1 Introduction
A drawing of a graph is, typically, defined as a mapping of each vertex to a distinct point on the plane and of each edge to a simple Jordan arc with endpoints at and . When edges are drawn as straightline segments the corresponding drawings are referred to as straightline drawings and the associated graphs are referred to as geometric graphs. Drawing algorithms are used to generate the mapping of vertices and edges to points and Jordan arcs on the plane, respectively. The produced drawings follow conventions, or drawing styles, which dictate the characteristic features of the drawing, for example, whether edges are allowed to cross each other, whether edges have to be drawn as a single (straightline) segment or are allowed to have “bends”, whether vertex placement has to follow a pattern (e.g., drawn on a circle, or on several parallel lines as a hierarchy), etc. The drawing algorithms usually aim to optimize some characteristic attributes of the drawing, having as ultimate goal to produce aesthetically pleasing and useful drawings, i.e., drawings that reveal properties of the underlying graphs and/or facilitate their exploratory analysis. Drawing characteristics that we typically attempt to optimize include the number of edge crossings, the area of the drawing (assuming vertices at integer coordinates), the angular resolution and the total number of bends (if they are allowed). Introductory as well as in depth coverage of graph drawing algorithms under several drawing styles is provided in [9, 24, 29].
Common to almost every drawing style, we find two restrictions that aim to eliminate any ambiguity on the drawn graph, and thus, to improve the readability of its drawing. These two conditions state that “edges cannot intersect (or pass over) vertices of the graph” and that “edges cannot overlap each other”. Fig. 1 demonstrates that when a drawing does not respect these restrictions we cannot interpret it in an unambiguous way.
When the “edgevertex intersection” restriction is formalized, we require that the line segment which corresponds to an arbitrary edge does not contain any point , where . Thus, when trying to enforce these restrictions, edges are treated as line segments of zero width, and vertices as points. However, in reality, in order to be able to identify the vertices we draw them as “thick” objects; typically in the shape of a disk or a square/rectangle. These objects can be of “unit” size (e.g., for vertex , a disk with center at and diameter equal to one), or have size that depends, for example, on the length of the label contained in it. Thus, when vertices are drawn as thick objects, we have to make sure that no edge intersects the area occupied by the vertex object and not just its centerpoint.
Reality dictates another restriction. Graph drawings typically are either displayed on a drawing canvas, where the centers of the vertex objects are being placed at grid positions, i.e., they have integer coordinates. So, when combined with the requirement that vertices are of at least unit size, we are left with following generic drawing problem: Given a graph , produce a grid drawing of where the vertices are represented by unitsized disks the edges as (zerowidth) linesegments and no edge intersects any vertex disk. The drawing problem is generic in the sense that the produced drawing has to also satisfy additional restrictions dictated by the drawing style (e.g., planar drawing, with bends allowed, etc). For simplicity, we concentrate on unit disk vertex objects. The size of the object as well as its shape can be treated as parameters of the drawing.
By assuming that our vertex objects are disks, we call the grid drawings that have no overlaps between edges and vertex objects disklink drawings (see Section 2 for a formal definition). Graph editors typically create grid drawings which have unitsized disk vertices but, they do not necessarily respect the “no intersection between edge and vertex objects” restriction. Fig. 2 shows such a drawing. However, as it is demonstrated in Fig. 2, by zooming out the problem is resolved, but, at the cost of increasing the area of the drawing. In this paper, we address precisely this problem. We design algorithms that compute disklink drawings in small area (smaller than the ones obtained by simple zoomout). As pointed out in the next paragraph, our research is related to other interesting problems studied in Graph Drawing and Computational Geometry.
Related Work. The problem caused by overlaps of vertex objects with other vertices and edges, leading to cluttered drawings, has been recognized from the early years of graph drawing. Davidson and Harel[5] back in 1996 presented a method to draw graphs nicely based on simulated annealing. The energy function they tried to minimize, among other terms, included a term which captured vertexedge distances, penalizing for edges that are too close to vertices in the drawing. It should be noted that the produced drawings, still contained edgevertex overlaps. A few years later, Gansner and North [21] and Dobkin et al. [12]
used two postprocessing heuristics to improve drawings by reducing clutter while conserving area. They firstly eliminated the overlapping of vertex objects by using Voronoi diagrams to reposition the vertices away from each other while maintaining (roughly) the original layout and, secondly, they redrew edges as smooth splines avoiding overlaps with vertex objects (but introducing some edgeedge overlaps).
