An Omega(n^2) Lower Bound for Random Universal Sets for Planar Graphs
AAn Ω( n ) Lower Bound for Random UniversalSets for Planar Graphs
Alexander Choi
Department of Computer Science, University of California, [email protected]
Marek Chrobak
Department of Computer Science, University of California, [email protected]
Kevin Costello
Department of Mathematics, University of California, [email protected]
Abstract
A set U ⊆ R is n -universal if all n -vertex planar graphs have a planar straight-line embedding into U . We prove that if Q ⊆ R consists of points chosen randomly and uniformly from the unit squarethen Q must have cardinality Ω( n ) in order to be n -universal with high probability. This showsthat the probabilistic method, at least in its basic form, cannot be used to establish an o ( n ) upperbound on universal sets. Mathematics of Computing – Discrete Mathematics – GraphTheory
Keywords and phrases graph theory, planar graphs, universal sets
Digital Object Identifier
Funding
Alexander Choi : Research supported by NSF grant CCF-1536026.
Marek Chrobak : Research supported by NSF grant CCF-1536026.
Planar universal sets.
Let n ≥
4. A set U of points in R is called n -universal if eachplanar graph G with n vertices has an embedding into R that maps one-to-one the verticesof G into points of U and maps each edge of G into a straight-line segment connectingits endpoints, in such a way that these segments do not intersect (except of course at theendpoints). It is easy to see that for U to be universal, it is sufficient that the requiredembedding exists only for maximal planar graphs, namely n -vertex planar graphs with 3 n − triangulated ) and that addingany 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 smallcardinality. That finite universal sets exist is trivial as, according to Fáry’s theorem [7], eachplanar graph has a straight-line embedding in R , so we can simply embed each n -vertexplanar graph into a separate set of points. For each n = 4 , n points are sufficient. This was extended by Cardinal et al. [3],who showed the existence of n -universal sets of cardinality n for n ≤
10, as well as theirnon-existence for n ≥
15. As shown recently by Scheucher et al. [10] with a computer-assistedproof, for n = 11 at least 12 points are necessary in a 11-universal set. This leaves thequestion of existence of n -universal sets of size n open only for n = 12 , , © Alexander Choi, Marek Chrobak, and Kevin Costello;licensed under Creative Commons License CC-BYLeibniz International Proceedings in InformaticsSchloss Dagstuhl – Leibniz-Zentrum für Informatik, Dagstuhl Publishing, Germany a r X i v : . [ c s . D M ] S e p X:2 Random Universal Sets Lower Bound
For arbitrary values of n , various algorithmic upper bounds for n -universal sets of size O ( n ) have been described that make use of points on an integer lattice [1, 2, 5, 6, 11]. Thebest current upper bound of n + O ( n ) was given by Bannister et al. [1]. The techniquein [1] involves reducing the problem to a combinatorial question about superpatterns ofinteger permutations.Very little is known about lower bounds for universal sets. Following an earlier sequenceof papers [6, 4, 9], recently Scheucher et al. [10] proved that (1 . − o (1)) n points arerequired for a set to be universal.There is also some research on constructing universal sets for some sub-classes of planargraphs. For example, Bannister et al. [1] describe a tight asymptotic bound of Θ( n log n ) forthe size of n -universal sets for a specific type of “simply-nested” graphs.Summarizing, in spite of 30 years of research, the gap between the lower and upperbounds for the size of n -universal sets is still very large, between linear and quadratic in n . Our contribution.
The probabilistic method is a powerful tool for proving existence ofvarious combinatorial structures with desired properties. In its standard form, it worksby establishing a probability distribution on these structures and showing that they havenon-zero probability (typically, in fact, large) of having the required property. We addresshere the question whether this approach can work for showing existence of smaller n -universalsets. Our result is, unfortunately, negative; namely we give an Ω( n ) lower bound foruniversal sets constructed in this way, as summarized by the following theorem. (cid:73) Theorem 1.
Let Q ⊆ R be a set of m random points chosen uniformly from the unitsquare. If m ≤ (cid:0) n e (cid:1) then with probability at least − · − n/ set Q is not n -universal. The constants in Theorem 1 are not optimized; they were chosen with the simplicity ofthe proof in mind.The idea behind the proof of Theorem 1 is to reduce the problem to the well-studiedproblem of estimating longest increasing sub-sequences in random permutations. We presentthis proof in the next section.We stress that Theorem 1 applies only to the probabilistic method in its basic form. Itdoes not preclude the possibility that, for some m = o ( n ), there exist some other probabilitydistribution 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. We now prove Theorem 1, starting with a lemma showing that a universal set must containa large monotone subset, as defined below.Let U = { ( x , y ) , ( x , y ) , ..., ( x m , y m ) } ⊆ R be a set of m points in the plane, where x < x < ... < x m and y i = y j for all i = j . Consider a subset S of U , say S = { ( x j , y j ) , ( x j , y j ) , ..., ( x j ‘ , y j ‘ ) } , with j < j < ... < j ‘ . We say that S is increasing iff y j < y j < ... < y j ‘ and we say that it’s decreasing iff y j > y j > ... > y j ‘ . If S is eitherincreasing or decreasing, we call it monotone . (cid:73) Lemma 2. If U is an n -universal set then U has a monotone subset S of cardinality b n/ c . Proof.
