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The difference between the PKCS#5 and PKCS#7 padding mechanisms is the block size; PKCS#5 padding is defined for 8-byte block sizes, PKCS#7 padding would work for any block size from 1 to 255 bytes.
This is the definition of PKCS#5 padding (6.2) as defined in the RFC:
The padding string PS shall consist of 8 - (||M|| mod 8) octets all
having value 8 - (...
answered Jul 4 '13 at 21:18
86
GCM and CBC modes internally work quite differently; they both involve a block cipher and an exclusive-or, but they use them in different ways.
In CBC mode, you encrypt a block of data by taking the current plaintext block and exclusive-oring that wth the previous ciphertext block (or IV), and then sending the result of that through the block cipher; the ...
55
SHA-256(SHA-256(x)) was proposed by Ferguson and Schneier in their excellent book "Practical Cryptography" (later updated by Ferguson, Schneier, and Kohno and renamed "Cryptography Engineering") as a way to make SHA-256 invulnerable to "length-extension" attack. They called it "SHA-256d". We started using SHA-256d for everything when we launched the Tahoe-...
39
Short: CBC mode in context of TLS protocol has had security issues, and would have had to be reworked.
AES-CBC mode combined with decent HMAC can be as secure as AES-GCM. However, combining the cipher and MAC securely has been in practice found to be much easier said than done. Also, padding that is required by AES-CBC mode complicates things.
In ...
33
XTS vs. Undiffused CBC. The issue here is malleability. Both XTS and CBC prevent an attacker from learning information about encrypted data. However, neither one completely succeeds in preventing an attacker from tampering with encrypted data.
However, it's arguably easier to tamper with an (undiffused) CBC ciphertext than it is to tamper with an XTS ...
30
In comparison against CBC mode and HMAC, GCM mode is quite commonly better alternative. But, I'll go to detail where it neccessarily is not. Just like Richie Frame, I also do not agree that CBC + HMAC is always the best comparison target. I've added few other details. Hope you find them useful.
Against CBC and HMAC
I'll discuss downsides first.
The ...
28
The crucial difference between plain encryption and authenticated encryption (AE) is that AE additionally provides authenticity, while plain encryption provides only confidentiality. Let's investigate in detail these two notions.
In the further text, we assume $K$ to be a secret key, which is known to authorized parties, but unknown to attackers.
Goals
...
24
With CBC (Cipher block chaining) mode, before encryption, each block is XOR-ed with the ciphertext of the previous block, to randomize the input to the block cipher (and avoid encrypting the same block twice with the same key, as this would give the same output, and tell the attacker something about the plaintext).
As the first block has no previous block, ...
23
You say that a random IV "would also be unique", but really that is the crux of the problem. The problem with counter mode is that it is secure unless the same counter is used twice; if it is, it is likely that an attacker will be able to recover both plaintext messages. This contrasts with CBC mode, which if you repeat an IV, it has the relatively benign ...
22
Contrary to what Stephen says, you absolutely can compute the tag in parallel.
Here's how it works; the tag computation is essentially "assemble the AAD, data, the length field and $Encr(Nonce)$ into a series of values $x_n, x_{n-1}, x_{n-2}, ..., x_0$", and then "compute the polynomial $x_nh^n + x_{n-1}h^{n-1} + x_{n-2}h^{n-2} + ... + x_0h^0$
This ...
21
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 ...
20
The paper you cite (Deterministic Authenticated-Encryption...) gives quite a bit of useful information (but I'm assuming you already knew that). It looks like a pretty good read (I'll let you know if that assumption holds after I finish it).
For why simpler constructions (CBC/CTR with a MAC or even AEX mode) don't satisfy (emphasis added):
A key-wrap ...
18
If you look closely at the definition of authenticated encryption modes, you will see they all are, actually, the combination of symmetric encryption and a MAC.
Using traditional encryption and an independent MAC has a few tricky points, none of them being unsolvable:
The encryption mode will use a key, and the MAC will also use a key; using the same key ...
18
There are some serious problems with this design that would preclude it from being standardized, so it probably does not have a name.
The 2 visibly main flaws are as follows:
If the plaintext follows a pattern similar to the block counter, the
block cipher inputs may repeat, exposing information about the
plaintext (exact same issue as reuse of nonce, but ...
17
The security of that approach is equivalent to that of normal CBC.
