Equality cases for a bound on the chromatic number

TL;DR

This paper characterizes the graphs that achieve equality in a known bound relating the chromatic number, vertices, and edges, showing they are composed of complete graphs or odd cycles with attached trees.

Contribution

It provides a complete description of all graphs attaining equality in the bound, extending understanding of the structure of such graphs.

Findings

Equality holds for graphs made of complete graphs or odd cycles with attached trees.

The characterization includes all such graphs with these specific structures.

The result clarifies the extremal cases for the chromatic number bound.

Abstract

It is known that the inequality $$ \frac{\chi(G)(\chi(G)-1)}{2} + |V| - \chi(G) \leq |E|$$ holds for all connected graphs, where $\chi(G)$ denotes the chromatic number of $G$. We prove that equality holds whenever the graph consists of a complete graph or an odd cycle, together with finitely many trees attached to its vertices.

Equations

$$\frac{\chi (G)(\chi (G)-1)}{2}+|V|-\chi (G)\le |E|$$

$$\frac{\chi (G)(\chi (G)-1)}{2}+|V|-\chi (G)\le |E|.$$

$$\frac{\chi (G)(\chi (G)-1)}{2}+|V|-\chi (G)=|E|$$

$$\frac{\chi (G_n)(\chi (G_n)-1)}{2}+n-\chi (G_n)+c=m_n.$$

$$\frac{\chi (G_{n+1})(\chi (G_{n+1})-1)}{2}+(n+1)-\chi (G_{n+1})=m_{n+1}.$$

$$d_{G_{n+1}}(v)=m_{n+1}-m_n=1-c.$$

$$d_{G_{n+1}}(v)\ge \chi (G_n).$$

$$\frac{\chi (G_n)(\chi (G_n)+1)}{2}+n-\chi (G_n)+c=m_n+\chi (G_n).$$

$$m_{n+1}$$

$$=\frac{\chi (G_n)(\chi (G_n)+1)}{2}+n-\chi (G_n)$$

$$=m_n+\chi (G_n)-c,$$

$$d_{G_{n+1}}(v)=m_{n+1}-m_n=\chi (G_n)-c.$$

$$\frac{\chi (G)(\chi (G)-1)}{2}-\chi (G)=\frac{m(m-1)}{2}-m=|E(G)|-|V(G)|$$

$$\frac{\chi (G)(\chi (G)-1)}{2}-\chi (G)=\frac{3(3-1)}{2}-3=0=|E(G)|-|V(G)|.$$

Full text

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holds for all connected graphs [1], where $\chi(G)$ denotes the chromatic number of $G$. We prove that equality holds whenever the graph consists of a complete graph or an odd cycle, together with finitely many trees attached to its vertices.

Throughout the article, we will only consider simple graphs (without loops or multiple edges). We begin by stating Lemma 1 of [1].

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We wish to investigate graphs for which (1) is an equality. We begin with a definition.

1. A.

A complete graph $K_{n}$ where $n\geq 1$ with finitely many trees attached to each vertex, or 2. B.

An odd cycle $C_{2m+1}$ where $m\geq 1$ with finitely many trees attached to each vertex

is said to be of type A or B respectively. If a graph $G_{A}$ or $G_{B}$ is of type A or B, we will occasionally refer to the complete graph or odd cycle it contains as its underlying graph.

We have the following result:

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if and only if it is of type A or B.

We will need the following lemma in the proof of Theorem 1.

1. (I)

At least one of the following statements are true:

$d_{G_{n}}(v)=1$* and $G_{n}$ is of type A or B, or* 2. 2.

$d_{G_{n}}(v)=\chi(G_{n-1})$* and $G_{n-1}$ is of type A or B;* 2. (II)

If $d_{G_{n}}(v)=\chi(G_{n-1})\leq 2$, then $G_{n}$ is of type A or B.

