问题描述
给定公钥的 x 和 y 坐标以及曲线名称,我需要确定这些坐标是否代表曲线上的有效点。如果是,则测试通过。如果否,则测试失败。
到目前为止我的代码是:
String curve = (String) testGroupHeaders.get("curve");
String curve_num = curve.split("-")[1];
String specName = "secp" + curve_num + "r1";
String qx = (String) testsData.get("qx");
String qy = (String) testsData.get("qy");
BigInteger x = new BigInteger(qx,16);
BigInteger y = new BigInteger(qy,16);
ECPoint ecPoint = new ECPoint(x,y);
if (ecPoint.equals(ECPoint.POINT_INFINITY)) {
testResultsObject.put("testPassed",false);
} else {
try {
AlgorithmParameters parameters = AlgorithmParameters.getInstance("EC");
parameters.init(new ECGenParameterSpec(specName));
ECParameterSpec ecParameters = parameters.getParameterSpec(ECParameterSpec.class);
EllipticCurve ellCurve = ecParameters.getCurve();
EcpublicKeySpec keySpec = new EcpublicKeySpec(ecPoint,ecParameters);
KeyFactory keyFactory = KeyFactory.getInstance("EC");
EcpublicKey publicKey = (EcpublicKey) keyFactory.generatePublic(keySpec);
testResultsObject.put("testPassed",true);
} catch (Exception e) {
System.out.println(e);
testResultsObject.put("testPassed",false);
}
我希望如果点无效,公钥的构建会失败,但是当输入已知的无效点时,这段代码仍然会导致测试通过。
任何帮助将不胜感激。
我发现这个问题 How to determine whether EC Point is on curve? 有一个模糊的答案:
“可能有更直接的方法,但如果您有 EllipticCurve 对象,您总是可以将点的坐标替换为曲线方程。”
EllipticCurve ellC = publicKey.getParams().getCurve();
希望它不会构建曲线,但它仍然成功。
请注意 BouncyCastle 不是我的选择,我必须使用 java.security 提供程序。
我也找到了这个答案Elliptic curve point
但我也无法从中得到任何东西。
编辑
好的,在从上面最后一个链接答案中的 Bouncy Castle 代码改编后,我非常接近此代码:
EllipticCurve ellCurve = ecParameters.getCurve();
BigInteger a = ellCurve.getA();
BigInteger b = ellCurve.getB();
ECField ecField = ellCurve.getField();
ECFieldFp fp = (ECFieldFp) ecField;
BigInteger p = ((ECFieldFp) ecField).getP();
BigInteger rhs = x.multiply(x).multiply(x).add(a.multiply(x)).add(b).mod(p);
BigInteger lhs = y.multiply(y).mod(p);
System.out.println(rhs);
System.out.println(lhs);
这适用于大多数情况,除了 secp521r1 的特殊情况:
{
"tcId": 8,"qx": "02ED3613D77D9096DC0545F3EEC44270762BF52E63044D29FC880556114F4FC176DBE20A9BC717C58FA63BB3308D3136355A072704C0B8BE9A6B3CFFFE36467722C0","qy": "00E857AE0D11FB1B79AC9531980A3E3BD1AC9E457E39A107D0AF5C9E09F2D6243F31C697C74CFD6A80F1CA5FCDA5950754DCC8724B9B21ED3A6705EBA1B9D2926B8F"
},
当我应用上述代码并将 lhs 与 rhs 进行比较时,我得到相同的输出,这表明这是曲线上的一个有效点。
rhs: 1634177850809752323211745106139613474061093186782816308703279366414782386677365518237753355552098931968864191666289807321598625802278923850729659158219971712
lhs: 1634177850809752323211745106139613474061093186782816308703279366414782386677365518237753355552098931968864191666289807321598625802278923850729659158219971712
解决方法
找出答案,稍后将其张贴在这里供我自己或遇到此问题的任何其他人使用。除了主帖中的编辑之外,我还没有检查仿射 x 和 y 是否在 [0,p-1] 范围内。参考:https://neilmadden.blog/2017/05/17/so-how-do-you-validate-nist-ecdh-public-keys/
这是没有充气城堡的完整工作代码:
String curve = (String) testGroupHeaders.get("curve");
String curve_num = curve.split("-")[1];
String specName = "secp" + curve_num + "r1";
JSONArray tests = (JSONArray) testGroupHeaders.get("tests");
JSONArray testResultsArray = new JSONArray();
for (int k = 0; k < tests.size(); k++){
JSONObject testResultsObject = new JSONObject();
JSONObject testsData = (JSONObject) tests.get(k);
long tcId = (long) testsData.get("tcId");
testResultsObject.put("tcId",tcId);
String qx = (String) testsData.get("qx");
String qy = (String) testsData.get("qy");
BigInteger x = new BigInteger(qx,16);
BigInteger y = new BigInteger(qy,16);
ECPoint ecPoint = new ECPoint(x,y);
if (ecPoint.equals(ECPoint.POINT_INFINITY)) {
testResultsObject.put("testPassed",false);
} else {
try {
AlgorithmParameters parameters = AlgorithmParameters.getInstance("EC");
parameters.init(new ECGenParameterSpec(specName));
ECParameterSpec ecParameters = parameters.getParameterSpec(ECParameterSpec.class);
EllipticCurve ellCurve = ecParameters.getCurve();
BigInteger a = ellCurve.getA();
BigInteger b = ellCurve.getB();
ECField ecField = ellCurve.getField();
BigInteger p = ((ECFieldFp) ecField).getP();
BigInteger rhs = x.multiply(x).multiply(x).add(a.multiply(x)).add(b).mod(p);
BigInteger lhs = y.multiply(y).mod(p);
// Do this part to try and generate an exception
ECPublicKeySpec keySpec = new ECPublicKeySpec(ecPoint,ecParameters);
KeyFactory keyFactory = KeyFactory.getInstance("EC");
ECPublicKey publicKey = (ECPublicKey) keyFactory.generatePublic(keySpec);
BigInteger affineX = publicKey.getW().getAffineX();
BigInteger affineY = publicKey.getW().getAffineY();
if (affineX.compareTo(BigInteger.ZERO) < 0 || affineX.compareTo(p) >= 0
|| affineY.compareTo(BigInteger.ZERO) < 0 || affineY.compareTo(p) >= 0) {
testResultsObject.put("testPassed",false);
} else {
if (rhs.equals(lhs)) {
testResultsObject.put("testPassed",true);
} else {
testResultsObject.put("testPassed",false);
}
}
} catch (Exception e) {
testResultsObject.put("testPassed",false);
}
请注意,这只是 POC 代码,可以进行大量清理/优化。所有 .get 调用都来自我正在解析的 JSON 请求,因此获取字符串值/输入的方式可能会有所不同。