A Note on the Gyárfás–Sumner Conjecture

Abstract

The Gyárfás–Sumner conjecture says that for every tree T and every integer $t\ge 1$, if G is a graph with no clique of size t and with sufficiently large chromatic number, then G contains an induced subgraph isomorphic to T. This remains open, but we prove that under the same hypotheses, G contains a subgraph H isomorphic to T that is “path-induced”; that is, for some distinguished vertex r, every path of H with one end r is an induced path of G.

1 Introduction

The Gyárfás–Sumner conjecture says [2, 14]:

Conjecture 1.1

For every tree T, and every integer $t\ge 1$, if G is a graph with no clique of size t, and with no induced subgraph isomorphic to T, then its chromatic number is bounded.

This has been proved for a few families of trees (for instance [1, 3,4,5, 7, 8, 10], and see [9] for a survey), but remains open in general. However, if in its statement, we replace “no induced subgraph isomorphic to T” with “no subgraph isomorphic to T” then the statement becomes true and easy, since every such (non-null) graph G has a vertex of degree less than |T| (the number of vertices of T), and so has chromatic number at most |T|. We are concerned with something between these two.

Let T be a tree and let r be a vertex of T (we call (T, r) a rooted tree). A path-induced copy of (T, r) in G is an isomorphism from T to a subgraph of a graph G, such that, for every path P of T with one end r, the image of P under $\phi$ is an induced path in G. So, admitting a path-induced copy of (T, r) lies partway between having T as a subgraph and having T as an induced subgraph.

We will show here that Conjecture 1.1 is true if instead of excluding an induced tree we exclude a path-induced tree. Our main result is the following.

Theorem 1.2

For every rooted tree (T, r), and every integer $t\ge 1$, if G is a graph with no clique of size t, that admits no path-induced copy of (T, r), then its chromatic number is bounded.

We prove Theorem 1.2 in the next section, and give some further discussion in the final section.

2 The Proof

We denote the chromatic number of G by $\chi (G)$, and for $X\subseteq V(G)$, we write $\chi (X)$ for $\chi (G[X])$. We need the following lemma. It is a special case of a theorem of [1], but we give the proof since it is short. For $c\ge 1$, we say $X\subseteq V(G)$ is a c-creature of G if every vertex in X has fewer than c neighbours in $V(G)\setminus X$.

Lemma 2.1

For all integers $a,c\ge 0$, if G is a graph with $\chi (G)>ac$, and $X\subseteq V(G)$ is a c-creature with $\chi (X)\le a$, then $\chi (G\setminus X)=\chi (G)$.

Proof

Suppose that $\chi (G\setminus X)<\chi (G)$, and let $\kappa :V(G)\setminus X\rightarrow \{1,\ldots ,\chi (G)-1\}$ be a colouring of $G\setminus X$ with $\chi (G)-1$ colours. Since $\chi (X)\le a$, there is a partition $X_1,\ldots ,X_a$ of X into a stable sets. For $1\le i\le a$, let

$$\begin{aligned} J_i=\{(i-1)c+1,(i-1)c+2,\ldots ,ic\}. \end{aligned}$$

For each $v\in X_i$, choose $\kappa (v)\in J_i$ different from $\kappa (u)$ for each neighbour $u\in V(G)\setminus X$ of v (this is possible since $|J_i|=c$ and v has fewer than c neighbours in $V(G)\setminus X$). Thus, we have extended $\kappa$ to a $(\chi (G)-1)$-colouring of G, a contradiction. This proves Lemma 2.1. $\square$

If $\phi$ is a path-induced copy of some (T, r), we denote by $V(\phi )$ the vertex set of its image, that is, the set of all vertices $\phi (h)\;(h\in V(T))$. For integers $d\ge 2$ and $k\ge 1$, let $(T^k_d, r)$ be the rooted tree in which the root r has degree d, every vertex has degree d or 1, and every path from r to a leaf has length k. If u is a vertex, N(u) denotes the set of its neighbours. We need:

Lemma 2.2

Let $k\ge 1$, $d\ge 2$, and $\tau \ge 0$ be integers. Then there exists an integer K with the following property. Let G be a graph, such that $\chi (N(u))\le \tau$ for every $u\in V(G)$, and let $v\in V(G)$. Then either G admits a path-induced copy $\phi$ of $(T_d^k,r)$ with $\phi (r) = v$, or there exists a $(1+d+d^2+\cdots +d^{k-1})$-creature X of G with $v\in X$ such that $\chi (X)\le K$.

Proof

For each $j\ge 1$, define $c(j) = 1+d+d^2+\cdots +d^{j-1}$. Define $f(1) = 1$, and inductively for $j\ge 2$, define $f(j) = f(j-1)c(j-1)+\tau$. We will prove by induction on k that $K=f(k)$ satisfies the theorem.

