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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 ...


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/...


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 ...


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

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.


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 ...



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