Connected non-isomorphic graphs and edge-Ramsey classes

PDF p. 1: Introduction, question attributed to Fox et al. [5]; PDF p. 4: References [5]

For a finite simple graph $H$ with at least one vertex, let its two-colour edge-Ramsey class consist of the finite simple graphs $G$ with at least one vertex such that every function $E(G)\to \{1,2\}$ has an edge-monochromatic ordinary copy of $H$. An ordinary copy is a subgraph isomorphic to $H$; it need not include every host edge between its vertices. A copy is edge-monochromatic if all its edges have the same colour, with the condition vacuous when it has no edges.

A graph is connected if every two of its vertices are joined by a path. Two graphs are isomorphic if a bijection of their vertex sets preserves and reflects adjacency; they are non-isomorphic if no such bijection exists. Does there exist a pair of connected non-isomorphic graphs whose two-colour edge-Ramsey classes are equal? A negative answer is equivalent to the assertion that every two connected non-isomorphic graphs have distinct two-colour edge-Ramsey classes.

Fox and coauthors posed this existence question in Question 4.3 using two-colour Ramsey equivalence. Their abstract reports the partial result that a clique has no other connected graph in its Ramsey-equivalence class. The article under study, dated October 5, 2026, reports the general question as open. No solution is asserted here; later literature has not been systematically reviewed, so this is a dated report rather than a certification of current worldwide status.

The article's introduction does not specify an edge-colour count in its question. The two-colour formulation above follows the definition in the cited Fox et al. paper and makes that question precise. Distinct classes means that some host belongs to one class and not the other.