Help splitting brute-force attack - Printable Version +- hashcat Forum (https://hashcat.net/forum) +-- Forum: Deprecated; Previous versions (https://hashcat.net/forum/forum-29.html) +--- Forum: Old oclHashcat Support (https://hashcat.net/forum/forum-38.html) +--- Thread: Help splitting brute-force attack (/thread-3386.html) |
Help splitting brute-force attack - ner0 - 05-14-2014 Hi there. I've read some topics and wiki references but I don't seem to be able to get this right. Let's assume I'm trying to crack an MD5 hash and it's password is comprised of 8 characters, for ex: fvA#K2%h How would I set my command-line arguments to distribute this into 8 workloads that might be executed on 8 separate machines? Is it even possible what I'm asking? I've read this but it seems deprecated: https://hashcat.net/wiki/doku.php?id=distributing_workload_in_oclhashcat RE: Help splitting brute-force attack - epixoip - 05-14-2014 This is covered in the release notes for 1.20. See the "Improved distributed cracking support" section of http://hashcat.net/forum/thread-3323.html RE: Help splitting brute-force attack - ner0 - 05-14-2014 (05-14-2014, 08:46 PM)epixoip Wrote: This is covered in the release notes for 1.20. See the "Improved distributed cracking support" section of http://hashcat.net/forum/thread-3323.html Thanks epixoip. I've now understood how it works, my example: Quote:Hashcat -mNUM -aNUM --outfile=cracked_pwd.file --outfile-format=NUM hash_or_hashfile.hash ?a?a?a?a?a?a?a?a --keyspace ?a?a?a?a?a?a?a?a - being the length and charset used (8 chars). In my particular case, it gave me the number of keyspace as 7737809375 Then, I divided that number by how many individual machines to run the attack on, for example if it where 8 machines: 7737809375 / 8 = 967226171,875 = 967226172 (round number) Then we create 8 specific commands, one for each machine, like so: Quote:MACHINE 1: I think that's roughly it. Thanks for the help! |