TypescriptClasses/NumberAlgorithms.ts

//typescriptClasses/NumberAlgorithms.ts
//Matthew Ellison
// Created: 07-13-21
//Modified: 07-13-21
//Algorithms for 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/>.
*/


import { InvalidResult } from "./InvalidResult";


//Generate an infinite sequence of prime numbers using the Sieve of Eratosthenes
export function* sieveOfEratosthenes(){
	//Return 2 the first time, this lets us skip all even numbers later
	yield 2;

	//Dictionary to hold the primes we have already found
	let dict = new Map<number, number[]>();

	//Start checking for primes with the number 3 and skip all even numbers
	for(let possiblePrime = 3;true;possiblePrime += 2){
		//If possiblePrime is not in the dictionary it is a new prime number
		//Return it and mark its next multiple
		if(!dict.has(possiblePrime)){
			yield possiblePrime;
			dict.set(possiblePrime * possiblePrime, [possiblePrime]);
		}
		//If possiblePrime is in the dictionary it is a composite number
		else{
			//Move each witness to its next multiple
			for(let num of dict.get(possiblePrime)){
				let loc: number = possiblePrime + num + num;
				if(dict.has(loc)){
					dict.get(loc).push(num);
				}
				else{
					dict.set(loc, [num]);
				}
				//We no longer need this, free the memory
				dict.delete(possiblePrime);
			}
		}
	}
}
export function* sieveOfEratosthenesBig(){
	//Return 2 the first time, this lets us skip all even numbers later
	yield 2n;

	//Dictionary to hold the priems we have already found
	let dict = new Map<bigint, bigint[]>();

	//Start checking for primes with the number 3 and skip all even numbers
	for(let possiblePrime = 3n;true;possiblePrime += 2n){
		//If possiblePrime is not in the dictionary it is a new prime number
		//Return it and mark its next multiple
		if(!dict.has(possiblePrime)){
			yield possiblePrime;
			dict.set(possiblePrime * possiblePrime, [possiblePrime]);
		}
		//If possiblePrime is in the dictionary it is a composite number
		else{
			//Move each witness to its next multiple
			for(let num of dict.get(possiblePrime)){
				let loc: bigint = possiblePrime + num + num;
				if(dict.has(loc)){
					dict.get(loc).push(num);
				}
				else{
					dict.set(loc, [num]);
				}
				//We no longer need this, free the memory
				dict.delete(possiblePrime);
			}
		}
	}
}

//This returns the sqrt of the bigint passed in
export function sqrtBig(value: bigint): bigint{
	if(value < 0n){
		throw "Negative numbers are not supported";
	}

	let k = 2n;
	let o = 0n;
	let x = value;
	let limit = 100;

	while(x ** k !== k && x !== o && --limit){
		o = x;
		x = ((k - 1n) * x + value / x ** (k - 1n)) / k;
	}

	return x;
}

//Returns an array with all Fibonacci numbers up to goalNumber
export function getAllFib(goalNumber: number): number[]{
	//Setup the variables
	let fibNums: number[] = [];

	//If the number is <= 0 return an empty list
	if(goalNumber <= 0){
		return fibNums;
	}
	else if(goalNumber == 1){
		fibNums.push(1);
		return fibNums;
	}

	//This means that at least 2 1's are elements
	fibNums.push(1);
	fibNums.push(1);

	//Loop to generate the rest of the Fibonacci numbers
	while(fibNums[fibNums.length - 1] <= goalNumber){
		fibNums.push((fibNums[fibNums.length - 1]) + (fibNums[fibNums.length - 2]));
	}

	//At this point the most recent number is > goalNumber, so remove it and return the rest of the list
	fibNums.pop();
	return fibNums;
}
export function getAllFibBig(goalNumber: bigint): bigint[]{
	//Setup the variables
	let fibNums:bigint[] = [];

	//If the number is <= 0 return an empty list
	if(goalNumber <= 0){
		return fibNums;
	}
	else if(goalNumber == 1n){
		fibNums.push(1n);
		return fibNums;
	}

	//This means that at least 2 1's are elements
	fibNums.push(1n);
	fibNums.push(1n);

	//Loop to generate the rest of the Fibonacci numbers
	while(fibNums[fibNums.length - 1] <= goalNumber){
		fibNums.push((fibNums[fibNums.length - 1]) + (fibNums[fibNums.length - 2]));
	}

	//At this point the most recent number is > goalNumber, so remove it and return the rest of the list
	fibNums.pop();
	return fibNums;
}

//Returns an array with all primes up to goalNumber
export function getPrimes(goalNumber: number): number[]{
	let primes: number[] = [];	//Holds the prime numbers
	let foundFactor: boolean = false;	//A flag for whether a factor of the current number has been found

