If $G$ is a random graph with $n$ vertices and each edge included independently with probability $1/2$ then is it true that almost surely \[\chi(G) - \zeta(G) \to \infty\] as $n\to \infty$?

OPEN - $1000

The cochromatic number of $G$, denoted by $\zeta(G)$, is the minimum number of colours needed to colour the vertices of $G$ such that each colour class induces either a complete graph or empty graph. Let $\chi(G)$ denote the chromatic number.

If $G$ is a random graph with $n$ vertices and each edge included independently with probability $1/2$ then is it true that almost surely \[\chi(G) - \zeta(G) \to \infty\] as $n\to \infty$?

A problem of Erdős and Gimbel (see also [Gi16]). At a conference on random graphs in Poznan, Poland (most likely in 1989) Erdős offered \$100 for a proof that this is true, and \$1000 for a proof that this is false (although later told Gimbel that \$1000 was perhaps too much).

It is known that almost surely \[\frac{n}{2\log_2n}\leq \zeta(G)\leq \chi(G)\leq (1+o(1))\frac{n}{2\log_2n}.\] (The final upper bound is due to Bollobás [Bo88]. The first inequality follows from the fact that almost surely $G$ has clique number and independence number $< 2\log_2n$.)

Heckel [He24] and, independently, Steiner [St24b] have shown that it is not the case that $\chi(G)-\zeta(G)$ is bounded with high probability, and in fact if $\chi(G)-\zeta(G) \leq f(n)$ with high probability then $f(n)\geq n^{1/2-o(1)}$ along an infinite sequence of $n$. Heckel conjectures that, with high probability, \[\chi(G)-\zeta(G) \asymp \frac{n}{(\log n)^3}.\]

SOLVED

The cochromatic number of $G$, denoted by $\zeta(G)$, is the minimum number of colours needed to colour the vertices of $G$ such that each colour class induces either a complete graph or empty graph. Let $z(n)$ be the maximum value of $\zeta(G)$ over all graphs $G$ with $n$ vertices.

Determine $z(n)$ for small values of $z(n)$. In particular is it true that $z(12)=4$?

A question of Erdős and Gimbel, who knew that $4\leq z(12)\leq 5$ and $5\leq z(15)\leq 6$. The equality $z(12)=4$ would follow from proving that if $G$ is a graph on $12$ vertices such that both $G$ and its complement are $K_4$-free then either $\chi(G)\leq 4$ or $\chi(G^c)\leq 4$.

In fact there do exist such graphs - Bhavik Mehta found computationally that there is exactly one (up to taking the complement) graph on $12$ vertices such that both $G$ and its complement are $K_4$-free with chromatic number $\geq 5$. This graph was explicitly checked to have cochromatic number $4$, and hence this proves that indeed $z(12)=4$.

The values of $z(n)$ are now known for $1\leq n\leq 19$: \[1,1,2,2,3,3,3,3,4,4,4,4,5,5,5,6,6,6,6.\] (The only significant difficulty here is proving $z(12)=4$ - the others follow from easy inductive arguments and the facts that $R(3)=6$ and $R(4)=18$.) It is unknown whether $z(20)=6$ or $7$.

Gimbel [Gi86] has shown that $z(n) \asymp \frac{n}{\log n}$.

SOLVED

The cochromatic number of $G$, denoted by $\zeta(G)$, is the minimum number of colours needed to colour the vertices of $G$ such that each colour class induces either a complete graph or empty graph.

Let $z(S_n)$ be the maximum value of $\zeta(G)$ over all graphs $G$ which can be embedded on $S_n$, the orientable surface of genus $n$. Determine the growth rate of $z(S_n)$.

A problem of Erdős and Gimbel. Gimbel [Gi86] proved that
\[\frac{\sqrt{n}}{\log n}\ll z(S_n) \ll \sqrt{n}.\]
Solved by Gimbel and Thomassen [GiTh97], who proved
\[z(S_n) \asymp \frac{\sqrt{n}}{\log n}.\]

SOLVED

The cochromatic number of $G$, denoted by $\zeta(G)$, is the minimum number of colours needed to colour the vertices of $G$ such that each colour class induces either a complete graph or empty graph.

If $G$ is a graph with chromatic number $\chi(G)=m$ then must $G$ contain a subgraph $H$ with \[\zeta(H) \gg \frac{m}{\log m}?\]

A problem of Erdős and Gimbel, who proved that there must exist a subgraph $H$ with
\[\zeta(H) \gg \left(\frac{m}{\log m}\right)^{1/2}.\]
The proposed bound would be best possible, as shown by taking $G$ to be a complete graph.

The answer is yes, proved by Alon, Krivelevich, and Sudakov [AKS97].

OPEN

The cochromatic number of $G$, denoted by $\zeta(G)$, is the minimum number of colours needed to colour the vertices of $G$ such that each colour class induces either a complete graph or empty graph. The dichromatic number of $G$, denoted by $\delta(G)$, is the minimum number $k$ of colours required such that, in any orientation of the edges of $G$, there is a $k$-colouring of the vertices of $G$ such that there are no monochromatic oriented cycles.

Must a graph with large chromatic number have large dichromatic number? Must a graph with large cochromatic number contain a graph with large dichromatic number?

The first question is due to Erdős and Neumann-Lara. The second question is due to Erdős and Gimbel. A positive answer to the second question implies a positive answer to the first via the bound mentioned in [760].

SOLVED

The cochromatic number of $G$, denoted by $\zeta(G)$, is the minimum number of colours needed to colour the vertices of $G$ such that each colour class induces either a complete graph or empty graph.

Is it true that if $G$ has no $K_5$ and $\zeta(G)\geq 4$ then $\chi(G) \leq \zeta(G)+2$?

A conjecture of Erdős, Gimbel, and Straight [EGS90], who proved that for every $n>2$ there exists some $f(n)$ such that if $G$ contains no clique on $n$ vertices then $\chi(G)\leq \zeta(G)+f(n)$.

This has been disproved by Steiner [St24b], who constructed a graph $G$ with $\omega(G)=4$, $\zeta(G)=4$, and $\chi(G)=7$.