# Tag Info

10

Yes, you are remembering correctly. Yes, this is a reasonable method to find the key length. The reason why this works is because, typically, the plaintext is not uniformly random. For instance, rather than a random bit-string, the plaintext might be some English text, encoded in ASCII. If $X,Y$ represent two random English letters, encoded in ASCII, ...

9

Within the DES block cipher itself, the XOR operation is used at two different places: On the input of S-boxes, XOR-ing 48 bits per round: 48 bits from a subkey (extracted from the 56-bit key), and 48 bits that are the output of expansion E. The 48-bit result forms the eight 6-bit inputs of the S-boxes. On the output of S-boxes, XOR-ing 32 bits per round: ...

6

The encryption scheme seems to be: re-use an existing 128-bit secret, originally used to unlock a read-prevention mechanism, as the 128-bit key; split the plaintext (data to protect from prying eyes) into 128-bit blocks; XOR each block with that 128-bit key. That approach is flawed. Two cardinal mistakes are made: Use XOR with a keystream that repeats. ...

5

What do you mean 'a one time pad with a password'? One time pads don't take passwords, they take samples of truly random data as long as the message. If what you're doing is taking the password, repeating it N times, and using that as if it were a random one-time pad, well, that can usually be broken even if you don't send a second message. If what you're ...

5

To answer your question: no this is not homomorphic encryption because one of the plaintexts is used unencrypted. There may be times when it is a useful property, but the only uses I know of it are to demonstrate the malleability of xor ciphers. To be a homomorphic encryption function, it should be possible to calculate the encryption of some function of ...

5

The inequality is obtained by a distance argument. Consider two points $X,Y$ on the real line. Taking another point $Z$, you have $|X-Z| + |Y-Z| \geq |X-Z+Z-Y| = |X-Y|$. Applying this "triangle" inequality to your equality 1, we have for any $z \in \mathbb{R}$, $\begin{array}{l} \bigl\lvert\Pr[A(x\oplus g(U_n))=1] - z\bigr\rvert + \bigl\lvert ... 4 XOR operations, fixed bit movements (as in taking the 2 topmost bits or concatenating bits etc.) and data dependent rotations form a functional complete set of operations. This means that you can realize any function between fixed length binary strings, including all possible blockciphers, using them. To show that these operations form a functional complete ... 4 We typically refer to a homomorphic cipher if we can take two ciphertexts and combine them in a way that has a predictible result on the plaintexts. In your example you have taken one ciphertext and one plaintext. Using a stream cipher correctly you should never have 2 ciphertexts encrypted with the same portion of a keystream. So, combining two ciphertexts ... 3 Occasionally, for instance in very constrained environment, it can be useful to use only a few cryptographic primitives for all processing. (When you only have a hammer everything looks like a nail.) In such environments, it may be useful to use key derivation function to derive stream to be used as a stream cipher, or use hash function as cipher and so on. ... 3 This is called an Even-Mansour cipher. Actually, for the differential cryptanalysis it does not matter what sort of difference you use, you only need that it propagates deterministically through linear transformations (whatever linearity means). In this case you use a difference modulo$2^{32}$: $$A \boxminus B \equiv (A-B)\pmod{2^{32}}.$$ You compute ... 3 For most purposes, you probably don't need to provide confidentiality protection for the version field. There may be some settings where the version field allow traffic analysis, and to analyze that, we'd need to know so much detailed information about your particular protocol that the question would probably become too localized to be a good fit for this ... 2 If you have known plaintext, namely one input file that is known in its entirety, this is trivial to break. So I'll explore methods that might lead to a break, if you don't know what's in the input file that was compressed. I suggest that you start by analyzing the DEFLATE stream format carefully (see also these handy notes). This will probably help you ... 2 So a fundamental property of multi-party anonymous systems is you are only anonymous out of the number of honest participants in the system. If the Stasi control everyone else at the dinner table and know they didn't send the message, then they know you did no matter what protocol you use. In your case with this ring topology, because only your two ... 2 The Hamming distance is more effective when you suspect that the plaintext has been XORed with some repeating keystream. That's because XOR works at a bit level, as does the Hamming distance. The Index of Coincidence is more effective when you suspect that the plaintext has been combined with some repeating keystream, where the combiner works ... 2 Apart from the obvious security vulnerability from using a simple XOR operation to convert between plaintext and ciphertext you also run into the massive security-flaw of each identical plaintext generates identical ciphertext. In any given message, if a constant cipher operation is used along with a constant key, the resultant ciphertext will be same for ... 2 A better way to solve your problem is: on the server, encrypt the document under a document encryption key (a unique key that's specific to that document). The document remains encrypted in storage encrypted under this document key. When a client requests the document, send the client a copy of the encrypted document, as well as an encryption of the ... 1 Collusion is a concern but unlikely in very large ad-hoc ring networks where each ring is a one-shot random walk of a suitably large and mostly trustworthy membership pool. Collision and congestion are problems though; read below. If the opponent(s) can determine where the message or the key to decipher the message came from, the poster is done for. ... 1 The answer is almost definitely no for what I think you want (question is still unclear). I was hoping someone else could give a definite no, but since they haven't I figured I'd write this up. Your exact secret sharing method probably needs to be fully described in the question. I'm going to assume that$a=a_1\oplus a_2\oplus\dots\oplus a_n$and similarly ... 1 There is an entire class of ciphers that do exactly this. They're called stream ciphers. At their most basic, they pretend to be a one-time pad. You give them a small key and they generate a stream of seemingly random bits. You then XOR these bits with your plain-text to make ciphertext. You can add bells and whistles to this basic behaviour but what I've ... 1 To allow effective cryptanalysis of a scheme, two things must be clear: first, our goal (e.g., what we want to keep secret, what we want to preserve the integrity of, etc.), and second, what the attacker(s) are allowed to know or do. So, our goal is to preserve the secrecy of$m_1$, I take it. So, the attacks will try to find it. Now, what can the ... 1 Once you've XORed two messages with the same secret value, the net result is the same as if you had XORed them with each other without using the secret at all. Given$plaintext_1$⊕$key$=$cyphertext_1$and$plaintext_2$⊕$key$=$cyphertext_2$, then$cyphertext_1$⊕$cyphertext_2$==$plaintext_1$⊕$plaintext_2\$. Because ...

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