# Detecting *K*\_2, 3 as an Induced Minor

Clément Dallard, Maël Dumas, Claire Hilaire, Martin Milanič, Anthony Perez, Nicolas Trotignon

► To cite this version:

Clément Dallard, Maël Dumas, Claire Hilaire, Martin Milanič, Anthony Perez, et al. Detecting *K*\_2, 3 as an Induced Minor. IWOCA 2024, Jul 2024, Ischia, Italy. pp.151-164, 10.1007/978-3-031-63021-7_12. hal-04659728

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<figure><img src="image_1.png" /></figure>
Detecting $K_{2,3}$ as an induced minor

Clément Dallard¹, Maël Dumas², Claire Hilaire³, Martin Milanič³, Anthony Perez², and Nicolas Trotignon⁴

¹Department of Informatics, University of Fribourg, Switzerland
firstname.lastname@unifr.ch

²Université d'Orléans, INSA Centre Val de Loire, LIFO EA 4022, Orléans, France
firstname.lastname@univ-orleans.fr

³FAMNIT and IAM, University of Primorska, Slovenia
firstname.lastname@upr.si

⁴LIP, CNRS, École Normale Supérieure de Lyon, France
firstname.lastname@ens-lyon.fr

We consider a natural generalization of chordal graphs, in which every minimal separator induces a subgraph with independence number at most 2. Such graphs can be equivalently defined as graphs that do not contain the complete bipartite graph $K_{2,3}$ as an induced minor, that is, graphs from which $K_{2,3}$ cannot be obtained by a sequence of edge contractions and vertex deletions.

We develop a polynomial-time algorithm for recognizing these graphs. Our algorithm relies on a characterization of $K_{2,3}$-induced minor-free graphs in terms of excluding particular induced subgraphs, called Truemper configurations.

# 1 Introduction

Chordal graphs have been extensively studied in the literature, mainly owing to their well-understood structure, their multiple characterizations and linear-time recognition algorithms (see, e.g. [1]). Among these characterizations, one states that chordal graphs are exactly the graphs in which every minimal separator forms a clique, that is, has independence number at most 1. In this paper, we investigate the complexity of recognizing graphs whose minimal separators have independence number at most 2. More specifically, we design a polynomial-time algorithm that decides whether each minimal separator of a given graph induces a subgraph with independence number at most 2.

Another way to generalize the above characterization of chordal graphs is to require that each minimal separator is a union of two cliques (rather than having independence number at most 2). Graphs satisfying this condition are called *bisimplicial* and were recently studied by Milanič, Penev, Pivač, and Vušković [18]. This result together with [11, Theorem 6.3] implies that the class of bisimplicial graphs forms a generalization of the
class of 1-perfectly orientable graphs (see, e.g., [11, 12]), which in turn form a common
generalization of the classes of chordal graphs and circular-arc graphs.

All these graph classes have the nice property that they are closed under vertex deletions
and edge contractions; in other words, they are closed under induced minors. This is
also the case for several geometrically defined graph classes such as the class of planar
graphs, and for intersection graph classes. More precisely, for any graph class $\mathcal{G}$, the class
of graphs defined as intersection graphs of connected subgraphs of graphs in $\mathcal{G}$ is closed
under induced minors. Among others, this framework captures the classes of chordal
graphs and circular-arc graphs, corresponding respectively to the case when $\mathcal{G}$ is the class
of all trees and all cycles. The connection between induced minors and the property
that each minimal separator of a graph induces a graph with independence number at
most 2 comes from the following result of Dallard, Milanič, and Štorgel [5, Lemma 3.2]:
For any positive integer $q$, a graph $\mathcal{G}$ does not contain the complete bipartite graph
$K_{2,q}$ as an induced minor if and only if each minimal separator of $\mathcal{G}$ induces a graph
with independence number less than $q$. This implies, in particular, that each minimal
separator of a graph $\mathcal{G}$ induces a graph with independence number at most 2 if and only
if $\mathcal{G}$ does not contain $K_{2,3}$ as an induced minor. As shown in [18], the aforementioned
class of bisimplicial graphs is characterized by an infinite family of forbidden induced
minors, that is, graphs that do not belong to the class but for which deleting any vertex
or contracting any edge results in a graph in the class. The smallest forbidden induced
minor of bisimplicial graphs happens to be $K_{2,3}$.

Every induced-minor closed class can be characterized by a family $\mathcal{F}$ of forbidden
induced minors. The set $\mathcal{F}$ may be finite or infinite (as in the case of bisimplicial graphs).<fnref n="1" />
However, contrary to case of the induced subgraph relation, finiteness of the family of
forbidden induced minors does not imply polynomial-time recognition (unless $P = NP$).
Indeed, as shown by Fellows, Kratochvíl, Middendorf, and Pfeiffer in 1995 [8], there exists
a graph $H$ such that determining if a given graph $G$ contains $H$ as an induced minor is
NP-complete. Recently, Korhonen and Lokshtanov showed that this can happen even if
$H$ is a tree [13].

It is thus a natural question to determine for which graphs $H$ determining if a given
graph $G$ contains $H$ as an induced minor can be done in polynomial time. This is the
case for all graphs $H$ on at most four vertices. The problem is particularly simple if $H$ is
a forest on at most four vertices or the disjoint union a vertex and a triangle. Otherwise,
excluding $H$ always leads to a well structured class that is easy to recognize. When $H$ is
the 4-cycle, that is, $K_{2,2}$, the class of graphs excluding $H$ as an induced minor is precisely
the class of chordal graphs. For $H = K_4$, excluding $H$ as an induced minor is equivalent
to excluding it as a minor, which leads to the class of graphs with treewidth at most
two. For the diamond, that is, the graph $H$ obtained from $K_4$ by removing one edge,
excluding $H$ leads to graphs whose blocks are all complete graphs or chordless cycles
(see [10, Theorem 4.6]). For the paw, that is, the graph $H$ obtained from $K_4$ by removing
two adjacent edges, a direct algorithm is easy. In this latter case, we note that excluding

<footnote n="1">¹For a simpler example, consider the case when $\mathcal{F}$ consists of all complements of cycles.</footnote>
$H$ as an induced minor is equivalent to excluding $H$ as an induced topological minor, and this leads to a very simple class [3].

## Results and organization of the paper

Our main result is a polynomial-time algorithm to decide whether an input graph contains $K_{2,3}$ as an induced minor. Our method relies on the study of the induced subgraphs that must be present whenever the input graph contains $K_{2,3}$ as an induced minor.

