Drawing planar graphs of bounded degree with few slopes
aa r X i v : . [ m a t h . C O ] S e p Drawing planar graphs of bounded degreewith few slopes
Balázs Keszegh ∗ János Pach † Dömötör Pálvölgyi ‡ November 3, 2018
Abstract
We settle a problem of Dujmović, Eppstein, Suderman, and Wood by showing that there existsa function f with the property that every planar graph G with maximum degree d admits a drawingwith noncrossing straight-line edges, using at most f ( d ) different slopes. If we allow the edges tobe represented by polygonal paths with one bend, then d slopes suffice. Allowing two bends peredge, every planar graph with maximum degree d ≥ can be drawn using segments of at most ⌈ d/ ⌉ different slopes. There is only one exception: the graph formed by the edges of an octahedronis 4-regular, yet it requires 3 slopes. These bounds cannot be improved. A planar layout of a graph G is called a drawing if the vertices of G are represented by distinctpoints in the plane and every edge is represented by a continuous arc connecting the corresponding pairof points and not passing through any other point representing a vertex [3]. If it leads to no confusion,in notation and terminology we make no distinction between a vertex and the corresponding point andbetween an edge and the corresponding arc. If the edges are represented by line segments, the drawingis called a straight-line drawing . The slope of an edge in a straight-line drawing is the slope of thecorresponding segment.In this paper, we will be concerned with drawings of planar graphs. Unless it is stated otherwise,all drawings will be noncrossing , that is, no two arcs that represent different edges have an interiorpoint in common.Every planar graph admits a straight-line drawing [9]. From the practical and aesthetical point ofview, it makes sense to minimize the number of slopes we use [24]. The planar slope number of a planargraph G is the smallest number s with the property that G has a straight-line drawing with edges ofat most s distinct slopes. If G has a vertex of degree d , then its planar slope number is at least ⌈ d/ ⌉ ,because in a straight-line drawing no two edges are allowed to overlap.Dujmović, Eppstein, Suderman, and Wood [4] raised the question whether there exists a function f with the property that the planar slope number of every planar graph with maximum degree d canbe bounded from above by f ( d ) . Jelinek et al. [13] have shown that the answer is yes for outerplanar graphs, that is, for planar graphs that can be drawn so that all of their vertices lie on the outer face.In Section 2, we answer this question in full generality. We prove the following. ∗ EPFL, Lausanne and Rényi Institute, Budapest † EPFL, Lausanne and Rényi Institute, Budapest. Supported by Grants from NSF, SNF, NSA, PSC-CUNY, andOTKA. ‡ EPFL, Lausanne and Eötvös University, Budapest heorem 1. Every planar graph with maximum degree d admits a straight-line drawing, using segmentsof O ( d (3 + 2 √ d ) ≤ K d distinct slopes. The proof is based on a paper of Malitz and Papakostas [18], who used Koebe’s theorem [14] ondisk representations of planar graphs to prove the existence of drawings with relatively large angularresolution. As the proof of these theorems, our argument is nonconstructive; it only yields a nondeter-ministic algorithm with running time O ( dn ) . However, if one combines our result with a polynomialtime algorithm that computes the ǫ -approximation of the disk representation (see e.g. Mohar [19]),then one can obtain a deterministic algorithm running in time exponential in d but polynomial in n .For d = 3 , much stronger results are known than the one given by our theorem. Dujmović at al.[4] showed that every planar graph with maximum degree 3 admits a straight-line drawing using atmost 3 different slopes, except for at most 3 edges of the outer face, which may require 3 additionalslopes. This complements Ungar’s old theorem [23], according to which 3-regular, 4-edge-connectedplanar graphs require only 2 slopes and 4 extra edges.The exponential upper bound in Theorem 1 is probably far from being optimal. However, we wereunable to give any superlinear lower bound for the largest planar slope number of a planar graph withmaximum degree d . The best constructions we are aware of are presented in Section 5.It is perhaps somewhat surprising that if we do not restrict our attention to planar graphs, then noresult similar to Theorem 1 holds. For every d ≥ , Barát, Matoušek, and Wood [1] and, independently,Pach and Pálvölgyi [21] constructed graphs with maximum degree d with the property that no matterhow we draw them in the plane with (possibly crossing) straight-line edges, we must use an arbitrarilylarge number of slopes. (See also [5].) The case d ≤ is different: Keszegh et al. [15] proved that everygraph with maximum degree 3 can be drawn with 5 slopes. Moreover, Mukkamala and Szegedy [20]showed that 4 slopes suffice if the graph is connected. The case d = 4 remains open.Returning to planar graphs, we show that significantly fewer slopes are sufficient if we are allowedto represent the edges by short noncrossing polygonal paths. If such a path consists of k + 1 segments,we say that the edge is drawn by k bends . In Section 3, we show if we allow one bend per edge, thenevery planar graph can be drawn using segments with O ( d ) slopes. Theorem 2.
