Polynomial Bounds for Chromatic Number. IV: A Near-polynomial Bound for Excluding the Five-vertex Path

Abstract

A graph G is H-free if it has no induced subgraph isomorphic to H. We prove that a $P_5$-free graph with clique number $\omega \ge 3$ has chromatic number at most $\omega ^{\log _2(\omega )}$. The best previous result was an exponential upper bound $(5/27)3^{\omega }$, due to Esperet, Lemoine, Maffray, and Morel. A polynomial bound would imply that the celebrated Erdős-Hajnal conjecture holds for $P_5$, which is the smallest open case. Thus, there is great interest in whether there is a polynomial bound for $P_5$-free graphs, and our result is an attempt to approach that.

1 Introduction

If G, H are graphs, we say G is H-free if no induced subgraph of G is isomorphic to H; and for a graph G, we denote the number of vertices, the chromatic number, the size of the largest clique, and the size of the largest stable set by $|G|, \chi (G), \omega (G),\alpha (G)$ respectively.

The k-vertex path is denoted by $P_k$, and $P_4$-free graphs are well-understood; every $P_4$-free graph G with more than one vertex is either disconnected or disconnected in the complement [24], which implies that $\chi (G)=\omega (G)$. Here we study how $\chi (G)$ depends on $\omega (G)$ for $P_5$-free graphs G.

The Gyárfás-Sumner conjecture [10, 25] says:

1.1 Conjecture: For every forest H there is a function f such that $\chi (G)\le f(\omega (G))$ for every H-free graph G.

This is open in general, but has been proved [10] when H is a path, and for several other simple types of tree ([3, 11,12,13,14, 17, 19]; see [18] for a survey). The result is also known if all induced subdivisions of a tree are excluded [17].

A class of graphs is hereditary if the class is closed under taking induced subgraphs and under isomorphism, and a hereditary class is said to be $\chi$-bounded if there is a function f such that $\chi (G)\le f(\omega (G))$ for every graph G in the class (thus, the Gyárfás-Sumner conjecture says that, for every forest H, the class of H-free graphs is $\chi$-bounded). Louis Esperet [8] made the following conjecture:

1.2 (False) Conjecture: Let $\mathcal G$ be a $\chi$-bounded class. Then there is a polynomial function f such that $\chi (G)\le f(\omega (G))$ for every $G\in \mathcal G$.

Esperet’s conjecture was recently shown to be false by Briański, Davies and Walczak [2]. However, this raises the further question: which $\chi$-bounded classes are polynomially $\chi$-bounded? In particular, the two conjectures 1.1 and 1.2 would together imply the following, which is still open:

1.3 Conjecture: For every forest H, there exists $c>0$ such that $\chi (G)\le \omega (G)^c$ for every H-free graph G.

This is a beautiful conjecture. In most cases where the Gyárfás-Sumner conjecture has been proved, the current bounds are very far from polynomial, and 1.3 has been only been proved for a much smaller collection of forests (see [5, 15, 16, 20,21,22,23]). In [22] we proved it for any $P_5$-free tree H, but it has not been settled for any tree H that contains $P_5$. In this paper we focus on the case $H=P_5$.

The best previously-known bound on the chromatic number of $P_5$-free graphs in terms of their clique number, due to Esperet, Lemoine, Maffray, and Morel [9], was exponential:

1.4   If G is $P_5$-free and $\omega (G)\ge 3$ then $\chi (G)\le (5/27)3^{\omega (G)}$.

Here we make a significant improvement, showing a “near-polynomial” bound:

1.5   If G is $P_5$-free and $\omega (G)\ge 3$ then $\chi (G)\le \omega (G)^{\log _2(\omega (G))}$.

(The cycle of length five shows that we need to assume $\omega (G)\ge 3$. Sumner [25] showed that $\chi (G)\le 3$ when $\omega (G)=2$.) Conjecture 1.3 when $H=P_5$ is of great interest, because of a famous conjecture due to Erdős and Hajnal [6, 7], that:

1.6 Conjecture: For every graph H there exists $c>0$ such that $\alpha (G)\omega (G)\ge |G|^c$ for every H-free graph G.

This is open in general, despite a great deal of effort; and in view of [4], the smallest graph H for which 1.6 is undecided is the graph $P_5$. Every forest H satisfying 1.3 also satisfies the Erdős-Hajnal conjecture, and so showing that $H=P_5$ satisfies 1.3 would be a significant result. (See [1] for some other recent progress on this question.)

