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9

You would not just need a mode of operation for what you're asking. What you need is a secure transport protocol. Probably the best well known one for TCP connections is TLS of course. For UDP connections you could use DTLS. If you have a shared key you could use one of the pre-shared key (PSK) variants. If you want to create your own transport protocol you ...


9

I would pick e) none of the above. None of those modes offers integrity protection, so unless integrity is handled elsewhere, your application is wildly insecure. An attacker could modify bits in transit and do nefarious things. Of the three, CFB and CTR are the worst for the application and should be very easy for an attacker to mount successful attacks, ...


6

the key and plaintext is the same. The attacker knows this and the IVs used but doesn't know the plaintext. Is there anything to learn about the plaintext when multiple ciphertexts are available instead of only one? No. Giving the attacker multiple encryptions of a single plaintext using a randomly chosen IV and a fixed key with AES-CBC does not leak ...


3

If you have a nonrepeating (but possibly predictable) value, you can convert that into an unpredictable CBC-mode IV at fairly minimal cost. Here's how: Prepend the 128 bit nonrepeating value to the message CBC mode encrypt the (value, message), using any IV that's not correlated to the nonrepeating value (all 0's work) Use the first 16 bytes of the ...


2

If you use a random IV each time you encrypt a file, the result will be different. $$\text{AES-CBC}(IV_1, Key, M) = C_1$$ $$\text{AES-CBC}(IV_2, Key, M) = C_2$$ you do not gain any pieces of informations by having $C_1$ and $C_2$. Even knowing that they share the same key and should they share the same length, an attacker can not even know whether or not ...


2

Scheme is not IND-CPA for any message longer than one block. I'll include a image of CBC mode below for reference (Source: Wikipedia). Suppose instead of block cipher encryption we have plaintext xor-ed with the key as you propose. You'll note that for message block 1, $M_1$, the ciphertext block $C_1 = M_1 \oplus IV \oplus Key$. Similarly $C_2 = M_2 ...


1

It depends on the position of the two interchanged bits. Assume they have different values, i.e. one is $0$ and the other is $1$. Indeed, if they had the same value, interchanging would not affect plaintext decryption. Let's say $n$ is the length of every block. In CBC, every ciphertext block is involved in two plaintext blocks decryption: its own and ...



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