A note on triangle-free graphs

We show that if $G$ is a simple triangle-free graph with $n\geq 3$ vertices, without a perfect matching, and having a minimum degree at least $\frac{n-1}{2}$, then $G$ is isomorphic either to $C_5$ or to $K_{\frac{n-1}{2},\frac{n+1}{2}}$.

💡 Research Summary

The paper investigates the structure of simple triangle‑free graphs under a strong degree condition combined with the absence of a perfect matching. The main theorem states that if a graph G on n ≥ 3 vertices satisfies three properties—(a) minimum degree δ(G) ≥ (n − 1)/2, (b) G does not contain a perfect matching, and (c) G is triangle‑free—then G must be isomorphic to one of two very specific graphs: the 5‑cycle C₅ or the complete bipartite graph K_{(n‑1)/2,(n+1)/2}.

The proof begins by invoking Ore’s theorem, which guarantees a Hamiltonian path whenever the minimum degree is at least (n − 1)/2. Since a perfect matching is missing, n must be odd. The authors then bound the maximum degree Δ(G). Assuming Δ(G) ≥ (n + 3)/2 leads to a contradiction: a vertex of maximum degree u would have at least (n + 3)/2 neighbours, leaving at most (n − 5)/2 other vertices. Because each neighbour also has degree at least (n − 1)/2, some two neighbours of u must be adjacent, forming a triangle with u, which violates condition (c). Hence Δ(G) ≤ (n + 1)/2.

Two cases are examined.

1. Δ(G) = (n + 1)/2. Let u be a vertex of maximum degree, V the set of its neighbours (size (n + 1)/2) and U the remaining vertices (size (n − 3)/2). The minimum‑degree condition forces every vertex in V to be adjacent to every vertex in U. Condition (c) forbids edges inside V and inside U. Consequently the graph is exactly the complete bipartite graph K_{(n‑1)/2,(n+1)/2}.

2. Δ(G) = (n − 1)/2. Here G is regular of degree r = (n − 1)/2. Since n is odd, r is even. The authors show that r cannot be 4 or larger. Selecting any edge uv, the remaining neighbours of u and v are disjoint because of the triangle‑free property. The single vertex w not in these neighbour sets must connect to a subset of the neighbours of u and v. Careful case analysis (whether w is adjacent to more neighbours of u or of v) always yields a forced triangle, contradicting (c). Therefore r must be 2, which forces n = 5 and makes G the 5‑cycle C₅.

Thus the combination of a high minimum degree and the lack of a perfect matching forces a triangle‑free graph into one of two extremal configurations. This result complements classical theorems such as Mantel’s bound on the number of edges in triangle‑free graphs and the Andr'asfai‑Erdős‑Sós theorem that links minimum degree to bipartiteness. The paper highlights how the additional restriction of “no perfect matching” dramatically narrows the possible structures, offering a clean classification that may inspire further investigations into degree‑constrained graphs with other forbidden subgraphs.