Java ECAlgorithms类代码示例

import org.spongycastle.math.ec.ECAlgorithms; //导入依赖的package包/类
/**
 * <p>Given the components of a signature and a selector value, recover and return the public key
 * that generated the signature according to the algorithm in SEC1v2 section 4.1.6.</p>
 *
 * <p>The recId is an index from 0 to 3 which indicates which of the 4 possible keys is the correct one. Because
 * the key recovery operation yields multiple potential keys, the correct key must either be stored alongside the
 * signature, or you must be willing to try each recId in turn until you find one that outputs the key you are
 * expecting.</p>
 *
 * <p>If this method returns null it means recovery was not possible and recId should be iterated.</p>
 *
 * <p>Given the above two points, a correct usage of this method is inside a for loop from 0 to 3, and if the
 * output is null OR a key that is not the one you expect, you try again with the next recId.</p>
 *
 * @param recId Which possible key to recover.
 * @param sig the R and S components of the signature, wrapped.
 * @param messageHash Hash of the data that was signed.
 * @return 65-byte encoded public key
 */
@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);
}
 

import org.spongycastle.math.ec.ECAlgorithms; //导入依赖的package包/类
/**
 * <p>Given the components of a signature and a selector value, recover and return the public key
 * that generated the signature according to the algorithm in SEC1v2 section 4.1.6.</p>
 *
 * <p>The recId is an index from 0 to 3 which indicates which of the 4 possible keys is the correct one. Because
 * the key recovery operation yields multiple potential keys, the correct key must either be stored alongside the
 * signature, or you must be willing to try each recId in turn until you find one that outputs the key you are
 * expecting.</p>
 *
 * <p>If this method returns null it means recovery was not possible and recId should be iterated.</p>
 *
 * <p>Given the above two points, a correct usage of this method is inside a for loop from 0 to 3, and if the
 * output is null OR a key that is not the one you expect, you try again with the next recId.</p>
 *
 * @param recId Which possible key to recover.
 * @param sig the R and S components of the signature, wrapped.
 * @param messageHash Hash of the data that was signed.
 * @return 65-byte encoded public key
 */
@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);
  Fp q = (Fp) ECAlgorithms.sumOfTwoMultiplies(CURVE.getG(), eInvrInv, R, srInv);
  return q.getEncoded(/* compressed */ false);
}
 

import org.spongycastle.math.ec.ECAlgorithms; //导入依赖的package包/类
/**
 * <p>Given the components of a signature and a selector value, recover and return the public key
 * that generated the signature according to the algorithm in SEC1v2 section 4.1.6.</p>
 *
 * <p>The recId is an index from 0 to 3 which indicates which of the 4 possible keys is the correct one. Because
 * the key recovery operation yields multiple potential keys, the correct key must either be stored alongside the
 * signature, or you must be willing to try each recId in turn until you find one that outputs the key you are
 * expecting.</p>
 *
 * <p>If this method returns null it means recovery was not possible and recId should be iterated.</p>
 *
 * <p>Given the above two points, a correct usage of this method is inside a for loop from 0 to 3, and if the
 * output is null OR a key that is not the one you expect, you try again with the next recId.</p>
 *
 * @param recId Which possible key to recover.
 * @param sig the R and S components of the signature, wrapped.
 * @param message Hash of the data that was signed.
 * @param compressed Whether or not the original pubkey was compressed.
 * @return An ECKey containing only the public part, or null if recovery wasn't possible.
 */
@Nullable
public static ECKey recoverFromSignature(int recId, ECDSASignature sig, Sha256Hash message, boolean compressed) {
    Preconditions.checkArgument(recId >= 0, "recId must be positive");
    Preconditions.checkArgument(sig.r.signum() >= 0, "r must be positive");
    Preconditions.checkArgument(sig.s.signum() >= 0, "s must be positive");
    Preconditions.checkNotNull(message);
    // 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 = message.toBigInteger();
    //   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);
    if (compressed) {
        // We have to manually recompress the point as the compressed-ness gets lost when multiply() is used.
        q = new ECPoint.Fp(curve, q.getX(), q.getY(), true);
    }
    return new ECKey((byte[])null, q.getEncoded());
}
 

