# Tag Info

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In the substitution step of a typical Substitution-Permutation Network (e.g. in AES SubBytes), the whole state is broken in parts and each part substituted. That's not the case in (the core of) a Feistel cipher, where at each step/round some sizable part of the state is bound to remain unchanged (in order that each step be reversible).

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If you can generate uniform random numbers, you can use a variant of Fisher-Yates. //given an array s with the elements to be permuted for i from n-1 to 1: t = rand(0, i) # inclusive swap(s[i], s[t])

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Assuming that the probability distributions of $\pi_{k_1}$ and $\pi_{k_2}$ are both uniform (that is, each permutation can take on any particular setting with probability $1/n!$), then no, adversary does not have enough information to learn anything about the original positions. This remains true even if we assume the adversary can perform unbounded ...

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Are there many permutation polynomials in a field? For a field $F$ of order $q$, every function from $F$ to $F$ is expressable (uniquely) as a polynomial of order $q-1$. $q!$ of these will represent permutation polynomials. Is there anyway to pick uniformly at random permutation polynomial in a field of prime order? Given a function from $f$ with $f(x_i) ... 2 Hint: you can notice that$n! > 2^n$(except for very small$n$). 1 A while ago, I spent time playing with modern field/pen & paper ciphers, especially with Card-Chameleon. Card-Chameleon needs two separate full alphabet permutations as a key. As it's a field/pen & paper cipher, I tried to find a computer-less, math-less, way to generate these permutations from passwords. My solution is a two steps process. Let's ... 1 What is meant by vectors here? Just the inputs and outputs of the function. The function$f\$ takes, as input, a sequence of bits (for example, 1010), and returns a sequence of bits (for example, 1100); the text refers to such a sequence of bits as a vector Can someone explain this? Well, the idea is that every output bit depends on all input ...

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AES transformation can be viewed as a sequence of invertible transformations each processing only a small part of the state. All these transformations would be even, and so is the entire AES for any key (see also this question).

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In general, the key length and number of rounds are the dominant factors in deciding cipher strength. But you need to consider how the rounds are constructed and how the key is used. Substitution and permutation are the bread and butter of DES. That's literally all it is - substitution, permutation, and XOR. Here is a diagram of the DES fiestel function ...

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