Timeline for Encrypted text - C# SHA256, implementation
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2021 at 7:48 | comment | added | RemarkLima | Is that a reference to the knights who say... Ni! | |
| Apr 12, 2021 at 21:11 | comment | added | Maarten Bodewes | Unrelated, I seem to have a problem with a word today, unable to spell "its". Ni! | |
| Apr 12, 2021 at 14:08 | comment | added | Maarten Bodewes | @RemarkLima No, the calculations are over the number of digits. So if e+16 then you have slightly over 16 digits, then (16 / 3) x 10 = 53 bits and cryptographers do use caution so we guess to the low end. KeePass doesn't know that they are diceware rolls so it's guess is way too high, this kind of knowledge should be included. A key size of 32 bytes is 256 bit, which gives you enough protection even against quantum computers. And as you may notice, the 51 bit password is what you need to protect - not the 256 bit key. | |
| Apr 12, 2021 at 13:41 | comment | added | RemarkLima | One more thing - I've had a look at the AES-NI in C# and it only supports smaller key sizes, and the key and IV is only 32 and 16 bytes. It "feels" safer to use the Bouncy Castle with the full width key and IV...?? | |
| Apr 12, 2021 at 10:28 | vote | accept | RemarkLima | ||
| Apr 12, 2021 at 10:28 | comment | added | RemarkLima | Hi @maarten-bodewes, I've put a few 4 word passwords into KeePass and always get over 110 bits of entropy? Not sure how to get an absolute on this | |
| Apr 10, 2021 at 20:15 | comment | added | RemarkLima | From 4 rolls, and your sums I get: (3,656,158,440,062,976 / 3) * 10 = 1.2187195e+16 ... Is that right? 4 rolls gives over 3.5 trillion (??? Quadrillion??) combinations right? Or are my maths way off? On my phone, hence a bit gammy with the numbers! | |
| Apr 10, 2021 at 19:58 | comment | added | Maarten Bodewes | Heh, no, it is the total of 4 rolls. I simply asked for a 4 word roll on the site, got the answer back + the number of possibilities. Then it is simply a question of taking the $\log_2$ of that number. Or counting the number of digits, and divide by 3 and multiply by 10 to get an approximation (and you're welcome :) ). | |
| Apr 10, 2021 at 19:53 | comment | added | RemarkLima |
Quick question before I digest the rest of the review (thanks BTW) - the 51 bits for a diceware, is that 4 times diceware? i.e. fish-sock-cheese-river as 4 separate rolls to create the 4 words? Or is it 51 bits of entropy per roll? And so 51 bits of entropy multiplied by 4?
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| Apr 10, 2021 at 15:42 | history | answered | Maarten Bodewes | CC BY-SA 4.0 |