Drawing Shortest Paths in Geodetic Graphs
Sabine Cornelsen, Maximilian Pfister, Henry Förster, Martin Gronemann, Michael Hoffmann, Stephen Kobourov, Thomas Schneck
DDrawing Shortest Paths in Geodetic Graphs (cid:63)
Sabine Cornelsen − − − X ] ,Maximilian Pfister − − − , Henry F¨orster − − − ,Martin Gronemann − − − X ] ,Michael Hoffmann − − − ,Stephen Kobourov − − − , andThomas Schneck − − − University of Konstanz, Germany [email protected] University of T¨ubingen, Germany { pfister,foersth,schneck } @informatik.uni-tuebingen.de University of Osnabr¨uck, Germany [email protected] Department of Computer Science, ETH Z¨urich, Switzerland [email protected] Department of Computer Science, University of Arizona, USA [email protected]
Abstract.
Motivated by the fact that in a space where shortest pathsare unique, no two shortest paths meet twice, we study a question posedby Greg Bodwin: Given a geodetic graph G , i.e., an unweighted graph inwhich the shortest path between any pair of vertices is unique, is therea philogeodetic drawing of G , i.e., a drawing of G in which the curvesof any two shortest paths meet at most once? We answer this questionin the negative by showing the existence of geodetic graphs that requiresome pair of shortest paths to cross at least four times. The bound onthe number of crossings is tight for the class of graphs we construct.Furthermore, we exhibit geodetic graphs of diameter two that do notadmit a philogeodetic drawing. Keywords:
Edge crossings · Unique Shortest Paths · Geodetic graphs.
Greg Bodwin [1] examined the structure of shortest paths in graphs with edgeweights that guarantee that the shortest path between any pair of vertices isunique. Motivated by the fact that a set of unique shortest paths is consistent in (cid:63) This research began at the Graph and Network Visualization Workshop 2019(GNV’19) in Heiligkreuztal. S. C. is funded by the German Research FoundationDFG – Project-ID 50974019 – TRR 161 (B06). M. H. is supported by the SwissNational Science Foundation within the collaborative DACH project
Arrangementsand Drawings as SNSF Project 200021E-171681. S. K. is supported by NSF grantsCCF-1740858, CCF-1712119, and DMS-1839274. a r X i v : . [ c s . D M ] A ug S. Cornelsen et al. v v v v v Fig. 1: A drawing of the geodetic graph K . It has a crossing formed by edges v v and v v . In addition, edges v v and v v meet but do not cross since their meetincludes vertex v . Finally, edges v v and v v meet twice violating the property ofphilogeodetic drawings. the sense that no two such paths can “intersect, split apart, and then intersectagain”, he conjectured that if the shortest path between any pair of vertices in agraph is unique then the graph can be drawn so that any two shortest paths meetat most once. Formally, a meet of two Jordan curves γ , γ : [0 , → R is a pairof maximal intervals I , I ⊆ [0 ,
1] for which there is a bijection ι : I → I so that γ ( x ) = γ ( ι ( x )) for all x ∈ I . A crossing is a meet with ( I ∪ I ) ∩ { , } = ∅ .Two curves meet k times if they have k pairwise distinct meets. For example,shortest paths in a simple polygon (geodesic paths) have the property that theymeet at most once [6].A drawing of a graph G in R maps the vertices to pairwise distinct pointsand maps each edge to a Jordan arc between the two end-vertices that is dis-joint from any other vertex. Drawings extend in a natural fashion to paths:Let ϕ be a drawing of G , and let P = v , . . . , v n be a path in G . Then let ϕ ( P ) denote the Jordan arc that is obtained as the composition of the curves ϕ ( v v ) , . . . , ϕ ( v n − v n ). A drawing ϕ of a graph G is philogeodetic if for everypair P , P of shortest paths in G the curves ϕ ( P ) and ϕ ( P ) meet at most once.An unweighted graph is geodetic if there is a unique shortest path betweenevery pair of vertices. Trivial examples of geodetic graphs are trees and completegraphs. Observe that any two shortest paths in a geodetic graph are eitherdisjoint or they intersect in a path. Thus, a planar drawing of a planar geodeticgraph is philogeodetic. Also every straight-line drawing of a complete graphis philogeodetic. Refer to Fig. 1 for an illustration of a drawing of a completegraph that is not philogeodetic; this example also highlights some of the conceptsdiscussed above. It is a natural question to ask whether every (geodetic) graphadmits a philogeodetic drawing. Results.
