# Tag Info

1

When you are e.g. sending TLS encrypted data over a SSH tunnel, there are two things in particular that should be noted: The TLS handshake will only commence, once the SSH connection has been established. The bulk encryption keys of TLS will be completely independent of the SSH encryption keys. Since the handshakes and keys are completely independent, ...

1

Any good software should use PBKDF (a password based key derivation function) that uses a random salt. This salt is stored with the ciphertext and should be different for each ciphertext. As long as this is the case they key will be different for each ciphertext. The best way an attacker can then attack your ciphertext (when stored on disk) is to iterate ...

0

It really depends on you block cipher mode of operation and likeliness between your files. With any proper implementation, it should really not, but since you don't really give any details about your encryption scheme, it's hard to tell. (side-note: how do you derive the key from the password?)

1

It's a hardware implementation of something that typically needs to be written in software. Imagine if nobody had hardware multiplication circuits and everybody had to write software implementations of multiplication. Then a new processor came out that had a dedicated circuit to perform multiplication. Obviously a circuit for multiplying numbers would be ...

4

AES-NI is just a fast way for the processor to execute the calculations of AES. Normally the computer has to calculate every single step of the AES key schedule and the rounds as a single instruction: Substitute it with the S-boxes, shift the rows, mix the columns, XOR the round key. This is called a software implementation. Every instruction has to be done ...

1

If you start with a random key and zero counter, there's 128 bits of entropy in the system state. If you start with a random key and random counter value, there's 256 bits of entropy. Whether that matters depends on what you are using the PRNG output for. If you are using the output for anything where 256 bits of entropy would be an asset – say random ...

1

When using counter mode you can start at any value, it doesn't matter. The only important thing is that you never use the same counter value twice for the lifetime of the key. So, as long as your key is actually random and, as you say, you don't use it for more than 1 MB of data, then your generator should be fine.

1

There's no particular computation you have to do. Since all other tables could be computed just by rotations of Te0 what you need is just to perform rotations, xors and table lookups. Here's what usually done when you have only Te0. You read the state column-wise, keeping in one 32 bit variable/register the value of one column. (As convention hereafter, ...

0

Currently you seem to be using a Password Based KDF (PBKDF); you are using PBKDF2, as defined in Rfc2898. You don't need to do this as randomly generated data is already fine for creating an AES key. So - as you don't need a PBKDF - you don't actually need a salt. If you need more keys or key data then what is actually required is a Key Based KDF (KBKDF), ...

4

Like the other answers say, it does not always have to be the case. One other case where it is often not stored is when you have a single use key, for example as part of some hybrid encryption scheme. Then there is no need to use a nonce at all and it is usually taken to have zero value.

5

If you want strict indistinguishability, then yes, you need to store the IV (initial counter) somewhere. However, there are some relaxed modes that are used in practice for things like disk encryption, where it is often very useful to decrypt things "in the middle" like you say. For instance, XEX uses a counter which is derived from the sector and offset ...

7

I would like to ask if that is true for every AES CTR mode implementation?, Doesn't have to be. You can store the nonce anywhere. You could even send it to the recipient via a different channel (e.g., email the ciphertext and use SMS to transmit the nonce). Storing it at the beginning has its advantages. For example, if streaming the data, you can ...

1

As you can make up from this encryption scheme, I'm using the encrypt-then-authenticate approach to enforce ciphertext integrity. In step 2 of the decryption process I perform this authentication step. If the calculated HMAC turns out to be equal to the HMAC in the file, does this mean, apart from the implied ciphertext integrity, that the supplied ...

1

CCM (Counter with CBC-MAC) Message authentication (via CBC-MAC) is done on the plaintext not the ciphertext. (This is generally not a desireable feature.) On the encrypt operation, the encryption and MAC could happen in parallel, but generally do not (typically because there is just one AES engine in a chip, just one AES thread at a time, etc.). Similar ...

3

I am little curious about how do we calculate hardness proof of any cryptography algorithm? This is typically done by assuming some problem is hard (e.g., solving discrete log). Then proving that if someone can break the cryptography algorithm (e.g., diffie-hellman) that they can also break the hard problem. Once this relationship is established, we ...

14

CBC does not perform authentication This property makes it less suitable for places where authentication is required, basically any transport protocol. TLS uses CBC, but by default performs authentication over the plain text instead of the ciphertext, which opened up a host of attacks. CBC can be used here, but it is error prone and may require an ...

7

There are several scenarios where you wouldn’t want to use AES in CBC mode. In CBC mode, each block is dependent on a previous one. As @fgrieu nicely hinted at in his comment, using CBC means that if you have a large, encrypted file and you only want to update/change/modify a tiny fraction of it, you would have to follow the decrypt-modify-encrypt path each ...

-2

CBC is one of the five confidentiality modes of block cipher. It needs one IV (Initialization Vector). The IV must be same length to the blocksize of the algorithm (e.g. 128bit for AES). The IV is done exclusive-ORed with the first plaintext block P1. The result of the first XOR operation is C1. Next C1 is exclusive-ORed with the second plaintext block P2. ...

0

There are alternative representations of $GF(2^8)$ where the multiplicative inverse can be computed by computing the inverse in $GF(2^4)$, along with a handful of other operations (several $GF(2^4)$ multiplies, additions and a square). Does your $\delta$ convert from the standard AES representation into such an alternative representation of $GF(2^8)$? ...

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