CPPClasses/headers/mee/numberAlgorithms.hpp

//myClasses/headers/mee/numberAlgorithms.hpp
//Matthew Ellison
// Created: 07-02-21
//Modified: 07-02-21
//This file contains declarations of functions I have created to manipulate numbers
/*
	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/>.
*/
#ifndef MEE_NUMBER_ALGORITHMS_HPP
#define MEE_NUMBER_ALGORITHMS_HPP


#include <algorithm>
#include <bitset>
#include <cinttypes>
#include <cmath>
#include <map>
#include <string>
#include <unordered_map>
#include <vector>
#include "Generator.hpp"


namespace mee{


//This function determines whether the number passed into it is a prime
template <class T>
bool isPrime(T possiblePrime){
	if(possiblePrime <= 3){
		return possiblePrime > 1;
	}
	else if(((possiblePrime % 2) == 0) || ((possiblePrime % 3) == 0)){
		return false;
	}
	for(T cnt = 5;(cnt * cnt) <= possiblePrime;cnt += 6){
		if(((possiblePrime % cnt) == 0) || ((possiblePrime % (cnt + 2)) == 0)){
			return false;
		}
	}
	return true;
}

//This is a function that returns all the primes <= goalNumber and returns a vector with those prime numbers
template <class T>
std::vector<T> getPrimes(T goalNumber){
	std::vector<T> primes;
	bool foundFactor = false;

	//If the number is 1, 0, or a negative number return an empty vector
	if(goalNumber <= 1){
		return primes;
	}
	else{
		primes.push_back(2);
	}

	//We can now start at 3 and skip all of the even numbers
	for(T possiblePrime = 3;possiblePrime <= goalNumber;possiblePrime += 2){
		//Step through every element in the current primes. If you don't find anything that divides it, it must be a prime itself
		uint64_t topPossibleFactor = ceil(sqrt(possiblePrime));
		for(uint64_t cnt = 0;(cnt < primes.size()) && (primes.at(cnt) <= topPossibleFactor);++cnt){
			if((possiblePrime % primes.at(cnt)) == 0){
				foundFactor = true;
				break;
			}
		}
		//If you didn't find a factor then it must be prime
		if(!foundFactor){
			primes.push_back(possiblePrime);
		}
		//If you did find a factor you need to reset the flag
		else{
			foundFactor = false;
		}
	}

	std::sort(primes.begin(), primes.end());
	return primes;
}

//This function returns a vector with a specific number of primes
template <class T>
std::vector<T> getNumPrimes(T numberOfPrimes){
	std::vector<T> primes;
	primes.reserve(numberOfPrimes);	//Saves cycles later
	bool foundFactor = false;

	//If the number is 1, 0, or a negative number return an empty vector
	if(numberOfPrimes <= 1){
		return primes;
	}
	//Otherwise 2 is the first prime number
	else{
		primes.push_back(2);
	}

	//Loop through every odd number starting at 3 until we find the requisite number of primes
	//Using possiblePrime >= 3 to make sure it doesn't loop back around in an overflow error and create an infinite loop
	for(T possiblePrime = 3;(primes.size() < numberOfPrimes) && (possiblePrime >= 3);possiblePrime += 2){
		//Step through every element in the current primes. If you don't find anything that divides it, it must be a prime itself
		uint64_t topPossibleFactor = ceil(sqrt(possiblePrime));
		for(uint64_t cnt = 0;(cnt < primes.size()) && (primes.at(cnt) <= topPossibleFactor);++cnt){
			if((possiblePrime % primes.at(cnt)) == 0){
				foundFactor = true;
				break;
			}
		}
		//If you didn't find a factor then it must be prime
		if(!foundFactor){
			primes.push_back(possiblePrime);
		}
		//If you did find a factor you need to reset the flag
		else{
			foundFactor = false;
		}
	}

	//The numbers should be in order, but sort them anyway just in case
	std::sort(primes.begin(), primes.end());
	return primes;
}

//This function returns all prime factors of a number
template <class T>
std::vector<T> getFactors(T goalNumber){
	//Get all the prime numbers up to sqrt(number). If there is a prime < goalNumber it will have to be <= sqrt(goalNumber)
	std::vector<T> primes = getPrimes((T)ceil(sqrt(goalNumber)));	//Make sure you are getting a vector of the correct type
	std::vector<T> factors;

	//Need to step through each prime and see if it is a factor of the number
	for(int cnt = 0;cnt < primes.size();){
		if((goalNumber % primes[cnt]) == 0){
			factors.push_back(primes[cnt]);
			goalNumber /= primes[cnt];
		}
		else{
			++cnt;
		}
	}

	//If it didn't find any factors in the primes the number itself must be prime
	if(factors.size() == 0){
		factors.push_back(goalNumber);
		goalNumber /= goalNumber;
	}

