Coloring ($P_6$, diamond, $K_4$)-free graphs

TL;DR

This paper proves that all ($P_6$, diamond, $K_4$)-free graphs can be colored with six colors and provides an example of such a graph requiring exactly six colors, extending previous results.

Contribution

It establishes the exact chromatic number for ($P_6$, diamond, $K_4$)-free graphs and presents a minimal example demonstrating this bound.

Findings

Every ($P_6$, diamond, $K_4$)-free graph is 6-colorable.

Existence of a ($P_6$, diamond, $K_4$)-free graph with chromatic number 6.

Abstract

We show that every ($P_6$, diamond, $K_4$)-free graph is $6$-colorable. Moreover, we give an example of a ($P_6$, diamond, $K_4$)-free graph $G$ with $\chi(G) = 6$. This generalizes some known results in the literature.

Equations

$$W_i$$

$$Y_i$$

$$A_i$$

$$A_{jk}$$

$$A_{p14}$$

$$S_5:=\{\begin{array}{l}{A}_5^{\prime }\cup A_{15}\cup \{v_3\},\text{mbox}if{A}_{14}=\varnothing \text{mbox}or{A}_{45}=\varnothing ,\\ A_5^{\prime }\cup A_{45}\cup \{v_3\},\text{mbox}if{A}_{15}=\varnothing ,\end{array}$$

$$S_6:=\{\begin{array}{l}{A}_5^{\prime \prime }\cup A_{14}\cup A_{45}\cup A_{145},\text{mbox}if{A}_{14}=\varnothing \text{mbox}or{A}_{45}=\varnothing .\\ A_5^{\prime \prime }\cup A_{14}\cup A_{145},\text{mbox}if{A}_{15}=\varnothing .\end{array}$$

Full text

1 Introduction

We consider simple, finite, and undirected graphs. For notation and terminology not defined here we refer to [22]. Let $P_{n}$, $C_{n}$, $K_{n}$ denote the induced path, induced cycle and complete graph on $n$ vertices respectively. If $G_{1}$ and $G_{2}$ are two vertex disjoint graphs, then their union $G_{1}\cup G_{2}$ is the graph with $V(G_{1}\cup G_{2})$ $=V(G_{1})\cup V(G_{2})$ and $E(G_{1}\cup G_{2})$ $=$ $E(G_{1})\cup E(G_{2})$. Similarly, their join $G_{1}+G_{2}$ is the graph with $V(G_{1}+G_{2})$ $=$ $V(G_{1})\cup V(G_{2})$ and $E(G_{1}+G_{2})$ $=$ E(G_{1})\cup E(G_{2})$$\cup\{(x,y)\mid x\in V(G_{1}),~{}y\in V(G_{2})\}. For any positive integer $k$, $kG$ denotes the union of $k$ graphs each isomorphic to $G$. If $\cal{F}$ is a family of graphs, a graph $G$ is said to be $\cal{F}$-free if it contains no induced subgraph isomorphic to any member of $\cal{F}$. A clique (independent set) in a graph $G$ is a set of vertices that are pairwise adjacent (non-adjacent) in $G$. The clique number of $G$, denoted by $\omega(G)$, is the size of a maximum clique in $G$.

A $k$-coloring of a graph $G=(V,E)$ is a mapping $f:V\rightarrow\{1,2,\ldots,k\}$ such that $f(u)\neq f(v)$ whenever $uv\in E$. We say that $G$ is $k$-colorable if $G$ admits a $k$-coloring. That is, a partition of the vertex set $V(G)$ into $k$ independent sets. The chromatic number of $G$, denoted by $\chi(G)$, is the smallest positive integer $k$ such that $G$ is $k$-colorable. It is well known that a graph is $2$-colorable if and only if it is bipartite. Given an integer $k$, the $k$-Coloring problem is that of testing whether a given graph is $k$-colorable. The $k$-Coloring problem is $NP$-complete for every fixed $k\geq 3$ [11, 14]. The problem of finding the maximum chromatic number of graphs without forbidden induced subgraphs from some finite/infinite set and with small clique number is well studied and still receives much attention. We refer to [19] for a survey and we give some of them that are not in [19]. We note that some of the cited results are consequences of much stronger results available in the literature. Mycielski [16] showed that for any integer $k$, there exists a triangle-free graph with chromatic number $k$. Fan et al. [10] showed that every (fork, $K_{3}$)-free graph with odd-girth at least $7$ is $3$-colorable. Pyatkin [18] showed that every ($2P_{3},K_{3}$)-free graph is 4-colorable. Esperet et al. [9] showed that every ($P_{5},K_{4}$)-free graph is $5$-colorable. It follows from a result of Gravier, Hoáng and Maffray [12] that every ($P_{6},K_{3}$)-free graph is $4$-colorable (see also [19]), and that every ($P_{6},K_{4}$)-free graph is $16$-colorable. Randerath et al. [20] showed that every ($P_{6},K_{1,3},W_{5},DD,K_{4}$)-free graph is $3$-colorable, where $DD$ denotes the double-diamond graph, and $W_{5}$ is the 5-wheel. It follows from a result of [4] that every ($P_{2}\cup P_{3},C_{4},K_{4}$)-free graph is $4$-colorable, and from a result of [13] that every ($P_{5}$, diamond, $K_{4}$)-free graph is $4$-colorable. Chudnovsky et al. [7] showed that every (odd hole, $K_{4}$)-free graph is 4-colorable. This implies that every ($P_{6},C_{5},K_{4}$)-free graph is $4$-colorable. Addario-Berry et al. [1] showed that every (even hole, $K_{4}$)-free graph is 5-colorable, and Kloks, Müller and Vuskovic [15] showed that every (even-hole, diamond, $K_{4}$)-free is $4$-colorable.

