Simple Topological Drawings of k -Planar Graphs
Michael Hoffmann, Chih-Hung Liu, Meghana M. Reddy, Csaba D. Tóth
SSimple Topological Drawings of k -Planar Graphs (cid:63) Michael Hoffmann , Chih-Hung Liu , Meghana M. Reddy , andCsaba D. T´oth , Department of Computer Science, ETHZ, Z¨urich, Switzerland { hoffmann,chih-hung.liu,meghana.mreddy } @inf.ethz.ch Department of Mathematics, Cal State Northridge, Los Angeles, CA, USA Department of Computer Science, Tufts University, Medford, MA, USA [email protected]
Abstract.
Every finite graph admits a simple (topological) drawing , thatis, a drawing where every pair of edges intersects in at most one point.However, in combination with other restrictions simple drawings do notuniversally exist. For instance, k -planar graphs are those graphs that canbe drawn so that every edge has at most k crossings (i.e., they admit a k -plane drawing ). It is known that for k ≤
3, every k -planar graph admitsa k -plane simple drawing. But for k ≥
4, there exist k -planar graphsthat do not admit a k -plane simple drawing. Answering a question bySchaefer, we show that there exists a function f : N → N such that every k -planar graph admits an f ( k )-plane simple drawing, for all k ∈ N . Notethat the function f depends on k only and is independent of the sizeof the graph. Furthermore, we develop an algorithm to show that every4-planar graph admits an 8-plane simple drawing. Keywords:
Topological graphs · local crossing number · k -planar graphs A topological drawing of a graph G in the plane is a representation of G inwhich the vertices are mapped to pairwise distinct points in the plane and edgesare mapped to Jordan arcs that do not pass through (the images of) vertices.Moreover, no three Jordan arcs pass through the same point in the plane, andevery pair of Jordan arcs has finitely many intersection points, each of which iseither a common endpoint or a crossing , where the two arcs cross transversally.A graph is k -planar if it admits a topological drawing in the plane where everyedge is crossed at most k times, and such a drawing is called a k -plane drawing . (cid:63) This work was initiated at the 17 th Gremo Workshop on Open Problems (GWOP)2019. The authors thank the organizers of the workshop for inviting us and providinga productive working atmosphere. M. H. and M. M. R. are supported by the SwissNational Science Foundation within the collaborative DACH project
Arrangementsand Drawings as SNSF Project 200021E-171681. Research by C. D. T. was supportedin part by the NSF award DMS-1800734. a r X i v : . [ c s . C G ] A ug Michael Hoffmann, Chih-Hung Liu, Meghana M. Reddy, and Csaba D. T´oth A simple topological drawing of a graph refers to a topological drawing whereno two edges cross more than once and no two adjacent edges cross. We studysimple topological drawings of k -planar graphs.It is well known that drawings of a graph G that attain the minimum numberof crossings (i.e., the crossing number of G ) are simple topological drawings [4,p. 18]. However, a drawing that minimizes the total number of crossings neednot minimize the maximum number of crossings per edge; and a drawing thatminimizes the maximum number of crossings per edge need not be simple. A k -plane simple topological drawing is a simple topological drawing where everyedge is crossed at most k times. We study the simple topological drawings of k -planar graphs and prove that there exists a function f : N → N such that every k -planar graph admits an f ( k )-plane simple topological drawing by designingan algorithm to obtain the simple topological drawing from a k -plane drawing.The function f in our bound is exponential in k , more precisely f ( k ) ∈ O ∗ (3 k ).It remains open whether this can be improved to a bound that is polynomial in k . We also present a significantly better bound for 4-planar graphs.In a k -plane drawing adjacent edges may cross, and two edges may cross manytimes. To obtain a simple topological drawing, we need to eliminate crossingsbetween adjacent edges and ensure that any two edges cross at most once. Related Work.
It is easy to see that every 1-planar graph admits a 1-plane simpletopological drawing [3]. Pach et al. [2, Lemma 1.1] proved that every k -planargraph for k ≤ k -plane simple topological drawing. However, theseresults do not extend to k -planar graphs, for k >
3. In fact, Schaefer [4, p. 57]constructed k -planar graphs that do not admit a k -plane simple topologicaldrawing for k = 4. The construction idea can be extended to all k >
4. The local crossing number lcr( G ) of a graph G is the minimum integer k such that G admits a drawing where every edge has at most k crossings. The simple localcrossing number lcr ∗ ( G ) minimizes k over all simple topological drawings of G .Schaefer [4, p. 59] asked whether the lcr ∗ ( G ) can be bounded by a function oflcr( G ). We answer this question in the affirmative and show that there exists afunction f : N → N such that lcr ∗ ( G ) ≤ f (lcr( G )).The family of k -planar graphs, for small values of k , was instrumental inproving the current best bounds on the multiplicative constant in the CrossingLemma and the Szemer´edi-Trotter theorem on point-line incidences [1,2]. Ack-erman [1] showed that every graph with n ≥ m ≤ n −
12 edges, and claims that this boundholds for all 4-planar graphs. Pach et al. [2, Conjecture 5.4] conjectured thatfor all k, n ≥
1, the maximum number of edges in a k -planar n -vertex graph isattained by a graph that admits a simple k -plane drawing. Lenses in topological drawings.