Your scheme with first plaintext block $IV^\prime$ is clearly identical to normal CBC with $IV=AES(IV^\prime)$. Since a block cipher is a permutation over a block, a uniformly random first plaintext block will lead to a uniformly random IV for normal CBC.
A ciphertext produced with your ...
17
A block cipher is an invertible transformation that maps an $n$ bit block of bits to an $n$ bit block of bits, under the control of a key (and where $n=128$ in the case of AES)
Now, we most often need to do things other than mapping blocks of $n$ bits; how we do that is using the block cipher within a Mode of Operation. A mode of operation is just a way to ...
14
Thomas is correct; there's no attack on CFB mode if you can predict the IV; NIST is just being cautious.
With CBC, the value of the first encrypted block $C_0 = E_k( IV \oplus P_0)$, where $IV$ is the IV used for that packet, $P_0$ is the value of the first plaintext block, and $E_k$ is the evaluation of the block cipher.
If an attacker can predict the ...
14
The algorithm (now reasonably clear) is reminiscent of a block cipher in CFB mode, with $random$ as the IV (which can be public), $secret$ as the key, and MD5 used as keystream generator instead of the block cipher.
Decryption works as in CFB:
\begin{align*}
M_1 &= C_1 \oplus \operatorname{MD5}( secret\mathbin\|random )\\
M_n &= C_n \oplus \...
14
Short answer:
There would be nothing (that isn't already wrong with TLS) necessarily wrong with a CTR + HMAC cipher suite, but the technical merits are only one factor in a technical feature getting to RFC status in the TLS working group.
Without being discourteous to the TLS Working Group (WG) participants or process, other reasons can be: political (...
14
TLS 1.3 is a reboot of the TLS protocol which focused on up to date cryptography rather than backwards compatibility.
Now CBC is not as secure as you make it to be, and the way that it was used in TLS made it particularly vulnerable. To note: in TLS the HMAC authentication tag was created over the plaintext rather than the ciphertext. This made TLS ...
answered Oct 26 '17 at 23:53
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Given only what you've said, and assuming the keys are created and stored in a strong manner, using a different key to encrypt database entries mitigates the problem of ECB mode. Namely that identical plaintext, when encrypted with the same key, always outputs the same ciphertext. No security is gained by switching to CBC mode (assuming you can easily store ...
13
For block ciphers, it depends on which mode of operation you're using — nobody uses just a plain block cipher for anything, at least not unless all their messages are shorter than a single cipher block (typically 8 or 16 bytes).
ECB mode, which just amounts to chopping the message up into blocks and feeding each block through the cipher, does not use ...
13
Do not use a fixed IV. It can have seriously negative consequences. This is especially true for CBC mode.
That said, a random 128-bit IV stored in plaintext is typically what you want. The IV can be known to an attacker without breaking security.
13
While you do operate block-by-block when generating the pseudorandom stream, the actual encryption step (i.e., the XOR) is bitwise, and therefore does not require the message to be padded.
For example, the message "Hello" will be processed as follows (pseudocode):
byte stream[16] = AES(Key, Nonce);
byte plaintext[5] = "Hello";
byte ciphertext[5];
for i ...
12
After reading the paper How to Break XML Encryption (thanks to Krzysztof for the link), here are my two cents.
This attack relies on the fact that a CBC-ciphertext C = (IV, C1, ... Cd) can be decomposed into pairs of (IV, C1), (C1, C2), (C2, C3), ... (C(d-1), Cd), each of which is also a valid CBC ciphertext for the same key, relating to the corresponding ...
12
Free space and used space look exactly the same to someone who only sees one version of the ciphertext.
First, the basic idea of a secure block cipher is that you learn nothing about the plaintext block simply by observing the ciphertext block. You may be able to learn something about the plaintext from the surrounding context, such as by collecting more ...
12
The MAC value should be calculated over all of the input, not just the first block. The chaining of CBC makes sure that the bits in the last block of ciphertext depends on all the previous blocks.
answered Apr 16 '16 at 9:17
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Each mode of operation has its own IV requirements. Some need uniform, unpredictable randomness. Other are equally happy with just uniqueness.
CBC is well-known for its need of an IV chosen randomly and uniformly among the possible IV values, and such that an attacker who can choose the text to encrypt may not predict the IV value before submitting the said ...
11
It is an IV and it is safe to transmit with the ciphertext (if it wasn't, we would call it a key).
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