Thank you very much for your answer, it helps a lot!



________________________________
From: Sascha Krissler <[email protected]>
To: [email protected]
Sent: Fri, May 14, 2010 1:56:02 AM
Subject: Re: [A51] rainbow table

The size of an LAPDm frame is always 23 bytes, but most messages are
shorter. A good candidate is the CIPHER MODE ENABLED message which is
the first encrypted message sent in a connection and only uses 3 bytes.
So 20 unused bytes are padded with a constant (0x2b).
Another source of known plaintext are messages with predictable content,
for example SYSTEM INFORMATION messages, which are
the same for all subscribers of a cell and are periodically sent to each
connected MS. Among those known bits of information are frequency
allocation of neighbouring cells, "location area identification" or
"cell identity".
In theory from analysis of the voice codec and the forward error correction
there may be sources of known plaintext bits found in the bursts transmitting
voice data. AFAIK nobody from the project looked into this by now.

---------------------
Von: MIHAITA ADELA <[email protected]>
Gesendet: 13.05.2010 19:08:29
An: [email protected]
Betreff: [A51] rainbow table




Hello,

I'm interested to learn more about the project you're working on and I would 
kindly ask you to help me with an answer to a simple question.
As far as I understood, the rainbow tables provide, for 64 bits known of 
keystream, the initial internal state that generated those bits. Once you have 
the initial internal state, you can  find the secret key used. My question is 
how do you   have access to  few samples of 64 bits of keystream, how do you 
get to find them out?  Is it because of that padding with known plaintext? but 
then, I don't get it with that known plaintext, who add is there?

I tried to follow the explanations given in the "GSM Basic" section but it's 
not clear for me. 

Thank you very much,
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