If I understood this correctly, I think there is a way to remove the already checked keyspace.
I think mathematically one would say there is an intersection between ?l?u?d and ?l?d and the new set of (sub-)masks (=> keyspaces) you want to check now is basically:
Code:
?l?u?d - (?l?u?d ∩ ?l?d) = ?l?u?d - ?l?d
This means you want to remove from the new keyspace the keyspace that intersects the new and old one, basically this one:
It is possible for instance to generate a mask-file (.hcmask,
http://hashcat.net/wiki/doku.php?id=mask...mask_files ) that contains all the (sub-)masks which are not in (?l?u?d ∩ ?l?d), for instance:
Code:
./mp64.bin -1 lud ???1???1???1???1???1???1???1???1 > all.txt
./mp64.bin -1 ld ???1???1???1???1???1???1???1???1 > ld.txt
while read i; do fgrep $i ld.txt > /dev/null || echo $i; done < all.txt > mymask.hcmask
(Note: mp64.bin is the maskprocessor:
http://hashcat.net/wiki/doku.php?id=maskprocessor )
basically the while loop removes from the full list of (sub-)mask of ?l?u?d, all masks that intersect (I use while combined w/ fgrep here, but other tools can do this too).
It would be interesting to know how much faster it is to use this mask-file approach vs full ?l?u?d (I mean if the overhead of mask-files, should be minimal, does add much to the overall time, i.e. if the time of running ?l?u?d is almost the same as running ?l?d + intersection above).
There may be other approaches too, maybe w/o the use of maskprocessor.... but this seems to work very well...
It would be great to hear other ideas too.
UPDATE: corrected image since ?l?d is a subset of ?l?u?d