Tree independence number
II. Three-path-configurations.

Let $X=((N(s_{1})\cup N(s_{2}))\setminus W)\cup N(v)$. Suppose for a contradiction that there is a component of $G\setminus X$ intersecting both $S^{*}$ and $W\setminus S$. It follows that there is a path $P=p_{1}\hbox{-}\dots\hbox{-}p_{k}$ in $G\setminus X$, possibly with $k=1$, such that $p_{1}$ has a neighbor in $S^{*}$ and $p_{k}$ has a neighbor in $W\setminus S$. In particular, $P$ is disjoint from and anticomplete to $\{s_{1},s_{2},v\}$.

Choose $P$ with $|P|=k$ as small as possible. It follows that

• •

  we have $P\subseteq G\setminus(W\cup X)$;

• •

  $P^{*}$ is anticomplete to $W\cup\{v\}$;

• •

  if $k>1$, then $p_{1}$ is anticomplete to $W\setminus S^{*}$ and $p_{k}$ is anticomplete to $S$.

Let $t_{1}$ and $t_{2}$ be the (unique) neighbors of $s_{1}$ and $s_{2}$ in $W\setminus S^{*}$, respectively. Since $(W,v)$ has at least two long sectors, it follows that $s_{1},s_{2},t_{1},t_{2}$ are all distinct, and that $W\setminus N[S]\neq\emptyset$. In particular, since $W\cup\{v\}$ is not a pyramid, pinched prism, or theta, it follows that $v$ has a neighbor in $w\in W\setminus N[S]$.

Traversing $S$ from $s_{1}$ to $s_{2}$, let $u_{1}$ and $u_{2}$ be the first and the last neighbor of $p_{1}$ in $S$, respectively. It follows that $u_{1},u_{2}\in S^{*}$. We deduce:

Suppose not. Then there is a path $Q$ in $G$ from $p_{k}$ to $v$ such that $Q^{*}\subseteq W\setminus N[S]$. Assume that $u_{1}=u_{2}$. Then there is a theta in $G$ with ends $u_{1},v$ and paths $u_{1}\hbox{-}S\hbox{-}s_{1}\hbox{-}v,u_{1}\hbox{-}S\hbox{-}s_{2}\hbox{-}v,u_{1}\hbox{-}p_{1}\hbox{-}P\hbox{-}p_{k}\hbox{-}Q\hbox{-}v$. Next, assume that $u_{1}$ and $u_{2}$ are distinct and non-adjacent. Then there is a theta in $G$ with ends $p_{1},v$ and paths $p_{1}\hbox{-}u_{1}\hbox{-}S\hbox{-}s_{1}\hbox{-}v,p_{1}\hbox{-}u_{2}\hbox{-}S\hbox{-}s_{2}\hbox{-}v,p_{1}\hbox{-}P\hbox{-}p_{k}\hbox{-}Q\hbox{-}v$. Since $G$ is theta-free, it follows that $u_{1},u_{2}$ are distinct and adjacent. But now there is a pyramid in $G$ with apex $v$, base $p_{1}u_{1}u_{2}$ and paths $p_{1}\hbox{-}P\hbox{-}p_{k}\hbox{-}Q\hbox{-}v,u_{1}\hbox{-}S\hbox{-}s_{1}\hbox{-}v,u_{2}\hbox{-}S\hbox{-}s_{2}\hbox{-}v$, a contradiction. This proves (2).

