Coloring graphs with no induced subdivision of $K_4^+$

TL;DR

This paper proves that graphs with a large gap between chromatic number and clique number necessarily contain an induced subdivision of a specific graph called $K_4^+$, advancing understanding of graph coloring constraints.

Contribution

It establishes a new structural property linking high chromatic number to the presence of an induced subdivision of $K_4^+$ in graphs.

Findings

Graphs with high chromatic number relative to clique number contain an induced $K_4^+$ subdivision.

Provides a structural characterization connecting coloring and induced subgraphs.

Enhances understanding of the relationship between chromatic number and graph subdivisions.

Abstract

Let $K_4^+$ be the 5-vertex graph obtained from $K_4$, the complete graph on four vertices, by subdividing one edge precisely once (i.e. by replacing one edge by a path on three vertices). We prove that if the chromatic number of some graph $G$ is much larger than its clique number, then $G$ contains a subdivision of $K_4^+$ as an induced subgraph.

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Full text

The authors are partially supported by ANR Project STINT (anr-13-bs02-0007), and LabEx PERSYVAL-Lab (anr-11-labx-0025).

Given a graph $H$, a subdivision of $H$ is a graph obtained from $H$ by replacing some edges of $H$ (possibly none) by paths. We say that a graph $G$ contains an induced subdivision of $H$ if $G$ contains a subdivision of $H$ as an induced subgraph.

A class of graphs $\mathcal{F}$ is said to be $\chi$-bounded if there is a function $f$ such that for any graph $G\in\mathcal{F}$, $\chi(G)\leqslant f(\omega(G))$, where $\chi(G)$ and $\omega(G)$ stand for the chromatic number and the clique number of $G$, respectively.

Scott [7] conjectured that for any graph $H$, the class of graphs without induced subdivisions of $H$ is $\chi$-bounded, and proved it when $H$ is a tree. But Scott’s conjecture was disproved in [6]. Finding which graphs $H$ satisfy the assumption of Scott’s conjecture remains a fascinating question. It was proved in [1] that every graph $H$ obtained from the complete graph $K_{4}$ by subdividing at least 4 of the 6 edges once (in such a way that the non-subdivided edges, if any, are non-incident), is a counterexample to Scott’s conjecture. On the other hand, Scott proved that the class of graphs with no induced subdivision of $K_{4}$ has bounded chromatic number (see [5]). Le [4] proved that every graph in this class has chromatic number at most 24. If triangles are also excluded, Chudnovsky et al. [2] proved that the chromatic number is at most 3.

In this paper, we extend the list of graphs known to satisfy Scott’s conjecture. Let $K_{4}^{+}$ be the 5-vertex graph obtained from $K_{4}$ by subdividing one edge precisely once.

We will need the following result of Kühn and Osthus [3].

Proof of Theorem 1. Let $k$ be an integer, let $d(\cdot,\cdot)$ be the function defined in Theorem 2, and let $R(s,t)$ be the Ramsey number of $(s,t)$, i.e. the smallest $n$ such that every graph on $n$ vertices has a stable set of size $s$ or a clique of size $t$.

We will prove that every graph $G$ with no induced subdivision of $K_{4}^{+}$, and with clique number at most $k$, is $d$-colorable, with $d=\max(k,d(K_{4}^{+},R(4,k)))$. The proof proceeds by induction on the number of vertices of $G$ (the result being trivial if $G$ has at most $k$ vertices). Observe that all induced subgraphs of $G$ have clique number at most $k$ and do not contain any induced subdivision of $K_{4}^{+}$. Therefore, by the induction, we can assume that all induced subgraphs of $G$ are $d$-colorable. In particular, we can assume that $G$ is connected.

Assume first that $G$ does not contain $K_{s,s}$ as a subgraph, where $s=R(4,k)$. Then by Theorem 2, $G$ has average degree less than $d$, and hence contains a vertex of degree at most $d-1$. By the induction, $G-v$ has a $d$-coloring and this coloring can be extended to a $d$-coloring of $G$, as desired.

We can thus assume that $G$ contains $K_{s,s}$ as a subgraph. Since $G$ has clique number at most $k$, it follows from the definition of $R(4,k)$ that $G$ contains $K_{4,4}$ as an induced subgraph. Let $M$ be a set of vertices of $G$ inducing a complete multipartite graph with at least two partite sets containing at least 4 vertices. Assume that among all such sets of vertices of $G$, $M$ is chosen with maximum cardinality. Let $V_{1},V_{2},\ldots,V_{t}$ be the partite sets of $M$.

Let $v$ be a vertex of $G$, and $S$ be a set of vertices not containing $v$. The vertex $v$ is complete to $S$ if $v$ is adjacent to all the vertices of $S$, anticomplete to $S$ if $v$ is not adjacent to any of the vertices of $S$, and mixed to $S$ otherwise. Let $R$ be the vertices of $G$ not in $M$. We can assume that $R$ is non-empty, since otherwise $G$ is clearly $k$-colorable and $k\leqslant d$. We claim that:

[TABLE]

Assume for the sake of contradiction that $v$ has two neighbors $a,b$ in $V_{i}$ and a neighbor $c$ and a non-neighbor $d$ in $V_{j}$, with $j\neq i$. Then $v,a,b,c,d$ induce a copy of $K_{4}^{+}$, a contradiction. This proves ([math]).

