Diffie-Hellman -- Primitive root mod n -- cryptography question.
- by somewhat confused
In the below snippet, please explain starting with the first "for" loop what is happening and why. Why is 0 added, why is 1 added in the second loop. What is going on in the "if" statement under bigi. Finally explain the modPow method. Thank you in advance for meaningful replies.
public static boolean isPrimitive(BigInteger m, BigInteger n) {
BigInteger bigi, vectorint;
Vector<BigInteger> v = new Vector<BigInteger>(m.intValue());
int i;
for (i=0;i<m.intValue();i++)
v.add(new BigInteger("0"));
for (i=1;i<m.intValue();i++)
{
bigi = new BigInteger("" + i);
if (m.gcd(bigi).intValue() == 1)
v.setElementAt(new BigInteger("1"), n.modPow(bigi,m).intValue());
}
for (i=0;i<m.intValue();i++)
{
bigi = new BigInteger("" + i);
if (m.gcd(bigi).intValue() == 1)
{
vectorint = v.elementAt(bigi.intValue());
if ( vectorint.intValue() == 0)
i = m.intValue() + 1;
}
}
if (i == m.intValue() + 2)
return false;
else
return true;
}
