Symmetry

An embarrassing challenge for me. The solution to this challenge is quite simple. The symmetry of the xor operation enables us to decrypt the message/plaintext PP from the IVIV and ciphertext CC. Denote the output after running block cipher encryption with key k EkE_k to be OO, then we have: O=Ek(IV) O = E_k(IV) C=PO C = P \oplus O xor both sides of the above equation, we have:...

December 6, 2022 · 2 min · qvinhprolol