1 Introduction
This paper studies questions about duality between crossings and noncrossings in graph drawings. This idea is best illustrated by an example. A graph is planar if it has a drawing with no crossings, while a thrackle is a graph drawing in which every pair of edges intersect exactly once. So in some sense, thrackles are the graph drawings with the most crossings. Yet thracklehood and planarity appear to be related. In particular, a widely believed conjecture would imply that every thrackleable graph is planar. Loosely speaking, this says that a graph that can be drawn with the maximum number of crossings has another drawing with no crossings. This paper explores this seemingly counterintuitive idea through the notions of thickness and antithickness. First we introduce the essential definitions.
A (topological) drawing of a graph^{1}^{1}1We consider undirected, finite, simple graphs with vertex set and edge set . The number of vertices and edges of are respectively denoted by and . Let denote the subgraph of induced by a set of vertices . Let and . is a function that maps each vertex of to a distinct point in the plane, and maps each edge of to a simple closed curve between the images of its endvertices, such that:

the only vertex images that an edge image intersects are the images of its own endvertices (that is, an edge does not ‘pass through’ a vertex),

the images of two edges are not tangential at a common interior point (that is, edges cross ‘properly’).
Where there is no confusion we henceforth do not distinguish between a graph element and its image in a drawing. Two edges with a common endvertex are adjacent. Two edges in a drawing cross if they intersect at some point other than a common endvertex. Two edges that do not intersect in a drawing are disjoint. A drawing of a graph is noncrossing if no two edges cross. A graph is planar if it has a noncrossing drawing.
In the 1960s John Conway introduced the following definition. A drawing of a graph is a thrackle if every pair of edges intersect exactly once (either at a common endvertex or at a crossing point). A graph is thrackeable if it has a drawing that is a thrackle; see [75, 49, 88, 22, 92, 13, 15, 52, 72, 89, 14, 12, 84, 17, 83, 47, 5]. Note that in this definition, it is important that every pair of edges intersect exactly once since every graph has a drawing in which every pair of edges intersect at least once^{2}^{2}2Proof: Let . Position each vertex at . Define a relation on where if and only if . Observe that is a partial order of . Let , where implies that . Draw each edge as the 1bend polyline . Then every pair of edges intersect at least once..
A drawing is geometric if every edge is a straight line segment. A geometric drawing is convex if every vertex is on the convex hull of the set of vertices. A track drawing is a convex drawing of a bipartite graph in which the two colour classes are separated in the ordering of the vertices around the convex hull. For the purposes of this paper, we can assume that the two colour classes in a track drawing are on two parallel lines (called tracks). The notion of a convex thrackle is closely related to that of outerplanar thrackle, which was independently introduced by Cairns and Nikolayevsky [17].
1.1 Thickness and Antithickness
The thickness of a graph is the minimum such that the edge set can be partitioned into planar subgraphs. Thickness is a widely studied parameter; see the surveys [60, 79]. The thickness of a graph drawing is the minimum such that the edges of the drawing can be partitioned into noncrossing subgraphs. Equivalently, each edge is assigned one of colours such that crossing edges receive distinct colours.
Every planar graph can be drawn with its vertices at prespecified locations [64, 56, 85]. It follows that a graph has thickness if and only if it has a drawing with thickness [64, 56]. However, in such a representation the edges might be highly curved^{3}^{3}3In fact, Pach and Wenger [85] proved that for every planar graph that contains a matching of edges, if the vertices of are randomly assigned prespecified locations on a circle, then edges of have bends in every polyline drawing of .. The minimum integer such that a graph has a geometric / convex / track drawing with thickness is called the geometric / book / track thickness of . Book thickness is also called pagenumber and stacknumber in the literature; see the surveys [7, 34]^{4}^{4}4In the context of this paper it would make sense to refer to book thickness as convex thickness, and to refer to thickness as topological thickness, although we refrain from the temptation of introducing further terminology.
The following results are well known for every graph :
The antithickness of a graph is the minimum such that can be partitioned into thrackeable subgraphs. The antithickness of a graph drawing is the minimum such that the edges of the drawing can be partitioned into thrackles. Equivalently, each edge is assigned one of colours such that disjoint edges receive distinct colours. The minimum such that a graph has a topological / geometric / convex / 2track drawing with antithickness is called the topological / geometric / convex / 2track antithickness of . Thus a graph is thrackeable if and only if it has antithickness .
