Induced colorful trees and paths in large chromatic graphs

In a proper vertex coloring of a graph a subgraph is colorful if its vertices are colored with different colors. It is well-known that in every proper coloring of a k-chromatic graph there is a colorful path Pk on k vertices. If the graph is k-chromatic and triangle-free then in any proper coloring there is also a path Pk which is an induced subgraph. N.R. Aravind conjectured that these results can be put together: in every proper coloring of a k-chromatic triangle-free graph, there is an induced colorful Pk. Here we prove the following weaker result providing some evidence towards this conjecture. For a suitable function f(k), in any proper coloring of a {C3, C4}-free f(k)chromatic graph there is an induced colorful path on k vertices. A special case of a result of the first author in [7] says that every triangle-free kchromatic graph G contains an induced path on k vertices. The following more general conjecture is attributed to N.R. Aravind in [2]. A path (or more generally a ∗Research supported in part by grant (no. K104343) from the National Development Agency of Hungary, based on a source from the Research and Technology Innovation Fund. †Research supported in part by grant (no. K104343) from the National Development Agency of Hungary, based on a source from the Research and Technology Innovation Fund.


Introduction
A special case of a result of the first author in [7] says that every triangle-free k-chromatic graph G contains an induced path on k vertices.The following more general conjecture is attributed to N.R. Aravind in [2].A path (or more generally a subgraph) in a proper coloring of G is called colorful if its vertices are colored with distinct colors.
Conjecture 1.In any proper coloring of any triangle free k-chromatic graph G there is an induced colorful path on k vertices.
The main result of [2] is the proof of Conjecture 1 for the case when G has girth k.One can easily see that Conjecture 1 cannot be extended from paths to other trees.Indeed, the following example shows that there are graphs of arbitrary large chromatic number with proper colorings that contain no colorful K 1,3 .For other similar problems on colorful paths see [3].
Example 1. ( [5,10]) Let SH n be the graph whose vertex set is the set of n 3 triples of [n] and where for 1 i < j < k < n, vertex (i, j, k) is adjacent to (j, k, ).Coloring (i, j, k) with j, we have a proper coloring containing no colorful K 1,3 and the chromatic number of SH n is unbounded.
However, if we drop the colorful condition then (according to a well-known conjecture of the first author and Sumner [6,12]) the existence of any induced subtree might be guaranteed in triangle-free graphs of sufficiently large chromatic number.If the trianglefree condition is strengthened, considering the family G 5 of graphs with no cycles of length three and four, then the induced tree conjecture becomes easy, in fact large minimum degree can replace the chromatic bound.
Theorem 1. (Gyárfás, Szemerédi, Tuza [8]).Let T k be a tree on k vertices.Then every graph in G 5 with minimum degree at least k − 1 contains T k as an induced subgraph.
Assume we have a proper coloring on G.The color degree cod G (v) is the number of distinct colors appearing on the neighbors of v and cod Our first result is the following "colorful" variant of Theorem 1.
Theorem 2. Let T k be a tree on k 4 vertices.Then every proper coloring of G ∈ G 5 with cod(G) 2k − 5 contains T k as an induced colorful subgraph.
A related subject is to find induced subgraphs in oriented large chromatic trianglefree graphs, for old and new results see [1].By a result of Chvátal [4], acyclic digraphs with no induced subgraph with edges (1, 2), (2, 3), (4, 3) are perfect.On the other hand, triangle-free digraphs with no induced subgraph with edges (1, 2), (3, 2), (3,4) exist with an arbitrary large chromatic number (see [9]).In [9] it was asked what happens for the directed P 4 = (1, 2), (2, 3), (3,4)?This was answered by Kierstead and Trotter [10] by constructing arbitrary large chromatic triangle-free oriented graphs without induced directed P 4 .They also proved that if the clique size of a graph is fixed and its chromatic number is large then in every proper coloring and with orienting edges from smaller to larger color, there is either an induced colorful star S k (a vertex with outdegree k) or an induced colorful directed path P k .Here we present a result in a similar vein.Theorem 3. Let k be a positive integer and T k be a tree on k vertices.There exists a function f (k) such that the following holds.If G ∈ G 5 with χ(G) f (k) then in any proper coloring of G and in any acyclic orientation of G there is either an induced colorful T k or an induced directed path P k .
Note that in Theorem 3 the orientation of T k is not prescribed (but P k is the directed path).Also, P k is induced but not necessarily colorful.However, if G is oriented so that for c(v) < c(w) we have (v, w) ∈ E(G), P k must be colorful as well.Selecting this acyclic orientation and T k = P k , we get from Theorem 3 the following weakened form of Conjecture 1.
Corollary 1.In any proper coloring of an f (k)-chromatic graph G ∈ G 5 , G contains an induced colorful path on k vertices.
To get closer to Conjecture 1 it would be very desirable to forbid only triangles (and allow four-cycles) in Corollary 1.It is worth considering the following problem.
Problem 1.Let k be a positive integer and T k be a tree on k vertices.Is there a function f (k) such that the following holds?If G is a triangle-free graph with χ(G) f (k) then in any proper coloring of G with χ(G) colors, there is an induced colorful T k .
Problem 1 seems certainly difficult since it contains the Gyárfás -Sumner conjecture.The case when T k is a path should be easier, it is weaker than Conjecture 1.Note that the condition that the proper coloring of G must use χ(G) colors, eliminates Example 1.In fact, Problem 1 has an affirmative answer for any k-vertex star with f (k) = k, since a k-chromatic graph must contain a vertex adjacent to all other color classes in any k-coloring and these neighbors form an independent set since G is triangle-free.

