Triangle-Free Geometric Intersection Graphs with No Large Independent Sets

Abstract

It is proved that there are triangle-free intersection graphs of line segments in the plane with arbitrarily small ratio between the maximum size of an independent set and the total number of vertices.

1 Introduction

Pawlik et al. [7] proved that there are triangle-free intersection graphs of line segments in the plane with arbitrarily large chromatic number. The graphs they construct have independent sets containing more than $1/3$ of all the vertices. It has been left open whether there is a constant $c>0$ such that every triangle-free intersection graph of $n$ segments in the plane has an independent set of size at least $cn$. Fox and Pach [3] conjectured a much more general statement, that $K_k$-free intersection graphs of curves in the plane have linear-size independent sets, for every $k$. This would imply a well-known conjecture that $k$-quasi-planar graphs (graphs drawn in the plane so that no $k$ edges cross each other) have linearly many edges [5], which is proved up to $k=4$ [1].

In this note, I resolve the independent set problem in the negative, proving the following strengthening of the result of Pawlik et al.:

Theorem

There are triangle-free segment intersection graphs with arbitrarily small ratio between the maximum size of an independent set and the total number of vertices.

The constructions presented in the next two sections give rise to triangle-free intersection graphs of $n$ segments in the plane with maximum independent set size $\Theta (n/\log \log n)$.

2 Construction

Pawlik et al. [7] construct, for $k\geqslant 1$, a triangle-free graph $G_k$ and a family $\mathcal {P}_k$ of subsets of $V(G_k)$, called probes, with the following properties:

• (i) $|\mathcal {P}_k|={2^{2^{k-1}-1}}$,

• (ii) every member of $\mathcal {P}_k$ is an independent set of $G_k$,

• (iii) for every proper coloring of the vertices of $G_k$, there is a probe $P\in \mathcal {P}_k$ such that at least $k$ colors are used on the vertices in $P$.

They are built by induction on $k$, as follows. The graph $G_1$ has just one vertex $v$, and $\mathcal {P}_1$ has just one probe $\{v\}$. For $k\geqslant 2$, first, a copy $(G,\mathcal {P})$ of $(G_{k-1},\mathcal {P}_{k-1})$ is taken. Then, for every probe $P\in \mathcal {P}$, another copy $(G_P,\mathcal {P}_P)$ of $(G_{k-1},\mathcal {P}_{k-1})$ is taken. There are no edges between vertices from different copies. Finally, for every probe $P\in \mathcal {P}$ and every probe $Q\in \mathcal {P}_P$, a new vertex $d_Q$ connected to all vertices in $Q$, called the diagonal of $Q$, is added. The resulting graph is $G_k$. The family of probes $\mathcal {P}_k$ is defined by

$$\begin{aligned} \mathcal {P}_k=\big \{P\cup Q:P\in \mathcal {P}\text { and }Q\in \mathcal {P}_P\big \}\cup \big \{P\cup \{d_Q\}:P\in \mathcal {P}\text { and }Q\in \mathcal {P}_P\big \}. \end{aligned}$$

It is easy to check that the graph $G_k$ is indeed triangle-free and the conditions (i)–(iii) are satisfied for $(G_k,\mathcal {P}_k)$—see [7] for details. It is also shown in [7] how the graph $G_k$ is represented as a segment intersection graph.

I will show that there is an assignment $w_k$ of positive integer weights to the vertices of $G_k$ with the following properties:

• (iv) the total weight of $G_k$ is $\frac{k+1}{2}\cdot {2^{2^{k-1}-1}}$,

• (v) for every independent set $I$ of $G_k$, the number of probes $P\in \mathcal {P}_k$ such that $P\cap I\ne \emptyset$ is at least the weight of $I$.

Once this is achieved, the proof of the theorem of this paper follows easily. Namely, it follows from (i) and (v) that every independent set $I$ of $G_k$ has weight at most ${2^{2^{k-1}-1}}$. We can take the representation of $G_k$ as a segment intersection graph and replace every segment representing a vertex $v\in V(G_k)$ by $w_k(v)$ parallel segments lying very close to each other, so as to keep the property that any two segments representing vertices $u,v\in V(G_k)$ intersect if and only if $uv\in E(G_k)$. It follows from (iv) that the family of segments obtained this way has size $\frac{k+1}{2}\cdot {2^{2^{k-1}-1}}$, while every independent set of its intersection graph has size at most ${2^{2^{k-1}-1}}$.