In this note, we first give an alternative and simple proof to the fact that every ($P_{2}\cup P_{3}$, diamond, $K_{4}$)-free graph is $6$-colorable given in [2]. Then we show that the conclusion holds even for a general class of graphs, namely ($P_{6}$, diamond, $K_{4}$)-free graphs. That is, we show that every ($P_{6}$, diamond, $K_{4}$)-free graph is $6$-colorable. Moreover, we give an example of a ($P_{6}$, diamond, $K_{4}$)-free graph $G$ with $\chi(G)=6$. This generalizes the aforementioned results for ($P_{5}$, diamond, $K_{4}$)-free graphs, ($P_{6},K_{3}$)-free graphs and ($P_{2}\cup P_{3}$, diamond, $K_{4}$)-free graphs. The proof of our results depend on a sequence of partial results given below, and we give tight examples for each of them. See Figure 1.

• •

Let $G$ be a ($P_{2}\cup P_{3}$, diamond, $K_{4}$)-free graph that contains a non-dominating $K_{3}$. Then $G$ is $4$-colorable.

• •

Let $G$ be a ($P_{2}\cup P_{3}$, diamond, $K_{4}$)-free graph such that every triangle in $G$ dominates $G$. Then $G$ is $6$-colorable.

• •

Let $G$ be a ($P_{6}$, diamond, bull, $K_{4}$)-free graph. Then $G$ is 4-colorable.

• •

Let $G$ be a ($P_{6}$, diamond, $K_{4}$)-free graph that contains an induced bull. Then $G$ is $6$-colorable.

Note that the class of ($H$, diamond)-free graphs, for various $H$, is well studied in variety of contexts in the literature. Chudnovsky et al. [5] showed that there are exactly six 4-critical ($P_{6}$, diamond)-free graphs. Tucker [21] gave an $O(kn^{2})$ time algorithm for $k$-Coloring perfect diamond-free graphs. It is also known that $k$-Coloring is polynomial-time solvable for (even-hole, diamond)-free graphs [15] as well as for (hole, diamond)-free graphs [3]. Dabrowski et al. [8] showed that if $H$ is a graph on at most five vertices, then $k$-Coloring is polynomial time solvable for ($H$, diamond)-free graphs, whenever $H$ is a linear forest and NP-complete otherwise. However, the computational complexity of the $k$-Coloring problem for ($P_{6}$, diamond)-free graphs is open. It is also known that the Maximum Weight Independent Set problem is solvable in polynomial time for ($P_{6}$, diamond)-free graphs [17] as well as for (hole, diamond)-free graphs [3].

We devote the rest of this section to the notations and terminologies used in this paper. For any integer $k$, we simply write $[k]$ to denote the set $\{1,2,\ldots,k\}$. Let $G$ be a graph, with vertex-set $V(G)$ and edge-set $E(G)$. A diamond or a $K_{4}-e$ is the graph with vertex set $\{a,b,c,d\}$ and edge set $\{ab,bc,cd,ad,bd\}$.

A graph $G$ is called perfect if $\chi(H)=\omega(H)$, for every induced subgraph $H$ of $G$. By the Strong Perfect Graph Theorem [6], A graph is perfect if and only if it contains no odd hole (chordless cycle) of length at least $5$ and no odd antihole (complement graph of a hole) of length at least $5$.

For $x\in V(G)$, $N(x)$ denotes the set of all neighbors of $x$ in $G$. The neighborhood $N(X)$ of a subset $X\subseteq V(G)$ is the set $\{u\in V(G)\setminus X\mid u\mbox{~{}is adjacent to a vertex of }X\}$. For any two disjoint subsets $S,~{}T\subseteq V(G)$, $[S,T]$ denotes the edge-set $\{uv\in E(G)\mid u\in S,\ v\in T\}$. The set $[S,T]$ is said to be complete if every vertex in $S$ is adjacent to every vertex in $T$. Also, for $S\subseteq V(G)$, let $G[S]$ denotes the subgraph induced by $S$ in $G$, and for convenience we simply write $[S]$ instead of $G[S]$. The length of a path is the number of edges in it. The length of a shortest path between two vertices $x$ and $y$ is denoted by $dist(x,y)$. For $S\subseteq V(G)$ and $x\in V(G)\setminus S$, we define dist$(x,S):=\min\{\mbox{dist}(x,y)\mid y\in S\}$. We say that a subgraph $H$ of $G$ is dominating if every vertex in $V(G)\setminus V(H)$ has a neighbor in $H$; otherwise, it is a non-dominating subgraph. We say that a graph $H$ is obtained from $G$ by duplication if it can be obtained from $G$ by substituting independent sets for some of the vertices in $G$.

2 Our Results

We use the following preliminary results often. Let $G$ be a (diamond, $K_{4}$)-free graph. Then the following hold:

1. (R1)

If $T$ is a triangle in $G$, then any vertex $p\in V(G)\setminus V(T)$ has at most one neighbor in $T$. 2. (R2)

For any $v\in V(G)$, $N(v)$ induces a $P_{3}$-free graph, and hence $[N(v)]$ is a union of $K_{2}$’s and $K_{1}$’s. 3. (R3)

For any two non-adjacent vertices $x$ and $y$ in $G$, the set of common neighbors of $x$ and $y$ is an independent set. 4. (R4)

For any two adjacent vertices $x$ and $y$ in $G$, the number of common neighbors of $x$ and $y$ is at most one. 5. (R5)

Let $G$ be a diamond-free graph with $n$ vertices. Let $v_{1},v_{2},\ldots,v_{k}$ $(1\leq k\leq n)$ be vertices in $G$ such that $N(v_{i})$ is an independent set, for each $i\in\{1,2\ldots,k\}$. Then the duplicated graph $H$ obtained from $G$ by substituting an independent set for each $v_{i}$, $i\in\{1,2\ldots,k\}$ is also diamond-free.

Theorem 1

Let $G$ be a connected ($P_{2}\cup P_{3}$, diamond, $K_{4}$)-free graph that contains a non-dominating $K_{3}$. Then $G$ is $4$-colorable.

Proof of Theorem 1. Let $T$ be a non-dominating triangle in $G$ induced by the vertex-set $N_{0}:=\{v_{1},v_{2},v_{3}\}$. For $j\geq 1$, let $N_{j}$ denote the set $\{y\in V(G)\setminus N_{0}\mid d(y,N_{0})=j\}$. Then by (R1), every vertex of $N_{1}$ has at most one neighbor in $N_{0}$; more precisely:

1. (1)

If $x\in N_{1}$, then $[N(x)\cap N_{0}]$ is isomorphic to $K_{1}$.