Proofs
Proof of Theorem 2. We construct an induced colorful T k by induction.For k = 4 we have two trees to construct from the condition that cod(G) 3.For each of the two trees the proof is obvious and we omit the details.
For the inductive step, assume v is a leaf of a tree T k with neighbor w in T k .Let T * be the tree T k − v.By induction we find T * as an induced colorful subgraph of G so w ∈ V (T * ) ⊆ V (G).By the condition on the color degree, w is adjacent to a set No edge of G goes from S to any vertex of T * − {w} that is at distance one or two from w in T * since G ∈ G 5 .There are at most k − 4 vertices of T * that are at distance at least three from w in T * and each of them sends at most one edge to S since G is C 4 -free.Thus at least one vertex in S is nonadjacent to any vertex of T * − {w} and it extends T * to a tree isomorphic to T k and it is an induced colorful subgraph of G.
Proof of Theorem 3. Let c be a proper coloring of G, G * is an orientation of G. Assume that we have an ordering < on V (G), then the forward color degree f cod G (v) is the number of distinct colors appearing on the neighbors of v that are larger than v, i.e.
We shall prove that the following function f (k) is suitable.
the electronic journal of combinatorics 23(4) (2016), #P4.46 where For k 3 the theorem holds with the first alternative: a 1-chromatic graph has a vertex, a 2-chromatic graph has an edge, 3-chromatic graphs without triangles have odd induced cycles of length at least 5 which must contain colorful induced P 3 .Thus we assume k 4.
Assume first that G has a subgraph G such that cod G (v) 2k − 5 for every v ∈ G .By Theorem 2 we find a colorful induced T k = P k in G .Now we may assume that G has no subgraph G such that cod G (v) 2k − 5 = d for every v ∈ G .Thus we can define an ordering π on V (G) by repeatedly selecting vertices v 1 , . . ., v i , . . .such that f cod G i (v i ) < d for all v i ∈ V (G), where G i is the subgraph of G induced by the vertices (V (G) \ {v 1 , . . ., v i−1 }).
The oriented graph G * can be written as G 1 ∪ G 2 where both graphs have vertex set V (G) and G 1 contains the edges of G * oriented forward with respect to π and G 2 contains the edges of G * oriented backward with respect to π.By the Gallai -Vitaver -Roy theorem ( [11], Ex. 9.9) we can find a directed path P t in one of the two G i -s with t = g(k).Indeed, for k = 4, g(4 for some i ∈ {1, 2}.Assume that P t is oriented forward in the ordering π.Lemma 1. P = P t contains an induced P k starting from the first vertex of P . Proof.For k = 4 the lemma is obvious since in a graph G ∈ G 5 any P 4 is induced.Assuming that it is true for some k − 1 4, consider the at most d − 1 = 2k − 6 forward edges from the starting point v ∈ P to other points w 1 , . . ., w s of P where s 2k −6.This partitions P − v into at most 2k − 6 disjoint paths, Q 1 = w 1 . . ., Q 2 = w 2 . . ., Q s = w s . . ., one of them, Q j , must contain at least g(k)−1 2k−6 = g(k − 1) vertices.By induction, Q j contains an induced P k−1 from its first vertex w j .No edge of G is oriented from Q j to v since the orientation is acyclic.Also, apart from vw j , no edge of G is oriented from v to Q j by the definition of Q j .Thus vP k−1 is the required induced P k .This proves the lemma.
The proof of Theorem 3 is now finished, observing that if P t is oriented backward in the ordering π, Lemma 1 should be used in "backward" version, stating that P = P t contains an induced P k ending in the first vertex of P .