如何从曲线上的点（SECP256K1或P-256）导出tezos公钥

- 7
- 219 2019-02-27
如何从曲线上的x,y点(SECP256K1或P-256)得出Tezos公钥?

例如,对于以太坊,您可以使用Keccak-256上的[x,y]进行计算. 然后,通过获取最后40个字节(20个十六进制字符)并以0x作为前缀,总共42个字节来获取地址.

How can I derive Tezos public key from x,y points on the curve (either SECP256K1 or P-256)?

For example, for Ethereum you can compute this using Keccak-256 on the [x,y]. The address is then obtained by taking the last 40 bytes (20 hex chars) and prefixing it with 0x for a total of 42 bytes.

address

1 个回答

投票数

- 8
- 2019-02-27
Tezos中的公钥以压缩形式存储:

<1 byte prefix>|X

如果Y为偶数,则前缀为b'\ x02',否则为b'\ x03',现在您的压缩密钥长度为33个字节；

为secp256k1添加tezos前缀b'\ x03 \ xfe \ xe2V',为p256添加b'\ x03 \ xb2 \ x8b \ x7f';

使用带有校验和的Base58算法进行编码.

检出pytezos.crypto 源以获取实施示例.
Public keys in Tezos are stored in compressed form:

<1 byte prefix>|X

Prefix is b'\x02' if Y is even, b'\x03' otherwise, now you have a compressed key of 33 bytes long;

Add tezos prefix b'\x03\xfe\xe2V' for secp256k1, b'\x03\xb2\x8b\x7f' for p256;

Encode with Base58 algo with checksum.

Checkout pytezos.crypto sources for implementation examples.
- 这里的" Y"是什么?(一个错字,应该是X吗?)
  
  What is `Y` here? (A typo, should be X?)
  - 0
  - 2020-01-17
  - arvidj
- X,Y来自椭圆曲线方程,对于secp256k1,它是y ^ 2=x ^ 3 + 7
  
  X, Y are from the elliptic curve equation, for secp256k1 it's `y^2 = x^3 + 7`
  - 0
  - 2020-01-17
  - Michael Zaikin