--ProjectEuler/lua/Problem14.lua
--Matthew Ellison
-- Created: 02-07-19
--Modified: 06-19-20
--[[
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?
]]
--All of my requires, unless otherwise listed, can be found at https://bitbucket.org/Mattrixwv/luaClasses
--[[
	Copyright (C) 2020  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"


local timer = Stopwatch:create();
timer:start();

TOP_NUM = 1000000	--The largest number that you will check against the chain

--This function returns a table of numbers created by the chain
local function getChain(startNum)
	--Put the starting number in the list
	local chain = {};
	chain[#chain+1] = startNum;

	--Starting with the current number perform the correct opperations on the numbers until that number reaches 1
	while(startNum > 1) do
		--Determine if the number is odd or even and perform the correct operations and add the new number to the list
		if((startNum % 2) == 0) then
			startNum = startNum / 2;
		else
			startNum = (3 * startNum) + 1;
		end
		--Add the new number to the chain
		chain[#chain+1] = startNum;
	end

	--Return the list
	return chain;
end

--Setup your variables
local largestChain = {};

--Start at 1 and run up to TOP_NUM checking how long the chain is when started with each number
for startingNumber=1,TOP_NUM do
	local currentChain = getChain(startingNumber);
	--If the new chain is longer than the current longest chain replace it
	if(#currentChain > #largestChain) then
		for location=1,#currentChain do
			largestChain[location] = currentChain[location];
		end
	end
end

timer:stop();

--Print the results
print("The longest chain with a starting number < " .. TOP_NUM .. " starts with " .. largestChain[1] .. " with a length of " .. #largestChain);
print("It took " .. timer:getSeconds() .. " seconds to run this algorithm")

--[[Results:
The longest chain with a starting number < 1000000 starts with 837799 with a length of 525
It took 13.704 seconds to run this algorithm
]]