3 ms·

Nope. No dictionary words is a non-starter. If I tried to enforce this as policy I'd get zero compliance.

rmrfstar 6y ago
It's a good point. One thing that would have been nice to see is a graph with several lines. X-axis is bits of entropy, Y-axis is recall rate. The individual lines reflect a single combination of (a) formatting scheme (e.g. diceware) and (b) reinforcement mechanism (e.g. type 20x once per day). I've run this experiment on myself and found that diceware with intra-word punctuation gives the highest entropy at the lowest effort.
stordoff 6y ago
Every password I need to remember has a dictionary word in it, because it's easy _additional_ entropy. Once I've got a password that's borderline too long to remember, I can't really add more random characters to it without risking forgetting it. Concatenating it with a dictionary word ("password-dictword") is much easier to remember, but (I believe) still adds a meaningful amount of security.
- rmrfstar 6y ago
  It removes you from the list of people who get pwned with a rainbow table from leaked passwords, but that's not much security. English has only got ~1.2 bits of entropy per character. [1] [1] Table 1 in https://arxiv.org/pdf/1808.04444.pdf https://arxiv.org/pdf/1808.04444.pdf
  - ravi-delia 6y ago
    I was going to snarkily make a point about that figure only making sense in the context of grammatical english text, with the example of a randomly chosen 7 letter word being one of 32,909 options. log(32,909) is about 15. Which is of course 1.5 bits per character, and that's generously assuming that a person will actually randomly pick from all those potential options. So sorry for what I almost commented, but now I have to wonder why the entropy of a word on its own is so close to the entropy of a word in a sentence. Am I missing something? Is it just coincidence?
    - deleted 6y ago
      [deleted]
  - schoen 6y ago
    But entropy is relative to an attacker (it measures the attacker's uncertainty about your password). The 1.2 bits/character measurement is about grammatical English text that occurs in a communicative setting. That means that before you get there, you probably already know what word goes at the end of this sentence. You can measure this with human native speakers, or with a compression algorithm, or with a naive letter n-gram or word n-gram model, and get different statistics -- each of which is correct in the sense of each model's uncertainty about how the text continues. In the password setting, words can be used in all sorts of ways, some of which are very unpredictable to an attacker, so it's plausible that each word adds at least log₂(len(dictionary)) to the password entropy when it's chosen at random from a dictionary, even when the attacker expects that the password selection method could have used that dictionary.
    - deleted 6y ago
      [deleted]

It's a good point. One thing that would have been nice to see is a graph with several lines. X-axis is bits of entropy, Y-axis is recall rate. The individual lines reflect a single combination of (a) formatting scheme (e.g. diceware) and (b) reinforcement mechanism (e.g. type 20x once per day). I've run this experiment on myself and found that diceware with intra-word punctuation gives the highest entropy at the lowest effort.

stordoff 6y ago

Every password I need to remember has a dictionary word in it, because it's easy _additional_ entropy. Once I've got a password that's borderline too long to remember, I can't really add more random characters to it without risking forgetting it. Concatenating it with a dictionary word ("password-dictword") is much easier to remember, but (I believe) still adds a meaningful amount of security.

rmrfstar 6y ago
It removes you from the list of people who get pwned with a rainbow table from leaked passwords, but that's not much security. English has only got ~1.2 bits of entropy per character. [1] [1] Table 1 in https://arxiv.org/pdf/1808.04444.pdf https://arxiv.org/pdf/1808.04444.pdf
- ravi-delia 6y ago
  I was going to snarkily make a point about that figure only making sense in the context of grammatical english text, with the example of a randomly chosen 7 letter word being one of 32,909 options. log(32,909) is about 15. Which is of course 1.5 bits per character, and that's generously assuming that a person will actually randomly pick from all those potential options. So sorry for what I almost commented, but now I have to wonder why the entropy of a word on its own is so close to the entropy of a word in a sentence. Am I missing something? Is it just coincidence?
  - deleted 6y ago
    [deleted]
- schoen 6y ago
  But entropy is relative to an attacker (it measures the attacker's uncertainty about your password). The 1.2 bits/character measurement is about grammatical English text that occurs in a communicative setting. That means that before you get there, you probably already know what word goes at the end of this sentence. You can measure this with human native speakers, or with a compression algorithm, or with a naive letter n-gram or word n-gram model, and get different statistics -- each of which is correct in the sense of each model's uncertainty about how the text continues. In the password setting, words can be used in all sorts of ways, some of which are very unpredictable to an attacker, so it's plausible that each word adds at least log₂(len(dictionary)) to the password entropy when it's chosen at random from a dictionary, even when the attacker expects that the password selection method could have used that dictionary.
  - deleted 6y ago
    [deleted]