We show that there exist geodetic graphs that require some pair ofshortest paths to meet at least four times (Theorem 1). The idea is to start witha sufficiently large complete graph and subdivide every edge exactly twice. TheCrossing Lemma [8] can be used to show that some pair of shortest paths mustcross at least four times. By increasing the number of subdivisions per edge, wecan reduce the density and obtain sparse counterexamples. The bound on the rawing Shortest Paths in Geodetic Graphs 3 number of crossings is tight because any uniformly subdivided K n can be drawnso that every pair of shortest paths meets at most four times (Theorem 2).On one hand, our construction yields counterexamples of diameter five. Onthe other hand, the unique graph of diameter one is the complete graph, whichis geodetic and admits a philogeodetic drawing (e.g., any straight-line drawingsince all unique shortest paths are single edges). Hence, it is natural to ask whatis the largest d so that every geodetic graph of diameter d admits a philogeodeticdrawing. We show that d = 1 by exhibiting an infinite family of geodetic graphsof diameter two that do not admit philogeodetic drawings (Theorem 3). Theconstruction is based on incidence graphs of finite affine planes. The proof alsorelies on the crossing lemma. Geodetic graphs.
Geodetic graphs were introduced by Ore who asked for a char-acterization as Problem 3 in Chapter 6 of his book “Theory of Graphs” [7, p. 104].An asterisk flags this problem as a research question, which seems justified, asmore than sixty years later a full characterization is still elusive.Stemple and Watkins [14,15] and Plesn´ık [10] resolved the planar case byshowing that a connected planar graph is geodetic if and only if every block is(1) a single edge, (2) an odd cycle, or (3) stems from a K by iteratively choosinga vertex v of the K and subdividing the edges incident to v uniformly. Geodeticgraphs of diameter two were fully characterized by Scapellato [12]. They includethe Moore graphs [3] and graphs constructed from a generalization of affineplanes. Further constructions for geodetic graphs were given by Plesn´ık [10,11],Parthasarathy and Srinvasan [9], and Frasser and Vostrov [2].Plesn´ık [10] and Stemple [13] proved that a geodetic graph is homeomorphicto a complete graph if and only if it is obtained from a complete graph K n byiteratively choosing a vertex v of the K n and subdividing the edges incident to v uniformly. A graph is geodetic if it is obtained from any geodetic graph byuniformly subdividing each edge an even number of times [9,11]. However, thegraph G obtained by uniformly subdividing each edge of a complete graph K n anodd number of times is not geodetic: Let u, v, w be three vertices of K n and let x be the middle subdivision vertex of the edge uv . Then there are two shortest x - w -paths in G , one containing v and one containing u . The complete graph K n is geodetic and rather dense. However, all shortest pathsare very short, as they comprise a single edge only. So despite the large numberof edge crossings in any drawing, every pair of shortest paths meets at mostonce, as witnessed, for instance, by any straight-line drawing of K n . In order tolengthen the shortest paths it is natural to consider subdivisions of K n .As a first attempt, one may want to “take out” some edge uv by subdividingit many times. However, Stemple [13] has shown that in a geodetic graph everypath where all internal vertices have degree two must be a shortest path. Thus,it is impossible to take out an edge using subdivisions. So we use a differentapproach instead, where all edges are subdivided uniformly. S. Cornelsen et al.
Theorem 1.