Let G be a graph, such that $\chi (N(u))\le \tau$ for every $u\in V(G)$, and let $v\in V(G)$. If $k=1$, and v has at least d neighbours in G, then there is a path-induced copy $\phi$ of $(T_d^k,r)$ with $\phi (r) = v$; and otherwise $\{v\}$ is a d-creature. Thus, we may assume that $k\ge 2$ and that the result holds for $k-1$. Let $M(v)=V(G)\setminus (N(v)\cup \{v\})$. Choose $A\subseteq N(v)$ maximal such that for each $a\in A$, there is a path-induced copy $\phi _a$ of $(T_d^{k-1},r)$ such that

• $\phi _a(r) = a$ and $V(\phi _a)\subseteq \{a\}\cup M(v)$, for each $a\in A$; and

• the sets $V(\phi _a)\; (a\in A)$ are pairwise disjoint.

If $|A|\ge d$, then G admits a path-induced copy $\phi$ of $(T_d^k,r)$ with $\phi (r) = v$, as required, so we may assume that $|A|<d$. Consequently the union of the sets $V(\phi _a) \; (a\in A)$ has cardinality at most $(d-1)(1+d+d^2+\cdots d^{k-1})=d^k-1$. Let us denote this union by W.

For each $u\in N(v)\setminus A$, from the inductive hypothesis applied to the subgraph induced on $(\{u\}\cup M(v))\setminus W$, there is a $c(k-1)$-creature $X_u$ of this subgraph with $u\in X_u$ and with $\chi (X_u)\le f(k-1)$. Let X be the union of $\{v\}$ and all the sets $X_u\; (u\in N(v)\setminus A)$. We claim that X satisfies the theorem.

(1) X is a c(k)-creature.

First, since X contains all vertices of $N(v)\setminus A$, it follows that v has $|A|<d\le c(k)$ neighbours in $V(G){\setminus } X$. Every other vertex $x\in X$ belongs to one of the sets $X_u$ where $u\in N(v)\setminus A$, and so has fewer than $c(k-1)$ neighbours in $M(v)\setminus (X_u\cup W)$, and consequently has fewer than $c(k-1)$ in $V(G)\setminus (X\cup W)$. Moreover, it has at most $d^k$ neighbours in W because $|W|\le d^k$, and so has in total fewer than $c(k-1)+d^k=c(k)$ neighbours in $V(G)\setminus X$. This proves (1).

(2) $\chi (X)\le f(k)$.

For each $B\subseteq N(v)\setminus A$, let $X_B$ be the union of the sets $(X_b\setminus \{b\})\;(b\in B)$. Thus, $X_B\subseteq M(v){\setminus } W$. Suppose first that $\chi (X_{N(v){\setminus } A})> f(k-1)c(k-1)$, and choose $B\subseteq N(v)\setminus A$ minimal such that $\chi (X_B))> f(k-1)c(k-1)$. Since $B\ne \emptyset$, there exists $b\in B$; but $X_b\setminus \{b\}$ is a $c(k-1)$-creature of $G[X_B]$ with chromatic number at most $f(k-1)$, contrary to Lemma 2.1 and the minimality of B. This proves that $\chi (X_{N(v)\setminus A)})\le f(k-1)c(k-1)$. Since $\chi (N(v))\le \tau$, it follows that $\chi (X)\le f(k-1)c(k-1)+\tau =f(k)$. This proves (2).

From (1) and (2), X satisfies the theorem. This proves Lemma 2.2. $\square$

Now we can deduce Theorem 1.2, which we restate:

Theorem 2.3

For every rooted tree (T, r) and every integer $t\ge 1$, if G is a graph with no clique of size t and with sufficiently large chromatic number, then G admits a path-induced copy of (T, r).

Proof

We may assume that (T, r) equals $(T_d^k,r)$ for some choice of $d\ge 2$ and $k\ge 1$. We proceed by induction on t; and so may assume that there exists $\tau \ge 0$ such that if G is a graph with no clique of size $t-1$, and $\chi (G)>\tau$, then G admits a path-induced copy of (T, r). Choose f(k) as in the proof of Lemma 2.2. We claim that if G is a graph with no clique of size t that does not admit a path-induced copy of (T, r)i, then $\chi (G)\le (1+d+d^2+\cdots +d^{k-1})f(k)$. We may assume that G is critical with its chromatic number; that is, for every nonempty subset $X\subseteq V(G)$, $\chi (G\setminus X)<\chi (G)$. By Lemma 2.2, there is a $(1+d+d^2+\cdots +d^{k-1})$-creature X of G with $v\in X$ such that $\chi (X)\le f(k)$. Since G has no clique of size t, for every vertex v, G[N(v)] has no clique of size $t-1$, and so $\chi (G[N(v)])\le \tau$. Since $\chi (G\setminus X)<\chi (G)$, Lemma 2.1 implies that

$$\begin{aligned} \chi (G)\le (1+d+d^2+\cdots +d^{k-1})f(k). \end{aligned}$$

This proves Theorem 2.3. $\square$

3 Strengthenings

In this final section, we discuss possible strengthenings of Theorem 1.2.