Another related research direction considers drawings where vertices are objects with integer coordinates and the edges are fat segments. Barequet et al. [3] in an attempt to visualize weighted graphs study drawings where the width of each edge is proportional to its weight and the width of each vertex is proportional to the sum of the weights of its incident edges. If denotes the maximum edge weight, then for an vertex maximal planar graph, a drawing of area is produced. Also, these drawings are not straightline, as edge segments attach around the objects representing the vertices. Barequet et al. use diamond shaped vertex objects however, the use of disks is also possible. For edges with “zero” width, the drawings appear to be of similar style to the ones we consider in this paper, however one significant difference remain; in the drawing produced in [3] the edges do not connect the centers of the incident vertexdisks but rather simply enter these disks through varying angles. Duncan et al. [15] also use fat edges but, in contrast to the work of Barequet et al. [3], they do not compute a drawing from scratch but rather they try to extend an existing one without modifying the area of the layout. Given a planar weighted graph of maximum degree one and an embedding for (given as a set of homotopic shortest paths), they indentify in time a planar drawing such that all edges are drawn as thick as possible and proportional to their corresponding edge weights where is the number of paths and is the maximum of input and outut complexities of the wiring. They also show how to extend their result to general planar graphs.
Van Kreveld [30] introduced and studied bold drawings. In a bold drawing, vertices are drawn as disks of radious and edges as rectangles of width , where . They concentrated on good bold drawings of planar graphs, defined (informally) as bold drawing which have no vertexvertex ore edgevertex intersections, having all of its vertices and edges at least partially visible and having being completelly exposed in the sense that the area covered by overlapping edges in not sufficient to hide any vertex disk or and edgerectangle. They showed that if a typical graph drawing (i.e., with point vertices and zero width edges) is in nondegenerate position (i.e., no edge intersects a nonincident vertex and no three edges pass through a common point), then there exist positive values and that will turn it into a bold drawing. They also presented algorithms for i) deciding whether for given and values a drawing is bold, and ii) for maximizing and/or for a given drawing so that it is turned to a bold one. Pach [25] answered one question posed by Van Kreveld in [30]. More specifically, he showed that every graph admits a bold drawing in which the region occupied by the union of disks and rectangles representing the vertices and edges does not contain any disk or radius other than the ones representing the graph’s vertices (i.e., no vertices can be hidden).
When the input graph is a complete graph, our problem can be regarded as a generalization of the nothreeinline problem[14], which asks for the maximum number of points that can be placed on an grid such that no three points are collinear. In this regard, a result by Wood [31] states that the balanced complete partite graph , where is the number of vertices in each partition, admits a drawing on a grid of size , where is the minimum prime number such that . In our case, we seek for the maximum value such that has a disklink drawing on an grid.
We finally remark that a straightline drawing on an integer grid using only the horizontal, vertical and slopes is a disklink drawing. For this reason, triconnected cubic planar graphs admit a disklink drawing on a grid of quadratic size [11]. For biconnected graphs, the problem is still open.
Contribution and paper organization. We tackle the problem of designing algorithms that produce disklink drawings in compact area. Our contribution is as follows.

We first give some basic results (Section 2), in particular, we give an upper bound on the stretching factor that turns any grid drawing into a disklink drawing. This result immediately implies some area upper bounds for disklink drawings of certain graph classes.

We then study improved area bounds for nonplanar graphs (Section 3). We show that proper level graphs and bounded bandwidth graphs admit disklink drawings in quadratic area and linear area, respectively. The latter result is obtained by exploiting a construction of Erdős [16] for the nothreeinline problem. Moreover, as the main result of this section, we prove that every complete graph has a convex disklink drawing in quartic area. This is obtained by using the corners of a regular gon as an initial placement of the vertices, and by suitably rounding the coordinates of each vertex to enforce integer coordinates.

Afterwards, we turn our attention to crossingfree disklink drawings (Section 4). Our main result states that every vertex planar graph admits a planar disklink drawing in area. This is obtained by extending a central technique by de Fraysseix, Pach and Pollack [8], which draws an vertex planar graph on a grid of size . We also show that every outerplanar graph admits an outerplanar disklink drawing in precisely area. Such disklink drawings are computed through an inductive algorithm that exploits a BFStraversal of the graph. In addition, we prove a basic superlinear lower bound for the area requirement of disklink drawings of star graphs; since star graphs obviously admit grid drawings in linear area, this last result corroborates the fact that computing disklink drawings in compact area may be a challenging task already for trivial classes of graphs.
