Date: Sun, 12 May 2019 23:25:14 +0300 From: Aleksey Cherepanov <lyosha@...nwall.com> To: john-users@...ts.openwall.com Subject: Re: team john-users write-up for CracktheCon contest at CypherCon 2019 Aleksey Cherepanov <lyosha@...nwall.com> writes: > Duplicate character 3 times for every position, for positions 0-35: [...] > Just to demonstrate endless possibilities of rules, there is the same > rule for positions 0-124, i.e. up to max length (but position 124 is not > meaningful for duplication): > $ echo 123 | john --pipe --stdout 2>/dev/null \ > --rules=': vc0V vccV vd0c vccV vc0c vb0[0Vdc] va[0-9A-V]b >a Xa1a Xa1a' > 11123 > 12223 > 12333 Minor mistake in this rule: there is 1 position overlap between internal packs (e.g. 0+V vs V+0), but sequential dupe suppression hides it, so there are no dupes during hashing. So the rule should be the following: --rules=': vc0V vccV vd0c vccV vc0c vb0[0Vdc] va[0-9A-U]b >a Xa1a Xa1a' The change is 'U' in 'va[0-9A-U]b'. It covers positions 0-123. To debug manipulations on variables, I used the following command: $ python -c "print '_' * 125" | john 2>/dev/null --pipe --stdout \ --rules=': ... oaA' \ | perl -pe 's/^(_*)A_*$/length $1/e' Positions 0-30: va[0-9A-U]0 Positions 31-61: vb0V va[0-9A-U]b Merge the ranges using preprocessor: Positions 0-61: vb0[0V] va[0-9A-U]b Positions 62-92: vc0V vccV vd0c vb0d va[0-9A-U]b Positions 0-92: vc0V vccV vd0c vb0[0Vd] va[0-9A-U]b Positions 93-123: vc0V vccV vd0c vccV vc0c vb0c va[0-9A-U]b Positions 0-123: vc0V vccV vd0c vccV vc0c vb0[0Vdc] va[0-9A-U]b So there are base values 0, 31, 62 and 93. All base values are computed in every rule (after preprocessing), but only one value is used. Then 0-30 is added. Lack of gaps may be checked with the following additional command in the pipe: ... | perl -ne 'print "failure\n" if $. != $_ + 1' Thanks! -- Regards, Aleksey Cherepanov
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