We use standard notation throughout. When $X\subseteq V(G)$, G[X] denotes the subgraph induced on X. We write $\chi (X)$ for $\chi (G[X])$ when there is no ambiguity.

2 The Main Proof

We denote the set of nonnegative real numbers by $\mathbb {R}_+$, and the set of nonnegative integers by $\mathbb {Z}_+$. Let $f:\mathbb {Z}_+\rightarrow \mathbb {R}_+$ be a function. We say

• f is non-decreasing if $f(y)\ge f(x)$ for all integers $x,y\ge 0$ with $y>x\ge 0$;

• f is a binding function for a graph G if it is non-decreasing and $\chi (H)\le f(\omega (H))$ for every induced subgraph H of G; and

• f is a near-binding function for G if f is non-decreasing and $\chi (H)\le f(\omega (H))$ for every induced subgraph H of G different from G.

In this section we show that if a function f satisfies a certain inequality, then it is a binding function for all $P_5$-free graphs. Then at the end we will give a function that satisfies the inequality, and deduce 1.5.

A cutset in a graph G is a set X such that $G\setminus X$ is disconnected. A vertex $v\in V(G)$ is mixed on a set $A\subseteq V(G)$ or a subgraph A of a graph G if v is not in A and has a neighbour and a non-neighbour in A. It is complete to A if it is adjacent to every vertex of A. We begin with the following:

2.1   Let G be $P_5$-free, and let f be a near-binding function for G. Let G be connected, and let X be a cutset of G. Then

$$\begin{aligned} \chi (G\setminus X)\le f(\omega (G)-1)+ \omega (G) f(\lfloor \omega (G)/2\rfloor ). \end{aligned}$$

Proof

We may assume (by replacing X by a subset if necessary) that X is a minimal cutset of G; and so $G\setminus X$ has at least two components, and every vertex in X has a neighbour in V(B), for every component B of $G\setminus X$. Let B be one such component; we will prove that $\chi (B)\le f(\omega (G)-1)+ \omega (G) f(\lfloor \omega (G)/2\rfloor )$, from which the result follows.

Choose $v\in X$ (this is possible since G is connected), and let N be the set of vertices in B adjacent to v. Let the components of $B\setminus N$ be $R_1,\ldots ,R_k, S_1,\ldots ,S_\ell$, where $R_1,\ldots ,R_k$ each have chromatic number more than $f(\lfloor \omega (G)/2\rfloor )$, and $S_1,\ldots ,S_\ell$ each have chromatic number at most $f(\lfloor \omega (G)/2\rfloor )$. Let S be the union of the graphs $S_1,\ldots ,S_\ell$; thus, $\chi (S)\le f(\lfloor \omega (G)/2\rfloor )$. For $1\le i\le k$, let $Y_i$ be the set of vertices in N with a neighbour in $V(R_i)$, and let $Y=Y_1\cup \cdots \cup Y_k$.

(1) For $1\le i\le k$, every vertex in $Y_i$ is complete to $R_i$.

Let $y\in Y_i$. Thus, y has a neighbour in $V(R_i)$; suppose that y is mixed on $R_i$. Since $R_i$ is connected, there is an edge ab of $R_i$ such that y is adjacent to a and not to b. Now v has a neighbour in each component of $G\setminus X$, and since there are at least two such components, there is a vertex $u\in V(G)\setminus (X\cup V(B))$ adjacent to v. But then $u\hbox {-}v\hbox {-}y\hbox {-}a\hbox {-}b$ is an induced copy of $P_5$, a contradiction. This proves (1).

