you can tell the keyspace covered by a table from the sorted.table
file in the torrent. if it lists 32 rounds then the average chain length
is 1 million, with 8 rounds it is 256 thousand.
Look at the startbits value and calculate the number of chains:
filesize(sorted.start.tbl) * 8 / 53.
Then you have the value for a single table. multiply that by the number
of tables available and you have the size of the covered keyspace.
as a rule of thumb: 8 GB torrents are 8 rounds and 760M chains, so
760M * 256k = 2^47.5
2GB torrents are 32 rounds and 190M chains, so
190M * 1M = 2^47.5
The initial table parameters said to generate 380M chains and
would cover 380M * 1M = 2^48.5 and be about 4GB in size.

i know of at least 32 tables lying around which would provide a coverage
of 2^52.5. so with only those tables you need 2^63 / 2^52.5 = 2^10.5
64bit samples for 50% success. 2^10.5 samples can be gathered from 28
fully known GSM bursts (7 fully known plaintext LAPDm messages).

On Mon, Dec 28, 2009 at 08:57:20AM +0000, p q wrote:
> do you have estimate how much tables have been generated successfully ? how
> far we are from the point that having 50% success is guaranteed?
> 
> 
> From: sascha <[email protected]>
> > Date: Mon, Dec 28, 2009 at 8:44 AM
> > Subject: [A51] A51 torrents
> > To: [email protected]
> >
> >
> > The torrents for completed and sorted tables are being made available
> > at http://reflextor.com/torrents
> > More will be added as they come in.
> > Of course they are fastest to access at the congress building, but i
> > suggest you download what you want to have in the next few days to
> > take advantage of the big congress uplink capacity.
> > _______________________________________________
> > A51 mailing list
> > [email protected]
> > http://lists.lists.reflextor.com/cgi-bin/mailman/listinfo/a51
> >
> >

> _______________________________________________
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