Suppose not. By (2) and without loss of generality, we may assume that $p_{k}$ is adjacent to $t_{1}$. Assume first that $t_{1}$ and $v$ are not adjacent. If $u_{1}$ and $u_{2}$ are not adjacent either, then there is a theta in $G$ with ends $p_{1},s_{1}$ and paths $p_{1}\hbox{-}u_{1}\hbox{-}S\hbox{-}s_{1},p_{1}\hbox{-}u_{2}\hbox{-}S\hbox{-}s_{2}\hbox{-}v\hbox{-}s_{1},p_{1}\hbox{-}P\hbox{-}p_{k}\hbox{-}t_{1}\hbox{-}s_{1}$, and if $u_{1},u_{2}$ are adjacent, then there is a pyramid in $G$ with apex $s_{1}$, base $p_{1}u_{1}u_{2}$ and paths $p_{1}\hbox{-}P\hbox{-}p_{k}\hbox{-}t_{1}\hbox{-}s_{1},u_{1}\hbox{-}S\hbox{-}s_{1},u_{2}\hbox{-}S\hbox{-}s_{2}\hbox{-}v\hbox{-}s_{1}$. Since $G$ is (theta, pyramid)-free, it follows that $t_{1}$ and $v$ are adjacent. Assume that $u_{1}$ and $u_{2}$ are not adjacent. Then there is a pyramid in $G$ with apex $p_{1}$, base $s_{1}t_{1}v$ and paths paths $s_{1}\hbox{-}S\hbox{-}u_{1}\hbox{-}p_{1},t_{1}\hbox{-}p_{k}\hbox{-}P\hbox{-}p_{1},v\hbox{-}s_{2}\hbox{-}S\hbox{-}u_{2}\hbox{-}p_{1}$. Again, since $G$ is pyramid-free, it follows that $u_{1},u_{2}$ are adjacent. But now there is a prism in $G$ with triangles $u_{1}p_{1}u_{2},s_{1}t_{1}v$ and paths $u_{1}\hbox{-}S\hbox{-}s_{1},p_{1}\hbox{-}P\hbox{-}p_{k}\hbox{-}t_{1},u_{2}\hbox{-}S\hbox{-}s_{2}\hbox{-}v$, a contradiction. This proves (2).

Henceforth, let $u=u_{1}=u_{2}$. It follows that:

Suppose that $k>1$. Since $W\cup\{p_{k}\}$ is not a theta with ends $t_{1},t_{2}$, we may assume by (2) and without loss of generality, that $N(p_{k})\cap(W\setminus S)=\{t_{1}\}$. But then $W\cup P$ is a theta in $G$ with ends $u,t_{1}$, a contradiction. This proves (2).

Henceforth, let $p=p_{1}=p_{k}$. Since $W\cup\{p\}$ is not a theta with one end $u$ and the other end in $\{t_{1},t_{2}\}$, (2) implies that $N(p)\cap W=\{t_{1},t_{2},u\}$. Recall that $v$ has a neighbor $w\in W\setminus N[S]$.

Suppose not. Then we may assume, without loss of generality, that either $t_{1}$ and $w$ are not adjacent, or $t_{2}$ and $v$ are adjacent. In either case, it follows that there is a path $Q$ in $G$ from $p$ to $v$ such that $t_{2}\in P^{*}\subseteq W\setminus(S\cup N(t_{1}))$. Now there is a theta in $G$ with ends $p,s_{1}$ and paths $p\hbox{-}u\hbox{-}S\hbox{-}s_{1},p\hbox{-}t_{1}\hbox{-}s_{1},p\hbox{-}Q\hbox{-}v\hbox{-}s_{1}$, a contradiction. This proves (2).

We will now finish the proof. From (2), it follows that $W\setminus S=t_{1}\hbox{-}w\hbox{-}t_{2}$ and $N(v)\cap W=\{s_{1},s_{2},w\}$. Recall also that $N(p)\cap W=\{t_{1},t_{2},u\}$. Since $|W|>6$, it follows either $s_{1}\hbox{-}S\hbox{-}u$ or $s_{2}\hbox{-}S\hbox{-}u$, say the former, has non-empty interior. But then there is a theta in $G$ with ends $s_{1},u$ and paths $s_{1}\hbox{-}S\hbox{-}u,s_{1}\hbox{-}t_{1}\hbox{-}p\hbox{-}u$ and $s_{1}\hbox{-}v\hbox{-}s_{2}\hbox{-}S\hbox{-}u$, a contradiction. ∎

For every vertex $y\in Y$, let $P_{y}$ be the path in $D_{1}$ such that $y$ is complete to the ends of $P_{y}$ and anticomplete to $D_{1}\setminus P_{y}$. Let $I$ be the graph with $V(I)=Y$, such that two distinct vertices $y,y^{\prime}\in Y$ are adjacent in $I$ if and only if $P_{y}\cap P_{y^{\prime}}\neq\emptyset$. Then $I$ is an interval graph, and so by [9], $I$ is perfect. Since $|V(I)|=(3x+2)(c+2)$, we deduce that $I$ contains either a clique of cardinality $c+2$ of a stable set of cardinality $3x+2$.