Every planar graph G with maximum degree d can be drawn with at most bend per edge,using at most d slopes. Allowing two bends per edge yields an optimal result: almost all planar graphs with maximumdegree d can be drawn with ⌈ d/ ⌉ slopes. In Section 4, we establish Theorem 3.
Every planar graph G with maximum degree d ≥ can be drawn with at most 2 bendsper edge, using segments of at most ⌈ d/ ⌉ distinct slopes. The only exception is the graph formed bythe edges of an octahedron, which is 4-regular, but requires 3 slopes. These bounds are best possible. It follows from the proof of Theorem 3 that in the cyclic order of directions, the slopes of theedges incident to any given vertex form a contiguous interval. Moreover, the ⌈ d/ ⌉ directions we usecan be chosen to be equally spaced in [0 , π ) . We were unable to guarantee such a nice property inTheorem 2: even for a fixed d , as the number of vertices increases, the smallest difference between the d − slopes we used tends to zero. We suspect that this property is only an unpleasant artifact of ourproof technique. Note that it is sufficient to prove the theorem for triangulated planar graphs, because any planargraph can be triangulated by adding vertices and edges so that the degree of each vertex increases onlyby a factor of at most three [22], so at the end we will lose this factor.2e need the following result from [18], which is not displayed as a theorem there, but is statedright above Theorem 2.2.
Lemma 4. (Malitz-Papakostas)
The vertices of any triangulated planar graph G with maximum degree d can be represented by nonoverlapping disks in the plane so that two disks are tangent to each other ifand only if the corresponding vertices are adjacent, and the ratio of the radii of any two disks that aretangent to each other is at least α d − , where α = √ ≈ . . Lemma 4 can be established by taking any representation of the vertices of G by tangent disks, asguaranteed by Koebe’s theorem, and applying a conformal mapping to the plane that takes the diskscorresponding to the three vertices of the outer face to disks of the same radii. The lemma now followsby the observation that any internal disk is surrounded by a ring of at most d mutually touching disks,and the radius of none of them can be much smaller than that of the central disk.The idea of the proof of Theorem 1 is as follows. Let G be a triangulated planar graph withmaximum degree d , and denote its vertices by v , v , . . . . Consider a disk representation of G meetingthe requirements of Lemma 4. Let D i denote the disk that represents v i , and let O i be the center of D i .By properly scaling the picture if necessary, we can assume without loss of generality that the radiusof the smallest disk D i is sufficiently large. Place an integer grid on the plane, and replace each center O i by the nearest grid point. Connecting the corresponding pairs of grid points by segments, we obtaina straight-line drawing of G . The advantage of using a grid is that in this way we have control of theslopes of the edges. The trouble is that the size of the grid, and thus the number of slopes used, is verylarge. Therefore, in the neighborhood of each disk D i , we use a portion of a grid whose side length isproportional to the radius of the disk. These grids will nicely fit together, and each edge will connecttwo nearby points belonging to grids of comparable sizes. Hence, the number of slopes used will bebounded. See Figure 1.Now we work out the details. Let r i denote the radius of D i ( i = 1 , . . . ) , and suppose without lossof generality that r ∗ , the radius of the smallest disk is r ∗ = min i r i = √ /α d − > , where α denotes the same constant as in Lemma 4.Let s i = ⌊ log d ( r i /r ∗ ) ⌋ ≥ , and represent each vertex v i by the integer point nearest to O i suchthat both of its coordinates are divisible by d s i . (Taking a coordinate system in general position, wecan make sure that this point is unique.) For simplicity, the point representing v i will also be denotedby v i . Obviously, we have that the distance between O i and v i satisfies O i v i < d s i √ . Since the centers O i of the disks induce a (crossing-free) straight-line drawing of G , in order toprove that moving the vertices to v i does not create a crossing, it is sufficient to verify the followingstatement. Lemma 5.
