On the Maximum Number of Crossings in Star-Simple Drawings of K n with No Empty Lens
Stefan Felsner, Michael Hoffmann, Kristin Knorr, Irene Parada
OOn the Maximum Number of Crossings inStar-Simple Drawings of K n with No Empty Lens (cid:63) Stefan Felsner , Michael Hoffmann , Kristin Knorr , and Irene Parada Institute of Mathematics, Technische Universitt Berlin, Germany [email protected] Department of Computer Science, ETH Zrich, Switzerland [email protected] Department of Computer Science, Freie Universitt Berlin, Germany [email protected] Department of Mathematics and Computer Science,TU Eindhoven, The Netherlands [email protected]
Abstract.
A star-simple drawing of a graph is a drawing in which adja-cent edges do not cross. In contrast, there is no restriction on the numberof crossings between two independent edges. When allowing empty lenses(a face in the arrangement induced by two edges that is bounded by a2-cycle), two independent edges may cross arbitrarily many times in astar-simple drawing. We consider star-simple drawings of K n with noempty lens. In this setting we prove an upper bound of 3(( n − n !. Keywords: star-simple drawings · topological graphs · edge crossings. A topological drawing of a graph G is a drawing in the plane where vertices arerepresented by pairwise distinct points, and edges are represented by Jordanarcs with their vertices as endpoints. Additionally, edges do not contain anyother vertices, every common point of two edges is either a proper crossing or acommon endpoint, and no three edges cross at a single point. A simple drawing is a topological drawing in which adjacent edges do not cross, and independentedges cross at most once. (cid:63) This research started at the 3rd Workshop within the collaborative DACH project
Arrangements and Drawings , August 19–23, 2019, in Wergenstein (GR), Switzerland,supported by the German Research Foundation (DFG), the Austrian Science Fund(FWF), and the Swiss National Science Foundation (SNSF). We thank the partic-ipants for stimulating discussions. S.F. is supported by DFG Project FE 340/12-1.M.H. is supported by SNSF Project 200021E-171681. K.K. is supported by DFGProject MU 3501/3-1 and within the Research Training Group GRK 2434
Facets ofComplexity . I.P. was supported by FWF project I 3340-N35. a r X i v : . [ c s . C G ] A ug S. Felsner et al.
We study a broader class of topological drawings, which are called star-simple drawings, where adjacent edges do not cross, but independent edges maycross any number of times; see Fig. 1 for illustration. In such a drawing, forevery vertex v the induced substar centered at v is simple, that is, the drawingrestricted to the edges incident to v forms a plane drawing. In the literature(e.g., [1,2]) these drawings also appear under the name semi-simple , but weprefer star-simple because the name is much more descriptive. (a) simple (b) star-simple but not simple (c) not star-simple Fig. 1: Topological drawings of K and a (nonempty) lens (shaded in (b)).In contrast to simple drawings, star-simple drawings can have regions or cellswhose boundary consists of two continuous pieces of (two) edges. We call sucha region a lens ; see Fig. 1b. A lens is empty if it has no vertex in its interior.If empty lenses are allowed, the number of crossings in star-simple drawings ofgraphs with at least two edges is unbounded (twisting), as illustrated in Fig. 2a.We restrict our attention to star-simple drawings with no empty lens. This re-striction is—in general—not sufficient to guarantee a bounded number of cross-ings (spiraling), as illustrated in Fig. 2b. However, we will show that star-simpledrawings of the complete graph K n with no empty lens have a bounded numberof crossings. (a) twisting (b) spiraling Fig. 2: Constructions to achieve an unbounded number of crossings.Empty lenses also play a role in the context of the crossing lemma for multi-graphs [5]. This is because a group of arbitrarily many parallel edges can bedrawn without a single crossing. Hence, for general multigraphs there is no hope rossings in Star-Simple Drawings of K n with No Empty Lens 3 to get a lower bound on the number of crossings as a function of the numberof edges. However, if we forbid empty lenses, we cannot draw arbitrarily manyparallel edges.Kynˇcl [3, Section 5 ”Picture hanging without crossings”] proposed a construc-tion of two edges in a graph on n vertices with an exponential number (2 n − )of crossings and no empty lens; see Fig. 3. This configuration can be completedto a star-simple drawing of K n , cf. [6]. For n = 6 it is possible to have one morecrossing while maintaining the property that the drawing can be completed to astar-simple drawing of K ; see Fig. 4. Repeated application of the doubling con-struction of Fig. 3 leads to two edges with 2 n − + 2 n − crossings in a graph on n vertices. This configuration can be completed to a star-simple drawing of K n .We suspect that this is the maximum number of crossings of two edges in astar-simple drawing of K n .Fig. 3: The doubling construction yields an exponential number of crossings. (a) 5 crossings (b) star-simple completion (c) the stars of the drawing Fig. 4: Two edges with 2 n − + 2 n − crossings in a star-simple drawing of K n , for n = 6. S. Felsner et al.
