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7

Security is clearly broken if there is a polynomial-length period with non-negligible probability (where by this I mean if a random point falls in a cycle with a poly-length period with non-negligible probability). In order to find a preimage, just go forward until you get back to the starting point, keeping the previous value each time.


6

Let us first consider the problem without involving Shamir secret-sharing at all. Suppose that $n = 140$ and that the secret $\sigma$ is a 140-byte Twitter message. The space is thus restricted considerably, from all possible $256$ byte values to the printable characters permitted to be used in Twitter messages, and the distribution in this restricted space ...


4

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])


4

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).


3

I'm not sure I understand your question entirely. If there is only one possible message, then the ciphertext can be trivially decrypted simply by choosing this message. I'll assume instead that the ciphertext contains the shuffled bit pattern of a name chosen from a set of more than one name. The problem with bit shuffling is that the number of set bits ...


3

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 ...


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 ...


1

Ryan, the permutation IP is fixed so it is a table of only 64 entries, mapping the $i$-th position to the $P(i)$th position. Each entry $i$ (and $P(i)$) of the table is in the range $$0,1,\ldots,63$$ so 6 bits are enough to represent each, but a byte can also be used.



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