Blurgh

@Nullable
    public static byte[] recoverPubBytesFromSignature(int recId, ECDSASignature sig, byte[] messageHash) {
        check(recId >= 0, "recId must be positive");
        check(sig.r.signum() >= 0, "r must be positive");
        check(sig.s.signum() >= 0, "s must be positive");
        check(messageHash != null, "messageHash must not be null");
        // 1.0 For j from 0 to h   (h == recId here and the loop is outside this function)
        //   1.1 Let x = r + jn
        BigInteger n = CURVE.getN();  // Curve order.
        BigInteger i = BigInteger.valueOf((long) recId / 2);
        BigInteger 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.
        ECCurve.Fp curve = (ECCurve.Fp) CURVE.getCurve();
        BigInteger prime = curve.getQ();  // Bouncy Castle is not consistent about the letter it uses for the prime.
        if (x.compareTo(prime) >= 0) {
            // Cannot have point co-ordinates larger than this as everything takes place modulo Q.
            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.
        ECPoint 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.
        BigInteger e = new BigInteger(1, messageHash);
        //   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.
        BigInteger eInv = BigInteger.ZERO.subtract(e).mod(n);
        BigInteger rInv = sig.r.modInverse(n);
        BigInteger srInv = rInv.multiply(sig.s).mod(n);
        BigInteger eInvrInv = rInv.multiply(eInv).mod(n);
        ECPoint.Fp q = (ECPoint.Fp) ECAlgorithms.sumOfTwoMultiplies(CURVE.getG(), eInvrInv, R, srInv);
        return q.getEncoded(/* compressed */ false);
    }