# Tag Info

21

Let's assume that the plaintexts consist only of spaces and ASCII letters. Given the hint, that seems like a reasonable assumption to start with, even if it might turn out to be only mostly correct. Now, take one of the ciphertexts and XOR it with each of the others. Of course, the XOR operation cancels out the keystream, so you end up with the plaintext ...

6

I was/am assuming that for public key encryption, COA means "other than the public key, ciphertext only". Otherwise, any secure symmetric cipher with the key published becomes a "COA resistant" PKE scheme. With that in mind, access to an encryption oracle cannot possibly help an attacker, since the attacker can already encrypt any plaintext using the ...

6

If I understand right, your operation effectively is $$\forall i: c_i = p_i \oplus k_0 \oplus k_1 \oplus k_2 \oplus \dots \oplus k_n,$$ whith $c_i$ the ciphertext bits, $p_i$ the plaintext bits, and $k_j$ the key bits. As $\oplus$ (this is XOR) is associative, this is equivalent to $$k^* := k_0 \oplus k_1 \oplus k_2 \oplus \dots \oplus k_n,$$ \forall ...

5

Unless you are badly (and I mean truly badly) misrepresenting the idea, it is one of the worse ideas I've seen in crypto in quite some time. The first bit is effectively exclusive or'ed with the parity of the key; the ciphertext bit will be either the plaintext bit (if the key has an even number of '1' bits) or the complement of the first plaintext bit (if ...

5

Well, from your previous questions, I'm assuming that your writing a utility to brute-force decrypt a password protected file (encrypted with a certain encryption utility), and you're looking for a way to determine whether your trial decryption is plausible. Normally, when an attacker attempts to decrypt something, he has some idea about what it is (why ...

4

The usual assumption is that the attacker knows a full plaintext block; that's what the EFF DES-cracking machine uses. That machine knows exactly 8 consecutive plaintext bytes and the corresponding ciphertext block; it stops when it finds a matching key. Since there are 256 possible DES keys, and 264 possible 8-byte blocks, chances are high that there is ...

4

If you don't know the system, you just check one after the other: frequency analysis of bigrams detects Ceasar and Playfair. Try Caesar first then Playfair. Auto correlation method for Vigenere (for each x: count the number of occurances, where letter at position i and i+x are equal. For the correct codeword length, it will spike) If you have a Hill ...

4

Since for any fixed key the encryption algorithm is a bijection from the set of $n$-bit strings onto itself, the set of all possible ciphertexts is regardless of the key and algorithm always the set of all $n$-bit strings and does not provide any information.

3

I'm not exactly certain what answer you're looking for; I tried to cover all the obvious bases. Actually, with RSA, we generally assume that the attacker knows the public key (the modulus $n$ and the public exponent $e$); with that, he can encrypt as much plaintext as he cares to (by selecting a value $M$ and computing $M^e \bmod n$). So, in that sense, the ...

3

Simplified DES is a toy Feistel cipher with 16-bit 8-bit block and 10-bit key, and only two rounds, intended for educational purposes. Here is a preview of the original paper, and an implementation; another; yet another. If one knows one block of ciphertext, but nothing about the plaintext and key, the plaintext can not be guessed entirely: each of the ...

2

No, that's not really possible without blatant flaws of the implementation. Modern modes of operation of ciphers are resistent to attacks even if you know many pairs of plaintext and ciphertext - and the IV is public knowledge. Knowing it is the normal case. You also didn't mention what operation mode was used. Well, of course you could brute force the key, ...

2

In an effort to guess what the poster is saying, I tried to develop a "secure-as-possible" (but not really secure at all) cipher using only minor modifications of the above: $Key = {k_0, k_1 ... k_m}$ $Plaintext = {p_0, p_1 ... p_n}$ $Ciphertext = {c_0, c_1 ... c_n}$ $\forall i: i \le m \rightarrow c_i = p_i \oplus k_i$ The first $m$ bits of ciphertext ...

2

When you have a RSA key pair, it means that you know the private key (otherwise this is not "your" key pair). The private key format, normally, contains the two factors $p$ and $q$ (at least so it goes with PKCS#1). Even if all you have are the modulus $n$, the public exponent $e$ and the private exponent $d$, the factorization of $n$ can still be worked out ...

2

The answer is yes, primarily because it has been done. Linear B, Akkadian, Sumerian, and hieroglyphics all had no persons with knowledge of them for centuries, and yet we can translate them today.

2

A core assumption of natural languages is that conceptual frequency and the grammar to construct and articulate concepts is not evenly and randomly distributed; so pattern discovery can occur. However, two problems can quickly arise: If the language sample (or collection of samples) is too small, the patterns may not be distinct and consistent. Cretan ...

2

I had a look at classic transpositioning and substitution methods which lead me to encryptions like Vigenére, Autokey, Beaufort and so on. However, those are designed to work with non-numeric alphabets as far as my understanding goes. While most classical ciphers are applied to the usual alphabet (without different cases), they are not limited to that. ...

1

If your fully homomorphic scheme is asymmetric and deterministic, then, it is already broken. One could just encrypt the value one, $c_1 = E(pk, 1)$, then, adds $c_1$ to itself until $c_1 + c_1 + ... + c_1 == E(pk, a)$ or $c_1 + c_1 + ... + c_1 == E(pk, b)$ etc...

1

This is an interesting question. If the input set does not have any identity element (for both mul and add operations) then one cannot construct a field and subsequently there cannot be homomorphic operations. More clearly, if the input plaintext elements do not form a field (and their respective ciphertext elements do not form a field either) then there ...

1

I have a definition of COA-security in my head but I cannot find this definition (applied to public key cryptography) in the literature or reference books. Under it, textbook RSA is an example of a public key cryptosystem that is COA-secure but not CPA-secure. To be CPA-secure, a necessary but not sufficient property is that encrypting the same message ...

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