BlockChain-Account_TakeOver
题目描述
ECDSA
签名
假设我们的私钥为 d A d_A dA而公钥为 Q A Q_A QA， Q A = d A ⋅ G Q_A=d_Acdot G QA=dA⋅G，接下来就是签名的过程，要签名的消息为 m m m
- 取 e = H A S H ( m ) e = HASH(m) e=HASH(m)
- 取 e e e的左边的 L n L_n Ln个bit长度的值为 z z z， L n L_n Ln即为前面提到的参数里 n n n的比特长度
- 从 [ 1 , n − 1 ] [1, n-1] [1,n−1]范围内，随机选择一个整数 k k k
- 利用 k k k得到椭圆曲线上的一点 ( x 1 ， y 1 ) = k ⋅ G (x1，y1)=k cdot G (x1，y1)=k⋅G
- 然后计算 r ≡ x 1 ( m o d n ) r equiv x_1 (mod n) r≡x1(modn)，如果如果 r = 0 r=0 r=0则返回步骤3重新选择 k k k
- 计算 s = k − 1 ( z + r ⋅ d A ) ( m o d n ) s = k^{-1}(z + rcdot d_A) (mod n) s=k−1(z+r⋅dA)(modn)，如果 s = 0 s=0 s=0则返回步骤3重新选择 k k k
- 得到的数字签名即为 ( r ， s ) (r，s) (r，s)
验证
取点 P = ( X p , Y p ) = s − 1 ⋅ z ⋅ G + s − 1 ⋅ r ⋅ Q A P=(X_p,Y_p)= s ^ {-1} cdot z cdot G + s ^ {-1} cdot r cdot Q_A P=(Xp,Yp)=s−1⋅z⋅G+s−1⋅r⋅QA 若 X p = r X_p=r Xp=r，则签名有效，否则无效
Simple Proof P = ( X p , Y p ) = s − 1 ⋅ z ⋅ G + s − 1 ⋅ r ⋅ Q A P=(X_p,Y_p)= s ^ {-1} cdot z cdot G + s ^ {-1} cdot r cdot Q_A P=(Xp,Yp)=s−1⋅z⋅G+s−1⋅r⋅QA 因为 Q A = d A ⋅ G Q_A=d_A cdot G QA=dA⋅G，故有 P = s − 1 ⋅ z ⋅ G + s − 1 ⋅ r ⋅ Q A P= s ^ {-1} cdot z cdot G + s ^ {-1} cdot r cdot Q_A P=s−1⋅z⋅G+s−1⋅r⋅QA = s − 1 ⋅ z ⋅ G + s − 1 ⋅ r ⋅ d A ⋅ G = s ^ {-1} cdot z cdot G + s ^ {-1} cdot r cdot d_A cdot G =s−1⋅z⋅G+s−1⋅r⋅dA⋅G = s − 1 ⋅ G ⋅ ( z + r ⋅ d A ) = s ^ {-1} cdot G cdot {(z + r cdot d_A)} =s−1⋅G⋅(z+r⋅dA) 又有 s = k − 1 ( z + r ⋅ d A ) ( m o d n ) s = k^{-1}(z + rcdot d_A) (mod n) s=k−1(z+r⋅dA)(modn) 即 s − 1 = k ⋅ ( z + r ⋅ d A ) − 1 ( m o d n ) s^{-1} = kcdot(z + rcdot d_A)^{-1} (mod n) s−1=k⋅(z+r⋅dA)−1(modn) 代入得到 P = k ⋅ G P=k cdot G P=k⋅G
利用冲突的随机数恢复私钥
从上面的签名过程我们可以看到最关键的地方就在于随机数k，对于一个固定的椭圆曲线，一个确定的k就意味着一个确定的r，所以如果有两个相同的私钥签署的签名出现了相同的r就代表着在生成随机数时取到了相同的k，看到这里想必你也明白了我们题目的交易签名的问题出在哪了，这两笔交易的r值相同，代表在它们签名时使用的随机数k是相同的，而这就是我们恢复私钥的关键 我们不妨设这两个签名的 z z z与 s s s分别为 z 1 z_1 z1， z 2 z_2 z2与 s 1 s_1 s1， s 2 s_2 s2 则有 s 1 − s 2 = k − 1 ( z 1 + d A ⋅ r ) − k − 1 ( z 2 + d A ⋅ r ) s_1-s_2= k ^ {-1}(z_1 + d_A cdot r)-k ^ {-1}(z_2+ d_A cdot r) s1−s2=k−1(z1+dA⋅r)−k−1(z2+dA⋅r) = k − 1 ( z 1 − z 2 ) = k ^ {-1}(z_1 - z_2) =k−1(z1−z2) 那么 k = ( z 1 − z 2 ) ( s 1 − s 2 ) k = frac {(z_1-z_2)}{(s_1-s_2)} k=(s1−s2)(z1−z2) 通过 k k k，可以计算出 d A = ( s ⋅ k − z ) / r d_A=(s cdot k -z) /r dA=(s⋅k−z)/r
P2PKH
PubkeyScript是一张记录了交易记录的指令列表，它控制了下一名使用者如何解锁已接收的比特币并传送。收款人会制造一个signature script，而该文件必须满足最后一个发送者创建的PubkeyScript的参数。 PubkeyScript的参数：
- 公钥哈希(Public Key Hash) (比特币地址）
- 电子签署(ScriptSig: (r,s)+pubkey)
ASM是汇编代码
- OP_PUSHBYTES_71指压入栈中一个71字节大小的数据
- nSequence用以记录该笔交易是否可以上链
- ScriptPubKey是一个脚本，用以验证当前用户有能力使用这个UTXO中的比特币来支付，即证明身份 OP_DUP复制栈顶数据 OP_HASH160先后进行两种hash操作然后压入栈 OP_PUSHBYTES_20将签名的hash值压入栈 OP_EQUALVERIFY比较计算签名hash和刚压入栈的hash来验证有效性，有效返回1否则返回0 OP_CHECKSIG检测栈顶的2个元素，pub key和signature是否能对应的上。对应的上，说明这个签名的私钥，和收款人的公钥可以对上。有资格花这笔钱 OP_RETURN 用来当注释，携带一些信息 Transaction hex则包含所有信息连接在一起
