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171 lines
5.2 KiB
Lua
171 lines
5.2 KiB
Lua
--luaClasses/Algorithms.lua
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--Matthew Ellison
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-- Created: 2-4-19
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--Modified: 2-4-19
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--This is a file of algorithms that I have found it useful to keep around at all times
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--This function returns a list with all the primes numbers <= goalNumber
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function getPrimes(goalNumber)
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local primes = {}; --Holds the prime numbers
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local foundFactor = false;
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--If the number is 0 or negative return an empty table
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if(goalNumber <= 1) then
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return primes;
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--Otherwise the number is at lease 2, therefore 2 should be added to the list
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else
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primes[#primes + 1] = 2;
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end
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--We can now start at 3 and skip all even numbers, because they cannot be prime
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for possiblePrime=3,goalNumber,2 do
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--Check all current primes, up to sqrt(possiblePrime), to see if there is a divisor
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primesCnt = 1;
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--We can safely assume that there will be at least 1 element in the primes list because of 2 being added before the loop
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topPossibleFactor = math.ceil(math.sqrt(possiblePrime))
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while(primes[primesCnt] <= topPossibleFactor) do
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if((possiblePrime % primes[primesCnt]) == 0) then
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foundFactor = true;
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break;
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else
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primesCnt = primesCnt + 1;
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end
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--Check if the index has gone out of range
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if(primesCnt >= #primes) then
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break;
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end
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end
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--If you didn't find a factor then the current number must be prime
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if(not foundFactor) then
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primes[#primes + 1] = possiblePrime;
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else
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foundFactor = false;
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end
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end
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--Sort the array just to be neat and safe
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table.sort(primes);
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--Return the array
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return primes;
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end
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--This function gets a specific number of primes
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function getNumPrimes(numberOfPrimes)
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local primes = {}; --Holds the prime numbers
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local foundFactor = false;
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--If the number is 0 or negative return an empty table
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if(numberOfPrimes <= 1) then
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return primes;
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--Otherwise the number is at lease 2, therefore 2 should be added to the list
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else
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primes[#primes + 1] = 2;
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end
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--We can now start at 3 and skip all even numbers, because they cannot be prime
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possiblePrime = 3;
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while((#primes < numberOfPrimes) and (possiblePrime >= 0)) do
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--Check all current primes, up to sqrt(possiblePrime), to see if there is a divisor
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primesCnt = 1;
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--We can safely assume that there will be at least 1 element in the primes list because of 2 being added before the loop
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topPossibleFactor = math.ceil(math.sqrt(possiblePrime))
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while(primes[primesCnt] <= topPossibleFactor) do
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if((possiblePrime % primes[primesCnt]) == 0) then
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foundFactor = true;
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break;
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else
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primesCnt = primesCnt + 1;
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end
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--Check if the index has gone out of range
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if(primesCnt >= #primes) then
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break;
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end
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end
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--If you didn't find a factor then the current number must be prime
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if(not foundFactor) then
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primes[#primes + 1] = possiblePrime;
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else
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foundFactor = false;
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end
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--Advance to the next possible prime number
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possiblePrime = possiblePrime + 2;
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end
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--Sort the array just to be neat and safe
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table.sort(primes);
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--Return the array
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return primes;
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end
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--This is a function that returns all the factors of goalNumber
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function getFactors(goalNumber)
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local primes = getPrimes(math.ceil(math.sqrt(goalNumber))); --Get all the primes up the largest possible divisor
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local factors = {}; --Holds all the factors
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--You need to step through each prime and see if it is a factor of the number
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cnt = 1;
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while((cnt <= #primes) and (goalNumber > 1)) do
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--If the prime is a factor you need to add it to the factor list
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if((goalNumber % primes[cnt]) == 0) then
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factors[#factors + 1] = primes[cnt];
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goalNumber = goalNumber / primes[cnt];
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--Otherwise advance the location in the primes array you are looking at
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--By not advancing if the primes is a factor you allow for multiple of the same prime number as a factor
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else
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cnt = cnt + 1
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end
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end
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--If you didn't get any factors the number itself must be a prime number
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if(#factors == 0) then
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factors[#factors + 1] = goalNumber;
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goalNumber = 1
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end
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--If your some reason teh goalNumber is not 1 print an error message
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if(goalNumber > 1) then
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print("There was an error in getFactors(). A leftover of " .. goalNumber);
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end
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--Return the list of factors
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return factors;
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end
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function getDivisors(goalNumber)
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local divisors = {};
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--Start by checking that the number is positive
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if(goalNumber <= 0) then
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return divisors;
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--If the number is 1 return just itself
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elseif(goalNumber == 1) then
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divisors[#divisors+1] = 1;
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--Otherwise add 1 and itself to the list
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else
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divisors[#divisors+1] = 1;
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divisors[#divisors+1] = goalNumber;
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end
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--Start at 3 and loop through all numbers < (goalNumber / 2) looking for a number that divides it evenly
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local topPossibleDivisor = math.ceil(math.sqrt(goalNumber));
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for possibleDivisor=2,topPossibleDivisor do
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--If you find one add it and the number it creates to the list
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if((goalNumber % possibleDivisor) == 0) then
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divisors[#divisors+1] = possibleDivisor;
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--Account for the possibility of sqrt(goalNumber) being a divisor
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if(possibleDivisor ~= topPossibleDivisor) then
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divisors[#divisors + 1] =(goalNumber / possibleDivisor);
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end
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end
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end
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--Sort the list before returning for neatness
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table.sort(divisors);
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--Return the list
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return divisors;
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end
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