Re: When will md5crk complete?
From: WinTerMiNator (me_at_privacy.net)
Date: 05/23/04
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Date: Sun, 23 May 2004 21:03:10 +0200
Tom St Denis wrote:
> Gregory G Rose wrote:
>> I was just looking at the md5crk web site, stats
>> page, http://www.md5crk.com/stats/ , and realized
>> that the rate of total MD5s per second isn't
>> climbing very much, and therefore the probability
>> of completion in a given time isn't getting much
>> better. So then I asked myself, "When do we expect
>> it to complete?", and I didn't like the answer (or
>> the punctuation :-) ); I get about 222 years.
>>
>> This surprised me. Were there really that many
>> more machines on the RC5/64 project? Or is MD5
>> really that much slower? Or what?
>
> IIRC on my Athlon XP box I got around 6M/sec with dnet and on my P4 I
> get around 11.3M/sec with md5crk.
>
> I don't think that md5crk is any slower I just think there are far
> fewer active members. Just recently I've gone up from mid 500 to
> ~290 in the ranking in the span of a week.
>
> The best advice is to just keep plugging, hope that Jean-Luc got his
> coding right [hehehe just kidin JL] and try to recruit new volunteers.
>
> Tom
I have read md5crck FAQ carefully, and I have found strong contradictions:
- JL wants to show that the use of MD5 hash in digital signatures is
unsecure; he gives as examples the use of certificates by Paypal, merchants
sites... In that case, the signature is existing, and, using brute force,
there is a 0.5 probability to find a collision against a given signature
(that means, a forged certificate having same MD5 hash) after 2^127 trials,
provided MD5 is really a randomly distributed function.
- However, his attack follows the "birthday paradox" scheme: he just wants
to find, in a large set, two documents colliding. In that case, the order of
magnitude of trials will be proportional to the root of the number of
possible hashes, ~2^64 (and the birthday attack in signing a contract is
very easy to counter: just make a very slight change - adding a space
character - to any contract proposed to signature that you have not written
yourself!).
This is all the problem of md5crck: the attack is not against the threat
model! And an eventual success in finding a collision in a set will prove
nothing about forging a given signature: let's suppose as, like JL expects,
~2 years will be necessary, with thousands computers, to find a collision
like he intends. This will just prove that, with the same computer power,
one will need 2^63 more time, that is 2^64 years, or 18446744073709551616
years to forge a certificate...
Very counterproductive! This is not the way to convince certificate builders
to stop to use MD5...
-- Michel Nallino aka WinTerMiNator http://www.chez.com/winterminator (Internet et sécurité: comment surfer en paix) http://www.gnupgwin.fr.st (GnuPG pour Windows) Adresse e-mail: http://www.cerbermail.com/?vdU5HHs5WG
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