Limiting crossing numbers for geodesic drawings on the sphere
LLimiting crossing numbers for geodesic drawingson the sphere
Marthe Bonamy (cid:63) , Bojan Mohar (cid:63)(cid:63) , and Alexandra Wesolek (cid:63) (cid:63) (cid:63) CNRS, LaBRI, Universit´e de Bordeaux, France [email protected] Department of Mathematics, Simon Fraser University, Burnaby, BC, Canada [email protected]@sfu.ca
Abstract.
We introduce a model for random geodesic drawings of thecomplete bipartite graph K n,n on the unit sphere SS in R , where weselect the vertices in each bipartite class of K n,n with respect to two non-degenerate probability measures on SS . It has been proved recently thatmany such measures give drawings whose crossing number approximatesthe Zarankiewicz number (the conjectured crossing number of K n,n ).In this paper we consider the intersection graphs associated with suchrandom drawings. We prove that for any probability measures, the re-sulting random intersection graphs form a convergent graph sequencein the sense of graph limits. The edge density of the limiting graphonturns out to be independent of the two measures as long as they areantipodally symmetric. However, it is shown that the triangle densitiesbehave differently. We examine a specific random model, blow-ups of an-tipodal drawings D of K , , and show that the triangle density in thecorresponding crossing graphon depends on the angles between the greatcircles containing the edges in D and can attain any value in the interval (cid:0) , (cid:1) . Keywords:
Crossing Number · Graph Limits · Geodesic Drawing · Ran-dom Drawing · Triangle Density.
The crossing number cr ( G ) of a graph G is the minimum number of crossingsobtained by drawing G in the plane (or the sphere). In this paper we consider the (spherical) geodesic crossing number cr ( G ), for which we minimize the numberof crossings taken over all drawings of G in the unit sphere SS in R such thateach edge uv is a geodesic segment joining points u and v in SS . Recall that (cid:63) Supported in part by the ANR Project DISTANCIA (ANR-17-CE40-0015) operatedby the French National Research Agency (ANR). (cid:63)(cid:63)
Supported in part by the NSERC Discovery Grant R611450 (Canada), by the CanadaResearch Chairs program, and by the Research Project J1-8130 of ARRS (Slovenia). (cid:63) (cid:63) (cid:63)
Supported by the Vanier Canada Graduate Scholarships program. a r X i v : . [ c s . C G ] A ug Marthe Bonamy , Bojan Mohar, and Alexandra Wesolek geodesic segments (or geodesic arcs ) in SS are arcs of great circles whose lengthis at most π . Also note that cr ( G ) ≤ cr ( G ) for every graph G .Crossing number minimization has a long history and is used both in appli-cations and as a theoretical tool in mathematics. We refer to [13] for an overviewabout the history and the use of crossing numbers. Despite various breakthroughresults about crossing numbers, some of the very basic questions remain openas of today, two of the most intriguing being what are the crossing numbersof the complete graphs K n and what are the crossing numbers of the completebipartite graphs K n,n (the Tur´an Brickyard Problem). The asymptotic versionsof both problems are strongly related [12] and a lower bound for the limitingcrossing number of K n,n gives a related lower bound for K n . The asymptoticversion of the rectilinear crossing number of K n is related to Sylvester’s Fourpoint problem in the plane [15,14], see also [13] for recent results. The geodesicversion on the sphere, which we discuss in this paper, is a spherical version ofSylvester’s problem. In this paper we initiate the study of limiting properties of intersection graphsassociated with drawings of complete and complete bipartite graphs. We limitourselves to geodesic drawings on the unit sphere in R in which case the draw-ings are determined by the choice of the placements of the vertices on the sphere.The first main result of this work shows that whenever the vertices in each bi-partite class of K n,n are selected according to some (non-degenerate) probabilitymeasure on SS (where the two measures used for each class can be different),then, with probability 1, the intersection graphs form a convergent sequence ofgraphs in the sense of graph limits [6]. See Theorem 2.The basic combinatorial property of convergent graph sequences is that ofsubgraph densities. The density of edges in the crossing graphs corresponds tothe asymptotic crossing number. In addition to this, we examine one particu-lar related basic question: what is the density of triangles. We show that theirdensity can be substantially different among different randomized models. Al-though this result may be seen as “expected”, it is still somewhat surprising.Indeed, it shows that there is a large variety of drawings of K n,n , all attainingthe Zarankiewicz bound, in which the number of triples of mutually crossingedges varies significantly, and can attain any value in the interval (cid:0) , (cid:1) .See Theorems 4 and 6. We believe that further exploring of subgraph densitiesin crossing graphons may give a deeper insight into the basic Tur´an’s BrickyardProblem for geodesic drawings on the sphere. During World War II, Hungarian mathematician P´al Tur´an worked in a brickfactory near Budapest. There the bricks were transported on wheeled trucksfrom kilns to storage yards. It was difficult to push the trucks past the railcrossings and it would result in extra work if bricks fell of the trucks. Therefore imiting crossing numbers for geodesic drawings on the sphere 3
Tur´an wondered if there was a way of arranging the rails such that there wouldbe less crossings between them. Seeing the kilns and storage yards as parts of abipartite graph, this led to the more general question of the minimum number ofcrossings in drawings of complete bipartite graphs K n,n . Zarankiewicz [18] andUrbanik [16] suggested drawings that involved Z ( m, n ) = (cid:98) n (cid:99) (cid:98) n − (cid:99) (cid:98) m (cid:99) (cid:98) m − (cid:99) = n ( n − m ( m − , n, m are even; n ( n − m − , n is even, m is odd; ( n − ( m − , n, m are odd (1)crossings. Whether this value is the best possible remains unanswered to thisday despite numerous attacks using powerful machinery in trying to resolve thisconjecture.A general construction of drawings of complete bipartite graphs attaining theZarankiewicz bound was recently exhibited [9]. All of them are geodesic drawingsin SS and they show that cr ( K n,n ) ≤ cr ( K n,n ) ≤ Z ( n, n ) for every n ≥ . (2)It is not hard to see that the following limits exist: λ := lim n →∞ n − cr ( K n,n ) and λ := lim n →∞ n − cr ( K n,n ) . Clearly, (2) implies that λ ≤ λ ≤ . The asymptotic Zarankiewicz conjecture for the usual and the geodesic crossing number is also open. Conjecture 1. λ = λ = . In 1965, Moon [11] proved that a random set of n points on the unit sphere SS in R joined by geodesics gives rise to a drawing of K n whose number of crossingsasymptotically approaches the conjectured value. It was proved recently [10] thatthe same phenomenon appears in a much more general random setting. Theseresults can also be extended to random drawings of the complete bipartite graphs K n,n where it was shown that under a symmetry condition on the probabilitymeasures the crossings in such drawings converge to the Zarankiewicz value.A probability distribution µ on SS is nondegenerate if for every great circle Q ⊂ SS , µ ( Q ) = 0. It is antipodally-symmetric if for every measurable set A ⊆ SS the measure of its antipodal set A is the same, µ ( A ) = µ ( A ). Theorem 1 ([10]).