Suppose inductively that the result holds for all $k\leq n$, and let $G_{n+1}$ be a graph with $n+1$ vertices and $m_{n+1}$ edges such that equality in (2) holds. Suppose $v\in V(G_{n+1})$ is such that $G_{n}:=G_{n+1}-v$ is connected, and let $m_{n}$ denote the number of edges of $G_{n}$. Then, by Lemma 1, there exists a non-negative integer $c$ such that

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Since equality in (2) holds for $G_{n+1}$ by hypothesis, we have

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As the addition of a vertex may either increase the chromatic number by one or leave it unchanged, we must consider these cases separately.

Case 1: $\boldsymbol{\chi(G_{n+1})=\chi(G_{n})}$. Subtracting (3) from (4), we find that

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Since $G_{n+1}$ is connected, we must have $d_{G_{n+1}}(v)>0$, so that $c=0$ and $d_{G_{n+1}}(v)=1$. It follows that equality as in (2) is satisfied for $G_{n}$, so that it is of type A or B by hypothesis. Since $d_{G_{n+1}}(v)=1$, it follows from Lemma 2 (see Appendix) that $G_{n+1}$ is type A or B. That is, (I1) holds, so that (I) holds. Since $d_{G_{n+1}}(v)=1$, the antecedent in (II) holds iff $\chi(G_{n})=1$. But that means that $G_{n}$ is the graph on one vertex, so that $G_{n+1}$ must be the connected graph on two vertices. It follows that $\chi(G_{n+1})=2\neq\chi(G_{n})$, a contradiction. Thus (II) holds vacuously.

Case 2: $\boldsymbol{\chi(G_{n+1})=\chi(G_{n})+1}$. We claim that

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Indeed, suppose for contradiction that $d_{G_{n+1}}(v)<\chi(G_{n})$. Then, given a $\chi(G_{n})$-coloring of $G_{n}$, the vertices adjacent to $v$ would not use all $\chi(G_{n})$ colors, so there would be an unused color one may assign to $v$. But this gives us a $\chi(G_{n})$-coloring of $G_{n+1}$, a contradiction.

By adding $\chi(G_{n})$ to both sides of (3), we find that

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Since $\chi(G_{n+1})=\chi(G_{n})+1$ by assumption, it follows from (4) that

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so that

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Since $\chi(G_{n+1})>\chi(G_{n})$, it follows that $d_{G_{n+1}}(v)\geq\chi(G_{n})$. But since $\chi(G_{n})-d_{G}(v)=c\geq 0$, it follows that $d_{G_{n+1}}(v)=\chi(G_{n})$ and that $c=0$. Thus, equality as in (2) holds for $G_{n}$, so that $G_{n}$ is of type A or B by hypothesis. It follows that (I2) holds, so that (I) holds.

We now prove (II). If $d_{G_{n+1}}(v)=\chi(G_{n})\leq 2$, then there are two cases to consider.

Case 2.1: $\boldsymbol{\chi(G_{n})=1}$. In this case, $G_{n}$ must be the graph on one vertex, so that $G_{n+1}$ is the connected graph on two vertices. It follows that $G_{n+1}$ is of type A, so that (II) holds.

Case 2.2: $\boldsymbol{\chi(G_{n})=2}$. By (I), it follows that $G_{n}$ is of type A or B.

Case 2.2.1: $\boldsymbol{G_{n}}$ is of type A with underlying graph $\boldsymbol{K_{m}}$ with $\boldsymbol{m\leq 2}$. In this case, $G_{n}$ is a tree. In $G_{n+1}$, we claim that the vertex $v$ must be adjacent to exactly two vertices of $G_{n}$. Indeed, we have $\chi(G_{n+1})=\chi(G_{n})+1=3$ by hypothesis. If $v$ were adjacent to fewer vertices, then $\chi(G_{n+1})<3$, and similarly we would have $\chi(G_{n+1})>3$ if $v$ were adjacent to more than two vertices.

Since $v$ is attached to two vertices of $G_{n}$, exactly one cycle is present in $G_{n+1}$. If this cycle has odd length, then $G_{n+1}$ is of type B (with the odd cycle being the underlying graph), so we are done. If this cycle has even length, then $\chi(G_{n+1})=2$ by Lemma 1, so that $\chi(G_{n+1})\neq\chi(G_{n})+1$, a contradiction.

