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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?

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? Correct Answer c_i 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 = 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?
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