Java Code Examples for org.spongycastle.math.ec.ECAlgorithms#sumOfTwoMultiplies()

/**
 * 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>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>If this method returns null it means recovery was not possible and recId should be
 * iterated.
 *
 * <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.
 *
 * @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
 */
public 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);
}
 

/**
 * <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> <p>The recId is an index from 0 to 3 which indicates which of the 4 possible allKeys is the
 * correct one. Because the key recovery operation yields multiple potential allKeys, 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> <p>If this method returns null it means recovery was not possible and recId should be
 * iterated.</p>
 *
 * <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 allKeys 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);
}
 

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);
}
 

/**
 * <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> <p>The
 * recId is an index from 0 to 3 which indicates which of the 4 possible allKeys is the correct
 * one. Because the key recovery operation yields multiple potential allKeys, 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> <p>If this method returns
 * null it means recovery was not possible and recId should be iterated.</p> <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 allKeys 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);
}
 

/**
 * <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);
}
 

/**
 * <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));
}