[bitcoin-dev] Time to worry about 80-bit collision attacks or not?
watsonbladd at gmail.com
Fri Jan 8 14:34:09 UTC 2016
On Fri, Jan 8, 2016 at 4:38 AM, Gavin Andresen via bitcoin-dev
<bitcoin-dev at lists.linuxfoundation.org> wrote:
> On Fri, Jan 8, 2016 at 7:02 AM, Rusty Russell <rusty at rustcorp.com.au> wrote:
>> Matt Corallo <lf-lists at mattcorallo.com> writes:
>> > Indeed, anything which uses P2SH is obviously vulnerable if there is
>> > an attack on RIPEMD160 which reduces it's security only marginally.
>> I don't think this is true? Even if you can generate a collision in
>> RIPEMD160, that doesn't help you since you need to create a specific
>> SHA256 hash for the RIPEMD160 preimage.
>> Even a preimage attack only helps if it leads to more than one preimage
>> fairly cheaply; that would make grinding out the SHA256 preimage easier.
>> AFAICT even MD4 isn't this broken.
> It feels like we've gone over that before, but I can never remember where or
> when. I believe consensus was that if we were using the broken MD5 in all
> the places we use RIPEMD160 we'd still be secure today because of Satoshi's
> use of nested hash functions everywhere.
>> But just with Moore's law (doubling every 18 months), we'll worry about
>> economically viable attacks in 20 years.
>> That's far enough away that I would choose simplicity, and have all SW
>> scriptPubKeys simply be "<0> RIPEMD(SHA256(WP))" for now, but it's
>> not a no-brainer.
> 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.
> Instead they then generate about 2^81scripts that are some form of
> pay-to-attacker ....
> ... wait, no that doesn't work, because SHA256 is used as the inner hash
> function. They'd have to generate 2^129 to find a cycle in SHA256.
For 2^80 they simply generate 2^80 scripts that look innocent, and
2^80 that are not. With high probability there is a collision. I agree
that most cryptanalysis won't work because of the nesting, but 2^80 is
> Instead, they .. what? I don't see a viable attack unless RIPEMD160 and
> SHA256 (or the combination) suffers a cryptographic break.
> Gavin Andresen
> bitcoin-dev mailing list
> bitcoin-dev at lists.linuxfoundation.org
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