We start with definitions needed to describe thekey operations in our algorithms. In a topological drawing, we define a structure imple Topological Drawings of k -Planar Graphs 3 called lens . Consider two edges, e and f , that intersect in two distinct points, α and β (each of which is either a common endpoint or a crossing). Let e αβ (resp., f αβ ) denote the portion of e (resp., f ) between α and β . The arcs e αβ and f αβ together are called a lens if e αβ and f αβ do not intersect except at α and β . SeeFig. 1 for examples. The lens is denoted by L ( e αβ , f αβ ). A lens L ( e αβ , f αβ ) isbounded by independent arcs if both α and β are crossings, else (if α or β is avertex of G ) it is bounded by adjacent arcs . Lemma 1.
If a pair of edges e and f intersect in more than one point, thenthere exist arcs e αβ ⊂ e and f αβ ⊂ f that form a lens. u vs tα βef (a) u vs tα β ef (b) Fig. 1.
Lenses formed by two edges.
Operations.
We present algorithms that transform a k -plane drawing into asimple topological drawing by a sequence of elementary operations. Each oper-ation modifies one or two edges that form a lens so that the lens is eliminated.We use two elementary operations, Swap and
Reroute . Both have been usedpreviously (e.g., in [2, Lemma 1.1]); we describe them here for completeness.The common setup in both operations is the following. Let e = uv and f = st be edges that form a lens L ( e αβ , f αβ ), where α and β are each a crossing or acommon endpoint. Assume that the Jordan arc of e visits u , α , β , v , and theJordan arc of f visits s , α , β , and t in this order. Let α and β be sufficiently smalldisks centered at α and β , resp., so that their boundary circles each intersect e and f twice, but do not intersect any other edge. Swap operation.
We modify the drawing of e and f in three steps as follows.(1) Redraw e such that it follows its current arc from u to α , then continuesalong f αβ to β , and further to v along its original arc. Similarly, redraw f suchthat it follows its current arc from s to α , then continues along e αβ to β , andfurther to t along its original arc. (2) Replace the portion of e and f in α and β by straight line segments. (3) Eliminate self-crossings, if any is introduced, byremoving any loops from the modified arcs of e and f . The swap operation isdenoted by Swap ( e αβ , f αβ ); see Fig. 2 for illustrations. The swap operation fora lens bounded by adjacent arcs is defined similarly. Observation 1.
Let D be a topological drawing of a graph G , and let L ( e αβ , f αβ ) be a lens. Operation Swap ( e αβ , f αβ ) produces a topological drawing that has atleast one fewer crossing than D . Michael Hoffmann, Chih-Hung Liu, Meghana M. Reddy, and Csaba D. T´oth u vs tα βef (a) u vs tα βe f (b)
Fig. 2.
Swap ( e αβ , f αβ ) applied to the two lenses in Fig. 1 Reroute operation.
We modify the drawing of f in three steps as follows. (1) Re-draw f such that it follows its current arc from s to the first intersection with α , it does not cross e in α , and then it closely follows arc e αβ to β , and furtherfollows its original arc from β to t . (2) Replace the portion of f in the interiorof α and β by straight line segments. (3) Eliminate self-crossings, if any areintroduced, by removing any loops from the modified arc of f . The reroute oper-ation is denoted by Reroute ( e αβ , f αβ ); see Fig. 3 for illustrations. The rerouteoperation for a lens bounded by adjacent arcs is defined similarly. u vs tα βfe (a) u vs α β t fe (b) Fig. 3.
Reroute ( e αβ , f αβ ) operation on the two lenses in Fig. 1 Observation 2.
Let D be a topological drawing of a graph G , and let L ( e αβ , f αβ ) be a lens. Operation Reroute ( e αβ , f αβ ) produces a topological drawing. While a
Reroute ( e αβ , f αβ ) operation modifies only the edge f , it may in-crease the total number of crossings, as well as the number of crossings on f . Planarization.
Let D be a topological drawing of a graph G . Denote by N theplanarization of D (i.e., we introduce a vertex of degree four at every crossingin D ). We call this graph a network . We refer to the vertices and edges of N as nodes and segments , respectively, so as to distinguish them from the corre-sponding entities in G . Our algorithms in Section 3–4 use the planarization N ofa drawing D , then successively modify the drawing D , and ultimately return asimple topological drawing of G . We formulate invariants for these algorithms interms of the planarization N of the initial drawing. In other words, N remainsfixed (in particular, N will not be the planarization of the modified drawings).As Reroute operations redraw edges to closely follow existing edges, our algo-rithms will maintain the following invariants: imple Topological Drawings of k -Planar Graphs 5 (I1) Every edge in D closely follows a path in the network N ;(I2) every pair of edges in D cross only in a small neighborhood of a node of N ;(I3) every pair of edges crosses at most once in each such neighborhood. Length of an arc and number of crossings.
