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Your teacher is right, and here's why: What happens if you encrypt A with G and B with G? You can't decipher it, because you have no idea if the G in the ciphertext was an A or a B. So… For the plaintext letter A you can use the ciphertext letter A, B, C, … , X, Y, or Z. ($26$ possible letters.) For B you can use A, B, C, … , X, Y, or Z, but not the ...


Yes, you can do that. Start by generating a random key and encrypting the data under that key. Now, you in turn encrypt that key with your password-derived key. The twist is that you include the already encrypted file in your PBKDF. So you generate a random key $k$ and then encapsulate it as $\mathcal{E}_{k'}(k)$, where $k' = H(PBKDF(\text{password}) \| ...

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