Lemma 1.
Every thrackeable graph has a thrackled drawing with each vertex at a prespecified location.
Proof.
Consider a thrackled drawing of . Replace each crossing point by a dummy vertex. Let be the planar graph obtained. Let be a distinct prespecified point in the plane for each vertex of . For each vertex choose a distinct point . Every planar graph can be drawn with its vertices at prespecified locations [64, 56, 85]. Thus can be drawn planar with each vertex of at . This drawing defines a thrackled drawing of with each vertex of at , as desired. ∎
Corollary 2.
A graph has antithickness if and only if it has a drawing with antithickness .
Every graph satisfies
Moreover, if is bipartite, then
For the final equality, observe that a track layout of with antithickness is obtained from a track layout of with thickness by simply reversing one track, and vice versa.
1.2 An Example: Trees
Consider the thickness of a tree. Every tree is planar, and thus has thickness and geometric thickness . It is well known that every tree has track thickness at most . Proof: Orient the edges away from some vertex . Properly colour the vertices of black and white. Place each colour class on its own track, ordered according to a breadthfirst search of starting at . Colour each edge according to whether it is oriented from a black to a white vertex, or from a white to a black vertex. It is easily seen that no two monochromatic edges cross, as illustrated in Figure 1.
The 2claw is the tree with vertex set and edge set , as illustrated in Figure 2(a). The upper bound of on the track thickness of trees is best possible since Harary and Schwenk [57] proved that the 2claw has track thickness exactly , as illustrated in Figure 2(b).
What about the antithickness of a tree? Since every tree has track thickness at most , by reversing one track, every tree has track antithickness at most . And again the 2claw shows that this bound is tight. In fact:
Lemma 3.
The 2claw is not a geometric thrackle.
Proof.
Suppose to the contrary that the 2claw is a geometric thrackle, as illustrated in Figure 2(c). For at least one of the three edges incident to , say , the other two vertices adjacent to are on distinct sides of the line through . Thus can only intersect one of and , which is the desired contradiction. ∎
1.3 Main Results and Conjectures
A graph parameter is a function that assigns to every graph a nonnegative integer . Examples that we have seen already include thickness, geometric thickness, book thickness, antithickness, geometric antithickness, and convex antithickness. Let be a class of graphs. Let denote the function , where is the maximum of , taken over all vertex graphs . We say has bounded if (where is the hidden variable in ).
A graph parameter is bounded by a graph parameter (for some class ), if there exists a binding function such that for every graph (in ). If is bounded by (in ) and is bounded by (in ) then and are tied (in ). If and are tied, then a graph family has bounded if and only if has bounded . This definition is due to Ding and Oporowski [28] and Reed [90]. Note that ‘tied’ is a transitive relation. For and to be not tied means that for some class of graphs, is bounded but is unbounded (or vice versa). In this case, and are separated, which is terminology introduced by Eppstein [37, 38].
The central questions of this paper ask which thickness/antithickness parameters are tied. In Section 4 we prove that thickness and antithickness are tied—in fact we prove that these parameters are both tied to arboricity, and thus only depend on the maximum density of the graph’s subgraphs. (See Section 4 for the definition of arboricity.)
Eppstein [37] proved that book thickness and geometric thickness are separated. In particular, for every , there exists a graph with geometric thickness and book thickness at least ; see [9, 8] for a similar result. Thus book thickness is not bounded by geometric thickness. The example used here is , which is the graph obtained from by subdividing each edge exactly once. In Lemma 17 we prove that has geometric antithickness . At the end of Section 5 we prove that has convex antithickness at least (which is unbounded). Thus convex antithickness is not bounded by geometric antithickness, implying that convex antithickness and geometric antithickness are separated.
