>of the 2^64 keyspace, 2^53 data points are computed and 2^38 are stored >on disk. The 2^53 data points that are effectively contained in the tables >map about 2^56.5 data points of the keyspace. Since each chain has an >average length of 2^15 the number of data points stored is 2^38, even though >the table contains 2^53 data points.
>2^56.5 is enough because a single gsm burst lets you do 51 independent lookups >in the tables (Time Memory Data Tradeoff). 2^53 table values cover 2^56.5 >of the keyspace, because in gsm A5/1 is clocked 100 times before the keystream >is used to encrypt the plaintext stream and the A5/1 cipher by its structure >converges to a smaller subset of the state space when it is clocked. >Meaning that of the 2^64 possible states, 2^3 are unreachable after 100 >clocks and thus can be ignored. By keyspace you mean Kc with 56 bits as approx. the last 8 bits are zeroes right ? What do you mean by gsm brust do 51 independent lookups? What I understand is that you generate a key stream which is used to encrypt plain text and it will do it after the clock passed 100 times ? >>2^3 are unreachable after 100 clocks and thus can be ignored. Why is that ? where is the COUNT that has 22 bits should be fed into A5? I understand from the demo that you have 3 LFSR , and you generate keystream based on the clock right that key stream is XORED with plain text and generate the cipher text ? Thanks,, Omar Atia -----Original Message----- From: [email protected] [mailto:[email protected]] On Behalf Of [email protected] Sent: Tuesday, January 04, 2011 1:23 PM To: [email protected] Subject: Re: [A51] I'm new to A5/1 On Tue, Jan 04, 2011 at 12:57:28PM +0200, Omar Atia wrote: > Dears, > > Why you are creating rainbow tables ? I have read what is in Airprobe > sniffing ways it is missing how do you decrypt the cipher text with Kc ? do > you have this stage or you are only creating these tables ? The tables had been finished in last years summer and they work as intended. > > How can I start the study ? wikipedia: Rainbow Table http://reflextor.com/trac/a51/wiki/BackclockA51 > > Are you writing all possibilities on a disk ? of the 2^64 keyspace, 2^53 data points are computed and 2^38 are stored on disk. The 2^53 data points that are effectively contained in the tables map about 2^56.5 data points of the keyspace. Since each chain has an average length of 2^15 the number of data points stored is 2^38, even though the table contains 2^53 data points. 2^56.5 is enough because a single gsm burst lets you do 51 independent lookups in the tables (Time Memory Data Tradeoff). 2^53 table values cover 2^56.5 of the keyspace, because in gsm A5/1 is clocked 100 times before the keystream is used to encrypt the plaintext stream and the A5/1 cipher by its structure converges to a smaller subset of the state space when it is clocked. Meaning that of the 2^64 possible states, 2^3 are unreachable after 100 clocks and thus can be ignored. _______________________________________________ A51 mailing list [email protected] http://lists.lists.reflextor.com/cgi-bin/mailman/listinfo/a51 _______________________________________________ A51 mailing list [email protected] http://lists.lists.reflextor.com/cgi-bin/mailman/listinfo/a51