There exists an infinite family of sparse geodetic graphs for whichin any drawing in R some pair of shortest paths meets at least four times.Proof. Take an even number t and a complete graph K s for some s ∈ N . Subdi-vide each edge t times. The resulting graph K ( s, t ) is geodetic. See Fig. 4 for adrawing of K (8 , K ( s, t ) has n = s + t (cid:0) s (cid:1) vertices and m = ( t +1) (cid:0) s (cid:1) edges, with m ∈ O ( n ), for s fixed and t sufficiently large. Consider a drawing Γ of K ( s, t ).Let B denote the set of s branch vertices in K ( s, t ), which correspond to thevertices of the original K s . For two distinct vertices u, v ∈ B , let [ uv ] denotethe shortest uv -path in K ( s, t ), which corresponds to the subdivided edge uv ofthe underlying K s . As t is even, the path [ uv ] consists of t + 1 (an odd numberof) edges. For every such path [ uv ], with u, v ∈ B , we charge the crossings in Γ along the t + 1 edges of [ uv ] to one or both of u and v as detailed below; seeFig. 2 for illustration. – Crossings along an edge that is closer to u than to v are charged to u ; – crossings along an edge that is closer to v than to u are charged to v ; and – crossings along the single central edge of [ uv ] are charged to both u and v . u v Fig. 2: Every crossing is charged to at least one endpoint of each of the two involved(independent) edges. Vertices are shown as white disks, crossings as red crosses, andcharges by dotted arrows. The figure shows an edge uv that is subdivided four times,splitting it into a path with five segments. A crossing along any such segment is assignedto the closest of u or v . For the central segment, both u and v are at the same distance,and any crossing there is assigned to both u and v . Let Γ s be the drawing of K s induced by Γ : every vertex of K s is placed atthe position of the corresponding branch vertex of K ( s, t ) in Γ and every edgeof K s is drawn as a Jordan arc along the corresponding path of K ( s, t ) in Γ .Assuming (cid:0) s (cid:1) ≥ s (i.e., s ≥ (cid:0) s (cid:1) s = 1512 s ( s − ≥ c · s pairs of independent edges cross in Γ s , for some constant c . Every crossing in Γ s corresponds to a crossing in Γ and is charged to at least two (and up to four) rawing Shortest Paths in Geodetic Graphs 5 vertices of B . Thus, the overall charge is at least 2 cs , and at least one vertex u ∈ B gets at least the average charge of 2 cs .Each charge unit corresponds to a crossing of two independent edges in Γ s ,which is also charged to at least one other vertex of B . Hence, there is a vertex v (cid:54) = u so that at least 2 cs crossings are charged to both u and v . Note that thereare only s − u and v , and the common edge uv is notinvolved in any of the charged crossings (as adjacent rather than independentedge). Let E x , for x ∈ B , denote the set of edges of K s that are incident to x .We claim that there are two pairs of mutually crossing edges incident to u and v , respectively; that is, there are sets C u ⊂ E u \ { uv } and C v ⊂ E v \ { uv } with | C u | = | C v | = 2 so that e crosses e , for all e ∈ C u and e ∈ C v .Before proving this claim, we argue that establishing it completes the proof ofthe theorem. By our charging scheme, every crossing e ∩ e happens at an edgeof the path [ e ] in Γ that is at least as close to u as to the other endpoint of e .Denote the three vertices that span the edges of C u by u, x, y . Consider the twosubdivision vertices x (cid:48) along [ ux ] and y (cid:48) along [ uy ] that form the endpoint of themiddle edge closer to x and y , respectively, than to u ; see Fig. 3 for illustration. ux y (cid:124)(cid:123)(cid:122)(cid:125) t / v e r t i ce s (cid:124)(cid:123)(cid:122)(cid:125) t / v e r t i ce s (cid:124) (cid:123) (cid:122) (cid:125) t / v e r t i c e s (cid:124) (cid:123) (cid:122) (cid:125) t / v e r t i c e s x (cid:48) y (cid:48) Fig. 3: Two adjacent edges ux and uy , both subdivided t times, and the shortest pathbetween the “far” endpoints x (cid:48) and y (cid:48) of the central segments of [ ux ] and [ uy ]. The triangle uxy in K s corresponds to an odd cycle of length 3( t + 1) in K ( s, t ). So the shortest path between x (cid:48) and y (cid:48) in K ( s, t ) has length 2(1 + t/