For $i\in[3]$, let $A_{i}:=\{x\in N_{1}\mid N(x)\cap N_{0}=\{v_{i}\}\}$. Then since $G$ is $K_{4}$-free, by (R2), each $[A_{i}]$ is a union of $K_{2}$’s and $K_{1}$’s, and so $[A_{i}]$ is bipartite. Let $(A_{1}^{\prime},A_{1}^{\prime\prime})$ be a bipartition of $A_{1}$ such that $A_{1}^{\prime}$ is a maximal independent set of $A_{1}$.

1. (2)

$N_{j}=\emptyset$, for all $j\geq 3$.

Proof. Suppose not, and let $y\in N_{3}$. Then by the definition of $N_{3}$, there exist vertices $x_{1}\in N_{1}$ and $x_{2}\in N(x)\cap N_{2}$ such that $x_{1}-x_{2}-y$ is a $P_{3}$ in $G$. By (1), $x_{1}\in A_{i}$, for some $i\in[3]$, say $i=1$. But then $\{v_{2},v_{3},x_{1},x_{2},y\}$ induces a $P_{2}\cup P_{3}$ in $G$, which is a contradiction. $\diamond$

Now, since $T$ is non-dominating, $N_{2}\neq\emptyset$. Also, since $G$ is ($P_{2}\cup P_{3}$)-free, $[N_{2}]$ is a $P_{3}$-free graph, and hence $[N_{2}]$ is a union of complete graphs. Moreover:

1. (3)

If $C$ is a component of $[N_{2}]$, then there exists $x\in N_{1}$ such that $[N(x)\cap N_{2}]\cong C$.

Proof. Otherwise, $G$ induces a $P_{2}\cup P_{3}$ or diamond, a contradiction. $\diamond$

Furthermore, since $G$ is $K_{4}$-free, by (3), $[N_{2}]$ is a union of $K_{2}$’s and $K_{1}$’s. In particular, $[N_{2}]$ is bipartite. Let $(N_{2}^{\prime},N_{2}^{\prime\prime})$ be a bipartition of $N_{2}$.

1. (4)

Let $i,j\in[3]$. Let $x\in A_{i}$ be such that $N(x)\cap N_{2}\neq\emptyset$. Then $A_{j}$ ($j\neq i$) is an independent set.

Proof. We may assume that $i=1$. Let $z\in N(x)\cap N_{2}$. Suppose to the contrary that there are vertices $y_{1},y_{2}\in A_{2}$ such that $y_{1}y_{2}\in E$. Then since $\{y_{1},y_{2},v_{2},x\}$ does not induce a diamond in $G$, we have either $xy_{1}\notin E$ or $xy_{2}\notin E$. Assume $xy_{1}\notin E$. Now, since $\{x,z,y_{1},v_{2},v_{3}\}$ does not induce a $P_{2}\cup P_{3}$ in $G$, $zy_{1}\in E$. Then since $\{v_{1},v_{3},z,y_{1},y_{2}\}$ does not induce a $P_{2}\cup P_{3}$ in $G$, $zy_{2}\in E$. But, then $\{v_{2},y_{1},y_{2},z\}$ induces a diamond in $G$, a contradiction. So, $A_{2}$ is an independent set. Similarly, $A_{3}$ is also an independent set. $\diamond$

Now, we show that $G$ is 4-colorable using the above properties. Suppose that $K_{2}\sqsubseteq[N_{2}]$. Then by (3), there is a vertex $x\in A_{i}$ such that $[N(x)\cap N_{2}]\cong K_{2}$. We may assume that $i=1$, and let $y_{1},y_{2}\in N(x)\cap N_{2}$ be such that $y_{1}y_{2}\in E$. Then $A_{2}=\emptyset=A_{3}$. Otherwise, if $w\in A_{2}\cup A_{3}$, then since $\{w,v_{2},v_{3},y_{1},y_{2}\}$ or $\{v_{1},v_{3},w,y_{1},y_{2}\}$ or $\{v_{1},v_{2},w,y_{1},y_{2}\}$ does not induce a $P_{2}\cup P_{3}$ in $G$, we have $wy_{1},wy_{2}\in E$. But, then $\{x,w,y_{1},y_{2}\}$ induces a diamond or a $K_{4}$ in $G$, a contradiction. Then we define the sets $S_{1}:=\{v_{1}\}\cup N_{2}^{\prime},S_{2}:=\{v_{2}\}\cup A_{1}^{\prime},S_{3}:=\{v_{3}\}\cup A_{1}^{\prime\prime}$, and $S_{4}:=N_{2}^{\prime\prime}$. Clearly, $S_{j}$ is an independent set, for each $j\in\{1,2,3,4\}$. Hence $(S_{1},S_{2},S_{3},S_{4})$ is a $4$-coloring of $G$.

So, assume that $N_{2}$ $(\neq\emptyset)$ is an independent set. Then there exists $x\in A_{i}$ such that $N(x)\cap N_{2}\neq\emptyset$. We may assume that $i=1$, and let $z\in N(x)\cap N_{2}$. By (4), $A_{2}$ and $A_{3}$ are independent sets. Moreover, if $y\in A_{1}^{\prime\prime}$, then $N(y)\cap N_{2}=\emptyset$. (Otherwise, since $A_{1}^{\prime}$ is maximal, there exist $y^{\prime}\in A_{1}^{\prime}$ and $z^{\prime}\in N(y)\cap N_{2}$ such that $yy^{\prime},yz^{\prime}\in E$. Then since $\{v_{2},v_{3},y,y^{\prime},z^{\prime}\}$ does not induces a $P_{2}\cup P_{3}$ in $G$, $y^{\prime}z^{\prime}\in E$. But, then $\{v_{1},y,y^{\prime},z^{\prime}\}$ induces a diamond in $G$, a contradiction.) Now, we define $S_{1}:=\{v_{1}\}\cup A_{2},S_{2}:=\{v_{2}\}\cup A_{1}^{\prime},S_{3}:=\{v_{3}\}\cup A_{1}^{\prime\prime}\cup N_{2},$ and $S_{4}:=A_{3}$. Then $S_{j}$ is an independent set, for each $j\in\{1,2,3,4\}$. Hence $(S_{1},S_{2},S_{3},S_{4})$ is a $4$-coloring of $G$.

