5 ms·

> he can trivially test combinations of dictionary words in very short amount of time. Explain the reasoning behind this, please. Start with: You don't know t

> he can trivially test combinations of dictionary words in very short amount of time. Explain the reasoning behind this, please. Start with: You don't know the dictionary I used, but have to use one that seems 'good enough' (i.e. a superset of mine, if possible). How many words are in there? How many combinations can you create for 'two word phrases'? (You don't know the length of my phrase) How many for three? How many for four words?

hesselink 15y ago
The XKCD criticism of the 'bad' password has the same problems. How does the hacker know I have this kind of password (starting with a real english word, for example)?
- darklajid 15y ago
  No, it doesn't have the same problem. It criticizes a particular way to choose passwords, leading to a result that seems 'secure' and even quite good to lots of people. One that easily satisfies braindead corporate password rules. The entropy given there is based on that way to choose a password and even explained, graphically. If you choose your password totally different and from different sets of characters, then the number will be off. That's not a surprise though?
burgerbrain 15y ago
In fact, even if the attacker: 1) Knows this scheme was used. 2) Knows the list of words you used. 3) Knows how many words you used. You will still have that amount of entropy. Diceware entropy values are calculated with all three of these assumptions already made.
ErrantX 15y ago
> Start with: You don't know the dictionary I used, but have to use one that seems 'good enough' (i.e. a superset of mine, if possible). People are likely to use a standard English dictionary. In my experience (which is exactly within this field) people use a fairly tight subset of the English vocabulary. So I would be quite happy to test for a dictionary of, say, 100,000 words and be hopeful of a good hit rate (note that XKCD says common words, which is easily missed) Our software has a test (which runs about third in its list of tests) which does dictionary combinations up to three words (two words is quite a commonly used password based on our statistics) with a dictionary size of <s>175,000</s>17,500. (Edit: sorry, apparently it is an order of magnitude smaller, I checked with one of the engineers :)) This includes English words plus a few commonly used foreign/slang terms. The hit rate on this is fairly high. (we crack document/windows passwords mainly) You could of course choose deliberately obscure words to invalidate this - but they aren't so easy to remember (so people will tend not to). If someone is going out of their way to secure a password, sure, you're going to hit a brick wall. But what every password scheme tends to forget is the "human factor" whereby people not concentrating on being secure will introduce attack vectors.
- darklajid 15y ago
  I actually didn't ignore the 'common' limitation (and didn't downvote you - I'm actually interested how you come up with that). Follow-up questions: - What are the first tests, before this 3rd that tests for words? I assume tests for passwords of the first/left variety in the comic? Aren't they cheaper? - 'Up to three words' is reducing the exponent of possible combinations by one. Length/number of words is relevant Edit: Another issue. You say 'people forget the human factor', while you, yourself, propose something like '4 times Hack News with substitutions' as better. How is that including the 'human factor'?
  - ErrantX 15y ago
    <scrubbed answer> You know what; it's been so long since I played around with this stuff (it's even a separate company now, that we just consult for) that I'm way out of touch with my thought process :) You're right; there is nothing particularly wrong with the suggestion that makes it intrinsically very weak for most uses. I'd best stop commenting before I make a total mess :) Sorry.
  - deleted 15y ago
    [deleted]
- d0mine 15y ago
  100,000 words x 4 words per password ~ 66 bits of entropy XKCD suggests 2048 words x 4 words per password ~ 44 bits of entropy It could be good enough depending on a use case.
- demallien 15y ago
  If the dictionary really has 100 000 words, you're looking down the barrel of 52 bits of entropy for a three word phrase In a more likely dictionary of the 5000 most commonly used words in the English language, you still get a three word pass phrase of about 40bits of entropy. Make that a four word passphrase, and you're back up around 52 bits.
  - xkcdentropy 15y ago

hesselink 15y ago

The XKCD criticism of the 'bad' password has the same problems. How does the hacker know I have this kind of password (starting with a real english word, for example)?

darklajid 15y ago
No, it doesn't have the same problem. It criticizes a particular way to choose passwords, leading to a result that seems 'secure' and even quite good to lots of people. One that easily satisfies braindead corporate password rules. The entropy given there is based on that way to choose a password and even explained, graphically. If you choose your password totally different and from different sets of characters, then the number will be off. That's not a surprise though?

burgerbrain 15y ago

In fact, even if the attacker: 1) Knows this scheme was used. 2) Knows the list of words you used. 3) Knows how many words you used. You will still have that amount of entropy. Diceware entropy values are calculated with all three of these assumptions already made.

ErrantX 15y ago

> Start with: You don't know the dictionary I used, but have to use one that seems 'good enough' (i.e. a superset of mine, if possible). People are likely to use a standard English dictionary. In my experience (which is exactly within this field) people use a fairly tight subset of the English vocabulary. So I would be quite happy to test for a dictionary of, say, 100,000 words and be hopeful of a good hit rate (note that XKCD says common words, which is easily missed) Our software has a test (which runs about third in its list of tests) which does dictionary combinations up to three words (two words is quite a commonly used password based on our statistics) with a dictionary size of <s>175,000</s>17,500. (Edit: sorry, apparently it is an order of magnitude smaller, I checked with one of the engineers :)) This includes English words plus a few commonly used foreign/slang terms. The hit rate on this is fairly high. (we crack document/windows passwords mainly) You could of course choose deliberately obscure words to invalidate this - but they aren't so easy to remember (so people will tend not to). If someone is going out of their way to secure a password, sure, you're going to hit a brick wall. But what every password scheme tends to forget is the "human factor" whereby people not concentrating on being secure will introduce attack vectors.

darklajid 15y ago
I actually didn't ignore the 'common' limitation (and didn't downvote you - I'm actually interested how you come up with that). Follow-up questions: - What are the first tests, before this 3rd that tests for words? I assume tests for passwords of the first/left variety in the comic? Aren't they cheaper? - 'Up to three words' is reducing the exponent of possible combinations by one. Length/number of words is relevant Edit: Another issue. You say 'people forget the human factor', while you, yourself, propose something like '4 times Hack News with substitutions' as better. How is that including the 'human factor'?
- ErrantX 15y ago
  <scrubbed answer> You know what; it's been so long since I played around with this stuff (it's even a separate company now, that we just consult for) that I'm way out of touch with my thought process :) You're right; there is nothing particularly wrong with the suggestion that makes it intrinsically very weak for most uses. I'd best stop commenting before I make a total mess :) Sorry.
- deleted 15y ago
  [deleted]
d0mine 15y ago
100,000 words x 4 words per password ~ 66 bits of entropy XKCD suggests 2048 words x 4 words per password ~ 44 bits of entropy It could be good enough depending on a use case.
demallien 15y ago
If the dictionary really has 100 000 words, you're looking down the barrel of 52 bits of entropy for a three word phrase In a more likely dictionary of the 5000 most commonly used words in the English language, you still get a three word pass phrase of about 40bits of entropy. Make that a four word passphrase, and you're back up around 52 bits.
- xkcdentropy 15y ago