One can refine Theorem 1.2 a little, and we first give a sketch of this. Start with a graph G with bounded clique number and large (really really huge!) chromatic number. For fixed k, d, we can apply Theorem 1.2 to get a path-induced copy of $(T_D^k,r)$ in G for some huge value of D, and then use Ramsey arguments to get a path-induced copy of $T_d^k$ where we have some control over the edges that stop this subgraph being induced. For instance, since there is a bound on the maximum size of a clique, each vertex of the tree that has children has a large (say size $D'$) set of children that is stable in G; and the set of vertices such that all their ancestors belong to the selected stable subsets forms a path-induced copy of $(T_{D'}^k,r)$ in which for every vertex, its set of children is stable in G. We can also arrange, using the bipartite Ramsey theorem, that for every two vertices of the tree at the same height, the children of the first are either completely adjacent, or completely nonadjacent, to the children of the second. And then we can get a path-induced copy of $(T_{D''}^k,r)$, for some huge $D''$, such that for each vertex, its set of grandchildren is stable in G. And so on: we can arrange that for each i, the set of vertices at distance i from the root is a stable set. (Let us call this being level-stable.)

We can also arrange, using the bipartite Ramsey theorem, that for every two vertices of the tree that are not leaves, even if their height is different, the children of the first are either completely adjacent, or completely nonadjacent, to the children of the second. (The argument here is tricker: it is important to fix up pairs in the right order, but we omit the details.)

But we can go further. Say two vertices u, v of the tree are incomparable if neither is an ancestor of the other; and let d(u, v) denote the distance between u, v in the tree. If u, v belong to the tree, let w be their “join” (their common ancestor furthest from the root), and let $a=d(u,r), b=d(v,r)$ and $c=d(w,r)$. The triple (a, b, c) describes the pair u, v up to isomorphisms of the tree. But we need a little more information. For each vertex, choose a linear order of its set of children. So now, if u, v are incomparable, then they descend from different children (say $u', v'$ respectively) of their join w, and one of these is earlier than the other in the linear order of the children of w. If $u'$ is earlier than $v'$ we say u is earlier than v. Let us say that the type of an unordered pair $\{u,v\}$ (where u, v are incomparable) is the triple (a, b, c) defined as before, where u is earlier than v.

Let us say a path-induced copy of $T_d^r$ in G is type-uniform if the adjacency of each incomparable pair of vertices depends only on their type; in other words, if $\{u,v\}$ and $\{u',v'\}$ have the same type, then they are both adjacent or both nonadjacent pairs. One can use more Ramsey arguments (we omit the details, which are straightforward) to arrange, again by reducing D, that the adjacency of each pair of vertices depends only on their type; in other words, if $\{u,v\}$ and $\{u',v'\}$ have the same type, then they are both adjacent or both nonadjacent pairs. In conclusion, then, we deduce:

Theorem 3.1

For all $k,t\ge 1$ and $d\ge 2$, if G is a graph with no clique of size t and with sufficiently large chromatic number, then G admits a path-induced, level-stable, type-uniform copy of $(T_d^k,r)$.

Here is another way in which we might strengthen Theorem 1.2: can we obtain polynomial bounds? There is an analogous problem where we exclude the complete bipartite graph $K_{t,t}$ as a subgraph, instead of excluding $K_t$, and the following was shown in [11]:

Theorem 3.2

For every tree T, there is a polynomial f(t) such that for every integer $t\ge 1$, if G has no induced subgraph isomorphic to T and no subgraph isomorphic to $K_{t,t}$, then G has average degree at most f(t).

This is an improvement of a result of Kierstead and Penrice [5], who proved that there is a function f(t) as in Theorem 3.2, not necessarily a polynomial; and that in turn was an improvement of a theorem of Rödl (see [4,5,6]), who proved the same with average degree replaced by chromatic number. Is there any hope for a comparable strengthening of Theorem 1.2?

Theorem 3.2 assumes that G does not contain $K_{t,t}$ as a subgraph. In Theorem 1.2 we replace this by the much weaker hypothesis that the clique number of G is bounded, although in compensation we must weaken the conclusion, replacing the bound on average degree with a bound on chromatic number. This change is necessary: $K_{n,n}$ has large minimal degree, but no $K_3$ and no path-induced copy of $P_4$.

But we could still ask for a polynomial bound on chromatic number. Indeed, it is possible that the Gyárfás–Sumner conjecture holds with polynomial bounds (in other words, Conjecture 1.1 with a bound on chromatic number that is polynomial in t). This has recently been shown for a few trees, including every tree that does not contain the five-vertex path as an induced subgraph [12]. However, the five-vertex path appears intractable. Here, the best current bound is slightly superpolynomial (see [13] for this and for related discussion):

Theorem 3.3

If G does not contain the five-vertex path $P_5$ as an induced subgraph, and has clique number t, then $\chi (G)\le t^{\log _2 t}$.

If P is a path and r is one end of P, then a graph G contains a path-induced copy of (P, r) if and only if it contains an induced copy of P. Thus, obtaining polynomial bounds in Theorem 1.2 even for $P_5$ would also require an improvement of Theorem 3.3.