Let µ , µ be nondegenerate antipodally-symmetric proba-bility distributions on the unit sphere SS . Then a µ -random set of n pointson SS joined by geodesics (segments of great circles) to a µ -random set of n points gives rise to a drawing D n of the complete bipartite graph K n,n such that cr ( D n ) /Z ( n, n ) = 1 + o (1) a.a.s. The random drawing model in the theorem will be referred to as ( µ , µ ) -random drawing of the complete bipartite graph K n,n . Marthe Bonamy , Bojan Mohar, and Alexandra Wesolek
Let N = { n , n , n , . . . } be an infinite set of positive integers, where n < n Let µ and µ be nondegenerate probability measures on SS . Let A n and B n be a µ -random and a µ -random set of n points in SS , respectively,let D n be the corresponding ( µ , µ ) -random geodesic drawing of K n,n on parts A n and B n , and let X n be its crossing graph. The sequence of graphs ( X n ) isconvergent with probability and there is a graphon W = W ( µ , µ ) that is thelimit of this convergent sequence. Since the number of edges in the crossing graph corresponds to the numberof crossings in D n , we have t ( K , X n ) = 2 | E ( X n ) || X n | = 2 cr ( D n ) n . Thus, Theorem 1 shows a tight relationship with the asymptotic Zarankiewiczconjecture and can be expressed as follows. imiting crossing numbers for geodesic drawings on the sphere 5 Theorem 3. Let µ , µ be nondegenerate antipodally-symmetric probability mea-sures on SS . Let W = W ( µ , µ ) be the corresponding graphon of the sequence ( X n ) as defined above. Then t ( K , W ( µ , µ )) = 18 . We follow standard terminology from [1,4] for graph theory and from [13] fordrawings of graphs. A drawing of a graph is good if any two edges cross atmost once, no two edges with a common endvertex cross, and no three edgescross at the same point. The first two conditions are clear when we considergeodesic drawings, and the third condition can always be satisfied if we make aninfinitesimal perturbation.We say that a set of points on the unit sphere SS is in general position if notwo of the points are antipodal to each other, no three of them lie on the samegreat circle and no three geodesic arcs joining pairs of points cross at the samepoint. If µ is a nondegenerate probability distribution on SS , then randomlychosen vertices will be in general position with probability 1. In the following we want to draw a comparison of subgraph densities of thecrossing graphs X n to a concept similar to the Buffon Needle Problem (see,e.g. [5] or [17]). We pick endpoints of segments randomly w.r.t. some probabilitydistribution and consider the crossings formed by the segments. If the probabilitydistribution is uniform on the sphere, it is equivalent as throwing a (bended)needle onto the sphere, where the needle length varies. Now considering a smallnumber of such segments on the sphere we ask how they will cross each other.Let µ and µ be nondegenerate probability measures on SS . A ( µ , µ ) -random geodesic segment is a geodesic segment uv whose endpoints u, v arechosen randomly w.r.t. µ and µ , respectively. For a given graph H of order k = | H | , we pick k ( µ , µ )-random geodesic segments on the sphere and lookat the probability that H is a subgraph of their intersection graph. Let A = { a , . . . , a k } be a µ -random set of points in SS and B = { b , . . . , b k } be a µ random set of points in SS . The segments we are considering are a b , . . . , a k b k .Note that the probability that H is a subgraph of the intersection graph of a b , . . . , a k b k depends on µ and µ only. Definition 1. Let X be the intersection graph of k ( µ , µ ) -random geodesicsegments a b , . . . , a k b k and let H be a graph of order k . For a bijection φ : V ( H ) → V ( X ) we define p H := P r [ φ is a graph homomorphism ] . Marthe Bonamy , Bojan Mohar, and Alexandra Wesolek Observe that p H is independent of φ , since the segments a i b i ( i = 1 , . . . , k ) areselected independently.We want to compare the above model with another model where we pick n (cid:29) k points with respect to µ and µ each, and consider the correspondingcrossing graph X n of a drawing D n of K n,n . We will show that the models areclosely related: with growing n , picking k vertices from X n , they will with highprobability come from k independent geodesic segments and therefore represent( µ , µ )-random geodesic segments. In the following we fix a graph H and amapping φ : V ( H ) → V ( X n ). Definition 2. For given X n , let φ : V ( H ) → V ( X n ) and we define the randomvariable y H,φ on X n to be y H,φ ( X n ) = (cid:40) if φ is a graph homomorphism H → X n otherwiseand denote its expectation by E φ := E [ y H,φ ] . Note that E φ is not the same for every φ . For example, if H is a completegraph, then E φ = 0 whenever im ( φ ) contains edges that share a vertex, as thoseedges never cross and hence are not adjacent in the crossing graph. Lemma 1. Let ( X n ) be a sequence of the crossing graphs of ( µ , µ ) -randomgeodesic drawings D n of K n,n for n = 1 , , . . . , and let