bugtraq - Re: SHA-1 broken

lists.openwall.net		lists / announce owl-users owl-dev john-users john-dev passwdqc-users yescrypt popa3d-users / oss-security kernel-hardening musl sabotage tlsify passwords / crypt-dev xvendor / Bugtraq Full-Disclosure linux-kernel linux-netdev linux-ext4 linux-hardening linux-cve-announce PHC
Open Source and information security mailing list archives

Hash Suite: Windows password security audit tool. GUI, reports in PDF.

[<prev] [next>] [thread-next>] [day] [month] [year] [list]

Message-ID: <b841ffed05022013587114cb42@mail.gmail.com>
Date: Mon, 21 Feb 2005 08:58:54 +1100
From: Michael Silk <michaelsilk@...il.com>
To: bugtraq@...urityfocus.com, exon@...e.se
Subject: Re: SHA-1 broken


Inline. 

> -----Original Message-----
> From: exon [mailto:exon@...e.se] 
> Sent: Saturday, 19 February 2005 8:58 PM
> To: bugtraq@...urityfocus.com
> Subject: Re: SHA-1 broken
> 
> Michael Silk wrote:
> > Michael,
> > 
> >  But wouldn't it render a login-based hashing system 
> resistant to the 
> > current hashing problems if it is implemented something like:
> > 
> >  --
> >  result = hashFunc1( input + hashFunc2(input) + salt ) 
> > // 
> > // instead of
> > //
> > result = hashFunc1( input + salt )
> >  --
> > 
> 
> I assume you mean hashFUnc2 inside the parentheses 

Yes.

 
> No it won't, because if hashFunc2 has collisions the 
> resulting output will collide in hashFunc1 as well. 

How?

The attackers input is "input". He can only choose to enter a
collision for "hashFunc1" _OR_ "hashFunc2". He can't enter a
collision for both, but that is what he needs to pass this
function with a different string from the original.


> The 
> collision resistance in this case is somewhat less than that 
> of hashFunc2 (because two different outputs of hashFunc2 
> might collide in hashFunc1, 

Sure, hashFunc2 might give collisions, but it doesn't mean anything
unless _THOSE_ collisions are collisions in hashFunc1 that lead to the
original hash.


> but a 
> strong hash isn't supposed to depend on the algorithm not being known.

Obviously.

-- Michael