Structure and coloring of some ($P_{7}$, $C_{4}$)-free graphs

All graphs considered in this paper are finite and simple. We follow [1] for undefined notations and terminology. We use $P_{t}$ and $C_{t}$ to denote a path and a cycle on $t$ vertices, respectively. Let $G$ be a graph, let $v\in V(G)$, and let $X$ and $Y$ be two subsets of $V(G)$. We say that $v$ is complete to $X$ if $v$ is adjacent to all vertices of $X$, and say that $v$ is anticomplete to $X$ if $v$ is not adjacent to any vertex of $X$. We say that $X$ is complete (resp. anticomplete) to $Y$ if each vertex of $X$ is complete (resp. anticomplete) to $Y$.

A hole of $G$ is an induced cycle of length at least 4, and a $k$-hole is a hole of length $k$. Given a graph $H$, we say that $G$ is $H$-free if it dose not contain an induced subgraph isomorphic to $H$. Given a set $\{H_{1},H_{2},\cdots\}$ of graphs, we say that $G$ is $(H_{1},H_{2},\cdots)$-free if $G$ is $H_{i}$-free for each $i$.

A clique blowup of a graph $H$ with $V(H)=\{v_{1},v_{2},\cdots,v_{n}\}$ is any graph $G$ such that $V(G)$ can be partitioned into $n$ nonempty cliques $A_{i}$, $v_{i}\in V(H)$ such that $A_{i}$ is complete to $A_{j}$ if $v_{i}\sim v_{j}$, and $A_{i}$ is anticomplete to $A_{j}$ if $v_{i}\not\sim v_{j}$. Particularly, if $|A_{i}|=t$ for all $i\in\{1,\cdots,n\}$, then we call $G$ a $t$-size clique blowup of $H$.

For $v\in V(G)$, let $N_{G}(v)$ be the set of vertices adjacent to $v$, $d_{G}(v)=|N_{G}(v)|$, $N_{G}[v]=N_{G}(v)\cup\{v\}$, and $M_{G}(v)=V(G)\setminus N_{G}[v]$. For $X\subseteq V(G)$, let $N_{G}(X)=\{u\in V(G)\setminus X\;|\;u$ has a neighbor in $X\}$ and $M_{G}(X)=V(G)\setminus(X\cup N_{G}(X))$. Moreover, $G[X]$ denotes the subgraph of $G$ induced by $X$. If it does not cause any confusion, we usually omit the subscript $G$ and simply write $N(v)$, $d(v)$, $N[v]$, $M(v)$, $N(X)$ and $M(X)$. Let $\delta(G)$ denote the minimum degree of $G$. For $u$, $v\in V(G)$, we simply write $u\sim v$ if $uv\in E(G)$, and write $u\not\sim v$ if $uv\not\in E(G)$.

A clique (stable set) of $G$ is a set of mutually adjacent (non-adjacent) vertices in $G.$ The clique number of $G$, denoted by $\omega(G),$ is the maximum size of a clique in $G$. For positive integer $k$, a $k$-coloring of $G$ is a function $c:V(G)\rightarrow\{1,\cdots,k\}$, such that for each edge $uv$ of $G$, $c(u)\not=c(v)$. The chromatic number of $G$, denoted by $\chi(G)$, is the minimum number $k$ for which there exists a $k$-coloring of $G$. A graph is perfect if all its induced subgraphs $H$ satisfy $\chi(H)=\omega(H)$. A family ${\cal G}$ of graphs is said to be $\chi$-bounded [11] if there exists a function $f$ such that for every graph $G\in{\cal G}$, $\chi(G)\leq f(w(G))$. If such a function $f$ does exist to ${\cal G}$, then $f$ is called a $\chi$-binding function of ${\cal G}$.

A clique cutset of $G$ is a clique $K$ in $G$ such that $G-K$ has more components than $G$. A vertex $u$ is a cut vertex of $G$ if it is a clique cutset of size 1.