(2) $\chi (Y)\le (\omega (G)-1) f(\lfloor \omega (G)/2\rfloor )$.

Let $1\le i\le k$. Since $f(\lfloor \omega (G)/2\rfloor )<\chi (R_i)\le f(\omega (R_i))$, and f is non-decreasing, it follows that $\omega (R_i)>\omega (G)/2$. By (1), $\omega (G[Y_i])+\omega (R_i)\le \omega (G)$, and so $\omega (G[Y_i])<\omega (G)/2$. Consequently $\chi (Y_i)\le f(\lfloor \omega (G)/2\rfloor )$, for $1\le i\le k$. Choose $I\subseteq \{1,\ldots ,k\}$ minimal such that $\bigcup _{i\in I}Y_i=Y$. From the minimality of I, for each $i\in I$ there exists $y_i\in Y_i$ such that for each $j\in I\setminus \{i\}$ we have that $y_i\notin Y_j$; and so the vertices $y_i\;(i\in I)$ are all distinct. For each $i\in I$ choose $r_i\in V(R_i)$. For all distinct $i,j\in I$, if $y_i,y_j$ are nonadjacent, then $r_i\hbox {-}y_i\hbox {-}v\hbox {-}y_j\hbox {-}r_j$ is isomorphic to $P_5$, a contradiction. Hence the vertices $y_i\;(i \in I)$ are all pairwise adjacent, and adjacent to v; and so $|I|\le \omega (G)-1$. Thus, $\chi (Y)=\chi (\bigcup _{i\in I}Y_i)\le (\omega (G)-1)f(\lfloor \omega (G)/2\rfloor )$. This proves (2).

All the vertices in $N\setminus Y$ are adjacent to v, and so $\omega (G[N\setminus Y])\le \omega (G)-1$. Moreover, for $1\le i\le k$, each vertex of $R_i$ is adjacent to each vertex in $Y_i$, and $Y_i\ne \emptyset$ since B is connected, and so $\omega (R_i)\le \omega (G)-1$. Since there are no edges between any two of the graphs $G[N\setminus Y], R_1,\ldots ,R_k$, their union (Z say) has clique number at most $\omega (G)-1$ and so has chromatic number at most $f(\omega (G)-1)$. But V(B) is the union of Y, V(S) and V(Z); and so

$$\begin{aligned} \chi (B)\le f(\omega (G)-1)+(\omega (G)-1)f(\lfloor \omega (G)/2\rfloor ) +f(\lfloor \omega (G)/2\rfloor ). \end{aligned}$$

This proves 2.1. $\square$

2.2   Let $\Omega \ge 1$, and let $f:\mathbb {Z}_+\rightarrow \mathbb {R}_+$ be non-decreasing, satisfying the following:

• f is a binding function for every $P_5$-free graph H with $\omega (H)\le \Omega$; and

• $f(w-1) + (w+2) f(\lfloor w/2\rfloor ) \le f(w)$ for each integer $w> \Omega$.

Then f is a binding function for every $P_5$-free graph G.

Proof

We prove by induction on |G| that if G is $P_5$-free then f is a binding function for G. Thus, we may assume that G is $P_5$-free and f is near-binding for G. If G is not connected, or $\omega (G)\le \Omega$, it follows that f is binding for G, so we assume that G is connected and $\omega (G)>\Omega$. Let us write $w=\omega (G)$ and $m=\lfloor w/2\rfloor$. If $\chi (G)\le f(w)$ then f is a binding function for G, so we assume, for a contradiction, that:

(1) $\chi (G)>f(w-1) + (w+2)f(m)$.

We deduce that:

(2) Every cutset X of G satisfies $\chi (X)> 2f(m)$.

If some cutset X satisfies $\chi (X)\le 2f(m)$, then since $\chi (G\setminus X)\le f(w-1)+ w f(m)$ by 2.1, it follows that $\chi (G)\le f(w-1)+(w+2)f(m)$, contrary to (1). This proves (2).