For any three mutually adjacent vertices, v i , v j , v k in G , the orientation of the triangles O i O j O k and v i v j v k are the same.Proof. By Lemma 4, the ratio between the radii of any two adjacent disks is at least α d − . Supposewithout loss of generality that r i ≥ r j ≥ r k ≥ α d − r i . For the orientation to change, at least one of O i v i , O j v j , or O k v k must be at least half of the smallest altitude of the triangle O i O j O k , which is atleast r k . 3igure 1: Straight-line graph from disk representationOn the other hand, as we have seen before, each of these numbers is smaller than d s i √ ≤ r i /r ∗ √ α d − r i ≤ r k which completes the proof.Now we are ready to complete the proof of Theorem 1. Take an edge v i v j of G , with r i ≥ r j ≥ α d − r i .The length of this edge can be bounded from above by v i v j ≤ O i O j + O i v i + O j v j ≤ r i + r j + d s i √ d s j √ ≤ r i + √ d s i ≤ r i + √ r i /r ∗ ≤ r i /r ∗ (2 r ∗ + √ ≤ r j /r ∗ α d − (2 r ∗ + √ < d s j +1 α d − ( 2 √ α d − + √ . According to our construction, the coordinates of v j are integers divisible by d s j , and the coordinatesof v i are integers divisible by d s i ≥ d s j , thus also by d s j .Thus, shrinking the edge v i v j by a factor of d s j , we obtain a segment whose endpoints are integerpoints at a distance at most dα d − ( √ α d − + √ . Denoting this number by R ( d ) , we obtain that thenumber of possible slopes for v i v j , and hence for any other edge in the embedding, cannot exceed thenumber of integer points in a disk of radius R ( d ) around the origin. Thus, the planar slope number of4ny triangulated planar graph of maximum degree d is at most roughly R ( d ) π = O ( d /α d ) , whichcompletes the proof. (cid:3) Our proof is based on the result of Malitz and Papakostas that does not have an algorithmic version.However, with some reverse engineering, we can obtain a nondeterministic algorithm for drawing atriangulated planar graph of bounded degree with a bounded number of slopes. Because of the enormousconstants in our expressions, this algorithm is only of theoretical interest. Here is a brief sketch.
Nondeterministic algorithm.
First, we guess the three vertices of the outer face and their coordinatesin the grid scaled according to their radii. Then embed the remaining vertices one by one. For eachvertex, we guess the radius of the corresponding disk as well as its coordinates in the proportionallyscaled grid. This algorithm runs in nondeterministic O ( dn ) time. In this section, we represent edges by noncrossing polygonal paths, each consisting of at most twosegments. Our goal is to establish Theorem 2, which states that the total number of directions assumedby these segments grows at most linearly in d .The proof of Theorem 2 is based on a result of Fraysseix et al. [10], according to which everyplanar graph can be represented as a contact graph of T -shapes. A T -shape consists of a vertical anda horizontal segment such that the upper endpoint of the vertical segment lies in the interior of thehorizontal segment. The vertical and horizontal segments of T are called its leg and hat , while theirpoint of intersection is the center of the T -shape. The two endpoints of the hat and the bottom endpointof the leg are called ends of the T -shape.Two T -shapes are noncrossing if the interiors of their segments are disjoint. Two T -shapes are tangent to each other if they are noncrossing but they have a point in common. Lemma 6. (Fraysseix et al.)
The vertices of any planar graph with n vertices can be represented bynoncrossing T -shapes such that1. two T -shapes are tangent to each other if and only if the corresponding vertices are adjacent;2. the centers and the ends of the T -shapes belong to an n × n grid.Moreover, such a representation can be computed in linear time. The proof of the lemma is based on the canonical ordering of the vertices of a planar graph,introduced in [11].
Proof of Theorem 2.