We conclude with a brief discussion of our research and with open problems (Section 5).
2 Basic Results
We assume familiarity with basic graph theoretic concepts [23] and standard notions of graph drawing [24, 28]. In what follows, a grid drawing is always a straightline drawing whose vertices are at integer coordinates, whereas a disklink drawing is formally defined as follows.
Definition 1.
A disklink drawing of a graph maps each vertex of to a distinct open disk with radius and each edge of to a (zerowidth) straightline segment connecting the centers of the two disks corresponding to its endvertices, such that the center of each disk is at integer coordinates, no two disks intersect, and the edgevertex resolution is at least , that is, no edge segment intersects a vertexdisk except at its endpoints.
We say that a graph admits a disklink drawing (resp. a grid drawing) on a grid of size (or, equivalently, in area ), if the minimum axisaligned box containing it has side lengths and . In other words, the size of a disklink drawing (resp. a grid drawing) is the number of grid points in its bounding box. As already mentioned, we assume for simplicity that , even though our results carry over for any constant radius (up to some multiplicative constant factor for the area). We now introduce a central property, which we use to transform a grid drawing into a disklink drawing; the  and span of an edge whose endpoints are and in a grid drawing are the quantities and , respectively.
Lemma 1.
Let be a grid drawing of a graph and let be an edge of such that and . Let be the drawing obtained by mapping each vertex with coordinates in to the point , where and are integers such that and . Then, is a grid drawing of in which the minimum distance between any vertex and the edge segment representing is at least .
Proof.
Drawing is a grid drawing of , as it is obtained through an affine transformation of and both and are integers. We prove that the minimum distance between any vertex and the edge segment representing is at least (and thus at least ). To this aim, it suffices to consider the case in which and , as for larger values the distance between and any vertex in can only increase further. Up to a translation, we may assume that one endpoint of in is , which implies that its other endpoint is . Since and , the endpoints of in are and . Assume to the contrary that there is a vertex in , which is at a distance strictly less than from . It follows that must lie at a grid point either on line with slope through the point or on line with slope through the point . By symmetry, we may assume that the former situation applies. For some integer number , let be the grid point representing along in . By the stretching factors and , the position of in is , which must be a grid point since is a grid drawing. Since both and are even and either or
is odd, either
or is not integer, which contradicts the fact that is a grid drawing. ∎Theorem 2 (Stretching Theorem).
Every graph that admits a grid drawing also admits a disklink drawing on a grid of size .
Proof.
Let and be the maximum  and span over all edges in the grid drawing. Since and , the result follows by Lemma 1. ∎
Corollary 3 is obtained by combining Theorem 2 and a result by Wood [31], who proved that every vertex colorable graph has an grid drawing. Note that Corollary 3 applied to a planar graph yields a disklink drawing on a grid of quadratic size, which, however, is not necessarily planar. Corollary 4 is an immediate implication of Theorem 2 and the fact that every vertex planar graph has an grid drawing [7, 26]; a drastic improvement will be presented in Section 4.
Corollary 3.
Every colorable vertex graph admits a disklink drawing on a grid of size .
Corollary 4.
Every vertex planar graph admits a planar disklink drawing on a grid of size .
3 Nonplanar Drawings
In this section we study disklink drawings for families of nonplanar graphs. We begin with a simple result that turns a proper level drawing (see, e.g., [10]) into a disklink drawing. Proper level drawings have been intesively studied, with particular attention devoted to planar graphs (see, e.g., [2] for references). The idea of exploiting a proper leveling of a graph to compute a disklink drawing will be later reused for outerplanar graphs (Section 4).
3.1 Proper level graphs
We recall some notation that will be useful also in the next section. A leveling of a graph is a function for some integer number . For a vertex of , is referred to as the level of . A drawing of a graph is proper level with respect to a leveling , if the coordinate of each vertex of is and for each edge of it holds . A graph is proper level if it admits a proper level drawing. We show that proper level graphs admit disklink drawings in quadratic area.
Theorem 5.
Every proper level vertex graph admits a disklink drawing on a grid of size .
Proof.
Refer to Fig. 3 for an example of the construction. Let be a proper level drawing of a proper level graph with respect to a leveling . We process the vertices of based on their levels from to such that vertices with the same level are processed based on their lefttoright order in . Let be the th vertex according to this order, with . We place the center of the disk representing with coordinates if is even, and with coordinates if is odd.
