Initial commit with existing files

Problem23.lua

@@ -0,0 +1,91 @@
+--ProjectEuler/lua/Problem23.lua
+--Matthew Ellison
+-- Created: 03-22-19
+--Modified: 03-28-19
+--Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers
+--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"
+
+
+MAX_NUM = 28123;	--The highest number that will be evaluated
+
+
+function isSum(abund, num)
+	local sumOfNums = 0;
+	--Pick a number for the first part of the sum
+	for firstNum = 1, #abund do
+		for secondNum = firstNum, #abund do
+			sumOfNums = abund[firstNum] + abund[secondNum];
+			if(sumOfNums == num) then
+				return true;
+			elseif(sumOfNums > num) then
+				break;
+			end
+		end
+	end
+	return false;
+end
+
+
+--Setup the variables
+local timer = Stopwatch:create();
+local divisorSums = {};
+
+--Start the timer
+timer:start();
+
+--Get the sum of the divisors of all numbers < MAX_NUM
+for cnt = 1, MAX_NUM do
+	local div = getDivisors(cnt);
+	if(#div > 1) then
+		table.remove(div, #div);
+	end
+	divisorSums[cnt] = getSum(div);
+end
+
+--Get the abundant numbers
+local abund = {};
+for cnt = 1, #divisorSums do
+	if(divisorSums[cnt] > cnt) then
+		abund[#abund + 1] = cnt;
+	end
+end
+
+--Check if each number can be the sum of 2 abundant numbers and add to the sum if no
+local sumOfNums = 0;
+for cnt = 1, MAX_NUM do
+	if(not isSum(abund, cnt)) then
+		sumOfNums = sumOfNums + cnt;
+	end
+end
+
+--Stop the timer
+timer:stop();
+
+--Print the results
+io.write("The answer is " .. sumOfNums .. '\n');
+io.write("It took " .. timer:getSeconds() .. " seconds to run this algorithm\n");
+
+--[[Results:
+The answer is 4179871
+It took 393.139 seconds to run this algorithm
+]]