Consider a representation of G by T -shapes satisfying the conditions in the lemma.See Figure 2(a). For any v ∈ V ( G ) , let T v denote the corresponding T -shape. We define a drawing of G , in which the vertex v is mapped to the center of T v . To simplify the presentation, the center of T v is also denoted by v . For any uv ∈ E ( G ) , let p uv denote the point of tangency of T u and T v . Thepolygonal path up uv v consists of a horizontal and a vertical segment, and these paths together almostform a drawing of G with one bend per edge, using segments of two different slopes. The only problemis that these paths partially overlap in the neighborhoods of their endpoints. Therefore, we modifythem by replacing their horizontal and vertical pieces by almost horizontal and almost vertical ones,as follows.For any ≤ i ≤ d , let α i denote the slope of the (almost horizontal) line connecting the origin (0 , to the point (2 in, − . Analogously, let β i denote the slope of the (almost vertical) line passingthrough (0 , and (1 , in ) . 5ix a T -shape T v in the representation of G . It is tangent to at most d other T -shapes. Starting atits center v , let us pass around T v in the counterclockwise direction, so that we first visit the upper leftside of its hat, then its lower left side, then the left side and right side of its leg, etc. We number thepoints of tangencies along T v in this order. (Note that there are no points of tangencies on the lowerside of the hat.)Suppose now that the hat of a T -shape T u is tangent to the leg of T v , and let p uv be their pointof tangency. Assume that p uv was the number i point of tangency along T u and the number j pointof tangency along T v . Let p ′ uv denote the unique point of intersection of the (almost horizontal) linethrough u with slope α i and the (almost vertical) line through v with slope β j . In our drawing of G ,the edge uv will be represented by the polygonal path up ′ uv v . See Figure 2(c) for the resulting drawingand Figure 2(b) for a version distorted for the human eye to show the underlying structure. u vu u u u u u u (a) u vu u u u u u u (b) u vu u u u u u u (c) Figure 2: Representation with T -shapes and the drawing with one bend per edgeSince the segments we used are almost horizontal or vertical, the modified edges up ′ uv v are veryclose (within distance / ) of the original polygonal paths up uv v . Thus, no two nonadjacent edges cancross each other. On the other hand, the order in which we picked the slopes around each v guaranteesthat no two edges incident to v will cross or overlap. This completes the proof. In this section, we draw the edges of a planar graph by polygonal paths with at most two bends.Our aim is to establish Theorem 3.Note that the statement is trivially true for d = 1 and is false for d = 2 . It is sufficient to proveTheorem 3 for even values of d . For d = 4 , the assertion was first proved by Liu et al. [17] and later,independently, by Biedl and Kant [2] (also that the only exception is the octahedral graph). The latterapproach is based on the notion of st -ordering of biconnected ( -connected) graphs from Lempel et al.[16]. We will show that this method generalizes to higher values of d ≥ . As it is sufficient to provethe statement for even values of d , from now on we suppose that d ≥ even. We will argue that it isenough to consider biconnected graphs. Then we review some crucial claims from [2] that will enableus to complete the proof. We start with some notation.Take d ≥ lines that can be obtained from a vertical line by clockwise rotation by , π/d, π/d,. . . , ( d − π/d degrees. Their slopes are called the d regular slopes . We will use these slopes to draw G . Since these slopes depend only on d and not on G , it is enough to prove the theorem for connectedgraphs. If a graph is not connected, its components can be drawn separately.6n this section we always use the term “slope” to mean a regular slope. The directed slope of adirected line or segment is defined as the angle (mod π ) of a clockwise rotation that takes it to aposition parallel to the upward directed y -axis. Thus, if the directed slopes of two segments differ by π , then they have the same slope. We say that the slopes of the segments incident to a point p forma contiguous interval if the set ~S ⊂ { , π/d, π/d, . . . , (2 d − π/d } of directed slopes of the segmentsdirected away from p , has the property that for all but at most one α ∈ ~S , we have that α + π/d mod 2 π ∈ ~S (see Figure 5).Finally, we say that G admits a good drawing if G has a planar drawing such that every edge hasat most bends, every segment of every edge has one of the ⌈ d/ ⌉ regular slopes, and the slopes of thesegments incident to any vertex form a contiguous interval. If t is a vertex whose degree is at least twobut less than d , then we can define the two extremal segments at t as the segments corresponding to theslopes at the two ends of the contiguous interval formed by the slopes of all the segments incident to t . Also define the t -wedge as the infinite cone bounded by the extension of the two extremal segments,which contains all segments incident to t and none of the “missing” segments. See Figure 3. For a degreeone vertex t we define the t -wedge as the infinite cone bounded by the extension of the rotations of thesegment incident to t around t by ± π/ d . t (a) t (b) Figure 3: The t -wedgeTo prove Theorem 3, we show by induction that every connected planar graph with maximumdegree d ≥ with an arbitrary t vertex whose degree is strictly less than d admits a good drawing thatis contained in the t -wedge. Note that such a vertex always exist because of Euler’s polyhedral formula,thus Theorem 3 is indeed a direct consequence of this statement. First we show how the induction stepgoes for graphs that have a cut vertex, then (after a lot of definitions) we prove the statement also forbiconnected graphs (without the induction hypothesis). Lemma 7.