	///Should add some kind of error throwing inc ase the number != 1 after searching for all prime factors

	return factors;
}

//This is a function that gets all the divisors of num and returns a vector containing the divisors
template <class T>
std::vector<T> getDivisors(T num){
	std::vector<T> divisors;	//Holds the number of divisors
	//Ensure the parameter is a valid number
	if(num <= 0){
		return divisors;
	}
	else if(num == 1){
		divisors.push_back(1);
		return divisors;
	}
	//You only need to check up to sqrt(num)
	T topPossibleDivisor = ceil(sqrt(num));
	for(uint64_t possibleDivisor = 1;possibleDivisor <= topPossibleDivisor;++possibleDivisor){
		//Check if the counter evenly divides the number
		//If it does the counter and the other number are both divisors
		if((num % possibleDivisor) == 0){
			//We don't need to check if the number already exists because we are only checking numbers <= sqrt(num), so there can be no duplicates
			divisors.push_back(possibleDivisor);
			//We still need to account for sqrt(num) being a divisor
			if(possibleDivisor != topPossibleDivisor){
				divisors.push_back(num / possibleDivisor);
			}
			//Take care of a few occations where a number was added twice
			if(divisors.at(divisors.size() - 1) == (possibleDivisor + 1)){
				++possibleDivisor;
			}
		}
	}
	//Sort the vector for neatness
	std::sort(divisors.begin(), divisors.end());
	//Return the vector of divisors
	return divisors;
}

//These functions return the numth Fibonacci number
template <class T>
T getFib(const T num){
	//Make sure the number is within bounds
	if(num <= 2){
		return 1;
	}
	//Setup the variables
	T fib = 0;
	T tempNums[3];
	tempNums[0] = tempNums[1] = 1;

	//Do the calculation
	unsigned int cnt;
	for(cnt = 2;(cnt < num) && (tempNums[(cnt - 1) % 3] >= tempNums[(cnt - 2) % 3]);++cnt){
		tempNums[cnt % 3] = tempNums[(cnt + 1) % 3] + tempNums[(cnt + 2) % 3];
	}
	fib = tempNums[(cnt - 1) % 3];	//Transfer the answer to permanent variable. -1 to account for the offset of starting at 0

	return fib;
}

//This function returns a vector that includes all Fibonacci numbers <= num
template <class T>
std::vector<T> getAllFib(const T num){
	std::vector<T> fibList;
	//Make sure the number is within bounds
	if(num <= 1){
		fibList.push_back(1);
		return fibList;
	}
	else{	//Make sure to add the first 2 elements
		fibList.push_back(1);
		fibList.push_back(1);
	}

	//Setup the variables
	T fib = 0;
	T tempNums[3];
	tempNums[0] = tempNums[1] = 1;

	//Do the calculation and add each number to the vector
	for(T cnt = 2;(tempNums[(cnt - 1) % 3] <= num) && (tempNums[(cnt - 1) % 3] >= tempNums[(cnt - 2) % 3]);++cnt){
		tempNums[cnt % 3] = tempNums[(cnt + 1) % 3] + tempNums[(cnt + 2) % 3];
		fibList.push_back(tempNums[cnt % 3]);
	}

	//If you triggered the exit statement you have one more element than you need
	fibList.pop_back();

	//Return the vector that contains all of the Fibonacci numbers
	return fibList;
}

//This function converts a number to its binary equivalent
template <class T>
std::string toBin(T num){
	//Convert the number to a binary string
	std::string fullString = std::bitset<sizeof(T) * 8>(num).to_string();
	//Remove leading zeros
	int loc = 0;
	for(loc = 0;(loc < fullString.size()) && (fullString[loc] == '0');++loc);
	std::string trimmedString = fullString.substr(loc);
	if(trimmedString == ""){
		trimmedString = "0";
	}
	return trimmedString;
}

//Return the factorial of the number passed in
template <class T>
T factorial(T num){
	T fact = 1;
	for(T cnt = 1;cnt <= num;++cnt){
		fact *= cnt;
	}
	return fact;
}

//A generator for prime numbers
template <class T>
mee::Generator<T> sieveOfEratosthenes(){
	//Return 2 the first time, this lets us skip all even numbers later
	co_yield 2;

	int num = 0;

	//Dictionary to hold the primes we have already found
	std::unordered_map<T, std::vector<T>> dict;

	//Start checking for primes with the number 3 and skip all even numbers
	for(T possiblePrime = 3;true;possiblePrime += 2){
		//If possiblePrime is in the dictionary it is a composite number
		if(dict.contains(possiblePrime)){
			//Move each number to its next odd multiple
			for(T num : dict[possiblePrime]){
				dict[possiblePrime + num + num].push_back(num);
			}
			//We no longer need this, free the memory
			dict.erase(possiblePrime);
		}
		//If possiblePrime is not in the dictionary it is a new prime number
		//Return it and mark its next multiple
		else{
			co_yield possiblePrime;
			dict[possiblePrime * possiblePrime].push_back(possiblePrime);
		}
	}
}
//An alternate to sieveOfEratosthenes that uses map instead of unordered_map for greater compatibility but lower performance
template <class T>
mee::Generator<T> sieveOfEratosthenesAlt(){
	//Return 2 the first time, this lets us skip all even numbers later
	co_yield 2;

	int num = 0;

	//Dictionary to hold the primes we have already found
	std::map<T, std::vector<T>> dict;

	//Start checking for primes with the number 3 and skip all even numbers
	for(T possiblePrime = 3;true;possiblePrime += 2){
		//If possiblePrime is in the dictionary it is a composite number
		if(dict.contains(possiblePrime)){
			//Move each number to its next odd multiple
			for(T num : dict[possiblePrime]){
				dict[possiblePrime + num + num].push_back(num);
			}
			//We no longer need this, free the memory
			dict.erase(possiblePrime);
		}
		//If possiblePrime is not in the dictionary it is a new prime number
		//Return it and mark its next multiple
		else{
			co_yield possiblePrime;
			dict[possiblePrime * possiblePrime].push_back(possiblePrime);
		}
	}
}


}


#endif	//MEE_NUMBER_ALGORITHMS_HPP