[TABLE]

Assume for the sake of contradiction that some vertex $v\in R$ has two neighbors $a,b$ in some set $V_{i}$. Then by ([math]), $v$ is complete or anticomplete to each set $V_{j}$ with $j\neq i$. Let $\mathcal{A}$ be the family of sets $V_{j}$ to which $v$ is anticomplete, and let $\mathcal{C}$ be the family of sets $V_{j}$ to which $v$ is complete. If $\mathcal{A}$ contains at least two elements, i.e. if $v$ is anticomplete to two sets $V_{j}$ and $V_{j^{\prime}}$ then by taking $u\in V_{j}$ and $u^{\prime}\in V_{j^{\prime}}$, we observe that $v,a,b,u,u^{\prime}$ induces a copy of $K_{4}^{+}$, a contradiction. It follows that $\mathcal{A}$ contains at most one element.

Next, we prove that $v$ is complete to $V_{i}$. Assume instead that $v$ is mixed to $V_{i}$. If $v$ is complete to some set $V_{\ell}$ containing at least two vertices, then we obtain a contradiction with ([math]). It follows that all the elements of $\mathcal{C}$ are singleton. By the definition of $M$, this implies that $\mathcal{A}$ contains exactly one set $V_{j}$, which has size at least 4. Let $c$ be a non-neighbor of $v$ in $V_{i}$, and let $d,d^{\prime}$ be two vertices in $V_{j}$. Then $v,a,b,c,d,d^{\prime}$ is an induced subdivision of $K_{4}^{+}$, a contradiction. We proved that $v$ is complete to $V_{i}$. Hence, every set $V_{j}$ is either in $\mathcal{A}$ or in $\mathcal{C}$. Since $\mathcal{A}$ contains at most one element, the graph induced by $M\cup\{v\}$ is a complete multipartite graph, with at least two partite sets containing at least 4 elements. This contradicts the maximality of $M$, and concludes the proof of ([math]).

[TABLE]

Assume for the sake of contradiction that some connected component of $G-M$ has at least two neighbors in some set $V_{i}$. Then there is a path $P$ whose endpoints $u,v$ are in $V_{i}$, and whose internal vertices are in $R$. Choose $P,u,v,V_{i}$ such that $P$ contains the least number of edges. Note that by ([math]), $P$ contains at least 3 edges. Observe also that by the minimality of $P$, the only edges in $G$ between $V_{i}$ and the internal vertices of $P$ are the first and last edge of $P$. Let $V_{j}$ be a partite set of $M$ with at least 4 elements, with $j\neq i$ (this set exists, by the definition of $M$). By ([math]) and the minimality of $P$, at most two vertices of $V_{j}$ are adjacent to some internal vertex of $P$. Since $V_{j}$ contains at least four vertices, there exist $a,b\in V_{j}$ that are not adjacent to any internal vertex of $P$. If $V_{i}$ has at least three elements then it contains a vertex $w$ distinct from $u,v$. As $w$ is not adjacent to any vertex of $P$, the vertices $w,a,b$ together with $P$ induce a subdivision of $K_{4}^{+}$, a contradiction. If $V_{i}$ has at most two elements, then there must be an integer $\ell$ distinct from $i$ and $j$ such that $V_{\ell}$ has at least four elements. In particular, $V_{\ell}$ contains a vertex $c$ that is not adjacent to any internal vertex of $P$. As a consequence, the vertices $a,c$ together with $P$ induce a subdivision of $K_{4}^{+}$, which is again a contradiction. This proves ([math]).

Recall that we can assume that $R$ is non-empty. An immediate consequence of ([math]) is that the neighborhood of each connected component of $R$ is a clique. Since $G$ is connected, it follows that it contains a clique cutset $K$ (a clique whose deletion disconnects the graph). Let $C$ be a connected component of $G-K$, let $G_{1}=G-C$, and let $G_{2}$ be the subgraph of $G$ induced by $C\cup K$. It follows from the induction that there exist $d$-colorings of $G_{1}$ and $G_{2}$. Furthermore, since $K$ is a clique, we can assume that the colorings coincide on $K$. This implies that $G$ is $d$-colorable and concludes the proof of Theorem 1. $\Box$

We remark that we could have used $K_{3,3}$ instead of $K_{4,4}$ in the proof, at the expense of a slightly more detailed analysis. The resulting bound on the chromatic number would have been $\max(k,d(K_{4}^{+},R(3,k)))$ instead of $\max(k,d(K_{4}^{+},R(4,k)))$.

Acknowledgement. The main result of this paper was proved in January 2016 during a meeting of the ANR project STINT at Saint Bonnet de Champsaur, France. We thank the organizers and participants for the friendly atmosphere. We also thank Alex Scott for spotting a couple of typos in a previous version of the draft.