(3) If P, Q are cliques of G, both of cardinality at least w/2, then $G[P\cup Q]$ is connected.

Suppose not; then there is a minimal subset $X\subseteq V(G){\setminus } (P\cup Q)$ such that P, Q are subsets of different components (A, B say) of $G\setminus X$. From the minimality of X, every vertex $x\in X$ has a neighbour in V(A) and a neighbour in V(B). If x is mixed on A and mixed on B, then since A is connected, there is an edge $a_1a_2$ of A such that x is adjacent to $a_1$ and not to $a_2$; and similarly there is an edge $b_1b_2$ of B with x adjacent to $b_1$ and not to $b_2$. But then $a_2\hbox {-}a_1\hbox {-}x\hbox {-}b_1\hbox {-}b_2$ is an induced copy of $P_5$, a contradiction; so every $x\in X$ is complete to at least one of A, B. The set of vertices in X complete to A is also complete to P, and hence has clique number at most m, and hence has chromatic number at most f(m); and the same for B. Thus, $\chi (X)\le 2f(m)$, contrary to (2). This proves (3).

If $v\in V(G)$, we denote its set of neighbours by N(v), or $N_G(v)$. Let $a\in V(G)$, and let B be a component of $G{\setminus } (N(a)\cup \{a\})$; we will show that $\chi (B)\le (w-m+2)f(m)$.

A subset Y of V(B) is a joint of B if there is a component C of $B{\setminus } Y$ such that $\chi (C)> f(m)$ and Y is complete to C. If $\emptyset$ is not a joint of B then $\chi (B)<f(m)$ and the claim holds, so we may assume that $\emptyset$ is a joint of B; let Y be a joint of B chosen with Y maximal, and let C be a component of $B\setminus Y$ such that $\chi (C)> f(m)$ and Y is complete to C.

(4) If $v\in N(a)$ has a neighbour in V(C), then $\chi (V(C)\setminus N(v))\le f(m)$.

Let $N_C(v)$ be the set of neighbours of v in V(C), and $M=V(C)\setminus N_C(v)$; and suppose that $\chi (M)> f(m)$. Let $C'$ be a component of G[M] with $\chi (C')> f(m)$, and let Z be the set of vertices in $N_C(v)$ that have a neighbour in $V(C')$. Thus, $Z\ne \emptyset$, since $N_C(v),V(C')\ne \emptyset$ and C is connected. If some $z\in Z$ is mixed on $C'$, let $p_1p_2$ be an edge of $C'$ such that z is adjacent to $p_1$ and not to $p_2$; then $a\hbox {-}v\hbox {-}z\hbox {-}p_1\hbox {-}p_2$ is an induced copy of $P_5$, a contradiction. So every vertex in Z is complete to $V(C')$; but also every vertex in Y is complete to V(C) and hence to $V(C')$, and so $Y\cup Z$ is a joint of B, contrary to the maximality of Y. This proves (4).

(5) $\chi (Y)\le f(m)$ and $\chi (C)\le (w-m+1) f(m)$.

Let X be the set of vertices in N(a) that have a neighbour in V(C). Since C is a component of $B\setminus Y$ and hence a component of $G{\setminus } (X\cup Y)$, and a belongs to a different component of $G\setminus (X\cup Y)$, it follows that $X\cup Y$ is a cutset of G. By (2), $\chi (X\cup Y)>2f(m)$. Since $\omega (C)\ge m+1$ (because $\chi (C)>f(m)$, and f is near-binding for G) and every vertex in Y is complete to V(C), it follows that $\omega (G[Y])\le w-m-1\le m$, and so has chromatic number at most f(m) as claimed; and so $\chi (X)>f(m)$. Consequently there is a clique $P\subseteq X$ with cardinality $w-m$. The subgraph induced on the set of vertices of C complete to P has clique number at most m, and so has chromatic number at most f(m); and for each $v\in P$, the set of vertices of C nonadjacent to v has chromatic number at most f(m) by (4). Thus, $\chi (C)\le (|P|+1)f(m)= (w-m+1)f(m)$. This proves (5).

