this post was submitted on 19 Jun 2024
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This is really good for encrypted content such as a hard drive, password vault or gpg key

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[–] catalog3115@lemmy.world -1 points 5 months ago (8 children)

Doesn't using a particular wordlist limits choice, gives attackers a wordlist to generate the password from

[–] MalReynolds@slrpnk.net 13 points 5 months ago (1 children)

Nope, As long as you're not as uncreative as to use Correct Horse Battery Staple.

[–] narc0tic_bird@lemm.ee 2 points 5 months ago (1 children)

Well it still technically does give them a list, it's just that even with the list there are a lot of combinations.

[–] laughterlaughter@lemmy.world 6 points 5 months ago

Technically, yes. But the article already mentioned the amount of effort for the brute force to succeed (that is, practically never, if the phrase is truly random.)

But anyway. With regular passwords, the attackers already have a list: the alphabet plus numbers and symbols. Not really that different.

[–] MajorSauce@sh.itjust.works 5 points 5 months ago

Even a "traditional" password would have a "list" that attackers could know (all the possible characters that can be used in a password), now compare this set of +-150 characters with the set of possible words that can be used (probably close to 250k per language if you take out some similarities).

Even with only 4 words, the number of possibilities is astounding.

[–] kevincox@lemmy.ml 3 points 5 months ago

Technically yes. But the method is by far strong enough that this isn't an issue. This is sort of always the issue with calculating entropy. We say that password has less entropy than 8(A>Ni'[. But that is baking in assumptions about the search space. If password is a randomly generated string of lower, upper, numbers and symbols it is just as secure as the latter. (80^8^ ≈ 10^15^ candidates) but if password was generated as just lowercase characters it is far less secure (26^8^ ≈ 10^11^ candidates) but if it was a random dictionary word it is not very secure at all (≈ 10^5^ candidates) and if it was chosen as one of the most popular passwords it is even less secure. How can one password have different entropy?

The answer is basically it matters how the attacker searches. But in practice the attacker will search the more likely smaller sets first, then expand to the larger. So the added time to search the smaller sets is effectively negligible.

What may be more useful is the "worst case" entropy. Basically the attacker knows exactly what set you picked. For the password case that is 1 because I just picked the most common password. For the rolling method described above it is 6^5^6^ ≈ 10^23^ because even if they know the word list they don't know the rolls. You may be able to go slightly higher by building your own word list, but the gains will probably be fairly small and you would likely get far more value just by rolling one more word on the existing list than spending the time to generate your own.

[–] possiblylinux127@lemmy.zip 3 points 5 months ago* (last edited 5 months ago)

1200*1200*1200*1200*1200*1200 is a lot of combinations

[–] ShortN0te@lemmy.ml 2 points 5 months ago

Yes, but when the list is long enough and you have enough words, it is to difficult to guess.

Think about it. The list of all possible characters is also known, still with enough length and randomness it becomes too difficult to guess too.

[–] AHemlocksLie@lemmy.zip 2 points 5 months ago

It does limit choice, but so long as you aren't retaining the generation list somewhere an attacker can find it, how are they to know your list? As long as your list is incorporating less common words, your attacker can't even simplify the problem by focusing on the most common words. Just one rare word can expand the list they need to use by tens of thousands of words.

[–] hushable@lemmy.world 1 points 5 months ago

Just mispel a word, replace a vowel with a numb3r and the wordlist is useless