Eppstein [38] also proved that geometric thickness and thickness are separated. In particular, for every , there exists a graph with thickness and geometric thickness at least . Thus geometric thickness is not bounded by thickness. (Note that it is open whether every graph with thickness has bounded geometric thickness.) Eppstein [38] used the following graph to establish this result. Let be the graph having as its vertices the singleton and tripleton subsets of an element set, with an edge between two subsets when one is contained in the other. (Note that can be analogously defined—just replace tripleton by doubleton.) Then has thickness , and for all there is an for which has geometric thickness at least . We expect that an analogous separation result holds for antithickness and geometric antithickness. Since has an edgepartition into three starforests, has antithickness . We conjecture that for all there is an for which has geometric antithickness at least . This would imply that geometric antithickness is not bounded by antithickness.
In the positive direction, we conjecture the following dualities:
Conjecture 4.
Geometric thickness and geometric antithickness are tied.
Conjecture 5.
Book thickness and convex antithickness are tied.
In Theorem 15 we prove that convex antithickness and queuenumber (defined in Section 2) are tied. Thus the truth of creftype 5 would imply that book thickness and queuenumber are tied. This would imply, since planar graphs have bounded book thickness [100, 11], that planar graphs have bounded queuenumber, which is an open problem due to Heath et al. [59, 58]; see [23, 31, 29] for recent progress. Thus a seemingly easier open problem is to decide whether planar graphs have bounded geometric antithickness.
Lovász et al. [72]
proved two related results. First they proved that every bipartite thrackleable graph is planar. And more generally, they proved that a bipartite graph has a drawing in which every pair of edges intersect an odd number of times if and only if the graph is planar. In their construction, nonadjacent edges cross once, and adjacent edges intersect three times.
1.4 Other Contributions
In addition to the results discussed above, this paper makes the following contributions. In Section 2 we prove that convex antithickness is tied to queuenumber and tracknumber. Several interesting results follow from this theorem. Section 3 surveys the literature on the problem of determining the thickness or antithickness of a given (uncoloured) drawing of a graph. Sections 4 and 5 respectively prove two results discussed above, namely that thickness and antithickness are tied, and that convex antithickness and geometric antithickness are separated. Section 6 studies natural extremal questions for all of the above parameters. Finally, Section 7 considers the various antithickness parameters for a complete graph.
2 Stack, Queue and Track Layouts
This section introduces track and queue layouts, which are well studied graph layout models. We show that they are closely related to convex antithickness.
A vertex ordering of an vertex graph is a bijection . We write to mean that . Thus is a total order on . We say or is ordered by . Let and denote the endvertices of each edge such that . At times, it will be convenient to express by the list , where . These notions extend to subsets of vertices in the natural way. Suppose that are disjoint sets of vertices, such that each is ordered by . Then denotes the vertex ordering such that whenever and with , or , , and . We write .
Let be a vertex ordering of a graph . Consider two edges with no common endvertex. There are the following three possibilities for the relative positions of the endvertices of and in . Without loss of generality .

and cross: .

and nest and is nested inside :

and are disjoint:
A stack (respectively, queue) in is a set of edges such that no two edges in are crossing (nested) in . Observe that when traversing , edges in a stack (queue) appear in LIFO (FIFO) order—hence the names.
A linear layout of a graph is a pair where is a vertex ordering of , and is a partition of . A stack (queue) layout of is a linear layout such that each is a stack (queue) in . At times we write (or ) if .
A graph admitting a stack (queue) layout is called a stack (queue) graph. The stacknumber of a graph , denoted by , is the minimum such that is a stack graph. The queuenumber of , denoted by , is the minimum such that is a queue graph. See [34] for a summary of results and references on stack and queue layouts.
A stack layout of a graph defines a convex drawing of with thickness , and vice versa. Thus the stacknumber of equals the book thickness of .
Lemma 6.
For every graph , the queuenumber of is at most the convex antithickness of .
Proof.
Consider a convex drawing of a graph with convex antithickness . Let be the underlying circular ordering and let be the corresponding edgepartition. Any two edges in cross or intersect at a common endvertex with respect to the vertex ordering . Thus each is a queue, and has queuenumber at most . ∎
We now set out to prove a converse to Lemma 6. A key tool will be track layouts, which generalise the notion of 2track drawings, and have been previously studied by several authors [30, 26, 32, 35, 78, 77, 76, 25, 24, 29, 31].
