3 ms·

remember though, the l33t letter replacements are in the dictionaries for common passwords, and used in brute-force algorithms.

patio11 16y ago
True, but increasing the raw length of the password (which passphrases will give you in spades) makes those leetification tricks combinatorially explosive. "apple" has 3 characters (2^3 = 8 variations) for easy leetspeak. "apples" has 2^4 = 16. "applestastegreat" has 2^12 = 4,096. Assuming you don't totally fubar your entropy in the underlying passphrase, you'll end up with something both memorable and infeasible to brute force.
- ominous_prime 16y ago
  Oh yes, and I do use this in my own passphrases for the added combinational complexity. I just wanted to point out to the naive that using `passw0rd` or `s3cr3t`, they are in password lists that a brute force attack will use. My point that crackers will programaticaly try this was less relevant.
  - nostrademons 16y ago
    You can choose to l33tify different combinations of letters in the password. passw0rd may be in the dictionary, but is p4ssword, or pa$$word? Patrick's point is that there are combinatorially many different ways that you can 133t!fy a password, so they don't really help in dictionary attacks.
    - JoeAltmaier 16y ago
      I know I am annoyed by "keyspace rules" that actually reduce the keyspace (rant). So, I capitalize the 1 letter and put 0 or 1 at the end - voila! IT appeased. Am I the only one? No. So the 'keyspace rules' don't accomplish anything, and may actually reduce the keyspace.
    - redthrowaway 16y ago
      >passw0rd may be in the dictionary, but is p4ssword, or pa$$word? I'm guessing yes. If I were building a dictionary, I would add every password that's been dumped from sites like Gawker, etc. With hundreds of thousands of passwords floating around in the ether, you can quite quickly cover common leetifications. Also, it's trivially easy to leetify algorithmically, so padding a dictionary wouldn't be difficult. You'll run into space issues eventually, but you can easily split the dictionary over several machines. The best solution for the user is to pick a fairly complex, long, and unique password. Server-side, assume your database will be compromised use something like bcrypt to make cracking it computationally expensive.
    - tripzilch 16y ago
      > You can choose to l33tify different combinations of letters in the password. passw0rd may be in the dictionary, but is p4ssword, or pa$$word? yes they are. well, actually, password cracking software (most used is Jack the Ripper) comes with a bunch of commandline tools that allow you to do substitutions and common variations to your password lists. generally, when trying to crack a specific target, you start with a few "seed" password lists (common passwords, but also lists of target-specific jargon), and then apply the transformation tools on those to expand your lists. say you'd want to crack accounts at an aviator forum, you'd include a list of airplane types and aviation jargon. then you'd apply the transformation tools, and you'll end up with boeing Boeing b0eing Bo3!ng, etc. check this guy's blog, he writes about how it's (usually) done: http://www.skullsecurity.org/blog/2010/the-ultimate-faceoff-between-password-lists http://www.skullsecurity.org/blog/2010/the-ultimate-faceoff-... (analysis of leaked password lists) http://www.skullsecurity.org/blog/?s=ripper http://www.skullsecurity.org/blog/?s=ripper (describes Jack the Ripper in these articles)
iuguy 16y ago
If you want to cheat at complexity you can look at the most common mechanisms for brute force and their defaults. For example, the publicly available rainbow tables are mainly built for US character sets. If you're using characters that aren't in those character sets the rainbow table lookup will fail (as the hash wasn't precomputed). Likewise, if dictionaries (and the permutation rules) lack a non-standard character, using that character will defeat the dictionary attack. If you use a straight brute force then there's a trade off between the characters you use and the amount of time it'll take to exhaust a particular set of characters. It's only if the attacker is determined (and has usually exhausted alphanumeric character set brute forces) that you'll start to see other characters being added to the brute force. The great thing of course is that the attacker has to pretty much start from scratch with another set of permutations for each additional character.

patio11 16y ago

True, but increasing the raw length of the password (which passphrases will give you in spades) makes those leetification tricks combinatorially explosive. "apple" has 3 characters (2^3 = 8 variations) for easy leetspeak. "apples" has 2^4 = 16. "applestastegreat" has 2^12 = 4,096. Assuming you don't totally fubar your entropy in the underlying passphrase, you'll end up with something both memorable and infeasible to brute force.

ominous_prime 16y ago
Oh yes, and I do use this in my own passphrases for the added combinational complexity. I just wanted to point out to the naive that using `passw0rd` or `s3cr3t`, they are in password lists that a brute force attack will use. My point that crackers will programaticaly try this was less relevant.
- nostrademons 16y ago
  You can choose to l33tify different combinations of letters in the password. passw0rd may be in the dictionary, but is p4ssword, or pa$$word? Patrick's point is that there are combinatorially many different ways that you can 133t!fy a password, so they don't really help in dictionary attacks.
  - JoeAltmaier 16y ago
    I know I am annoyed by "keyspace rules" that actually reduce the keyspace (rant). So, I capitalize the 1 letter and put 0 or 1 at the end - voila! IT appeased. Am I the only one? No. So the 'keyspace rules' don't accomplish anything, and may actually reduce the keyspace.
  - redthrowaway 16y ago
    >passw0rd may be in the dictionary, but is p4ssword, or pa$$word? I'm guessing yes. If I were building a dictionary, I would add every password that's been dumped from sites like Gawker, etc. With hundreds of thousands of passwords floating around in the ether, you can quite quickly cover common leetifications. Also, it's trivially easy to leetify algorithmically, so padding a dictionary wouldn't be difficult. You'll run into space issues eventually, but you can easily split the dictionary over several machines. The best solution for the user is to pick a fairly complex, long, and unique password. Server-side, assume your database will be compromised use something like bcrypt to make cracking it computationally expensive.
  - tripzilch 16y ago
    > You can choose to l33tify different combinations of letters in the password. passw0rd may be in the dictionary, but is p4ssword, or pa$$word? yes they are. well, actually, password cracking software (most used is Jack the Ripper) comes with a bunch of commandline tools that allow you to do substitutions and common variations to your password lists. generally, when trying to crack a specific target, you start with a few "seed" password lists (common passwords, but also lists of target-specific jargon), and then apply the transformation tools on those to expand your lists. say you'd want to crack accounts at an aviator forum, you'd include a list of airplane types and aviation jargon. then you'd apply the transformation tools, and you'll end up with boeing Boeing b0eing Bo3!ng, etc. check this guy's blog, he writes about how it's (usually) done: http://www.skullsecurity.org/blog/2010/the-ultimate-faceoff-between-password-lists http://www.skullsecurity.org/blog/2010/the-ultimate-faceoff-... (analysis of leaked password lists) http://www.skullsecurity.org/blog/?s=ripper http://www.skullsecurity.org/blog/?s=ripper (describes Jack the Ripper in these articles)

iuguy 16y ago

If you want to cheat at complexity you can look at the most common mechanisms for brute force and their defaults. For example, the publicly available rainbow tables are mainly built for US character sets. If you're using characters that aren't in those character sets the rainbow table lookup will fail (as the hash wasn't precomputed). Likewise, if dictionaries (and the permutation rules) lack a non-standard character, using that character will defeat the dictionary attack. If you use a straight brute force then there's a trade off between the characters you use and the amount of time it'll take to exhaust a particular set of characters. It's only if the attacker is determined (and has usually exhausted alphanumeric character set brute forces) that you'll start to see other characters being added to the brute force. The great thing of course is that the attacker has to pretty much start from scratch with another set of permutations for each additional character.