你好，
我尝试从cookie中恢复公钥，这是我所做的，不确定那是否正确...
1）注册并登录以获取cookie：
user: test
pass: asd
cookie: dGVzdC0tMEYCIQDkk9vxwQ2A81geSQSTCxQEzGwTkA7gdYR0%2BpSr6MTNEwIhAJLLemlZFZZT6unXBg4i1WvdJy6tKVJrLQmj%2FM8SujPA
2）将Cookie解码为网址：
dGVzdC0tMEYCIQDkk9vxwQ2A81geSQSTCxQEzGwTkA7gdYR0+pSr6MTNEwIhAJLLemlZFZZT6unXBg4i1WvdJy6tKVJrLQmj/M8SujPA
3）将cookie解码为base64
test--0F\x02!\x00\xE4\x93\xDB\xF1\xC1\r\x80\xF3X\x1EI\x04\x93\v\x14\x04\xCCl\x13\x90\x0E\xE0u\x84t\xFA\x94\xAB\xE8\xC4\xCD\x13\x02!\x00\x92\xCBziY\x15\x96S\xEA\xE9\xD7\x06\x0E\"\xD5k\xDD'.\xAD)Rk-\t\xA3\xFC\xCF\x12\xBA3\xC0
4）解码cookie中“-”之后的部分，因为ECDSA（DER）说出它的“信号”。
ECDSA::Format::SignatureDerString.decode(sig)
<ECDSA::Signature:0x000056532e7cc928 @s=66397190700537287645903651815357348182011798486667182586289641565984306901952, @r=103388573995635080359749164254216598308788835304023601477803095234286494993683>
5）检查代码：（这现在不重要*） * ================================================== ==========
$group = Secp256k1
$private_key = UNKNOWN
def sign(str)
digest = Digest::SHA256.digest(str)
temp_key = str.size
signature = ECDSA.sign($group, $private_key, digest, temp_key)
end
temp_key = 'test'.size ==> 4
================================================ ============= *
6）公共密钥恢复：
1）检查s和r是否为整数：
signature.r.class
signature.s.class
2）计算曲线点R（x1，y1），其中x（h）x可以是r，r + n（r + 2n等）取决于$ group中的h值。
从十六进制转换n
n = 0xFFFFFFFF_FFFFFFFF_FFFFFFFF_FFFFFFFE_BAAEDCE6_AF48A03B_BFD25E8C_D0364141.ord
n = 115792089237316195423570985008687907852837564279074904382605163141518161494337
3）当h = 1 ==> j <= h
时计算xj所以我得到x0 = r和x1 = r + n
x0 = 103388573995635080359749164254216598308788835304023601477803095234286494993683
x1 = 219180663232951275783320149262904506162058819969664165517260679242195329665346```
# y² = x³ + 7 ==> x³ + 7 - y² = 0 ==> y = sqrt(x³ + 7)
So there are Y that can be -Y.
y0 = (Math.sqrt(x0**3 + 7)).to_i
y0 = 33243658930125968263381409750062872803328956523994685349884711523646619725501695561221713273581676204319400941060096
y1 = 102613229253348668425867949464551151644184036880118254846069064957181439439331464178252481009905814258844193460846592`
R0 = (x0, y0) or R0 = (x0, -y0) ==> R0 = x0,y0 R0m = x0,-y0 (arrays)
R1 = (x1, y1) or R1 = (x1, -y1) ==> R0 = x1,y1 R1m = x1,-y1 (arrays)
3）计算e = HASH（m）
m = "test"
e = Digest::SHA256.digest(m)
"\x9F\x86\xD0\x81\x88L}e\x9A/\xEA\xA0\xC5Z\xD0\x15\xA3\xBFO\e+\v\x82,\xD1]l\x15\xB0\xF0\n\b"
e.unpack("H*")
"9f86d081884c7d659a2feaa0c55ad015a3bf4f1b2b0b822cd15d6c15b0f00a08"
==> IN BINARY "1001111110000110110100001000000110001000010011000111110101100101100110100010111111101010101000001100010101011010110100000001010110100011101111110100111100011011001010110000101110000010001011001101000101011101011011000001010110110000111100000000101000001000"
4）z =（Ln-e的第一个len（n）n = 256）的最左位
z = 1000000000001000100010001000100010000000000000000000100010000000100010000000100000000000000000001000000000000000000000000000100010000000000000001000000000000000000010000000000010001000000000000000100010001000100010000000100000001000100000000000100000001000
==> IN DECIMAL 57911121712712222031468434730037426356855313884841210343156710740660779485192
5）计算u1 = -z * r ^（-1）mod n和u2 = s * r ^（-1）mod n
u1 = (-z * r**(-1)) % n
=> 11971578986221445867450691250469536884332685577127711573660715696043877254293372045122426347501727868950695322967745945196078037375262636091467191875787979/103388573995635080359749164254216598308788835304023601477803095234286494993683
u2 = (-z * s**(-1)) % n
=>(66397190700537287645903651815357348182011798486667182586289641565984306901952/103388573995635080359749164254216598308788835304023601477803095234286494993683)
6) Calculate curve point Qa = (Xa, Ya) = u1 *G + u2 * R
G = [55066263022277343669578718895168534326250603453777594175500187360389116729240, 32670510020758816978083085130507043184471273380659243275938904335757337482424]
u1.to_int = 115792089237316195423570985008687907852837564279074904382605163141518161494336
u2.to_int = 115792089237316195423570985008687907852837564279074904382605163141518161494336
现在是时候计算每个R. R0，R0m，R1和R1m的Qa了，但是我觉得我浪费了时间……有人可以看一下这个，并帮助我了解如何从cookie中恢复公钥吗？
我使用了维基百科[ecdsa恢复公用密钥]
https://en.m.wikipedia.org/wiki/Elliptic_Curve_Digital_Signature_Algorithm
和这些网站：
参考文献：
1）https://www.instructables.com/id/Understanding-how-ECDSA-protects-your-data/
2）https://www.johndcook.com/blog/2018/08/14/bitcoin-elliptic-curves/