# Tag Info

0

There is no such thing as an AES encryption standard. There is the block cipher, which can be used for a mode of operation (and possibly a padding method). Such a mode of operation however requires a keyed block cipher. And a password is not a key - even the PGP mcrypt guys have started to see this now. So you need (at least) some kind of Password Based Key ...

1

That's because AES is not a password-based encryption algorithm. It's a block cipher. It may seem like a detail, but such details matter. In cryptography, and in security in general, details often matter. AES is a pair of functions, each of which takes a key and a 128-bit message and produces a 128-bit message. The two functions are called encryption and ...

0

In the event this is a homework question, I'll give an answer that would be improbably correct so you may not use it. ;) But maybe people won't accuse you of asking a homework question if you would source where you got the example? The password and the key are two different things. Early cryptographic implementations used the password directly. This is ...

4

There are two things here: Encryption uses mode of operation, and not "AES alone". Some of them are randomized by an initialization vector - that means the encryption of the same text under the same algorithm is still randomized and not deterministic. The encryption methods take care of that. You only need the correct key to decrypt. Passwords are not ...

2

The first (and hardest) step is to factor $n$; the easiest way to do this (given $e$ and $d$) is with this randomized procedure: Select a random value $z$ from the range $(2, n-2)$ Compute the value $\lambda = (ed-1)/2^k$, where $k$ is that integer that makes $\lambda$ an odd integer. Compute $t = z^\lambda \bmod n$. If $t = 1$ or $t = n-1$, we fail on ...

2

Not quite, if all possible permutations are allowed. There are $8! = 40320$ permutations over 3 bits; 15 bits of key allows you to specify $2^{15} = 32768$ of them; hence any mapping of 15 bits will necessarily $40320-32768 = 7552$ of permutations unexpressable. It is doable if you don't allow every single permutation (e.g. allow only even permutations), ...

2

PKCS10 looks like relevant industry practice for private keys. See "Note 2" at page 4, Certification Request Syntax Specification - RFC 2986: The signature on the certification request prevents an entity from requesting a certificate with another party's public key. That is, soneone requesting a certificate on a public key demonstrates his knowledge of ...

1

XORing a key and message is called a one time pad. It is perfectly secure, providing confidentiality, when used correctly. That last part is the hard part, along with finding a situation in which you only need confidentiality.

1

Most ciphers — both classical and modern — will work just fine with any key. It's just that, if the key used to dechipher the message does not match1 the key used to encipher it, the output will be essentially nonsense, and the actual intended message will not be revealed. (Some encryption systems may then detect that the decrypted text is ...

1

I personally have always seen a key used for encryption as a key used in a door, I never compared it to a keystone. But i think it is fair to compare it to a key with which you open a door. Since a key in cryptographic sense give you access to data or even to complete systems. Further more a keystone in the sense that a key is needed to make it work is not ...

1

fkraiem's definition is too narrow. $\:$ "In the context of encryption schemes," keys are "whatever piece of information the legitimate recipient of an encrypted message possesses, which allows him to decrypt the ciphertext" and any information related to keys of the type mentioned above, which allows its possessor to encrypt the plaintext .

3

In the context of encryption schemes, the key is whatever piece of information the legitimate recipient of an encrypted message possesses, which allows him to decrypt the ciphertext efficiently. Hence, the key must be kept hidden from an attacker, since otherwise the attacker could decrypt efficiently just as the legitimate recipent does.

Top 50 recent answers are included