	//If the number is 0 or a negative return an empty list
	if(goalNumber <= 1){
		return primes;
	}
	//Optherwise the number is at least 2, so 2 should be added to the list
	else{
		primes.push(2);
	}

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

		//If you didn't find a factor then the current number must be prime
		if(!foundFactor){
			primes.push(possiblePrime);
		}
		else{
			foundFactor = false;
		}
	}

	//Sort the list before returning it
	primes = primes.sort((n1, n2) => n1 - n2);
	return primes;
}
export function getPrimesBig(goalNumber: bigint): bigint[]{
	let primes: bigint[] = [];	//Holds the prime numbers
	let foundFactor: boolean = false;	//A flag for whether a factor of the current number has been found

	//If the number is 0 or a negative return an empty list
	if(goalNumber <= 1){
		return primes;
	}
	//Optherwise the number is at least 2, so 2 should be added to the list
	else{
		primes.push(2n);
	}

	//We can now start at 3 and skip all even numbers, because they cannot be prime
	for(let possiblePrime: bigint = 3n;possiblePrime <= goalNumber;possiblePrime += 2n){
		//Check all current primes, up to sqrt(possiblePrime), to see if there is a divisor
		let topPossibleFactor: bigint = sqrtBig(possiblePrime);
		//We can safely assume that there will be at least 1 element in the primes list because of 2 being added before this
		for(let primesCnt: number = 0;primes[primesCnt] <= topPossibleFactor;){
			if((possiblePrime % primes[primesCnt]) == 0n){
				foundFactor = true;
				break;
			}
			else{
				++primesCnt;
			}
			//Check if the index has gone out of range
			if(primesCnt >= primes.length){
				break;
			}
		}

		//If you didn't find a factor then the current number must be prime
		if(!foundFactor){
			primes.push(possiblePrime);
		}
		else{
			foundFactor = false;
		}
	}

	//Sort the list before returning it
	primes = primes.sort(function(n1, n2){
		if(n1 > n2){
			return 1;
		}
		else if(n1 < n2){
			return -1;
		}
		else{
			return 0;
		}
	});
	return primes;
}

//Returns an array with numberOfPrimes prime elements
export function getNumPrimes(numberOfPrimes: number): number[]{
	let primes: number[] = [];	//Holds the prime numbers
	let foundFactor: boolean = false;	//A flag for whether a factor of the current number has been found

	//If the number is 0 or negative return an empty list
	if(numberOfPrimes <= 1){
		return primes;
	}
	//Otherwise the number is at least 2, so 2 should be added to the list
	else{
		primes.push(2);
	}

	//We can now start at 3 and skip all even number, because they cannot be prime
	for(let possiblePrime: number = 3;primes.length < numberOfPrimes;possiblePrime += 2){
		//Check all the current primes, up to sqrt(possiblePrime), to see if there is a divisor
		let topPossibleFactor: number = Math.ceil(Math.sqrt(possiblePrime));
		//We can safely assume that there will be at least 1 element in the primes list because of 2 being added by default
		for(let primesCnt: number = 0;primes[primesCnt] <= topPossibleFactor;){
			if((possiblePrime % primes[primesCnt]) == 0){
				foundFactor = true;
				break;
			}
			else{
				++primesCnt;
			}
			//Check if the index has gone out of bounds
			if(primesCnt >= primes.length){
				break;
			}
		}

		//If you didn't find a factor then the current number must be prime
		if(!foundFactor){
			primes.push(possiblePrime);
		}
		else{
			foundFactor = false;
		}
	}

	//Sort the list before returning it
	primes = primes.sort((n1, n2) => n1 - n2);
	return primes;
}
export function getNumPrimesBig(numberOfPrimes: bigint): bigint[]{
	let primes: bigint[] = [];	//Holds the prime numbers
	let foundFactor: boolean = false;	//A flag for whether a factor of the current number has been found

	//If the number is 0 or negative return an empty list
	if(numberOfPrimes <= 1){
		return primes;
	}
	//Otherwise the number is at least 2, so 2 should be added to the list
	else{
		primes.push(2n);
	}

	//We can now start at 3 and skip all even number, because theyu cannot be prime
	for(let possiblePrime: bigint = 3n;primes.length < numberOfPrimes;possiblePrime += 2n){
		//Check all the current primes, up to sqrt(possiblePrime), to see if there is a divisor
		let topPossibleFactor: bigint = sqrtBig(possiblePrime);
		//We can safely assume that ther ewill be at least 1 element in the primes list because of 2 being added by default
		for(let primesCnt: number = 0;primes[primesCnt] <= topPossibleFactor;){
			if((possiblePrime % primes[primesCnt]) == 0n){
				foundFactor = true;
				break;
			}
			else{
				++primesCnt;
			}
			//Check if the index has gone out of bounds
			if(primesCnt >= primes.length){
				break;
			}
		}