import org.spongycastle.math.ec.ECAlgorithms; //导入依赖的package包/类
/**
 * <p>Given the components of a signature and a selector value, recover and return the public key
 * that generated the signature according to the algorithm in SEC1v2 section 4.1.6.</p>
 *
 * <p>The recId is an index from 0 to 3 which indicates which of the 4 possible keys is the correct one. Because
 * the key recovery operation yields multiple potential keys, the correct key must either be stored alongside the
 * signature, or you must be willing to try each recId in turn until you find one that outputs the key you are
 * expecting.</p>
 *
 * <p>If this method returns null it means recovery was not possible and recId should be iterated.</p>
 *
 * <p>Given the above two points, a correct usage of this method is inside a for loop from 0 to 3, and if the
 * output is null OR a key that is not the one you expect, you try again with the next recId.</p>
 *
 * @param recId Which possible key to recover.
 * @param sig the R and S components of the signature, wrapped.
 * @param message Hash of the data that was signed.
 * @param compressed Whether or not the original pubkey was compressed.
 * @return An ECKey containing only the public part, or null if recovery wasn't possible.
 */
@Nullable
public static ECKey recoverFromSignature(int recId, ECDSASignature sig, Sha256Hash message, boolean compressed) {
    Preconditions.checkArgument(recId >= 0, "recId must be positive");
    Preconditions.checkArgument(sig.r.compareTo(BigInteger.ZERO) >= 0, "r must be positive");
    Preconditions.checkArgument(sig.s.compareTo(BigInteger.ZERO) >= 0, "s must be positive");
    Preconditions.checkNotNull(message);
    // 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 = message.toBigInteger();
    //   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);
    if (compressed) {
        // We have to manually recompress the point as the compressed-ness gets lost when multiply() is used.
        q = new ECPoint.Fp(curve, q.getX(), q.getY(), true);
    }
    return new ECKey((byte[])null, q.getEncoded());
}
 

import org.spongycastle.math.ec.ECAlgorithms; //导入依赖的package包/类
/**
 * <p>Given the components of a signature and a selector value, recover and return the public key
 * that generated the signature according to the algorithm in SEC1v2 section 4.1.6.</p>
 *
 * <p>The recId is an index from 0 to 3 which indicates which of the 4 possible keys is the correct one. Because
 * the key recovery operation yields multiple potential keys, the correct key must either be stored alongside the
 * signature, or you must be willing to try each recId in turn until you find one that outputs the key you are
 * expecting.</p>
 *
 * <p>If this method returns null it means recovery was not possible and recId should be iterated.</p>
 *
 * <p>Given the above two points, a correct usage of this method is inside a for loop from 0 to 3, and if the
 * output is null OR a key that is not the one you expect, you try again with the next recId.</p>
 *
 * @param recId Which possible key to recover.
 * @param sig the R and S components of the signature, wrapped.
 * @param messageHash Hash of the data that was signed.
 * @param compressed Whether or not the original pubkey was compressed.
 * @return An ECKey containing only the public part, or null if recovery wasn't possible.
 */
@Nullable
public static ECKey recoverFromSignature(int recId, ECDSASignature sig, byte[] messageHash, boolean compressed) {
    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 ECKey.fromPublicOnly(q.getEncoded(compressed));
}
 

import org.spongycastle.math.ec.ECAlgorithms; //导入依赖的package包/类
/**
 * <p>Given the components of a signature and a selector value, recover and return the public key
 * that generated the signature according to the algorithm in SEC1v2 section 4.1.6.</p>
 *
 * <p>The recId is an index from 0 to 3 which indicates which of the 4 possible keys is the correct one. Because
 * the key recovery operation yields multiple potential keys, the correct key must either be stored alongside the
 * signature, or you must be willing to try each recId in turn until you find one that outputs the key you are
 * expecting.</p>
 *
 * <p>If this method returns null it means recovery was not possible and recId should be iterated.</p>
 *
 * <p>Given the above two points, a correct usage of this method is inside a for loop from 0 to 3, and if the
 * output is null OR a key that is not the one you expect, you try again with the next recId.</p>
 *
 * @param recId Which possible key to recover.
 * @param sig the R and S components of the signature, wrapped.
 * @param message Hash of the data that was signed.
 * @param compressed Whether or not the original pubkey was compressed.
 * @return An ECKey containing only the public part, or null if recovery wasn't possible.
 */
@Nullable
public static ECKey recoverFromSignature(int recId, ECDSASignature sig, Sha256Hash message, boolean compressed) {
    Preconditions.checkArgument(recId >= 0, "recId must be positive");
    Preconditions.checkArgument(sig.r.signum() >= 0, "r must be positive");
    Preconditions.checkArgument(sig.s.signum() >= 0, "s must be positive");
    Preconditions.checkNotNull(message);
    // 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 = message.toBigInteger();
    //   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 new ECKey((byte[])null, q.getEncoded(compressed));
}