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The modern trend for encryption-only modes is clearly CTR, which has a number of advantages over other modes: no padding is needed (contrary to CBC); the computationally-intensive part can be efficiently performed with the IV (and key) only, before the plaintext or ciphertext is available (contrary to CBC, CFB); the computationally-intensive part can be ...


3

It is certainly wrong to state that "MAC can only be produced with AES in CBC and CFB mode", but there seems to be a simple reason that people were inspired by these modes when thinking up possible MAC constructions: They carry along some state that incorporates information from the message while traversing the input blocks. In both modes, encrypting a block ...


3

When seeing clearly through all the cube "magic", one recognizes the following: All the cube operations are just key-dependent bit permutations. Therefore, the whole cipher is a sequence of key-dependent permutations and XORs with key bits. This admits an algebraic description: For all keys, there is a permutation matrix $A\in\mathbb F_2^{512\times512}$ and ...


3

In your example, $Encryption_1$ is $\textsf{AES}_{CTR}$ and $Encryption_2$ is $\textsf{Salsa20}$. Then, the encryption method you are proposing is $Encryption_1(Encryption_2(plaintext))$, which is in fact a cascade of stream ciphers. Note that, because you simply XOR the streams, this cascade cipher commutes, that is, you will have the same result if you use ...


3

Asmuth and Blakley provided a proof that, assuming the keys for each cryptosystem are chosen independently, breaking their composite cryptosystem is at least as hard as breaking the hardest part of either. [1] Building on their work, cascade ciphers have been shown to in fact be harder to break than the hardest part of either. Admittedly, what you're ...


3

Under your scheme, the keystream generated by the stream cipher will be the same for each message enciphered with the same key. While your scheme does not encrypt identical blocks of plaintext in the same message to the same value, it does encrypt identical blocks of plaintext in different messages to the same value, meaning that XORing two ciphertexts ...


2

With concatenation the caller only has to ensure the nonce is unique. For example they can use a counter that increments for each message. Incrementing a counter for each message is convenient in many scenarios, including encrypted network transports like TLS. If you use xor or add nonce and counter you get overlaps, so a counter as nonce would be fatally ...


2

Aren't $IV_1$ and $IV_2$ public in TLS 1.2 as well? $IV_1$ certainly is (as that's just the ciphertext block in front of the block we're attacking); however the IV that the TLS 1.2 sender will use for the next message ($IV_2$) is not. In fact, the sender might not know it yet, as it might not have not picked it yet. But doesn't this mean that BEAST ...


2

Your scheme is indeed an instance of output feedback mode (OFB), using $$(\mathit{key},\mathit{pad}) \mapsto H(\mathit{key}\oplus\mathit{pad})\text,$$ where $\mathit{key}$ corresponds to keyhash and $\mathit{pad}$ to hash, as the "block cipher". (It is very likely not really a block cipher due to lack of bijectivity, but that's not needed for output feedback ...


1

For CBC encryption, you would store the initialization vector as the beginning of the ciphertext, regardless of whether you read in an IV or generated it yourself. The CBC decryption algorithm would do what you described at the end of your post. Your implementation should also offer access to the block cipher itself, without any mode of operation.


1

For question (1): This page gives some hints on IVs and CBC: https://defuse.ca/cbcmodeiv.htm I copy-paste the part about IVs "predictability" Chosen-Plaintext Attacks Randomness is not enough, though. IVs have to be unpredictable, too[2]. Suppose there is a CBC-mode encryption system that selects a random IV, publishes it, asks the user ...


1

You say are asking as a learning exercise, to learn how to invent ciphers. The way to learn that is not to try to invent some block cipher and then ask others to break it. The way to learn is to learn cryptanalysis, by breaking other ciphers. See Schneier's self-study course on cryptanalysis for one good resource.



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