我正在使用Bouncy Castle在智能卡相关软件中使用ECDH协议执行Elliptic Curve Key Aggreement,如BSI-TR-03111规范中所定义,§3.4
目的是在PACE协议中执行nonce的通用映射,如ICAO SAC技术报告,§3.4.2.1.1
中所定义(我正在使用Java,但我认为语言并不重要)
KeyAgreement
类使它变得非常简单(例如here),但它只允许输出生成的ECPoint
的X坐标,即Z AB (通常,但并不总是需要的。)
有没有办法让实际的ECPoint
(即S AB )返回而不必重新实现算法?即使这个公式本身很简单,也必须检查错误和异常并考虑变化(例如辅助因子乘法)
答案 0 :(得分:1)
我完全忘记了这个问题。这是我实施的解决方案:
package com.arjowiggins.arjonaut.crypto.ec;
import org.bouncycastle.crypto.BasicAgreement;
import org.bouncycastle.crypto.CipherParameters;
import org.bouncycastle.crypto.params.ECPrivateKeyParameters;
import org.bouncycastle.crypto.params.ECPublicKeyParameters;
import org.bouncycastle.math.ec.ECPoint;
import java.math.BigInteger;
/**
* Created by Salvatore on 22/04/2015.
*/
public class ECDHPointAgreement implements BasicAgreement {
private ECPrivateKeyParameters key;
public void init(CipherParameters key) {
this.key = (ECPrivateKeyParameters) key;
}
public int getFieldSize() {
return (key.getParameters().getCurve().getFieldSize() + 7) / 8;
}
public BigInteger calculateAgreement(CipherParameters pubKey) {
ECPoint P = calculatePoint(pubKey);
return P.getAffineXCoord().toBigInteger();
}
public ECPoint calculatePoint(CipherParameters pubKey) {
ECPublicKeyParameters pub = (ECPublicKeyParameters) pubKey;
ECPoint P = pub.getQ().multiply(key.getD()).normalize();
if (P.isInfinity()) {
throw new IllegalStateException("Infinity is not a valid agreement value for ECDH");
}
return P;
}
}
大多数代码来自BouncyCastle的KeyAgreement
类。
答案 1 :(得分:0)
解决Y的曲线方程正如我所做的那样。
public static BigInteger computeAffineY(BigInteger affineX, ECParameterSpec params) {
ECCurve bcCurve = toBouncyCastleECCurve(params);
ECFieldElement a = bcCurve.getA();
ECFieldElement b = bcCurve.getB();
ECFieldElement x = bcCurve.fromBigInteger(affineX);
ECFieldElement y = x.multiply(x).add(a).multiply(x).add(b).sqrt();
return y.toBigInteger();
}