Timeline for What is the entropy of a password made from the initial letters from an English sentence
Current License: CC BY-SA 4.0
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| Nov 17, 2021 at 1:47 | comment | added | Adam Katz | That log(13**n) is an estimate given the fact that the top ten letters represent 66.5% of words and the top half of letters represent 70.4%. You're also still assuming there is no relationship between the words that the letters represent. If the words are fully random, then you could indeed eliminate that variable with more words, but why not just have the whole words present? Yes, this is a scheme people use and experts used to suggest. It is not strong. | |
| Nov 17, 2021 at 0:18 | comment | added | hugomg | Of course this is not supposed to replace a password manager, which can store a long 100% random password. The point of the question is, how long does the sentence needs to be to have enough entropy? Is that log(13) an actual estimate or just a guess? Even if it is a lower amount per letter, that would just mean you'd need more characters. And on a character-by-character basis, I'd expect it would be higher entropy than the wordlist-based password... Not to mention, this is an actual password generation scheme that some people use. I think it is valueble to be able to estimate the entropy. | |
| Nov 16, 2021 at 23:57 | history | answered | Adam Katz | CC BY-SA 4.0 |