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13

We can't implement "AES 512 key size" because AES is defined for key sizes $k\in\{128,192,256\}$ bits only; much like we can't make a bicycle with 3 wheels. I see no reason why we would want to define an AES variant with 512-bit key size (since AES-128 is safe enough for anything foreseeable most current applications except those that require huge security ...


4

You could do a brute force attack where you simply try the keys with highest Hamming weight first (those with the most ones). I am not sure if you would call this attack practical but at least it would be much more likely to succeed quickly than brute force when the key is selected uniformly at random. Just consider the key of all ones, and assume we are ...


3

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 ...


3

There are two major problems with this method. The first problem is that Susan is likely to be able to recover significant amount of data from a series of such blocks. For example, if Susan knows $subkey_1$, then she could recover the value $subblock_1 \oplus subblock_2 \oplus subkey_2$; if a single block is encrypted with this key, she can't deduce ...


3

To answer your question, the cryptographic complexity will reduce by $2^{k-n}$ where $k$ is the key length and $n$ is the number of bits known. A slightly better scheme that doesn't use a cryptographic key sharing scheme can be as follows: Generate three random numbers $B_1$, $B_2$, $B_3$ with bit length equal to the size of your key $K_p$. Generate each ...


3

Just use Shamir's Secret sharing: wiki link It's designed by Adi Shamir, who is the "S" in RSA. It's fairly simple to use and there are no known weaknesses in it. While it doesn't split the data, splitting the key may work just as well (unless you absolutely must split the data.)


2

What they are trying to say that an effective way to multiply $a$ and $b$ in $GF(2^8)$ is to first check if either $a$ or $b$ are zero (if either are zero, the result of the multiplication is zero), and if neither are zero, then: $$a \otimes b = E( L(a) + L(b) \bmod 255 )$$ where $E$ and $L$ are lookups in the ETable, LTable given, and the addition is over ...


2

You have more than one question in your... question. user13741 already did answer the "is 2x AES-128 as secure as 1x AES-256?" part, so only a small summary: Plaintext is the message before it was encryptet, and ciphertext after it was encryptet. You can encrypt a known plaintext with all possible keys and save the result. Now you decrypt the ciphertext with ...


2

I believe this question is only answerable if $F_k$ is easily invertible. In other words, if you can compute $M=F^{-1}_k(F_k(M))$. Then a standard meet-in-the-middle attack applies. Given message $M$, ciphertext $C = E_k(M)$ for unknown $k \in \{0,1\}^{128}$, an efficiently-computable function $X$ such that $k = X(k_1, k_2)$ for some $k_1, k_2 \in ...


2

You misunderstand $(02) \cdot 10000100$; it is not integer multiplication (resulting in a 9 bit integer); instead, it is multiplication in $GF(2^8)$ (which results in an element in $GF(2^8)$, which can be represented in 8 bits). AES uses a polynomial representation of $GF(2^8)$, using the polynomial $x^8 + x^4 + x^3 + x + 1$; what this means is that ...


1

AES is secure in such a way that you cannot find the key even if you know (part of) the plaintext. So it is not possible (feasible) to deduce the key based on the file prefix to decrypt the complete file. I suspect this is the attack that you anticipate. Like woliveirajr said, you can just use the existing filename with a suffix like .enc. What you need to ...


1

If you are encrypting one file to one file, simply save the correct extension. Example: open test.pdf, encrypt the content, and save as test1.pdf, or test1.pdf.enc (so that you know that the file is encrypted and any pdf won't try to open it when you double click). If you are encrypting more than one file together (and, in the end, you have one big chunk of ...


1

There is nothing in the GCM cipher that prevents it's use it in streaming mode. You should however not use the resulting plaintext during decryption for anything that requires security before you have verified the authentication tag. The authentication tag is not to prevent you from decrypting the ciphertext. It is there to provide for integrity and ...


1

In both options, if the adversary has a way to check either AES key, then a brute password guessing attack can be attempted, and BCrypt is the main line of defense against that. For constant effort, option 2 force to halve the cost parameter in BCrypt, and is thus twice more vulnerable to password guessing than option 1 is. BCrypt's output is described as ...


1

This is pretty common. The method of encryption is public information, so it is safe to leave in the clear. In your post you made no mention of authentication. This is something you absolutely must use, otherwise you have no way of knowing if your data has been modified and leaves you open to many attacks. Consider using HMAC with SHA-256, or a block ...


1

Encrypting once with a 256-bit key is stronger than encrypting twice with two 128-bit keys. This is because of the Meet in the middle Attack. Encrypting once with a 256-bit key gives you 256 bits of security. Encrypting twice with two 128-bit keys gives you 129 bits of security.


1

This question has nothing to do with AES; instead, it has to do with an encoding that someone picked for this example (and the fact that you have questions about it indicates that it wasn't a great choice by whoever came up with the example). The encoding is simple; we translate the letters 'A' through 'Z' to the integers 0 to 25, which is in hexidecimal ...



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