求解过程
提取r,s,z
# -*-coding:utf-8-*-
"""
@author: iceland
"""
import sys
import hashlib
import argparse
from urllib.request import urlopen
# # ==============================================================================
# parser = argparse.ArgumentParser(
# description=This tool helps to get ECDSA Signature r,s,z values from Bitcoin rawtx or txid,
# epilog=Enjoy the program! :) Tips BTC: bc1q39meky2mn5qjq704zz0nnkl0v7kj4uz6r529at)
#
# parser.add_argument("-txid", help="txid of the transaction. Automatically fetch rawtx from given txid", action="store")
# parser.add_argument("-rawtx", help="Raw Transaction on the blockchain.", action="store")
#
# if len(sys.argv) == 1:
# parser.print_help()
# sys.exit(1)
# args = parser.parse_args()
# # ==============================================================================
#
# txid = args.txid if args.txid else
# rawtx = args.rawtx if args.rawtx else
#
# if rawtx == and txid == :
# print(One of the required option missing -rawtx or -txid);
# sys.exit(1)
# ==============================================================================
def get_rs(sig):
rlen = int(sig[2:4], 16)
r = sig[4:4 + rlen * 2]
# slen = int(sig[6+rlen*2:8+rlen*2], 16)
s = sig[8 + rlen * 2:]
return r, s
def split_sig_pieces(script):
sigLen = int(script[2:4], 16)
sig = script[2 + 2:2 + sigLen * 2]
r, s = get_rs(sig[4:])
pubLen = int(script[4 + sigLen * 2:4 + sigLen * 2 + 2], 16)
pub = script[4 + sigLen * 2 + 2:]
assert (len(pub) == pubLen * 2)
return r, s, pub
# Returns list of this list [first, sig, pub, rest] for each input
def parseTx(txn):
if len(txn) < 130:
print([WARNING] rawtx most likely incorrect. Please check..)
sys.exit(1)
inp_list = []
ver = txn[:8]
if txn[8:12] == 0001:
print(UnSupported Tx Input. Presence of Witness Data)
sys.exit(1)
inp_nu = int(txn[8:10], 16)
first = txn[0:10]
cur = 10
for m in range(inp_nu):
prv_out = txn[cur:cur + 64]
var0 = txn[cur + 64:cur + 64 + 8]
cur = cur + 64 + 8
scriptLen = int(txn[cur:cur + 2], 16)
script = txn[cur:2 + cur + 2 * scriptLen] # 8b included
r, s, pub = split_sig_pieces(script)
seq = txn[2 + cur + 2 * scriptLen:10 + cur + 2 * scriptLen]
inp_list.append([prv_out, var0, r, s, pub, seq])
cur = 10 + cur + 2 * scriptLen
rest = txn[cur:]
return [first, inp_list, rest]
# ==============================================================================
def get_rawtx_from_blockchain(txid):
try:
htmlfile = urlopen("https://blockchain.info/rawtx/%s?format=hex" % txid, timeout=20)
except:
print(Unable to connect internet to fetch RawTx. Exiting..)
