椭圆曲线点 [英] Elliptic curve point

问题描述

目前，iam正在开发一个使用椭圆曲线的项目。请给我一个解决方案，确定一个点是在椭圆曲线上吗？以及如何在椭圆曲线上获得点

Presently iam working on a project that uses elliptic curve. Please provide me a solution that determines whether a point is on the elliptic curve or not? and also how to get a point on the elliptic curve

推荐答案

检查点是否在椭圆曲线上很容易。只需检查您的点（x，y）是否满足定义椭圆曲线的方程： y ^ 2 = x ^ 3 + ax + b （记得在正确的字段中执行计算）。

Checking whether a point is on the elliptic curve is easy. Just check whether your point (x,y) fulfills the equation defining your elliptic curve : y^2 = x^3 + ax + b (remember to perform the calculation in the correct field).

使用Bouncycastle你可以这样做：

Using Bouncycastle you can do it like this:

ECCurve curve = //...
ECFieldElement x = //...
ECFieldElement y = //...

ECFieldElement a = curve.getA();
ECFieldElement b = curve.getB();
ECFieldElement lhs = y.multiply(y);
ECFieldElement rhs = x.multiply(x).multiply(x).add(a.multiply(x)).add(b);

boolean pointIsOnCurve = lhs.equals(rhs);

您已使用加密技术标记问题，因此我假设您要询问有限域上的椭圆曲线。该曲线将具有生成器 g 和订单。
要获取随机点，只需生成一个随机整数 x ，范围在0到（order -1）之间，然后选择 x * g 。

You have tagged the question with cryptography, so I assume you are asking about elliptic curves over a finite field. The curve will have a generator, g and an order. To get a random point, just generate a random integer, x, between 0 and (order - 1), and choose x * g.

您可以使用Bouncycastle这样做：

You can do it using Bouncycastle like this:

X9ECParameters x9 = NISTNamedCurves.getByName("P-224"); // or whatever curve you want to use
ECPoint g = x9.getG();
BigInteger n = x9.getN();
int nBitLength = n.bitLength();
BigInteger x;
do
{
    x = new BigInteger(nBitLength, random);
}
while (x.equals(ZERO)  || (x.compareTo(n) >= 0));
ECPoint randomPoint = g.multiply(x); 

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