# Tag Info

## Hot answers tagged cbc

20

It's meaningless nonsense. I would be inclined to avoid spending any money with these people. If you scroll down on this page, you'll find a table labelled key size vs. time to crack, according to which their $2 \times 256$ bit encryption takes $3.31 \times 10^{112}$ years to crack, making it (apparently) superior to ordinary $256$-bit encryption (which can ...

8

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 ...

7

In the first block, the IV provides the "randomness", and in subsequent blocks you just use the previous block of ciphertext instead. Based on the assumption, that the cipher is not weak and behaves like a pseudorandom permutation, this is basically the same: You XOR something unpredictable on the plaintext, and then encrypt. As long as the IV is chosen ...

5

The modes you are referencing are specifically modes of operations for block ciphers, and therefore are not directly applicable to hash functions. Block cipher operations take 2 inputs, the key and a block-sized input value, and output a block-sized keyed permutation of the input. Hash functions take a variable length input, and output a fixed length value. ...

5

At a high level, the major flaw is that you are rolling your own crypto protocol. You should strongly consider using a standardized protocol like DTLS. Some specific problems: Symmetric key distribution is left unspecified. Keys must be changed occasionally to thwart distinguishers. No way to recover from symmetric key compromise. Your message ...

4

AES CBC usually requires padding, such as PKCS#7 padding. This padding is 1 to 16 bytes, 16 being the block size of AES. The HMAC will add 256 / 8 = 32 bytes to the total. Usually you will need to store the randomized IV as well with ciphertext, to allow for reuse of the key, adding another 16 bytes (the block size again). So the total overhead will be about ...

4

What is this method/algorithm/construction called? Dunno; this is a new one on me. Is it as secure as CBC implemented the normal way? Should be. Modeled as an abstract 'take plaintext, output ciphertext' model, this method (with a random last ciphertext bits) has precisely the same ciphertext output distribution as CBC mode (with a random IV), ...

4

First, the advice: What are the best-practices to store the message length / strip away padding? Use standard padding, like PKCS#7 padding. It handles finding the length uniquely for you. Use encrypt-then-MAC to prevent padding oracle attacks. (Or better yet, don't use CBC. Use an authenticated encryption mode like GCM, or use CTR+MAC which doesn't ...

3

Rejecting replays is the duty of a higher level protocol. Simple authenticated encryption will accept any message with a valid MAC, even if you receive it several times. Decryption is a stateless process, but you need state to keep track of messages you already received. For example you could associate an increasing counter for each message you send. The ...

3

Better is a subjective term. However for the choice between ECB and CBC, the choice should be CBC for almost all situations. Although ECB and CBC are modes of operation of a block cipher, you could also turn this way of thinking around and see the block cipher as a configuration option for the mode of operation. The mode of operation has a big influence on ...

3

As long as the IV is chosen correctly, every individual block of the encrypted output will be uniformly random over the set of all bit-patterns of the given size. Each block is independent from the clear text, but they are not independent from each other. The first block contains the IV itself, which by construction is uniformly random and independent from ...

3

Yes there is a significant difference concerning brute-force. ECB suffers from multi target attacks whenever you encrypt the same message block. This is always possible in a chosen-plaintext attack and often possible in practice with a known-plaintext attack. With CBC the IV means that the plaintexts passed to the blockcipher are almost certainly unique ...

3

As Maarten Bodewes already wrote in a comment, if you ignore the computational overhead of XOR, then there is essentially no difference in CBC and ECB for a bruteforce attack. However, the question is actually mixing oranges and apples (and it is not obvious), because the security weakness of modes of operation has nothing to do with the underlying ...

3

The "interesting" part of your encryption is here: Therefore, I prepend a block at the beginning of my packet. Its content goes as follows: First four bytes: current timestamp in seconds Next 12 bytes: zeros I compute the sha256 hash of the message (32 bytes) I xor the timestamp + zeros block with the first half of the hash I xor the ...

3

Your mode is essentially equivalent to CFB mode, except that: you've reversed the order of the blocks in the message, and you're using the block cipher in the opposite direction than usual. Neither of those differences should have any direct security implications (since all standard block ciphers have the same security properties in both directions), ...