In Section 3, we prove that all of them are particular instances of the so-called Truemper configurations. Fortunately, the detection of Truemper configurations is a well-studied question (see [6] for the most recent survey). For instance, detecting $K_{2,3}$ as an induced topological minor can be equivalently stated as detecting a Truemper configuration called the *theta*, a problem that can be solved in polynomial time [4, 14]. It turns out that by relying on several such previous works, we can easily detect three of the four Truemper configurations identified in Section 3 (namely the 3-path-configurations known as the *long prisms*, *pyramids*, and *thetas*, the definitions are given later). This is all done in Section 4.

In Section 5, we give an algorithm to detect the last configuration identified in Section 3 (namely the so-called *broken wheel*) provided that the graph does not contain any of the 3-path-configuration detected in Section 4, thus completing our algorithm. Our approach relies on what is called a *shortest path detector*, a method originally invented for the detection of pyramids in [2].

Observe that detecting $K_{2,3}$ as an induced minor might look easy, in particular because the topological version (that is, detecting a *theta*) is known already. We might indeed have overlooked a simpler method to solve it. But several NP-completeness results for questions similar to the ones that we address here (namely the detection of prisms and broken wheel, see below for the details) suggest that a too naive attempt is likely to fail.

All our algorithms are written for the decision version of the problem they address, but writing an algorithm that actually outputs the structure whose existence is decided is straightforward. We omit it for the sake of readability.

We conclude the paper with some open questions, see Section 6.

## 2 Preliminaries

All graphs considered in this paper are finite, simple, and undirected, with $n$ vertices and $m$ edges. Given a graph $G$ and a set $S \subseteq V(G)$, the *subgraph of G induced by S* is the graph denoted by $G[S]$ with vertex set $S$ in which two vertices are adjacent if and only if they are adjacent in $G$. A *subdivision* of a graph $G$ is any graph obtained from $G$ by a sequence of edge subdivisions. Given two graphs $G$ and $H$, the graph $H$ is an *induced subgraph of G* if $H$ can be obtained from $G$ by a sequence of vertex deletions, an *induced minor of G* if $H$ can be obtained from $G$ by a sequence of vertex deletions or edge contractions, and an *induced topological minor* of $G$ if some subdivision of $H$ is an induced subgraph of $G$. Note that $H$ is an induced minor of $G$ if and only if $G$ admits an *induced minor model* of $H$, that is, a set $\{X_v: v \in V(H)\}$ of pairwise disjoint nonempty
subsets of $V(G)$ indexed by the vertices of $H$, each inducing a connected subgraph of $G$
and such that for any two distinct vertices $u, v \in V(H)$, there is an edge in $G$ between a
vertex in $G[X_u]$ and a vertex in $G[X_v]$ if and only if $u$ and $v$ are adjacent in $H$. Given a
graph $G$ and a set $S \subseteq V(G)$, we denote by $N(S)$ the set of vertices in $V(G) \setminus S$ having a
neighbor in $S$ and by $N[S]$ the set $S \cup N(S)$. For $S = \{v\}$ for some $v \in V(G)$, we write
$N(v)$ for $N(\{v\})$ and similarly for $N[v]$. Given two graphs $G$ and $H$, we say that $G$ is
$H$-free if $G$ contains no induced subgraph isomorphic to $H$.

A *path* in a graph *G* is a sequence *v*₁...*v*ₖ of pairwise distinct vertices of *G* such that *v*ᵢ is adjacent to *v*ᵢ₊₁, for every integer *i* such that 0 ≤ *i* ≤ *k* − 1. We also view the empty set as an empty path. Given a nonempty path *P* = *v*₁...*v*ₖ, we define the *length* of *P* as *k* − 1, its *endpoints* as the vertices *v*₁ and *v*ₖ, and its *interior* as the subpath *v*₂...*v*ₖ₋₁ (which is empty if *k* ≤ 2). A *chord* of a path in *G* is an edge in *G* whose endpoints are non-consecutive vertices of the path. A path is *chordless* if it has no chord. Given a path *P* = *v*₁...*v*ₖ and two vertices *a* and *b* on *P* such that *a* = *v*ᵢ and *b* = *v*ⱼ where 1 ≤ *i* ≤ *j* ≤ *k*, we denote by *aPb* the subpath of *P* from *a* to *b*, that is, the path *v*ᵢ...*v*ⱼ. Given an integer *k* ≥ 3, a *cycle of length *k* in a graph *G* is a sequence *v*₁...*v*ₖ*v*₁ of vertices of *G* such that the vertices *v*₁, ..., *v*ₖ are pairwise distinct, *v*ᵢ is adjacent to *v*ᵢ₊₁ for all *i* ∈ {1, ..., *k* − 1}, and *v*ₖ is adjacent to *v*₁. A *chord* of a cycle in *G* is an edge in *G* whose endpoints are non-consecutive vertices of the cycle. A cycle is *chordless* if it has no chord. For convenience, and when there is no ambiguity, we will not distinguish between a path and its vertex set (and the same for a cycle), nor between a set of vertices and the subgraph it induces. For example, for a vertex *v* in a path *P*, we write *P* \ *v* to denote the subgraph induced by *V*(*P*) \ {*v*}, etc.

An *independent set* in a graph *G* is a set of pairwise nonadjacent vertices. A *clique* in *G* is a set of pairwise adjacent vertices. A *triangle* is a clique of size three. The graph *K*<sub>*p*,*q*</sub> is a complete bipartite graph with parts of size *p* and *q*, respectively.

**3 Reducing to Truemper configurations**

In this section, we show that detecting $K_{2,3}$ as an induced minor is in fact equivalent to detecting whether some specific graphs are contained as induced subgraphs. First, we give some definitions.

A *prism* is a graph made of three vertex-disjoint chordless paths $P_1 = a_1...b_1$, $P_2 = a_2...b_2$, $P_3 = a_3...b_3$ of length at least 1, such that $\{a_1, a_2, a_3\}$ and $\{b_1, b_2, b_3\}$ are triangles and no edges exist between the paths except those of the two triangles. A prism is *long* if at least one of its three paths has length at least 2.

A *pyramid* is a graph made of three chordless paths $P_1 = a...b_1$, $P_2 = a...b_2$, $P_3 = a...b_3$ of length at least 1, two of which have length at least 2, vertex-disjoint except at *a*, and such that $\{b_1, b_2, b_3\}$ is a triangle and no edges exist between the paths except those of the triangle and the three edges incident to *a*.

A *theta* is a graph made of three internally vertex-disjoint chordless paths $P_1 = a...b$, $P_2 = a...b$, $P_3 = a...b$ of length at least 2 and such that no edges exist between the
<figure><img src="image_2.png" /><figcaption>Figure 1: A schematic representation of a prism, a long prism, a pyramid, a theta, and a wheel. Dashed edges represent paths of length at least 1. Dotted edges may be present or not.</figcaption></figure>

paths except the three edges incident to $a$ and the three edges incident to $b$. Prisms, pyramids, and thetas are referred to as *3-path-configurations*.