		//If you didn't find a factor then the current number must be prime
		if(!foundFactor){
			primes.push(possiblePrime);
		}
		else{
			foundFactor = false;
		}
	}

	//Sort the list before returning it
	primes = primes.sort(function(n1, n2){
		if(n1 > n2){
			return 1;
		}
		else if(n1 < n2){
			return -1;
		}
		else{
			return 0;
		}
	});
	return primes;
}

//Returns true if possiblePrime is prime
export function isPrime(possiblePrime: number): boolean{
	if(possiblePrime <= 3){
		return possiblePrime > 1;
	}
	else if(((possiblePrime % 2) == 0) || ((possiblePrime % 3) == 0)){
		return false;
	}
	for(let cnt: number = 5;(cnt * cnt) <= possiblePrime;cnt += 6){
		if(((possiblePrime % cnt) == 0) || ((possiblePrime % (cnt + 2)) == 0)){
			return false;
		}
	}
	return true;
}
export function isPrimeBig(possiblePrime: bigint): boolean{
	if(possiblePrime <= 3n){
		return possiblePrime > 1n;
	}
	else if(((possiblePrime % 2n) == 0n) || ((possiblePrime % 3n) == 0n)){
		return false;
	}
	for(let cnt : bigint = 5n;(cnt * cnt) <= possiblePrime;cnt += 6n){
		if(((possiblePrime % cnt) == 0n) || ((possiblePrime % (cnt + 2n)) == 0n)){
			return false;
		}
	}
	return true;
}

//Returns an array with all the factors of goalNumber
export function getFactors(goalNumber: number): number[]{
	//You need to get all the primes that could be factors of this number so you can test them
	let topPossiblePrime: number = Math.ceil(Math.sqrt(goalNumber));
	let primes: number[] = getPrimes(topPossiblePrime);
	let factors: number[] = [];

	//You need to step through each prime and see if it is a factor in the number
	for(let cnt: number = 0;cnt < primes.length;){
		//If the prime is a factor you need to add it to the factor list
		if((goalNumber % primes[cnt]) == 0){
			factors.push(primes[cnt]);
			goalNumber /= primes[cnt];
		}
		//Otherwise advance the location in primes you are looking at
		//By not advancing f the prime is a factor you allow for multiple of the same prime number as a factor
		else{
			++cnt;
		}
	}

	//If you didn't get any factors the number itself must be a prime
	if(factors.length == 0){
		factors.push(goalNumber);
		goalNumber /= goalNumber;
	}

	//If for some reason the goalNumber is not 1 throw an exception
	if(goalNumber != 1){
		throw new InvalidResult("The factor was not 1: " + goalNumber);
	}

	//Return the list of factors
	return factors;
}
export function getFactorsBig(goalNumber: bigint): bigint[]{
	//You need to get all the primes that could be factors of this number so you can test them
	let topPossiblePrime: bigint = sqrtBig(goalNumber);
	let primes: bigint[] = getPrimesBig(topPossiblePrime);
	let factors: bigint[] = [];

	//You need to step through each prime and see if it is a factor in the number
	for(let cnt: number = 0;cnt < primes.length;){
		//If the prime is a factor you need to add it to the factor list
		if((goalNumber % primes[cnt]) == 0n){
			factors.push(primes[cnt]);
			goalNumber /= primes[cnt];
		}
		//Otherwise advance the location in primes you are looking at
		//By not advancing f the prime is a factor you allow for multiple of the same prime number as a factor
		else{
			++cnt;
		}
	}

	//If you didn't get any factors the number itself must be a prime
	if(factors.length == 0){
		factors.push(goalNumber);
		goalNumber /= goalNumber;
	}

	//If for some reason the goalNumber is not 1 throw an exception
	if(goalNumber != 1n){
		throw new InvalidResult("The factor was not 1: " + goalNumber);
	}

	//Return the list of factors
	return factors;
}

//Returns an array with the prime divisors of goalNumber
export function getDivisors(goalNumber: number): number[]{
	let divisors: number[] = [];

	//Start by checking that the number is positive
	if(goalNumber <= 0){
		return divisors;
	}
	//If the number is 1 return just itself
	else if(goalNumber == 1){
		divisors.push(1);
		return divisors;
	}

	//Start at 3 and loop through all numbers < sqrt(goalNumber) looking for a number that divides it evenly
	let topPossibleDivisor: number = Math.ceil(Math.sqrt(goalNumber));
	for(let possibleDivisor = 1;possibleDivisor <= topPossibleDivisor;++possibleDivisor){
		//If you find one add it and the number it creates to the list
		if((goalNumber % possibleDivisor) == 0){
			divisors.push(possibleDivisor);
			//Account for the possibility of sqrt(goalNumber) being a divisor
			if(possibleDivisor != topPossibleDivisor){
				divisors.push(goalNumber / possibleDivisor);
			}
			if(divisors[divisors.length - 1] == (possibleDivisor + 1)){
				++possibleDivisor;
			}
		}
	}