Assume that $I$ contains a clique of cardinality $c+2$. Then there exists $C\subseteq Y$ with $|C|=c+2$ and $d\in D_{1}$ such that $d\in P_{y}$ for every $y\in C$. It follows that for every $y\in C$, either $y$ is adjacent to $d$, or $D_{1}\setminus d$ has two components and $y$ has a neighbor in each of them. Since $|N(d)\cap Y|\leq c$, we deduce that there are two vertices $y,y^{\prime}\in C\subseteq Y$ as well as two paths $P_{1}$ and $P_{2}$ from $y$ to $y^{\prime}$ with disjoint and anticomplete interiors contained in $D_{1}$. On the other hand, since $D_{2}$ is connected and $N(D_{2})=Y$, it follows that there is a path $P_{3}$ in $G$ from $y$ to $y^{\prime}$ whose interior is contained in $D_{2}$. But now there is a theta in $G$ with ends $y,y^{\prime}$ and paths $P_{1},P_{2},P_{3}$, a contradiction.

We deduce that $I$ contains a stable set $S$ of cardinality $3x+2$. From the definition of $I$, it follows that there is a subpath $H_{1}$ of $D_{1}$ such that $(H_{1},S)$ is an alignment. Hence, there exists $X\subseteq S\subseteq Y$ with $|X|=x$ such that $(H_{1},X)$ is a consistent alignment. This completes the proof of Lemma 3.4. ∎

Let $\sigma(x)=18\mu(3((x+2)^{2}+1)(x+1))$, where $\mu(\cdot)$ comes from Theorem 3.3. We begin with the following:

Suppose for a contradiction that for some $i\in\{1,\dots,k\}$, there is a subset $Z\subseteq Y$ of cardinality five such that $d_{i}$ is complete to $Z$. It follows that for every $y\in Z$, we have $i(y)\neq i$, and so there is $3$-subset $T$ of $Z$ such that either $i(y)<i$ for all $y\in T$ or $i<i(y)$ for all $y\in T$. Consequently, there are two distinct vertices $y,y^{\prime}\in T\subseteq Z\subseteq Y$ for which $d_{i}$ is disjoint from and anticomplete to $d_{i(y)}\hbox{-}D_{1}\hbox{-}d_{i(y^{\prime})}$. On the other hand, since $D_{2}$ is connected and $N(D_{2})=Y$, it follows that there is a path $Q$ in $G$ from $y$ to $y^{\prime}$ whose interior is contained in $D_{2}$. But now there is a theta in $G$ with ends $y,y^{\prime}$ and paths $y\hbox{-}d_{i(y)}\hbox{-}D_{1}\hbox{-}d_{i(y^{\prime})}\hbox{-}y^{\prime},y\hbox{-}d_{i}\hbox{-}y^{\prime},Q$, a contradiction. This proves (3).

From (3), Lemma 3.4 and the choice of $\sigma(x)$, it follows that:

Henceforth, let $Y_{1}$ be as in (3). Since $G_{1}=G[Y_{1}\cup D_{2}]$ and $G_{1}\setminus Y_{1}=D_{2}$ are both connected, we can apply Theorem 3.3 to $G_{1}$ and $Y_{1}$. It follows that there is a set $Y^{\prime}\subseteq Y_{1}$ with $|Y^{\prime}|=3((x+2)^{2}+1)(x+1)$ and an induced subgraph $H$ of $G_{1}\setminus Y_{1}=D_{2}$ for which one of the following holds.

• •

  $H$ is a path and every vertex of $Y^{\prime}$ has a neighbor in $H$.

• •

  $H$ is a caterpillar, or the line graph of a caterpillar, or a subdivided star or the line graph of a subdivided star. Moreover, every vertex of $Y^{\prime}$ has a unique neighbor in $H$ and every vertex of $H\cap N(Y^{\prime})$ is simplicial in $H$.

Assume that the second bullet above holds. By (3), $(H_{1},Y^{\prime})$ is a consistent alignment. But then it is straightforward to observe that $G$ contains either a theta, a prism or a pyramid, a contradiction. It follows that $H$ is indeed a path and every vertex of $Y^{\prime}$ has a neighbor in $H$.

