BigInteger 作为中性符号

Question

我目前正在研究与密码学相关的算法。更具体地说，在椭圆曲线上添加点。有一个选项，我必须处理这样的情况，比如添加一个点，例如 P(x,y) = (1,4) 和一些中性点的符号，例如Q=(e, e)。这种“相加”的结果应该是点(1,4)。它 (e) 不能为零值，因为那时点将是 Q(qx,qy)=(0,0) 并且将激活另一个功能，因此结果也会不同。你能为 BigInteger 分配一个符号吗？我需要类似

的东西

if(qx == e){
  BigInteger r1 = x1;
  BigInteger r2 = x2;
}

这是完整的功能：

static void addPoints(BigInteger x1, BigInteger y1, BigInteger x2, BigInteger y2, BigInteger a, BigInteger b, BigInteger p) throws Exception {
        BigInteger lambda = null;
        BigInteger x3 = null;
        BigInteger y3 = null;
       
        if (x1.compareTo(x2) == 0 && y1.compareTo(y2) == 0) { //same points
            lambda = (((new BigInteger("3").multiply(x1.pow(2))).add(a))
                    .multiply(Modul1.getReverseNumber(
                            (new BigInteger("2").multiply(y1)), p)))
                    .mod(p);

            x3 = ((lambda.pow(2)).subtract(new BigInteger("2").multiply(x1))).mod(p);
            y3 = ((lambda.multiply(x1.subtract(x3))).subtract(y1)).mod(p);

        } else if (x1.compareTo(x2) != 0) { //points are diffrent
            lambda = ((y2.subtract(y1)).multiply(
                    Modul1.getReverseNumber(x2.subtract(x1), p)
            )).mod(p);

            x3 = (((lambda.pow(2)).subtract(x1)).subtract(x2)).mod(p);
            y3 = ((lambda.multiply(x1.subtract(x3))).subtract(y1)).mod(p);

        } else if (x1.compareTo(x2) == 0 && y1.compareTo(p.subtract(y2)) == 0) { //y2 is negate
            System.out.println(O);
        } else { //Point add Neutral Point
            System.out.println("Punkt P + neutral : (" + x1 + "," + y1 + ")");
        }
    }

Answer 1

我稍微解决了一下。我使用 String 作为函数参数有一点。如果它是无穷大符号，则结果是第一个 Point，否则 null BigInteger 将被赋予此 String 的值。

 static void addPoints(BigInteger x1, BigInteger y1, String e1, String e2, BigInteger a, BigInteger b, BigInteger p) throws Exception {
        BigInteger lambda = null;
        BigInteger x3 = null;
        BigInteger y3 = null;

        if (e1.equals("e") || e2.equals("e")) {
            System.out.println("Punkt P + O to: (" + x1 + "," + y1 + ")");
        } else {
            BigInteger x2 = new BigInteger(e1);

            BigInteger y2 = new BigInteger(e2);


            String O = "symbol O";
            if (x1.compareTo(x2) == 0 && y1.compareTo(y2) == 0) {
                lambda = (((new BigInteger("3").multiply(x1.pow(2))).add(a))
                        .multiply(Modul1.getReverseNumber(
                                (new BigInteger("2").multiply(y1)), p)))
                        .mod(p);

                x3 = ((lambda.pow(2)).subtract(new BigInteger("2").multiply(x1))).mod(p);
                y3 = ((lambda.multiply(x1.subtract(x3))).subtract(y1)).mod(p);

                System.out.println("lamda to: " + lambda);
                System.out.println("x3: " + x3);
                System.out.println("y3: " + y3);
                System.out.println("Punkt P+P = (" + x3 + "," + y3 + ")");
            } else if (x1.compareTo(x2) != 0) {
                lambda = ((y2.subtract(y1)).multiply(
                        Modul1.getReverseNumber(x2.subtract(x1), p)
                )).mod(p);

                x3 = (((lambda.pow(2)).subtract(x1)).subtract(x2)).mod(p);
                y3 = ((lambda.multiply(x1.subtract(x3))).subtract(y1)).mod(p);

                System.out.println("Punkt P+Q = (" + x3 + "," + y3 + ")");
            } else if (x1.compareTo(x2) == 0 && y1.compareTo(p.subtract(y2)) == 0) {
                System.out.println("Infinity,Infinity");
            }

        }
    }

BigInteger 作为中性符号

BigInteger as neutral symbol

java

biginteger