Thus, Theorem 1 is proved. $\Box$

The bound given in Theorem 1 is tight. For example, consider the duplicated graph $G$ obtained from $G_{1}$ (shown in Figure 2) by substituting each vertex indicated in circle by an independent set (of order $\geq 1$). As the graph $G$ is highly symmetric, using (R5), there are not too many cases to directly verify that $G$ is ($P_{2}\cup P_{3}$, diamond, $K_{4}$)-free. Also, it is easy to see that $G$ contains a non-dominating triangle $T$ with vertices $\{v_{1},v_{2},v_{3}\}$, and that $\chi(G)=4$.

Theorem 2

Let $G$ be a connected ($P_{2}\cup P_{3}$, diamond, $K_{4}$)-free graph such that every triangle in $G$ dominates $G$. Then $G$ is $6$-colorable.

Proof of Theorem 2. Let $K$ be a $K_{3}$ in $G$ with vertices $\{x_{1},x_{2},x_{3}\}$ that dominates $G$. Since $G$ is ($K_{4}$, diamond)-free and $K$ dominates $G$, every vertex in $V(G)\setminus V(K)$ has exactly one neighbor in $K$. For $i\in[4]$, let $A_{i}:=\{x\in V(G)\setminus V(K)\mid N(x)\cap V(K)=\{x_{i}\}\}$. Then by (R2), each $[A_{i}]$ is a union of $K_{2}$’s and $K_{1}$’s. Now, it is easy to check that $G$ is $6$-colorable. $\Box$

The bound given in Theorem 2 is tight. For example, consider the graph $G$ which is isomorphic to the complement of the 16-regular Schläfli graph on 27 vertices. It is verified that $G$ is ($P_{2}\cup P_{3}$, diamond)-free and it is well known that $\chi(G)=6$ and $\omega(G)=3$; see (https://hog.grinvin.org/ViewGraphInfo.action?id=19273).

Theorem 3

Let $G$ be a ($P_{2}\cup P_{3}$, diamond, $K_{4}$)-free graph. Then $G$ is $6$-colorable.

Proof. Follows by Theorems 1 and 2. $\Box$

Theorem 4

Let $G$ be a ($P_{6}$, diamond, bull, $K_{4}$)-free graph. Then $G$ is 4-colorable.

Proof. If $G$ is perfect, then $G$ is $3$-colorable and the theorem holds. So we may assume that $G$ is connected and $G$ is not perfect. Since $G$ is $P_{6}$-free, $G$ contains no hole of length at least $7$, and since $G$ is diamond-free, $G$ contain no anti-hole of length at least $7$. Thus, it follows from the Strong Perfect Graph Theorem [6] that $G$ contains a $5$-hole (hole of length $5$), say $N_{0}$ with vertex-set $\{v_{1},v_{2},v_{3},v_{4},v_{5}\}$, and edge-set $\{v_{1}v_{2},v_{2}v_{3},v_{3}v_{4},v_{4}v_{5},v_{5}v_{1}\}$. Throughout this proof, we take all the subscripts of $v_{i}$ to be modulo $5$. For any integer $j\geq 1$, let $N_{j}$ denote the set $\{x\in V(G)\mid d(x,N_{0})=j\}$.

1. (1)

If $x\in N_{1}$, then $[N(x)\cap N_{0}]$ is isomorphic to either $K_{1}$ or $2K_{1}$.

Proof. Suppose not. Then there exists an $i\in[5]$ such that $\{v_{i},v_{i+1}\}\subseteq N(x)$. Then since $\{v_{i-1},v_{i},v_{i+1},v_{i+2},x\}$ does not induce a bull in $G$, either $v_{i-1}\in N(x)$ or $v_{i+2}\in N(x)$. But then either $\{v_{i},v_{i+1},v_{i+2},x\}$ or $\{v_{i-1},v_{i},v_{i+1},x\}$ induces a diamond in $G$, a contradiction. So $(1)$ holds. $\diamond$

By (1), we partition $N_{1}$ as follows: For any $i\in[5]$, $i$ mod $5$, let:

[TABLE]

Moreover, let $W:=W_{1}\cup\cdots\cup W_{5}$ and $Y:=Y_{1}\cup\cdots\cup Y_{5}$.

1. (2)

For $i\in[5]$, $i$ mod $5$, the following hold:

1. (i)

$[W_{i}]$ is union of $K_{2}$’s and $K_{1}$’s. 2. (ii)

$[W_{i},W_{i+1}]=\emptyset$. 3. (iii)

$[W_{i},W_{i+2}]$ is complete. In particular, if $W_{i}\neq\emptyset$, then $W_{i+2}$ and $W_{i-2}$ are independent sets. 4. (iv)

$Y_{i}\cup Y_{i+2}$ is an independent set. 5. (v)

$[W_{i},Y_{i+1}]=\emptyset$

Proof of $(2)$. We prove for $i=1$.

$(i)$: Follows by the definition of $W_{1}$, and by (R2).

$(ii)$: Suppose not. Then there exist vertices $x\in W_{1}$ and $y\in W_{2}$ such that $xy\in E$. But, then $\{x,y,v_{2},v_{3},v_{4},v_{5}\}$ induces a $P_{6}$ in $G$, which is a contradiction. So (ii) holds.

$(iii)$: Let $x\in W_{1}$. We show that $W_{3}$ is an independent set. If not, then there exist adjacent vertices $y$ and $z$ in $W_{3}$. Since $\{x,v_{1},v_{5},v_{4},v_{3},y\}$ and $\{x,v_{1},v_{5},v_{4},v_{3},z\}$ do not induce a $P_{6}$ in $G$, we have $xy\in E$ and $xz\in E$. But, then $\{x,y,z,v_{3}\}$ induces a diamond in $G$, which is a contradiction. So, $W_{3}$ is an independent set. Similarly, $W_{4}$ is also an independent set. Thus (iii) holds.

