# Tag Info

43

In the "ideal cipher" model, the block cipher is a permutation of the space of input blocks, chosen uniformly among all such permutations. A plaintext that gets encrypted to itself is a fixed point for the permutation; it is expected that about 63.21% of all permutations have at least one fixed point (a permutation with no fixed point is called a derangement)...

21

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 ...

16

TL;DR No, the approach is not secure. Use a standard like CMAC instead. Or even better, check your AES accelerator module to see if it supports any AEAD modes of encryption like GCM, CCM, EAX. Long Version In order for a message authentication code (MAC) to be secure, an adversary with oracle access to the MAC (basically this means the adversary can send ...

16

Even if the 32 characters are completely random, they won't contain non-printable characters. Actually, there are only about 107 printable characters in ASCII (out of 256 values for a full byte) and that even includes the space character. So if all the printable characters are used, it would result to a security level of about $log_2(107^{32}) = 215$ bits, ...

16

There are two important differences between AES-128 and AES-256: AES-128 has 10 rounds, AES-256 has 14 The key expansion process (that is, how they generate subkeys) is different If your AES-128 encryption hardware just takes a plaintext block and a 128 bit key, and produces a ciphertext block, well, no, there's not much you can do. In this case, the ...

16

You have clarified the question as asking about whether replacing ShiftRows with a random byte permutation would strengthen AES against differential attacks. It would not. ShiftRows and MixColumns were carefully selected to work in tandem, such that every byte affects every other byte in the state within just two rounds. MixColumns ensures that every ...

16

If you "crack the AES", a very hypothetical assumption, then you could simply parse the WiFi packets and decrypt the messages that are being send. Cracking AES is however considered impossible. AES-256 is even considered secure against attacks that involve (somewhat less hypothetical but still unavailable) quantum computers. Note that even RC4, which was ...

15

No, this isn't an oversight. AES is a block cipher, which is a keyed permutation. Now if you have a permutation of, say, three elements there are e few permutations possible: a -> a b -> b c -> c but also: a -> b b -> c c -> a and a -> c b -> a c -> b and a -> c b -> b c -> a (there should be $6$ for $3!$, the ...

13

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.

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It actually leaks information. You are sending: Encrypted IV: $AES(k,IV)$ First ciphertext block of CBC: $AES(k, M_1 \oplus IV)$ Eavesdropper can observe whether the two blocks are equal, which happens iff $M_1$ is all zeroes.

11

In general the AAD itself is not required or won't change the security of the GCM mode of operation itself. It may however directly influence the security of the protocol in which GCM is deployed. For instance, you may have specific configurable parameters outside the ciphertext itself. These parameters may very well include: version number of the ...

10

There is no such thing as a 16 bit AES key. AES is a block cipher with a block size of 128 bits and a key size of 128, 192 or 256 bits. As a block cipher, AES can only encrypt 16 bytes (128) bits at a time. AES in itsef is not (CPA) secure as repetition of the plaintext would lead to repetitions of the ciphertext. To encrypt larger amounts of data, AES ...

9

What makes crypto code vulnerable to timing attacks is data dependent timing variations. Branching according to a round counter, or to the key size, does not create a vulnerability. Most implementations of AES make no branch according to key or data value, and supressing other branches won't help. The main source of data-dependent timing variations in AES ...

9

AES has a block-size of 128 bits in all its variants. The number in AES-128/192/256 is the key-size. Rijndael, the block-cipher that became AES, also supports 256 bit blocks, but that part was not standardized as AES. Since the block-size is 128 bits, GCM works exactly the same way for AES-256 as it does for AES-128.

9

CBC mode decryption for the first block is defined as: $$P_0 = IV \oplus D_k( C_0 )$$ where $P_0$ is the first plaintext block, $C_0$ is the first ciphertext block, and $D_k$ is the decryption by the block cipher using the key $k$. This can be rearranged as: $$IV = P_0 \oplus D_k( C_0 )$$ which allows you to reconstruct the IV, if you know the key, the ...

9

I assume that you mean the S-box. The answer is NO! Randomly chosen S-boxes are not good choices for differential and linear cryptanalysis. When Biham and Shamir presented differential attacks on DES, one of the things that they showed was that if you replace the S-boxes in DES with randomly chosen ones, then the differential attack becomes much more ...

