# Tag Info

3

Anon2000 - as currently constructed your mode is fatally flawed. Given two known messages encrypted with the same key (i.e. where the attacker knows the plaintexts) that are each at least two blocks in length (not counting the IV or final validation block), the attacker can trivially forge at least two other 'valid' messages (and many more than that if the ...

2

This does not meaningfully authenticate the ciphertext. Your encryption is the same as OFB, meaning no block depends on the previous plaintext; for instance, $$C_2=P_2\oplus E(P_1\oplus (P_1\oplus E(IV\oplus 0)))=P_2\oplus E(E(IV)).$$ That means confidentiality should be fine; however, it provides no authentication except of the length of the message.

1

Consider what would happen if an attacker altered one of the ciphertext blocks, but kept the IV and all previous ciphertext blocks identical. i.e. Xor some difference $\Delta$ to ciphertext block $C_x$, so that the altered block $C''_x = C_x \oplus \Delta$. This will produce an altered corresponding Plaintext block, like so: P''_x = C''_x \oplus ...

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