Journal of Graph Algorithms and Applications Drawing Graphs with Few Arcs 394 A. Schulz Drawing Graphs with Few Arcs

Let G = (V, E) be a planar graph. An arrangement of circular arcs is called a composite arc-drawing of G, if its 1-skeleton is isomorphic to G. Similarly, a composite segment-drawing is described by an arrangement of straight-line segments. We ask for the smallest possible ground set of arcs/segments for a composite arc/segment-drawing. We present algorithms for constructing composite arc-drawings with a small ground set for trees, series-parallel graphs, planar 3-trees and general planar graphs. In the case where G is a tree, we also introduce an algorithm that realizes the vertices of the composite drawing on a O(n 1.81) × n grid. For each of the graph classes we provide a lower bound for the maximal size of the arrangement's ground set.


Introduction
There exists a large number of design criteria for good drawings of planar graphs such as small area, good vertex and angular resolution, or a small number of edge crossings.All these measures assure that vertices and edges in a drawing are distinguishable for the observer.In this paper we propose a novel criterion for aesthetic and readable graph drawings.Our goal is to generate drawings that are easy to perceive by the viewer.When reading a drawing the human mind decomposes the received picture into geometric entities such as lines, segments, arcs, disks, circles, and so on.By interpreting the relationship between these entities an understanding of the drawing is obtained.We refer to the number of entities used in the drawing as its visual complexity.
Straight edges and the absence of crossings are desirable features for a drawing.A straight edge would be considered as one single entity, whereas, for example, a polygonal chain might be considered as a combination of several geometric entities.Something similar is true for edge crossings.If two edges cross, they introduce a new perceptional feature in the drawing -the crossing point.Therefore, a noncrossing straight-line drawing is formed by a combination of |V | + |E| geometric entities.In this paper we want to reduce the number of geometric entities even further.To make this possible, we group edges, such that they form a new entity.For example, if we are able to draw a path of the graph as a single straight-line segment in the drawing (with vertices in its interior), the visual complexity of the drawing is reduced.More formally, we define: Definition 1 (Composite drawing) Let A be an arrangement of simple geometric bounded 1d objects in the plane.The objects might be subdivided by placing additional vertices on them.Let G be the 1-skeleton of the subdivided arrangement.The arrangement A is called a composite drawing of G.If A contains only line segments it is called a composite segment-drawing, if A contains also circular arcs it is called a composite arc-drawing.The number of arcs/segments of A refers to the cardinality of the ground set of A.
Fig. 1 shows examples of composite arc-drawings.
Our motivation for the perception based approach stems partially from the work of the artist Mark Lombardi.Lombardi's visual art was focused on graph drawings of social networks within the political and financial sector [10].The drawings of Lombardi have a unique style.Maybe the most characteristic feature is the use of circular arcs to represent consecutive edges.These circular-arc paths kept the visual complexity of the drawings low.By aligning edges Lombardi enhanced his drawings with additional information.For example, these alignments were used to decode temporal or sequential dependencies of events represented by the vertices.
In this work, we focus on the combinatorial aspects of drawings with low visual complexity.As simple geometric objects for composite drawings we consider (straight-line) segments and circular arcs.Using straight-line segments graph class upper bounds lower bounds segments [6] arcs segments is the most natural way for drawing edges, but also circular arcs have been proposed as "edge shapes" before [1,5].In our understanding, a line segment is a degenerated circular arc, and by a suitable Möbius transformation these segments can be converted to proper circular arcs.We present bounds on the maximal number of arcs/segments necessary in a composite drawing.Our approach cannot handle edge crossings, since every crossing defines a subdivision of geometric objects and hence introduces an additional vertex in the composite drawing.Therefore we only study (noncrossing) drawings of planar graphs, such as trees, series-parallel graphs, and planar 3-trees.Moreover, all graphs that we consider are simple, which means that we forbid parallel edges and self-loops.The results of this paper are listed in Tab. 1.All lower bounds presented in this paper are due to the following simple lemma [6].
Lemma 1 Let G be a graph with N vertices of odd degree.Every composite arc-drawing or segment-drawing of G requires at least N/2 arcs.
Proof.In every odd degree vertex at least one arc/segment has an endpoint.
Hence we have at least N endpoints of arcs.Since the number of odd vertices is an even number the statement of the lemma follows.User studies comparing straight-line drawings with circular-arc drawing have been conducted only recently [12,16].Both studies showed that certain tasks are easier to carry out by the observer, when straight edges are used.On the other hand, users preferred the aesthetics of circular arc drawings over straight-line drawings in one of the studies [12].Note that these studies have not considered drawings with low visual complexity, but only drawings with circular arcs.The hypothesis that drawings with low visual complexity are indeed easier to perceive still needs to be checked empirically, which is work in progress.
Recently, so-called smooth orthogonal layouts have been studied [2,3].Instead of polyline edges, which are typically used in orthogonal layouts, chains of smooth circular arcs are proposed as edge shapes.The goal is to optimize the maximal edge complexity (number of circular arc pieces per edge) and not the visual complexity of the whole drawing.