	//Sort the list before returning it for neatness
	divisors.sort((a, b) => a - b);
	//Return the list
	return divisors;
}
export function getDivisorsBig(goalNumber: bigint): bigint[]{
	let divisors: bigint[] = [];

	//Start by checking that the number is positive
	if(goalNumber <= 0n){
		return divisors;
	}
	//If the number is 1 return just itself
	else if(goalNumber == 1n){
		divisors.push(1n);
		return divisors;
	}

	//Start at 3 and loop through all numbers < sqrt(goalNumber) looking for a number that divides it evenly
	let topPossibleDivisor: bigint = sqrtBig(goalNumber);
	for(let possibleDivisor = 1n;possibleDivisor <= topPossibleDivisor;++possibleDivisor){
		//If you find one add it and the number it creates to the list
		if((goalNumber % possibleDivisor) == 0n){
			divisors.push(possibleDivisor);
			//Account for the possibility of sqrt(goalNumber) being a divisors
			if(possibleDivisor != topPossibleDivisor){
				divisors.push(goalNumber / possibleDivisor);
			}
			if(divisors[divisors.length - 1] == (possibleDivisor + 1n)){
				++possibleDivisor;
			}
		}
	}

	//Sort the list before returning it for neatness
	divisors.sort((a, b) => {
		if(a > b){
			return 1;
		}
		else if(a < b){
			return -1;
		}
		else{
			return 0;
		}
	});
	//Return the list
	return divisors;
}

//Returns F[goalSubscript]
export function getFib(goalSubscript: number): number{
	//Setup the variables
	let fibNums: number[] = [1, 1, 0];	//A list to keep track of the Fibonacci numbers. It need only be 3 long because we only need the one we are working on and the last 2

	//If the number is <= 0 return 0
	if(goalSubscript <= 0){
		return 0;
	}

	//Loop through the list, generating Fibonacci numbers until it finds the correct subscript
	let fibLoc: number = 2;
	for(fibLoc = 2;fibLoc < goalSubscript;++fibLoc){
		fibNums[fibLoc % 3] = fibNums[(fibLoc - 1) % 3] + fibNums[(fibLoc - 2) % 3];
	}

	//Return the proper number. The location counter is 1 off of the subscript
	return fibNums[(fibLoc - 1) % 3];
}
export function getFibBig(goalSubscript: bigint): bigint{
	//Setup the varibles
	let fibNums: bigint[] = [1n, 1n, 0n];	//A list to keep track of the Fibonacci numbers. It need only be 3 long because we only need the one we are working on and the last 2

	//If the number is <= 0 return 0
	if(goalSubscript <= 0n){
		return 0n;
	}

	//Loop through the list, generating Fibonacci numbers until it finds the correct subscript
	let fibLoc: bigint = 2n;
	for(fibLoc = 2n;fibLoc < goalSubscript;++fibLoc){
		fibNums[Number(fibLoc % 3n)] = fibNums[Number((fibLoc - 1n) % 3n)] + fibNums[Number((fibLoc - 2n) % 3n)];
	}

	//Return the proper number. The location counter is 1 off of the subscript
	return fibNums[Number((fibLoc - 1n) % 3n)];
}

//This function returns the GCD of the two numbers sent to it
export function gcd(num1: number, num2: number){
	while((num1 != 0) && (num2 != 0)){
		if(num1 > num2){
			num1 %= num2;
		}
		else{
			num2 %= num1;
		}
	}
	return num1 | num2;
}
export function gcdBig(num1: bigint, num2: bigint){
	while((num1 != 0n) && (num2 != 0n)){
		if(num1 > num2){
			num1 %= num2;
		}
		else{
			num2 %= num1;
		}
	}
	return num1 | num2;
}

//Return the factorial of the number passed in
export function factorial(num: number): number{
	let fact: number = 1;
	for(let cnt = 1;cnt <= num;++cnt){
		fact *= cnt;
	}
	return fact;
}
export function factorialBig(num: bigint): bigint{
	let fact: bigint = 1n;
	for(let cnt = 1n;cnt <= num;++cnt){
		fact *= cnt;
	}
	return fact;
}

//Converts a number to its binary equivalent
export function toBin(num: number): string{
	return (num >>> 0).toString(2);
}
export function toBinBig(num: bigint): string{
	let binNum = "";
	while(num > 0n){
		let rest = num % 2n;
		if(rest == 1n){
			binNum += "1";
		}
		else{
			binNum += "0";
		}
		num = (num - rest) / 2n;
	}
	binNum = binNum.split("").reverse().join("");
	if(binNum == ""){
		binNum = "0";
	}
	return binNum;
}