In AES, to make the s-box, we apply the transformation – b’i = bi XOR b(i+4) XOR b(i+5) XOR b(i+6) XOR b(i+7) XOR ci What is ci in this transformation?

In AES, to make the s-box, we apply the transformation – b’i = bi XOR b(i+4) XOR b(i+5) XOR b(i+6) XOR b(i+7) XOR ci What is ci in this transformation? Correct Answer ci is the ith bit of byte c with value 0x63

ci is the ith bit of byte c with value 0x63 i.e, c = 01100011

Related Questions

In AES, to make the s-box, we apply the transformation b’_i = b_i XOR b_(i+4) XOR b(i+5) XOR b_(i+6) XOR b_(i+7) XOR c_i What is c_i in this transformation?
The inverse s-box permutation follows, b’_i = b_(i+2) XOR b(i+5) XOR b_(i+7) XOR d_i Here d_i is
The inverse s-box permutation follows, b’i = b(i+2) XOR b(i+5) XOR b(i+7) XOR di Here di is –
On comparing AES with DES, which of the following functions from DES does not have an equivalent AES function?