Let a be a Jordan arc that closelyfollows a path in N such that its endpoints are in the small neighborhoods ofnodes of N . The length of a , denoted by (cid:96) ( a ), is the graph-theoretic length ofthe path of N that a closely follows. Let x ( a ) denote the number of crossingson the arc a in a drawing D . Note that the length (cid:96) ( a ) is measured in terms ofthe (fixed) network N , and x ( a ) is measured in terms of the (varying) drawing D . For instance, in Fig. 3(b) we have (cid:96) ( f ) = 3 both before and after rerouting,whereas x ( f ) = 2 before and x ( f ) = 1 after rerouting. k -Planar Graphs In this section we describe and analyze an algorithm to transform a topolog-ical drawing into a simple topological drawing whose local crossing number isbounded by a function of the local crossing number of the original drawing.
Algorithm 1.
Let D be a topological k -plane drawing of a graph G = ( V, E ). Let N be theplanarization of D . Let D := D .While there exists a lens in D , do the following.Let L ( e αβ , f αβ ) be a lens so that w.l.o.g. (cid:96) ( e αβ ) < (cid:96) ( f αβ ), or (cid:96) ( e αβ ) = (cid:96) ( f αβ )and x ( e αβ ) ≤ x ( f αβ ). Modify D by applying Reroute ( e αβ , f αβ ).When the while loop terminates, return the drawing D . Observation 3.
Algorithm 1 maintains invariants (I1)–(I3), and the length ofevery edge decreases or remains the same.
Corollary 1.
Algorithm 1 maintains the following invariant:(I4) The length of every edge in D is at most k + 1 . Lemma 2.
Algorithm 1 terminates and transforms a k -plane topological draw-ing into a simple topological drawing of G .Proof. Let the sum of lengths of all edges in the drawing be defined as the totallength of the drawing (recall that the length of an edge is the length of the cor-responding path in N ). By Observation 3, the total length of the drawing mono-tonically decreases. If the total length remains the same in one iteration of thewhile loop, then (cid:96) ( e αβ ) = (cid:96) ( f αβ ) and x ( e αβ ) ≤ x ( f αβ ). Since Reroute ( e αβ , f αβ )eliminates a crossing at α or β , the total number of crossings strictly decreasesin this case. Thus, the algorithm terminates. By Observations 1–2, the algorithmmaintains a topological drawing. The drawing D (cid:48) returned by the algorithm doesnot contain lenses. By Lemma 1, any two edges in D (cid:48) intersect in at most onepoint. Consequently, D (cid:48) is a simple topological drawing of G . Michael Hoffmann, Chih-Hung Liu, Meghana M. Reddy, and Csaba D. T´oth
Lemma 3 (Crossing Lemma [1, Theorem 6]).
Let G be a graph with n vertices and m edges and D be a topological drawing of G . Let cr ( D ) be definedas the total number of crossings in D , and cr ( G ) be defined as the minimum ofcr ( D ) over all drawings D of G . If m ≥ . n , then cr ( G ) ≥ m n . Theorem 1.
There exists a function f ( k ) such that every k -planar graph admitsan f ( k ) -plane simple topological drawing, and there exists an algorithm to obtainan f ( k ) -plane simple topological drawing from a given k -plane drawing of a graph.Proof. The statement holds for k ≤ f ( k ) = k [2, Lemma 1.1]. Hence wemay suppose that k ≥
4. Consider the drawing D (cid:48) returned by Algorithm 1,and a node γ of the network N that corresponds to a crossing. We analyse thesubgraph G γ of G formed by the edges of G that in D (cid:48) pass through a smallneighborhood γ of γ . Let n γ and m γ be the number of vertices and edges of G γ ,respectively. By (I4), every edge in D (cid:48) corresponds to a path of length at most k + 1 in N . If an edge uv passes through γ in D (cid:48) , then N contains a path oflength at most k from γ to u (resp., v ) in which internal vertices correspond tocrossings in D . Every node in N that corresponds to a crossing has degree 4.Hence the number of vertices reachable from γ on such a path is n γ ≤ · k − .We apply Lemma 3 to the graph G γ , and distinguish between two cases:Either m γ < . n , otherwise m γ ≥ . n and then cr( G γ ) ≥ m γ /n γ .Since G γ has m γ edges and each edge has at most k crossings in D , we ob-tain m γ /n γ ≤ m γ k/
2, which implies m γ ≤ (cid:112) k/ n γ . The combination ofboth cases yields an upper bound m γ ≤ max { . n γ , (cid:112) k/ n γ } . So, for k ≥ m γ ≤ (cid:112) k/ n γ .Since m γ edges pass through γ , by invariant (I3) every edge passing through γ has at most m γ − γ . By invariant (I4), every edge in G passesthrough (the neighborhood of) at most k nodes of N . By (I2), an edge passingthrough γ , . . . , γ k crosses at most (cid:80) ki =1 ( m γ i −
1) edges in D (cid:48) . Combining theupper bounds on m γ and n γ , we obtain that every edge in the output drawing D (cid:48) has at most (cid:112) k/ · k · k − = √ · k / · k crossings, for k ≥ The function f from our proof of Theorem 1 yields f (4) = 23 √ · / · ≈ . Theorem 2.