sys.exit(1)
else:
res = htmlfile.read().decode(utf-8)
return res
# =============================================================================
def getSignableTxn(parsed):
res = []
first, inp_list, rest = parsed
tot = len(inp_list)
for one in range(tot):
e = first
for i in range(tot):
e += inp_list[i][0] # prev_txid
e += inp_list[i][1] # var0
if one == i:
e += 1976a914 + HASH160(inp_list[one][4]) + 88ac
else:
e += 00
e += inp_list[i][5] # seq
e += rest + "01000000"
z = hashlib.sha256(hashlib.sha256(bytes.fromhex(e)).digest()).hexdigest()
res.append([inp_list[one][2], inp_list[one][3], z, inp_list[one][4], e])
return res
# ==============================================================================
def HASH160(pubk_hex):
return hashlib.new(ripemd160, hashlib.sha256(bytes.fromhex(pubk_hex)).digest()).hexdigest()
# ==============================================================================
txn = 010000000153db4e56f159c0679818ef8ce814ce8fcaad12b854da7e582fb5f19266945f63000000006a47304402200e1e942f62d61cc25117d71bc2da4b523bd720dc7feec77551a0b152eb042cd7022030d7d78612b765dff96dd14fc5d723e06a8fa61b42a93410236273baf82f7f15012102572263bbac032e37cf96fe7664fb799f56353108c032807cc23ca557fb60b394ffffffff02d0070000000000001976a914b1c75a61c0461cd92c124e00ee275a600aa096b288ac0000000000000000056a0343544600000000
# if rawtx == :
# rawtx = get_rawtx_from_blockchain(txid)
print(
Starting Program...)
m = parseTx(txn)
e = getSignableTxn(m)
for i in range(len(e)):
print(= * 70,
f
[Input Index #: {
i}]
R: {
e[i][0]}
S: {
e[i][1]}
Z: {
e[i][2]}
PubKey: {
e[i][3]})
求dA
#-*-coding:utf-8-*-
r1 = 0x0e1e942f62d61cc25117d71bc2da4b523bd720dc7feec77551a0b152eb042cd7
s1 = 0x5611099541793c7681a9f8b48364d6e7088e16afe7b7b6244c52e94d28252a3b
z1 = 0x5ecd4154a2db20480d7715d6e47a772aaf596e11e6f16a4d58e9f0d260294660
r2 = 0x0e1e942f62d61cc25117d71bc2da4b523bd720dc7feec77551a0b152eb042cd7
s2 = 0x30d7d78612b765dff96dd14fc5d723e06a8fa61b42a93410236273baf82f7f15
z2 = 0x50f1c6205aab5f8dac7b505f91dfc437b1b13cd00f12a492570a040a27c38e25
assert r1 == r2
r = r1
def inverse_mod( a, m ):
"""Inverse of a mod m."""
if a < 0 or m <= a: a = a % m
c, d = a, m
uc, vc, ud, vd = 1, 0, 0, 1
while c != 0:
q, c, d = divmod( d, c ) + ( c, )
uc, vc, ud, vd = ud - q*uc, vd - q*vc, uc, vc
assert d == 1
if ud > 0: return ud
else: return ud + m
def derivate_privkey(p, r, s1, s2, z1, z2):
z = z1 - z2
s = s1 - s2
r_inv = inverse_mod(r, p)
s_inv = inverse_mod(s, p)
k = (z * s_inv) % p
d = (r_inv * (s1 * k - z1)) % p
return d, k
p = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141
privatekey,k=derivate_privkey(p,r,s1,s2,z1,z2)
利用私钥解密AES
import base64 cipher = base64.b64decode(b"4w/VLHqPZi/epoOGvjoY9TZWhDtYpL3iLsUTyvzghJM=") privatekey = 0xF41AA419CB6BD43F322D403F40728CE9784CD0B465F409322A76A3DF0A984A29 from Crypto.Cipher import AES from Crypto.Util.number import * key = long_to_bytes(privatekey)[0:16] iv = long_to_bytes(privatekey)[16:] aes = AES.new(key=key,iv=iv,mode=AES.MODE_CBC) flag = aes.decrypt(cipher.encode()) print(flag)