public static bool GetQuadraticCharacter(Relation rel, FactorPair quadraticFactor)
{
BigInteger ab = rel.A + rel.B;
BigInteger abp = BigInteger.Abs(BigInteger.Multiply(ab, quadraticFactor.P));
int legendreSymbol = Legendre.Symbol(abp, quadraticFactor.R);
return(legendreSymbol != 1);
}
public List <FactorPair> GetFactors(int n)
{
var primeFactors = this.GetPrimeFactors(n).ToArray();
var factors = new List <FactorPair>();
if (primeFactors.Length == 0)
{
return(factors);
}
// in order to find the factors, we create all subsets of prime factors and put them in one of two sets
// the values in the first set will form the first factor, and the second set will form the second set
// since each pair of factors will be represented twice, we will assume the first value will always be in the first set
// we've also already added the value of (1, n).
//
// if we represent these by bit strings, we need to iterate from i=2^(n-1) (the most significant bit is a 1)
// to i=(i^2n) - 1
//
// it's also possible for the same value to be added multiple times.
// for instance for n=16, the prime factorization is (2, 2, 2, 2)
// bitstrings 1010 and 1001 create the same value. because of that, we need to implement equality and make sure that it doesn't already exist
var numPrimesInPrimeFactorization = primeFactors.Length;
for (int i = (int)Math.Pow(2, numPrimesInPrimeFactorization - 1); i <= (int)Math.Pow(2, numPrimesInPrimeFactorization) - 1; i++)
{
var bitString = Convert.ToString(i, 2).PadLeft(numPrimesInPrimeFactorization, '0').ToCharArray().Select(a => a - '0').ToArray();
var factor1 = 1;
var factor2 = 1;
for (int prime = 0; prime < numPrimesInPrimeFactorization; prime++)
{
if (bitString[prime] == 1)
{
factor1 *= primeFactors[prime];
}
else
{
factor2 *= primeFactors[prime];
}
}
var newFactor = new FactorPair(factor1, factor2);
if (!factors.Any(f => f.SmallerFactor == newFactor.SmallerFactor && f.LargerFactor == newFactor.LargerFactor))
{
factors.Add(newFactor);
}
}
return(factors);
}