$(iv)$: Suppose not. Then there exist adjacent vertices $x$ and $y$ in $Y_{1}\cup Y_{3}$. If both $x$ and $y$ are in $Y_{1}$ or in $Y_{3}$, then either $\{v_{1},x,v_{5},y\}$ or $\{v_{2},x,v_{4},y\}$ induces a diamond in $G$, a contradiction. So, we may assume that $x\in Y_{1}$ and $y\in Y_{3}$. But, then $\{v_{1},v_{2},y,v_{4},x\}$ induces a bull in $G$, which is a contradiction. So (iv) holds.

$(v)$: Suppose not. Then there exist vertices $x\in W_{1}$ and $y\in Y_{2}$ such that $xy\in E$. But, then $\{v_{5},v_{1},y,v_{3},x\}$ induces a bull in $G$, which is a contradiction. So (v) holds. $\diamond$

By (2)(i), for each $i\in[5]$, $W_{i}$ is bipartite. Let $(W_{i}^{\prime},W_{i}^{\prime\prime})$ is a bipartition of $W_{i}$.

1. (3)

The following hold:

1. (i)

$[W,N_{2}]=\emptyset$. 2. (ii)

If $xx^{\prime}$ is an edge in $[N_{2}]$, then $N(x)\cap N_{1}=N(x^{\prime})\cap N_{1}$. Moreover, $|N(x)\cap N_{1}|=1$. 3. (iii)

If $C$ is a component of $[V(G)\setminus(N_{0}\cup N_{1})]$, then there exists a vertex $y\in Y$ such that $N(y)\cap V(C)=V(C)$.

Proof of $(3)$. $(i)$: Suppose not. Then there exist vertices $x\in W$ and $y\in N_{2}$ such that $xy\in E$. We may assume that $x\in W_{1}$. But, then $\{y,x,v_{1},v_{2},v_{3},v_{4}\}$ induces a $P_{6}$ in $G$ which is a contradiction. So (i) holds.

$(ii)$: Let $y\in N(x)\cap N_{1}$. Then by (i), $y\in Y_{i}$, for some $i$. Say $y\in Y_{1}$. Since $\{x^{\prime},x,y,v_{2},v_{3},v_{4}\}$ does not induce a $P_{6}$ in $G$, $x^{\prime}y\in E$. So, $N(x)\cap N_{1}\subseteq N(x^{\prime})\cap N_{1}$. Similarly, $N(x^{\prime})\cap N_{1}\subseteq N(x)\cap N_{1}$. Hence, $N(x)\cap N_{1}=N(x^{\prime})\cap N_{1}$. Moreover, if $|N(x)\cap N_{1}|>1$, and if $y_{1},y_{2}\in N(x)\cap N_{1}$, then $y_{1},y_{2}\in N(x^{\prime})$, and then $\{x,x^{\prime},y_{1},y_{2}\}$ induces either a diamond or a $K_{4}$ in $G$, a contradiction. So, $|N(x)\cap N_{1}|=1$.

$(iii)$: Since $C$ is a component of $[V(G)\setminus(N_{0}\cup N_{1})]$, by (i), there exists a vertex $y\in Y$ such that $N(y)\cap V(C)\neq\emptyset$. We may assume that $y\in Y_{1}$. Now, we show that $N(y)\cap V(C)=V(C)$. Suppose not. Then there exist vertices $x,z\in V(C)$ such that $yx\in E$ and $yz\notin E$. Then since $C$ is connected, there exists a path joining $x$ and $z$ in $C$, say $P$. But, then $V(P)\cup\{y,v_{2},v_{3},v_{4}\}$ will induce a $P_{6}$ in $G$, a contradiction. So (iii) holds. $\diamond$

By (3)(iii), we see that for each $j\geq 3$, $N_{j}=\emptyset$. So, $V(G)=N_{0}\cup W\cup Y\cup N_{2}$.

Also, by (3)(iii) and (R2), $[N_{2}]$ is a union of $K_{2}$’s and $K_{1}$’s, and hence $[N_{2}]$ is bipartite. Let $(N_{2}^{\prime},N_{2}^{\prime\prime})$ be a bipartition of $[N_{2}]$ such that $N_{2}^{\prime}$ is a maximal independent set of $N_{2}$.

1. (4)

For each $i\in[5]$, $i$ mod $5$, we have: (i) $[Y_{i}\cup Y_{i+2},N(Y_{i+1}\cup Y_{i+3}\cup Y_{i+4})\cap N_{2}^{\prime\prime}]=\emptyset$, and (ii) $[Y_{i+1}\cup Y_{i+3},N(Y_{i}\cup Y_{i+2})\cap N_{2}^{\prime\prime}]=\emptyset$.

Proof of $(4)$. $(i)$: We prove for $i=1$. Suppose to the contrary that there exist adjacent vertices $y\in Y_{1}\cup Y_{3}$ and $x\in N(Y_{2}\cup Y_{4}\cup Y_{5})\cap N_{2}^{\prime\prime}$. Since $N_{2}^{\prime}$ is a maximal independent set of $N_{2}$, there exists $x^{\prime}\in N_{2}^{\prime}$ such that $xx^{\prime}\in E$. Also, since $x\in N(Y_{2}\cup Y_{4}\cup Y_{5})\cap N_{2}^{\prime\prime}$, there exists a vertex $y^{\prime}\in Y_{2}\cup Y_{4}\cup Y_{5}$ such that $xy^{\prime}\in E$. But, then $\{y,y^{\prime}\}\subseteq N(x)\cap N_{1}$, a contradiction to (3)(ii). So (i) holds.

$(ii)$: Similar to the proof of $(i)$. $\diamond$

Now, by using the above properties, we prove the theorem in three cases as follows:

Case 1. $W_{i}=\emptyset$, for each $i\in[5]$.

