thanks . sorry this is probably my fault but i couldn't clearly
understand what would be the success chance of this attack , roughly
and why . is there any detailed paper on this so i can put some time
and read to finally understand it or maybe you can elaborate more ?

thank you in advance

>
> 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 ?
>> >
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