To see that the computed representation is a disklink drawing of , consider an edge of . Then, either is represented as a horizontal segment, or . In the former case, by construction, there is no vertex horizontally aligned between and . In the latter case, the absolute value of the slope of is at least . In both cases, does not intersect any disk representing a vertex . ∎
3.2 Bounded bandwidth graphs
A graph has bandwidth if there is a total ordering of the vertices of , denoted by , such that for every edge with , the cardinality of the set is at most (see, e.g., [13, 17]). We show that the graphs with bounded bandwidth admit disklink drawings in linear area.
Theorem 6.
Every vertex graph of bandwidth admits a disklink drawing on a grid of size , where is the minimum prime number such that .
Proof.
Let be an vertex graph of bandwidth , which we assume to be maximal (i.e., no edge can be added without increasing its bandwidth). At a high level, we first construct a grid drawing of in which no three vertices forming a cycle in are collinear, and for any edge it holds and . Applying Lemma 1 to yields the desired disklink drawing. To construct , we make use of a result by Erdős [16], who showed that for every prime , there do not exist three collinear points in the set consisting of the points
(1) 
Let be the vertices of according to . Let be the minimum prime number such that . We partition the vertices of into groups of vertices, and for each group of vertices, we obtain a drawing using the aforementioned results by Erdős. In the construction, we ensure that any two groups of vertices are separated by units horizontally (see Fig. 4). More precisely, for and , we draw vertex of with at the point that is obtained by shifting point of Eq. (1) by units along the horizontal direction. Namely, we set
(2) 
We say that two vertices and are in the same group if there exist three indices , and such that and , where and . We claim that no three vertices forming a cycle in are collinear in the constructed drawing. To see this, consider any three vertices , and that form a cycle in , and assume w.l.o.g. that . Observe first that at least two of them belong to the same group. Otherwise, , and would belong to three different groups, and thus the distance in between and would be greater than . Since has bandwidth , vertices and would not be adjacent in , a contradiction.
If , and all belong to the same group, the noncollinearity is guaranteed by Erdős’s construction. Consider now the case in which two vertices belong to the same group, say and , while the third vertex belongs to a different group. In this case, the three vertices cannot be horizontally aligned. Namely, suppose for a contradiction that , and are horizontally aligned and observe that, by construction, two vertices that belong to two distinct groups and have the same coordinate are at distance at least in . Then, by this observation, and are at distance at least , and thus and are at distance at least , which contradicts the fact that and are connected by an edge. Thus, to prove noncollinearity, we can restrict to the case in which and are not horizontally aligned. Consider the line that passes through them. By Eq. (2), the absolute value of the slope of ranges in . On the other hand, vertex is at horizontal distance at least and at vertical distance at most , hence does not pass through .
Putting all together, we constructed a grid drawing of on a grid of size , which is . Additionally, for any edge it holds and . The result follows by Lemma 1.
∎
3.3 Complete graphs
Corollary 3 implies that the complete graph admits a disklink drawing on a grid of size . We conclude this section by strengthening this result. Namely, the next theorem shows that the same area bound can be obtained by disklink drawings that are also convex. Here, a convex drawing is a grid drawing where the vertices of the graph are placed at the corners of a convex polygon. We remark that, in contrast to Corollary 3, the next theorem cannot be obtained by exploiting Theorem 2. This is because of a known (super quadratic) lower bound on the area required to produce a convex grid drawing of a complete graph, given by Acketa and Zunic [1].
Theorem 7.
The complete vertex graph admits a convex disklink drawing on a grid of size .
Proof.
Denote by the vertices of . Let be a regular gon centered at point such that the distance between its center and any of its vertices is , where is a positive integer that we will define below. For , we place vertex at the th corner of and obtain a drawing of , which is not necessarily a grid drawing. For , denote by the distance between vertex and edge , where the indices are taken modulo . It follows that . Observe that the edgevertex resolution of equals to .
We now claim that if , then is at least (a suitable value greater than one, as it will become clear below). To prove the claim, for , denote by the smallest of the two angles between the line segments that connect the center of with the vertices and . Since is a regular gon, it follows that . Since the edge is perpendicular to the line segment connecting the center of with vertex , it follows that . Hence, the goal that we set above is equivalent to . Since and , what we have to prove is that for every . To see this, let be such that . Clearly, if for , then our claim follows. Using elementary properties of trigonometric functions, we can rewrite as . Since , is equivalent to . Let be such that . The first derivative of is . Hence, if and only if , which holds for all . The fact the first derivative of is positive implies that is increasing. Hence, for all , or equivalently for . The latter implies that for , as desired.