In this section we study the induced drawing D ( e, e (cid:48) ) of two independent edges e and e (cid:48) in a star-simple drawing D of the complete graph. We start by observingthat the endpoints of e and e (cid:48) must lie in the same region of D ( e, e (cid:48) ). This factwas also used in earlier work by Aichholzer et al. [1] and by Kynˇcl [4]. Lemma 1.
The four vertices incident to e and e (cid:48) belong to the same regionof D ( e, e (cid:48) ) .Proof. Assuming that the two edges cross at least two times, the drawing D ( e, e (cid:48) )has at least two regions. Otherwise, the statement is trivial. If the four verticesdo not belong to the same region of D ( e, e (cid:48) ), then there is a vertex u of e anda vertex v of e (cid:48) that belong to different regions. Now consider the edge uv inthe drawing D of the complete graph. This edge has ends in different regionsof D ( e, e (cid:48) ), whence it has a crossing with either e or e (cid:48) . This, however, makesa crossing in the star of u or v . This contradicts the assumption that D is astar-simple drawing. (cid:117)(cid:116) Lemma 1 implies that the deadlock configurations as shown in Fig. 5a donot occur in star-simple drawings of complete graphs. Formally, a deadlock is apair e, e (cid:48) of edges such that not all incident vertices lie in the same region of thedrawing D ( e, e (cid:48) ).Now suppose that D is a star-simple drawing of a complete graph with noempty lens. In this case we can argue that e and e (cid:48) do not form a configurationas the black edge e and the red edge e (cid:48) in Fig. 5b. Indeed, that configurationhas an interior lens L and by assumption this lens is non-empty, i.e., L containsa vertex x . Let e and e (cid:48) be the black and the red edge in Fig. 5b, respectively,and let u be a vertex of e . The edge xu (the green edge in the figure) has nocrossing with e , hence it follows the ”tunnel” of the black edge. This yields adeadlock configuration of the edges xu and e (cid:48) . Note that if in Fig. 5b instead ofdrawing the green edge xu we connect x with an edge f to one of the vertices ofthe red edge e (cid:48) such that f and the red edge have no crossing, then f and theblack edge e form a deadlock. e e (cid:48) e e (cid:48) (a) deadlocks x ue e (cid:48) (b) spiral Fig. 5: Constructions to achieve an unbounded number of crossings. rossings in Star-Simple Drawings of K n with No Empty Lens 5 We use this intuition to formally define a spiral. Two edges e, e (cid:48) form a spiral if they form a lens L such that if we place a vertex x in L and draw a curve γ connecting x to a vertex u of e so that γ does not cross e , then γ and e (cid:48) form adeadlock. The discussion above proves the following lemma: Lemma 2.
A star-simple drawing of a complete graph with no empty lens hasno pair e , e (cid:48) of edges that form a spiral. In this section we derive an upper bound for the number of crossings of twoedges in a star-simple drawing of K n with no empty lens. Theorem 1.
Consider a star-simple drawing of K n with no empty lens. If C ( k ) is the maximum number of crossings of a pair of edges that (a) form no deadlockand no spiral and such that (b) all lenses formed by the two edges can be hit by k points, then C ( k ) ≤ e · k ! , where e ≈ . is Euler’s number.Proof. Due to Lemma 1 we can assume that all four vertices of e and e (cid:48) are onthe outer face of the drawing D ( e, e (cid:48) ). We think of e (cid:48) as being drawn red andhorizontally and of e as being a black meander edge. Let p , . . . , p k be pointshitting all the lenses of the drawing D ( e, e (cid:48) ). Let u be one of the endpoints of e .For each i = 1 , . . . , k we draw an edge e i connecting p i to u such that e i has nocrossing with e and, subject to this, the number of crossings with e (cid:48) is minimized.Fig. 6 shows an example.Note that we do not claim that all these edges e , . . . , e k together with e and e (cid:48) can be extended to a star-simple drawing of a complete graph. Therefore,we cannot use Lemma 2 directly but state the assumption (a) instead. p i e i e ue (cid:48) Fig. 6: The drawing D ( e, e (cid:48) ) and an edge e i connecting p i to u .We claim the following three properties:(P1) The edges e i and e (cid:48) form no deadlock and no spiral. S. Felsner et al. (P2) All lenses of e i and e (cid:48) are hit by the k − p , . . . , p i − , p i +1 , . . . , p k . (P3) Between any two crossings of e and e (cid:48) from left to right, i.e., in the orderalong e (cid:48) , there is at least one crossing of e (cid:48) with one of the edges e i .Before proving the properties, we show that they imply the statement of thetheorem by induction on k . The base case 1 = C (0) ≤ e ·
0! = e is obvious.From (P1) and (P2) we see that the number X i of crossings of e i and e (cid:48) is upperbounded by C ( k − C ( k ) ≤ (cid:80) i X i . Combiningthese we get C ( k ) ≤ k · C ( k −
1) + 1 ≤ k ! · k (cid:88) s =0 s ! ≤ k ! · e . (cid:117)(cid:116) For the proof of the three claims we need some notation. Let ξ , ξ , . . . , ξ N be the crossings of e and e (cid:48) indexed according to the left to right order along thehorizontal edge e (cid:48) . Let g i and h i be the pieces of e (cid:48) and e , respectively, betweencrossings ξ i and ξ i +1 . The bounded region enclosed by g i ∪ h i is the bag B i and g i is the gap of the bag. In the drawing D ( e, e (cid:48) ) the bags B i where h i is acrossing free piece of e are exactly the inclusion-wise minimal lenses formed by e and e (cid:48) . From now on when referring to a lens we always mean such a minimallens. Indeed if there is no empty minimal lens, then there is no empty lens. Thefollowing observation is crucial. Observation 2
For two bags B i and B j the open interiors are either disjointor one is contained in the other.Proof. Every bag is bounded by a closed Jordan curve, and the boundaries oftwo distinct bags do not cross (at most they may touch at a single point that isone of ξ , ξ , . . . , ξ N ). (cid:117)(cid:116) Observation 2 implies that the containment order on the bags is a downwardsbranching forest. The minimal elements in the containment order are the lenses.Consider a lens L and the point p i inside L . Since the vertex u of e is in the outerface of D ( e, e (cid:48) ), the edge e i has to leave each bag that contains L . Furthermore,by definition e i does not cross e and therefore it has to leave a bag B containing L through the gap g of B . We now reformulate and prove the third claim (P3).(P3’) For each pair ξ i , ξ i +1 of consecutive crossings on e (cid:48) there is a lens L and apoint p j ∈ L such that e j crosses e (cid:48) between ξ i and ξ i +1 . Proof sketch for (P3’).