Composite drawings of trees
Let T = (V, E) be a tree that we want to realize as a composite segment-drawing.Drawings with |E|/2 segments can be constructed by a greedy algorithm [6], which is optimal.

Grid drawings of trees with few arcs
In this subsection we show how to draw an unordered tree as a composite arcdrawing with few arcs and the additional constraint that all vertices lie on the Z 2 grid.Our objective is to obtain a drawing that uses few arcs but also requires a small grid.Note that the greedy algorithm in [6] yields an embedding on a grid exponential in |V |.
To obtain a drawing on a small grid we do not aim at drawings with the lowest visual complexity.We believe that both grid size, and visual complexity cannot be optimized at the same time.As an easy example, the reader might consider the realization of a simple cycle.Obviously this graph can be drawn with only one circle.However realizing a circle such that it contains many grid points is a highly nontrivial task.To our knowledge the best method uses a grid of size O(5 n/4 ) [13].

Heavy path decomposition
The drawing algorithm is based on a decomposition scheme for trees, called the heavy path decomposition [15], which works as follows.We root the tree T = (V, E) at some vertex r.Let u be a node of T , then T u denotes the subtree rooted at u, and N (u) denotes the size of this subtree.For every non-leaf u we select a child v, for which N (v) is maximal (with respect to the size of the subtrees of the other children).The edge (u, v) is called a heavy edge and all edges that are not heavy are called light edges.A maximal connected component of heavy edges is called a heavy path.The tree T decomposes into heavy paths and light edges.Note that every path in T to the root visits at most log |V | light edges.
For the drawing algorithm it is convenient to introduce the following definitions.We call the node on a heavy path that is closest to r its top node.The subtree rooted at the top node of a heavy path its called the heavy path subtree of this path.The light edge that links the top node of a heavy path P with its parent in T is called light parent edge of P .The depth of a heavy path P is defined as follows: If P is not incident to light parent edges of other heavy paths, it has depth one.Otherwise we obtain the depth of P by adding one to the maximal depth of a heavy path linked to P via its light parent edge.Note that the subtrees of heavy paths of a fixed depth are all disjoint.Figure 2 shows an example of a heavy path decomposition with annotated depth of the heavy paths.

Algorithm outline
The drawing algorithm works (high-level) as follows.We draw all subtrees of heavy paths with increasing order of their depth.Furthermore, we associate every subtree of a heavy path with an axis-aligned rectangle called its safe box.As an invariant the drawing of a subtree is exclusively contained inside its safe box, and the root of the subtree is placed on the top edge of its safe box, but not on its corners.For convenience we require that every safe box has width at least 3, in particular every leaf is placed inside a 3 × 1 safe box.When going from a depth k to a depth k + 1 subtree we arrange the drawings of the subtrees whose heavy paths have smaller depth to a new drawing (details will be given later).The algorithm terminates when the heavy path subtree with the largest depth has been drawn.
Let us explain the recursive step of the algorithm, that is, how to build the subtrees of the heavy paths (see Fig. 3(a) for an illustration).The heavy path is drawn as a single vertical segment.The only exception might be its edge (u, v) incident to the leaf v.Note that every subtree incident to u has to be a leaf as well, otherwise (u, v) would not be a heavy edge.Hence all k children of u are leaves, a node with this property is called a k-fork in the following.The children of u are placed on the line y = 0 and u is placed on (0, 1).In the case that k is even, we place the children of u symmetrically around the y-axis such that they have Two vertices are joined by an arc through u when they have the same absolute x-coordinate (see Fig. 3(a)).In case that k is odd we place the light edges as in the even case and realize the heavy edge (u, v) by extending the vertical segment that contains the remaining heavy path (see Fig. 3(b)).
Assume now that u is not a fork.All safe boxes of subtrees incident to u will be drawn, such that their roots lie on the same horizontal line which is one unit below u.Moreover, they will be distributed, such that two of them are connected by a single arc running through u.Note that if we have an odd number of light edges for u, one of the safe boxes does not have a sibling to pair with.In this case we draw the arc as if there would be a sibling (leaf) but we draw only the half of the arc that connects to v. The location of the safe boxes incident to u needs vertical space, which is determined by the safe box with the largest height.The smallest horizontal strip containing all safe boxes incident to u is called a row.The tree is constructed such that all of its rows are separated vertically by one unit.The child w following u on the heavy path is placed at the bottom boundary of the row directly below u.

