Timeline for How to solve MixColumns
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Oct 23, 2021 at 23:38 | history | suggested | Matthias Braun | CC BY-SA 4.0 | add link, try to make "exclusive or" more succinct and readable by replacing it with "XOR" |
| Oct 21, 2021 at 13:14 | review | Suggested edits | |||
| S Oct 23, 2021 at 23:38 | |||||
| Mar 30, 2020 at 3:31 | comment | added | poncho♦ | @ShantanuShinde: the most obvious way to compute $4 \times X$ is $4 \times X = 2 \times (2 \times X)$, that is, you apply the above double step twice. Or, am I missing something. | |
| Mar 30, 2020 at 1:56 | comment | added | Shantanu Shinde | how would you multiply 4 with X then? 4 = 2xor6 so (2 xor 6) x X. then what? | |
| Oct 22, 2017 at 12:40 | comment | added | poncho♦ | @LoicVerrall: xor is both associative and commutative, and so you can use whatever order is convenient. However, if you decide to xor all 4 together at once, you don't "output '1' if there is 1 in all 4 inputs", instead, it's "output 1 if there is a 1 in an odd number of the numbers" | |
| Oct 22, 2017 at 11:16 | comment | added | Loic Verrall | How do you XOR the 4 values at the end together? Do you XOR the first pair, then the result with the third input, and that result XOR the fourth input? Or do you do them all in one go (i.e. output '1' if there is literally one '1' in all 4 of the inputs)? | |
| Aug 9, 2012 at 17:57 | vote | accept | goldroger | ||
| Apr 20, 2012 at 20:24 | history | edited | Paŭlo Ebermann | CC BY-SA 3.0 | use mathtt formatting like in the question for better legibility. |
| Apr 20, 2012 at 1:58 | history | edited | poncho♦ | CC BY-SA 3.0 | Gave explicit evaluation of the example the submitter gave |
| Apr 19, 2012 at 19:32 | history | edited | poncho♦ | CC BY-SA 3.0 | Extended the answer to cover the next thing the OP is likely to stumble against |
| Apr 19, 2012 at 18:15 | history | answered | poncho♦ | CC BY-SA 3.0 |