A note on chromatic number and induced odd cycles

An odd hole is an induced odd cycle of length at least 5. Scott and Seymour confirmed a conjecture of Gyárfás and proved that if a graph G has no odd holes then χ(G) 6 22 ω(G)+2 . Chudnovsky, Robertson, Seymour and Thomas showed that if G has neither K4 nor odd holes then χ(G) 6 4. In this note, we show that if a graph G has neither triangles nor quadrilaterals, and has no odd holes of length at least 7, then χ(G) 6 4 and χ(G) 6 3 if G has radius at most 3, and for each vertex u of G, the set of vertices of the same distance to u induces a bipartite subgraph. This answers some questions in [17].


Introduction
Let G be a graph, and let k be an integer.A k-coloring of G is an assignment of k colors to the vertices of G such that adjacent vertices receive distinct colors.The chromatic number χ(G) of G is the minimum integer k such that G admits a k-coloring.We use ω(G) to denote the clique number of G which is the largest integer l such that G contains the complete graph K l as a subgraph.It is certain that χ(G) ω(G).But the difference χ(G) − ω(G) may be arbitrarily large as there are triangle-free graphs with arbitrary large chromatic number (see [7,23,15]), and furthermore, Erdős [8] showed that for every positive integers k and l there exists a graph G with χ(G) k whose shortest cycle has length at least l.
There are still quite a lot of families of graphs whose chromatic numbers are bounded by a function of their clique numbers.For instance, the Strong Perfect Graph Theorem [2] asserts that χ(G) = ω(G) if G contains neither odd cycles of length at least 5 nor their complements as induced subgraphs, and Vizing's theorem [24] together with Beineke's characterization [1] shows that χ(G) ω(G) + 1 if G contains none of the nine given graphs as induced subgraphs.As a natural question, one may ask, for a given family of graphs G, whether there is a function f such that χ(G) f (ω(G)) for each graph G ∈ G? If such a function f does exist, then we say that the family G is χ-bounded, and call f a binding function of G.In this literature, the family of perfect graphs is χ-bounded with a binding function f (x) = x, and the family of line graphs is χ-bounded with a binding function f (x) = x + 1.
For convenience, we say that a graph G induces a graph H if H is isomorphic to an induced subgraph of G. Let F be a family of graphs.A graph G is said to be F-free if it induces no member of F. For a finite family F, Erdős [8] shows that if F-free graphs are χ-bounded then F must contain a tree.Then, Gyárfás [10], and Sumner [22] independently, conjectured that F -free graphs are χ-bounded for every forest F .There are some partial results about this conjecture (see [4,12,11,13,14,18]).
A hole of a graph is an induced cycle of length at least four.Gyárfás [12] also proposed three conjectures on the relation between chromatic number and induced cycles in graphs.Let l be an integer of at least 4.
Scott [19] proved that for each l > 0, the family of graphs with neither odd holes nor hole of length at least l is χ-bounded.Chudnovsky, Robertson, Seymour and Thomas [3] confirmed Conjecture 1 on graphs of clique number at most 3 and showed that if G is {K 4 , odd holes}-free, then χ(G) 4 (note that if G is {K 3 , odd holes}-free, then it is bipartite).
In 2016, Scott and Seymour [20] proved that χ(G) 2 2 ω(G)+2 for {odd holes}-free graph G and thus confirmed Conjecture 1.As to Conjectures 2 and 3, Scott and Seymour [21] first proved that for each l > 0, every triangle-free graph with sufficiently large chromatic number contains holes of l consecutive lengths, and thus confirmed them on trianglefree graphs.Finally, Chudnovsky, Scott and Seymour [5] confirmed Conjecture 2, and Chudnovsky, Scott, Seymour and Spirkl [6] confirmed Conjecture 3.
We use G to denote the family of graphs that have neither triangles nor quadrilaterals, and have no odd holes of length larger than 5.
Robertson conjectured (see [16]) that the only 3-connected, internally 4-connected graph in G is the Petersen graph.This conjecture is true if the requirement on internally 4connectivity is replaced by cubic [16].Plummer and Zha [17] presented a counterexample to Robertson's conjecture, and posed a few new questions including (1) whether every such graph has bounded chromatic number?and (2) how close to being perfect are these graphs?
In this note, we prove that, for every graph G ∈ G, χ(G) 4 and χ(G) 3 if G has radius at most 3 (Theorem 6), which answers the first question of Plummer and Zha, and for each vertex u of G, the set of vertices of the same distance to u induces a bipartite subgraph (Lemma 4), which answers the second question in some sense.
We introduce some notations.Let S be a subset of V (G), and let x be a vertex of G.We use G[S] to denote the subgraph of G induced by S, and let N S (x) be the neighbors of x in S. Let x and y be two vertices of G.An xy-path refers to a path from x to y.We use d G (x, y) to denote the length of a shortest xy-path which is referred to as the distance between the two vertices.A cycle of length k is simply called a k-cycle.