The assignment $w_k$ of weights to the vertices of $G_k$ is defined by induction on $k$, following the inductive construction of $(G_k,\mathcal {P}_k)$. The weight of the only vertex of $G_1$ is set to $1$. This clearly satisfies (iv) and (v). For $k\geqslant 2$, let $G$, $\mathcal {P}$, $G_P$, $\mathcal {P}_P$ and $d_Q$ be defined as in the inductive step of the construction of $(G_k,\mathcal {P}_k)$. Let $p=|\mathcal {P}_{k-1}|={2^{2^{k-2}-1}}$. The weights $w_k$ of the vertices of $G$ are their original weights $w_{k-1}$ in $G_{k-1}$ multiplied by $p$. The weights $w_k$ of the vertices of every $G_P$ are equal to their original weights $w_{k-1}$ in $G_{k-1}$. The weight $w_k$ of every diagonal $d_Q$ is set to $1$. It remains to prove that (iv) and (v) are satisfied for $(G_k,\mathcal {P}_k,w_k)$ assuming that they hold for $(G_{k-1},\mathcal {P}_{k-1},w_{k-1})$.

The proof of (iv) is straightforward:

$$\begin{aligned} w_k(G_k)=w_k(G)+\sum _{P\in \mathcal {P}}\bigl (w_k(G_P)+|\mathcal {P}_P|\bigr )=2pw_{k-1}(G_{k-1})+p^2=\tfrac{k+1}{2}\cdot 2^{2^{k-1}-1}. \end{aligned}$$

For the proof of (v), let $I$ be an independent set in $G_k$. Let $\mathcal {I}=\{P\in \mathcal {P}:P\cap I\ne \emptyset \}$. For every probe $P\in \mathcal {P}$, define

$$\begin{aligned} \mathcal {I}_P&=\{Q\in \mathcal {P}_P:Q\cap I\ne \emptyset \},&\mathcal {P}'_P&=\{P\cup Q:Q\in \mathcal {P}_P\}\cup \{P\cup \{d_Q\}:Q\in \mathcal {P}_P\},\\ D_P&=\{d_Q:Q\in \mathcal {P}_P\},&\mathcal {I}'_P&=\{P'\in \mathcal {P}'_P:P'\cap I\ne \emptyset \}. \end{aligned}$$

By the induction hypothesis, we have

$$\begin{aligned} w_k(V(G)\cap I)\leqslant p|\mathcal {I}|,\qquad \qquad w_k(V(G_P)\cap I)\leqslant |\mathcal {I}_P|. \end{aligned}$$

Suppose $P\in \mathcal {I}$. It follows that $(P\cup Q)\cap I\ne \emptyset$ and $(P\cup \{d_Q\})\cap I\ne \emptyset$ for every $Q\in \mathcal {P}_P$. Hence $|\mathcal {I}'_P|=|\mathcal {P}'_P|=2p$. Moreover, we have $d_Q\notin I$ whenever $Q\in \mathcal {I}_P$, because $d_Q$ is connected to all vertices in $Q$, one of which belongs to $I$. Hence

$$\begin{aligned} w_k(V(G_P)\cap I)+w_k(D_P\cap I)\leqslant |\mathcal {I}_P|+|\mathcal {P}_P\backslash \mathcal {I}_P|=|\mathcal {P}_P|=p. \end{aligned}$$

Now, suppose $P\in \mathcal {P}\backslash \mathcal {I}$. If $Q\in \mathcal {I}_P$, then $(P\cup Q)\cap I\ne \emptyset$, $d_Q\notin I$ (by the same argument as above), and $(P\cup \{d_Q\})\cap I=\emptyset$. If $Q\in \mathcal {P}_P\backslash \mathcal {I}_P$, then $(P\cup Q)\cap I=\emptyset$, and $(P\cup \{d_Q\})\cap I\ne \emptyset$ if and only if $d_Q\in I$. Hence

$$\begin{aligned} w_k(V(G_P)\cap I)+w_k(D_P\cap I)\leqslant |\mathcal {I}_P|+|D_P\cap I|=|\mathcal {I}'_P|. \end{aligned}$$