We now prove that the drawing obtained from by rounding each vertex in to its nearest grid point in has edgevertex resolution at least , that is, by replacing each vertex with a disk centered at that point we obtain a disklink drawing. Consider the effect of this rounding operation on the edgevertex resolution of . In particular, consider vertex and the edge for some . The rounding of vertex may result in bringing edge one unit closer to in the worst case. Similarly, in the worst case the same effect may be observed by the rounding of the vertices and . Hence, in the worst case the rounding may result in decreasing the edgevertex resolution of by three units in . This completes the proof. ∎
4 Planar and Outerplanar Drawings
In this section we study crossingfree disklink drawings of planar and outerplanar graphs. By Corollary 4 every planar graph admits a disklink drawing on a grid of quartic size; we reduce this upper bound to quadratic, which is asymptotically worstcase optimal even for planar grid drawings [6].
4.1 Planar graphs
We present an algorithm that builds upon the wellknown shiftmethod by de Fraysseix, Pach and Pollack [8], which we outline in the following. We first recall the notion of canonical ordering for maximal planar graphs [8] used by the shiftmethod. Let be a maximal planar graph and let be a permutation of . Assume that edges , and form a face of , which we assume w.l.o.g. to be its outerface. For , let be the subgraph induced by and denote by the outerface of . Then, is a canonical ordering of if for each the following hold: (i) is biconnected, (ii) all neighbors of in are (consecutive) on , and (iii) if , then has at least one neighbor , with . A canonical ordering of a maximal planar graph always exists and can be computed in time [6].
The shiftmethod [8] is an incremental algorithm, which constructs a planar drawing of a maximal planar graph ; in the following, we refer to the lineartime variant by Chrobak and Payne [4]. Drawing has integer grid coordinates and fits in a grid of size . More precisely, based on a canonical order of , drawing is constructed as follows. Initially, vertices , and are placed at points , and . For , assume that a planar grid drawing of has been constructed in which edges of are drawn as straightline segments with slopes , except for the edge , which is drawn as a horizontal line segment (contour condition; see Fig. LABEL:fig:shiftmethod1). Also, for vertex has been associated with a socalled shiftset . For , and , it holds that , and . Let be the vertices of from left to right in , where and . Let also , with be the neighbors of from left to right along in . To avoid edgeoverlaps, the algorithm first translates each vertex in one unit to the right and each vertex in two units to the right; see Fig. LABEL:fig:shiftmethod2. Then, the algorithm places vertex at the intersection of the line of slope through with the line of slope through (which is a grid point, since by the contour condition the Manhattan distance between and is even) and sets .
While constructing drawing , it is also possible to compute a coloring of the edges of , which is known as Schnyder realizer in the literature [19, 27]. In particular, color blue, green and when a vertex with is placed, color edge blue, edge green and the remaining edges incident to in red, that is, with . It follows that all edges that appear in the contour of are either blue or green, which further implies that all faces of (and thus of ) are either bichromatic or trichromatic. Since vertices in the same shiftset are always translated by the same amount, the red edges are rigid, i.e., neither the slope nor the length of a red edge incident to in can change due to a shift required by the placement of a vertex with . Consider now an edge in and let be the angle formed by and the axis. The construction ensures that if is blue, then ; if is green, then ; if is red, then .
Our algorithm works as follows. We start by placing , and as in the original shiftmethod. For placing , with , our algorithm shifts the vertices of in three “shifting phases”. First, each vertex in with is shifted by units to the right (instead of a single unit, as in the original shiftmethod). In the second phase, each vertex in is shifted by units to the right, where is either or so to guarantee that the Manhattan distance between and is even. In the final phase, each vertex in is moved by units to the right^{1}^{1}1Note that although the second and the third shifting phases shift the relevant vertices by the same amount , we distinguish the two phases for clarity of presentation.; see Fig. 6. After all three shifting phases have been executed, we have the final placement for the vertices of in . We complete the construction of by placing vertex at the intersection of the line of slope through with the line of slope through , as in the original shiftmethod. Hence, the contour condition is maintained, assuming that the coordinates of are integer (a property which is formally proven in the following).