A diamond is a graph obtained from two triangles that share exactly one edge. A kite is a graph consists of a diamond and another vertex adjacent to a vertex of degree 2 of the diamond. A gem is a graph that consists of a $P_{4}$ plus a vertex adjacent to all vertices of the $P_{4}$. A bull is a graph consisting of a triangle with two disjoint pendant edges. Let $C=v_{1}v_{2}v_{3}v_{4}v_{5}v_{6}v_{7}v_{1}$ be a 7-hole. We use $F$ to denote a graph obtained from $C$ by adding a stable set $\{y_{1},y_{2},y_{3}\}$ such that $N(y_{1})\cap V(C)=\{v_{1},v_{4},v_{5}\}$, $N(y_{2})\cap V(C)=\{v_{2},v_{5},v_{6}\}$ and $N(y_{3})\cap V(C)=\{v_{3},v_{6},v_{7}\}$. (See Figure 1).

Gyárfás [11] showed that $\chi(G)\leq(t-1)^{\omega(G)-1}$ for all $P_{t}$-free graphs. This upper bound was improved to $\chi(G)\leq(t-2)^{\omega(G)-1}$ in [10]. However, this $\chi$-binding function is exponential in $\omega(G)$. It is natural to ask that is it possible to improve the exponential bound for $P_{t}$-free graphs to a polynomial bound [18]. Interested readers are referred to [17, 19, 21] for more results and still open problems on this topic.

It is known that $P_{3}$-free graphs are complete graphs and $P_{4}$-free graphs are perfect [22]. But up to now, no polynomial binding function for $P_{t}$-free graphs has been found, when $t\geq 5$. We refer the interested readers to [9] for results of $P_{5}$-free graphs, and list here some conclusions on binding functions of some $P_{6}$-free or $P_{7}$-free graphs.

Choudum, Karthick and Shalu [5] proved that every ($P_{6}$, gem)-free graph $G$ satisfies $\chi(G)\leq 8\omega(G)$. Cameron, Huang and Merkel [2] proved that every ($P_{6}$, diamond)-free graph $G$ satisfies $\chi(G)\leq\omega(G)+3$. Gasper and Huang [8] proved that every $(P_{6},C_{4})$-free graph $G$ satisfies $\chi(G)\leq\lfloor\frac{3}{2}\omega(G)\rfloor$. This bound was recently improved by Karthick and Maffray [13] to $\chi(G)\leq\lceil\frac{5}{4}\omega(G)\rceil$. Mishra [15] proved that every ($P_{6}$, diamond, bull)-free graphs $G$ satisfies $\chi(G)\leq 4$ when $\omega(G)=2$ and $\chi(G)=\omega(G)$ when $\omega(G)\neq 2$, and every ($P_{7}$, diamond, bull)-free graph $G$ satisfies $\chi(G)\leq\max\{7,\omega(G)\}$. Cameron, Huang, Penev and Sivaraman [3] showed that every $(P_{7},C_{4},C_{5})$-free graph satisfies $\chi(G)\leq\frac{3}{2}\omega(G)$. This upper bound was recently improved by Huang [12] to $\chi(G)\leq\lceil\frac{11}{9}\omega(G)\rceil$.

In 2021, Choudum, Karthick and Belavadi [4] proved that $\chi(G)\leq 2\omega(G)-1$ if $G$ is $(P_{7},C_{7},C_{4}$, gem)-free, and $\chi(G)\leq\max\{3,\omega(G)\}$ if $G$ is $(P_{7},C_{7},C_{4}$, diamond)-free. In 2022, Lan, Zhou and Liu [14] proved that $\chi(G)\leq\max\{3,\omega(G)\}$ if $G$ is $(P_{6},C_{4}$, diamond)-free.

In this paper, we study ($P_{7},C_{4},H)$-free graphs for $H\in$ {diamond, kite, gem}, and get some structural properties of these graphs. For graphs $G$ and $H$, we use $G+H$ to denote a graph with vertex set $V(G)\cup V(H)$ and edge set $E(G)\cup E(H)\cup\{uv\;|\;u\in V(G),v\in V(H)\}$. A bisimplicial vertex is one whose neighborhoods is the union of two cliques.