3

In RFC2246, if you need 12 bytes of padding total, that means that you have 11 padding bytes, followed by a padding length field. So, each padding byte has a value of 11 (0x0b), as well as the padding length field. This is implied by the requirement that the total TLSCiphertext.length must be a multiple of the block size, and this TLSCiphertext.length ...

3

POODLE is primarily a padding oracle attack against SSLv3.0, which is inherently vulnerable to the attack due to the protocol design. The "on downgraded legacy encryption" part of POODLE's name comes from the fact that most SSL/TLS client implementations will allow a TLS connection to downgrade to SSLv3.0 if the handshake fails - see this answer for more ...

2

The schemas from the relevant Wikipedia page really explain it all: As you see in the decryption schema, the IV is used for a single XOR that yields the first plaintext block; it is obvious that the IV impacts only that block. When encrypting, though, modifying the IV alters the first ciphertext block, then the second ciphertext block, and so on. The ...

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

That's a lot of questions, I'll try and answer in order. A hash or message digest alone is not secure because anybody can calculate and thus substitute a hash value. If you (correctly) add a key to the mix then you get a HMAC, which can be used. Nowadays often a HMAC is used, or an authenticated mode of authentication such as GCM, CCM (for packet ...

2

Yes, we always have to pad the message. The reason is simple: How do we know if the message has a padding or not if we don't always pad? Let's say we pad with adding only $0$ bits. We got the (after padding) message $0101\,1100\,0000\,0000$ and a block size of 2 bytes (16 bits). Well, what was the original message? Was it $0101\,11$? Or was it $0101\,1100$? ...

2

SSL padding always pads, using 1..blocksize bytes (8 bytes for triple DES, 16 for AES). This padding makes it deterministic independently of the value of the plaintext. It's a padding mode similar to ISO 10126 (only the last padding byte is one less). Other padding values - such as the zero padding performed by PHP's mcrypt library - are also ...

2

To show that a family of functions is not a PRP, you have to either show that the functions are not permutations or that they do not behave pseudo-randomly. As it is already established that the functions are in fact permutation you need to show the latter. For a family of permutations to be a PRP means that it is computationally infeasible to distinguish a ...

2

Patterns aren't hidden. Assume you've got a pair of plaintext blocks $p_1, p_2$ then the encryption of $p_2$ will always be the same: $c_2=E_k(p_1 \oplus p_2)$ as it only depends on the plaintext. So you haven't basically solved the basic problem of ECB (patterns remain) but rather moved it (to the more rare case) that two blocks repeat. Plaintext isn't ...

2

Normal CBC mode cannot be parallelized during encryption; that's because CBC mode encryption is defined as: $$C_i = E_k(C_{i-1} + P_i)$$ That is, what you encrypt during the processing of block $i$ depends on ciphertext of block $i-1$; hence you can't start the next block until you've completed processing of the previous block. You can't do any ...

1

If the last byte in the block is larger than the block size, that would generate a padding error in SSL 3.0. SSL 3.0 padding is up to $BlockSize-1$ bytes of unspecified data, and one byte with the length of the padding (not including the length byte itself). TLS 1.0 and above allow padding longer than the block size, and requires that each byte contains ...

1

For example, let say that you have a message: "100 dollars should be moved." and you encrypt it with OTP. Then everybody can just take the first character "1" and change it to a 9 by XOR the 1 from the cipher-text and then XOR a "9" with the key you got. What you are describing is what happens if an attacker introduces changes in the ciphertext by a ...

1

Is CBC mode in OTP more secure? No. If your one time pad satisfies the required properties (it's truly random, the attacker has no information about it, and it's only used once), then OTP already has perfect secrecy; playing around with how it works can't make things better. If your one time pad doesn't satisfy the required properties, then all bets ...

1

During CBC mode decryption, each ciphertext block is first decrypted with the block cipher, and then the first decrypted block is XORed with the IV, while later blocks are XORed with the previous ciphertext block: Thus, the first block of plaintext is computed as: $$P_1 = IV \oplus D_K(C_1).$$ Equivalently, we may solve this equation for $IV$ to obtain: ...

1

If the MAC'ing is done right (after the encryption) and if you pad the CBC data correctly there's no security risk. Howver changing to CTR or even better to GCM would be better, as both don't operate on whole blocks and GCM even provides authentication.

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