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1

Assuming our domain and codomain both have 4 elements, the uniform distribution of inputs to outputs means the function is injective. You seem to be using the standard technical term ‘uniform distribution’ in a confusing way. Normally the uniform distribution on a finite set $A$ means the probability distribution $P$ with $P(x) = 1/\#A$ for all $x \in A$, ...

-1

Assuming our domain and codomain both have 4 elements, the uniform distribution of inputs to outputs means the function is injective. No. This is not a cryptographic hash (what I call a pseudo random function). Simplistically: Hash -> avalanche effect -> bin collisions -> 37% rate -> non injective -> codomain =/= domain. You may be over thinking this. ...

1

It's simply due to a hash function acting (within it's block size $n$) as:- You'll notice that there is no possible 'A' output as there was a collision at 'C'. It's a version of the pigeon hole principle meaning that when two birds occupy one hole, you must have an empty hole remaining. Some of the output bins of a pseudo random function (hash extractor) ...

3

I didn't follow all of the question, but let's take a small example. Suppose $X$ is uniformly distributed in $\{1,2,3\}$, so that $\Pr[X = x] = 1/3$ for any $x \in \{1,2,3\}$. If we define \begin{equation*} f(1) = 2, \qquad f(2) = 3, \qquad f(3) = 2, \end{equation*} then we have \begin{align*} \Pr[f(X) = 1] &= 0, \\ \Pr[f(X) = 2] &= \...

0

You should pick a key-based key derivation function, such as HKDF-SHA256. Then: Generate a master key $k$ uniformly at random. Derive an Ed25519 seed $k_0$ from it by $$k_0 = \operatorname{HKDF-Expand}_k(\text{info=‘Ed25519 signing key’},\, \text{size=32 bytes}).$$ Derive an AES key $k_1$ from it by k_1 = \operatorname{HKDF-Expand}_k(\text{info=‘AES ...

1

Does deriving the public key from an RSA private key always yield the same result? Well, if $e$ is the correct public exponent, then any value of the form $e + k \cdot \text{lcm}(p-1, q-1)$ (for any integer $k$) is also a correct public exponent (in the sense that it would function equivalently when used with the same modulus). Of course, we usually select ...

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