--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
]]