A *hole* in a graph is a chordless cycle of length at least 4. Observe that the lengths of the paths in the three definitions above are designed so that the union of any two of the paths induces a hole. A *wheel* $W = (H, x)$ is a graph formed by a hole $H$ (called the *rim*) together with a vertex $x$ (called the *center*) that has at least three neighbors in the hole. Prism, pyramids, thetas, and wheels are referred to as *Truemper configurations*; see Fig. 1.

A *sector* of a wheel $(H, x)$ is a path $P$ that is contained in $H$, whose endpoints are neighbors of $x$ and whose internal vertices are not. Note that $H$ is edgewise partitioned into the sectors of $W$. Also, every wheel has at least three sectors. A wheel is *broken* if at least two of its sectors have length at least 2.

We denote by $\mathcal{S}$ the class of subdivisions of the graph $K_{1,3}$. The vertex with degree 3 in a graph $G \in \mathcal{S}$ is called the *apex* of $G$. The class $\mathcal{T}$ is the class of graphs that can be obtained from three paths of length at least one by selecting one endpoint of each path and adding three edges between those endpoints so as to create a triangle. The class $\mathcal{M}$ is the class of graphs $H$ that consist of a path $P$ and a vertex $a$, called the *center* of $H$, such that $a$ is nonadjacent to the endpoints of $P$ and $a$ has at least two neighbors in $P$; see Fig. 2. Given a graph $H \in \mathcal{S} \cup \mathcal{T} \cup \mathcal{M}$, the *extremities* of $H$ are the vertices of degree one as well as the center of $H$ in case $H \in \mathcal{M}$. Observe that any $H \in \mathcal{S} \cup \mathcal{T} \cup \mathcal{M}$ has exactly three extremities.

The key ingredient in our proof of Theorem 3.2 is the following sufficient condition for the presence of an induced subgraph from $\mathcal{S} \cup \mathcal{T} \cup \mathcal{M}$ (which is also used later, in the proof of Lemma 4.3).

**Lemma 3.1.** Let $G$ be a graph and $I$ be an independent set in $G$ with $|I| = 3$. If there exists a component $C$ of $G \setminus I$ such that $I \subseteq N(C)$, then there exists an induced subgraph $H$ of $G[N[C]]$ such that $H \in \mathcal{S} \cup \mathcal{T} \cup \mathcal{M}$ and $I$ is exactly the set of extremities of $H$.

*Proof.* Let $I = \{a_1, a_2, a_3\}$ and assume that there exists a component $C$ of $G \setminus I$ such that $I \subseteq N(C)$.

Let $P$ be a chordless path from $a_1$ to $a_2$ in $G[C \cup \{a_1, a_2\}]$. Consider a chordless path $P'$ in $G[C \cup \{a_3\}]$ from $a_3$ to some vertex $u$ adjacent to some internal vertex of $P$
<figure><img src="image_3.png" /><figcaption>Figure 2: A schematic representation of graphs in $\mathcal{S}$, $\mathcal{T}$, and $\mathcal{M}$. Dashed edges represent paths of length at least 1. Dotted edges may be present or not.</figcaption></figure>

(possibly $u = a_3$), chosen so that no vertex of $P' \setminus u$ has a neighbor in the interior of $P$. Let $H$ be the subgraph of $G$ induced by $P \cup P'$. Note that $H$ is an induced subgraph of $G[C \cup \{a_1, a_2, a_3\}]$. We assume that $P$ and $P'$ are chosen subject to the minimality of $|V(H)|$.

Let $v_1$ (resp. $v_2$) be the neighbor of $u$ in $P$ closest to $a_1$ (resp. $a_2$) along $P$.

Assume first that $u = a_3$. Then, $H$ belongs to $\mathcal{S}$ (if $v_1 = v_2$) or to $\mathcal{M}$ (if $v_1 \neq v_2$), in which case the center of $H$ is $a_3$. Note that in both cases the set of extremities of $H$ is $\{a_1, a_2, a_3\} = I$. Hence, from here on we may assume that $u \neq a_3$, so $P'$ has length at least 1.

If both $a_1$ and $a_2$ have neighbors in $P'$, then some chordless path from $a_1$ to $a_2$ with interior in $P'$ and some subpath of $P'$ contradict the minimality of $|V(H)|$. Hence, from here on, we assume that $a_1$ has no neighbors in $P'$.

Assume that $a_2$ has neighbors in $P'$, and let $u'$ be the neighbor of $a_2$ in $P'$ closest to $u$ along $P'$. If $v_1a_2 \notin E(G)$, then the paths $a_1Pv_1uP'u'a_2$ and $(a_3P'u') \setminus u'$ contradict the minimality of $|V(H)|$. If $v_1a_2 \in E(G)$, then $H$ belongs to $\mathcal{M}$, with the set of extremities equal to $I$ (with center $a_2$). Hence, from here on, we assume that $a_2$ has no neighbor in $P'$.

If $v_1 = v_2$, then $H$ belongs to $\mathcal{S}$ and $I$ is the set of its extremities. If $v_1v_2 \in E(G)$, then $H$ belongs to $\mathcal{T}$, with triangle $\{u, v_1, v_2\}$ and the set of extremities $I$. So, we may assume that $v_1 \neq v_2$ and $v_1v_2 \notin E(G)$. Since $u \neq a_3$, the path $a_1Pv_1uv_2Pa_2$ and $P' \setminus u$ contradict the minimality of $|V(H)|$. $\square$

**Theorem 3.2.** A graph contains $K_{2,3}$ as an induced minor if and only if it contains a long prism, a pyramid, a theta, or a broken wheel as an induced subgraph.

*Proof.* It is easy to check that a long prism, a pyramid, a theta, or a broken wheel all contain $K_{2,3}$ as an induced minor.

Conversely, assume that a graph $G$ contains $K_{2,3}$ as an induced minor. Let us denote the vertices of $K_{2,3}$ as $\{u, v, a, b, c\}$ so that $\{u, v\}$ and $\{a, b, c\}$ form a bipartition of
<figure><img src="image_4.png" /><figcaption>Figure 3: Illustration of two cases in the proof of Theorem 3.2: on the left, $H \in \mathcal{M}$ and $H' \in \mathcal{S}$; on the right, $H, H' \in \mathcal{M}$. Dashed edges represent paths of length at least 1.</figcaption></figure>

### Claim 1.
$X_a, X_b$, and $X_c$ each contain only one vertex.

*Proof of claim.* Let $a_u$ and $a_v$ be two vertices of $X_a$ with a neighbor respectively in $X_u$ and $X_v$. By minimality of $|X_a \cup X_b \cup X_c|$, the graph $G[X_a]$ is a chordless path $P = a_u \dots a_v$. If $a_u \neq a_v$, then $\{X_u \cup \{a_u\}, X_v, P \setminus a_u, X_b, X_c\}$ is an induced minor model of $K_{2,3}$ that contradicts the minimality of $|X_a \cup X_b \cup X_c|$. So, $X_a$ consists of one vertex. The proof is the same for $X_b$ and $X_c$.