Proof of the main results
Recall that G denotes the family of graphs that have neither C 3 nor C 4 , and have no odd holes of length at least 7. First, we have the following lemma on the structure of graphs in G.
Lemma 4. Let G be a graph in G, let u be an arbitrary vertex of G, and let Proof.Since G has no C 3 , L 0 = {u}, and G[L 1 ] is an independent set.So the conclusion holds for i = 0, 1. Suppose that, G[L i ] is bipartite for each 0 i h for some h 1.
Let H = G[L h+1 ], and suppose that H is not bipartite.Then, H has an odd cycle and thus a 5-cycle, say Among all paths of length h from u to v 1 or v 3 , we choose P to be a uv 1 -path, and P to be a uv 3 -path, such that P and P have the most common vertices.Let w ∈ L j be the last common vertex of P and P , let P w = x j x j+1 . . .x h be the segment of P from w to v 1 , and let P w = y j y j+1 . . .y h be the segment of P from w to v 3 , where x j = y j = w, x h = v 1 and y h = v 3 .Then, j h − 1, and thus C = wP w v 1 u 1 u 5 u 4 u 3 v 3 P w w is an odd cycle of length 2(h − j) + 5 7. Therefore, C has chords.
By the choice of P and P , each chord of C must be of the form x i y i for some j + 2 i h.Let i 0 be the largest index such that is an odd hole of length at least 7. Therefore, i 0 = h, i.e., v 1 v 3 ∈ E(G).
With the same arguments, we can show that a contradiction.Thus, Lemma 4 holds.
To prove our theorem, we need the following generalization of Brook's theorem by Gallai [9].
Theorem 5. ( [9]) Let G be a k-vertex-critical graph, and let V 1 be the set of vertices of degree k − 1 in G. Then every 2-connected induced subgraph of G[V 1 ] is either a complete graph or an odd hole.Now, we are ready to state and prove our theorem.Theorem 6.Let G be a graph in G.Then, χ(G) 4, and χ(G) 3 if G has radius at most 3.
Proof.Let u be an arbitrary vertex of G, and let L i = {u : d G (u, v) = i} for an integer i 0. By Lemma 4, the vertices with even distance to u induce a bipartite subgraph, and the vertices with odd distance to u induce a bipartite subgraph too.Therefore, χ(G) 4.
Next, we suppose that G has radius at most 3. Suppose that the conclusion does not hold.Let G be a counterexample in G with the smallest order, i.e., every proper subgraph of G is 3-colorable, and let k = max{i : L i = ∅}.Then, k 3, G is 4-vertex-critical, and δ(G) 3.
If k = 2, then χ(G) 3 as we may color L 2 ∪ {u} with two colors by Lemma 4, and color L 1 with the third color.So, we suppose that k = 3.
We will show that Then, we may color L 2 ∪ L 3 ∪ {u} with two colors, and color L 1 with the third color, and thus χ(G) 3. Suppose to the contrary that G[L 2 ∪ L 3 ] is not bipartite.Then, G[L 2 ∪ L 3 ] has odd cycles, and thus has 5-cycles.We choose C to be a We may assume that V (C) ∩ L 2 does not contain two nonadjacent vertices.For otherwise, let u 1 , u 3 ∈ L 2 by symmetry.Let P 1 = uw 1 u 1 and P 2 = uw 2 u 3 be two paths.
It follows that m 2, and if m = 2 then the two vertices in V (C) ∩ L 2 must be adjacent.
Suppose that m = 2, and let u 1 , u 2 ∈ L 2 by symmetry.Let x be a neighbor of u 4 in L 2 .Let P 1 = uw 1 u 1 , P 2 = uw 2 u 2 , and P 3 = ux x.It is clear that w 1 = w 2 , and so we suppose by symmetry that x = w 1 .Now, uw 1 u 1 u 5 u 4 xx u is an odd hole.
Finally, the only remaining case is that m = 1, i.e., For convenience, we relabel the 5-cycle C as v 0 u 0 u 1 u 2 u 3 v 0 , and let v 0 ∈ L 2 by symmetry.Let P = uw 0 v 0 , and let Recall that for a subset S of vertices and a vertex z, N S (z) denotes the neighbors of z in S.
Let B be the component of G[L 3 ] that contains u 0 .We will show that B is a cycle that can be labelled as u 0 u 1 . . .u 3q−1 for some integer q. (2) and for each integer i and each integer j ∈ {0, 1, 2}, We proceed to prove (2) and (3), and prove (3) by induction first.

By applying the same arguments as that used on
We have proved (3) for i ∈ {0, 1, 2, 3}.Suppose that (3) holds for i ∈ {0, 1, . . ., m}, where m 3. Let i = m + 1, and suppose, without loss of generality, that m ≡ 1 (mod 3).Then, Since G is finite, there must be an integer j such that u 0 u j ∈ E(G).Suppose that j ≡ r (mod 3).By (3), v 0 u 0 u j v 0 would be a C 3 if r = 0, and u 0 u 1 v 1 u j u 0 would be a C 4 if r = 1.Therefore, r = 2, and so B is a cycle of length j + 1 which is a multiple of 3.This completes the proof of (2).
Let H be the subgraph of G induced by the vertices of degree 3. Now, B is a 2connected subgraph of H, which is neither complete nor an odd hole.This contradiction to Theorem 5 shows that G[L 2 ∪ L 3 ] is bipartite.Now, by coloring L 2 ∪ L 3 ∪ {u} with two colors and coloring L 1 with the third color, we can complete the proof of Theorem 6.
Finally, we would like to mention that in [17], Plummer and Zha also conjectured that the graphs in G may have chromatic number at most 3.This problem is still open in general.
and N L 2 (u m−1 ) = {v 0 } by the inductive hypothesis.By applying the same argument to v 1 u m−3 u m−2 u m−1 u m v 1 as that used on C, we see thatN L 2 (u m+1 ) = {v 2 } and d(u m+1 ) = 3.Therefore, (3) holds for all i.