# Tag Info

## New answers tagged permutation

1

As DES doesn't have a random number generator as input (or anything else to draw entropy from), it cannot perform anything at random. DES itself is fully deterministic, i.e. for identical key and block of plaintext, it will always create an identical block of ciphertext. Fully deterministic doesn't mean that it cannot lose information or contain ...

0

Yes, a function $f$ is said to be negligible if for every polynomial function $p(n)$ there exits some constant $N$ such that $f(n) < \frac{1}{p(n)}$ for all $n > N$. If $\frac{1}{n!} < \frac{1}{p(n)}$ then $n! > p(n)$, for all polynomials $p(n)$ and suitable $N$ such that $n>N$. Thus, you'd only have to prove the second segment of the ...

2

Hint: you can notice that $n! > 2^n$ (except for very small $n$).

2

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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