(08-21-2020, 07:09 AM)royce Wrote: Only one way to find out.

(08-21-2020, 10:08 PM)philsmd Wrote:

Code:

hashcat --stdout -a 3 -o dict.txt Password?d?d?d
hashcat --stdout -a 3 -o dict.txt My?d?d?dPassword
hashcat -m 14800 -a 0 -w 3 hash.txt dict.txt

(08-22-2020, 10:11 AM)philsmd Wrote: I don't think it's practical, you would need to combine rules with each and every combination (10x -r multi-rule replaces within the command, rule-chaining, to change mutliple digits at the same time).
Let's say you have one rule that leaves the digit as is (: rule) and then for every digit (0-9) you replace them with another digit (10 * 9 replaces, because you don't need for instance "s00"), you get 1 + 90 = 91 replaces to just replace one unique digit in a password, but the password could also contain 0123456789, so you need to almost combine it with each and every other combination to make out all the 10 different digits replaced in the same password (almost 91 ^ 10), approximated : 91 * 90 * 89 * 88 * 87 * 86 * 85 * 84 * 83 * 82 = 23321927500622150400 combinations of replace (sxy) rules. That's not very practical if you want to do it that way.

It would make sense that you also mention the hash type, to make sure a good/clever strategy is feasible here.

You could for instance try to do the same as mentioned above (similar strategy, but not using a large dict file), but with a .hcmask file (no rules, but mask attack):

Code:

sed 's/[0-9]/?d/g' dict.txt > my.hcmask

sed here replaces every digits, with a ?d built-in charset and creates a mask. I would even filter it like this to make sure you do not run meaningless masks:

Code:

grep '[0-9]' dict.txt | sed 's/[0-9]/?d/g' > my.hcmask

this could produce several identical masks, so you should sort and unique these lines:

Code:

grep '[0-9]' dict.txt | sed 's/[0-9]/?d/g' | sort -u > my.hcmask

now you either could run it like this:

Code:

hashcat -m 18400 -a 3 -w 3 hash.txt my.hcmask

or filter and optimize it again, e.g. every masks that only ends or starts with at least one ?d, you could replace that with a -a 6 or -a 7 attack:

Code:

grep '^?d' my.hcmask | grep -v '[^?][^d]?d' | sed 's/?d//g' | sort -u > prepend_d_dict.txt

run this dict with -a 7 -i hash.txt ?d?d?d?d?d prepend_d_dict.txt

Code:

grep '?d$' my.hcmask | grep -v '?d[^?][^d]' | sed 's/?d//g' | sort -u > append_d_dict.txt

run this dict with -a 6 -i hash.txt append_d_dict.txt ?d?d?d?d?d

of course you could then eliminate these -a 6 and -a 7 compatible "masks" from the my.hcmask file and only leave the ones that replace something in the middle or on both sides/ends:

Code:

grep '^?d' my.hcmask | grep '[^?][^d]?d' > my_mod.hcmask

Code:

grep '?d$' my.hcmask | grep '?d[^?][^d]' >> my_mod.hcmask

Code:

grep -v '^?d' my.hcmask | grep -v '?d$' >> my_mod.hcmask

and uniq it:

Code:

sort -u my_mod.hcmask -o my_mod.hcmask

There are probably many more optimizations possible, but the -a 6 and -a 7 optimization seems to me the most important one (it doesn't mean that the speed will be much faster or you are guaranteed to complete the job much sooner, especially if you need to use it, as mentioned above, with -i  length increment etc).