Every -planar graph admits an -plane simple topological draw-ing. Given a -plane drawing of a graph with n vertices, an -plane simple topo-logical drawing can be computed in O ( n ) time. imple Topological Drawings of k -Planar Graphs 7 The proof of Theorem 2 is constructive: Let D be a 4-plane drawing of a 4-planar graph G = ( V, E ) with n = | V | vertices and m = | E | edges. The CrossingLemma implies that a k -planar graph on n vertices has at most 3 . √ kn edges.For k = 4, this implies m ≤ . n . (We note that Ackerman [1] proved a bound m ≤ n −
12 for 4-plane simple topological drawings with n ≥ L ( e αβ , f αβ ) is – a if e αβ has no crossings; – a quasi-0-lens if the arc e αβ has exactly one crossing γ , where e crosses anedge h , the edges h and f have a common endpoint s , and the arcs f sα and h sα cross the same edges in the same order (see Fig. 5(a) for an example); – a if x ( e ) = 4, x ( e αβ ) = 1, and x ( f αβ ) = 3; see Fig. 4(a).We show that all lenses other than 0-lenses and 1-3-lenses can be eliminatedby swap operations while maintaining a 4-plane drawing (Lemma 6). And 0-lenses can easily be eliminated by reroute operations (Lemma 4). The same holdsfor quasi-0-lenses (Lemma 5), which are of no particular concern in the initialdrawing but are important for the analysis of the last phase of our algorithm.The main challenge is to eliminate 1-3-lenses, which we do by rerouting the arcwith 3 crossings along the arc with 1 crossing.Our algorithm proceeds in three phases: Phase 1 eliminates all lenses otherthan 1-3-lenses. We show that it maintains a 4-plane drawing (Lemma 8). Phase 2eliminates every 1-3-lens using reroute operations. We show that this phase pro-duces an 8-plane drawing. Phase 2 may also create new lenses, but only 0- andquasi-0-lenses, which are eliminated in Phase 3 without creating any new lenses. feβαu wv fe βαu wv Fig. 4.
Reroute ( e uβ , f uβ ) applied to a 1-3-lens L ( e uβ , f uβ ). The initial 4-plane drawing has O ( n ) crossings since the graph has O ( n )edges and each edge has at most four crossings. The set of lenses in the initialdrawing can be identified in O ( n ) time. Due to the elimination of a single lens,a constant number of other lenses can be affected, which can be computed inconstant time. Further, each elimination operation strictly decreases the totalnumber of crossings in the drawing. Consequently, Algorithm 2 performs O ( n )elimination operations and can be implemented in O ( n ) time. Michael Hoffmann, Chih-Hung Liu, Meghana M. Reddy, and Csaba D. T´oth
Lemma 4.
Let L ( e αβ , f αβ ) be a 0-lens. Then operation Reroute ( e αβ , f αβ ) de-creases the total number of crossings and does not create any new crossing. Fur-ther, if any two edges have at most two points in common, then it does not createany new lens.Proof. The operation
Reroute ( e αβ , f αβ ) modifies only the edge f , by reroutingthe arc f αβ to closely follow e αβ . Since the arc e αβ is crossing-free, the edge f loses one of its crossings and no edge gains any new crossing. Overall, the totalnumber of crossings decreases, as claimed.Assume that any two edges have at most two points in common before theoperation. Consider a lens L ( g γδ , h γδ ) in the drawing after the operation. As nonew crossings are created, γ and δ are already common points of g and h beforethe operation. Since g and h have no other common points by assumption, thelens L ( g γδ , h γδ ) is already present before the operation.For quasi-0-lenses we define the operation Quasi-0-Reroute ( e αβ , f αβ ) asfollows; see Fig. 5. Let h be the edge that crosses e αβ at γ and shares an endpoint s with f . Redraw f such that it closely follows h from s to γ , it does not cross e in γ , and then it closely follows arc e αβ to β , and further follows its original arcfrom β to t . The analogue of Lemma 4 reads as follows. efα β tv us hγ eα β tv us hγ f Fig. 5.
Quasi-0-Reroute ( e αβ , f αβ ) applied to a quasi-0-lens L ( e αβ , f αβ ). Lemma 5.