There exist graphs showing that all the requirements in our theorems are necessary. We will give these examples separately after completing their proofs. As immediate consequences of these theorems, we have the following corollaries that generalize some results of Choudum, Karthick and Belavadi [4], and of Lan, Zhou and Liu [14].

We will prove Theorem 1.1 in Section 2, prove Theorem 1.2 in Section 3, and prove Theorem 1.3 in Section 4.

In this section, we focus on $(P_{7},C_{4}$, diamond)-free graphs, and prove Theorem 1.1 and Corollary 1.1. Theorem 1.1 generalizes the following result from [4].

It is certain that $X\cup Y\subseteq N(A)$. Since $G$ is diamond-free, we have that no vertex of $N(A)$ may have three consecutive neighbors in $A$. Let $x\in N(A)$. By symmetry, we may assume $x\sim v_{1}$. If $N(x)\cap A=\{v_{1}\}$, then $G[\{x,v_{1},v_{2},v_{3},v_{4},v_{5},v_{6}\}]=P_{7}$, a contradiction. Therefore $|N(x)\cap A|\geq 2$.

Suppose $x\sim v_{2}$. Then, $x$ has neighbors in $\{v_{3},\cdots,v_{7}\}$ as otherwise $G[\{x,v_{2},v_{3},v_{4},v_{5},v_{6},v_{7}\}]=P_{7}$, and $x\not\sim v_{3}$ and $x\not\sim v_{7}$ as otherwise $x$ has three consecutive neighbors in $A$. To forbid a 4-hole on $\{x,v_{2},v_{3},v_{4}\}$ or $\{x,v_{1},v_{6},v_{7}\}$, we have that $x\not\sim x_{4}$ and $x\not\sim x_{6}$. Therefore $x\sim v_{5}$, which implies that $x\in Y$. By symmetry, we have $x\in Y$ when $x\sim v_{7}$. So, we way assume that $x\not\sim v_{2}$ and $x\not\sim v_{7}$, and thus $x\not\sim v_{3}$ and $x\not\sim v_{6}$ to forbid a 4-hole on $\{x,v_{1},v_{2},v_{3}\}$ or $\{x,v_{1},v_{6},v_{7}\}$.

Now, $x\in X$ if $x$ has exactly one neighbor in $\{v_{4},v_{5}\}$, and $x\in Y$ if $x$ is complete to $\{v_{4},v_{5}\}$. This proves (1).

Suppose to its contrary, we may assume by the symmetry that $X_{1}$ or $Y_{1}$ has two distinct vertices $x$ and $x^{\prime}$. Then $G[\{x,x^{\prime},v_{1},v_{4}\}]$ is a diamond if $x\sim x^{\prime}$, and $xv_{1}x^{\prime}v_{4}x$ is a 4-hole if $x\not\sim x^{\prime}$. Both are contradictions. Therefore, (2) holds.

For $i\in\{1,\cdots,7\}$, let $X_{i}=\{x_{i}\}$ if $X_{i}\neq\mbox{{\rm\O}}$, and $Y_{i}=\{y_{i}\}$ if $Y_{i}\neq\mbox{{\rm\O}}$.

Suppose to its contrary that $N(A)$ has two adjacent vertices $u$ and $v$. First we assume that $\{u,v\}\cap X\neq\mbox{{\rm\O}}.$ Without loss of generality, suppose that $u\in X_{1}.$ Then $v\not\sim v_{j}$ for $j\in${2, 7} as otherwise $G[\{u,v,v_{1},v_{j}\}]$ is a diamond if $v\sim v_{1}$, and $uvv_{j}v_{1}u$ is a 4-hole if $v\not\sim v_{1}$. Moreover, $v\not\sim v_{k}$ for $k\in\{3,5\}$ as otherwise $G[\{u,v,v_{4},v_{k}\}]$ is a diamond if $v\sim v_{4}$, and $uvv_{k}v_{4}u$ is a 4-hole if $v\not\sim v_{4}$. Therefore $N(v)\cap A\subseteq\{v_{1},v_{4},v_{6}\}$. Since $|N(v)\cap A|\geq 2$ and $|X_{1}|\leq 1$, we have that $N(v)\cap A\in\{\{v_{1},v_{6}\},\{v_{4},v_{6}\},\{v_{1},v_{4},v_{6}\}\}$. Then $vv_{6}v_{7}v_{1}v$ is a 4-hole if $N(v)\cap A\in\{\{v_{1},v_{6}\},\{v_{1},v_{4},v_{6}\}\}$, and $vv_{4}v_{5}v_{6}v$ is a 4-hole if $N(v)\cap A=\{v_{4},v_{6}\}$, both are contradictions.