Box displacement
We now discuss how to arrange the safe boxes within each row.Let u be a node on the heavy path P (not a leaf or fork) and let v 1 , v 2 , . . ., v k be the k children of u not on P .By recursion, the subtrees rooted at the v i s have already been drawn, so we have for every v i a safe box B i with width w i and height h i .Recall that v i is placed on the top edge of B i .We will arrange all safe boxes B i such that their top edges lie on a common horizontal line, the node v i has x-coordinate x i , and the node u is placed one unit above at x = 0. To draw multiple light edges with a single arc, we pair two children, say v i and v j , and connect both by an arc running through u.This implies that x i = −x j for every Figure 3: (a) A drawing of a heavy path's subtree.Rows are drawn shaded and the heavy path is drawn thick.(b) An example of a composite arc-drawing of a tree.The safe boxes used in the algorithm are indicated by dashed rectangles.such pair of vertices.We determine the location of the safe boxes by a greedy strategy (see Fig. 4).Let i be the distance from v i to the top left corner of B i , and similarly, let r i be the distance from v i to the top right corner of B i .We first orient all boxes such that i ≥ r i (it is valid to reflect the whole safe box including the drawing).Then we sort the boxes by i in increasing order, and finally we flip all boxes with an even index vertically, such that r i ≥ i .
Assume for now that k is an even number.We place the safe boxes in rounds.In round t we place the safe boxes B 2t−1 and B 2t , t = 1, 2, . . .k/2, and connect them by an arc passing through u.For convenience we introduce the following notation: If a box B i is placed left of the heavy edge, then c i := r i and b i := i , otherwise c i := i and b i := r i .This implies that b i ≥ c i for i ≤ k.
In the first round we place B 1 and B 2 .Without loss of generality we assume that c 1 ≤ c 2 (otherwise the strategy is symmetric).We place B 2 as close as possible to the line x = 0. Since no safe boxes have been placed before, we only have to avoid the heavy edge emanating from u. Hence, the safe box is placed such that x 2 = c 2 + 1. Next, we place B 1 .The location of x 1 is already determined since we have fixed x 2 .Note that B 1 is guaranteed to be to the left of the heavy edge emanating from u, since c 2 ≥ c 1 .Let S be vertical strip with smallest width centered at the y-axis that contains the safe boxes placed so far.In the following rounds we place the remaining safe boxes such that they are separated from S by one unit and update S after every round.
In case k is odd, only one safe box needs to be placed in the final round.We draw the final safe box on the left side, such that it is separated from S by one unit.When all safe boxes have been arranged we determine the width of the displacement ∆, that is the distance between the most extreme top corners.The only exception is when k = 1; in this case ∆ equals the width of the only safe box plus 2 as shown in Fig. 5. Lemma 2 Assume we carried out the box displacement for the safe boxes incident to some u on P by the greedy strategy as explained above.We have to our preprocessing step we know that (i) b i ≥ c i , and (ii) the safe boxes are sorted such that b i+1 ≥ b i for all i < k.As a consequence of (i) we have that Let us first check the case k = 1.To maintain the invariant, that the root does not lie on a corner of the safe box, we have to extend the width ∆ (see Fig. 5).We obtain ∆ = 2 + w 1 , which is smaller than 7/4w 1 for w 1 ≥ 3. Since we placed all leaves inside 3 × 1 boxes the statement of the lemma holds for this case.
Assume now that k is even.In every nonterminal round i the width of the strip S increases by 2(1+max{b 2i−1 , b 2i }+max{c 2i−1 , c 2i }).In the last round we determine the width of the displacement ∆ by extending the width of the current strip S by 2