A vertex colouring of a graph is a partition of such that for every edge , if and then . The elements of are colours, and each set is a colour class. Suppose that is a total order on each colour class . Then each pair is a track, and is an track assignment of . To ease the notation we denote track assignments by when the ordering on each colour class is implicit.
An Xcrossing in a track assignment consists of two edges and such that and , for distinct colours and . An edge colouring of is simply a partition of . A track layout of consists of a track assignment of and an edge colouring of with no monochromatic Xcrossing. A graph admitting a track layout is called a track graph. The tracknumber of a graph is the minimum such that is a track graph.
The next two lemmas give a method that constructs a convex drawing from a track layout.
Lemma 7.
Suppose that has a convex drawing with antithickness , in which each thrackle is a matching. Then every track graph has convex antithickness at most .
Proof.
In the given convex drawing of , say the vertices are ordered around a circle, and is an edgepartition into thrackled matchings. Let be the track assignment and be the edge colouring in a track layout of . Let be a circular vertex ordering of . For each and , let . We now show that each set is a convex thrackle in , as illustrated in Figure 3.
Consider two edges with no common endvertex. If the endvertices of and belong to four distinct tracks, then and cross in , since the edges in pairwise cross. The endvertices of and do not belong to three distinct tracks, since is a matching. If the endvertices of and belong to two distinct tracks, then and cross in , as otherwise and form a monochromatic crossing in . Thus is a convex thrackle in , and has convex antithickness at most . ∎
The following results show that track graphs have bounded convex antithickness (for bounded and ). We start by considering small values of .
Lemma 8.
(a) Every track graph has convex antithickness at most .
(b) Every track graph has convex antithickness at most .
(c) Every track graph has convex antithickness at most .
Proof.
By Lemma 7, it is enough to show that , and admit convex drawings with their edges partitioned into 3, 5, and 8 thrackled matchings, respectively. The first claim is trivial. For the second claim, position the vertices of around a circle. Colour the two crossing edges blue. Colour each of the four other edges by a distinct colour. We obtain a convex drawing of with its edges partitioned into five thrackled matchings. Finally, say . Position the vertices of around a circle in the order . Then is a partition of into 8 thrackled matchings. ∎
Dujmović et al. [32] proved that every outerplanar graph has a track layout. Thus Lemma 8(c) with implies:
Corollary 9.
Every outerplanar graph has convex antithickness at most .
Let be the natural logarithm of . Let denote the th harmonic number. It is wellknown that
(1) 
The constructions in the proof of Lemma 8 generalise as follows.
Lemma 10.
The complete graph has a convex drawing with antithickness in which every thrackle is a matching, for some integer . That is, there is an edge colouring, such that edges that are disjoint or have a vertex in common receive distinct colours.
Proof.
By the above constructions, we may assume that . Let be the vertices of in order around a circle. For each and , let be the set of edges
As illustrated in Figure 4, is a thrackle and a matching.
Theorem 11.
Every track graph has convex antithickness at most .
A similar result was proved by Dujmović et al. [32], who showed that a track graph has geometric thickness at most . It is interesting that track layouts can be used to produce graph drawings with small geometric thickness, and can be used to produce graph drawings with small convex antithickness.
Dujmović et al. [32] proved that every queue colourable graph has a track layout. Thus Theorem 11 implies:
Corollary 12.
Every queue colourable graph has convex antithickness at most .
Dujmović and Wood [34] proved that every queue graph is colourable, and thus has a track layout. Thus Theorem 11 implies:
Corollary 13.
Every queue graph has convex antithickness at most .
A graph is series parallel if it has no minor.
Theorem 14.
Every series parallel graph has convex antithickness at most .
Proof.
Dujmović et al. [32] proved that queuenumber and tracknumber are tied. Thus Lemma 6 and Corollary 13 imply:
Theorem 15.
Queuenumber, tracknumber and convex antithickness are tied.