So, we may assume that $u\in Y$ and $v\in Y$. Without loss of generality, suppose that $u\in Y_{1}$. Then $v\not\sim v_{j}$ for $j\in\{2,7\}$ as otherwise $G[\{u,v,v_{1},v_{j}\}]$ is a diamond if $v\sim v_{1}$, and $uvv_{j}v_{1}u$ is a 4-hole if $v\not\sim v_{1}$. Moreover, $v\not\sim v_{3}$ to forbid a diamond on $\{u,v,v_{3},v_{4}\}$ if $v\sim v_{4}$ or a 4-hole on $\{u,v,v_{3},v_{4}\}$ if $v\not\sim v_{4}$. If $v\sim v_{5}$, then $v\not\sim v_{4}$ as otherwise $v\in Y_{1}$ and thus $|Y_{1}|\geq 2$, a contradiction. But now $G[\{u,v,v_{4},v_{5}\}]$ is a diamond. Therefore, $N(v)\cap A=\{v_{1},v_{4},v_{6}\}$ since $|N(v)\cap A|=3$. However, $vv_{4}v_{5}v_{6}v$ is a 4-hole, a contradiction. This proves (3).

If this is not true, we may assume by symmetry that $X_{1}\neq\mbox{{\rm\O}}$ and $X_{3}\neq\mbox{{\rm\O}}$, then $x_{1}\not\sim x_{3}$ by (3), which implies that $G[\{x_{1},x_{3},v_{1},v_{2},v_{4},v_{5},v_{6}\}]=P_{7}$, a contradiction. This proves (4).

If (5) does not hold, we may assume by symmetry that $Y_{1}\neq\mbox{{\rm\O}}$ and $Y_{4}\neq\mbox{{\rm\O}}$, then $y_{1}\not\sim y_{4}$ by (3), which implies that $y_{1}v_{4}y_{4}v_{1}y_{1}$ is a 4-hole, a contradiction. This proves (5).

If it is not the case, we may assume by symmetry that $X_{1}\neq\mbox{{\rm\O}}$ and $Y_{1}\cup Y_{2}\cup Y_{3}\cup Y_{4}\neq\mbox{{\rm\O}}$. Let $y\in Y_{1}\cup Y_{2}\cup Y_{3}\cup Y_{4}$. Then $x_{1}\not\sim y$ by (3), which implies that $x_{1}v_{4}yv_{1}x_{1}$ is a 4-hole if $y\in Y_{1}\cup Y_{4}$, $G[\{x_{1},y,v_{2},v_{3},v_{4},v_{6},v_{7}\}]=P_{7}$ if $y\in Y_{2}$, and $G[\{x_{1},y,v_{1},v_{2},v_{4},v_{5},v_{6}\}]=P_{7}$ if $y\in Y_{3}$, all of which are contradictions. This proves (6).

Suppose that $Y=\mbox{{\rm\O}}.$ By (4), we have that $\{X_{1},X_{2},\cdots,X_{7}\}$ has at most three nonempty elements, and thus $d(v_{i})=2$ for some $i\in\{1,\cdots,7\}$. If $\omega(G)=2$, then $Y=\mbox{{\rm\O}}$, and thus $\delta(G)\leq 2$.

Suppose that $\omega(G)\geq 4$ and suppose by symmetry that $Y_{1}\neq\mbox{{\rm\O}}$. Then, $Y_{4}\cup Y_{5}\cup X_{1}\cup X_{7}\cup X_{6}\cup X_{5}=\mbox{{\rm\O}}$ by (5) and (6), and one can verify easily that $d(v_{1})=3\leq\omega(G)-1\leq\max\{2,\omega(G)-1\}$.

