[jcifs] Quick LM Hash question.
crh at ubiqx.mn.org
Fri Apr 5 16:33:27 EST 2002
I am working on figuring out the authentication schemes used by Windows,
both for my book and for a presentation I'm doing. I've got a quick
question. My hope is that someone (Mike?) can point me to the specific
place where this is handled in jCIFS. I've been through the DES and MD4
code, but that's at a slightly lower level.
Here's the deal. All of the docs I've seen show the LanMan (LM) alorithm
E(K0K1,D0D1) = E(K0,D0)E(K0,D1)E(K1,D0)E(K1,D1)
This is what Leach wrote, but it doesn't make sense. The result of the
above is the same as:
E(K0K1,D0D1) = E(K0,D0D1)E(K1,D0D1)
Which just gives two concatonated strings, each the result of encrypting
the same data with two different keys.
I *think* that what should happen is:
E(K0K1,D0D1) = E(K1, E(K0,D0D1))
So, I'm digging through to find the code that actually does LanMan
Mike: Where do I look? NegProt?
Samba Team -- http://www.samba.org/ -)----- Christopher R. Hertel
jCIFS Team -- http://jcifs.samba.org/ -)----- ubiqx development, uninq.
ubiqx Team -- http://www.ubiqx.org/ -)----- crh at ubiqx.mn.org
OnLineBook -- http://ubiqx.org/cifs/ -)----- crh at ubiqx.org
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