Several upper bounds on convex antithickness immediately follow from known upper bounds on queuenumber. In particular, Dujmović et al. [30] proved that graphs with bounded treewidth have bounded tracknumber (see [24, 98] for quantitative improvements to the bounds). Thus Theorem 11 with implies that graphs with bounded treewidth have bounded convex antithickness. Improving on a breakthrough by Di Battista et al. [23], Dujmović [29] proved that vertex planar graphs have queuenumber. More generally, Dujmović et al. [31] proved that vertex graphs with Euler genus have queuenumber. Since such graphs are colourable, Corollary 12 implies a upper bound on the convex antithickness. Most generally, for fixed , Dujmović et al. [31] proved that minorfree graphs have queuenumber, which implies a bound on the convex antithickness (since such graphs are colourable).
3 Thickness and Antithickness of a Drawing
This section considers the problem of determining the thickness or antithickness of a given drawing of a graph. We employ the following standard terminology. For a graph , a clique of is a set of pairwise adjacent vertices of . The clique number of , denoted by , is the maximum number of vertices in a clique of . The cliquecovering number of , denoted by is the minimum number of cliques that partition . An independent set of is a set of pairwise nonadjacent vertices of . The independence number of , denoted by , is the maximum number of vertices in an independent set of . The chromatic number of , denoted by , is the minimum number of independent sets that partition . Obviously and for every graph . Let be a family of graphs. is bounded if is bounded by in , and is bounded if is bounded by in .
Now let be a drawing of a graph . Let be the maximum number of pairwise crossing edges in , and let be the maximum number of pairwise disjoint edges in . Then is a lower bound on the thickness of , and is a lower bound on the antithickness of . Our interest is when the thickness of is bounded from above by a function of , or the antithickness of is bounded from above by a function of .
Let be the graph with such that two vertices of are adjacent if and only if the corresponding edges cross in . Let be the graph with such that two vertices of are adjacent if and only if the corresponding edges cross in or have an endvertex in common. Note that is a spanning subgraph of . By definition, the thickness of equals , and the antithickness of equals .
A string graph is the intersection graph of a family of simple curves in the plane; see [63, 45, 46, 44, 74, 93] for example. If we consider edges in as curves, then is a string graph. And deleting a small disc around each vertex in , we see that is also a string graph. Moreover, if is geometric, then both and are intersection graphs of sets of segments in the plane. If is convex, then both and are intersection graphs of sets of chords of a circle, which is called a circle graph; see [55, 65] for example. If is a track drawing, then both and are permutation graphs, which are perfect; see [50] for example.
Whether the thickness / antithickness of an (unrestricted) drawing is bounded by the maximum number of pairwise crossing / disjoint edges is equivalent to whether string graphs are bounded / bounded. Whether the thickness / antithickness of a geometric drawing is bounded by the maximum number of pairwise crossing / disjoint edges is equivalent to whether intersection graphs of segments are bounded / bounded. For many years both these were open; see [69, 68]. However, in a recent breakthrough, Pawlik et al. [86, 87] constructed set of segments in the plane, whose intersection graph is trianglefree and with unbounded chromatic number. Thus the thickness of a drawing is not bounded by any function of the maximum number of pairwise crossing edges, and this remains true in the geometric setting.
For convex drawings, more positive results are known. Gyárfás [53, 54] proved that the family of circle graphs is bounded, the best known bound being due to Černý [20] (slightly improving an earlier bound by Kostochka and Kratochvíl [66]). This implies that a convex drawing with at most pairwise crossing edges has thickness less than . For small values of much better bounds are known [1, 2]. Kostochka [67] proved that for every circle graph ; also see [66]. Thus a convex drawing with at most pairwise disjoint edges has antithickness at most .
Now consider a 2track drawing . Then and are permutation graphs, which are perfect. Thus and . This says that if has at most pairwise crossing edges, and at most pairwise disjoint edges, then has thickness at most and antithickness at most . There is a very simple algorithm for computing these partitions. First we compute the partition into 2track thrackles. For each edge , if is a set of maximum size of pairwise disjoint edges such that in one layer and in the other layer, then assign to the ()th set. Consider two disjoint edges and . Without loss of generality, and . Suppose that by the above rule is assigned to the ()th set and is assigned to the (th set. Let be a set of maximum size of pairwise disjoint edges such that in one layer and in the other layer. Then is a set of pairwise disjoint edges such that in one layer and in the other layer. Thus . That is, two edges that are both assigned to the same set are not disjoint, and each such set is a 2track thrackle. In the above rule, . Thus this procedure partitions the edges into 2track thrackles. To partition the edges into 2track noncrossing subdrawings, simply reverse the order of the vertices in one track, and apply the above procedure.