9

The fastest block cipher is identity, which leaves input blocks completely unchanged. This is infinitely fast on all platforms; however, it is not secure. So maybe you want the fastest block cipher that still offers some given non-trivial level of security? Then it depends a lot on what you want to implement the block cipher on. With recent PC, you would ...

8

The Simplified AES uses the Galois field GF(16), which can be represented by polynomials over the F_2 field: hex ... bin ... polynomial 1 ... 0001 ... 1 2 ... 0010 ... x 3 ... 0011 ... x + 1 4 ... 0100 ... x^2 ... ... n ... abcd ... ax^3 + bx^2 + cx + d ... ... F ... 1111 ... x^3 + x^2 + x + 1 with multiplication defined as ...

8

Is there a key that makes AES the identity function? No, probably not. That would mean that exactly the right permutation would be chosen out of the almost infinite set of permutations that are possible. Is there a key that makes AES the identity function for certain inputs? That's more likely, but it won't be easy to find it. Is it known whether ...

8

You're missing a piece in your understanding of modern encryption. AES is a symmetrical block encryption cipher. It describes how to use a key (which can be 128, 192 or 256 bits) long to encrypt and decrypt a single block of fixed size (128 bits) of data. That's it. In order to have a complete encryption/decryption system, you need to couple it with ...

8

It depends how the “AES-128 encryption hardware units” you mention are actually defined. I've already encountered processors that allow to independently compute AES operations such as $\texttt{SubBytes}$ and $\texttt{MixColumns}$ – which are the same regardless the key size involved (128 or 256 bits). In that case: yes, it can speed up the calculation for ...

8

AAD has nothing to do with making it "more secure". The aim of AAD is to attach information to the ciphertext that is not encrypted, but is bound to the ciphertext in the sense that it cannot be changed or separated. (Conceptually, the MAC is computed over the AAD and the ciphertext together.)

8

It would appear that this is as a consequence of the Wassenaar Arrangement (more detailed explanation). Basically, because cryptography falls under certain laws regarding munitions and arms trafficking there are restrictions on the strength of the cryptography you are allowed to sell (if the product is not a mass market product). It would appear that the ...

7

I'll answer in order: Output size = input size That's correct, GCM uses CTR internally. It encrypts a counter value for each block, but it only uses as many bits as required from the last block. CTR turns the block cipher into a stream cipher. IV of any size For GCM a 12 byte IV is strongly suggested as other IV lengths will require additional calculations....

7

Yes, this is exactly what a message authentication code is for. Its job is to prevent an attacker from tampering with your message, or from forging completely bogus messages. For a secure MAC, it should not matter what these messages contain. (And no, a secure MAC cannot compromise your key; if it did, it would by definition not be secure, since an ...

7

Yes, AES-128 is intended to be the standard block cipher for building a secure and efficient symmetric cryptosystem using some block cipher operating mode, like CTR for encryption or GCM for authenticated encryption; efficiency can be particularly good when there is hardware support for AES and GCM. There might be better choices in the case at hand, like ...

7

If your software needs to decrypt the data and you want to prevent even those with physical access from decrypting without your software, you are basically out of luck. It is impossible to achieve purely in software, since even if a good white-box algorithm existed, an attacker could copy it into their software and be able to decrypt (without directly ...

7

If you aim to show the effect of the encryption, meaning a scrabled image you have to encrypt only the "image data" and keep the file structure unchaned. A simple tutorial can be found here: https://blog.filippo.io/the-ecb-penguin/. Author uses ppm files, but you can easily adapt to your file format. # First convert the Tux to PPM with Gimp # Then take the ...

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Is AES solvable in this way? In other words, will the algorithm eventually complete, producing the correct key? Almost yes. It will produce some correct key — there might be more than one. (It should quite plausibly be unique given "enough" plaintext-ciphertext samples, but this need not be the case in general.) Generally, computing the key in a known-...

7

The comments already have covered the two main points, but let me try to put it in the form of an answer. There are not (that we know) weak keys in AES, in the sense that you cannot formulate a routine $isWeak(key)$. However, there are weak ways to generate an AES key (i.e. bad randomness). An AES key is just a bit string of length $n$. That means that ...

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