-1
$\begingroup$

I found this binary operation, however It is the first time that I found this operator '.' :

That is the equation:

02 • 63 = 00000010 • 01100011 = 11000110

I don't understand how 0 • 0=1

I would be very grateful if you could help me.

$\endgroup$
  • $\begingroup$ This operator $\cdot$, like multiplication? Check the wiki for how binaray multiplication works. en.wikipedia.org/wiki/Binary_multiplier Though binary multiplication by 2 is simply a left shift. $\endgroup$ – Fleeep Jul 20 '16 at 14:12
  • $\begingroup$ " 00000010 • 01100011 = 11000110 " stands for a variety of operators • . Examples include ordinary multiplication when values are in binary; ordinary multiplication modulo $2^8$; carry-less multiplication; and multiplication in $GF(2^8)$ for any polynomial. We have no way to guess which without context. Regarding " 0 • 0 = 1 ", that seems to be an error. $\endgroup$ – fgrieu Jul 20 '16 at 14:16
  • 2
    $\begingroup$ I'm voting to close this question as off-topic because it has nothing to do with cryptography at all. It is just regular multiplication, where numbers are written in binary. $\endgroup$ – tylo Jul 20 '16 at 14:17
  • $\begingroup$ @tylo, I found this operation in the description of an example of AES.iu.edu.jo/files/FacultyIT/Computer-Science/Courses/… page 7 $\endgroup$ – user6594048 Jul 20 '16 at 14:32
  • 1
    $\begingroup$ I'd guess it's multiplication in the binary finite field GF(2^8). $\endgroup$ – CodesInChaos Jul 20 '16 at 14:48
3
$\begingroup$

This '.' is not 'AND' operation , its modular multiplication in the Galois Field $$GF(2^8)$$ and other operation $$\oplus$$ used in Mix Column Step of AES encrytion is modular addition in the Galois Field $$GF(2^8)$$. Here is a link to youtube channel by Christoff Paar where you can understand well, Watch lecture 7 and 8.

Hope it helps.

| improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.