◇

Observe that, by Claim 1, $I = X_a \cup X_b \cup X_c$ is an independent set of size 3 in the subgraph $G'$ of $G$ induced by the vertices of the model $\mathcal{H}$. Moreover, $I$ is contained in the neighborhood of each of the two connected components $X_u$ and $X_v$ of $G' \setminus I$. Consider the graphs $H = G[X_u \cup I]$ and $H' = G[X_v \cup I]$. By Lemma 3.1 and minimality of $\mathcal{H}$, the graphs $H$ and $H'$ belong to $S \cup T \cup \mathcal{M}$, since otherwise we could find a proper induced subgraph of one of them that is in $S \cup T \cup \mathcal{M}$, yielding an induced minor model of $K_{2,3}$ in $G$ smaller than $\mathcal{H}$, contradicting the minimality of $\mathcal{H}$. Moreover, the extremities of both $H$ and $H'$ are exactly $I = \{a_1, a_2, a_3\}$.

### Claim 2.
The graph $X = G[V(H) \cup V(H')]$ contains a long prism, a pyramid, a theta, or a broken wheel as an induced subgraph.

*Proof of claim.* Assume first that $H \in \mathcal{M}$. W.l.o.g., we may assume that $a_2$ is the center of $H$. Let $P$ be the path from $a_1$ to $a_3$ in $H$ that does not contain $a_2$. Let $c_1$ and $c_3$ be the vertices of $P$ adjacent to $a_2$ that are closest to $a_1$ and $a_3$ along $P$, respectively.

If $H' \in S$, then let $u$ be the unique vertex of degree 3 in $H'$; see Fig. 3. Observe that either $X$ is a pyramid (if $c_1c_3 \in E(G)$) or a broken wheel (if $c_1c_3 \notin E(G)$ and $a_2u \in E(G)$), or $X$ contains a theta with apices $a_2$ and $u$ (whenever $c_1c_3 \notin E(G)$ and $a_2u \notin E(G)$).

If $H' \in T$, then notice that either $X$ is a long prism (if $c_1c_3 \in E(G)$) or $X$ contains a pyramid with apex $a_2$ (if $c_1c_3 \notin E(G)$).
If $H' \in \mathcal{M}$ and $H'$ is centered at $a_2$, then $X$ is a broken wheel. So, assume w.l.o.g. that $H'$ is centered at $a_1$; see Fig. 3. Let $P'$ be the path from $a_2$ to $a_3$ in $H'$ that does not contain $a_1$. Let $c'_2$ and $c'_3$ be the vertices of $P'$ adjacent to $a_1$ that are closest to $a_2$ and $a_3$ along $P'$, respectively. If $c_1c_3$ and $c'_2c'_3$ are both edges of $G$, then $X$ is a prism with triangles $\{a_2, c_1, c_3\}$ and $\{a_1, c'_2, c'_3\}$. So we may assume w.l.o.g. that $c_1c_3 \notin E(G)$. If $c'_2c'_3 \in E(G)$, then $X$ contains a pyramid with apex $a_2$ and triangle $\{a_1, c'_2, c'_3\}$. If $c'_2c'_3 \notin E(G)$, then $X$ contains a theta with apices $a_1$ and $a_2$.

Hence, we may assume that $H \notin \mathcal{M}$ and symmetrically $H' \notin \mathcal{M}$.

Assume that $H \in \mathcal{S}$. If $H' \in \mathcal{S}$, then $X$ is a theta and if $H' \in \mathcal{T}$, then $X$ is a pyramid. Hence, we may assume that $H \notin \mathcal{S}$ and symmetrically $H' \notin \mathcal{S}$.

We are left with the case where both $H$ and $H'$ belong to $\mathcal{T}$, which implies that $X$ is a long prism. $\diamond$

This completes the proof of Theorem 3.2. $\square$

# 4 Detecting 3-path-configurations

Detecting a pyramid in polynomial time was first done by Chudnovsky and Seymour [2]. A faster method was later discovered by Lai, Lu, and Thorup [14]. The $\tilde{O}(\cdot)$ notation suppresses polylogarithmic factors.

**Theorem 4.1.** *Determining whether a given graph contains a pyramid as an induced subgraph can be done in time $\tilde{O}(n^5)$.*

Detecting a theta in polynomial time was first done by Chudnovsky and Seymour [4]. A faster method was later discovered by Lai, Lu, and Thorup [14].

**Theorem 4.2.** *Determining whether a given graph contains a theta as an induced subgraph can be done in time $\tilde{O}(n^6)$.*

We now explain how a long prism can be detected in a graph with no pyramid. Note that detecting a prism is known to be an NP-complete problem [17]. Our algorithm is inspired from the algorithm in [17] to detect a pyramid or a prism. Adapting it to detect a long prism in a pyramid-free graph is straightforward. We include all details just for the sake of completeness. For the reader's convenience, we tried to optimize the length of the proof instead of the running time of the algorithm, since anyway the complexity bottleneck is in the next section.

**Lemma 4.3.** *Determining whether a given pyramid-free graph contains a long prism as an induced subgraph can be done in time $O(n^6(n+m))$.*

*Proof.* We call co-domino the graph with vertex set $\{a_1, a_2, v_1, v_2, v_3, v_4\}$ and edges $v_1v_2, v_2v_3, v_3v_4, v_4v_1, a_1v_1, a_1v_2, a_2v_3$, and $a_2v_4$. We call net the graph with vertex set $\{a_1, a_2, a_3, v_1, v_2, v_3\}$ and edges $v_1v_2, v_2v_3, v_3v_1, a_1v_1, a_2v_2$, and $a_3v_3$. See Fig. 4 for a representation of the co-domino and the net.

Consider the following algorithm whose input is a pyramid-free graph $G$.
<figure><img src="image_5.png" /><figcaption>Figure 4: The co-domino and the net.</figcaption></figure>

1. Enumerate all the induced subgraphs of $G$ isomorphic to a co-domino (with notation as in the definition). For each of them compute the connected components of $G \setminus (N(v_1) \cup N(v_2) \cup N(v_3) \cup N(v_4))$. If one of them contains neighbors of both $a_1$ and $a_2$, output that some long prism exists in $G$ and stop the algorithm.

