public static byte[] RecoverPubBytesFromSignature(int recId, ECDSASignature sig, byte[] message, bool compressed)
{
if (recId < 0)
{
throw new ArgumentException("recId should be positive");
}
if (sig.R.SignValue < 0)
{
throw new ArgumentException("r should be positive");
}
if (sig.S.SignValue < 0)
{
throw new ArgumentException("s should be positive");
}
if (message == null)
{
throw new ArgumentNullException(nameof(message));
}
var curve = CURVE;
// 1.0 For j from 0 to h (h == recId here and the loop is outside this function)
// 1.1 Let x = r + jn
var n = curve.N;
var i = BigInteger.ValueOf((long)recId / 2);
var x = sig.R.Add(i.Multiply(n));
// 1.2. Convert the integer x to an octet string X of length mlen using the conversion routine
// specified in Section 2.3.7, where mlen = ⌈(log2 p)/8⌉ or mlen = ⌈m/8⌉.
// 1.3. Convert the octet string (16 set binary digits)||X to an elliptic curve point R using the
// conversion routine specified in Section 2.3.4. If this conversion routine outputs “invalid”, then
// do another iteration of Step 1.
//
// More concisely, what these points mean is to use X as a compressed public key.
//using bouncy and Q value of Point
var prime = new BigInteger(1,
Org.BouncyCastle.Utilities.Encoders.Hex.Decode(
"FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFC2F"));
if (x.CompareTo(prime) >= 0)
{
return(null);
}
// Compressed keys require you to know an extra bit of data about the y-coord as there are two possibilities.
// So it's encoded in the recId.
var R = DecompressKey(x, (recId & 1) == 1);
// 1.4. If nR != point at infinity, then do another iteration of Step 1 (callers responsibility).
if (!R.Multiply(n).IsInfinity)
{
return(null);
}
// 1.5. Compute e from M using Steps 2 and 3 of ECDSA signature verification.
var e = new BigInteger(1, message);
// 1.6. For k from 1 to 2 do the following. (loop is outside this function via iterating recId)
// 1.6.1. Compute a candidate public key as:
// Q = mi(r) * (sR - eG)
//
// Where mi(x) is the modular multiplicative inverse. We transform this into the following:
// Q = (mi(r) * s ** R) + (mi(r) * -e ** G)
// Where -e is the modular additive inverse of e, that is z such that z + e = 0 (mod n). In the above equation
// ** is point multiplication and + is point addition (the EC group operator).
//
// We can find the additive inverse by subtracting e from zero then taking the mod. For example the additive
// inverse of 3 modulo 11 is 8 because 3 + 8 mod 11 = 0, and -3 mod 11 = 8.
var eInv = BigInteger.Zero.Subtract(e).Mod(n);
var rInv = sig.R.ModInverse(n);
var srInv = rInv.Multiply(sig.S).Mod(n);
var eInvrInv = rInv.Multiply(eInv).Mod(n);
var q = ECAlgorithms.SumOfTwoMultiplies(curve.G, eInvrInv, R, srInv);
q = q.Normalize();
if (compressed)
{
q = CURVE.Curve.CreatePoint(q.XCoord.ToBigInteger(), q.YCoord.ToBigInteger());
return(q.GetEncoded(true));
}
var xBytes = q.XCoord.ToBigInteger().ToByteArray(); //.BigIntegerToBytes(32);
var yBytes = q.YCoord.ToBigInteger().ToByteArray(); //.BigIntegerToBytes(32);
return(q.GetEncoded(false)); //xBytes.Concat(yBytes).ToArray();
}
public static byte[] RecoverPubBytesFromSignature(ECDSASignature sig, byte[] message, bool compressed)
{
return(RecoverPubBytesFromSignature(sig.V[0], sig, message, compressed));
}