Let L ( e αβ , f αβ ) be a quasi-0-lens, where h denotes the edge thatcrosses e αβ at γ and shares an endpoint s with f . Then operation Quasi-0-Reroute ( e αβ , f αβ ) decreases the total number of crossings, and does not in-crease the number of crossings between any pair of edges. Further, if any twoedges have at most two points in common, then it does not create any new lens.Proof. The operation
Quasi-0-Reroute ( e αβ , f αβ ) modifies only the edge f , byrerouting the arc f sβ to closely follow first h from s to γ and then e αβ to β .Let f (cid:48) denote the new drawing of f . Since (1) e has at least one fewer crossingwith f (cid:48) than with f , and (2) every crossing of f (cid:48) along the arc between s and γ corresponds to a crossing of f along the arc from s to α , the total number ofcrossings strictly decreases, and for each pair of edges the number of crossingsbetween them does not increase, as claimed. imple Topological Drawings of k -Planar Graphs 9 Assume that any two edges have at most two points in common before theoperation. Suppose
Quasi-0-Reroute ( e αβ , f αβ ) creates a new lens. This lensmust be formed by f (cid:48) and another edge, say g . Then f (cid:48) and g must have atleast two points in common, and g must cross f αβ , implying that f and g haveat least three points in common before the operation. However, by assumption,edges f and g have at most two points in common, which is a contradiction.Consequently, every lens in the resulting drawing corresponds to a lens in theoriginal drawing, where the arc f sα is shifted to the arc of f (cid:48) from s to γ . Lemma 6.
Let L ( e αβ , f αβ ) be a lens either bounded by nonadjacent arcs with x ( e αβ ) ≤ x ( f αβ ) ≤ x ( e αβ ) + 2 , or by adjacent arcs with x ( e αβ ) ≤ x ( f αβ ) ≤ x ( e αβ ) + 1 . Then the operation Swap ( e αβ , f αβ ) produces a drawing in which thetotal number of crossings on each edge does not increase, and the total numberof crossings decreases.Proof. The operation
Swap ( e αβ , f αβ ) modifies only the edges e and f , by ex-changing arcs e αβ and f αβ , and eliminating any crossing at the endpoints of thesearcs. In particular, the number of crossings on other edges cannot increase. Thisalready implies that the total number of crossings decreases.Let e (cid:48) and f (cid:48) denote the new drawing of e and f . If both α and β are crossings,then both crossings are eliminated, hence x ( e (cid:48) ) = x ( e ) − x ( f αβ ) − x ( e αβ )) ≤ x ( e ) and x ( f (cid:48) ) = x ( f ) − x ( e αβ ) − x ( f αβ )) ≤ x ( f ) −
2. If α or β is a vertex of G ,then only one crossing is eliminated, hence x ( e (cid:48) ) = x ( e ) − x ( f αβ ) − x ( e αβ )) ≤ x ( e ) and x ( f (cid:48) ) = x ( f ) − x ( e αβ ) − x ( f αβ )) ≤ x ( f ) −
1, as required.
Lemma 7.
Let D be a 4-plane drawing of a graph, and let L ( e αβ , f αβ ) be a lenswith x ( e αβ ) ≤ x ( f αβ ) .1. If x ( f αβ ) − x ( e αβ ) ≥ , then L ( e αβ , f αβ ) is either a 0-lens or x ( e αβ ) = 1 and x ( f αβ ) = 3 .2. If x ( e αβ ) = 1 and x ( f αβ ) = 3 , then e αβ and f αβ are adjacent arcs.Proof. As D is 4-plane, we have x ( e ) ≤ x ( f ) ≤
4. Assume first thatboth α and β are crossings, and so x ( f αβ ) ≤ x ( f ) − ≤
2. Combined with x ( f αβ ) − x ( e αβ ) ≥
2, this implies x ( e αβ ) = 0, hence L ( e αβ , f αβ ) is a 0-lens.Assume next that α or β is a vertex in G . Then x ( f αβ ) ≤ x ( f ) − ≤
3. With x ( f αβ ) − x ( e αβ ) ≥
2, this implies x ( e αβ ) = 0, or x ( e αβ ) = 1 and x ( f αβ ) = 3. Algorithm 2.Input.
Let D be a 4-plane drawing of a graph G = ( V, E ). Phase 1.
While there is a lens L ( e αβ , f αβ ) that is not a 1-3-lens, do:If it is a 0-lens, then Reroute ( e αβ , f αβ ), else Swap ( e αβ , f αβ ). Phase 2.
Let L be the set of 1-3-lenses. For every L ( e αβ , f αβ ) ∈ L , if neither e αβ nor f αβ has been modified in previous iterations of Phase 2 (regardlessof whether x ( e αβ ) or x ( f αβ ) has changed), apply Reroute ( e αβ , f αβ ). Phase 3.
While there is a 0-lens L ( e αβ , f αβ ), do: Reroute ( e αβ , f αβ ).While there is a quasi-0-lens L ( e αβ , f αβ ), do: Quasi-0-Reroute ( e αβ , f αβ ). For i ∈ { , , } , let D i denote the drawing obtained at the end of Phase i .We analyse the three phases separately. Lemma 8.