Let G be a connected planar graph of maximum degree d , let t ∈ V ( G ) be a vertex whosedegree is strictly smaller than d , and let v ∈ V ( G ) be a cut vertex. Suppose that for any connected planargraph G ′ of maximum degree d , which has fewer than | V ( G ) | vertices, and for any vertex t ′ ∈ V ( G ′ ) whose degree is strictly smaller than d , there is a good drawing of G ′ that is contained in the t ′ -wedge.Then G also admits a good drawing that is contained in the t -wedge.Proof. Let G , G , . . . denote the connected components of the graph obtained from G after the removalof the cut vertex v , and let G ∗ i be the subgraph of G induced by V ( G i ) ∪ { v } .If t = v is a cut vertex, then by the induction hypothesis each G ∗ i has a good drawing in the v -wedge . After performing a suitable rotation for each of these drawings, and identifying their vertices Of course the v -wedges for the different components are different. v , the lemma follows because the slopes of the segments incident to v form a contiguousinterval in each component.If t = v , then let G j be the component containing t . Using the induction hypothesis, G ∗ j has agood drawing. Also, each G ∗ i for i ≥ has a good drawing in the v -wedge. As in the previous case, thelemma follows by rotating and possibly scaling down the components for i = j and again identifyingthe vertices corresponding to v .In view of Lemma 7, in the sequel we consider only biconnected graphs. We need the followingdefinition. Definition 8.
An ordering of the vertices of a graph, v , v , . . . , v n , is said to be an st -ordering if v = s , v n = t , and if for every < i < n the vertex v i has at least one neighbor that precedes it and aneighbor that follows it. In [16], it was shown that any biconnected graph has an st -ordering, for any choice of the vertices s and t . In [2], this result was slightly strengthened for planar graphs, as follows. Lemma 9. (Biedl-Kant)
Let D G be a drawing of a biconnected planar graph, G , with vertices s and t on the outer face. Then G has an st -ordering for which s = v , t = v n and v is also a vertex of theouter face and v v is an edge of the outer face. We define G i to be the subgraph of G induced by the vertices v , v , . . . , v i . Note that G i is con-nected. If i is fixed, we call the edges between V ( G i ) and V ( G ) \ V ( G i ) the pending edges . For a drawingof G , D G , we denote by D G i the drawing restricted to G i and to an initial part of each pending edgeconnected to G i . Proposition 10.
In the drawing D G guaranteed by Lemma 9, v i +1 , . . . v n and the pending edges arein the outer face of D G i .Proof. Suppose for contradiction that for some i and j > i , v j is not in the outer face of D G i . Weknow that v n is in the outer face of D G i as it is on the outer face of D G , thus v n and v j are in differentfaces of D G i . On the other hand, by the definition of st -ordering, there is a path in G between v j and v n using only vertices from V ( G ) \ V ( G i ) . The drawing of this path in D G must lie completely in oneface of D G i . Thus, v j and v n must also lie in the same face, a contradiction. Since the pending edgesconnect V ( G i ) and V ( G ) \ V ( G i ) , they must also lie in the outer face.By Lemma 9, the edge v v lies on the boundary of the outer face of D G i , for any i ≥ . Thus,we can order the pending edges connecting V ( G i ) and V ( G ) \ V ( G i ) by walking in D G from v to v around D G i on the side that does not consist of only the v v edge, see Figure 4(a). We call this the pending-order of the pending edges between V ( G i ) and V ( G ) \ V ( G i ) (this order may depend on D G ).Proposition 10 implies Proposition 11.
