Ya know that’s what they say, but I’m not so sure - is your dictionary-based brute-forcer doing strings of three words together? Allowing for interspersed special characters between? The sheer character length of three truly random dictionary words in a row is already staggeringly high amounts of entropy - I’m not sure I need to be worried about an attacker capable of that kind of sheer number-buggery.
I’m sure they are ever since this comic came out. It is a very large amount of entropy, but it is still far, far less than an equivalent amount of random characters. Honestly, you do you. If passwords are properly hashed and password attempts are not unlimited and as fast as possible, you’re basically fine.
Good info! And that’s exactly what I meant - a word is weak, but several randomized words together is pretty crazy strong. Slightly less than random letters, but much easier to type in memorize when the situation calls for it.
On your bikelock you have a 3 character code with and alphabet of 0-9. So 10^3 = 1000 possible combinations.
If you pick 3 random words out of a dictionary with 40k words, there are 40000^3 possible combinations. (64 000 000 000 000).
Depending on how the password is hashed a 1000$ machine might be able to test anywhere from like 10 to 10 000 000 000 000 hashes per second. (100 billion hashes per second are more realistic)
So a 3 word password might be safe for a very very long time or cracked in seconds.
even 4 words with one common substitution fucks up dictionary attack, cause you expand the search space 3-10 fold per word. although one have to have at least one relatively uncommon word (from 20k vocabulary, not 2k vocabulary)
Weak to dictionary attacks. better make it six words
Ya know that’s what they say, but I’m not so sure - is your dictionary-based brute-forcer doing strings of three words together? Allowing for interspersed special characters between? The sheer character length of three truly random dictionary words in a row is already staggeringly high amounts of entropy - I’m not sure I need to be worried about an attacker capable of that kind of sheer number-buggery.
I’m sure they are ever since this comic came out. It is a very large amount of entropy, but it is still far, far less than an equivalent amount of random characters. Honestly, you do you. If passwords are properly hashed and password attempts are not unlimited and as fast as possible, you’re basically fine.
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Good info! And that’s exactly what I meant - a word is weak, but several randomized words together is pretty crazy strong. Slightly less than random letters, but much easier to type in memorize when the situation calls for it.
If you pick 3 randon words out of 40000 that’s less entropy than a 8 random character password with only letters and numbers.
It’s about combinatorics.
On your bikelock you have a 3 character code with and alphabet of 0-9. So 10^3 = 1000 possible combinations.
If you pick 3 random words out of a dictionary with 40k words, there are 40000^3 possible combinations. (64 000 000 000 000).
Depending on how the password is hashed a 1000$ machine might be able to test anywhere from like 10 to 10 000 000 000 000 hashes per second. (100 billion hashes per second are more realistic)
So a 3 word password might be safe for a very very long time or cracked in seconds.
A 4 word password will take 40000 times as long.
What if you just add a letter after each word. Like correct a horse b battery c staple d.
I imagine single letters are in the dictionary, but that would double the amount of words
or proper nouns that arent in dictionaries
What if you mix languages or do they brute force multiple languages at the same time?
I’ll be real I tried looking up dictionaries used in attacks, ran into the slightest amount of resistance, and immediately stopped caring. Wikipedia has a list of attacks with links if you want to find out yourself
even 4 words with one common substitution fucks up dictionary attack, cause you expand the search space 3-10 fold per word. although one have to have at least one relatively uncommon word (from 20k vocabulary, not 2k vocabulary)
What if I add Tr0ub4dor&3 to my dictionary?