2. Enumerate all induced subgraphs of $G$ isomorphic to a net (with notation as in the definition). For each of them compute the connected components of $G \setminus (N(v_1) \cup N(v_2) \cup N(v_3))$. If one of them contains neighbors of $a_1$, $a_2$, and $a_3$, output that some long prism exists in $G$ and stop the algorithm.

3. Output that $G$ contains no long prism.

The algorithm relies on brute force enumeration of sets on six vertices and computation of connected components, so it can be implemented to run in time $O(n^6(n+m))$. Also, it clearly outputs that $G$ contains a long prism or that it does not. It therefore remains to prove that the algorithm indeed gives the answer that $G$ contains a long prism if and only if it does.

If $G$ contains a long prism $H$, then assume first that two of the paths of $H$ have length 1. Then $H$ contains an induced co-domino that will be detected in the Step 1 of the algorithm. Because of the third path of the prism, at least one of the computed connected components contains neighbors of both $a_1$ and $a_2$, so the algorithm gives the correct answer. Assume now that at least two paths of $H$ have length at least 2. Then $H$ contains a net that will be detected in Step 2 of the algorithm (unless the algorithm stopped before, which is fine since $G$ contains a long prism by assumption). At least one of the computed connected components (the one that contains the rest of the prism) contains neighbors of $a_1$, $a_2$, and $a_3$, so the algorithm gives the correct answer and stops.

Conversely, assume that the algorithm gives the answer that $G$ contains a long prism.

If such an answer is given in Step 1, this is because of some co-domino and some connected component $X$ that contains neighbors of $a_1$ and $a_2$. A shortest path from $a_1$ to $a_2$ with interior in $X$ together with the co-domino form a long prism, so a long prism does exist in the graph.

If such an answer is given in Step 2, it is because of some connected component $X$ of $G \setminus (N(v_1) \cup N(v_2) \cup N(v_3))$ that contains neighbors of $a_1$, $a_2$, and $a_3$. In $G' =$
G[X ∪ {a₁, a₂, a₃}], let us apply Lemma 3.1 to obtain an induced subgraph H that belongs to S ∪ T ∪ M and has {a₁, a₂, a₃} as the set of extremities. If H ∈ S then G contains a pyramid with triangle {v₁, v₂, v₃}; this cannot happen since G is pyramid-free. If H ∈ T, then G contains a long prism and the output of the algorithm is correct. If H ∈ M then G contains a pyramid if the center of H has two non-adjacent neighbors in H or a long prism otherwise. The first case cannot happen by assumption and hence the output of the algorithm is correct. □

5 Detecting a broken wheel

We give here a polynomial-time algorithm to detect a broken wheel in a graph with no long prism, no pyramid, and no theta. Note that detecting a broken wheel in general is an NP-complete problem, even when restricted to bipartite graphs. The proof can be found in [7]. Note that this article addresses the question of the detection of wheels (broken or not), however, the proof (of [7, Theorem 4]) that detecting a wheel is NP-complete proves without a single modification that detecting a broken wheel is NP-complete.

Let W = (H, x) be a broken wheel. Since W is broken, it contains at least two sectors of length at least 2, say P = a...b and R = c...d. We orient H clockwise and assume that a, b, c, and d appear clockwise in this order along H. We denote by Q the path of H from b to c that does not contain a and d. We denote by S the path of H from d to a that does not contain b and c. Note that

$$
\text{possibly } b = c \text{ or } d = a, \text{ but not both,} \tag{1}
$$

since W is a wheel by assumption. Note also that a ≠ b, c ≠ d, ab / ∈ E(G), and cd / ∈ E(G).

For every vertex v in H, we denote by v⁺ its neighbor in H in the clockwise direction, and by v⁻ its neighbor in the counter-clockwise direction. Note that possibly a⁺ = b⁻, a⁺b⁻ ∈ E(G), c⁺ = d⁻, or c⁺d⁻ ∈ E(G).

We say that the 13-tuple F = (x, a, b, c, d, a⁺, b⁺, c⁺, d⁺, a⁻, b⁻, c⁻, d⁻) is a frame for W. Every broken wheel has at least one frame. In what follows, once a frame is fixed, we use the notation for the paths P, Q, R, S without recalling it; see Fig. 5.

If W is a broken wheel in a graph G with frame F, we say that (W, F) is optimal if:

• no broken wheel of G has fewer vertices than W, and

• among all broken wheels with the same number of vertices as W and their frames, the sum of the lengths of S and Q is minimal.

The cube is the graph with vertex set {v₁, v₂, v₃, v₄, v₅, v₆, x, y} such that v₁v₂v₃v₄v₅v₆v₁ is a hole, N(x) = {v₁, v₃, v₅}, and N(y) = {v₂, v₄, v₆}; see Fig. 6. Observe that removing any vertex from the cube yields a broken wheel on seven vertices. Hence, every graph that does not contain a broken wheel is cube-free.

Given a graph G, a wheel W = (H, x) contained in G as an induced subgraph, and a vertex y ∈ G \ W, we denote by N_H(y) the set of vertices in H that are adjacent to y.
<figure><img src="image_6.png" /><figcaption>Figure 5: A frame for a broken wheel. Dotted edges may be present or not.</figcaption></figure>

<figure><img src="image_7.png" /><figcaption>Figure 6: The cube.</figcaption></figure>

We start with three lemmas analyzing the structure around an optimal pair $(W, F)$ of a broken wheel and a frame.

**Lemma 5.1.** Let $G$ be a (long prism, pyramid, theta, cube)-free graph. Let $W = (H, x)$ be a broken wheel in $G$ with frame $F = (x, a, b, c, d, a^+, b^+, c^+, d^+, a^-, b^-, c^-, d^-)$ such that $(W, F)$ is optimal. Let $y$ be a vertex of $G \setminus W$ that is not adjacent to $a, b, c, \text{ or } d$. Then $N_H(y)$ induces a (possibly empty) path of length at most 2. In particular, $N_H(y)$ is contained in the interior of one of $P, Q, R$, or $S$.

*Proof.* Assume first that $y$ has neighbors in at most one path among $P, Q, R$, and $S$. If $N_H(y)$ is not contained in some path of $H$ of length at most 2, it can be used to replace some subpath of $H$ to obtain a broken wheel with fewer vertices than $W$, a contradiction (note that this is correct if $xy \in E(G)$ and if $xy \notin E(G)$). Hence, $N_H(y)$ is contained in some path $T$ of $H$ of length at most 2, and it must be equal to $T$, because otherwise $T$ has length 2 and $y$ is non adjacent to its middle vertex, and $y$ and $H$ form a theta. So, the conclusion of the lemma holds. Hence, from now on, we assume that $y$ has neighbors in at least two paths among $P, Q, R$, and $S$.