Define $S_{1}:=\{v_{1},v_{3}\}\cup Y_{1}\cup Y_{3}\cup(N(Y_{2}\cup Y_{4}\cup Y_{5})\cap N_{2}^{\prime\prime})$, $S_{2}:=\{v_{2},v_{4}\}\cup Y_{2}\cup Y_{4}\cup(N(Y_{1}\cup Y_{3})\cap N_{2}^{\prime\prime})$, $S_{3}:=\{v_{5}\}\cup Y_{5}$, and $S_{4}:=N_{2}^{\prime}$. Then by (2)(iv) and (4), $S_{1},S_{2}$, $S_{3}$ and $S_{4}$ are independent sets. So, $(S_{1},S_{2},S_{3},S_{4})$ is a $4$-coloring of $G$.

Case 2. $W_{i}\neq\emptyset$, for every $i\in[5]$.

By (2)(iii), $W_{i}$ is an independent set, for each $i$. Now, we define $S_{1}:=\{v_{1},v_{3}\}\cup W_{2}\cup Y_{1}\cup Y_{3}\cup(N(Y_{2}\cup Y_{4}\cup Y_{5})\cap N_{2}^{\prime\prime})$, $S_{2}:=\{v_{2},v_{4}\}\cup W_{3}\cup Y_{2}\cup Y_{4}\cup(N(Y_{1}\cup Y_{3})\cap N_{2}^{\prime\prime})$, $S_{4}:=\{v_{5}\}\cup W_{1}\cup Y_{5}$, and $S_{5}:=W_{4}\cup W_{5}\cup N_{2}^{\prime}$. Then by the above properties, we see that $(S_{1},S_{2},S_{3},S_{4})$ is a $4$-coloring of $G$.

Case 3. $W_{i}\neq\emptyset$ and $W_{i-1}=\emptyset$, for some $i\in[5]$, $i$ mod $5$.

Up to symmetry, we may assume that $i=1$. Then by (2)(iii), $W_{3}$ and $W_{4}$ are independent sets.

(a) Suppose that $W_{4}=\emptyset$. Then we define $S_{1}:=\{v_{1},v_{3}\}\cup W_{2}^{\prime}\cup Y_{1}\cup Y_{3}\cup(N(Y_{2}\cup Y_{5})\cap N_{2}^{\prime\prime})$, $S_{2}:=\{v_{2},v_{5}\}\cup W_{1}^{\prime}\cup Y_{2}\cup Y_{5}\cup(N(Y_{1}\cup Y_{3}\cup Y_{4})\cap N_{2}^{\prime\prime})$, $S_{3}:=\{v_{4}\}\cup W_{1}^{\prime\prime}\cup W_{2}^{\prime\prime}\cup N_{2}^{\prime}$, and $S_{4}:=W_{3}\cup Y_{4}$. Then by the above properties, $(S_{1},S_{2},S_{3},S_{4})$ is a $4$-coloring of $G$.

(b) Suppose $W_{4}\neq\emptyset$, then by (2)(iii), $W_{1}$ and $W_{2}$ are independent sets. Now, we define $S_{1}:=\{v_{1},v_{3}\}\cup W_{2}\cup Y_{1}\cup Y_{3}\cup(N(Y_{2}\cup Y_{5})\cap N_{2}^{\prime\prime})$, $S_{2}:=\{v_{2},v_{5}\}\cup W_{1}\cup Y_{2}\cup Y_{5}\cup(N(Y_{1}\cup Y_{3}\cup Y_{4})\cap N_{2}^{\prime\prime})$, $S_{3}:=\{v_{4}\}\cup W_{3}\cup Y_{4}$, and $S_{4}:=W_{4}\cup N_{2}^{\prime}$. Then by the above properties, $(S_{1},S_{2},S_{3},S_{4})$ is a 4-coloring of $G$.

This completes the proof of the theorem. $\Box$

The bound given in Theorem 4 is tight. For example, consider the duplicated graph $H_{i}$ obtained from $G_{i}$, $i\in\{2,3\}$ (shown in Figure 2) by substituting each vertex indicated in circle by an independent set (of order $\geq 1$). Then it is verified that both $H_{2}$ and $H_{3}$ are ($P_{6}$, diamond, $K_{4}$, bull)-free (using (R5)), and $\chi(H_{2})=\chi(H_{3})=4$. Note that the graph $G_{2}$ is a Greenwood-Gleason graph or a Clebsch graph, and the graph $G_{3}$ is the Mycielski 4-chromatic graph or a Grötzsch graph.

Theorem 5

Let $G$ be a connected ($P_{6}$, diamond, $K_{4}$)-free graph that contains an induced bull. Then $G$ is $6$-colorable.

Proof. Let $G$ be a connected ($P_{6}$, diamond, $K_{4}$)-free graph that contains an induced bull, say $H$ with vertex-set $N_{0}:=\{v_{1},v_{2},v_{3},v_{4},v_{5}\}$, and edge-set $\{v_{1}v_{2},v_{2}v_{3},v_{3}v_{4},v_{2}v_{5},v_{3}v_{5}\}$. For any integer $j\geq 1$, let $N_{j}$ denote the set $\{y\in V(G)\mid d(y,N_{0})=j\}$. Then we have the following:

1. (1)

If $x\in N_{1}$, then $|N(x)\cap\{v_{2},v_{3},v_{5}\}|\leq 1$ and so $|N(x)\cap N_{0}|\leq 3$.

Proof. If $|N(x)\cap\{v_{2},v_{3},v_{5}\}|\geq 2$, then $\{v_{2},v_{3},v_{5},x\}$ induces a diamond or a $K_{4}$ in $G$, a contradiction. $\diamond$

By (1) and since $G$ is diamond-free, we partition $N_{1}$ as follows. Let:

[TABLE]

1. (2)

By (R2) and (R3), for any $i\in\{1,\ldots,5\}$, $j\in\{1,4\}$, $k\in\{2,3,4,5\}$ ($j\neq k$) and $p\in\{2,3,5\}$, we see that: $[A_{i}]$ is a union of $K_{2}$’s and $K_{1}$’s, and hence bipartite, and $[A_{jk}]$ and $[A_{p14}]$ are independent sets.

For each $i\in\{1,\ldots,5\}$, let $(A_{i}^{\prime},A_{i}^{\prime\prime})$ be a bipartition of $A_{i}$ such that $A_{i}^{\prime}$ is a maximal independent set.

1. (3)

We have either $A_{1}=\emptyset$ or $A_{4}=\emptyset$.

