A note on Brooks' theorem for triangle-free graphs

For the class of triangle-free graphs Brooks’ Theorem can be restated in terms of forbidden induced subgraphs, i.e. let G be a triangle-free and K1,r+1-free graph. Then G is r-colourable unless G is isomorphic to an odd cycle or a complete graph with at most two vertices. In this note we present an improvement of Brooks’ Theorem for triangle-free and rsunshade-free graphs. Here, an r-sunshade (with r ≥ 3) is a star K1,r with one branch subdivided. A classical result in graph colouring theory is the theorem of Brooks [2], asserting that every graph G is (∆(G))-colourable unless G is isomorphic to an odd cycle or a complete graph. Bryant [3] simplified this proof with the following characterization of cycles and complete graphs. Thereby he highlights the exceptional role of the cycles and complete graphs in Brooks’ Theorem. Here we give a new elementary proof of this characterization. Proposition 1 (Bryant [3]). Let G be a 2-connected graph. Then G is a cycle or a complete graph if and only if G − {u, v} is not connected for every pair (u, v) of vertices of distance two. Proof. Let G be a 2-connected graph of order n. If G is a cycle or a complete graph, then obviously G− {u, v} is not connected for every pair (u, v) of vertices of distance two. Hence, assume that G is neither a cycle nor a complete graph and that G − {u, v} is not connected for every pair (u, v) of vertices of distance two. Note that then there exists at least one vertex v of G with 2 < dG(v) < n − 1. Since G Australasian Journal of Combinatorics 26(2002), pp.3–9 is 2-connected, there exists at least one cycle in G. Now let C be a longest cycle in G. Assume C is not a Hamiltonian cycle of G. Since C is a longest cycle and G is connected, there exist vertices y, z of C and x ∈ V (G) − V (C), such that z is adjacent to x and y and x is not adjacent to y, i.e. distG(x, y) = 2. Now G−{x, y} is not connected and the 2-connectivity of G ensures besides the x− y-connecting path P1 via the remaining vertices of C the existence of a second x − y-connecting path P2, which is vertex disjoint from P1. But then by gluing the common end-vertices of P1 and P2 together we obtain a cycle C ′ of length greater than C — a contradiction to the special choice of C. Thus C = v0v1 . . . vn−1v0 is a Hamiltonian cycle. Now we consider a vertex vi with 2 < dG(vi) < n−1. Then there exists j ∈ {i−1, i+1} such that vi is without loss of generality adjacent to vj but not adjacent to vj−1. Since G− {vi, vj−1} is not connected, we obtain that vj is not adjacent to vj−2. Therefore G − {vj, vj−2} is not connected. Thus dG(vj−1) = 2, and in particular vj−1 is not adjacent to vj+1. But now finally, since G−{vj−1, vj+1} is connected, we immediately achieve a contradiction, which completes the proof of this proposition. Theorem 2 (Brooks [2]). Let G be neither a complete graph nor a cycle graph with an odd number of vertices. Then G is ∆(G)-colourable. In the recent book of Jensen and Toft [5], (Problem 4.6, p. 83), the problem of improving Brooks’ Theorem (in terms of the maximal degree ∆) for the class of triangle-free graphs is stated. The problem has its origin in a paper of Vizing [7]. The best known (non-asymptotic) improvement of Brooks’ Theorem in terms of the maximal degree for the class of triangle-free graphs is due to Borodin and Kostochka [1], Catlin [4] and Kostochka (personal communication mentioned in [5]). The last author proved that χ(G) ≤ 2/3(∆(G) + 3) for every triangle-free graph G. The remaining authors independently proved that χ(G) ≤ 3/4(∆(G) + 2) for every triangle-free graph G. For the class of triangle-free graphs, Brooks’ Theorem can be restated in terms of forbidden induced subgraphs, since triangle-free graphs G satisfy G[NG[x]] ∼= K1,dG(x) for every vertex x of G. Theorem 3 (Triangle-free Version of Brooks’ Theorem) Let G be a triangle-free and K1,r+1-free graph. Then G is r-colourable unless G is isomorphic to an odd cycle or a complete graph with at most two vertices. Our main theorem will extend this triangle-free version of Brooks’ Theorem. An r-sunshade (with r ≥ 3) is a star K1,r with one branch subdivided. The 3-sunshade is sometimes called a chair and the 4-sunshade a cross. Proposition 4 Let G be a triangle-free and chair-free graph; then χ(G) ≤ 3. Moreover if G is connected, then equality holds if and only if G is an odd hole. Proof. Let G be a triangle-free and chair-free graph. Without loss of generality let G be a connected graph. If G is bipartite, then χ(G) ≤ 2 and we are done. So let G be a non-bipartite graph. With a result of König that every non-bipartite graph contains an odd cycle and the clique size constraint we deduce that G contains an odd hole C. If G ∼= C, then χ(G) = 3. If G ∼= C, then there exists a vertex