LuaClasses/Algorithms.lua

--luaClasses/Algorithms.lua
--Matthew Ellison
-- Created: 2-4-19
--Modified: 2-7-19
--This is a file of algorithms that I have found it useful to keep around at all times


--This function returns a list with all the primes numbers <= goalNumber
function getPrimes(goalNumber)
	local primes = {};	--Holds the prime numbers
	local foundFactor = false;

	--If the number is 0 or negative return an empty table
	if(goalNumber <= 1) then
		return primes;
	--Otherwise the number is at lease 2, therefore 2 should be added to the list
	else
		primes[#primes + 1] = 2;
	end

	--We can now start at 3 and skip all even numbers, because they cannot be prime
	for possiblePrime=3,goalNumber,2 do
		--Check all current primes, up to sqrt(possiblePrime), to see if there is a divisor
		primesCnt = 1;
		--We can safely assume that there will be at least 1 element in the primes list because of 2 being added before the loop
		topPossibleFactor = math.ceil(math.sqrt(possiblePrime))
		while(primes[primesCnt] <= topPossibleFactor) do
			if((possiblePrime % primes[primesCnt]) == 0) then
				foundFactor = true;
				break;
			else
				primesCnt = primesCnt + 1;
			end
			--Check if the index has gone out of range
			if(primesCnt >= #primes) then
				break;
			end
		end

		--If you didn't find a factor then the current number must be prime
			if(not foundFactor) then
				primes[#primes + 1] = possiblePrime;
			else
				foundFactor = false;
			end
	end

	--Sort the array just to be neat and safe
	table.sort(primes);

	--Return the array
	return primes;
end

--This function gets a specific number of primes
function getNumPrimes(numberOfPrimes)
	local primes = {};	--Holds the prime numbers
	local foundFactor = false;

	--If the number is 0 or negative return an empty table
	if(numberOfPrimes <= 1) then
		return primes;
	--Otherwise the number is at lease 2, therefore 2 should be added to the list
	else
		primes[#primes + 1] = 2;
	end

	--We can now start at 3 and skip all even numbers, because they cannot be prime
	possiblePrime = 3;
	while((#primes < numberOfPrimes) and (possiblePrime >= 0)) do
		--Check all current primes, up to sqrt(possiblePrime), to see if there is a divisor
		primesCnt = 1;
		--We can safely assume that there will be at least 1 element in the primes list because of 2 being added before the loop
		topPossibleFactor = math.ceil(math.sqrt(possiblePrime))
		while(primes[primesCnt] <= topPossibleFactor) do
			if((possiblePrime % primes[primesCnt]) == 0) then
				foundFactor = true;
				break;
			else
				primesCnt = primesCnt + 1;
			end
			--Check if the index has gone out of range
			if(primesCnt >= #primes) then
				break;
			end
		end

		--If you didn't find a factor then the current number must be prime
		if(not foundFactor) then
			primes[#primes + 1] = possiblePrime;
		else
			foundFactor = false;
		end

		--Advance to the next possible prime number
		possiblePrime = possiblePrime + 2;
	end

	--Sort the array just to be neat and safe
	table.sort(primes);

	--Return the array
	return primes;
end

--This is a function that returns all the factors of goalNumber
function getFactors(goalNumber)
	local primes = getPrimes(math.ceil(math.sqrt(goalNumber)));	--Get all the primes up the largest possible divisor
	local factors = {};	--Holds all the factors

	--You need to step through each prime and see if it is a factor of the number
	cnt = 1;
	while((cnt <= #primes) and (goalNumber > 1)) do
		--If the prime is a factor you need to add it to the factor list
		if((goalNumber % primes[cnt]) == 0) then
			factors[#factors + 1] = primes[cnt];
			goalNumber = goalNumber / primes[cnt];
		--Otherwise advance the location in the primes array you are looking at
		--By not advancing if the primes is a factor you allow for multiple of the same prime number as a factor
		else
			cnt = cnt + 1
		end
	end

	--If you didn't get any factors the number itself must be a prime number
	if(#factors == 0) then
		factors[#factors + 1] = goalNumber;
		goalNumber = 1
	end

	--If your some reason teh goalNumber is not 1 print an error message
	if(goalNumber > 1) then
		print("There was an error in getFactors(). A leftover of " .. goalNumber);
	end

	--Return the list of factors
	return factors;
end

--This is a function the returns all divisors of the number passed to it
function getDivisors(goalNumber)
	local divisors = {};
	--Start by checking that the number is positive
	if(goalNumber <= 0) then
		return divisors;
	--If the number is 1 return just itself
	elseif(goalNumber == 1) then
		divisors[#divisors+1] = 1;
	end

	--Start at 3 and loop through all numbers < (goalNumber / 2) looking for a number that divides it evenly
	local topPossibleDivisor = math.ceil(math.sqrt(goalNumber));
	local possibleDivisor = 1;
	while(possibleDivisor <= topPossibleDivisor) do
		--If you find one add it and the number it creates to the list
		if((goalNumber % possibleDivisor) == 0) then
			divisors[#divisors+1] = possibleDivisor;
			--Account for the possibility of sqrt(goalNumber) being a divisor
			if(possibleDivisor ~= topPossibleDivisor) then
				divisors[#divisors + 1] =(goalNumber / possibleDivisor);
			end
			--Take care of a few occations where a number was added twice
			if(divisors[#divisors] == (possibleDivisor + 1)) then
				possibleDivisor = possibleDivisor + 1;
			end
		end
		possibleDivisor = possibleDivisor + 1;
	end
	--Sort the list before returning for neatness
	table.sort(divisors);
	--Return the list
	return divisors;
end

function getSum(ary)
	sum = 0;	--Holds the sum of all elements. Start at 0 because num+1 = num
	--Look through every element in the array and add the number to the running sum
	for location = 1, #ary do
		sum = sum + ary[location];
	end
	--Return the sum of all elements
	return sum;
end

function getProd(ary)
	prod = 1;	--Holds the product of all elements. Start at 1 because num*1 = num
	--Look through every element in the array and add the number to the running product
	for location = 1, #ary do
		prod = prod * ary[location];
	end
	--Return the product of all elements
	return prod;
end