The pair ξ i , ξ i +1 is associated with the bag B i . In thecontainment order of bags a minimal bag below B i is a lens, let L be any of theminimal elements below B i . By assumption, L contains a point p j . Since L ⊆ B i ,we have that also p j ∈ B i . Thus, it follows that e j has a crossing with the gap g i ,i.e., e j has a crossing with e (cid:48) between ξ i and ξ i +1 . (cid:117)(cid:116) Proof sketch for (P1).
We have to show that e i and e (cid:48) form no deadlock and nospiral. The minimality condition in the definition of e i implies that if L = B i ⊂ rossings in Star-Simple Drawings of K n with No Empty Lens 7 B i ⊂ . . . ⊂ B i t is the maximal chain of bags with minimal element L , then e i crosses the gaps of these bags in the given order and has no further crossingswith e (cid:48) . If γ is a curve from L to u that avoids e , then in the ordered sequence ofgaps crossed by γ we find a subsequence that is identical to the ordered sequenceof gaps crossed by e i . Since e and e (cid:48) form no spiral, there is such a curve γ thatforms no deadlock with e (cid:48) . Therefore, e i forms no deadlock with e (cid:48) , either.Now assume that e i and e (cid:48) form a spiral. Let B be the largest bag contain-ing p i . Think of B as a drawing of e i with a broad pen, which may also have someextra branches that have no correspondence in e i , see Fig. 7. The formalizationof this picture is that for every bag β formed by e i with e (cid:48) there is a bag B ( β )formed by e and e (cid:48) with B ( β ) ⊂ β . Now, if there is a lens λ formed by e i with e (cid:48) such that every e i -avoiding curve to u is a deadlock with e (cid:48) , then there is alens L ( λ ) formed by e and e (cid:48) with L ( λ ) ⊂ λ such that every e -avoiding curvefrom L ( λ ) to u is also B -avoiding and hence e i -avoiding. Thus, every such curvehas a deadlock with e (cid:48) , whence e and e (cid:48) form a spiral, contradiction. (cid:117)(cid:116) Proof sketch for (P2).
We know by (P1) that e i and e (cid:48) form no deadlock. There-fore, by Lemma 1, the vertices of e i and e (cid:48) belong to the same region of D ( e i , e (cid:48) ).All crossings of e i with e (cid:48) correspond to bags of e and e (cid:48) , therefore the verticesof e and e (cid:48) are in the outer face of D ( e i , e (cid:48) ). Together this shows that p i is also inthe outer face of D ( e i , e (cid:48) ). Since every lens of D ( e i , e (cid:48) ) contains a lens of D ( e, e (cid:48) ),it also contains one of the points hitting all lenses of D ( e, e (cid:48) ). Hence, all lensesof D ( e i , e (cid:48) ) are hit by the k − p , . . . , p i − , p i +1 , . . . , p k . (cid:117)(cid:116) e p i e i p j ue (cid:48) Fig. 7: An edge e i (green) that forms a spiral with e (cid:48) . The bag B in gray and thelens L ( λ ) marked with the vertex p j (blue). that is, disjoint from e i except for possibly a shared endpoint S. Felsner et al. Accounting for the four endpoints of the two crossing edges we have k ≤ n − K n without empty lens is upper bounded by 3( n − K n has at most n ! crossings. This is thefirst finite upper bound on the number of crossings in star-simple drawings of thecomplete graph K n . We know drawings of K n in this drawing mode that havean exponential number of crossings. Thus, it would be interesting to reduce thehuge gap between the upper and the lower bound. Specifically, can a star-simpledrawing of K n have two edges with more than 2 n − + 2 n − crossings? References
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