2) = t + 2 and passes through u , whereas the path from x (cid:48) via x and y to y (cid:48) has length3( t + 1) − ( t + 2) = 2 t + 1, which is strictly larger than t + 2 for t ≥
2. It followsthat the shortest path between x (cid:48) and y (cid:48) in K ( s, t ) is crossed by both edgesin C v . A symmetric argument yields two subdivision vertices a (cid:48) and b (cid:48) alongthe two edges in C v so that the shortest a (cid:48) b (cid:48) -path in K ( s, t ) is crossed by bothedges in C u . By definition of our charging scheme (that charges only “nearby”crossings to a vertex), the shortest paths x (cid:48) y (cid:48) and a (cid:48) b (cid:48) in K ( s, t ) have at leastfour crossings.It remains to prove the claim. To this end, consider the bipartite graph X on the vertex set E u ∪ E v where two vertices are connected if the correspondingedges are independent and cross in Γ s . Observe that two sets C u and C v ofmutually crossing pairs of edges (as in the claim) correspond to a 4-cycle C in X . So suppose for the sake of a contradiction that X does not contain C as asubgraph. Then by the K˝ov´ari-S´os-Tur´an Theorem [5] the graph X has O ( s / )edges. But we already know that X has at least 2 cs = Ω ( s ) edges, which yieldsa contradiction. Hence, X is not C -free and the claim holds. (cid:117)(cid:116) S. Cornelsen et al.
The bound on the number of crossings in Theorem 1 is tight.
Theorem 2.
A graph obtained from a complete graph by subdividing the edgesuniformly an even number of times can be drawn so that every pair of shortestpaths crosses at most four times.Proof (Sketch).
Place the vertices in convex position. Draw the subdivided edgesalong straight-line segments. For each edge, put half of the subdivision verticesvery close to one endpoint and the other half very close to the other endpoint(Fig. 4). As a result, all crossings fall into the central segment of the path. (cid:117)(cid:116)
Fig. 4: A drawing of K (8 , K where every edge is subdividedtwice, so that every pair of shortest paths meets at most four times. Two shortest pathsthat meet four times are shown bold and orange. In this section we give examples of geodetic graphs of diameter two that cannotbe drawn in the plane such that any two shortest paths meet at most once.An affine plane of order k ≥ k points, (ii) forany two points there is a unique line containing both, (iii) there are three pointsthat are not contained in the same line, and (iv) for any line (cid:96) and any point p not on (cid:96) there is a line (cid:96) (cid:48) that contains p , but no point from (cid:96) . Two lines that donot contain a common point are parallel . Observe that each point is containedin k + 1 lines. Moreover, there are k points and k + 1 classes of parallel lineseach containing k lines. The 2-dimensional vector space F over a finite field F of order k with the lines { ( x, mx + b ); x ∈ F } , m, b ∈ F and { ( x , y ); y ∈ F } , rawing Shortest Paths in Geodetic Graphs 7 x ∈ F is a finite affine plane of order k . Thus, there exists a finite affine planeof order k for any k that is a prime power (see, e.g., [4]).Scapellato [12] showed how to construct geodetic graphs of diameter two asfollows: Take a finite affine plane of order k . Let L be the set of lines and let P be the set of points of the affine plane. Consider now the graph G k with vertexset L ∪ P and the following two types of edges: There is an edge between twolines if and only if they are parallel. There is an edge between a point and a lineif and only if the point lies on the line; see Fig. 5. There are no edges betweenpoints. It is easy to check that G k is a geodetic graph of diameter two. . . . k points. . . k + 1 cliques of k parallel lines each G k : . . .. . . Fig. 5: Structure of the graph G k . Theorem 3.
