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Yes, all combinations of 0,1 of length 4. Interpret $(0,1,0,1)$ (as an example) as $z_0 = 0, z_1 = 1, z_2 = 0, z_3 = 1$. Then $z_4$ is given by the sum of these 4, so $z_4 = 0$ (as we work mod 2), so the sequence becomes $(0,1,0,1,0)$. Then $z_5 = z_4 + z_3 + z_2 + z_1 = 0$ mod 2, again, so we have $(0,1,0,1,0,0)$, etc. You go on till you find that the last ...

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Roughly your password is used to encrypt a MasterSecretKey. Then you use this MasterSecretKey with a symmetric algorithm to encrypt or decrypt your data (the disk sectors). They eight key slots in LUKS are eight different encryptions of the same MasterSecretKey under eight different passwords. See also this image: ...

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summmary In general, no. An attacker who has lots of ciphertext+plaintext pairs may never be able to reverse-engineer an algorithm from them. An attack may not even be able to distinguish which one of a large group of known encryption algorithms was used to generate those ciphertexts. However, various weaknesses in some algorithms and protocols are known. ...

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Key length is the length of the key. It's a term whose meaning has evolved over time; these days, it typically means length in bits. With digital symmetric ciphers, it's fairly simple, because those tend to have a key that's just a string of some number of bits, and any string of that length is a valid key. With RSA, it's more complicated - the key has a ...

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Did you know the algorithm ? This is the central question. In modern cryptography, the algorithm is public while the all secret lies in the key. Knowing the algorithm allows analyst to quantify the strength, or the intractability of the mathematical transformation. You alse have to make a distinction. Two main classes exist between Stream Cipher and Bloc ...

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if the adversary knows $m^{e_1}$, $m^{e_2}$ and $d_1e_2$, is it able to guess m? Obviously not; if he could, then they could break RSA. Here's the RSA problem, given $m^e$, $e$ (and the modulus $N$), recover $m$. Suppose that we can solve your problem, that is, we had a black box that, given $m^{e_1}$, $m^{e_2}$ and $d_1e_2$, could recover $m$. Then, ...

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