If $N_H(y) \subseteq Q \cup S$, then $y$ has neighbors in both $Q$ and $S$. Since $G$ is theta-free, $y$ has at least three neighbors in $H$. It follows that $y$ is the center of a broken wheel with
rim $H$, with the same number of vertices as $W$ and that has a frame contradicting the minimality of the sum of the lengths of $Q$ and $S$ (since $y$ is not adjacent to the endpoints of $Q$ and $S$). Hence, $y$ has neighbors in the interior of at least one of $P$ or $R$, say in $P$ up to symmetry.

If $y$ is adjacent to $x$, then $P$, $x$, and $y$ induce a theta or a broken wheel with center $y$ and with fewer vertices than $W$, a contradiction. So, $y$ is not adjacent to $x$.

Assume that $y$ has neighbors in $S$. Let $a'$ be the neighbor of $y$ in $P$ closest to $a$ along $P$. Let $b'$ be the neighbor of $y$ in $P$ closest to $b$ along $P$. Let $d'$ be the neighbor of $y$ in $S$ closest to $d$ along $S$. Let $T$ be a shortest path from $x$ to $d'$ in the subgraph of $G$ induced by the vertices of the path $xdSd'$. If $d'a \notin E(G)$, the paths $xaPa'$, $xbPb'$, and $xTd'y$ induce a theta or a pyramid (namely, if $a' = b'$ they form a theta with apices $a'$ and $x$, if $a'b' \in E(G)$ they form a pyramid, and otherwise they form a theta with apices $x$ and $y$). Hence, $d'a \in E(G)$. It follows that $d' = a^-$ is the unique neighbor of $y$ in $S$.

We claim that $a' = b'$ and $a'a \in E(G)$. Indeed, if $a' = b'$ and $a'a \notin E(G)$, then the three paths $aPa'$, $axbPa'$, and $ad'ya'$ form a theta with apices $a$ and $a'$; if $a'$ and $b'$ are adjacent, then the three paths $aPa'$, $axbPb'$, and $ad'y$ form a pyramid with apex $a$ and triangle $\{a', b', y\}$; and if $a'$ and $b'$ are distinct and nonadjacent, then the three paths $aPa'y$, $axbPb'y$, and $ad'y$ form a theta with apices $a$ and $y$. It follows that $a' = a^+$ is the unique neighbor of $y$ in $P$. Therefore, the hole $xTa^-ya^+Pbx$ is the rim of a broken wheel $W'$ centered at $a$. To avoid that the wheel $W'$ has fewer vertices than $W$, we infer that $b = c$, $R$ has length exactly 2, and $T = xdSa^-$ (which implies $xa^- \notin E(G)$). So, $Q = b$ and $R = bb^+d$. Vertex $y$ must be adjacent to $b^+$, since otherwise $H$ and $y$ form a theta with apices $a^-$ and $a^+$ (recall that $y$ is not adjacent to $a, b, c$ and $d$). We have $a^-d \in E(G)$, for otherwise the three paths $dSa^-$, $dxaa^-$, and $db^+ya^-$ form a theta with apices $d$ and $a^-$. By a symmetric argument, we must have $a^+b \in E(G)$. Hence, $\{x, y, a, b, d, a^-, a^+, b^+\}$ induces a cube, a contradiction.

We proved that $y$ cannot have neighbors in $S$. Symmetrically, it cannot have neighbors in $Q$. Hence, $y$ has neighbors in both $P$ and $R$ and no neighbor in $Q$ and $S$. Since $G$ contains no theta, $y$ must have at least two neighbors in either $P$ or $R$, say in $R$ up to symmetry. Up to symmetry, we may assume by (1) that $a \neq d$ (so possibly $b = c$). Let $a'$ be the neighbor of $y$ in $P$ closest to $a$ along $P$. Let $d'$ be the neighbor of $y$ in $R$ closest to $d$ along $R$. Let $c'$ be the neighbor of $y$ in $R$ closest to $c$ along $R$. Note that $c' \neq d'$. If $ca' \notin E(G)$, then the three paths $yd'Rdx$, $yc'Rcx$, and $ya'Pax$ induce a long prism, a pyramid, or a theta (namely, they induce a long prism if $c'd' \in E(G)$ and $ad \in E(G)$, a pyramid if exactly one of $c'd' \in E(G)$ and $ad \in E(G)$ happens, and a theta otherwise). So, $ca' \in E(G)$. Hence, $b = c$ and $a'$ is the unique neighbor of $y$ in $P$. Hence, since $c' \neq d'$, the three paths $yd'Rdb$, $yc'Rb$, and $ya'b$ induce a pyramid with triangle $\{y, c', d'\}$ or a theta with apices $y$ and $b$, a contradiction. $\square$

**Lemma 5.2.** Let $G$ be a (long prism, pyramid, theta, cube)-free graph. Let $W = (H, x)$ be a broken wheel of $G$ with frame $F = (x, a, b, c, d, a^+, b^+, c^+, d^+, a^-, b^-, c^-, d^-)$ such that $(W, F)$ is optimal. Let $P' = a^+...b^-$ be a shortest path from $a^+$ to $b^-$ in $G \setminus ((N[a] \cup N[b] \cup N[c] \cup N[d] \cup N[x]) \setminus \{a^+, b^-\})$. Then the induced subgraph of $G$ obtained from $W$ by removing the internal vertices of $P$ and adding the vertices of $P'$
instead is a broken wheel with frame F, and $(W', F)$ is optimal. A similar result holds for R.

**Proof.** If no vertex of $P'$ has a neighbor in the interior of $Q, R$, or $S$ (in which case $P'$ is vertex-disjoint from $Q \cup R \cup S$), then the conclusion of the lemma holds. Hence, suppose for a contradiction that some vertex $v$ of $P'$ has a neighbor in the interior of $Q, R$, or $S$, and choose $v$ closest to $a^+$ along $P'$. Note that $v$ is in the interior of $P'$, hence, $v \notin W$ and $v$ is not adjacent to any of $a, b, c, d$, and $x$. Let $w$ be the vertex closest to $v$ along $vP'a^+$ that has some neighbor in $P$. Note that $w$ exists because $a^+ \in P$; moreover, $w$ is in the interior of $P'$; hence, $w \notin W$ and $w$ is not adjacent to any of $a, b, c, d$, and $x$. Since $v$ has a neighbor in the interior of $Q, R$, or $S$, Lemma 5.1 implies that $N_H(v)$ induces a nonempty path of length at most 2 that is contained in the interior of either $Q, R$, or $S$. Similarly, $N_H(w)$ induces a nonempty path of length at most 2 that is contained in the interior of $P$. Note that $w \neq v$. We set $T = vP'w$. Now, we focus on $T$ and $H$. By construction, $T$ is disjoint from $H$ and no vertex in the interior of $T$ has a neighbor in $H$. If $vw \notin E(G)$ or if one of $|N_H(v)|$ or $|N_H(w)|$ has size at most 2, then $T \cup H$ induces a long prism, a pyramid, or a theta. Hence, $vw \in E(G)$, and both $N_H(v)$ and $N_H(w)$ induce a path of length 2. It follows that the subgraph of $G$ induced by $H \cup T \cup \{x\}$ contains a theta with apices $v$ and $x$, a contradiction. □

The next lemma might look identical to the previous one, but there is an important
difference: the vertices of the shortest path are allowed to be adjacent to $x$.