The edges connecting v i +1 to vertices preceding it form an interval of consecutiveelements in the pending-order of the edges between V ( G i ) and V ( G ) \ V ( G i ) . For an illustration see Figure 4(a).Two drawings of the same graph are said to be equivalent if the circular order of the edges incidentto each vertex is the same in both drawings. Note that in this order we also include the pending edges(which are differentiated with respect to their yet not drawn end).Now we are ready to finish the proof of Theorem 3, as the following lemma is the only missing step.
Lemma 12.
For any biconnected planar graph G with maximum degree d ≥ and for any vertex t ∈ V ( G ) with degree strictly less then d , G admits a good drawing that is contained in the t -wedge. v v p p p p p p p p p p p G i (a) The pending-order of the pendingedges in D G i p v v p p p p p p p p p p p G i v i +1 (b) The preceding neighbors of v i +1 areconsecutive in the pending-order Figure 4: Properties of the st -ordering Proof.
Take a planar drawing D G of G such that t is on the outer face and pick another vertex, s , fromthe outer face. Apply Lemma 9 to obtain an st -ordering with v = s, v , and v n = t on the outer faceof D G such that v v is an edge of the outer face. We will build up a good drawing of G by startingwith v and then adding v , v , . . . , v n one by one to the outer face of the current drawing. As soon aswe add a new vertex v i , we also draw the initial pieces of the pending edges, and we make sure thatthe resulting drawing is equivalent to the drawing D G i .Another property of the good drawing that we maintain is that every edge consists of preciselythree pieces. (Actually, an edge may consist of fewer than 3 segments, because two consecutive piecesare allowed to have the same slope and form a longer segment) The middle piece will always be vertical,except for the middle piece of v v .Suppose without loss of generality that v follows directly after v in the clockwise order of thevertices around the outer face of D G . Place v and v arbitrarily in the plane so that the x –coordinateof v is smaller than the x –coordinate of v . Connect v and v by an edge consisting of three segments:the segments incident to v and v are vertical and lie below them, while the middle segment has anarbitrary non-vertical regular slope. Draw a horizontal auxiliary line l above v and v . Next, draw theinitial pieces of the other (pending) edges incident to v and v , as follows. For i = 1 , , draw a shortsegment from v i for each of the edges incident to it (except for the edge v v , which has already beendrawn) so that the directed slopes of the edges (including v v ) form a contiguous interval and theircircular order is the same as in D G . Each of these short segments will be followed by a vertical segmentthat reaches above l . These vertical segments will belong to the middle pieces of the correspondingpending edges. Clearly, for a proper choice of the lengths of the short segments, no crossings will becreated during this procedure. So far this drawing, including the partially drawn pending edges between V ( G ) and V ( G ) \ V ( G ) , will be equivalent to the drawing D G . As the algorithm progresses, thevertical segments will be further extended above l , to form the middle segments of the correspondingedges. For an illustration, see Figure 5(a).The remaining vertices v i , i > , will be added to the drawing one by one, while maintaining theproperty that the drawing is equivalent to D G i and that the pending-order of the actual pending edgescoincides with the order in which their vertical pieces reach the auxiliary line l i . At the beginning ofstep i + 1 , these conditions are obviously satisfied. Now we show how to place v i +1 .Consider the set X of intersection points of the vertical (middle) pieces of all pending edges be-tween V ( G i ) and V ( G ) \ V ( G i ) with the auxiliary line l i . By Proposition 11, the intersection points9 v l (a) Drawing v , v and the edgesincident to them v i v i − v i − v i − v i − v i − l i − l i − l i − l i − l i − l i − (b) Adding v i ; partial edges added in thisstep are drawn with dashed lines Figure 5: Drawing with at most two bendscorresponding to the pending edges incident to v i +1 must be consecutive elements of X . Let m be(one of) the median element(s) of X . Place v i +1 at a point above m , so that the x -coordinates of v i +1 and m coincide, and connect it to m . (In this way, the corresponding edge has only one bend,because its second and third piece are both vertical.) We also connect v i +1 to the upper