numbers is " .. getSum(amicable) .. '\n');
+io.write("It took " .. timer:getMilliseconds() .. " milliseconds to run this algorithm\n");
+
+--[[ Results:
+All amicable numbers less than 10000 are
+220
+284
+1184
+1210
+2620
+2924
+5020
+5564
+6232
+6368
+The sum of all of these amicable numbers is 31626
+It took 46.0 milliseconds to run this algorithm
+]]