Consider the analogous question for queue layouts: Given a fixed vertex ordering of a graph , determine the minimum value such that admits a queue layout of . We can again construct an auxillary graph with , where two vertices are adjacent if and only if the corresponding edges of are nested in . Then admits a queue layout of if and only if . A classical result by Dushnik and Miller [36] implies that is a permutation graph, and is thus perfect. Hence . A clique in corresponds to a set of edges of that are pairwise nested in , called a rainbow. Hence admits a queue layout of if and only if has no ()edge rainbow, which was also proved by Heath and Rosenberg [59]. Dujmović and Wood [34] observed the following simple way to assign edges to queues: if the maximum number of edges that are pairwise nested inside an edge is , then assign to the ()th queue.
This procedure can also be used to prove that a convex drawing with at most pairwise disjoint edges has antithickness at most . This is equivalent to the result of Kostochka [67] mentioned above. Let be any vertex ordering obtained from the order of the vertices around the convex hull. Then has no ()edge rainbow. Assign edges to queues as described at the end of the previous paragraph. Partition the th queue into sets of pairwise nondisjoint edges as follows. For each edge in the th queue, if the maximum number of pairwise disjoint edges with both endvertices to the left of the left endvertex of is , then assign to the ()th set. Thus two edges in the th queue that are both assigned to the same set are not disjoint. Let be a maximum set of pairwise disjoint edges in the th queue. Then . Thus the th queue can be partitioned into sets of pairwise nondisjoint edges. Now we bound . Under each edge in is an edge rainbow. This gives a set of edges that are pairwise disjoint. Thus and . Thus we can partition the th queue into at most sets of pairwise nondisjoint edges. In total we have at most sets, each with no two disjoint edges, which is less than . Loosely speaking, this proof shows that a convex drawing and an associated edgepartition into convex thrackles can be thought of as a combination of a queue layout and an arch layout; see [34] for the definition of an arch layout.
4 Thickness and Antithickness are Tied
The arboricity of a graph is the minimum number of forests that partition . NashWilliams [80] proved that the arboricity of equals
(2) 
We have the following connection between thickness, antithickness, and arboricity.
Theorem 16.
Thickness, antithickness, and arboricity are pairwise tied. In particular, for every graph with thickness , antithickness , and arboricity ,
Proof.
Every forest is planar. Thus a partition of into forests is also a partition of into planar subgraphs. Thus .
Woodall [99] proved that every forest is thrackeable. Thus a partition of into forests is also a partition of into thrackeable subgraphs. Thus .
Every planar graph has arboricity at most by (2) and since . (Indeed, much more is known about edgepartitions of planar graphs into three forests [94].) Since every forest is thrackeable [99], every planar graph has antithickness at most . Thus a partition of into planar subgraphs gives a partition of into thrackeable subgraphs. Thus .
Theorem 16 with implies that every thrackle has arboricity at most . It is an open problem whether every thrackle is planar. It follows from a result by Cairns and Nikolayevsky [16] that every thrackle has a crossingfree embedding in the projective plane. Also note that the constant in Theorem 16 can be improved to using the result of Fulek and Pach [47].
5 Separating Convex Antithickness and Geometric Thickness
As discussed in Section 1.3, the following lemma is a key step in showing that convex antithickness and geometric antithickness are separated. Recall that is the graph obtained from by subdividing each edge exactly once.
Lemma 17.
has geometric antithickness .
Proof.
Let be the original vertices of . Position each at . For , let be the division vertex of the edge ; colour the edge blue, and colour the edge red. Orient each edge of from the original endvertex to the division endvertex. This orientation enables us to speak of the order of crossings along an edge.
We now construct a geometric drawing of , such that every pair of blue edges crosses, and every pair of red edges cross. Thus the drawing has antithickness . In addition, the following invariants are maintained for all and :
(1) No blue edge crosses after the crossing between and .
