the keyspace of A5/1 is 2^64. When you record a conversation you can guess a 
whole LAPDm frame,
so you get enough plaintext and thus A5/1 keystream that you can do 204 lookups 
in the database.
That is because, suppose you have 114 bits of known keystream, then you can try 
to reverse the
A5/1 register state that produced keystream 0..63 or you can try to reverse the 
state that produced
keystream bits 1..64 and so on. with 114 bits you would get 51 values to lookup 
in the table that
gives you a register state for a keystream. we get 4x114 bits of known 
plaintext.
So a table with 2^57 values stored in it, would give you a success probability 
of 1/(2^7) (because
57 + 7 = 64). The probability of a failed lookup is thus 1-1/(2^7).
calculate (1-1/2^7)^204 and you get the probability that 204 lookups fail. that 
would be 20%,
so the tables at this point of completion can decrypt 80% of the conversations.
50%, 80% ... actually there is no definitive point to say we are done it is 
more like
the chances of success are continually rising.

i'm not advanced on cryptography . would you tell me why 50% ?
> 
> From:  *Sascha Krissler*<[email protected]>
> Date: Fri, Sep 11, 2009 at 8:16 PM
> Subject: Re: [A51] sucess?
> To: [email protected]
> 
> When reach our initial goal of 2^8.5 tables with 2^28.5 unique chains 
> each, we would get a probability
> of around 50%, so at some point this would be the actual probability. 
> It is not unlikely that people will carry
> on generating, increasing this probability and also increasing the 
> lookup speed.
> 
> > suppose the community build all the tables then how much is our 
> success chance to crack a given recorded conversation ?
> >
> >
> 
> _______________________________________________ A51 mailing list A51@
> lists.reflextor.com http://lists.lists.reflextor.com/cgi-bin/mailman/
> 
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