On this weblog: https://internet.archive.org/internet/20160308014317/http://www.nilsschneider.internet/2013/01/28/recovering-bitcoin-private-keys.html
the creator confirmed a case that utilizing identical okay twice will leak personal key.
Many individuals know this methodology. However I discover typically, the components cannot give the correct reply(or I compute mistaken).
Take a look at this, you possibly can confirm signatures by public key:
public_key = 02a50eb66887d03fe186b608f477d99bc7631c56e64bb3af7dc97e71b917c5b364
msghash1 = 01b125d18422cdfa7b153f5bcf5b01927cf59791d1d9810009c70cd37b14f4e6
msghash2 = 339ff7b1ced3a45c988b3e4e239ea745db3b2b3fda6208134691bd2e4a37d6e1
sig1 = 304402200861cce1da15fc2dd79f1164c4f7b3e6c1526e7e8d85716578689ca9a5dc349d02206cf26e2776f7c94cafcee05cc810471ddca16fa864d13d57bee1c06ce39a3188
sig2 = 304402200861cce1da15fc2dd79f1164c4f7b3e6c1526e7e8d85716578689ca9a5dc349d02204ba75bdda43b3aab84b895cfd9ef13a477182657faaf286a7b0d25f0cb9a7de2
So enter information:
r=0861cce1da15fc2dd79f1164c4f7b3e6c1526e7e8d85716578689ca9a5dc349d
s1=6cf26e2776f7c94cafcee05cc810471ddca16fa864d13d57bee1c06ce39a3188
s2=4ba75bdda43b3aab84b895cfd9ef13a477182657faaf286a7b0d25f0cb9a7de2
z1=01b125d18422cdfa7b153f5bcf5b01927cf59791d1d9810009c70cd37b14f4e6
z2=339ff7b1ced3a45c988b3e4e239ea745db3b2b3fda6208134691bd2e4a37d6e1
I work out:
personal key = eaa57720a5b012351d42b2d9ed6409af2b7cff11d2b8631684c1c97f49685fbb
public key = 04e0e81185567ea58fc7e7258aa4d5c3e201a8d4ce2810c1007d87727a67eeb9a8c2ba06935280209f8bf42fc7603b65095f036044c4124ddf7c6a250cb450e4c8
Nevertheless, it is mistaken.
I am utilizing this python code to compute:
# this perform is from
# https://github.com/warner/python-ecdsa/blob/grasp/ecdsa/numbertheory.py
def inverse_mod( a, m ):
"""Inverse of a mod m."""
if a < 0 or m <= a: a = a % m
# From Ferguson and Schneier, roughly:
c, d = a, m
uc, vc, ud, vd = 1, 0, 0, 1
whereas c != 0:
q, c, d = divmod( d, c ) + ( c, )
uc, vc, ud, vd = ud - q*uc, vd - q*vc, uc, vc
# At this level, d is the GCD, and ud*a+vd*m = d.
# If d == 1, which means ud is a inverse.
assert d == 1
if ud > 0: return ud
else: return ud + m
def derivate_privkey(p, r, s1, s2, hash1, hash2):
z = hash1 - hash2
s = s1 - s2
r_inv = inverse_mod(r, p)
s_inv = inverse_mod(s, p)
okay = (z * s_inv) % p
d = (r_inv * (s1 * okay - hash1)) % p
return d, okay
p = 0xFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEBAAEDCE6AF48A03BBFD25E8CD0364141
# this case is correct
public_key=0x04dbd0c61532279cf72981c3584fc32216e0127699635c2789f549e0730c059b81ae133016a69c21e23f1859a95f06d52b7bf149a8f2fe4e8535c8a829b449c5ff
r =0xd47ce4c025c35ec440bc81d99834a624875161a26bf56ef7fdc0f5d52f843ad1
s1=0x44e1ff2dfd8102cf7a47c21d5c9fd5701610d04953c6836596b4fe9dd2f53e3e
s2=0x9a5f1c75e461d7ceb1cf3cab9013eb2dc85b6d0da8c3c6e27e3a5a5b3faa5bab
z1=0xc0e2d0a89a348de88fda08211c70d1d7e52ccef2eb9459911bf977d587784c6e
z2=0x17b0f41c8c337ac1e18c98759e83a8cccbc368dd9d89e5f03cb633c265fd0ddc
print "personal:%xn random:%x" % derivate_privkey(p,r,s1,s2,z1,z2)
print
# this case might be mistaken
public_key=0x02a50eb66887d03fe186b608f477d99bc7631c56e64bb3af7dc97e71b917c5b364
r =0x0861cce1da15fc2dd79f1164c4f7b3e6c1526e7e8d85716578689ca9a5dc349d
s1=0x6cf26e2776f7c94cafcee05cc810471ddca16fa864d13d57bee1c06ce39a3188
s2=0x4ba75bdda43b3aab84b895cfd9ef13a477182657faaf286a7b0d25f0cb9a7de2
z1=0x01b125d18422cdfa7b153f5bcf5b01927cf59791d1d9810009c70cd37b14f4e6
z2=0x339ff7b1ced3a45c988b3e4e239ea745db3b2b3fda6208134691bd2e4a37d6e1
print "personal:%xn random:%x" % derivate_privkey(p,r,s1,s2,z1,z2)
In actual fact, there have one other one met this downside:
https://crypto.stackexchange.com/questions/16615/ecdsa-how-to-retrieve-a-non-random-k
However he did not gave extra infomation, possibly he figured it out.
I’ve not discovered extra folks complaining about it, so, it is seemingly my fault by some means.
Are you able to level out my error? or simply level out the correct approach? Thanks.