Observe that the first shifting phase implies that the horizontal distance between any two consecutive vertices and in with gets increased by one unit in , while in the original shiftmethod this would only be the case for and . In the second shifting phase, the choice of guarantees that if is placed at the intersection of the line of slope through with the line of slope through , then its position coincides with a grid point. This is due to the fact that an even Manhattan distance between and implies that the two aforementioned lines intersect at a grid point [8]. The choice of further implies that the horizontal distance between and gets increased by either one or two units in , while in the original shiftmethod the corresponding increment is always one unit. Finally, notice that the third translation phase does not affect the horizontal distances of the involved vertices that are on , as in the original shiftmethod.
Since the contour condition is maintained in the course of the construction, the planarity of is implied as in the original shiftmethod. Assuming that is a disklink drawing of (i.e., its edgevertex resoltuion is at least ), we prove in the following that: (i) the drawing produced by applying the three shifting phases on the vertices of has edgevertex resolution at least , and (ii) the newly introduced edges of incident to leave the edgevertex resolution of at least .
To prove (i), we establish that there is no edgedisk intersection in each individual face of after the three shifting phases have been applied. In other words, stretching some triangular face of does not introduce edgedisk intersections. To see this, consider a face in . Since is triangular, we denote the vertices delimiting by , and . Since is a valid disklink drawing of , it follows that there is no edgedisk intersection in the drawing of in . If face is not stretched in some of the three shifting phases, then clearly there is no edgedisk intersection in the drawing of in . Hence, we may assume that has been stretched. As already mentioned, is either bichromatic or trichromatic in the Schnyder realizer. We consider these two cases separately.

Case A. Assume that is bichromatic. Here, we further distinguish cases based on the color that appears twice in .

Case A.1. Assume that this color is red. We can easily conclude that has not been stretched, since the red edges are rigid, i.e., their length stay unchanged in the course of the algorithm.

Case A.2. Consider now the case in which the dominant color in is blue; the case in which this color is green is symmetric. We further assume w.l.o.g. that and are the blue edges of , which implies that is either red or green. Assume first that is red; refer to Fig. 7. Since red edges are rigid and we assumed that is stretched, it follows and are in the same shiftset, while is in a different shiftset from the one of and . Since and are blue, , while the fact that is red implies that ; note that in Fig. 7 we have also assumed w.l.o.g. that is below . In this case, since is to the left of and , it follows that is shifted by a smaller amount than and . Since the angle is increased by the shift while the length of remains unchanged, the distance of to is increased after the shifting. Thus, the edgevertex resolution is not decreased, as desired. To complete the case in which is bichromatic, we consider the case where is green; refer to Fig. 7. Assume, w.l.o.g., is above . Then, is also above , since and . Moreover, appears between and in the horizontal direction. Consider now vertex . The lowest point of a disk with radius centered at has coordinate greater than the coordinate of , and thus it cannot intersect edge . Similar arguments can be made about vertices and and their opposite edges, respectively, which completes the case in which is bichromatic.


Case B. To complete the proof of (i), we now consider the case in which is trichromatic. W.l.o.g., let be the topmost vertex of in and let and appear in this order in a counterclockwise traversal of starting from . There exist two cases to consider illustrated in Figs. 7 and 7. We focus on the former; the latter is treated similarly. Consider now vertex . The highest point of a disk with radius centered at has coordinate smaller than the coordinate of , and thus it cannot intersect edge . Similar arguments can be made about vertices and and their opposite edges, respectively, which completes the case in which is trichromatic.
It remains to show property (ii) regarding the newly introduced edges of that are incident to . By the first and the second shifting phases, it follows that the horizontal and vertical distance of each vertex , for , to any edge with is at least one, which implies that the actual edgevertex resolution is at least and thus more than , as desired.
To estimate the area required by disklink drawing
, we make use of an important property of Schnyder realizers, namely, that each monochromatic subgraph of induces a tree with vertices [27]. By the contour condition, is contained in an isosceles right triangle. Hence, to determine its area, it is enough to determine its width. Our modification of the shiftmethod elongates some edges, which were not elongated by the original method. In particular, when placing , the edges in the path from to in are elongated by exactly unit in the horizontal direction. Since after the placement of these edges connect vertices in shiftset , they are not further elongated, that is, they are elongated exactly once in the course of the algorithm. Furthermore, in the original shiftmethod the edge is elongated by one unit in the horizontal direction during the placement of , while in our construction it might be necessary to be elongated by an extra unit. To estimate the width of it is enough to estimate the additional width that is due to our modified shiftmethod. Towards this, we observe that we can charge the elongation of each of the aforementioned edges to the red edge that is incident to and to its right endvertex. Hence, the additional width that is due to our modified shiftmethod is at most , since the red subgraph of is a tree with exactly edges. Given that the width of the drawings produced by the original shiftmethod is at most , it follows that the width of the drawings of our algorithm is at most . We summarize this result in Theorem 8.Theorem 8.
