Date: Wed, 4 Aug 2010 00:55:12 +0400 From: Solar Designer <solar@...nwall.com> To: john-users@...ts.openwall.com Subject: Re: 1337 aka Leet Rules On Wed, Aug 04, 2010 at 12:35:32AM +0400, Solar Designer wrote: > To produce all combinations of "e" to "3" and "o" to "0" replacements in > a word containing 2 instances of each of these characters, you can use: > > %2e vap0 %2o vbp0 /e vcp0 /o op[o0] oc[e3] ob[o0] oa[e3] Actually, this can be simplified to: %2e op[e3] %2o op[o0] /e op[e3] /o op[o0] which is also more similar to the last line of the four-line and five-line versions: > An alternative that does not require 2 instances of each letter may be: > > /e op[e3] /o op[o0] > %2[eo] op\p /\r[eeo] op\p[e30] > %2[eo] op\p /e op3 /o op0 > %2e op3 %2o op0 /e op[e3] /o op[o0] ... > To avoid producing an unmodified instance of the word, this may be > further revised to: > > /[eo] op\p > /e op3 /o op0 > %2[eo] op\p /\r[eeo] op\p[e30] > %2[eo] op\p /e op3 /o op0 > %2e op3 %2o op0 /e op[e3] /o op[o0] Basically, we have the single line for all 16 combinations, and we may split it into 4*4 = 16 or 2+1+4+4+4 = 15 if we need to treat some of the combinations specially (such as exclude the no-op one). Alexander
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