Proof. Suppose not. Let $x\in A_{1}$ and $y\in A_{4}$. Then since $\{x,v_{1},v_{2},v_{3},v_{4},y\}$ does not induce a $P_{6}$ in $G$, $xy\in E$. But then $\{v_{1},x,y,v_{4},v_{3},v_{5}\}$ induces a $P_{6}$ in $G$, which is a contradiction. $\diamond$

1. (4)

We have the following:

1. (i)

$[A_{1},A_{14}\cup A_{15}\cup A_{45}]$ is complete. 2. (ii)

$[A_{5},A_{15}\cup A_{45}]=\emptyset$. 3. (iii)

$[A_{2}^{\prime\prime},A_{12}\cup A_{24}\cup A_{124}]=\emptyset=[A_{3}^{\prime\prime},A_{13}\cup A_{34}\cup A_{134}]$. 4. (iv)

$A_{2}^{\prime\prime}\cup A_{12}\cup A_{24}\cup A_{124}$ and $A_{3}^{\prime\prime}\cup A_{13}\cup A_{34}\cup A_{134}$ are independent sets. 5. (v)

$[A_{5}^{\prime\prime},A_{14}\cup A_{145}]=\emptyset$. 6. (vi)

$[A_{14}\cup A_{15}\cup A_{45},A_{145}]=\emptyset$.

Proof. $(i)$: Suppose not. Then there exist vertices $x\in A_{1}$ and $y\in A_{14}\cup A_{15}\cup A_{45}$ such that $xy\notin E$. But, then $\{x,v_{1},y,v_{3},v_{4},v_{5}\}$ or $\{x,v_{1},v_{2},v_{3},v_{4},y\}$ induces a $P_{6}$ in $G$, a contradiction. So (i) holds.

$(ii)$: Suppose not. Then there exist vertices $x\in A_{5}$ and $y\in A_{15}\cup A_{45}$ such that $xy\in E$. But, then $\{y,x,v_{1},v_{2},v_{3},v_{4}\}$ induces a $P_{6}$ in $G$, a contradiction. So, (ii) holds.

$(iii)$: Follows by the definitions of $A_{2}^{\prime\prime}$ and $A_{3}^{\prime\prime}$, and by (R1).

$(iv)$: Follows by (2), (iii), (R1), (R3), and (R4).

$(v)$: $[A_{5}^{\prime\prime},A_{145}]=\emptyset$ follows by the definition of $A_{5}^{\prime\prime}$ and by(R1). Also, $[A_{5}^{\prime\prime},A_{14}]=\emptyset$. Otherwise, there exist $x\in A_{5}^{\prime\prime}$ and $y\in A_{14}$ such that $xy\in E$. Since $A_{5}^{\prime}$ is maximal, there exists $x^{\prime}\in A_{5}^{\prime}$ such that $xx^{\prime}\in E$. But, then by (R1), $\{x^{\prime},x,y,v_{1},v_{2},v_{3}\}$ induces a $P_{6}$ in $G$ which is a contradiction. So, $[A_{5}^{\prime\prime},A_{14}]=\emptyset$.

$(vi)$: Follows by (R3). $\diamond$

Define $S_{1}:=A_{2}^{\prime\prime}\cup A_{12}\cup A_{24}\cup A_{124}$ and $S_{2}:=A_{3}^{\prime\prime}\cup A_{13}\cup A_{34}\cup A_{134}$.

1. (5)

If $C$ is a component of $[V(G)\setminus(N_{0}\cup N_{1})]$, then there exists a vertex $x\in N_{1}$ such that $N(x)\cap V(C)=V(C)$.

Proof. Suppose not. Since $G$ is connected there exists a vertex $x\in N_{1}$ such that $N(x)\cap V(C)\neq\emptyset$ and $N(x)\cap V(C)\neq V(C)$. Then there exist vertices $y_{1}$ and $y_{2}$ in $C$ with $y_{1}y_{2}\in E$ such that $y_{1}\in N(x)$ and $y_{2}\notin N(x)$. Now, if both $v_{1}$ and $v_{4}$ are neighbors of $x$ or if both $v_{1}$ and $v_{4}$ are non-neighbors of $x$, then by (1), $N_{0}\cup\{x,y_{1},y_{2}\}$ induces a $P_{6}$ in $G$, a contradiction. So, we may assume, up to symmetry that $xv_{1}\in E$ and $xv_{4}\notin E$. Now, if $xv_{2}\in E$, then by (1), $\{y_{2},y_{1},x,v_{2},v_{3},v_{4}\}$ induces a $P_{6}$ in $G$, a contradiction. So, $xv_{2}\notin E$. Then since $\{y_{2},y_{1},x,v_{1},v_{2},v_{5}\}$ does not induce a $P_{6}$ in $G$, $xv_{5}\in E$. But, then $\{y_{2},y_{1},x,v_{5},v_{3},v_{4}\}$ induces a $P_{6}$ in $G$, a contradiction. So, (5) holds. $\diamond$

By (5) and by (R2), $[N_{2}]$ is a union of $K_{2}$’s and $K_{1}$’s, and hence bipartite. Let $(N_{2}^{\prime},N_{2}^{\prime\prime})$ be a bipartition of $N_{2}$ such that $N_{2}^{\prime}$ is a maximal independent set of $N_{2}$.

Also, since $G$ is connected, by (5) and by the definition of $N_{j}$’s, $N_{j}=\emptyset$, for all $j\geq 3$. Thus, $V(G)=N_{0}\cup N_{1}\cup N_{2}$.

1. (6)

$[A_{1}\cup A_{4}\cup A_{15}\cup A_{45},N_{2}]=\emptyset$.

Proof. Suppose not. Then there exist vertices $x\in A_{1}\cup A_{4}\cup A_{15}\cup A_{45}$ and $y\in N_{2}$ such that $xy\in E$. But, then $\{x,y,v_{1},v_{2},v_{3},v_{4}\}$ induces a $P_{6}$ in $G$, a contradiction. $\diamond$

By (4:(iv)), it enough to show that $G-(S_{1}\cup S_{2})$ is $4$-colorable, and we do this in two cases using (3).