Every vertex planar graph admits a planar disklink drawing on a grid of size .
Pseudocode for a lineartime implementation of the algorithm supporting Theorem 8 is provided in Fig. 8. The pseudocode is based on the lineartime implementation of the shiftmethod by Chrobak and Payne [4].
The shiftmethod can easily be implemented to run in quadratic time by updating the coordinates of all vertices contained in the shiftsets explicitly at every vertex addition. In the original work of de Fraysseix, Pach and Pollack [8] a rather involved approach is used to achieve a runtime of . Later, Chrobak and Payne described a lineartime algorithm whose key ingredient is to store only relative coordinates rather than absolute values. This method required them to change the definition of the shiftset. The proof of Theorem 8 uses their definition of shiftset which enables us to adapt their approach for our needs.
In the pseudocode we use, besides the already introduced notation, some more variables. For a vertex that is part of the contour, denotes the horizontal distance to its predecessor. Furthermore, the shiftsets are stored as a forest of trees induced by the red edges. For every vertex we store its link to the parent in the corresponding variable. The relative horizontal offset of a vertex to its parent is denoted by .
4.2 Outerplanar graphs
In this subsection, we turn our attention to outerplanar graphs. Recall that a graph is outerplanar if it admits a planar drawing in which all vertices belong to the outerface, or equivalently, if it excludes and as minors. Grid outerplanar drawings in small area motivated a rich body of literature (see [20] for references), with the current best area upper bound being , as shown by Frati et al. [20]. Applying Theorem 2 to this result would give us disklink drawings of outerplanar graphs in nearquadratic area. On the other hand, Theorem 8 implies that every outerplanar graph admits a disklink drawing on a grid of size . We present a construction that results in more compact disklink drawings of area; furthermore, a superlinear lower bound is discussed in Section 5. Here, a disklink drawing is outerplanar if replacing each disk with its center yields an outerplanar grid drawing.
Theorem 9.
Every vertex outerplanar graph admits an outerplanar disklink drawing on a grid of size .
Proof.
Let be an outerplanar graph and let be an outerplanar drawing of ; for each vertex of , defines a counterclockwise ordering of the edges incident to . Let be a leveling of whose levels correspond to the levels of a breadth first search (BFS) traversal of (starting at any vertex of ). When the BFS visits a vertex, it scans its edges in the counterclockwise order defined by . Felsner et al. [18] proved that is a proper leveling of and that a proper level outerplanar drawing of can be computed by placing the vertices within each level one next to the other from left to right based on the order they have been visited in the BFS; see Fig. 9. From now on, we assume that comes with the outerplanar embedding defined by .
Let be the rooted tree formed by the tree edges defined by the BFS traversal of . For each vertex of , we denote by the subgraph of induced by the vertices in the subtree of rooted at . Furthermore, we write , if is different from , it holds , and is to the right of in . Let be the children of in (in the left to right order they appear in ). We describe an algorithm that computes a grid drawing of such that for each vertex of , the subdrawing of has the following properties: (P.1) Drawing is a grid outerplanar drawing. (P.2) Any two vertices share neither the same coordinate nor the same coordinate in . (P.3) There is no free column and no free row in the minimumarea grid that contains . (P.4) For any edge , the bounding box of does not contain any other vertex in its interior. (P.5) Consider any two vertices and with . If , then is above and to the left of . If and , then is below and to the left of .
If is the root of , and hence is a grid drawing of . Also, by P.1, P.2 and P.3, is outerplanar on a grid of size . Finally, by P.4, it is immediate to see that we can replace each point representing a vertex in with a disk centered at that point, and obtain the desired disklink drawing of .
The algorithm traverses bottomup; let be the current node visited by the algorithm. If is a leaf, it is drawn at point and P.1P.5 trivially hold. If is an internal node of , we draw it at and we combine the drawings , , , (for which P.1P.5 hold) by suitably translating them. For , denote by and the width and the height of , respectively. We translate such that has coordinates , where
For a schematic illustration refer to Fig. 10; Fig. 10 illustrates a complete disklink drawing obtained by applying this construction.
