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4

Another, more indirect take on this: because of the semantic security requirement, we evaluate ciphers by their ability to resist an adaptive chosen-plaintext attack—where the attack not only sees some plaintext/ciphertext pairs, but also: Gets to choose which plaintexts they wish to see ciphertexts for; Gets to use the knowledge they gain from earlier ...

16

If the user changes the key for every message sent, then what use is a known-plaintext attack? Stop right there. This is not what we are trying to prove when conducting a known-plaintext attack. A known-plaintext attack is one where we are given a bunch of ciphertexts, all stemming from encryption using a fixed key. We are then given one plaintext/...

20

if a cipher has a known-plaintext attack, then it is considered completely broken. Yes, pretty much... [Paraphrased] But can't we come up with a case where this isn't true, such as a One Time Pad? Yes, we can come up with cases like that; however the requirements of such a case (key as long as the plaintext, no key reuse) make such a cipher ...

1

Here's an analogue of your question: What is the use of cars which can seat six people when there are only four people in my family? If a key is not used more than once, then you do not need security against chosen plaintext attacks. But other (most) people do need it.

0

Well, as the author of the answer you cited, I can't much improve on tylo's answer. However, perhaps a couple of examples will be an useful supplement. First, note that for $a,b \in \mathbb{Z}_4$: $$a \oplus b \equiv a + b + 2ab \pmod 4$$ Where the $\oplus$ on the left side means XOR and on the right side are addition and multiplication modulo 4. In ...

2

Sicne your question is refering to an answer on this site, I will not quote the entire answer here. But the crucial point is there in the last paragraph: If none of that made any sense to you, that's OK. You just need to read up about Abstract Algebra, and Fields, Rings and Groups. It's a fascinating and beautiful area of mathematics, and much of ...

1

Actually, as you note, XOR doesn't satisfy the bilinear condition. What does is multiplication $\otimes$ in $\mathbb{F}_2$, that is: $$\otimes(u \oplus v, w) = \otimes(u,w) \oplus \otimes(v,w)$$ or, as more traditionally expressed (using $+$ and $\times$ as the field operations): $$(u+v) \times w = (u \times w) + (v \times w)$$ This is one of the ...

1

If solutions of more or fewer than $m$ vectors are also admissible, this is easy: XOR of $n$-bit strings can be rephrased as addition in the vector space $(\mathbb Z/2\mathbb Z)^n$. You are looking for a solution to the linear equation $$a_1v_1+\dots+a_mv_m=y$$ where $v_1,\dots,v_m\in (\mathbb Z/2\mathbb Z)^n$ are all the public bit vectors, \$y\in(\mathbb ...

1

To answer your question, I obfuscated 1MB of data that consisted of a single 1 followed by all 0's, using your technique, and fed the results to ent: Entropy = 0.000039 bits per byte. Optimum compression would reduce the size of this 1048576 byte file by 99 percent. Chi square distribution for 1048576 samples is 267385856.00, and randomly ...

-1

I would agree with @Biv regarding the strength of XOR encryptions are generally linked to the key. i.e. they are only as strong as the key stream.Conventionally, XOR encryption, relies on bitwise exclusive OR operation to generate the ciphertext since it is hardware efficient. In LTE, ciphering of user data takes place in the Packet Data Convergence ...

1

Some vocabulary (to answer your comments): Obfuscation is used in computer science to hide source code while maintaining it executable see here. The idea is to hide the source code and make it hard to copy, disassemble. Steganography is to hide a message such as the attacker does not know its existence. By having a encrypted file, this defeat this purpose ...

2

It sound like you're using something like an XOR cipher to obfuscate your code. It will appear to be encrypted, but this can still be broken by frequency analysis since the use of a constant shift means that the encryption effectively has no key. An example of how to break a similar XOR cipher can be found here. As @iismathwizard mentioned, decryption ...

3

All of the mathematical operations within the s-boxes, shift row, and mix column should be known to the attacker, correct? Yes, see Kerckhoff's principle I understand they could be hard to calculate but aren't they static operations for the most part? I'm interpreting "static operations" to mean subBytes + shiftRows + mixColumns + addRoundKey, and that ...

4

The XOR state is irreversible without the proper key which is what I understand, so whats the point of all of the other operations that happen on the key? Suppose all we had was secret keys and the XOR operation. Well, actually, it is possible to build a secure cipher out of that, called the one time pad. One time pads offer perfect secrecy, but suffer ...

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