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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 ...


9

CBC mode encryption is defined as: $C_i = E_k(P_i\oplus C_{i-1})$ (with $P_i$ being the $i$th plaintext block, and $C_{i-1}, C_i$ being the ciphertext blocks. What might happen if we have a lot of ciphertext encrypted with the same key is if two ciphertexts happen to be the same, that is: $C_i = C_j$ If we see that, we can then immediately deduce that: ...


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 ...


5

Let me see if I have this right (and please correct me if I misunderstand; my conclusions depend on the details of this); you distribute images for your firmware device; these images are encrypted with a secret AES key (using AES in CBC mode); the device decrypts the image, and then runs that decrypted image. The sole check to make sure that the image ...


5

The reason why CBC is considered better than ECB has nothing to do with situations involving an attacker with a partial ciphertext; we always assume that any attacker has full access to the ciphertext. Instead, the problem with ECB is that it leaks information. Specifically, if you encrypt two messages which has two blocks of plaintexts in common, then ...


5

Absolutely. The key point is that, whilst in CBC mode, the encryption can be thought of as using the previous ciphertext as the IV - have a look at this diagram from wikipedia: I assume from what you've said that you have a function that will "do" AES-CBC decryption on large amounts of data, and you wish to use this. So, you simply run: $$ D_k^{IV}(c_1\ ...


5

With CBC mode the initialization vector is referred to as IV, because it is not nonce. There are ways to construct nonce so that it does not meet the needs of CBC mode. Random IV is one generation choice which is usually fine. Nonce can also be a counter, which is not ok here. Definitions Nonce means number used once. IV means initialization vector. CBC ...


4

Here's a nice paper I came across a while ago: Wooding, Mark (2008), "New proofs for old modes", Cryptology ePrint Archive, report 2008/121: "Abstract: We study the standard block cipher modes of operation: CBC, CFB, and OFB and analyse their security. We don't look at ECB other than briefly to note its insecurity, and we have no new results on counter ...


4

Well, there is no really good way; the encryption of the plaintext is $E_k( Plaintext \oplus IV)$ (followed by 16 bytes which are a deterministic function of the first ciphertext block). The AES function $E_k$ is designed to be totally unpredictable if you don't know the key, there's nothing to leverage there. The only thing that allows you to gain any ...


4

Yes. Assume that the attacker knows the ciphertext $c = c_1 \mathbin\| c_2$, the initialization vector $v$ and the plaintext $m = m_1 \mathbin\| m_2$. This tells them that $D_k(c_1) = m_1 \oplus v$ and $D_k(c_2) = m_2 \oplus c_1$, where $D_k(\cdot)$ denotes block cipher decryption under the (unknown) key $k$. In particular, this implies that, if the ...


4

Normally you don't want to reverse the encryption used within a DRBG. Schemes like PKCS#7 padding and CTS are required to deterministically reverse the padding during decryption. If you just want to have a final block, you should be fine by using zero padding until the end of the block. Of course, if this zero padding is actually harming the output of your ...


4

From the sound of your questions, it almost appears that you have some confusion between the CBC-MAC key and the CBC-MAC tag. The CBC-MAC algorithm takes the message (in this case, most likely the ciphertext) and a secret key; it outputs a tag (which can be public). The security property of CBC-MAC is that someone who does not know the key cannot generate ...


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 ...


3

Remark: in One Time Pad the pad is used once, thus this is not OTP, since here $k$ is reused. Hint for part 1: Write the relations between $k$, the message blocks $m_i$, the ciphertext blocks $\small C_i$ with the convention $\small\text{IV}=\small C_0$. Then, find equations that allow computing the desired $m_3⊕m_4$ from known quantities. Hint for part 2, ...


3

First of all, I suggest you to try use as IV first 16 bytes of encrypted file. Because in general IV is the first block of ciphertext. But if that doesn't work, then – of course – you can decrypt all message except first block. Just use first block as IV, and start to decrypt from second block. That will work because CBC does not provide integrity, and ...


3

You should use random IV even when unique keys are used. This prevents key-collision attack where the attacker collects number of cryptograms that have been encrypted with unique keys and brute-forces for key. Using predictable IV will reduce security of your cryptosystem by a factor of N (where N is the number of ciphertexts created). The attack recovers ...