We now prove that P.1P.5 hold for , assuming that P.1P.5 recursively hold for , , , . Our construction immediately guarantees P.2, P.3, and P.5. Concerning P.1, we first observe that is a grid drawing. Since the counterclockwise circular order of the edges around is preserved, the algorithm computes a drawing that maintains the embedding in the plane of . Thus, is outerplanar. It remains to prove P.4. Consider an edge of . We distinguish three cases: (i) either or is vertex , (ii) both and belong to , for some , and (iii) and belong to and , respectively, where and . Clearly, P.4 holds for Cases (i) and (ii); the former by construction, while the latter by recursion. It remains to consider Case (iii). W.l.o.g., assume that ; under this assumption, P.5 guarantees that is above and to the right of . Moreover, by outerplanarity, we know , as otherwise there would be another drawing , with , whose vertices do not belong to the outer face, which is impossible by P.1. Consider the rectangle having as topright corner and as bottomleft corner. Let be the lowest common ancestor in of and . Any vertex in the unique path of from to is strictly to the left of this rectangle by P.5. Any other vertex of in this region would not belong to the outer face, which is not possible by P.1. Similarly, any vertex in the unique path of from to is strictly above this rectangle by P.5, and any other vertex of in this region would not belong to the outer face, which is again not possible by P.1. This shows that P.4 holds and concludes the proof. ∎
4.3 A lower bound on the area of star graphs
In the traditional straightline drawing model, an vertex star admits a planar drawing on a grid of size , e.g., by placing the center of the star at , its th leaf at , where , and its th leaf at . We prove in the next theorem that disklink drawing of vertex stars require asymptotically more area.
Theorem 10.
Any disklink drawing of the vertex star requires a grid of size .
Proof.
For any , let be any disklink drawing of the vertex star , and denote by the vertex of with degree ; refer to Fig. 11 for an illustration. For ease of description, we assume that (after possibly scaling up the drawing by a factor of ) the edgevertex resolution of is at least (rather than ); this does not change the asymptotic area requirement of . Moreover, up to a translation of , we can assume w.l.o.g. that the center of the disk representing is at point .
By the pigeonhole principle, we know that in there exist vertices that are drawn in one of the four quadrants of the Cartesian system centered at , say w.l.o.g. the topright one. Denote by the set of these vertices (which by definition includes vertex ). Let be the minimum axisaligned rectangle that contains all vertices of . For each edge with , consider the halfline starting at point and containing edge , which intersects in two points, namely at and . Observe that each point lies on either the top or the right side of . Since no edge intersects a nonadjacent disk, it follows that any two points and along the top (right) side of have horizontal (vertical) distance at least . Since at least one of the two sides of contains at least such points, it follows that either the width or the height of the grid supporting , say the width, is at least .
Suppose, for a contradiction, that the grid supporting has size , which implies that its height is , for some constant that does not depend on . By the pigeonhole principle, again we know that there exist vertices whose disk centers share the same coordinate, which we denote by ; refer to Fig. 11 for an illustration. Let be such vertices and let be their corresponding coordinates such that , for any . Let , for . We know (by similar triangles, see Fig. 11) that
(3) 
where is the length of the vertical segment having and an internal point of edge as endpoints. By assumption the edgevertex resolution of is at least , thus , and it holds that
(4) 
On the other hand, the horizontal distance between any two consecutive vertices and is at least one, which implies that and therefore
(5) 
By a simple manipulation of Equations 4 and 5, we have
(6) 
Note that, as soon as (which holds because ), we have (as otherwise an edge would intersect a disk), which guarantees that Equation 6 is welldefined. We finally observe that the sum over all is upper bounded by the width of :
(7) 
Equation 7 implies that , which contradicts our assumption that the area of is . ∎
5 Discussion and Open Problems
We remark that our results are all proved via constructive techniques, and it is possible to show that each of them can be implemented to run in linear time in the number of edges of the graph. The only exception is Theorem 6, which requires a linear ordering of the vertices with minimum bandwidth. Determining the bandwidth of a graph is NPhard [22], even to approximate within a constant factor [13]; nonetheless there are classes of graphs for which the problem becomes tractable or it can be approximated (see [13, 17] for references). Our research raises several interesting questions.

The main problem is to establish nontrivial lower bounds for disklink drawings. We conjecture that every vertex star requires area, which would be a drastic improvement over Theorem 10.

Establishing improved area bounds for specific classes of planar graphs (e.g, seriesparallel, bounded degree) is a natural direction.

Also, one could extend the proposed model by allowing bends along the edges.
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