(2) No red edge crosses after the crossing between and .
The drawing is constructed in two stages. First, for , position at , as illustrated in Figure 5.
Observe that all the blue segments intersect at , and all the red segments intersect at . Thus the invariants hold in this subdrawing. Moreover, every blue edge intersects every red edge (although this property will not be maintained).
For and (in an arbitrary order) position as follows. The blue segment and the red segment were drawn in the first stage, and thus cross at some point . In the arrangement formed by the drawing produced so far, let be the face that contains , such that the blue segment is on the left of , and the red segment is on the right of . Position in the interior of , as illustrated in Figure 6.
By invariant (1), no blue edge crosses after the red edge . It follows that the new blue edge crosses every blue edge already drawn, and invariant (1) is maintained. By invariant (2), no red edge crosses after the blue edge . It follows that the new red edge crosses every red edge already drawn, and invariant (2) is maintained. ∎
Dujmović and Wood [35, Lemma 10] proved that has queuenumber at least . Since the queuenumber of a graph is at most its convex antithickness (Lemma 6), has convex antithickness at least . This proves the claim in Section 1 that implies that convex antithickness is not bounded by geometric antithickness.
6 Extremal Questions
This section studies the maximum number of edges in an vertex graph with topological (or geometric or convex or track) thickness (or antithickness) . The results are summarised in Table 1. First we describe results from the literature, followed by our original results.
For book thickness and track thickness the maximum number of edges is known. Bernhart and Kainen [6] proved that the maximum number of edges in an vertex graph with book thickness equals . Dujmović and Wood [34] proved that the maximum number of edges in an vertex graph with track thickness equals .
Determining the maximum number of edges in a thrackle is a famous open problem proposed by John Conway, who conjectured that every thrackle on vertices has at most edges. Improving upon previous bounds by Lovász et al. [72] and Cairns and Nikolayevsky [15], Fulek and Pach [47] proved that every thrackle has at most edges. Thus every graph with antithickness has at most edges. For , it is easy to construct an vertex graph consisting of edgedisjoint copies of . Thus this graph has antithickness and edges.
Many authors have proved that every geometric thrackle has at most edges [99, 39, 61, 82]. Thus every graph with geometric antithickness has at most edges. For convex antithickness, FabilaMonroy and Wood [41] improved this upper bound to , and FabilaMonroy et al. [40] established a matching lower bound. This lower bound is the best known lower bound in the geometric setting. It is an open problem to determine the maximum number of edges in an vertex graph with geometric antithickness .
We also mention that many authors have considered graph drawings, with at most pairwise crossing edges or at most pairwise disjoint edges (instead of thickness or antithickness ). These weaker assumptions allow for more edges. See [18, 71, 70, 19, 3, 48, 96, 48, 95, 43].
6.1 Thickness
Since every planar graph with vertices has at most edges, every graph with vertices and thickness has at most edges. We now prove a lower bound.
Theorem 18.
For all and infinitely many there is a graph with vertices, thickness , and exactly edges.
Let be a graph. Let be a bijection of . Let be the graph with vertex set and edge set , where is an abbreviation for the edge . Bijections and of are compatible if and are edgedisjoint. By taking a union, the next lemma implies Theorem 18.
Lemma 19.
For each integer there are infinitely many edgemaximal planar graphs that admits pairwise compatible bijections.
Proof.
Let be a prime number greater than . Let be the graph with vertex set
and edge set , where
Then is edgemaximal planar, as illustrated in Figure 7.
For each , let be the function defined by
where vertex indices are always in the cyclic group . Thus is a bijection.
Suppose to the contrary that and have an edge in common, for some distinct . Since is mapped to by both and , and similarly for the and , the following cases suffice. All congruences are modulo .
Case 1. is from in both and :
Case 1a. for some : Thus . Then and (implying ), or and (implying ), which is a contradiction since .
Case 1b. for some : Thus . If and , then , which is a contradiction since . Otherwise and , implying , which is a contradiction since .
Case 1c. for some : Thus . If and , then , which is a contradiction since . Otherwise and , implying , which is a contradiction since .
Case 2. is from in both and :
Case 2a. for some . Thus
Comments
There are no comments yet.