This is a follow up on an older question.

We construct graph on the vertex set $\{0,1\}^n$ where $n$ is a positive integer. For $x,y \in \{0,1\}^n$ the Hamming distance of $x,y$ is the cardinality of the set $\{ i \in \{0, ..., n-1\} : x(i) \neq y(i)\}$ (i.e. we count the positions on which $x$ and $y$ do not agree).

Fix a positive integer $k \leq n$. Two distinct elements of $\{0,1\}^n$ form an edge if their Hamming distance is at most $k$ (so they are in some sense "close" to each other). We denote the resulting graph on $\{0,1\}^n$ by $H(n,k)$.

We say that a finite simple, undirected graph $G=(V,E)$ is *Hamming-representable* if there are positive integers $k\leq n$ such that $G$ is isomorphic to an induced subgraph of $H(n,k)$.

**Question.** Is every finite graph Hamming-representable?