# Conjecture on representing graphs within $\{0,1\}^n$

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 $$H(n,k)$$.

Conjecture: If $$G=(V,E)$$ is a simple, undirected subgraph with $$|V|=n$$, then $$G$$ is isomorphic to some induced subgraph of $$H(n,1)$$ or $$H(n,2)$$.

Is this conjecture true?

For $$n\geq 8$$, $$K_n$$ with one edge removed is not such an induced subgraph. It's clearly not a subgraph of $$H(n,1)$$ since the latter is bipartite.
Suppose that graph is an induced subgraph of $$H(n,2)$$, say it's spanned on strings $$x_1,\dots,x_n$$ with $$x_{n-1},x_n$$ not connected. Observe we may assume $$x_1$$ is the zero sequence, so the strings hence can be identified with at-most-two-element subsets of $$[n]$$ with pairwise symmetric differences of size at most two. First note that at least one of the sets has two elements (otherwise the induced graph would be complete), say $$x_2$$. But then at most two sets have exactly one element, ones corresponding to elements of $$x_2$$, let's say those are $$x_3,x_4$$ (we should also check for the case when $$x_{n-1}$$ or $$x_n$$ is one of those sets, but that case is easier). A little case work shows that if we have any four or more two-element sets which are pairwise intersecting, then they necessarily have a common element. Applying this to $$x_2,x_5,\dots,x_{n-2},x_{n-1}$$ and $$x_2,x_5,\dots,x_{n-2},x_n$$ have a common element. It follows $$x_{n-1},x_n$$ have a common element too, so they are connected in the graph. We need $$n\geq 8$$ here for there to be enough vertices, but this can be easily improved.
I believe a similar reasoning can be used to show that for a fixed $$k$$ and large enough $$n$$, $$K_n$$ without an edge is not an induced subgraph of $$H(n,k)$$ -- I can show this for $$k=3$$ (I won't bother writing that down in detail since it's a lot of case work), but for general $$k$$ we need some general kind of general result that if we have a family of set with large pairwise intersections, then there is some "core" contained in all of them.