Without loss of generality we can assume that n is a multiple of 12, for otherwise wecan apply the argument below to n = n − ( n mod 12). . Choi, M. Chrobak, and K. Costello XX:3 Let k = n/
6, and let G be a maximal planar graph that consists of a sequence of 2 k = n/ C , C , ..., C k with each consecutive pair of 3-cycles connected by 6 edges in sucha way that for 1 ≤ j ≤ k − C k and C k +1 is 2-regular. This graph is3-connected (removing two vertices can only destroy 4 edges between consecutive layers), soit follows from a result of Whitney [12] that G has a topologically unique embedding in theplane up to choice of the external face.Since U is universal, G has a planar straight-line embedding into U . In this embedding,no matter what face of G is selected as the external face, in the sequence C , C , ..., C k ,either the first k or the last k will be embedded into nested triangles in the plane. Denotethese nested triangles by T , T , ..., T k , listed in order of decreasing area. In other words,each T i is inside T i − , for i >
1. By definition, the corner points of each T i are in U . Foreach i , let B i denote the bounding box (an axis-parallel rectangle) of T i . By straighforwardgeometry, these bounding boxes are also nested, that is each B i is inside B i − , for i > T T T B B B Figure 1
Triangles T , T , T and their bounding boxes. We claim that, for each i , at least one corner of B i is also a corner of T i , and thus alsobelongs to U . This is straighforward: each of the four (axis-parallel) sides of B i must touchone of the three corners of T i . So there must be a corner of T i that is touched by two sidesof B i . This corner of T i is then also a corner of B i , proving our claim.Let S be the set of the corners of the bounding boxes B i that are in U . By the claim inthe above paragraph, we have | S | ≥ k . Partition S into two subsets S and S , where S isthe set of points in S that are either bottom-left corners or top-right corners of the boundingboxes, and S is the set of points in S that are either top-left corners or bottom-right corners.Since the bounding boxes B i are nested, we obtain that S is increasing and S is decreasing.To complete the proof, take S to be either S or S , whichever set is larger, breaking thetie arbitrarily. Then S is monotone. Furthermore, since | S | ≥ k , S ∪ S = S , and S and S are disjoint, we conclude that | S | ≥ k/ n/
12, completing the proof. (cid:74)
Let U and S be as defined before the statement of Lemma 2. With U we can associate apermutation π of { , , ..., m } determined by having π ( i ) be the rank of y i in the set of all X:4 Random Universal Sets Lower Bound y-coordinates of U , that is y π − (1) < y π − (2) < ... < y π − ( m ) . We denote this permutation π by perm ( U ). Then S naturally induces a subsequence π ( j ) π ( j ) ...π ( j ‘ ) of perm ( U ). Obviously, S is increasing (resp. decreasing or monotone) iffits induced subsequence of perm ( U ) is increasing (resp. decreasing or monotone).We are now ready to prove Theorem 1. Suppose now that Q is a random set of m pointschosen uniformly from the unit square. The probability that any two points have an equalcoordinate is 0, and thus this event can be simply neglected. Further, this distribution on sets Q induces a uniform distribution on the associated permutations perm ( Q ) of { , , ..., m } .Therefore, by Lemma 2, to prove Theorem 1 it is sufficient to show the following claim: (cid:66) Claim 3. If π is a random permutation of { , , ..., m } and m ≤ ( n e ) , then the probabilitythat π contains a monotone subsequence of length b n/ c is at most 8 · − n/ . Proof.
The proof of Claim 3 can be derived from the the argument given by Alan Friezein [8]. We include it here for the sake of completeness.Without loss of generality, we can assume that n ≥
24. Otherwise, m = 0 and the claimis trivially true.Let ‘ = b n c ≥ e √ m , where the inequality follows from the bound on m . Let L bethe family of ‘ -element subsets of { , , ..., m } . If L ∈ L , say L = { a , a , ..., a ‘ } where a < a < ... < a ‘ , we will say that L is monotone in a permutation π if π ( a ) π ( a ) . . . π ( a ‘ )defines a monotone sequence. Each monotone subsequence of length ‘ can be either increasingor decreasing, so each L ∈ L has probability 2 /‘ ! of being monotone. Using these observations,the union bound, and the inequality ‘ ! ≥ ( ‘/e ) ‘ (that follows from Stirling’s formula), weobtain: P ( ∃ monotone L ∈ L ) ≤ P L ∈L P ( L is monotone )= (cid:18) m‘ (cid:19) · ‘ != 2 · m · ( m − · . . . · ( m − ‘ + 1)( ‘ !) ≤ · m ‘ ( ‘/e ) ‘ = 2 · (cid:18) me ‘ (cid:19) ‘ ≤ · me (2 e √ m ) ! ‘ = 2 · − ‘ ≤ · − n/ , since ‘ ≥ n/ −
1. This completes the proof of Claim 3 and Theorem 1. (cid:74)
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