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Apr 30, 2020 at 12:04 comment added Graham Hill @Freiheit the paper was at research.microsoft.com/apps/pubs/default.aspx?id=227130
Apr 29, 2020 at 10:44 comment added gbjbaanb @Scz yes... i even wrote the words out right and got the numbers wrong, how daft!
Apr 28, 2020 at 21:54 comment added Scz @gbjbaanb shouldn't it be 52^30 ≙ 171 bit vs. 8000^5 ≙ 65 bit?
Apr 28, 2020 at 18:13 comment added gbjbaanb @JorgeLeitao calculate the entropy of 30^52 (assuming upper and lower case) to 5^8000 (assuming 8000 words in a dictionary). The number of combinations are set of symbols ^ number of them which works out as:3e+51 v 2e+19. Entropy is log2 of these values.
Apr 28, 2020 at 3:36 comment added Jorge Leitao Does 5 random words have the equivalent entropy of 30 random characters?
Apr 27, 2020 at 17:06 comment added Freiheit This article is also in a similar vein. techcommunity.microsoft.com/t5/azure-active-directory-identity/… .
Apr 27, 2020 at 17:05 comment added Freiheit I tried to locate the paper, I recall hearing similar advice from MS or some other authoritative source. I think this might be it but MS has published quite a few papers on password strength and mangement - microsoft.com/en-us/research/wp-content/uploads/2016/02/…
Apr 27, 2020 at 15:19 history answered Graham Hill CC BY-SA 4.0