(6) $\chi (B)\le (w-m+2) f(m)$.

By (3), every clique contained in $V(B){\setminus } (V(C)\cup Y)$ has cardinality less than w/2 (because it is anticomplete to the largest clique of C) and so

$$\begin{aligned} \chi (B\setminus (V(C)\cup Y))\le f(m); \end{aligned}$$

and hence $\chi (B\setminus Y)\le (w-m+1)f(m)$ by (5), since there are no edges between C and $V(B)\setminus (V(C)\cup Y)$. But $\chi (Y)\le f(m)$ by (5), and so $\chi (B)\le (w-m+2) f(m)$. This proves (6).

By (6), $G\setminus N(a)$ has chromatic number at most $(w-m+2)f(m)$. But G[N(a)] has clique number at most $w-1$ and so chromatic number at most $f(w-1)$; and so $\chi (G)\le f(w-1)+(w-m+2)f(m)$, contrary to (1). This proves 2.2. $\square$

Now we deduce 1.5, which we restate:

2.3   If G is $P_5$-free and $\omega (G)\ge 3$ then $\chi (G)\le \omega (G)^{\log _2(\omega (G))}$.

Proof

Define $f(0)=0$, $f(1)=1$, $f(2)=3$, and $f(x)=x^{\log _2(x)}$ for every real number $x\ge 3$. Let G be $P_5$-free. If $\omega (G)\le 2$ then $\chi (G)\le 3=f(2)$, by a result of Sumner [25]; if $\omega (G) = 3$ then $\chi (G)\le 5\le f(3)$, by an application of the result 1.4 of Esperet, Lemoine, Maffray, and Morel [9]; and if $\omega (G)=4$ then $\chi (G)\le 15\le f(4)$, by another application of 1.4. Consequently every $P_5$-free graph G with clique number at most four has chromatic number at most $f(\omega (G))$.

We claim that

$$\begin{aligned} f(x-1) + (x+2)f(\lfloor x/2 \rfloor ) \le f(x) \end{aligned}$$

for each integer $x>4$. If that is true, then by 2.2 with $\Omega =4$, we deduce that $\chi (G)\le f(\omega (G))$ for every $P_5$-free graph G, and so 1.5 holds. Thus, it remains to show that

$$\begin{aligned} f(x-1) + (x+2)f(\lfloor x/2 \rfloor ) \le f(x) \end{aligned}$$

for each integer $x>4$. This can be verified by direct calculation when $x=5$, so we may assume that $x\ge 6$.

The derivative of $f(x)/x^4$ is

$$\begin{aligned} (2\log _2(x)-4)x^{\log _2(x)-5}, \end{aligned}$$

and so is nonnegative for $x\ge 4$. Consequently

$$\begin{aligned} \frac{f(x-1)}{(x-1)^4}\le \frac{f(x)}{x^4} \end{aligned}$$

for $x\ge 5$. Since $x^2(x^2-2x-4)\ge (x-1)^4$ when $x\ge 5$, it follows that

$$\begin{aligned} \frac{f(x-1)}{x^2-2x-4}\le \frac{f(x)}{x^2}, \end{aligned}$$

that is,

$$\begin{aligned} f(x-1) + \frac{2x+4}{x^2}f(x) \le f(x), \end{aligned}$$

when $x\ge 5$. But when $x\ge 6$ (so that f(x/2) is defined and the first equality below holds), we have

$$\begin{aligned} f(\lfloor x/2\rfloor )\le f(x/2)=(x/2)^{\log _2(x/2)}=(x/2)^{\log _2(x) - 1}=(2/x)(x/2)^{\log _2(x)} = (2/x^2)f(x), \end{aligned}$$

and so

$$\begin{aligned} f(x-1) + (x+2)f(\lfloor x/2\rfloor ) \le f(x) \end{aligned}$$

when $x\ge 6$. This proves 2.3. $\square$