H be a fixed graph oforder k . Then lim n →∞ | X n | k (cid:88) φ : V ( H ) → V ( X n ) E φ = p H . Proof. Let im ( φ ) = { v w , . . . , v k w k } . Then if |{ v , . . . , v k , w , . . . , w k }| = 2 n weare in the setup of Definition 1 and E [ y H,φ ] = p H . Moreover, there are O ( n k − )choices for φ for which |{ v , . . . , v k , w , . . . , w k }| < n and the result follows. (cid:117)(cid:116) Let us now consider the sum of the above defined random variables Y H := (cid:88) φ : V ( H ) → V ( X n ) y H,φ , (3)and note that Y H ( X n ) = hom ( H, X n ) and E [ Y H ] = (cid:80) φ : V ( H ) → V ( X n ) E φ . Theaim is to show that Y H is in general not far from its expectation. This then givesus the tool to show the existence of lim n →∞ | Y H || X n | k = t ( H ) with probability 1. Proposition 1. Let Y H be defined as in (3). Then we have var ( Y H ) = O ( n k − ) . The proof of the proposition is in the appendix. imiting crossing numbers for geodesic drawings on the sphere 7 Proof (of Theorem 2). By Proposition 1 and Chebyshev’s inequality there existsa constant C such that P r (cid:2) | Y H − E [ Y H ] | ≥ kCn k − (cid:3) ≤ k . Now if we choose k = k ( n ) appropriately such that k ( n ) n − converges to zeroand the sum (cid:80) ∞ n =1 1 k ( n ) is finite we can use the Borel-Cantelli Lemma. Forexample, we can choose k = n / and using Lemma 1 we get P r (cid:20)(cid:12)(cid:12)(cid:12)(cid:12) | Y H || X n | k − p H (cid:12)(cid:12)(cid:12)(cid:12) − (cid:12)(cid:12)(cid:12)(cid:12) p H − E [ Y H ] | X n | k (cid:12)(cid:12)(cid:12)(cid:12) ≥ Cn k − / | X n | k (cid:21) ≤ n / = ⇒ P r (cid:20)(cid:12)(cid:12)(cid:12)(cid:12) | Y H || X n | k − p H (cid:12)(cid:12)(cid:12)(cid:12) ≥ C (cid:48) n / (cid:21) ≤ n / . for some constant C (cid:48) . Then the Borel-Cantelli Lemma implies the following. Claim. For each fixed H , | Y H || X n | k → p H := t ( H ) with probability 1.Given that for each H , t ( H, X n ) → t ( H ) with probability 1, and since the proba-bilities are countably additive, it follows with probability 1 that t ( H, X n ) → t ( H )for every H . Consequently, the sequence of random crossing graphs ( X n ) is con-vergent with probability 1. (cid:117)(cid:116) K , In the previous sections, we have established the existence of crossing graphonsand determined densities t ( H ) for H = K if our measures µ , µ are antipo-dally symmetric. Somewhat surprisingly, these edge densities are the same forany “suitable” measures µ , µ . It is natural to ask what happens with otherhomomorphism densities in these crossing graphons. The purpose of this sectionis to show that the homomorphism densities of triangles behave differently. Tous, this was not a priori clear. We study a particular case of ( µ , µ )-randomdrawings of complete bipartite graphs and determine t ( K ) for the correspondinggraphon W ( µ , µ ).In the following we fix a drawing D of the complete bipartite graph K , where each part consists of two antipodal pairs of vertices on SS as in Figure1. We will be considering a blowup drawing D ( n )4 of D for which we replaceeach vertex from D with a circle of some small radius r = r ( n ) that is centeredat that vertex, and position n evenly spaced vertices on that circle. These n vertices will be referred to as the node of the corresponding vertex of K , . Wealso assume that all 8 n vertices obtained in this way are in general position. Inthat way, each edge of K , is replaced by a complete bipartite graph between thecorresponding nodes which we call the edge bundle . This means for N = 4 n that Marthe Bonamy , Bojan Mohar, and Alexandra Wesolek v w v v w w αβ γ δ Fig. 1. The left part shows a drawing D of a K , on parts { v , v , v , v } and { w , w , w , w } . The angles α and β are in the triangle formed by w , v and acrossing, whereas γ and δ are in a triangle formed by v , w and the same crossing.The right-hand side shows part of a D (3)4 drawing with the circles of w and v eachcontaining 3 vertices and with nine edges for each incident bundle emanating fromthese two nodes. D ( n )4 is a drawing of K N,N . In what follows, we discuss the number of trianglesin the intersection graph (of edges in D ( n )4 ) when n grows large. To simplify ourdiscussion about triangles, we first classify the crossings in D ( n )4 . D ( n )4 In the blowup drawing D ( n )4 , we distinguish three types of crossings, dependingon what they stem from, as depicted in Figure 2. (C) (B) (N) Fig. 2. Possible crossings in the blow up: Bundle-bundle crossings (C), bundle crossings(B) and node crossings (N).imiting crossing numbers for geodesic drawings on the sphere 9 Let us define these types (B), (C), and (N) more precisely and state theircount. The corresponding counting process is described in the appendix.(C) Two edge-bundles cross in a small neighborhood of a previous crossing in D .We call these bundle-bundle crossings (C). Since each edge-bundle consistsof n edges, this gives n bundle-bundle crossings for each crossing in D .