Phase 1 terminates, and D is a 4-plane drawing in which everylens is a 1-3-lens, and any two edges have at most two points in common.Proof. By Lemma 4 and Observation 1, each iteration of the while loop reducesthe total number of crossings. Since D has at most · m ∈ O ( n ) crossings, thewhile loop terminates after O ( n ) iterations. By Lemma 7 all lenses satisfy theconditions of Lemma 6, except for 0-lenses and lenses L ( e αβ , f αβ ) with x ( e αβ ) =1 and x ( f αβ ) = 3. Each lens of the latter type is either a 1-3-lens, which remainsuntouched, or x ( e ) < e increases, it does not exceedfour and the total number of crossings in the drawing strictly decreases. In allother cases we can apply either Lemma 4 or Lemma 6 to conclude that eachiteration maintains a 4-plane drawing. By the end condition of the while loop,all lenses other than 1-3-lenses are eliminated.To prove the final statement, suppose to the contrary, two edges e and f in D have three or more points in common. By Lemma 1, there exist arcs e αβ ⊂ e and f αβ ⊂ f such that L ( e αβ , f αβ ) is a lens, which is necessarily a 1-3-lens. We mayassume without loss of generality that x ( e ) = 4, x ( e αβ ) = 1, and x ( f αβ ) = 3.Denote by γ a common point of e and f other than α and β . Since D is a4-plane drawing and x ( f αβ ) = 3, we may assume that α is common endpoint of e and f , furthermore γ is a crossing in the interior of f αβ . Since e αβ and f αβ form a lens, the arc e αβ cannot pass through γ . Hence γ is a crossing between f αβ and e \ e α,β . By Lemma 1, e βγ and f βγ form a lens, which is necessarily a1-3-lens. However, x ( e βγ ) ≤ x ( f βγ ) ≤
2, which is a contradiction.For the analysis of Phases 2 and 3, we introduce some notation. Let N denotethe planarization of D . Note that N is a simple graph, since a double edge wouldcorrespond to a lens whose arcs are crossing-free (i.e., a 0-lens). Phases 2 and3 apply only Reroute and
Quasi-0-Reroute operations. Hence the resultingdrawings satisfy invariants (I1)–(I3). For a node α of N , we denote by α a smallneighborhood of α . Recall that the length (cid:96) ( a ) of an arc a along an edge of G isthe combinatorial length of the path in N that the arc closely follows. Lemma 9. D has the following properties: (i) the length of every edge is atmost five; (ii) at most two edges of G pass along every segment of N ; (iii) throughevery node ν of N , at most two rerouted edges of G pass through ν ; and (iv) ateach node α of N , an edge passing through α crosses at most two edges in α ;(v) any two edges have at most two points in common.Proof. (i) By Lemma 8, the drawing D is a 4-plane drawing. Therefore, everyedge in D passes through at most 4 crossings, hence its length is at most 5.Each Reroute operation in Phase 2 replaces an edge of length 5 with an edgeof length 3 (cf. Fig. 4(a)). Property (i) follows. (ii)
Each
Reroute ( e αβ , f αβ ) operation in Phase 2 reroutes the longer arc alongthe shorter arc of a 1-3-lens in L . Let A be the set paths of length 2 in N imple Topological Drawings of k -Planar Graphs 11 that correspond to shorter arcs e αβ in some 1-3-lens L ( e αβ , f αβ ) ∈ L . By thedefinition of 1-3-lenses, (cid:96) ( e ) = 5 and e αβ consists of the first two segments of N along e . Thus every segment γδ of N is contained in at most one path in A .Consequently, at most one new edge can pass along γδ due to reroute operations. (iii) Let γ be a node in N that corresponds to a crossing in the drawing D .Then γ is incident to at most two paths in A (at most one along each of the twoedges that cross at γ ). Hence at most two rerouted edges can pass through γ . (iv) Let γ be a node of N , and let e be an edge that passes through γ in D . Byproperty (ii), at most 4 edges pass through γ . If at most 3 edges pass through γ , then it is clear that e crosses at most two edges in γ . Suppose that four edgespass through γ . Then the four segments of N incident to γ are each containedin the shorter arc of some 1-3-lens in D . Consequently, γ is the middle vertexof two distinct arcs in A . In the drawing D (after Reroute operations), twoedges run in parallel in each of these shorter arcs. Hence each edge that passesthrough γ crosses at most two other edges in γ , as claimed. (v) Suppose f and f have three points in common in D . By Lemma 8, we mayassume that f has been rerouted in Phase 2, and f follows a path ( v, γ, δ, u ) in N and ( v, γ, δ ) ∈ A . Since f and f have at most one common endpoint, theycross in both γ and δ . After the rerouting operation, f does not cross any edgeof D in δ , which implies that f has also been rerouted in Phase 2. Since both f and f have length three and pass through γ and δ , and N is a simple graph,both f and f pass along segment γδ , which contradicts the fact that at mostone new edge can pass along γδ (see the proof of (ii) above). Corollary 2. D is an 8-plane drawing of G .Proof. Every edge of G passes through the small neighborhood of at most fournodes of N by Lemma 9(i). In each such neighborhood, it crosses at most twoother edges by Lemma 9(iv), and it has at most one crossing with each by (I3).Overall, every edge has at most eight crossings in D .Unfortunately, Phase 2 may create new lenses, but only of very specific types.We analyze these types and argue that all remaining lenses are removed. Lemma 10.
