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1

According to Handbook of Applied Cryptography (15.3.2, ii), ANSI X9.9 (which SEJPM mentioned in the comments but I have no access to) defined CFB-MAC only as a compatible alternative to CBC-MAC: The X9.9 MAC algorithm may be implemented using either the cipher-block chaining (CBC) or 64-bit cipher feedback (CFB-64) mode, initialized to produce the same ...


2

Encryption does not imply authentication. With most unauthenticated encryption algorithms an attacker can cause at least random changes to the plaintext by modifying the ciphertext. Some algorithms are malleable, which means the attacker can even cause (some) changes they want. Therefore, even if you could find a secure encryption algorithm that works ...


5

First of all, there's no such thing as a secure 8-bit block cipher, at least not as such things are conventionally used. That's because there are only 256 possible values for an 8-bit byte, and a block cipher will map each of these values to a different fixed value. Thus, an attacker who can even occasionally guess the unencrypted value of some bytes (or, ...


3

An 8-bit block cipher would be practically unusable. A $b$-bit PRP can be used fewer than $2^{b/2}$ times before it's distinguishable from PRF and this limit carries over approximately as is to block ciphers. So you could use it (less than) 16 times per key, i.e. for 128 bits of data. If you had a 128-bit key you might as well use it as a one-time pad. If ...


1

The answer is that it depends very much exactly on what you are considering. However, better bounds can be achieved by using a 96 bit nonce and a 32 bit counter. This is certainly true for GCM as was proved in this paper (Breaking and Repairing GCM Security Proofs). Note that GCM uses CTR inside, so this is relevant.


1

I don't understand the difference between the split nonce/counter design and simply using a random value and incrementing. Why is using nonce +/⊕ counter insecure whereas nonce || counter is secure? Here's the context of your Wikipedia quote (my bold): If the IV/nonce is random, then they can be combined together with the counter using any lossless ...


4

Suppose you do CTR mode as: $E(k,nonce+1) \oplus m_1$, $E(k,nonce+2) \oplus m_2$, $E(k,nonce+3) \oplus m_3$, etc. The wikipedia page is talking about a non-random nonce, with a specific example of a packet counter. So suppose $nonce$ is a packet counter and in each packet you encrypt several blocks. You might end up with the following: In packet #$p$: ...


0

If each round key is of the same size of the block to be encrypted and all round keys are truly randomly generated independent from each other, then we have OTP. This is similar to using a truly randomly generated key of the same size of the plaintext to be encrypted. Each bit of this key is used only one-time to encrypt one plaintext bit.



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