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1

During CBC mode decryption, each ciphertext block is first decrypted with the block cipher, and then the first decrypted block is XORed with the IV, while later blocks are XORed with the previous ciphertext block: Thus, the first block of plaintext is computed as: $$P_1 = IV \oplus D_K(C_1).$$ Equivalently, we may solve this equation for $IV$ to obtain: ...


-2

This is NOT secure. I have found the weakness with mathematics. So this is our two messages: Sending the first message: $character_1 ⊕ key_1 ⊕ IV_1 = cipher_1$ $character_2 ⊕ key_2 ⊕ IV_1 = cipher_2$ Sending the second message: $character_3 ⊕ key_1 ⊕ IV_2 = cipher_3$ $character_4 ⊕ key_2 ⊕ IV_2 = cipher_4$ Now see what happens if we XOR all of ...


5

This is not a mathematical proof. A notable place it fails to be a proof is here: Pay attention to which cipher text I use, look up to match the message with the cipher-number below. $$ cipher1⊕cipher3=character1⊕character3⊕IV1⊕IV2$$ (Note that the cipher BOTH use the SAME KEY, but they remain secure because of the two different IV) This line is ...


3

Simply put: No. First recall that this is a mis-use of the term "One Time Pad" So lets call it a vigenere cipher instead. You can determine this is insecure with a simple algebraic combination: $ \text{attack} = cipher_1 + cipher_2 + cipher_3 + cipher_4 \\ \text{Simplify: } \\ \text{attack} = character_1 + key_1 + IV_1 + character_2 + key_2 + IV_1 + ...


2

You can use your HardwareID as basis for the encryption key. If the ID provides enough entropy it'll work. However, if anyone can somehow obtain the ID (which might be quite easy to do) one can decrypt the file. For CFB-Mode the IV must indeed be unpredictable (but need not be secret), so random is just fine, but DO NOT REUSE AN IV. Encryption large ...



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