JGAA, 19(1) 393-412 (2015) 401
Summing up all rounds yields The second to last inequality follows by the monotonicity of (b 1 , b 2 , . . ., b k ).For the last transition we have used the facts that X b + X c = i w i and due to the Finally, if k is odd and k = 1 we have to change the above estimation only marginally.For convenience, we assume that k is still an even number, but we place only k − 1 safe boxes.We obtain Since there are only k−1 safe boxes, we have to use the weaker bound b k−2 ≤ X b /2 for the last transition.2 Let t be the top node of a heavy path P .After carrying out the box displacements for all rows we can define the safe box for the subtree of P .Its width is determined by the row with the maximal displacement width.By construction, t lies on the top edge, but not on a corner of the new safe box.Fig. 4 shows an example of the greedy strategy.Lemma 3 By inductively laying out the safe boxes with the greedy strategy explained above, the heavy path decomposition yields a drawing where every vertex is placed on a O(n 1.81 ) × n grid.
Proof.In every inductive step we construct a drawing of a subtree and its safe box out of smaller safe boxes.Assume we have k such safe boxes B 1 , . . ., B k .Due to Lemma 2 the width of the new safe box is at most 7/4 k i=1 w i , since it might happen that all safe boxes are placed in one row.On the other hand, at least one box is placed in every row, and these rows are vertically separated by one unit.This shows that the height of the new safe box is at most m+ k i=1 h i , for m being the number of rows plus one.
The claim of the lemma follows by induction.We first discuss the height.When a small box contains only a single vertex, its height is one.When combining the small boxes to a new subtree, we obtain as new height m + k i=1 h i .This new subtree, however, has at least the vertices contained in the smaller safe boxes and the m vertices on its heavy path.Hence the height of its safe box is at most the number of its vertices.
For the width we notice that due to the heavy path decomposition the recursion depth is at most log n .By induction a subtree of a heavy path with depth d and n vertices is contained inside a safe box of width at most 3 • (7/4) d • n .Hence the whole tree is contained in a box of width 3n • (7/4) log n which is upper bounded by O(n 1.81 ). 2

Analysis
A node is called a heavy even-prefork (short hep), if its heavy edge child is a k-fork, with k even.Fig. 3(b) illustrates the definition.A charging scheme for the "saved edges" in forks and heps leads to the following lemma.
Lemma 4 Let T = (V, E) be a tree drawn as a composite arc-drawing with the algorithm based on the heavy path decomposition.Then the drawing uses at most 3|E|/4 arcs.
Proof.For technical reasons we introduce a virtual edge, from the root of the tree T to an imaginary father, such that every node has a father.This only increases the number of edges by 1.Let F k be the set of k-forks in T , and let E k denote the (k + 1) edges incident to a k-fork.We define E odd := k odd E k and E even := E \ E odd .The edges for all forks are disjoint and hence we have Finally, we denote by h the number of heps.For every subtree S we define a potential Φ(S) based on the composite drawing restricted to S. We set Φ(S) := 2(#arcs in S) − (#edges in S).
If the root s of S is a k-fork, then S is a star with center s, and we have If S is a single node (e.g., a leaf of T ) we have Φ(S) = 0. Assume now that s is neither a fork nor a leaf.Let S 1 , S 2 , . . ., S t be the subtrees rooted at its children.There are t − 1 light edges to children of s, which are drawn using t/2 arcs.Let e be the heavy edge that joins s with some S i .If s is not a hep, then e can be drawn for free, because the vertical segment that represents the heavy path (except maybe for the last edge) was already partially drawn in S i .Thus, e can be realized by extending this "heavy path part".On the other hand, if s is a hep, we have to draw e as a new arc.Let χ(S) be 1 if S is a hep, and 0 otherwise.We obtain This implies We now assign edges to a hep as follows: for every hep we select the two heavy edges in its subtree, one of the light edges of the fork, and the parent edge, which gives 4 edges in total.Note that the edges associated to a hep are neither assigned to another hep nor contained in a k-fork with odd k.As a consequence we have h ≤ |E even |/4.We apply our edge charging scheme to Eq. ( 1) and observe that since and therefore no more than 3|E|/4 arcs are used in the composite drawing.Note that we had introduced an additional edge to T .If we subtract this edge, we obtain that there are at most 3|E|/4 arcs in the composite drawing for T .
2 Note that the bound stated in Lemma 4 is tight for trees that are stars with subdivided edges as shown in Fig. 6.Combining Lemma 2, 3, 4 yields Theorem 1.