So, we suppose that

• •

  $\omega(G)=3$ and $\delta(G)\geq 3$,

• •

  $G$ is not isomorphic to $F$ and has no clique cutsets, and

• •

  $Y_{i_{0}}=\{y_{i_{0}}\}\neq\mbox{{\rm\O}}$ for some integer $i_{0}$.

By (5) and (6), we have that $Y_{i_{0}+3}\cup Y_{i_{0}+4}\cup X_{i_{0}}\cup X_{i_{0}-1}\cup X_{i_{0}-2}\cup X_{i_{0}-3}=\mbox{{\rm\O}}$, and so

We will show that

For simplicity, we may take $i_{0}=1$ by symmetry to illustrate the procedure. Before proving (8), we firstly prove that

Notice that we take $i_{0}=1$. Suppose $X_{2}\neq\mbox{{\rm\O}}$ and $X_{3}\neq\mbox{{\rm\O}}$. By (4) and (6), we have that $X_{4}\cup X_{7}\cup Y_{2}\cup Y_{3}\cup Y_{4}\cup Y_{5}\cup Y_{6}=\mbox{{\rm\O}}$, and so $X=X_{2}\cup X_{3}$ and $Y=Y_{1}\cup Y_{7}$. Since $Y_{7}=\mbox{{\rm\O}}$ implies that $d(v_{7})=2$, we have that $Y_{7}=\{y_{7}\}\neq\mbox{{\rm\O}}$.

Since $\delta(G)\geq 3$ and $N(A)$ is a stable set, both $x_{2}$ and $x_{3}$ have neighbors in $R$. Let $u$ be a neighbor of $x_{2}$ in $R$ and $v$ be a neighbor of $x_{3}$ in $R$, and let $S$ and $T$ be the components of $G[R]$ which contains $u$ and $v$, respectively. Since $G[\{x_{2},v_{1},v_{2},v_{6},v_{7}\}]=P_{5}$ and $G$ is $P_{7}$-free, we have that $x_{2}$ must be complete to $V(S)$, and thus $\omega(S)\leq 2$ as $\omega(G)=3$. Moreover, $S$ is $P_{3}$-free as $G$ is diamond-free, and thus $S$ is a $K_{1}$ or a $K_{2}$. Similarly, $x_{3}$ must be complete to $V(T)$, and $T$ is a $K_{1}$ or a $K_{2}$.

Suppose that $S=T$. Then $|V(S)|=|V(T)|=1$ as otherwise $G[\{x,y,x_{2},x_{3}\}]$ is a diamond for any edge $xy$ of $S$, and so $V(S)=V(T)=\{u\}=\{v\}$. If $u\sim y_{7}$, then $v_{3}x_{3}uy_{7}v_{3}$ is a 4-hole, a contradiction. So, $u\not\sim y_{7}$. By symmetry, $u\not\sim y_{1}$. But now $d(u)=2$, a contradiction. Therefore, $S\neq T$.

If $V(S)=\{u\}$, then we have that $u\not\sim y_{1}$ to forbid a 4-hole on $\{u,y_{1},x_{2},v_{5}\}$, and thus $u\sim x_{3}$ and $u\sim y_{7}$ since $\delta(G)\geq 3$, which implies that there is a 4-hole on $\{u,x_{3},v_{3},y_{7}\}$, a contradiction. So, $S$ is a $K_{2}$. Similarly, $T$ is a $K_{2}$. Let $V(S)=\{u,u^{\prime}\}$ and $V(T)=\{v,v^{\prime}\}$.

To forbid a 4-hole on $\{y_{1},x_{2},v_{5},u\}$ or $\{y_{1},x_{2},v_{5},u^{\prime}\}$, $y_{1}$ must be anticomplete to $V(S)$. Similarly, $y_{7}$ must be anticomplete to $V(T)$. Since $\delta(G)\geq 3$, we have that each vertex of $S$ has neighbors in $\{x_{3},y_{7}\}$. Since $G$ is diamond-free, we may assume, without loss of generality, that $u\sim x_{3}$ and $u^{\prime}\sim y_{7}$. Similarly, we may suppose that $v\sim x_{2}$ and $v^{\prime}\sim y_{1}$. But now, $ux_{2}vx_{3}u$ is a 4-hole, a contradiction. This proves (9).

