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Given: The attacker can call PRP() and the inverse function prp() on any message of his choosing. PRP is a pseudorandom permutation indistinguishable to the attacker from a random permutation. Assuming R and K are "sufficiently large", perfectly random, and never leaked to the attacker -- in particular, during a chosen-ciphertext attack, the decryptor only ...

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The scheme is secure against chosen-plaintext attacks up to $2^{|R|/2}$ queries. Indeed, given this number of queries, it is likely that every encryption call yields a new value $R$, which has never used as part of the permutation input. However, when this bound is reached, some problems occur. Suppose you encrypt the same message $M$ as many as ...

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Notice that $N_b$ has not been seen until step 4, so it is generated by B. In step 5, A sends $N_b-1$ back to B. This proves to B that A knows $K_{ab}$ otherwise A could not have recovered $N_b$. This is done to prevent A from replaying the message in step 3 and therefore authenticating. For example, suppose Eve sees Alice send the message in step 3 ...

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The part of this answer that talks about key storage is at the end, the first part is about implementing a cascade. There are 2 main methods for cascading block ciphers, inside of the mode and outside of the mode. Within the encryption you have your mode of operation, and you have your block cipher cascade. The first cipher in the cascade will be considered ...

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My own symmetric cipher also has this property. Here's what I can say about the general meaning of it: It means that the amount of preprocessing of a key is small. So the amount of time from generating/importing a new key to actually starting encrypting is neglegible. It means that the amount of state a cipher uses is small. The state of a cipher generally ...

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This depends on the encryption algorithms in use. If the encryption algorithm is a bijection, then every possible $C$ must have a preimage. Examples of this include block ciphers like AES. In these cases, the decryption algorithm will output a message $m$ such that $E_k(m)=c$. Note that if $c$ has been tampered with by an adversary but is still the same ...

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$V_{1}$ and $V_{2}$ should never be equal when using correct implementation of cbc by using the same input $(a,b,c)$. See following construction scheme: Even though you have two distinct encryption processes, namely one for $V_{1}$and another for $V_{2}$, the correct implementation of CBC uses an initialization Vector IV which has to be random. By xoring ...

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For AES-128, the block cipher works on 128 bits at a time. Whichever block cipher mode you use (ECB, CBC, CTR, etc.), the encrypting will always be done on 128-bit blocks. The assumption is also made that padding is being used. Let's assume that $m = (a||b||c)$ and that $m' = (c||a||b)$. That gives us two separate messages, each 900 bits. Using ...

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I'm going to assume that the comma $,$ operator used in your question means 'concatenate' (normally written $a||b||c$). Moreover, I'm assuming that $a,b,c$ are distinct. In that case, With incredibly high probability, No: $V_1$ and $V_2$ will not be equal. Think of it this way: if they were equal, then what would $D_k(V_1)$ be? Supposing $V_1=V_2$, we ...

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I've thought quite a lot about this, and I think in general the answer is no, it would not be a good idea to use a KCV for those kind of situations. Using a hash or even better a MAC (using the key as MAC Key) would be a much better idea if a KCV is required. Instead of zero's, it would be much better to use a block of bytes that is not likely to be of use ...

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It looks like what you are describing is comparable to IGE (Infinite Garble Extension) and especially biIGE mode of encryption. So I guess my question and the answer on my question here is of relevance.

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You need to use different IV for every message you encrypt. Thus rather than the process: encrypt plaintext reverse the ciphertext continue encrypting (now from finish to start) You need to generate IV each time. I.e.: generate IV encrypt the plaintext using the IV store/send the ciphertext and the materials required to recreate IV The ...

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