(B) Two edges cross within a bundle. We call these bundle crossings (B). Here wehave (cid:0) n (cid:1) crossings per bundle assuming r ( n ) (cid:28) n − and a suitable rotationof the circles.(N) Two edge-bundles cross at a node. We call these node crossings (N). Let α ∈ (0 , π ) be the angle between two incident edges e, f in D which wereblown up to the edge-bundles, and let cr α be the resulting number of nodecrossings between the edges in the corresponding edge-bundles. Then wehave: cr α + cr π − α = n ( n − . D ( n )4 The crossings in a triangle need to stem from bundle-bundle crossings (C), bun-dle crossings (B) or node crossings (N) as specified above. We first prove thefollowing lemma. Lemma 2. Let D be a spherical drawing of a K , where each part consists oftwo pairs of antipodal vertices. Then no edge in D is crossed twice.Proof. Let the parts of the K , be A = { v , v , v , v } and B = { w , w , w , w } .Note that the edge v w can only be crossed by an edge between the otherantipodal pairs, i.e. v w , v w , v w , v w . All of them lie on the great circledefined by v w so in fact only one of these edges can cross v w . By symmetrythe same holds for the other edges. (cid:117)(cid:116) We classify the triangles in the intersection graph of the blowup drawing D ( n )4 as follows. We assign each crossing (which is an edge in the intersection graph) a type (C), (B), or (N) depending on whether it is a bundle-bundle, within bundleor a node crossing. We say a triangle c c c is of type ( l ( c ) l ( c ) l ( c )) where l ( c i )is the type of crossing c i .The above lemma shows that there are no (CCC) or (CCN) triangles in D ( n )4 .Also note that (CBB), (BBN) and (CBN) are not possible in general since BBsuggests that all edges are from the same bundle and the bundled edges in (CBN)cross the third edge either at a node or at a bundle-bundle crossing but not atboth. Triangles of type (NNN) either appear at three different nodes or at onenode. However, we can not have (NNN) triangles at three different nodes since K , is bipartite and hence triangle-free. By the following lemma, the numberof (NNN) triangles with all three crossings at one node is only of order rn andcan therefore be neglected. Lemma 3. The number of (NNN) triangles in D ( n )4 that correspond to threeedges at the same node is O ( rn ) . Moreover, if r ( n ) (cid:28) n − , there are no suchtriangles. ( CCC ) ( CCN ) Fig. 3. Triangles of type (CCC) and (CCN). x yAB B r rd d Fig. 4. Two edges from node A leading to antipodal nodes B and B can cross. If d = min { d , d } and r ≤ d , then | L x | = O ( rn ). Proof. Let us refer to Figure 4 and consider the possibility that an edge incidentwith a vertex y and leading to a node B crosses an edge incident with a vertex x that leads to the antipodal node B . If the geodesics from x to B intersect thecircle C A corresponding to A , we denote by L x the set of vertices in A that areon the smallest circular arc that contains those intersections.Then it is easy to see that either x ∈ L y or y ∈ L x (or both as shown inthe figure). It can be shown (details can be found in the full paper) that thenumber of cases where y ∈ L x or x ∈ L y is O ( rn ). In particular, if r (cid:28) n − then L x is empty. For each such pair x, y , the number of vertices z whose incidentedges leading to a node different from B and B make an (NNN) crossing trianglewith two edges incident with x and y , respectively, is O (( t + r ) n ), where t is thenumber of vertices on the arc between x and y . We define the parameter l whichis the number of vertices in the node A between x and the lowest point on thecircle of A (assuming that x is in the lower half of the circle and on the left side).Then t ∈ [2 l − Θ ( rn ) , l + Θ ( rn )]. This gives the following upper bound for thenumber of such triples ( x, y, z ):4 n/ (cid:88) l =1 O ( rn ) O (2 l + rn ) = O ( rn ) . Finally, since each such triple involves O ( n ) triples of mutually crossing edgesincident with x, y, z , we confirm that the number of considered (NNN) trianglesis O ( rn ). (cid:117)(cid:116) We are left with the following four cases. imiting crossing numbers for geodesic drawings on the sphere 11 (CNN) We consider pairwise crossings of three edges such that two cross at abundle-bundle crossing and the third edge crosses one edge each at one node each.These crossings depend on the angles α, β, γ, δ as depicted in Figure 1. By Section(N) in Appendix B the number of pairs of vertices x, y such that all edges at angle α incident to x cross all horizontal edges incident to y is π − α π n + O ( rn + n ).It is easy to see that the number of crossings we get in the triangle including α and β is (cid:0) π − α π (cid:1) (cid:16) π − β π (cid:17) n + O ( rn + n ). We have a similar count for the angles γ and δ . Then we have to add three other contributions corresponding to othercrossings in D . The antipodal crossing involves a triangles with α, β and γ, δ ,whereas the other two crossings involve triangles with α, γ and β, δ . Overall, thisgives n (cr α + cr δ )(cr γ + cr β ) + O ( rn + n ) triangles of this kind.