4 ms·

I have actually thought about this a lot and done some napkin calculations: using a set of about 93 characters the search space for 5 character passwords is abo

I have actually thought about this a lot and done some napkin calculations: using a set of about 93 characters the search space for 5 character passwords is about 7 gigabytes; combining all the possible passwords <= length 4 is only about 74 megabytes in comparison. I think that the entropy loss is insignificant.

FabHK 7y ago
Indeed, and the factor is about 95, which is close to 93, and that is not a coincidence, when you think about it. If you want to crack my password, and I tell you that my password is L characters long, by skipping all passwords of length < L you've only saved about 1/N of the required effort (where N is the size of the character set you choose from). In other words, telling someone the length of your password does not help them very much.
- _hyn3 7y ago
  The impact of this really depends on the cost of calculating hashes, which varies based on hash algorithm. It might even be the difference between being able to ever find the hash or not. Being able to constrain the search space is the ONLY tool you can count on. If you say "No passwords less than eight characters", most passwords will be clustered around 9 or 10 characters. That's human nature. This means that your entire search space is only the character set in that very limited set of strings. And, if you're limiting that character set, and also it's expensive for you to hash (ie scrypt or bcrypt with lots of rounds), then not having to go through the first 1-8 set will save you a tremendous amount of time...
  - FabHK 7y ago
    > then not having to go through the first 1-8 set will save you a tremendous amount of time... Well, not quite. If you allow say capital letters and numbers only (36 different characters), and you have to go through all 9 and 10 character passwords, skipping the first set of 1 to 8 character passwords saves you less than 0.1% of the work, if my rough back-of-the-envelope calculation is correct. That was precisely what my statement was about - longer passwords are just so much better. (Note that this assumes that the cost of computing the hash is by and large independent of the length of the password, which is a very good assumption for the lengths we are talking about here).

FabHK 7y ago

Indeed, and the factor is about 95, which is close to 93, and that is not a coincidence, when you think about it. If you want to crack my password, and I tell you that my password is L characters long, by skipping all passwords of length < L you've only saved about 1/N of the required effort (where N is the size of the character set you choose from). In other words, telling someone the length of your password does not help them very much.

_hyn3 7y ago
The impact of this really depends on the cost of calculating hashes, which varies based on hash algorithm. It might even be the difference between being able to ever find the hash or not. Being able to constrain the search space is the ONLY tool you can count on. If you say "No passwords less than eight characters", most passwords will be clustered around 9 or 10 characters. That's human nature. This means that your entire search space is only the character set in that very limited set of strings. And, if you're limiting that character set, and also it's expensive for you to hash (ie scrypt or bcrypt with lots of rounds), then not having to go through the first 1-8 set will save you a tremendous amount of time...
- FabHK 7y ago
  > then not having to go through the first 1-8 set will save you a tremendous amount of time... Well, not quite. If you allow say capital letters and numbers only (36 different characters), and you have to go through all 9 and 10 character passwords, skipping the first set of 1 to 8 character passwords saves you less than 0.1% of the work, if my rough back-of-the-envelope calculation is correct. That was precisely what my statement was about - longer passwords are just so much better. (Note that this assumes that the cost of computing the hash is by and large independent of the length of the password, which is a very good assumption for the lengths we are talking about here).