Subject: Re: [bitcoin-dev] Time to worry about 80-bit collision attacks or not?

From: Gavin Andresen <[email-masked]>
Date: Fri, 8 Jan 2016 07:38:50 -0500

On Fri, Jan 8, 2016 at 7:02 AM, Rusty Russell <[email-masked]> wrote:

> Matt Corallo <[email-masked]> 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.[1]

> 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.

Instead, they .. what? I don't see a viable attack unless RIPEMD160 and
SHA256 (or the combination) suffers a cryptographic break.

-- 
--
Gavin Andresen