(BBB) We consider pairwise crossings of three edges such that all edges arefrom one bundle. For each bundle we get (cid:0) n (cid:1) + O ( rn ) such triangles by Section(B) in Appendix B. There are 16 bundles so in total we have ∼ n + O ( rn )triangles of the type (BBB).(CCB) We consider pairwise crossings of three edges such that two edges arein one bundle and cross the third edge at a bundle-bundle crossing. There are2 (cid:0) n (cid:1) n + O ( rn ) triangles per each crossing in D . We have 4 crossings so intotal ∼ n + O ( rn ) triangles of this kind.(BNN) We consider pairwise crossings of three edges such that two are in thesame bundle and cross the third edge at a node. The argument is analogous tothe one for crossings of type ( N ). Starting at the top vertex, we enumerate thevertices clockwise along the cycle as in Figure 6. We consider an edge at angle α which ends in the i -th vertex in part ( A ) and its crossings to horizontal edges.From Section (N) in the Appendix B , we know that | S i | = 2 i + O ( rn ), where S i is as defined there. We can choose from (cid:0) i + O ( rn )2 (cid:1) pairs of left endpoints and (cid:0) n (cid:1) pairs of right endpoints for a triangle. The number of triangles with an edgeending in i and another edge ending in a vertex in W x = { y ∈ A | x ∈ L y } is oforder O ( rn ), where L y is defined as in the proof of Lemma 3. We consider nowedges at angle α ending in a vertex x in ( B ). Note that | S x | = π − απ n + O ( rn ).We can choose for any one of (cid:0) π − απ n + O ( rn )2 (cid:1) pairs of left endpoints (cid:0) n (cid:1) pairs ofright endpoints for a triangle. The number of triangles with another edge endingin a vertex in W x is of order O ( rn ). The contribution of triangles from edgesin ( C ) is the same as for edges in ( A ). Hence the number of triangles of type(BNN) is2 n ( π − α ) n/ π (cid:88) i =1 (cid:18) i (cid:19) · (cid:18) n (cid:19) + (cid:16) α π n (cid:17) · (cid:18) π − απ n (cid:19)(cid:18) n (cid:19) + O ( rn + n ) . For α and π − α added together, this gives112 n − α ( π − α )8 π n + O ( rn + n ) . Now note that for two bundles at angle α we can choose one of the bundles tocontain the bundled edges. This gives two options. At each node we have two pairs of bundles meeting at angle α and two pairs of bundles meeting at angle π − α . (In addition to these possibilities we get further (BNN) triangles fromtwo bundles at the same node that lead to antipodal nodes and correspond tothe value of α = π . They give only O ( rn + n ) triangles.) If α, β, γ, δ are theangles as in Figure 1, the overall number of (BNN) triangles is α ( α − π ) + β ( β − π ) + γ ( γ − π ) + δ ( δ − π ) π n + 83 n + O ( rn + n ) . If we leave out smaller order terms, the total number of triangles in theintersection graph by summing up the number of (CNN), (BBB), (CCB) and(BNN) triangles is α + β + γ + δ − π ( α + β + γ + δ ) π n + (2 π − α − δ )(2 π − γ − β )2 π n + 469 n . Theorem 4. Given a drawing D of a K , where each part has two antipodalpairs, let D ( n )4 be the blowup drawing, and let α, β, γ, δ be the angles definedabove. Then the limiting triangle density t ( K ) of the sequence D (1)4 , D (2)4 , . . . isequal to π (cid:16) (2 π − α − δ )(2 π − γ − β ) + 2( α + β + γ + δ ) − π ( α + β + γ + δ ) (cid:17) + 233 · + O ( r ) . Proof. We have determined the number of triangles in the intersection graphs.Dividing by the number of possible triangles in the intersection graph, (cid:0) n (cid:1) = n + O ( n ), gives the triangle density. (cid:117)(cid:116) Finally, let us show that the crossing graphs of drawings D ( n )4 can be interpretedas certain graphons. Theorem 5. For fixed r > let µ and µ be uniform distributions over twopairs of antipodal circles on SS of radius r each and let W ( µ , µ ) be the cross-ing graph limit of corresponding drawings. If we consider blow-up drawings D ( n )4 w.r.t. the centers of the circles of radius r , then the crossing graphs of D ( n )4 converge and their limit is the graphon W ( µ , µ ) .Proof. All we need to show is that the density t ( H ) in the random case limit andthe density t ( H ) of the blow-up drawing limit are the same for each graph H .Let k = | H | be the number of vertices of H and let φ : V ( H ) → [ k ] be a bijection.For distinct points x , . . . , x k , y , . . . , y k in SS , let X ( x , . . . , x k , y , . . . , y k ) be imiting crossing numbers for geodesic drawings on the sphere 13 the intersection graph of the geodesic segments x y , . . . , x k y k . Consider thefollowing function f ( x , . . . , x k , y , . . . , y k ) = (cid:40) , if v (cid:55)→ x φ ( v ) y φ ( v ) is a hom. H → X ( x , . . . , y k )0 , otherwise.Let S and S be the two circles on which µ and µ are defined, respectively.Since f as defined above is measurable because f − (1) is open, we can represent t ( H ) as t ( H ) = 1(8 πr ) k (cid:90) x ∈ S n × S n f ( x ) dx. In order to approximate t ( H ) consider a set C n which consists of n equidistantpoints on each of the cycles from S , S . Let