--ProjectEuler/ProjectEulerLua/Problem35.lua
--Matthew Ellison
-- Created: 06-05-21
--Modified: 06-05-21
--How many circular primes are there below one million?
--All of my requires, unless otherwise listed, can be found at https://bitbucket.org/Mattrixwv/luaClasses
--[[
	Copyright (C) 2021  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"


--Setup the variables
local timer = Stopwatch:create();
local MAX_NUM = 999999;	--The largest number that we are checking for primes
local primes = {};	--The primes below MAX_NUM
local circularPrimes = {};	--The circular primes below MAX_NUM

--Functions
--Returns a list of all rotations of a string passed to it
local function getRotations(str)
	local rotations = {};
	table.insert(rotations, str);
	for cnt = 1, string.len(str) - 1 do
		str = string.sub(str, 2) .. string.sub(str, 1, 1);
		table.insert(rotations, str);
	end
	return rotations;
end

--Start the timer
timer:start();

--Get all primes under 1,000,000
primes = getPrimes(MAX_NUM);
--Go through all primes, get all their rotations, and check if those numbers are also primes
for cnt = 1, #primes do
	local prime = primes[cnt];
	local allRotationsPrime = true;
	--Get all of the rotations of the prime and see if they are also prime
	local rotations = getRotations(tostring(prime));
	for rotCnt = 1, #rotations do
		local rotation = rotations[rotCnt];
		local p = tonumber(rotation, 10);
		if(not isFound(primes, p)) then
			allRotationsPrime = false;
			break;
		end
	end
	--If all rotations are prime add it to the list of circular primes
	if(allRotationsPrime) then
		table.insert(circularPrimes, prime);
	end
end

--Stop the timer
timer:stop();

--Print the results
io.write("The number of all circular prime numbers under " .. MAX_NUM .. " is " .. #circularPrimes .. "\n");
io.write("It took " .. timer:getString() .. " to run this algorithm\n");

--[[ Results:
The number of all circular prime numbers under 999999 is 55
It took 102.268 seconds to run this algorithm
]]