2 ms·
Really the key is using words that aren't in a dictionary, or at least not the same dictionary. It's pretty trivial for a brute force attack to just take a list
by jjk166 5y ago
Really the key is using words that aren't in a dictionary, or at least not the same dictionary. It's pretty trivial for a brute force attack to just take a list of English words and try them in a large number of combinations. But if you have some non-english words, as far as that attacker is concerned its random gibberish. Of course the attacker might use a few languages so just mixing say english and spanish won't do that much better, but if you got some navajo or some quenya you're likely immune to any attack that isn't specifically targeted at you. And it doesn't even have to be a foreign language, even many proper nouns won't show up in a dictionary. For a password like "PoggleExclaimsGeonosianBarahunde" or "Seanbeanfearsburzum-ishikrimpatul" it would take trillions of years to guess a password without very specific pop culture knowledge.
- vbrandl 5y agoHere is a small back of the envelope calculation. Websters dictionary includes 470.000 words [0]. 52 lower and uppercase letters + 10 digits + ~20 special chars = 82 possibilities. There are are 82^20 = 1.89 * 10^38 possible combinations for passwords consisting of 20 random characters. Picking 7 random words from Websters dictionary has 470.000^7 = 5.07 * 10^39 possible combinations. I'd argue that remembering 7 words is easier for most people, than remembering 20 random characters. [0]: https://www.merriam-webster.com/help/faq-how-many-english-words https://www.merriam-webster.com/help/faq-how-many-english-wo...
- jjk166 5y agoYeah, for a given level of entropy words are a better choice, but the fact is that dictionary attacks greatly reduce the efficacy of what could be an incredibly strong password if they can safely assume it is formed exclusively from concoctenated standard words. However if you break the validity of that assumption by using a word that's not on a list, then the dictionary attack must be combined with a brute force attack with the same string length. If we assume an average word length of 5 letters, that 7 word password jumps to 1.22x10^79 possible combinations if you randomly substitute one letter.