**Lemma 5.3.** Let $G$ be a (long prism, pyramid, theta, cube)-free graph. Let $W = (H, x)$ be a broken wheel of $G$ with frame $F = (x, a, b, c, d, a^+, b^+, c^+, d^+, a^-, b^-, c^-, d^-)$ such that $(W, F)$ is optimal. Let $Q' = b^+ \dots c^-$ be a shortest path from $b^+$ to $c^-$ in $G \setminus ((N[a] \cup N[b] \cup N[c] \cup N[d]) \setminus \{b^+, c^-\})$. Then the induced subgraph of $G$ obtained from $W$ by removing the internal vertices of $Q$ and adding the vertices of $Q'$ instead is a broken wheel with frame $F$, and $(W', F)$ is optimal. A similar result holds for $S$.

*Proof.* If no vertex of $Q'$ has a neighbor in the interior of $P, R$, or $S$ (in which case $Q'$ is vertex-disjoint from $P \cup R \cup S$), then the conclusion of the lemma holds. Hence, suppose for a contradiction that some vertex $v$ of $Q'$ has some neighbor in the interior of $P, R$, or $S$, and choose $v$ closest to $b^+$ along $Q'$. Let $w$ be the vertex closest to $v$ along $vQ'b^+$ that has some neighbor in $Q$.

The rest of the proof is similar to the proof of Lemma 5.2, except that $x$ might be
adjacent to $v$ or $w$. For clarity, we provide in this paragraph the details of the proof that
are similar to those in the proof of Lemma 5.2. Note that $v$ is in the interior of $Q'$, hence,
$v \notin W$ and $v$ is not adjacent to any of $a, b, c$, and $d$. Similarly, $w$ exists because $b^+ \in Q$;
moreover, $w$ is in the interior of $Q'$; hence, $w \notin W$ and $w$ is not adjacent to any of $a, b$
$c$, and $d$. Since $v$ has a neighbor in the interior of $P, R$, or $S$, Lemma 5.1 implies that
$N_H(v)$ induces a nonempty path of length at most 2 that is contained in the interior of
either $P, R$, or $S$. Similarly, $N_H(w)$ induces a nonempty path of length at most 2 that
is contained in the interior of $Q$. Since $v$ has no neighbors in $Q$, it is distinct from $w$
and nonadjacent to both $b^+$ and $c^-$. We set $T = vQ'w$. Now we focus on $T$ and $H$. By
construction, $T$ is disjoint from $H$ and no vertex in the interior of $T$ has a neighbor in $H$. If $vw \notin E(G)$ or if one of $|N_H(v)|$ or $|N_H(w)|$ has size at most 2, then $T \cup H$ induces a long prism, a pyramid, or a theta. Hence, $vw \in E(G)$, and both $N_H(v)$ and $N_H(w)$ induce a path of length 2.

If $xv \notin E(G)$, then the subgraph of $G$ induced by $H \cup T \cup \{x\}$ contains a theta with apices $x$ and $v$. If $xw \notin E(G)$, then the subgraph of $G$ induced by $H \cup T \cup \{x\}$ contains a theta with apices $x$ and $w$. Hence, $xv \in E(G)$ and $xw \in E(G)$.

Recall that $N_H(v)$ induces a path of length 2 that is contained in the interior of exactly one of $P$, $R$, or $S$. We now analyze these three cases, starting with $S$. If $N_H(v) \subseteq S$, then let $w'$ be the middle vertex of $N_Q(w)$. The hole $H'$ obtained from $H$ by replacing $w'$ by $w$ is the rim of a broken wheel $W'$ centered at $v$. This wheel has a frame that contradicts the optimality of $(W, F)$.

If $N_H(v) \subseteq P$, then some hole going through $a$, $v$, $w$, $c$, and $d$ is the rim of a broken wheel centered at $x$ that has fewer vertices than $W$, a contradiction to the optimality of $(W, F)$. The case if $N_H(v) \subseteq R$ is similar. $\square$

Now we have everything ready to prove the main result of this section.

**Lemma 5.4.** *Determining whether a given (long prism, pyramid, theta)-free graph contains a broken wheel as an induced subgraph can be done in time $O(n^{13}(n+m))$.*

*Proof.* Consider the following algorithm whose input is any (long prism, pyramid, theta)-free graph $G$.

1. Enumerate all 7-tuples of vertices $G$, and if one of them induces a broken wheel, output “$G$ contains a broken wheel” and stop.

2. Enumerate all 13-tuples $F = (x, a, b, c, d, a^+, b^+, c^+, d^+, a^-, b^-, c^-, d^-)$ of vertices of $G$. For each of them:

(i) Compute a shortest path $P'$ from $a^+$ to $b^-$ in $G \setminus ((N[a] \cup N[b] \cup N[c] \cup N[d] \cup N[x]) \setminus \{a^+, b^-\})$. If no such path exists, set $P' = \emptyset$.

(ii) If $b = c$, set $Q' = \emptyset$. Otherwise, compute a shortest path $Q'$ from $b^+$ to $c^-$ in $G \setminus ((N[a] \cup N[b] \cup N[c] \cup N[d]) \setminus \{b^+, c^-\})$. If no such path exists, set $Q' = \emptyset$.

(iii) Compute a shortest path $R'$ from $c^+$ to $d^-$ in $G \setminus ((N[a] \cup N[b] \cup N[c] \cup N[d] \cup N[x]) \setminus \{c^+, d^-\})$. If no such path exists, set $R' = \emptyset$.

(iv) If $d = a$, set $S' = \emptyset$. Otherwise, compute a shortest path $S'$ from $d^+$ to $a^-$ in $G \setminus ((N[a] \cup N[b] \cup N[c] \cup N[d]) \setminus \{d^+, a^-\})$. If no such path exists, set $S' = \emptyset$.

(v) If $P' \cup Q' \cup R' \cup S' \cup F$ induces a broken wheel centered at $x$, output “$G$ contains a broken wheel” and stop.

3. Output “$G$ contains no broken wheel”.

If the algorithm outputs that $G$ contains a broken wheel in Step 2 (v), it obviously does. Conversely, assume that $G$ contains a broken wheel, and let us check that the algorithm gives the correct answer.
If $G$ contains a broken wheel on 7 vertices, then it is detected in Step 1. So we may assume that $G$ contains no broken wheel on 7 vertices, which implies that $G$ contains no cube and that the algorithm continues with Step 2.

