# Tag Info

5

Sorry for the late answer, I got busy... So, you know that $\mathsf{Gen}$ is a probabilistic algorithm. What's a probabilistic algorithm? It's an algorithm which, during its execution, can make some random choices, which can be modeled as coin tosses. In programming terms, the algorithm can use a special coin-tossing function, which returns $0$ or $1$ each ...

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A keyspace is the set of all possible keys; it's a set. The cardinality of the keyspace is an integer, and is the number of elements in the keyspace. There is no possibility of confusion, because one is a set and the other is an integer.

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Simply put: No. First recall that this is a mis-use of the term "One Time Pad" So lets call it a vigenere cipher instead. You can determine this is insecure with a simple algebraic combination: $\text{attack} = cipher_1 + cipher_2 + cipher_3 + cipher_4 \\ \text{Simplify: } \\ \text{attack} = character_1 + key_1 + IV_1 + character_2 + key_2 + IV_1 + ... 2 Key generation and key scheduling are different things. The key scheduling is part of the cipher, the key generation generally isn't. Symmetric keys should be indistinguishable from random, so often they are the product of a secure random number generator. There are other ways as well, such as derivation from a password using a Password Based Key Derivation ... 1 The triple DES (3DES) block cipher works by essentially running the block through DES three times. Triple DES is also known as "DES EDE" (encrypt-decrypt-encrypt) and under the name given by the standard document: "TDEA". The TDEA algorithm is described in FIPS NIST Special Publication 800-67 Revision 1 where paragraph 3.2 describes the TDEA Keying Options. ... 1 It looks fine; whether you use the secret$S_0, S_1$as the HMAC key, or whether you use the random value$r$as the HMAC key; if$t' = t$, it implies that either$S_0 = S_1$, or we found a collision in the underlying hash function. I would personally suggest you use$S_0, S_1$as the key. With HMAC, it doesn't really matter; however if we extend this to ... 1 First, you wouldn't call it a one-time-pad any longer if you reused the key. Second the security of the OTP can only be proven if the key is as long as the message. (which wouldn't be the case if you'd reuse a key) Third, OTP usually use XOR operation to combine key and message. If the key is never reused, you're safe, but if an attacker can mount a known ... 1 Using simply a hash function is not strong enough, even if the key is not stored. We the users tend to choose very crappy passwords, such as "1234" or "password". If you only use a hash function for generating the key, then there are a lot of chances that the generated keys are SHA256("1234") or SHA256("password"). That is, this method is very vulnerable to ... 1 Consider what it means for a key$k$to be chosen according to some distribution over key space$\kappa$: Assume that$\mathsf{Gen}$picks key$k$from key space$\kappa$with probability$p$. Since$\mathsf{Gen}$is randomized, this means that a$p$-fraction of all the random tapes will lead it to generate$k\$ as the key. If we now conceptually redefine ...

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