We can now turn to prove (8). Recall that we take $i_{0}=1$ by symmetry.

Suppose to its contrary that $X_{2}\neq\mbox{{\rm\O}}$. Then $X_{3}=\mbox{{\rm\O}}$ by (9). By (4) and (6), we have that $X_{4}\cup X_{7}\cup Y_{2}\cup Y_{3}\cup Y_{4}\cup Y_{5}=\mbox{{\rm\O}}$, and thus $X=X_{2}$ and $Y=Y_{1}\cup Y_{6}\cup Y_{7}$. Since, for $i\in\{6,7\}$, $Y_{i}=\mbox{{\rm\O}}$ implies that $d(v_{i})=2$, we have that $Y_{6}\neq\mbox{{\rm\O}}$ and $Y_{7}\neq\mbox{{\rm\O}}$. Since $\delta(G)\geq 3$ and $N(A)$ is a stable set, we have that $x_{2}$ must have a neighbor, say $u$, in $R$. Let $B$ be the component of $G[R]$ that contains $u$.

Since $G$ is $P_{7}$-free and diamond-free, we have that $x_{2}$ must be complete to $V(B)$ and $B$ is a $K_{1}$ or a $K_{2}$. Suppose that $V(B)=\{u\}$. Then, $u$ has at least two neighbors in $N(A)\setminus\{x_{2}\}$ as $\delta(G)\geq 3$. Since $u\sim y_{1}$ implies that there is a 4-hole on $\{u,y_{1},x_{2},v_{5}\}$, we have that $u\not\sim y_{1}$, and thus $u\sim y_{6}$ and $u\sim y_{7}$. But then, $uy_{6}v_{3}y_{7}u$ is a 4-hole, a contradiction. So $B$ is a $K_{2}$. Let $V(B)=\{u,v\}$. To forbid a 4-hole on $\{a,y_{1},x_{2},v_{5}\}$ or $\{a,x_{2},v_{2},y_{6}\}$, where $a\in V(B)$, we have that $V(B)$ is anticomplete to {$y_{1}$, $y_{6}$}. Since $\delta(G)\geq 3$, each vertex in $B$ must have neighbor in $\{y_{7}\}$, which implies a diamond $G[\{u,v,x_{2},y_{7}\}]$, a contradiction. This proves (8).

Next, we prove that

We still take $i_{0}=1$ for simplicity. By (7) and (8), we have that $X=X_{3}\cup X_{4}$ and $Y=Y_{1}\cup Y_{2}\cup Y_{3}\cup Y_{6}\cup Y_{7}$.

Suppose $X_{3}\neq\mbox{{\rm\O}}$. By (6), we have that $Y_{3}\cup Y_{4}\cup Y_{5}\cup Y_{6}=\mbox{{\rm\O}}$, and so $X=X_{3}\cup X_{4}$ and $Y=Y_{1}\cup Y_{2}\cup Y_{7}$. Since $d(v_{2})\geq 3$, we have that $Y_{2}\neq\mbox{{\rm\O}}$, and consequently $X_{3}=\mbox{{\rm\O}}$ by taking $i_{0}=2$ in (8), a contradiction. Therefore, $X_{3}=\mbox{{\rm\O}}$, and $X=X_{4}$ and $Y=Y_{1}\cup Y_{2}\cup Y_{3}\cup Y_{6}\cup Y_{7}$.

Suppose that $X_{4}\neq\mbox{{\rm\O}}$. Then, $Y_{4}\cup Y_{5}\cup Y_{6}\cup Y_{7}=\mbox{{\rm\O}}$ by (6), and so $Y=Y_{1}\cup Y_{2}\cup Y_{3}$. Moreover, $Y_{2}\neq\mbox{{\rm\O}}$ and $Y_{3}\neq\mbox{{\rm\O}}$ as $d(v_{2})\geq 3$ and $d(v_{3})\geq 3$. Consequently, we have a contradiction as $X_{4}=\mbox{{\rm\O}}$ by taking $i_{0}=3$ in (8). Therefore, $X_{4}=\mbox{{\rm\O}}$, and thus (10) holds.

