[bitcoin-dev] Time to worry about 80-bit collision attacks or not?
aj at erisian.com.au
Fri Jan 8 15:33:29 UTC 2016
On Fri, Jan 08, 2016 at 07:38:50AM -0500, Gavin Andresen via bitcoin-dev wrote:
> Lets see if I've followed the specifics of the collision attack correctly,
> Ethan (or somebody) please let me know if I'm missing something:
> So attacker is in the middle of establishing a payment channel with
> somebody. Victim gives their public key, attacker creates the innocent
> fund-locking script '2 V A 2 CHECKMULTISIG' (V is victim's public key, A
> is attacker's) but doesn't give it to the victim yet.
Using Ethan Heilman's procedure, the attacker can create two scripts:
2 V __A1__ 2 CHECKMULTISIG
2 V __A2__ 2 CHECKMULTISIG
and find values A1 and A2 which hash the scripts to the same result
with under 3*2**80 work. I think you can do that by setting the next
private key as the result of RIPEMD(SHA256(script with pubkey)), so you
could still spend either. But it doesn't change the script, so it's not
*that* helpful -- you've just got two different keys you can use.
Ah, but you can make the form of the script be a function of your key, so:
if privkey % 2 == 0:
script = "2 V %s 2 CHECKMULTISIG" % (pubkey)
script = "%s CHECKSIG" % (pubkey)
hash = ripemd160(sha256(script))
nextprivkey = hash
Then you have a 50% chance of your cycle giving you a matching hash for
one script with A1 and the other script with A2, and you can find the
cycle with under 3*2**80 work. Doing five attempts should give you ~96%
chance of hitting a usable pair, and should take under 15*2**80 work ~=
2**84 work, with trivial memory use.
Trying that in python with a vastly weakened hash function (namely,
the first five bytes of ripemd160(sha256()), with 40 bits of security
and 3*2**20 work) works as expected -- I got a "useful" collision on my
second try in about 7 seconds, seeding with "grumpycat3" ("grumpycat2"
didn't work) with the result being:
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