Phase 3 terminates with an 8-plane simple topological drawing D .Proof. The while loops in Phase 3 terminate, as each iteration decreases thenumber of crossings by Lemmas 4 and 5. The drawing D at the beginning ofPhase 3 is 8-plane by Corollary 2, and remains 8-plane and no new lens is createdby Lemmas 4 and 5. It remains to show that Phase 3 eliminates all lenses of D .Every lens in D is a 1-3-lens by Lemma 8, and they are all in L . Phase 2modifies an arc in every lens in L . Thus the lenses of D are no longer presentin D . (The two edges that form a lens L ∈ L may still form a lens L (cid:48) in D ,but technically this is a new lens, that is, L (cid:54) = L (cid:48) , which is created in Phase 2and will be discussed next.)We classify the new lenses created in Phase 2. Assume that edges e and f = uw form a lens in D . Without loss of generality, the edge f was modified in Phase 2. Each iteration in Phase 2 applies a reroute operation on a 1-3-lens,which decreases the length of an edge from 5 to 3. Therefore Phase 2 modifiesevery edge at most once. The drawing of edge f in D was produced by a Reroute ( g uβ , f uβ ) operation, for some edge g , where u is a common endpoint of f and g . The resulting drawing of f in D closely follows a path ( u, α, β ) in N andthen the original arc (in D ) from β to w . After operation Reroute ( g uβ , f uβ ),edges f and g do not cross each other.Suppose first that f crosses e in β . Then e was redrawn in Phase 2 to closelyfollow f from w to β and beyond; as in Fig. 6. However, in this case, e and f have a common endpoint at w . No other edges follow segment βw in N byLemma 9(ii), hence e and f form a 0-lens. All such 0-lenses are eliminated inPhase 3, without creating any new lenses (cf. Lemma 4). Therefore, we mayassume that f does not cross any edge in β . vα βu wvα βwu vα βu wf f fe e e Fig. 6.
New 0-lens formed by e and f crossing in β . By Lemma 9(iv), the edge f crosses at most two other edges in α . If it crossesexactly one other edge, and f forms a lens L with that edge, then this crossingin α is the only crossing of f in D and, thus, L is a 0-lens. Otherwise, f crossestwo edges, denote them by e (for which we know that it crosses f ) and h ; oneof them was redrawn in a Reroute operation in Phase 2 to closely follow theother, which passes through α ; see Fig. 7 and 8. Therefore, e and h are adjacent,and they do not cross at the end of that operation. Thus, they do not cross in D , either; otherwise, three rerouted edges would pass through α , contradictingLemma 9(iii). As no new crossing is introduced in Phase 3, the edges e and h do not cross anytime during (and after) Phase 3, either. vα βu w vα βu w vα βu we e ef f fh h h Fig. 7. f crosses two edges at α and forms two 0-lenses.imple Topological Drawings of k -Planar Graphs 13 If the common endpoint of e and h is u or w (see Fig. 7), then both e and h form a lens with f : One of these lenses is a 0-lens, and when this lensis eliminated, the other lens either disappears, or it becomes a 0-lens as well.Hence Phase 3 eliminates both crossings.If e and h share distinct endpoints with f , without loss of generality e and f are adjacent at u and h and f are adjacent at w . As e and h do not cross, thecrossing e ∩ f is closer to u and the crossing h ∩ f is closer to w along f . Hence, e and h each form a 0-lens with f , both of which are eliminated in Phase 3. vα βu w vα βu w vα βu wf f fe e ehh h Fig. 8. f crosses two edges at α and forms a quasi-0-lens. It remains to consider the case that only e is adjacent to f (while h is not).Assume first that e and f are adjacent at w (see Fig. 8). If the crossing e ∩ f iscloser to w along f than the crossing h ∩ f , then the lens formed by e and f is a0-lens; else it forms a quasi-0-lens. In any case, the lens is eliminated in Phase 3.The same argument works in case that e and f are adjacent at u . References
1. Eyal Ackerman. On topological graphs with at most four crossings per edge.
Com-putational Geometry , 85:101574, 2019. doi:10.1016/j.comgeo.2019.101574 .2. J´anos Pach, Radoˇs Radoiˇci´c, G´abor Tardos, and G´eza T´oth. Improving the cross-ing lemma by finding more crossings in sparse graphs.
Discrete Comput. Geom. ,36(4):527–552, 2006. doi:10.1007/s00454-006-1264-9 .3. Gerhard Ringel. Ein Sechsfarbenproblem auf der Kugel.
