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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 ...


18

In short: You must authenticate the IV. Which particular attacks apply if you don't depends on the block cipher mode; I will give two common examples. In CTR mode, an attacker who fiddles with the IV can forge authenticated messages, but the content of the corresponding plaintext is beyond his control (since he doesn't know the key). Depending on the ...


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

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 ...


14

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 ...


12

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.


10

Both an AES-128 key (as defined by FIPS 197), and a TDES Keying Option 2 key (as defined by FIPS SP-800-67) are 128-bit bitstrings. Similarly, both an AES-192 key and a TDES Keying Option 1 key are 192-bit bitstrings. The differences are: In AES, all bits of a key matter to the result; in TDES, 1 bit out of 8 (the lower-order bit of each byte in ...


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 ...


8

None of Twofish, Serpent and AES are currently known as broken, so as far as security is concerned, you can use any of them. AES has a slight advantage because it's very widely used, so if it gets broken you're more likely to hear about it and get relevant software updates quickly. The Snowden postings haven't changed much as far as cryptography usage is ...


8

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 ...


8

It sounds like you're using a password-based key derivation function that accepts an optional salt input to convert a passphrase into an encryption key, which you then use to encrypt messages with a block cipher mode (or possibly some other type of stream cipher) that takes an IV or a nonce, and you want to know whether it's necessary to provide a salt to ...


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

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

I assume you mean AES-GCM. Nonces must be unique for any use of a key. Given that $n = H(k)$ is constant for constant key $k$, this implies that such a nonce may only be used once, ever. Nonce reuse is particularly catastrophic in GCM mode (as with any other CTR-based mode), as it causes the keystream to be identical. Essentially, you wind up with two (or ...


7

DES is slow in software because it was designed back in the early 70's even before the 8086 processor existed, and uses several bit oriented operations that are just not implemented efficiently in a processor with a word oriented instruction set. Its intended product was ASIC hardware designs, in which DES runs quickly. DES hardware processors are quite ...


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

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.


7

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 ...


6

SIV is a mode specially designed for this purpose. SIV-AES would be a good choice, but it has the same issues as AES-wrap; not many implementations. If you use a GCM you should make sure that the IV is unique (if your plaintext is ever not random you would otherwise be in problems). As for the password based key derivation function: yes, PBKDF2 is good, ...


6

What you are looking for is called white-box cryptography. In short white-box crypto aims to make an implementation of a cypher (for example AES) in such a way that it is impossible for an attacker to extract the key, even if the attacker (the user of the computer) has access to the source code and a debugger. Up till now all academic white-box ...


6

First, it's not said that AES is unbreakable, merely that none of the currently known attacks reduce the computational cost to a point where it's feasible. The current best attack on AES-128 takes 2^126.1 operations, if we had a computer (or cluster) several million times more efficient than any current computer and could operate at the thermodynamic ...


6

I add my whitebox AES implementation on GitHub in: C++ Java C++ version implements both Chow's (mixing bijections, input/output encodings, external encodings) and Karroumi's (dual AES in each column) whitebox AES scheme plus Billet's key recovery attack on both schemes. Java implements Chow's scheme only. PS: Due to low reputation I post links to ...


6

AES-CTR is a stream cipher, of a particular kind where the keystream is obtained by encryption of a counter. So the question reduces to: what are drawbacks of AES-CTR compared to other stream ciphers? The main ones compared to ChaCha20 are: Without hardware support, AES can fail to cache-timing attacks. Without hardware support, AES is slower. Without ...


6

DES is slow compared to AES including in hardware because for comparable security we must use 3DES, which triples the number of rounds per block, to 48 for 3DES versus 10, 12, or 14 for AES; DES's block size is 64 bits, half of AES's 128 bit; so when encrypting a sizable block of data, 3DES does more rounds that AES by a factor of 96/10, 96/12, or 96/14; ...


6

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 ...


6

Given the choice, it is preferable to use the block encryption operation of AES, since it often faster than block decryption (never slower AFAIK). For this reason, AES-CTR is defined to use the block encryption operation of AES exclusively; that's both for AES-CTR encryption and AES-CTR decryption, which are the same operation except for IV generation/input. ...


6

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 ...


6

In AES-256 and RSA-1024, the numbers refer to the key size. For SHA-256, it refers to the output size. Though there is no hard, fast rule. For example, in the SHA-3 family, there are hash functions with variable length output, SHAKE128 and SHAKE256. The numbers there refer to the security level. Then again you have secp256k1, the elliptic curve used, among ...


6

There is no NIST oversight here. The key size and the block size are two completely different parameters and issues. The only reason that you need a large block size is because bad things start to happen when you encrypt too many blocks. Specifically, for an $n$-bit block size, the birthday paradox kicks in at $\sqrt{2^n}$. So, for a 128-bit block, you need ...


6

In this answer I'm assuming that a key is used to encrypt more than one message. The first weakness is that CBC with fixed IV leaks if messages share a common prefix. The second weakness is that it makes a padding oracle attack much more severe. Consider a device that knows the key and decrypts a ciphertext you send to it. While it won't tell you the ...



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