Re: Salt in encrypted password in pg_shadow
От | Steve Atkins |
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Тема | Re: Salt in encrypted password in pg_shadow |
Дата | |
Msg-id | 20040907235424.GA18866@gp.word-to-the-wise.com обсуждение исходный текст |
Ответ на | Re: Salt in encrypted password in pg_shadow (Tom Lane <tgl@sss.pgh.pa.us>) |
Ответы |
Re: Salt in encrypted password in pg_shadow
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Список | pgsql-general |
On Tue, Sep 07, 2004 at 03:09:28PM -0400, Tom Lane wrote: > David Garamond <lists@zara.6.isreserved.com> writes: > > Tom Lane wrote: > >> Also, MD5 hashing is fast enough that I'm not sure the above is really > >> significantly cheaper than a straight brute-force attack, ie, you just > >> take your list of possible passwords and compute the hashes on the fly. > >> The hashes are going to be much longer than the average real-world > >> password, so reading in a list of hashes is going to take several times > >> as much I/O as reading the passwords --- seems to me that it'd be > >> cheaper just to re-hash each password. > > > Many people use short and easy-to-guess passwords (remember we're not > > talking about the superuser only here), so the dictionary attack can be > > more effective than people think. > > And that responds to the speed argument how? I quite agree that a > guessable password is risky, but putting in a random salt offers no > real advantage if the salt has to be stored in the same place as the > encrypted password. The usual attack against hashed passwords is to use a dictionary driven password generator to create a large number of passwords, find the hash of each of those, then store the passwords on disk indexed by the hash. That's a one-time effort that can then be used in the future to crack any number of password hashes extremely cheaply - given any hash you can find the corresponding password, if you have one, with one index lookup. A random salt stored with the hashed password increases the storage and precomputation time required by the size of the salt (so a 16 bit salt would increase the storage and precomputation time needed by a factor of 65536). That increase makes the pre-computed dictionary attack pretty much infeasible. Cheers, Steve
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