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Fixed bug in getting BigInteger sqrts

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Mattrixwv
2021-03-11 13:35:04 -05:00
parent 32a937f626
commit fa5b4d2f15
1 changed files with 8 additions and 2 deletions
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10
CSClasses/Algorithms.cs
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@@ -29,6 +29,12 @@ using System.Numerics;
namespace mee{ namespace mee{
public class Algorithms{ public class Algorithms{
//These are a few functions that help with some utility within the class
//Returns the sqrt of the BigInt passed to it
public static BigInteger BigIntegerSqrt(BigInteger num){
return (BigInteger)Math.Ceiling(Math.Exp(BigInteger.Log(num) / 2));
}
//These functions return a list of all Fibonacci numbers <= goalNumber //These functions return a list of all Fibonacci numbers <= goalNumber
public static List<int> GetAllFib(int goalNumber){ public static List<int> GetAllFib(int goalNumber){
//Setup the variables //Setup the variables
@@ -304,7 +310,7 @@ namespace mee{
//We can now start at 3 and skip all even numbers, because they cannot be prime //We can now start at 3 and skip all even numbers, because they cannot be prime
for(BigInteger possiblePrime = 3;possiblePrime <= goalNumber;possiblePrime += 2){ for(BigInteger possiblePrime = 3;possiblePrime <= goalNumber;possiblePrime += 2){
//Check all current primes, up to sqrt(possiblePrime), to see if there is a divisor //Check all current primes, up to sqrt(possiblePrime), to see if there is a divisor
BigInteger topPossibleFactor = (BigInteger)Math.Exp(BigInteger.Log(possiblePrime) / 2); BigInteger topPossibleFactor = BigIntegerSqrt(possiblePrime);
//We can safely assume that there will be at least 1 element in the primes list because of 2 being added before this //We can safely assume that there will be at least 1 element in the primes list because of 2 being added before this
for(int primesCnt = 0;primes[primesCnt] <= topPossibleFactor;){ for(int primesCnt = 0;primes[primesCnt] <= topPossibleFactor;){
if((possiblePrime % primes[primesCnt]) == 0){ if((possiblePrime % primes[primesCnt]) == 0){
@@ -440,7 +446,7 @@ namespace mee{
//We can now start at 3 and skip all even numbers, because they cannot be prime //We can now start at 3 and skip all even numbers, because they cannot be prime
for(BigInteger possiblePrime = 3;primes.Count < numberOfPrimes;possiblePrime += 2){ for(BigInteger possiblePrime = 3;primes.Count < numberOfPrimes;possiblePrime += 2){
//Check all current primes, up to sqrt(possiblePrime), to see if there is a divisor //Check all current primes, up to sqrt(possiblePrime), to see if there is a divisor
BigInteger topPossibleFactor = (BigInteger)Math.Exp(BigInteger.Log(possiblePrime) / 2); BigInteger topPossibleFactor = BigIntegerSqrt(possiblePrime);
//We can safely assume that there will be at least 1 element in the primes list because of 2 being added before this //We can safely assume that there will be at least 1 element in the primes list because of 2 being added before this
for(int primesCnt = 0;primes[primesCnt] <= topPossibleFactor;){ for(int primesCnt = 0;primes[primesCnt] <= topPossibleFactor;){
if((possiblePrime % primes[primesCnt]) == 0){ if((possiblePrime % primes[primesCnt]) == 0){