Is there an exponential lower bound for the chromatic number?

253 Views Asked by At

Let $n$ be a positive integer. Define the Hamming distance $d_H(x,y)$ of $x,y\in\{0,1\}^n$ by $$d_H(x,y)=|\big\{i\in\{0,\ldots,n-1\}:x(i)\neq y(i)\big\}|.$$

For integers $n>1$ and $k$ with $1\leq k<n$ let $G_{n,k}$ be the graph defined on the vertex set $\{0,1\}^n$ such that two nodes $x,y$ are connected by an edge if and only if $d_H(x,y) =k$.

From small scale numerical experiments it appears that for $k$ even and linear in $n$, the chromatic number of $G_{n,k}$ increases exponentially with $n$. For odd $k$ it is straightforward to prove that the chromatic number is $2$.

Is there an exponential lower bound for the chromatic number $\chi(G_{n,k})$ when $k$ is $\lfloor n/2 \rfloor$ and even?

1

There are 1 best solutions below

4
On BEST ANSWER

Yes, it grows exponentially.

The forbidden intersection theorem, conjectured by Erdős and proved by Frankl and Rödl, states that if $\mathscr F$ is a family of subsets of $\{1,\dots,n\}$ such that $F,F' \in \mathscr F \implies |F\cap F'| \ne \lfloor n/4 \rfloor$ then $|\mathscr F| \le (2-\epsilon)^n$ for some constant $\epsilon>0$.

Frankl, P., & Rödl, V. (1987). Forbidden Intersections. Transactions of the American Mathematical Society, 300(1), 259–286. https://doi.org/10.2307/2000598

Let $k = \lfloor n/2 \rfloor$ and assume $k$ is even. For $F \subset \{1, \dots,n\}$, let $1_F \in \{0,1\}^n$ denote the vector defined by $1_F(i) = 1 \iff i \in F$. For $F, F' \subset \{1,\dots,n\}$ such that $|F| = |F'| = k$, observe that $d_H(1_F, 1_{F'}) = 2k - 2 |F \cap F'|$. Therefore $$d_H(1_F, 1_{F'}) = k \iff |F \cap F'| = k / 2 = \lfloor n/4\rfloor.$$ It therefore follows from the forbidden intersection theorem that any independent set of $G_{n,k}$ intersects the set of vectors of total weight $k$ in a set of size $\le (2-\epsilon)^n$. The chromatic number of $G_{n,k}$ is therefore at least $\binom{n}{k} / (2-\epsilon)^n$, which grows exponentially since $\binom n k \sim c 2^n / \sqrt n$.