Case 2.2.2: $\boldsymbol{G_{n}}$ is of type A with underlying graph $\boldsymbol{K_{m}}$ with $\boldsymbol{m\geq 3}$. By Lemma 1, we find that $\chi(G_{n})=m\neq 2$, a contradiction.

Case 2.2.3: $\boldsymbol{G_{n}}$ is of type B. Similarly, we apply Lemma 1 to show that $\chi(G_{n})=3\neq 2$, a contradiction.

∎

We are now able to prove Theorem 1.

Suppose that the result holds for all $k\leq n$ for some $n\in\mathbb{N}$, and let $G_{n+1}$ be a graph with $n+1$ vertices and $m_{n+1}$ edges such that equality in (2) holds. If $G_{n+1}$ is of type A or B, we are done. If not, by Proposition 1.4.1 of [2], we may label the vertices $v_{1},v_{2},\dots,v_{n+1}$ in such a way that the subgraph consisting of vertices $v_{1},\dots,v_{k}$ is connected for $1\leq k\leq n+1$. It follows that the graph $G_{n}:=G_{n+1}-v_{n+1}$ with $m_{n}$ edges is a connected graph with $n$ vertices.

By Lemma 2, we have at least one of the following is true:

1. (11)

$d_{G_{n+1}}(v_{n+1})=1$ and $G_{n+1}$ is type A or B,

in which case we are done, or

1. (I2)

$d_{G_{n+1}}(v_{n+1})=\chi(G_{n})$.

In this case, we must consider various cases depending on $\chi(G_{n})$. If $\chi(G_{n})\leq 2$, then $G_{n+1}$ is of type A or B by (II) of Lemma 2, so we are done. It suffices to prove the theorem for $\chi(G_{n})\geq 3$.

Case 1: $\boldsymbol{G_{n}}$ is of type A with no trees attached. That is, $G_{n}=K_{n}$. In this case, $\chi(G_{n+1})=\chi(K_{n})+1=n+1$, so that $G_{n+1}=K_{n+1}$ (since every $k$-chromatic graph has a $k$-chromatic subgraph of minimum degree at least $k-1$ ([2], Lemma 5.2.3)). That is, $G_{n+1}$ is of type A.

Case 2: $\boldsymbol{G_{n}}$ is of type B with no trees attached. That is, $G_{n}=C_{n}$.

Case 2.1: $\boldsymbol{G_{n}=C_{3}}$. Since $C_{3}=K_{3}$, this is just Case 1.

Case 2.2: $\boldsymbol{G_{n}=C_{2m+1}}$ for $\boldsymbol{m\geq 2}$. In this case, there must exist some $v\in V(G_{n})$ that is not adjacent to $v_{n+1}$. Define a coloring such that $v$ and $v_{n+1}$ get mapped to the same color $1$, and color the rest of the vertices in the odd cycle by alternating between colors $2$ and $3$. This shows that $\chi(G_{n+1})\leq 3$, so that $\chi(G_{n+1})\neq\chi(G_{n})+1=4$, a contradiction.

Case 3: $\boldsymbol{G_{n}}$ contains trees. In this case, there exists some $v^{\prime}\in V(G_{n})$ with $d_{G_{n}}(v^{\prime})=1$. We claim that $G_{n}-v^{\prime}$ is connected. Indeed, suppose for contradiction that there exists two vertices $a$, $b\in G_{n}-v^{\prime}$ such that there is no path from $a$ to $b$. However, $G_{n}$ is connected, so there exists a path from $a$ to $b$ in $G_{n}$. It follows that this this path must pass through $v^{\prime}$, so that $d_{G_{n}}(v^{\prime})\geq 2$, a contradiction.