endpoints ofthe appropriately extended vertical segments passing through the remaining elements of X , so thatthe directed slopes of the segments leaving v i +1 form a contiguous interval of regular slopes. For anillustration see Figure 5(b). Observe that this step can always be performed, because, by the definitionof st -orderings, the number of edges leaving v i +1 is strictly smaller than d . This is not necessarily truein the last step, but then we have v n = t , and we assumed that the degree of t was smaller than d . Tocomplete this step, draw a horizontal auxiliary line l i +1 above v i +1 and extend the vertical portionsof those pending edges between V ( G i ) and V ( G ) \ V ( G i ) that were not incident to v i +1 until theyhit the line l i +1 . (These edges remain pending in the next step.) Finally, in a small vicinity of v i +1 ,draw as many short segments from v i +1 using the remaining directed slopes as many pending edgesconnect v i +1 to V ( G ) \ V ( G i +1 ) . Make sure that the directed slopes used at v i +1 form a contiguousinterval and the circular order is the same as in D G . Continue each of these short segments by addinga vertical piece that hits the line l i +1 . The resulting drawing, including the partially drawn pendingedges, is equivalent to D G i +1 .In the final step, if we place the auxiliary line l n − high enough, then the whole drawing will becontained in the v n -wedge and we obtain a drawing that meets the requirements. In this section, we construct a sequence of planar graphs, providing a nontrivial lower bound forthe planar slope number of bounded degree planar graphs. They also require more than the trivialnumber ( ⌈ d/ ⌉ ) slopes, even if we allow one bend per edge. Remember that if we allow two bends peredge, then, by Theorem 3, for all graphs with maximum degree d ≥ , except for the octahedral graph, ⌈ d/ ⌉ slopes are sufficient, which bound is optimal. Theorem 13.
For any d ≥ , there exists a planar graph G d with maximum degree d , whose planar lope number is at least d − . In addition, any drawing of G d with at most one bend per edge requiresat least ( d − slopes. ab ca ′ b ′ c ′ (a) A straight line drawing of G b a c (b) At most four segments startingfrom a, b, c can use the same slope ina drawing of G d with one bend peredge Figure 6: Lower bounds
Proof.
The construction of the graph G d is as follows. Start with a graph of vertices, consisting oftwo triangles, abc and a ′ b ′ c ′ , connected by the edges aa ′ , bb ′ , and cc ′ (see Figure 6(a)). Add to thisgraph a cycle C of length d − , and connect d − consecutive vertices of C to a , the next d − ofthem to b , and the remaining d − to c . Analogously, add a cycle C ′ of length d − , and connectone third of its vertices to a ′ , one third to b ′ , one third to c ′ . In the resulting graph, G d , the maximumdegree of the vertices is d .In any crossing-free drawing of G d , either C lies inside the triangle abc or C ′ lies inside the triangle a ′ b ′ c ′ . Assume by symmetry that C lies inside abc , as in Figure 6(a).If the edges are represented by straight-line segments, the slopes of the edges incident to a, b, and c are all different, except that aa ′ , bb ′ , and cc ′ may have the same slope as some other edge. Thus, thenumber of different slopes used by any straight-line drawing of G d is at least d − .Suppose now that the edges of G d are represented by polygonal paths with at most one bendper edge. Assume, for simplicity, that every edge of the triangle abc is represented by a path withexactly one bend (otherwise, an analogous argument gives an even better result). Consider the d − polygonal paths connecting a , b , and c to the vertices of the cycle C . Each of these paths has a segmentincident to a , b , or c . Let S denote the set of these segments, together with the segments of the pathsrepresenting the edges of the triangle abc . Claim 14.
The number of segments in S with any given slope is at most .Proof. The sum of the degrees of any polygon on k vertices is ( k − π . Every direction is covered byexactly k − angles of a k -gon (counting each side / times at its endpoints). Thus, if we take everyother angle of a hexagon, then, even including its sides, every direction is covered at most times. (SeeFigure 6(b).)The claim now implies that for any drawing of G with at most one bend per edge, we need at least (3( d −
3) + 6) / ( d − different slopes. 11 cknowledgements We are grateful to Günter Rote for pointing out a mistake in the original proof of Lemma 7 and toan anonymous referee for calling our attention to [19].