Since $G$ contains a broken wheel, it contains a broken wheel $W$ with a frame $F$ such that $(W, F)$ is optimal. We denote by $P, Q, R$, and $S$ the paths of the broken wheel as above. At some point in Step 2, the algorithm considers $F$. By Lemma 5.2 applied to $W$, the paths $P'$ (defined in Step 2 (i)), $Q, R$, and $S$ together with $F$ form a broken wheel $W_1$. By Lemma 5.3 applied to $W_1$ if $b \neq c$ (and trivially otherwise), the paths $P', Q'$ (defined in Step 2 (ii)), $R$, and $S$ together with $F$ form a broken wheel $W_2$. By Lemma 5.2 applied a second time to $W_2$, the paths $P', Q'$, $R'$ (defined in Step 2 (iii)), and $S$ together with $F$ form a broken wheel $W_3$. By Lemma 5.3 applied a second time to $W_3$ if $d \neq a$ (and trivially otherwise), the paths $P', Q'$, $R'$, and $S'$ (defined in Step 2 (iv)) together with $F$ form a broken wheel $W_4$. Hence, the algorithm actually finds a broken wheel, and therefore gives the correct answer.

The algorithm relies on a brute force enumeration of 13-tuples of vertices that can be implemented in time $O(n^{13})$, and a search of the graph for computing shortest paths $P'$, $Q'$, $R'$, $S'$, as well as testing if $P'$, $Q'$, $R'$, $S'$, and $F$ induce a broken wheel centered at $x$ can be implemented in time $O(n+m)$. $\square$

# 6 Conclusion

We can now state our main theorem.

**Theorem 6.1.** *Determining whether a given graph contains $K_{2,3}$ as an induced minor can be done in time $O(n^{13}(n+m))$.*

*Proof.* By Theorem 3.2, it is enough to decide whether $G$ contains a long prism, a pyramid a theta, or a broken wheel as an induced subgraph. Detecting a pyramid can be performed in time $\tilde{O}(n^5)$ by Theorem 4.1. Detecting a theta can be performed in time $\tilde{O}(n^6)$ by Theorem 4.2. Hence, it can be assumed that the input graph contains no pyramid and no theta, so that detecting a long prism can be performed in time $O(n^6(n+m))$ by Lemma 4.3. Hence, it can be assumed that the input graph contains no pyramid, no theta, and no long prism, so that detecting a broken wheel can be performed in time $O(n^{13}(n+m))$ by Lemma 5.4. $\square$

# Open questions

We wonder what is the complexity of detecting a broken wheel in a theta-free graph. Detecting a theta or a wheel (possibly not broken) in an input graph can be performed in polynomial time. However, the algorithm given in [19] relies on a complicated structural description of the class of (theta, wheel)-free graphs. We wonder whether a simpler approach, based on computing shortest paths as we do here, might work. Symmetrically, our polynomial-time algorithm suggests that a full structural description of graphs that do not contain $K_{2,3}$ as an induced minor may be worth investigating.

In particular, we propose the following.
**Conjecture 6.2.** There exists a polynomial $p$ and an integer $k$ such that if $G$ has no clique cutset and does not contain $K_{2,3}$ as an induced minor, then either $G$ has at most $p(|V(G)|)$ minimal separators or $G$ has no hole of length at least $k$.

If Conjecture 6.2 is true, it would give an alternative polynomial-time recognition algorithm for determining whether a given graph contains $K_{2,3}$ as an induced minor.

It is not difficult to construct examples showing that the condition that $G$ does not contain any hole of length at least $k$ cannot be replaced with the condition that $G$ has independence number at most $k$. For a positive integer $k$, let $G_k$ be the graph with $V(G_k) = A \cup B \cup C \cup D$ where $A, B, C$, and $D$ are pairwise disjoint sets with $k$ vertices each, $A = \{a_1, \dots, a_k\}$ is an independent set, each of $B = \{b_1, \dots, b_k\}$, $C = \{c_1, \dots, c_k\}$, and $D = \{d_1, \dots, d_k\}$ is a clique, every vertex in $A$ is adjacent to every vertex of $C$, every vertex in $B$ is adjacent to every vertex of $D$, for every $i, j \in \{1, \dots, k\}$, vertices $a_i$ and $b_j$ are adjacent if and only if $i = j$, vertices $b_i$ and $c_j$ are adjacent if and only if $i \neq j$, vertices $c_i$ and $d_j$ are adjacent if and only if $i = j$, and there are no other edges. Then, $A \cup \{d_1\}$ is an independent set in $G_k$ with cardinality $k+1$, and it can be verified that $G_k$ has no clique cutset, does not contain $K_{2,3}$ as an induced minor, and the number of minimal separators of graphs in the family $\{G_k: k \ge 1\}$ is not bounded by any polynomial function in the number of vertices.

Let us mention that it is easy to settle the complexity status of the most classical optimization problems when restricted to graphs that do not contain $K_{2,3}$ as an induced minor. The maximum independent set and $k$-coloring (for any fixed $k$) problems are polynomial-time solvable because the tree-independence number is bounded (see [5]). Computing the chromatic number is NP-hard, because it is NP-hard for the subclass of circular-arc graphs (see [9]). Many problems such as the maximum clique and coloring the complement are NP-hard because the class contains the complement of all triangle-free graphs.

Finally, note that the complexity of recognizing bisimplicial graphs, that is, graphs in which each minimal separator is a union of two cliques (see [18]) remains open. In this regard, our main result implies that the problem is solvable in polynomial time when restricted to the class of perfect graphs, since in a perfect graph, a set of vertices is a union of two cliques if and only if it induces a graph with independence number at most 2 (see [15, 16]).

## Acknowledgments

The authors are grateful to Nevena Pivač for helpful discussions. This work is supported in part by the *Slovenian Research and Innovation Agency* (I0-0035, research program P1-0285 and research projects J1-3001, J1-3002, J1-3003, J1-4008, J1-4084, N1-0102, and N1-0160, and BI-FR/22-23-PROTEUS-01), by the research program *CogniCom* (0013103) at the University of Primorska, by the French *Fédération de Recherche ICVL* (Informatique Centre-Val de Loire), by the LABEX MILYON (ANR-10-LABX-0070) of Université de Lyon, within the program Investissements d'Avenir (ANR-11-IDEX-0007) operated by the French National Research Agency (ANR), by Agence Nationale de la Recherche (France) under research grant ANR DIGRAPHS ANR-19-CE48-0013-01 and by H2020-MSCA-RISE project CoSP- GA No. 823748.
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