There are geodetic graphs of diameter two that cannot be drawnin the plane such that any two shortest paths meet at most once.Proof.
Let k ≥
129 be such that there exists an affine plane of order k (e.g., theprime k = 131). Assume there was a drawing of G k in which any two shortestpaths meet at most once. Let G be the bipartite subgraph of G k without edgesbetween lines. Observe that any path of length two in G is a shortest path in G k .As G has n = 2 k + k vertices and m = k ( k +1) > kn/ m > n ,for k ≥
8. Therefore, by the Crossing Lemma [8, Remark 2 on p. 238] there areat least m / n > k n/
512 crossings between independent edges in G .Hence, there is a vertex v such that the edges incident to v are crossed morethan k /
128 times by edges not incident to v . By assumption, (a) any two edgesmeet at most once, (b) any edge meets any pair of adjacent edges at most once,and (c) any pair of adjacent edges meets any pair of adjacent edges at mostonce. Thus, the crossings with the edges incident to v stem from a matching.It follows that there are at most ( n − / k + k − / k + k − / < k / k ≥ (cid:117)(cid:116) We conclude with two open problems: (1) Are there diameter-2 geodetic graphswith edge density 1 + ε that do not admit a philogeodetic drawing? (2) What isthe complexity of deciding if a geodetic graph admits a philogeodetic drawing? S. Cornelsen et al.
References
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15. Watkins, M.E.: A characterization of planar geodetic graphs. J. Combin. Theory (1), 102–103 (1967), https://doi.org/10.1016/S0021-9800(67)80118-2 rawing Shortest Paths in Geodetic Graphs 9 A Proof of Theorem 2
Proof.
Draw the graph as described on Page 6 and as illustrated in Fig. 4 for K (8 , B denote the set of branch vertices, and let S denote theset of subdivision vertices. Note that for every edge uv of K n , only the centralsegment of the subdivided path [ uv ] may have crossings in the drawing. We claimthat every shortest path in the graph contains at most two central segments inthe drawing, from which the theorem follows immediately. Consider a pair u, v of vertices. Case 1: { u, v } ∩ B (cid:54) = ∅ . Suppose without loss of generality that u ∈ B . If v ∈ B or v ∈ S subdivides an edge incident to u , then the shortest uv -pathcontains at most one central segment. Otherwise, v ∈ S subdivides an edge xy disjoint from u . One of x or y , without loss of generality x is closer to v . Then theshortest uv -path is [ vx ][ xu ], which contains exactly one central segment, [ xu ]. Case 2: u, v ∈ S . If u and v subdivide the same edge, then the shortest uv -path contains at most one central segment. If u and v subdivide distinct adjacentsegments, xy and xz , then the shortest uv -path is either [ ux ][ xv ], which containsat most two central segments. Or the sum of the length of [ uy ] and [ zv ] is at mosthalf of the number of subdivision vertices per edge and the shortest uv -path is[ uy ][ yz ][ zv ], which then contains at most one central segment. Otherwise, u and v subdivide disjoint segments, xy and wz , where without loss of generality x is closer to u than y and w is closer to v than z . Then the shortest uv -path is[ ux ][ xw ][ wv ], which contains exactly one central segment, [ xw ]. (cid:117)(cid:116) B Proof that G k (as Defined in Section 3) is Geodetic The following statement follows from Scapellato’s classification [12]. As we needmuch less than this classification in its full generality, we provide an easy proofof what we use, for the sake of self-containment.
Lemma 1. G k is a geodetic graph of diameter two.Proof. Two lines have distance one if they are parallel. Otherwise they shareexactly one vertex and, hence, are connected by exactly one path of length two.For any two points there is exactly one line that contains both. Given a line (cid:96) and a point p then either p lies on (cid:96) and, thus, p and (cid:96) have distance one. Orthere is exactly one line (cid:96) (cid:48) containing p that is parallel to (cid:96) and, thus, there isexactly one path of length two between (cid:96) and p ..