Recall that $Y_{i_{0}}\neq\mbox{{\rm\O}}$ by our assumption. By symmetry, we may set $i_{0}=1$ in (10). Then, we have that

We prove further that

Suppose that $Y_{2}\neq\mbox{{\rm\O}}$, $Y_{3}\neq\mbox{{\rm\O}}$, and $R\neq\mbox{{\rm\O}}$. By (5), we have that $Y_{5}\cup Y_{6}\cup Y_{7}=\mbox{{\rm\O}}$, and thus $Y=Y_{1}\cup Y_{2}\cup Y_{3}=\{y_{1},y_{2},y_{3}\}$. Let $Q$ be a component of $G[R]$. Since $G$ is connected, there must exist a path $P$ between $Q$ and $Y$. If we can show that $P$ must contain $y_{2}$, then $y_{2}$ is a cutvertex of $G$, which leads to a contradiction. Suppose that $V(P)\cap Y=\{y\}$. To prove (12), it suffices to prove the following Claim 2.1.

Suppose to the contrary that $N_{2}$ has two adjacent vertices $x^{\prime}$ and $y^{\prime}$. Let $S^{\prime}$ be the component of $G[N_{2}]$ which contains $x^{\prime}$ and $y^{\prime}$. Let $x^{\prime\prime}$ be a neighbor of $x^{\prime}$ in $N_{1}$. Since $d(u,y_{3})\leq 2$ for each vertex $u$ of $Q$, we have that $x^{\prime\prime}$ is complete to $V(S^{\prime})$, which implies that $\omega(S^{\prime})=2$ as $\omega(G)=3$. Moreover, $S^{\prime}$ is $P_{3}$-free as $G$ is diamond-free. Then, $S^{\prime}=x^{\prime}y^{\prime}$.

Let $S^{\prime\prime}$ be the component of $G[N_{1}]$ which contains $x^{\prime\prime}$. Note that $S^{\prime\prime}$ is a $K_{1}$ or a $K_{2}$. For each vertex $t_{1}\in N_{1}\setminus\{x^{\prime\prime}\}$ and $t_{2}\in V(S^{\prime})$, to avoid a 4-hole on $\{x^{\prime\prime},y_{3},t_{1},t_{2}\}$ if $t_{1}\not\sim x^{\prime\prime}$ or a diamond on $\{y_{3},x^{\prime\prime},t_{1},t_{2}\}$ if $t_{1}\sim x^{\prime\prime}$, $\{x^{\prime},y^{\prime}\}$ must be anticomplete to $N_{1}\setminus\{x^{\prime\prime}\}$, which implies that $x^{\prime}$ and $y^{\prime}$ both have a neighbor in $\{y_{1},y_{2}\}$ as $\delta(G)\geq 3$.

Suppose $x^{\prime}\sim y_{1}$ and $y^{\prime}\sim y_{1}$. Then $x^{\prime\prime}\sim y_{1}$ to forbid a diamond on $\{x^{\prime},y^{\prime},x^{\prime\prime},y_{1}\}$. If $y_{2}$ has a neighbor in $S^{\prime}$, say $x^{\prime}$ by symmetry, then $y_{2}\sim y^{\prime}$ to forbid an induced $P_{7}$ on $\{x^{\prime},y^{\prime},y_{2},y_{3},v_{1},v_{2},v_{7}\}$, and thus $G[\{x^{\prime},y^{\prime},y_{1},y_{2}\}]$ is a diamond, a contradiction. So, $y_{2}$ has no neighbors in $S^{\prime}$, and thus $\{x^{\prime\prime},y_{1}\}$ is a clique cutset, which leads to a contradiction. This proves that $x^{\prime}\not\sim y_{1}$ or $y^{\prime}\not\sim y_{1}$.