Tag Info

New answers tagged

4

Yes. For fixed $a\in\mathbb Z/n\mathbb Z$ and $b\in(\mathbb Z/n\mathbb Z)^\ast$ — note that $b$ must be invertible modulo $n$, which need not necessarily be a prime (but if it is, invertibility is equivalent to $b\neq0$), the map $$ f\colon\;\mathbb Z/n\mathbb Z\to\mathbb Z/n\mathbb Z,\; r \mapsto a+br $$ is a bijection, hence it preserves uniform ...


7

If anything can distinguish a blob of random data from an encryption of some (known) data (other than any intentionally added data formats, like a header or a length difference), the encryption algorithm is called broken. (This is one of the variants of the ciphertext indistinguishability criterion for ciphers.) As there are several encryption algorithms ...


2

Let's assume that randInt gives you cryptographically random (independent) numbers in the range $[0, N)$. Then you can generate a larger number by taking two of those, $x_1$ and $x_2$ and adding $x_1 + x_2 \cdot N$ (which is just bit concatenation if $N = 2^n$). That's a random number in the range $[0, N^2)$, assuming the original numbers were random and ...


2

There is no such thing like a Gaussian distribution over [0,1]; this doesn't make any sense. So it is not clear what you have to begin with. However, if you have independent random values, you can generate a random bit by taking two values A and B and comparing them. E.g., if $A<B$ you set the bit to 0, otherwise you set the bit to 1. A sequence of 8 ...



Top 50 recent answers are included