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You have a typo, $x$ is also in $\mathbb{F}_2^4,$ since the Present Sbox is $4\times 4$ bits. The quantities $S(x),a,b,x$ all lie in $\mathbb{F}_2^4,$ and the inner products such as $\langle b,S(x)\rangle$ etc lie in $\mathbb{F}_2.$ We now treat the inner product values as in $\mathbb{N},$ so we can calculate $(-1)^{\langle b,S(x)\rangle+\langle a,x\rangle}.... 0 I think I've gotten the solution After reading the following posts: is it possible to get key in One time pad if key is reused? How does one attack a two-time pad (i.e. one time pad with key reuse)? Taking advantage of one-time pad key reuse? and the answer of @Emme above/below, I came out with this solution: Let M1 and M2 be the original messages. S1 ... 0 You know that the two ciphertext you are given are formed as $$C_1 = S_1 \oplus R\\C_2 = S_2 \oplus R$$ with the same$R$in both cases. So you can get rid of$R$by calculating $$C^\ast = C_1 \oplus C_2 = S_1 \oplus S_2$$ Then you know that there are only two words,$111$and$0000$. Thus, in$C^\ast1$can only occur if one sentence contains$A$while ... 3 The$48$-bit state$s_i$evolves per a Linear Congruential Random Number Generator, with modulus$n=2^{48}$, multiplier$a=\mathtt{5DEECE66D}_{16}$, and additive constant$b=\mathtt B_{16}$; that is$s_i=(a\cdot s_{i-1}+b)\bmod2^{48}$. We are given the OR$r_i$of its two upper bits for$i\in[1,81]$, or equivalently the$81$booleans$s_i\ge2^{46}$. Our goal ... 1 This is the pre-image attack on hash functions, i.e. pre-image attack : given a hash value$y$find a pre-image$x$such that$y=h(x)$. Good designed cryptographic hash functions have$\mathcal{O}(2^{n})$classical pre-image resistance for$n$-bit output. The classical attack tries all possible input space to find a$x$that hashes the target hash value. ... 0 I'm surprised nobody has mentioned the theoretical connection here. I'm unfortunately unqualified to provide this precise answer, but can mention what I've heard from my colleagues. A significant development in the last roughly twenty years in complexity theory has been the "Hardness vs Pseudorandomness" paradigm. This is based on formalizing the following ... 1 For weak ciphers, sure, for somewhat modern ciphers, e.g enigma, short of possible but not as efficient as other methods. for modern cryptography? No Machine Learning is a very broad field so obviously I can't give conclusive statements about what can't be done. But in general if we think of gradient decent techniques we need a notion of getting close to a ... 4 The best example of black-box, end-to-end learning of the type you describe in the literature is probably Greydanus' work on Learning the Enigma With Recurrent Neural Networks. They achieve functional key recovery for the restricted version of Enigma they study, but require much more data and computing power than traditional cryptanalysis of the same ... 0 The vulnerability exists in the presumption that one would seek the plaintext not the key. How many possible sequences of$2^8\$ can be a result of a key generator with a given entropy factor? How many orderings of such bytes will conform to the parameters of a given key generator? Even CSPRNGs and 'truly random' sources are vulnerable to such method; though ...