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Usually the length of a one-time-pad key is the same length of the plain text and you can't decipher it (If you follow the instructions of the OTP: Random, used one time, and key length = plain text length). As Raoul722 said, it may be Vigenere Cipher or XOR. Maybe you could go for Index of Coincidence


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If you want the encryption to be information-theoretically secure, then you need an information-theoretically secure RNG. And therein lies the problem—how do you establish that a given RNG is information-theoretically secure? Science may say that there are some types of events that are physically unpredictable, but that by itself is insufficient to get a ...


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I am basing my answer on Cryptopals. The basic idea is that as {c0,c0+3,c0+6,…} have all been xor-ed with the same byte, the number of differing bits between c0 and c3 is the same as between p0 and p3. (this number is called the Hamming distance between two characters. Furthermore, the distance between [c0 c1 c2] and [c3 c4 c5] is the same as between [p0 ...


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The technique called "Angecryption" might be relevant to you. It allows you to encrypt a known plaintext to a given ciphertext under a given key. However, I'm not sure what file types it supports, and if the types supported are relevant to you. The example in the article is turns a picture of annakin skywalker into darth vader given the key "Anger = ...


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Short answer: @Stephan Touset's answer is of course correct: if this is just for your learning, then it doesn't matter, use whatever's convenient. Anything that's not this should be fine: *Mandatory XKCD The longer answer is that philosophers like to debate whether "truly random", as you put it, actually exists, and if it does, does it apply to ...


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You're never going to release this, right? If so, it doesn't matter. Use a pseudorandom stream from /dev/urandom and "pretend" that it's truly random for the sake of learning about the concepts involved.


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Thomas Pornin's last paragraph is right on, but this concept is still so frequently misunderstood that I wanted to add my 2 cents. From a layman's perspective, If you really had something that implemented a one-time-pad, what would it look like? It would necessarily have to involve a physical machine that generates a truly random key stream. That machine ...


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There is no difference. In fact, any quasigroup operation will do. Specifically, the only property we really need is that, for every ciphertext symbol $C$ and every plaintext symbol $P$, there exists one (and only one) key symbol $K$ such that encrypting $P$ with $K$ yields $C$. This implies that, as long as the actual key symbols are chosen uniformly at ...


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In a finite field $\mathbb{F}_q$, both maskings are perfectly secure, provided that $x \neq 0$ for the multiplicative masking. This is easy to see. The finite field $\mathbb{F}_q$ defines two groups: the additive group $\mathbb{F}_p^+$ (i.e., $\mathbb{F}_q$ equipped with addition) and the multiplicative group $\mathbb{F}_q^*$ (i.e., $\mathbb{F}_q \setminus ...



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