Case 3.1: $\boldsymbol{d_{G_{n+1}}(v_{n+1})=1}$. If $G_{n}$ is of type A or B, direct application of Lemma 2 implies that $G_{n+1}$ is of type A or B. Otherwise, by (I) of Lemma 2, we have $G_{n}$ or $G_{n+1}$ is of type A or B. If $G_{n}$ is of type A or B, we apply Lemma 2 again to conclude that $G_{n+1}$ is of type A or B, as desired.

Case 3.2: $\boldsymbol{d_{G_{n+1}}(v_{n+1})>1}$. We have that $v_{n+1}$ must be connected to some vertex of $G_{n}-v^{\prime}$ in $G_{n+1}$. It follows that $G:=G_{n+1}-v^{\prime}=(G_{n}-v^{\prime})\cup\{v_{n+1}\}$ is connected, so that we may apply Lemma 2 to $G$ to conclude that

1. (1)

$G_{n+1}$ is type A or B, in which case we are done; or 2. (2)

$d_{G_{n+1}}(v^{\prime})=\chi(G)$ and $G$ is type A or B.

However, in the second case, we know that $d_{G_{n}}(v^{\prime})=1$ implies that $d_{G_{n+1}}(v^{\prime})\leq 2$. By (II) of Lemma 2, $G_{n+1}$ is of type A or B.

In all cases, it follows that $G_{n+1}$ is of type A or B, so that the result follows from induction.

We now prove the converse. Let $G$ be a graph of type A. Then either its underlying graph is $K_{1}$, in which case $G$ is either the graph on one vertex for which the result is quickly established by a straightforward computation, or a tree with $\chi(G)=2$ for which the result is established by noting that trees always satisfy $|V|-|E|=1$; or that its underlying graph is $K_{m}$ for $m\geq 2$, in which case we can apply Lemma 1 to find that $\chi(G)=m$. Since $|E(K_{m})|-|V(K_{m})|=m(m-1)/2-m$, and since every vertex added afterwards also adds exactly one edge, it follows that

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as desired. On the other hand, given a graph $G$ of type B, its underlying odd cycle $C_{2m+1}$ has $|E(C_{2m+1})|-|V(C_{2m+1})|=0$. Since the addition of vertices that form trees adds exactly one edge for each added vertex, we find that $|E(G)|-|V(G)|=0$. By Lemma 1, $\chi(G)=3$, so that

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The converse follows.

∎

Appendix A Technical lemmas

Lemma 1**.**

Let $G_{A}$ and $G_{B}$ be type A and B graphs respectively, where $K_{m}$ is the underlying graph for $G_{A}$. Let $G_{C}$ be a graph defined similarly to a type B graph, except that its underlying cycle is even. Then the following statements hold:

1. (1)

$\chi(G_{A})=m$* if $m\geq 2$,* 2. (2)

$\chi(G_{B})=3$, and 3. (3)

$\chi(G_{C})=2$.

Proof.

Let $C_{2n+1}$ be the underlying graph of $G_{B}$. Since odd cycles (with $n\geq 1$) have chromatic number equal to $3$, it suffices to show that the addition of any number of trees to any of the vertices of $C_{2n+1}$ does not change its chromatic number. Indeed, since any tree with more than two vertices has a chromatic number equal to $2$, and since $\chi(G_{B})\geq\chi(C_{2n+1})$, the result follows. The proofs for statements (1) and (3) are very similar, and only require noting that $\chi(K_{m})=m$ and $\chi(C_{2n})=2$. ∎

Lemma 2**.**

Let $G$ be a graph. If $v\in V(G)$ has $d_{G}(v)=1$ and if $G-v$ is of type A, then $G$ is of type A. The result also holds if we replace type A with type B.

Proof.

Let $G$ be a graph, and let $v\in V(G)$. Suppose that $d_{G}(v)=1$ and $G-v$ is of type A. Since $d_{G}(v)=1$, it follows that $v$ is either adjacent to a vertex of the underlying graph or to a vertex of a tree. In the first case, $v$ forms a tree at that vertex so that $G$ is type A. In the second case, $v$ becomes part of a tree at a vertex of the underlying graph, so that $G$ is type A. The proof works exactly the same for type B graphs. ∎