π n : S ∪ S → C n be the functionthat maps a points from S ∪ S to its closest point in X . Let g n be a function g n : ( S ∪ S ) n → ( C n ) n that applies π n componentwise. Then f n = f ◦ g n converges pointwise to f on S n × S n . By the bounded convergence theorem t ( H ) = 1(8 πr ) k (cid:90) x ∈ S n × S n f ( x ) dx = 1(8 πr ) k lim n →∞ (cid:90) x ∈ S n × S n f n ( x ) dx = t ( H ) . (cid:117)(cid:116) The theorem shows that the same values for triangle densities in the ( µ , µ )-random setting hold as for the blow-up limit in Theorem 4. Theorem 6. For fixed r > let µ and µ be uniform distributions over twopairs of antipodal circles on SS of radius r each and let W ( µ , µ ) be the cross-ing graph limit of the corresponding drawings. Then · + O ( r ) ≤ t ( K , W ( µ , µ )) ≤ · + O ( r ) , and these bounds are best possible. The limiting triangle density t ( K ) dependson the angles α, β, γ, δ , and any value in the interval (cid:0) , (cid:1) is possible. The proof is in the appendix. It should be noted that the proofs of Theorem 2 and Theorem 3 also extend tothe case of the complete graph K n where we choose n points from the sphere withrespect to some antipodally symmetric probability measure µ . (Let us observethat antipodal symmetry is needed for such a result.) In Theorem 4 the value · = 0 . . t ( K ) = , so it would be of interest to find a closed expression for t ( K ). References 1. Bondy, J.A., Murty, U.S.R.: Graph theory, Graduate Texts in Mathematics,vol. 244. Springer, New York (2008). https://doi.org/10.1007/978-1-84628-970-52. Borgs, C., Chayes, J.T., Lov´asz, L., S´os, V.T., Vesztergombi, K.: Convergent se-quences of dense graphs. I. Subgraph frequencies, metric properties and testing.Adv. Math. (6), 1801–1851 (2008). https://doi.org/10.1016/j.aim.2008.07.0083. Borgs, C., Chayes, J., Lov´asz, L., S´os, V.T., Vesztergombi, K.: Counting graphhomomorphisms. In: Topics in discrete mathematics, Algorithms Combin., vol. 26,pp. 315–371. Springer, Berlin (2006). https://doi.org/10.1007/3-540-33700-8 184. Diestel, R.: Graph theory, Graduate Texts in Mathematics, vol. 173. Springer-Verlag, Berlin, third edn. (2005)5. Isokawa, Y.: Buffon’s short needle on the sphere. Bull. Fac. Ed. Kagoshima Univ.Natur. Sci. , 17–36 (2000)6. Lov´asz, L.: Large networks and graph limits, American Mathematical Society Col-loquium Publications, vol. 60. American Mathematical Society, Providence, RI(2012). https://doi.org/10.1090/coll/0607. Lov´asz, L., S´os, V.T.: Generalized quasirandom graphs. J. Combin. Theory Ser. B (1), 146–163 (2008). https://doi.org/10.1016/j.jctb.2007.06.0058. Lov´asz, L., Szegedy, B.: Limits of dense graph sequences. J. Combin. Theory Ser.B (6), 933–957 (2006). https://doi.org/10.1016/j.jctb.2006.05.0029. Mohar, B.: On a conjecture by Anthony Hill. arXiv September (2020)10. Mohar, B., Wesolek, A.: Random geodesic drawings. In preparation11. Moon, J.W.: On the distribution of crossings in random complete graphs. J. Soc.Indust. Appl. Math. , 506–510 (1965)12. Richter, R.B., Thomassen, C.: Relations between crossing numbers of complete andcomplete bipartite graphs. The American Mathematical Monthly (2), 131–137(1997)13. Schaefer, M.: Crossing numbers of graphs. Discrete Mathematics and its Applica-tions (Boca Raton), CRC Press, Boca Raton, FL (2018)14. Scheinerman, E.R., Wilf, H.S.: The rectilinear crossing number of a completegraph and Sylvester’s “four point problem” of geometric probability. Amer. Math.Monthly (10), 939–943 (1994). https://doi.org/10.2307/297515815. Sylvester, J.J.: On a special class of questions on the theory of probabilities. Birm-ingham British Assoc. Rept. pp. 8–9 (1865)16. Urbanik, K.: Solution du probl`eme pos´e par P. Tur´an. In: Colloq. Math. vol. 3,pp. 200–201 (1955)17. Wegert, E., Trefethen, L.N.: From the Buffon needle problem to theKreiss matrix theorem. Amer. Math. Monthly (2), 132–139 (1994).https://doi.org/10.2307/232436118. Zarankiewicz, K.: On a problem of P. Tur´an concerning graphs. Fundamenta Math-ematicae (41), 137–145 (1955)imiting crossing numbers for geodesic drawings on the sphere 15 A Proof of Proposition 1 Proof (of Proposition 1). By definition var ( Y H ) = E [( Y H − E [ Y H ]) ]= E (cid:88) φ : V ( H ) → V ( X n ) y H,φ − E φ = (cid:88) φ : V ( H ) → V ( X n ) (cid:88) φ (cid:48) : V ( H ) → V ( X n ) E [( y H,φ − E φ )( y H,φ (cid:48) − E φ (cid:48) )] . (4)For independent variables y H,φ and y H,φ (cid:48) the expectation E [( y H,φ − E φ )( y H,φ (cid:48) − E φ (cid:48) )] equals zero so we only need to consider those pairs φ and φ (cid:48) for which y H,φ and y H,φ (cid:48) are dependent.The events “ φ is a graph homomorphism H → X n ” and “ φ (cid:48) is a graph homo-morphism H → X n ” are independent if im ( φ ) = { e , . . . , e k } = { v w , . . . , v k w k } and im ( φ (cid:48) ) = { e (cid:48) , . . . , e (cid:48) k } = { v (cid:48) w (cid:48) , . . . , v (cid:48) k w (cid:48) k } satisfy |{ v , . . . , v k , w , . . . , w k } ∩ { v (cid:48) , . . . , v (cid:48) k , w (cid:48) , . . . , w (cid:48) k }| ≤ . But note that for these sets to share at least two points we have (cid:0) n (cid:1) choicesfor those two special points and at most ( n k − ) for the remaining ones. Thenumber of edges ( e , . . . , e k ) that can be formed by a set of vertices in X n { v , . . . , v k , w , . . . , w k } does not depend on n , so we have at most O ( n k − )pairs y H,φ and y H,φ (cid:48) that are dependent as φ and φ (cid:48) are defined by ( e , . . . , e k )and ( e (cid:48) , . . . , e (cid:48) k ) only. Note that for each pair | E [( y H,φ − µ φ )( y H,φ (cid:48) − µ φ (cid:48) )] | ≤ | y H,φ ( X n ) − µ φ | ≤ X n . Summing up those expectations over thedependent variables, (4) gives var ( Y H ) = O ( n k − ). (cid:117)(cid:116) B Counting crossings of types (C), (B), and (N) To help us with counting crossings of type ( N ) below, we first prove the followingLemma. Lemma 4. Let A and B be two nodes corresponding to adjacent vertices in D and let x ∈ A . If the geodesics from x to B intersect the circle C A correspondingto A , we denote by L x the set of vertices in A that are on the smallest circulararc that contains those intersections. Then | L x | = O ( rn ) . Moreover, if W y = { x ∈ A | y ∈ L x } , then | W y | = O ( rn ) .Proof. Note that the length of L x is O ( r /d ), where d is the distance from A to B (see Figure 5). As we consider the distance d to be constant, the number ofvertices y such that y ∈ L x is O ( r n/ (2 πr )) = O ( rn ). Moreover, since the anglesat y and x , as shown in Figure 5, are almost the same as d is large compared to r , we also have | W y | = O ( rn ). (cid:117)(cid:116) x yA B rd Fig. 5. L x are the vertices on the arc between the extremal two edges leading from x to B . The dashed arc in this figure contains vertices in W y . In the following we discuss and count the crossings of each type.(C) Two edge-bundles cross in a small neighborhood of a previous crossing in D . (We assume that r ( n ) is small.) We call these bundle-bundle crossings (C). Since each edge-bundle consists of n edges, this gives n bundle-bundlecrossings for each crossing in D .(B) Two edges cross within a bundle. We call these bundle crossings (B). Herewe have (cid:0) n (cid:1) crossings per bundle if r ( n ) (cid:28) n − and suitably rotated circlesconsidering the following elementary argument: Claim. Let D A,B be the subdrawing of D ( n )4 consisting of all edges betweentwo nodes A, B corresponding to two adjacent vertices of D . If r ( n ) (cid:28) n − and the cycles are suitably rotated, then cr( D ) = (cid:0) n (cid:1) . Proof. Any 4-tuple of two vertices from A and two vertices from B deter-mines precisely one crossing, and each crossing corresponds to precisely onesuch 4-tuple of vertices. (cid:117)(cid:116) If we drop the restriction r ( n ) (cid:28) n − and consider general r the picturelooks slightly different. Referring to Figure 5 we can see that if y ∈ L x thenthe pair x, y does not contribute (B) crossings with any pair of vertices in B . For another pair of vertices w, z in B we can see that the edges from x to w, z and from y to the antipodals w, z contribute two crossings. Hencegenerally we have (cid:0) n (cid:1) + O ( rn ) bundle crossings and O ( rn ) additionalnode crossings.(N) Two edge-bundles cross at a node. We call these node crossings (N). Let α ∈ (0 , π ) be the angle between two incident edges e, f in D which wereblown up to the edge-bundles, and let cr α be the resulting number of nodecrossings between the edges in the corresponding edge-bundles. We considerone bundle to be horizontal whereas the other bundle is counterclockwiseat angle α . We partition the edges from the bundle at angle α into foursets depending on which vertex in the node they are adjacent to. Startingat the top vertex, we enumerate the vertices clockwise along the cycle. The imiting crossing numbers for geodesic drawings on the sphere 17 first π − α π n vertices belong to part ( A ), the next α π n vertices belong to part( B ), then we have π − α π n vertices belonging to part ( C ) and the last α π n vertices belong to part ( D ) as in Figure 6. Assuming the circle radius r issmall enough, all edges in one bundle are almost parallel to each other up toan error term that depends on r . For each vertex x we introduce two sets ofvertices, S x and W x . Let y ∈ S x if all horizontal edges incident with y crossall edges at angle α that are incident with x , and let W x = { y ∈ A | x ∈ L y } where L y is defined as in Lemma 3 with respect to the horizontal edges. If x is the i -th vertex in ( A ) then | S x | = 2 i + O ( rn ) and | W x | = O ( rn ). If x is in ( B ) then | S x | = π − απ n + O ( rn ) and | W x | = O ( rn ). If x is in ( C ) thena similar count as in ( A ) applies if we enumerate those vertices starting atthe last vertex in ( C ). If x ∈ ( D ) then S x is empty and | W x | = O ( rn ). Notethat each pair x, y such that y ∈ S x contributes n crossing and if y ∈ W x then the contribution is O ( n ) crossing. Finally the number of crossings iscr α ( r ) = 2 ( π − α ) n/ π (cid:88) i =1 (2 i + O ( rn )) · n + (cid:16) α π n (cid:17) (cid:18) π − απ + O ( rn ) (cid:19) n = π − α π · n + O ( rn + n ) . ( A )( C ) ( B )( D ) α Fig. 6. ( A ), ( B ), ( C ) and ( D ) are special areas of vertices of a node. An illustrationwhere 0 < α ≤ π . The numbers of nodes in each part are rounded up or down, but these changes willmake our counts of crossings deviate only in a lower order term and can thus beneglected.8 Marthe Bonamy , Bojan Mohar, and Alexandra Wesolek xy α Fig. 7. An illustration of the antipodal argument for the total number of node crossingsat one of the nodes. The edges incident with any x and y ( x (cid:54) = y ) yield precisely n node crossings. Let us observe that cr α + cr π − α = n ( n − . This formula, which is exact,can be obtained directly by considering all four bundles arriving to a nodeas shown in Figure 7. It is apparent from the figure that the total numberof crossings, which is equal to 2 cr α +2 cr π − α , can be counted by consideringany ordered pair ( x, y ) of distinct vertices in the node and observing thatthe horizontal edges incident with x and the angle α edges incident with y leaving in both directions from each vertex yield n crossings. C Proof of Theorem 6 Lemma 5. Let α, β, γ, δ be as in Figure 1. Then α + β + γ + δ < π .Proof. We refer to Figure 1. Let T be the triangle formed by w , v and thecrossing c of the segments w v and v w . Let T (cid:48) be the triangle formed by v , w and the crossing c and note that T (cid:48) contains T . Note that the angle a at v and the angle b at w in T (cid:48) are a = π − γ and b = π − β . As T is within T (cid:48) its area is smaller and hence its angular defect is smaller which is proportionalto the angle sum. This tells us that α + β < a + b = 2 π − γ − δ which proves thelemma. (cid:117)(cid:116) Proof (of Theorem 6). By Theorem 5 we can refer to Theorem 4 to find theextremal bounds. We give a proof for the upper bound first, which is attainedif all angles are close to zero as in Figure 8. This is optimal since rewriting the imiting crossing numbers for geodesic drawings on the sphere 19 equation from Theorem 4 gives32 π (4 π − π ( α + β + γ + δ ) + ( α + δ )( γ + β ) + 2 α + 2 β + 2 γ + 2 δ )+ 233 · + O ( r ) ≤ π ( − π ( α + β + γ + δ ) + (2 π )( γ + β ) + 2 πα + 2 πβ + 2 πγ + 2 πδ )+ 13 · + O ( r ) ≤ · + O ( r ) . v w v v w w αβ γδ Fig. 8. If w , v approach w , v , respectively, then all angles α, β, γ, δ converge to zero. The claimed value in the lower bound in Theorem 4 is attained for α = β = γ = δ = π . To construct an example where all values are close to π , we exchange v and v with w and w in Figure 8, respectively. To show that we can not dobetter let α = π + a , β = π + b , γ = π + c and δ = π + d . Omitting O(r) terms,the associated triangle density is32 π (cid:0) ( π − a − d )( π − c − b ) + 2 a + 2 b + 2 c + 2 d − π (cid:1) + 233 · = 833 · + π ( − a − b − c − d )+ 12 ( a + b + c + d ) + 12 ( a − d ) + 12 ( b − c ) + a + b + c + d and it attains its global minimum at a = b = c = d = 0 as − a − b − c − d ≥ As we can come from any arrangement of four antipodal pairs of points toany other arrangement by continuously changing the points, and since r canbe arbitrary small, any triangular density in the interval (cid:0) , (cid:1) can beattained with one of these graphons. (cid:117)(cid:116) D Sketch of the proof of Theorem 1 As ref. [10] is not available in its final form at this time, we add a sketch of theproof of Theorem 1. It is based on the following result, see [9]. Theorem 7. Suppose that n > is an even integer and that P, Q are disjointsets, each containing n/ points, on the unit sphere in general position. Let D bethe geodesic drawing of K n,n , where points in P ∪ P and Q ∪ Q are the verticesof the bipartition of K n,n . Then cr ( D ) = Z ( n, n ) .Proof (of Theorem 1, sketch). By the nondegeneracy we can assume that D n has no antipodal vertices andlet P, Q be the vertices of the bipartition of the K n,n . Let D be the correspondingdrawing of K n, n with parts P ∪ P and Q ∪ Q . We can partition the set ofantipodal geodesic drawings of K n, n on the sphere into classes of equivalentdrawings, where two drawings are isomorphic if there exists a homeomorphismof the sphere which transforms one into the other. There are only finitely manyequivalence classes of geodesic drawings of the complete bipartite graph with 2 n vertices in each part. Let C , . . . , C m be those equivalence classes. Consideringdrawings D in C i , if we delete one vertex from each antipodal pair uniformly atrandom, we get a drawing D n . The drawing D has Z ( n, n ) crossings by Theorem7 and one of those crossings appears in D n if and only if all the involved verticesare in D n , which is with probability . By linearity of expectation E ( cr ( D n ) | D ∈ C i ) = 116 Z (2 n, n ) = 116 n ( n − . Using the law of total expectation we get E ( cr ( D n )) = m (cid:88) i =1 P ( D ∈ C i ) · E ( cr ( D n ) | D ∈ C i ) = 116 n ( n − . Since Z ( n, n ) ∼ n the theorem follows.the theorem follows.