Case 1. Suppose that $A_{1}\cup A_{4}\neq\emptyset$.

By (3) and by symmetry we may assume that $A_{1}\neq\emptyset$ and $A_{4}=\emptyset$. Then:

Claim 1

The following hold:

1. (i)

$[A_{1},A_{5}]=\emptyset$. 2. (ii)

$A_{14}\cup A_{15}\cup A_{45}$* is an independent set.* 3. (iii)

$[A_{3},N_{2}]=\emptyset=[A_{5},N_{2}]$.

Proof of Claim 1. $(i)$: Suppose not. Then there exist vertices $x\in A_{1}$ and $y\in A_{5}$ such that $xy\in E$. But, then $\{y,x,v_{1},v_{2},v_{3},v_{4}\}$ induces a $P_{6}$ in $G$, a contradiction. So (i) holds. $\diamond$

$(ii)$: Suppose not. Then there exist adjacent vertices, say $x$ and $y$ in $A_{14}\cup A_{15}\cup A_{45}$. Since $A_{1}\neq\emptyset$, let $z\in A_{1}$. Then by (4:(i)), $[\{z\},A_{14}\cup A_{15}\cup A_{45}]$ is complete. But, then $\{z,x,v_{4},y\}$ or $\{z,x,v_{5},y\}$ induce a diamond in $G$ or $\{z,x,v_{1},y\}$ induces a $K_{4}$ in $G$, a contradiction. So (ii) holds. $\diamond$

$(iii)$: Suppose not. Since $A_{1}\neq\emptyset$, let $z\in A_{1}$.

Let $x\in A_{3}$ and $y\in N_{2}$ be such that $xy\in E$. Then since $\{z,v_{1},v_{2},v_{3},x,y\}$ does not induce a $P_{6}$ in $G$, $xz\in E$. But, then $\{v_{5},v_{2},v_{1},z,x,y\}$ induces a $P_{6}$ in $G$, a contradiction. So, $[A_{3},N_{2}]=\emptyset$.

Again, let $x\in N_{2}$ and $y\in A_{5}$ be such that $xy\in E$. By $(6)$ and $(i)$, $xz,yz\notin E$. But, then $\{z,v_{1},v_{2},v_{5},y,x\}$ induces a $P_{6}$ in $G$, a contradiction. So, $[A_{5},N_{2}]=\emptyset$. $\diamond$

Hence we have proved Claim 1. $\Diamond$

Now, let us define $S_{3}:=A_{2}^{\prime}\cup\{v_{1},v_{5}\}$; $S_{4}:=A_{3}^{\prime}\cup N_{2}^{\prime}\cup\{v_{2},v_{4}\}$; $S_{5}:=A_{1}^{\prime}\cup A_{5}^{\prime}\cup N_{2}^{\prime\prime}\cup\{v_{3}\}$; and $S_{6}:=A_{1}^{\prime\prime}\cup A_{5}^{\prime\prime}\cup A_{14}\cup A_{15}\cup A_{45}\cup A_{145}$. Then by the above claim and by the above properties, we see that $(S_{3},S_{4},S_{5},S_{6})$ is a $4$-coloring of $G-(S_{1}\cup S_{2})$.

Case 2. Suppose that $A_{1}\cup A_{4}=\emptyset$.

If $[A_{2}\cup A_{3}\cup A_{5},N_{2}]\neq\emptyset$, then we find a suitable bull with $A_{1}\neq\emptyset$, and we proceed as in Case 1 to get a $4$-coloring of $G-(S_{1}\cup S_{2})$. Also, if $[A_{14},A_{15}]$ is not complete, then there exist $u\in A_{14}$ and $v\in A_{15}$ such that $uv\notin E$. Now, $\{v_{2},v_{3},v_{5},v_{4},v\}$ induces a bull with $A_{1}\neq\emptyset$. So, we proceed as in Case 1 to get a $4$-coloring of $G-(S_{1}\cup S_{2})$. By symmetry, the same holds if $[A_{14},A_{45}]$ is not complete. So, we may assume that $[A_{2}\cup A_{3}\cup A_{5},N_{2}]=\emptyset$, and that $[A_{14},A_{15}]$ and $[A_{14},A_{45}]$ are complete.

We define $S_{3}:=N_{2}^{\prime}\cup A_{2}^{\prime}\cup\{v_{1},v_{5}\}$ and $S_{4}:=N_{2}^{\prime\prime}\cup A_{3}^{\prime}\cup\{v_{2},v_{4}\}$. Clearly $S_{3}$ and $S_{4}$ are independent sets. Now it is enough to show that $G\setminus(S_{1}\cup S_{2}\cup S_{3}\cup S_{4})$ is bipartite. Then:

Claim 2

$A_{14}=\emptyset$* or $A_{15}=\emptyset$ or $A_{45}=\emptyset$.*

Proof of Claim 2. Suppose not, let $x\in A_{14},y\in A_{15}$ and $z\in A_{45}$. Then since $\{y,v_{1},v_{2},v_{3},v_{4},z\}$ does not induce a $P_{6}$ in $G$, $yz\in E$. But then since $[A_{14},A_{15}]$ and $[A_{14},A_{45}]$ are complete, $\{x,y,z,v_{5}\}$ induces a diamond in $G$, a contradiction. So the claim holds. $\diamond$

Now, we define

[TABLE]

and

[TABLE]

Then by the above claim and by (4), we see that $(S_{5},S_{6})$ is a bipartition of $G\setminus(S_{1}\cup S_{2}\cup S_{3}\cup S_{4})$.

This completes the proof of the theorem. $\Box$

The bound given in Theorem 5 is tight. For example, consider the graph $G$ which is isomorphic to the complement of the 16-regular Schläfli graph on 27 vertices. As mentioned earlier, $G$ is ($P_{6}$, diamond)-free, $\chi(G)=6$ and $\omega(G)=3$. Also, it is verified that $G$ contains a bull.

Theorem 6

Let $G$ be a ($P_{6}$, diamond, $K_{4}$)-free graph. Then $G$ is $6$-colorable.

Proof. Follows by Theorems 4 and 5. $\Box$