3

To see the problem, let's see how I would chain up a single function (call it $AES\,CBC_k(iv, m)$) which only encrypts a single block at a time into something that can encrypt "chunks" of any size. Let $m=m_1||m_2||m_3||m_4$ be the message I want to encrypt. Each $m_i$ is a single block (in AES it is 128 bits). I want to use $AES\,CBC_k$ to encrypt $m$. The ...


3

Just for completeness sake, CBC is defined as follows: The error you have made is that: $$M;N = (M_1, ..., M_n, N_1 ⊕ \mathbf{T_m}, N_2, ..., N_n)$$ (I've changed notation from M||N to M;N to reflect this isn't just concatenation) You need to cancel the tag from the message $M$, not the tag from the $N$ message. In that case, $T_{M;N}=T_N$ as required.


3

The usual mode for disk encryption is XTS (let's say the mode suggested by the NIST). AEAD cipher seems to be promissing but typically with the GCM you will have also to store an authentication tag per encrypted block which may lead to a complex implementation (but interesting). I believe that regarding at least integrity, there exist "new" file systems ...


3

The classic proof is contained in http://www.cs.ucdavis.edu/~rogaway/papers/sym-enc.pdf (1997), but it is not quite easy.


3

My first thought was that I could set the IV to the first 8 bytes of the CT [and] decode the rest[.] This is exactly how CBC works. For all blocks but the first, encryption is defined by $C_n=E_K(C_{n-1}\oplus P_n)$ and, therefore, decryption is achieved by $P_n=C_{n-1}\oplus D_K(C_n)$. Since there is no previous ciphertext for the first plaintext ...


3

If your IV is predictable this is as (in)secure as assuming that you have a zero vector IV. And a zero vector IV allows you to perform a so-called Adaptive Chosen Plaintext Attack (ACPA). Why? Assume that you have a encryption mechanism that works in CBC mode. This means, that on the first iteration the $IV$ is XORed with your input message (which is ...


3

You should read the wikipedia article about disk encryption. In short: for disk encryption, data is organized in sectors (for instance of size 512 bytes), and data may be encrypted with a chaining mode of operation such as CBC only inside these sectors, using a different initialization vector for each sector.


3

Well, if you have hardware which you can give it a long ciphertext block, and say "decrypt this block in one shot", well, one could argue that reusing the last ciphertext block as the next IV might give some minimal amount of gain; you would concatinate all the ciphertexts in order, and ask for the hardware to decrypt the entire thing -- the result will be ...


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 ...


3

I didn't find anything about the exact way Crashplan encrypts files, only that it uses Blowfish in CBC mode. The block size of Blowfish is 64 bit, so there are $2^{64}$ different input blocks and the same number of output blocks. All in all $147573953$ terabytes of different output data. The problem with this is the birthday attack. Summarized it says that ...


2

In the padding oracle attack you have an oracle that only tells you whether a particular chosen ciphertext decrypts to a correctly padded plaintext. That oracle is used to build a last word oracle, which used iteratively can reveal a whole message. The reason it works in CBC mode is that we can make predictable, arbitrary changes to the plaintext of the ...


2

…are any other modes of operation vulnerable to padding oracle attacks? Nope, it’s purely restricted to CBC. A padding oracle attack, also known as “Vaudenay attack” because it was originally published by Serge Vaudenay in 2002 and introduced at EUROCRYPT 2002, is an attack against cipher-block chaining. The attack works against any block cipher in ...


2

Yes, your understanding is correct. With the CBC-MAC's of $(a), (b)$ and $(a||b)$ you can forge the following new messages: $(a||b \oplus MAC(a)), \\ (b||a \oplus MAC(b)), \\ (b||b \oplus MAC(b)),\\ (a||a \oplus MAC(a)),\\ (a||b||a \oplus MAC(a||b)), \\ (a||b||b \oplus MAC(a||b)), \\ (a||b \oplus MAC(a)||a \oplus MAC(b))\\ (a||b \oplus MAC(a)||b \oplus ...


2

If the last 16 bytes of the ciphertext are the padding, then you actually have the simple ECB (Electronic CodeBook) mode. ECB is secure as long as all your plaintexts are 1 block long and never repeat.



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