# Tag Info

3

No, because CFB isn't commutative. You can see this by looking at the decryption of double-CFB encrypted ciphertext. Even assuming a constant IV (so single-use keys), if you decrypt in the wrong order it cannot work, since the ciphertext is used as input into the block cipher and will differ from what was used with that key when encrypting. The exception, ...

3

I can make a few comments regarding points 1 and 3: If you are going to encrypt only one block, your first assumption is not that misled. However, you will almost always need to encrypt a file longer (maybe way longer) than the key length (let's say 128 bits). Without considering encryption modes, that means that for every block of 128 bits, you will ...

3

This question is based on opinion. At least kind-of. But the variants from which one can choose are quite a few. As for general construction the sponge construction (like Keccak / SHA-3 uses) are very versatile and can be used for many purposes, for example hashing, authenticating (= "MAC'ing"), authenticated encryption (see “General Overview of the ...

2

Regarding points 2 and 3, cipher designers want to ensure that the relationship between the plaintext, the ciphertext, and the key are complex, so that no attacker can efficiently untangle them. If the ciphertext can be expressed as a linear or sufficiently low-degree system of functions of the plaintext and key then attackers can use efficient algebraic ...

2

I will specifically address your question 3; that is, quite a lot of block ciphers (and hash functions) consist of a regular round structure (where you repeatedly do the same thing over and over); why is this? Well, one incentive for doing that is that it makes the cipher easier to analyze; we can study the round function in depth; once we've done that, we ...

2

XTS has been designed for disk encryption, where an attacker typically has access to the disk only a single time (when they steal/confiscate the device). When an attacker sees several ciphertexts encrypted using the same key, they can tell which blocks differ between the versions, but not the content of the blocks. Compare this with CTR mode, which leaks ...

1

How the cipher key is scheduled if it is less than 128 bit (for AES 128), or less than 192 bit (for AES 192) or less than 256 (for AES 256)? That is undefined; the AES specification does not address that possibility. The AES 128 algorithm assumes that you give it 128 bits of key (and tells you exactly what to do with that key); it says nothing about ...

1

Yes, but. While EME is a secure block cipher, its security is not as good as a regular block cipher of that size would be. Theorem 1 of the linked paper shows that the adversary has an advantage after ~$2^{n/2}$ oracle queries, which is expected with an $n$-bit block cipher, rather than $2^{m/2}$ as we would desire for an $m$-bit wide block cipher. Since ...

1

To make notations simpler, I note $R_i = F(k_i, IV_i)$. Then: $$C_1 = P \oplus R_1$$ $$C_2 = P \oplus R_1 \oplus R_2$$ $$C_3 = P \oplus R_2$$ Therefore: $$C_1 \oplus C_2 \oplus C_3 = P \oplus R_1 \oplus P \oplus R_1 \oplus R_2 \oplus P \oplus R_2 = P$$ Your protocol looks like Shamir's three-pass protocol but it requires a bit more than mere commutativity, ...

1

Yes it is possible for a passive eavesdropper to recover the secret $P$. Here's how: The attacker observes $C_1,C_2,C_3$ and formes the XOR of all those values. That's it, the result of $C_1\oplus C_2 \oplus C_3=P\oplus F(K_1,IV_1)\oplus P \oplus F(K_1,IV_1) \oplus F(K_2,IV_2) \oplus P \oplus F(K_2,IV_2)=P$ yields the desired plaintext.

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