Abhandlungen ausdem Mathematischen Seminar der Universit¨at Hamburg , 29:107–117, 1965. doi:10.1007/BF02996313 .4. Marcus Schaefer. The graph crossing number and its variants: A survey.
TheElectronic Journal of Combinatorics , 20, 2013. Version 4 (February 14, 2020). doi:10.37236/2713 .4 Michael Hoffmann, Chih-Hung Liu, Meghana M. Reddy, and Csaba D. T´oth
A Proof of Lemma 1
Lemma 1.
If a pair of edges e and f intersect in more than one point, thenthere exist arcs e αβ ⊂ e and f αβ ⊂ f that form a lens.Proof. Suppose, to the contrary, that there are no two intersection points thatdetermine a lens. Then the arcs between any two intersection points cross eachother. Consider two intersection points, α and β , for which the arcs e αβ ∪ f αβ have the minimum number of crossings. Let γ be one of these crossings. Then thearcs e αγ ⊂ e αβ and f αγ ⊂ f αβ have fewer crossings, contradicting the minimalityassumption in the choice of α and β . B Proof of Observation 1
Observation 1.
Let D be a topological drawing of a graph G , and let L ( e αβ , f αβ ) be a lens. Operation Swap ( e αβ , f αβ ) produces a topological drawing that has atleast one fewer crossing than D .Proof. The swap operation modifies only two edges, and step (3) removes anyself-intersections introduced in previous steps. Consequently, the output is atopological drawing. The operation eliminates crossings at both α and β . Atleast one of α and β is a crossing (one of them may be a common vertex), soat least one crossing is eliminated. On the other hand, the operation does notcreate any new crossings, hence the total number of crossings drops by at leastone. C Proof of Observation 2
Observation 2.
Let D be a topological drawing of a graph G , and let L ( e αβ , f αβ ) be a lens. Operation Reroute ( e αβ , f αβ ) produces a topological drawing.Proof. Only one edge is modified and Step (3) removes any self-intersectionsintroduced in previous steps. Thus the output is a topological drawing.
D Proof of Observation 3
Observation 3.
Algorithm 1 maintains invariants (I1)–(I3), and the length ofevery edge decreases or remains the same.Proof.
By construction,
Reroute operations maintain invariants (I1)–(I3). Ini-tially, D is a k -plane drawing, so every edge corresponds to a path of length atmost k +1 in the planarization N of D . Each iteration of the while loop replacesan arc with another arc of the same or smaller length. The claim follows. imple Topological Drawings of k -Planar Graphs 15 E Remarks about Algorithm 1 (1)
Constant-factor improvements to the bound f ( k ) ≤ √ · k / · k are notdifficult to obtain, at the expense of making the algorithm and analysis morecomplicated. We mention two possible improvements.(a) In the bound n γ ≤ · k − , we assume that all endpoints of the edges thatpass through γ are at distance k from γ in N , which is impossible. However, onecould have 2 vertices at distance 1 from γ , and 2 · k − vertices at distance k . Amore careful analysis might save up to a factor of 2.(b) We could modify Algorithm 1 so that we apply operation Swap ( e αβ , f αβ )if (cid:96) ( e αβ ) = (cid:96) ( f αβ ), and operation Reroute ( e αβ , f αβ ) if (cid:96) ( e αβ ) = (cid:96) ( f αβ ) and x ( e αβ ) ≤ x ( f αβ ). Note that under Exchange operations, the number of edgespassing though a neighborhood γ does not increase. Using this modified algo-rithm, the length of every edge that has been rerouted is at most k . So γ is atdistance at most k − γ (withthe possible exception of the two edges that cross at γ in the original drawing).This would imply n γ ≤ · k − , improving the bound on f ( k ) by a factor of 3. (2) Algorithm 1 incrementally modifies the edges of a drawing to eliminatelenses. An alternative algorithm, which follows a global redrawing strategy, wouldyield essentially the same bound on f ( k ), and ensure that every edge in theresulting simple topological drawing closely follows a shortest path in N . Specif-ically, we could label the segments of N by s , . . . , s t , where t ∈ O ( kn ) is thenumber of segments in N , and assign a weight w ( s i ) = 2 i + 2 t +1 to every seg-ment. For each edge e = ( u, v ) ∈ E ( G ), consider the network N uv − ( V \ { u, v } ),and then draw e so that it closely follows the weighted path in N from u to v .The weights guarantee that this is also an unweighted path in N uv . Furthermore,it is not difficult to show that if L ( e αβ , f αβ ) is a lens, then both e αβ and f αβ follow the same path in N ; and all such lenses can successively be eliminated by Exchange operations. (3 ) We do not know whether the upper bound f ( k ) ∈ exp( O ( k )) can be im-proved to a bound polynomial in k . However, Algorithm 1 (as well as the global“shortest path” approach mentioned above) does not yield a sub-exponentialbound, as there exist k -plane drawings for which these algorithms return simpletopological graphs whose local crossing numbers are exponential in kk