Now, assume that some vertex in $z\in H$ has at least $x$ neighbors in $Y^{\prime}$. Choose $X\subseteq N(z)\cap Y^{\prime}\subseteq Y$ with $|X|=x$. Let $H_{2}=\{z\}$. By (3), $(H_{1},X)$ is a consistent alignment. Note that if $(H_{1},X)$ is spiky, then $H_{1}\cup X\cup\{z\}$ contains a theta, and if $(H_{1},X)$ is triangular, then $H_{1}\cup X\cup\{z\}$ contains a pyramid. Therefore, $(H_{1},X)$ is wide. But now $X$ and $H_{2}$ satisfy Theorem 3.2.

Therefore, we may assume that every vertex in $H$ has fewer than $x$ neighbors in $Y^{\prime}$. Let $H_{2}=H$. Since $|Y^{\prime}|=3((x+2)^{2}+1)(x+1)$, it follows from Lemma 3.4 that there exists $X^{\prime}\subseteq Y^{\prime}$ with $|X^{\prime}|=(x+2)^{2}$ such that $(H_{2},X^{\prime})$ is a consistent alignment. Also, by (3), $(D_{1},X^{\prime})$ is a consistent alignment. This, along with the Erdős-Szekeres Theorem [7], implies that there exists $X\subseteq X^{\prime}\subseteq Y^{\prime}\subseteq Y$ with $|X|=x+2$ such that both $(H_{1},X)$ and $(H_{2},X)$ are consistent alignments, and the orders given on $X$ by $(H_{1},X)$ and $(H_{2},X)$ are the same. Moreover, since $G$ is (theta, pyramid, pinched prism)-free, it follows that at least one of $(H_{1},X)$ and $(H_{2},X)$ is wide. Hence, $X$ and $H_{2}$ satisfy Theorem 3.2. This completes the proof. ∎

By Corollary 3.5, $\mathcal{C}$ is an amiable class. With same notation as in the definition of a $3$-amicable class, our goal is to show that there exists a subset $Z\subseteq D_{2}\cup\{d_{i+2},\dots,d_{j-2}\}\cup\{x_{4}\}$ with $|Z|\leq 3$ such that $N[Z]$ separates $d_{i}$ from $d_{j}$. Consequently, $N[Z]$ separates $d_{1}\hbox{-}D_{1}\hbox{-}d_{i}$ from $d_{j}\hbox{-}D_{1}\hbox{-}d_{k}$.

Let $l\in\{1,\ldots,k\}$ be minimum such that $x_{4}$ is adjacent to $d_{l}$, and let $m\in\{1,\ldots,k\}$ be maximum such that $x_{4}$ is adjacent to $d_{m}$. It follows that $i+2<l\leq m<j-2$.

Let $R$ be the (unique) path in $H_{2}$ with ends $r_{1},r_{2}$ (possibly $r_{1}=r_{2}$) such that $x_{1}$ is adjacent to $r_{1}$ and anticomplete to $R\setminus\{r_{1}\}$ and $x_{7}$ is adjacent to $r_{2}$ and anticomplete to $R\setminus\{r_{2}\}$. Then $x_{4}$ has a neighbor in $R$. Traversing $R$ from $r_{1}$ to $r_{2}$, let $z_{1}$ and $z_{2}$ be the first and the last neighbor of $x_{4}$ in $R$.

Let $W=d_{i}\hbox{-}D_{1}\hbox{-}d_{j}\hbox{-}x_{7}\hbox{-}r_{2}\hbox{-}R\hbox{-}r_{1}\hbox{-}x_{1}\hbox{-}d_{i}$. Then $W$ is a hole in $G$ and $|W|\geq 7$. It follows that $(W,x_{4})$ is a useful wheel in $G$. In particular, $S=d_{l}\hbox{-}D_{1}\hbox{-}d_{i}\hbox{-}x_{1}\hbox{-}r_{1}\hbox{-}R\hbox{-}z_{1}$ and $S^{\prime}=d_{m}\hbox{-}D_{1}\hbox{-}d_{j}\hbox{-}x_{7}\hbox{-}r_{2}\hbox{-}R\hbox{-}z_{2}$ are two long sectors of $(W,x_{4})$.