References [1] J. Barát, J. Matoušek, and D. Wood: Bounded-degree graphs have arbitrarily large geometricthickness,
Electronic J. Combinatorics /1 (2006), R3.[2] T. Biedl and G. Kant: A better heuristic for orthogonal graph drawings, Comput. Geom.
Graph Drawing , Prentice Hall, Upper SaddleRiver, N.J., 1999.[4] V. Dujmović, D. Eppstein, M. Suderman, and D. R. Wood: Drawings of planar graphs with fewslopes and segments,
Comput. Geom. (2007), no. 3, 194–212.[5] V. Dujmović, M. Suderman, and D. R. Wood: Graph drawings with few slopes, Comput. Geom. (2007), 181–193.[6] C. A. Duncan, D. Eppstein, and S. G. Kobourov: The geometric thickness of low degree graphs.In: Proc. 20th ACM Symp. on Computational Geometry (SoCG’04) , ACM Press, 2004, 340–346.[7] M. Engelstein: Drawing graphs with few slopes, Research paper submitted to the Intel Competitionfor high school students, New York, October 2005.[8] D. Eppstein: Separating thickness from geometric thickness,
Towards a Theory of Geometric Graphs(J. Pach, ed.), Contemporary Mathematics , Amer. Math. Soc, Providence, 2004, 75–86.[9] I. Fáry, On straight line representation of planar graphs,
Acta Univ. Szeged. Sect. Sci. Math. (1948), 229–233.[10] H. de Fraysseix, P. Ossona de Mendez, P. Rosenstiehl: On triangle contact graphs, Combinatorics,Probability and Computing (1994), 233–246.[11] H. de Fraysseix, J. Pach, and R. Pollack: How to draw a planar graph on a grid, Combinatorica (1) (1990), 41–51.[12] A. Garg and R. Tamassia: Planar drawings and angular resolution: Algorithms and bounds, SecondAnnual European Symposium, Lecture Notes in Computer Science , Springer, 1994, 12–23.[13] V. Jelinek, E. Jelinková, J. Kratochvíl, B. Lidický, M. Tesař, and T. Vyskočil: The planar slopenumber of planar partial 3-trees of bounded degree. In:
Graph Drawing, Lecture Notes in ComputerScience , Springer, Berlin, 2010, 304–315.[14] P. Koebe: Kontaktprobleme der konformen Abbildung,
Berichte Verhand. Sächs. Akad. Wiss. Leipzig,Math.-Phys. Klasse (1936) 141–164.[15] B. Keszegh, J. Pach, D. Pálvölgyi, and G. Tóth: Drawing cubic graphs with at most five slopes,Drawing cubic graphs with at most five slopes. Comput. Geom. (2008), no. 2, 138–147.1216] A. Lempel, S. Even, I. Cederbaum: An algorithm for planarity testing of graphs, Theory of Graphs (P. Rosenstiehl, ed.), Gordon and Breach, New York, 1967, 215–232 .[17] Y. Liu, A. Morgana, B. Simeone: General theoretical results on rectilinear embeddability of graphs,
Acta Math. Appl. Sinica (1991), 187–192.[18] S. Malitz and A. Papakostas: On the angular resolution of planar graphs, SIAM J. Discrete Math. (1994), 172-183.[19] B. Mohar, A polynomial time circle packing algorithm, Discrete Math., 117 (1993), 257-263.[20] P. Mukkamala and M. Szegedy: Geometric representation of cubic graphs with four directions, Comput. Geom. (2009), 842–851.[21] J. Pach and D. Pálvölgyi: Bounded-degree graphs can have arbitrarily large slope numbers, Elec-tronic J. Combinatorics /1 (2006), N1.[22] J. Pach and G. Tóth: Crossing number of toroidal graphs, Graph drawing, Lecture Notes in Comput.Sci. , Springer, Berlin, 2006, 334–342.[23] P. Ungar: On diagrams representing maps,
J. London Math. Soc. (1953), 336–342.[24] G. A. Wade and J. H. Chu: Drawability of complete graphs using a minimal slope set, The ComputerJ.37