# An Omega(n^2) Lower Bound for Random Universal Sets for Planar Graphs

A set U⊆^2 is n-universal if all n-vertex planar graphs have a planar straight-line embedding into U. We prove that if Q ⊆^2 consists of points chosen randomly and uniformly from the unit square then Q must have cardinality Ω(n^2) in order to be n-universal with high probability. This shows that the probabilistic method, at least in its basic form, cannot be used to establish an o(n^2) upper bound on universal sets.

## Authors

• 1 publication
• 11 publications
• 2 publications
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## 1 Introduction

Planar universal sets. Let . A set of points in is called -universal if each planar graph with vertices has an embedding into that maps one-to-one the vertices of into points of and maps each edge of into a straight-line segment connecting its endpoints, in such a way that these segments do not intersect (except of course at the endpoints). It is easy to see that for to be universal, it is sufficient that the required embedding exists only for maximal planar graphs, namely -vertex planar graphs with vertices. Maximal planar graphs have the property that in any planar embedding all their faces are -cycles (for this reason, they are also sometimes called triangulated) and that adding any edge to such a graph destroys the planarity property.

Past work. The main goal of research on universal sets is to construct such sets of small cardinality. That finite universal sets exist is trivial as, according to Fáry’s theorem [7], each planar graph has a straight-line embedding in , so we can simply embed each -vertex planar graph into a separate set of points. For each there is only one (non-isomorphic) maximal planar graph, so points are sufficient. This was extended by Cardinal et al. [3], who showed the existence of -universal sets of cardinality for , as well as their non-existence for . As shown recently by Scheucher et al. [10] with a computer-assisted proof, for at least points are necessary in a -universal set. This leaves the question of existence of -universal sets of size open only for .

For arbitrary values of , various algorithmic upper bounds for -universal sets of size have been described that make use of points on an integer lattice [1, 2, 5, 6, 11]. The best current upper bound of was given by Bannister et al. [1]. The technique in [1] involves reducing the problem to a combinatorial question about superpatterns of integer permutations.

Very little is known about lower bounds for universal sets. Following an earlier sequence of papers [6, 4, 9], recently Scheuer et al. [10] proved that points are required for a set to be universal.

There is also some research on constructing universal sets for some sub-classes of planar graphs. For example, Bannister et al. [1] describe a tight asymptotic bound of for the size of -universal sets for a specific type of “simply-nested” graphs.

Summarizing, in spite of 30 years of research, the gap between the lower and upper bounds for the size of -universal sets is still very large, between linear and quadratic in .

Our contribution.

The probabilistic method is a powerful tool for proving existence of various combinatorial structures with desired properties. In its standard form, it works by establishing a probability distribution on these structures and showing that they have non-zero probability (typically, in fact, large) of having the required property. We address here the question whether this approach can work for showing existence of smaller

-universal sets. Our result is, unfortunately, negative; namely we give an lower bound for universal sets constructed in this way, as summarized by the following theorem.

Let be a set of random points chosen uniformly from the unit square. If then with probability at least set is not -universal.

The constants in Theorem 1 are not optimized; they were chosen with the simplicity of the proof in mind.

The idea behind the proof of Theorem 1

is to reduce the problem to the well-studied problem of estimating longest increasing sub-sequences in random permutations. We present this proof in the next section.

We stress that Theorem 1 applies only to the probabilistic method in its basic form. It does not preclude the possibility that, for some

, there exist some other probability distribution on the unit square for which the probability of generating a universal set is large, or even that the uniform distribution has non-zero probability of generating such a set.

## 2 Lower Bound Proof

We now prove Theorem 1, starting with a lemma showing that a universal set must contain a large monotone subset, as defined below.

Let be a set of points in the plane, where and for all . Consider a subset of , say , with . We say that is increasing iff and we say that it’s decreasing iff . If is either increasing or decreasing, we call it monotone.

If is an -universal set then has a monotone subset of cardinality .

###### Proof.

Without loss of generality we can assume that is a multiple of , for otherwise we can apply the argument below to .

Let , and let be a maximal planar graph that consists of a sequence of 3-cycles with each consecutive pair of 3-cycles connected by edges in such a way that for the graph between and is -regular. This graph is -connected (removing two vertices can only destroy edges between consecutive layers), so it follows from a result of Whitney [12] that has a topologically unique embedding in the plane up to choice of the external face.

Since is universal, has a planar straight-line embedding into . In this embedding, no matter what face of is selected as the external face, in the sequence , either the first or the last will be embedded into nested triangles in the plane. Denote these nested triangles by , listed in order of decreasing area. In other words, each is inside , for . By definition, the corner points of each are in . For each , let denote the bounding box (an axis-parallel rectangle) of . By straighforward geometry, these bounding boxes are also nested, that is each is inside , for . (See Figure 1.)

We claim that, for each , at least one corner of is also a corner of , and thus also belongs to . This is straighforward: each of the four (axis-parallel) sides of must touch one of the three corners of . So there must be a corner of that is touched by two sides of . This corner of is then also a corner of , proving our claim.

Let be the set of the corners of the bounding boxes that are in . By the claim in the above paragraph, we have . Partition into two subsets and , where is the set of points in that are either bottom-left corners or top-right corners of the bounding boxes, and is the set of points in that are either top-left corners or bottom-right corners. Since the bounding boxes are nested, we obtain that is increasing and is decreasing.

To complete the proof, take to be either or , whichever set is larger, breaking the tie arbitrarily. Then is monotone. Furthermore, since , , and and are disjoint, we conclude that , completing the proof. ∎

Let and be as defined before the statement of Lemma 2. With we can associate a permutation of determined by having be the rank of in the set of all y-coordinates of , that is

 yπ−1(1)

We denote this permutation by . Then naturally induces a subsequence of . Obviously, is increasing (resp. decreasing or monotone) iff its induced subsequence of is increasing (resp. decreasing or monotone).

We are now ready to prove Theorem 1. Suppose now that is a random set of points chosen uniformly from the unit square. The probability that any two points have an equal coordinate is , and thus this event can be simply neglected. Further, this distribution on sets induces a uniform distribution on the associated permutations of . Therefore, by Lemma 2, to prove Theorem 1 it is sufficient to show the following claim:

If is a random permutation of and , then the probability that contains a monotone subsequence of length is at most .

###### Proof.

The proof of Claim 2 can be derived from the the argument given by Alan Frieze in [8]. We include it here for the sake of completeness.

Without loss of generality, we can assume that . Otherwise, and the claim is trivially true.

Let , where the inequality follows from the bound on . Let be the family of -element subsets of . If , say where , we will say that is monotone in a permutation if defines a monotone sequence. Each monotone subsequence of length can be either increasing or decreasing, so each has probability of being monotone. Using these observations, the union bound, and the inequality (that follows from Stirling’s formula), we obtain:

 P(∃monotoneL∈L) ≤∑L∈LP(Lis % monotone) =(mℓ)⋅2ℓ! =2⋅m⋅(m−1)⋅…⋅(m−ℓ+1)(ℓ!)2 ≤2⋅mℓ(ℓ/e)2ℓ =2⋅(me2ℓ2)ℓ ≤2⋅⎛⎝me2(2e√m)2⎞⎠ℓ =2⋅4−ℓ ≤8⋅4−n/12,

since . This completes the proof of Claim 2 and Theorem 1. ∎

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