Masks ordered by occurrences instead of by efficiency

[Sc00bz]

You should order masks by efficiency (occurrences/key_space) because this will lead to the less guesses to crack passwords.

Looking at the top 5:

Kaonashi/masks/kaonashi_masks_numbered.txt

Lines 1 to 5 in 5239bd3

	227251350:?l?l?l?l?l?l?l?l
	188463345:?d?d?d?d?d?d
	171728714:?l?l?l?l?l?l
	122109481:?d?d?d?d?d?d?d?d
	117878793:?l?l?l?l?l?l?l

And ordering just the 1000 in kaonashi_masks_numbered.txt by efficiency the top 5 are now:

673407:?u?u?u?u             // was #309
122109481:?d?d?d?d?d?d?d?d  // was #4
409898:?u?u?u               // was #470
149077:?l?d?d?d             // was #979
442599:?u?l?l               // was #444

You can use PACK to do this:

sed -e "s/^\(.*\):\(.*\)$/\2,\1/" kaonashi_masks_numbered.txt > input.txt
./maskgen.py input.txt -o output.hcmask

But first you'll need to add a line for ?b:

elif char == "b": count *= 256

But I'd suggest that you change ?b to ?1 and add -1 not_a.hcchr to hashcat options. Where "not_a.hcchr" is everything not in ?a which is what I assume ?b was used for. And obviously the line would be:

elif char == "1": count *= 161

Looking at the top 3 with ?b/?1, they change ranks quite a bit:

#315  ?l?l?l?l?l?b
#457  ?l?l?l?l?l?l?b
#459  ?d?d?d?d?d?d?b?b
---
#302  ?l?l?l?l?l?1
#432  ?d?d?d?d?d?d?1?1
#441  ?l?l?l?l?l?l?1

[Sc00bz]

P.S. I made a pull request (iphelix/pack#13) so if you use that code, you can just do:

./maskgen.py input.txt -o output.hcmask --custom-charset1-len 161