To conclude, we gather all the inequalities and obtain

$$\begin{aligned} \begin{aligned} w_k(I)&=w_k(V(G)\cap I)+\sum _{P\in \mathcal {P}}\bigl (w_k(V(G_P)\cap I)+w_k(D_P\cap I)\bigr )\\&\leqslant p|\mathcal {I}|+\sum _{P\in \mathcal {I}}p+\sum _{P\in \mathcal {P}\backslash \mathcal {I}}|\mathcal {I}'_P|=\sum _{P\in \mathcal {I}}|\mathcal {I}'_P|+\sum _{P\in \mathcal {P}\backslash \mathcal {I}}|\mathcal {I}'_P|=\sum _{P\in \mathcal {P}}|\mathcal {I}'_P|. \end{aligned} \end{aligned}$$

3 Improved Construction

Pawlik et al. [7] define also a graph $\tilde{G}_k$, which arises from $(G_k,\mathcal {P}_k)$ by adding, for every probe $P\in \mathcal {P}_k$, a diagonal $d_P$ connected to all vertices in $P$. This is the smallest triangle-free segment intersection graph known to have chromatic number greater than $k$. Define the assignment $\tilde{w}_k$ of weights to the vertices of $\tilde{G}_k$ so that $\tilde{w}_k$ is equal to $w_k$ on the vertices of $G_k$ and $\tilde{w}_k(d_P)=1$ for every $P\in \mathcal {P}_k$. Let $I$ be an independent set in $\tilde{G}_k$. Let $\mathcal {I}=\{P\in \mathcal {P}_k:P\cap I\ne \emptyset \}$. Hence $d_P\notin I$ for $P\in \mathcal {I}$. It follows that

$$\begin{aligned} \tilde{w}_k(I)&= w_k(V(G_k)\cap I)+|\{d_P:P\in \mathcal {P}_k\}\cap I|\\&\leqslant |\mathcal {I}|+|\mathcal {P}_k\backslash \mathcal {I}|=|\mathcal {P}_k|=2^{2^{k-1}-1},\\ \tilde{w}_k(\tilde{G}_k)&= w_k(G_k)+|\mathcal {P}_k|=\tfrac{k+3}{2}\cdot 2^{2^{k-1}-1}. \end{aligned}$$

The graph $\tilde{G}_k$ is the smallest one for which I can prove that it has a weight assignment such that the ratio between the maximum weight of an independent set and the total weight is at most ${\frac{2}{k+3}}$. It is not difficult to prove (e.g. using weak LP duality) that the assignment of weights $\tilde{w}_k$ to the vertices of $\tilde{G}_k$ is optimal (gives the least ratio) for this particular graph.

Both constructions give rise to triangle-free intersection graphs of $n$ segments in the plane with maximum independent set size $\Theta (n/\log \log n)$. On the other hand, it follows from the result of McGuinness [4] that every triangle-free intersection graph of $n$ segments has chromatic number $O(\log n)$ and maximum independent set size $\Omega (n/\log n)$.

4 Other Geometric Shapes

It is known that the graphs $G_k$ and $\tilde{G}_k$ have intersection models by many other geometric shapes, for example, L-shapes, axis-parallel ellipses, circles, axis-parallel square boundaries [6] or axis-parallel boxes in $\mathbb {R}^3$ [2]. The result of this paper can be extended to those models for which every geometric object $X$ representing a vertex of the intersection graph can be replaced by many pairwise disjoint objects approximating $X$. This is possible, for example, for intersection graphs of L-shapes, circles or axis-parallel square boundaries, but not for intersection graphs of axis-parallel ellipses or axis-parallel boxes in $\mathbb {R}^3$. The problem whether triangle-free intersection graphs of the latter kind of shapes have linear-size independent sets remains open.