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Re: Determining the encryption used
From: Tim <tim-pentest () sentinelchicken org>
Date: Sat, 13 May 2006 13:01:47 -0400

Hello Phoebe,

I'm don't know a lot about these matters, but I was under the
impression that if a password verification system is checking passwords
against a hash table, all you needed was a collision (as this would hash
to the correct value in the table and the comparison of the two would
return true).

Yes, any hash function will be subject to collisions on arbitrary
inputs.  However, the vulnerabilities found in MD5 and SHA1 don't
involve taking an existing hash and generating collisions against it.
They involve generating two seperate hashes which have a collision.
This seems like a very minor distinction at first, but it is actually a
very different type of attack.

Normally it should be very difficult to generate any collision at all
against secure hash functions, let alone using useful inputs.

Is this really naive?

Somewhat.  A summary of the three desired properties of a hash function
can be found here:
  http://en.wikipedia.org/wiki/Cryptographic_hash_function#Cryptographic_properties

The collision attacks found can break the security of cryptographic
signatures, since attackers potentially have control over multiple hash
inputs.  Where an attacker has control over only one input, (but knows
the value of it), collisions can only be generated by breaking the
second preimage property.  Reversing a hash to an original unknown
value, requires a (first) preimage attack.

Collision attacks are much easier to conduct due to the birthday
"paradox".  Just because this property of a hash has been broken,
doesn't mean the others have.

HTH,
tim

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