Theorem 1
The algorithm for realizing a tree G = (V, E) as composite arcdrawings uses at most 3|E|/4 arcs.The computed drawing realizes all vertices on a O(n 1.81 ) × n grid, for n = |V |.
3 Composite drawings of serial-parallel graphs and planar 3-trees In this section we study composite arc-drawings of series-parallel graphs and planar 3-trees.We start with series-parallel graphs.
A graph G = (V, E) is a series-parallel graph if it can be built from a sequence of serial or parallel combination steps [7].We do not give the details of this standard definition but use the following alternative definition instead.A graph is a series-parallel graph if it can be decomposed into a sequence of paths E 1 , E 2 , . . .E k , such that (1) the endpoints for every path not E 1 lie both on some path E j with smaller index (the path between the two endpoints on E j is called a nested interval ), (2) no interior point of a path is contained in a path with smaller index, and (3) all nested intervals are either disjoint or contain each other [7].Such a decomposition is called a nested open ear-decomposition.
Theorem 2 Let G = (V, E) be a series-parallel graph.Based on a nested open ear-decomposition we can obtain a composite arc-drawing with at most (|E| + 1)/2 arcs.For every n there is a series-parallel graph G = (V, E) with more than n vertices, whose composite segment-drawings need at least (|E| + 2)/4 segments.
Proof.Let E 1 , E 2 , . . .E k be a nested open ear-decomposition of G.As noted by Miller and Ramachandran [11], we have k = |E| − |V | + 2. We first draw E 1 as single segment and then draw the other paths in increasing order as circular arcs.Suppose that we draw the path E i with endpoints on E j with j < i.By construction, E j has been realized as a circular arc.Let E j be a copy of the part of E j that lies between the two endpoints of E i .We can draw E i "on top" of E j by slightly increasing the radius of E j .The angle of the tangents of E i and E j at their intersection can be made arbitrarily small.Hence, when executing the drawing process, we can assume that every circular arc has small curvature.Since all paths are "nested" we can finish the drawing without introducing an edge crossing.Fig. 7(a) shows a drawing obtained by the algorithm.
Any (simple) series-parallel graph has |E| ≤ 2|V | − 3 edges.The drawing uses exactly k arcs.This gives For the lower bound consider a graph as depicted in Fig. 7(b).More generally, we consider for some odd n > 3 the nested open ear-decomposition given by In such an n-vertex graph only v 2 has even degree and thus we have n − 1 = |E|/2 + 1 odd degree vertices.Lemma 1 implies that such a graph needs at least |E|/4 + 1/2 arcs in any composite arc-drawing.2 The next class of graphs we consider are the planar 3-trees.A planar 3tree is a triangulation that can be defined recursively as follows: Suppose G = ({v 1 , . . ., v n }, E) is a triangulation, we can pick one of its faces, say it is spanned by the vertices v i , v j , v k and add a new vertex u inside this face together with the three edges connecting v i , v j , v k with u.By this we remove one face and introduce 3 new faces.This operation is called a stacking operation.Any graph that can be generated from a triangle by a sequence of stacking operations is called a planar 3-tree.We say that a planar 3-tree is a k-fan if it has k + 3 vertices and it contains the triangle v 1 , v 2 , v 3 and for every 4 , and (v i , v i−1 ).
To develop an algorithm for a composite arc-drawing we first introduce a crucial lemma.For the lemma we need the following definitions.A triangle is called spherical if its edges are circular arcs that do not intersect.If furthermore every triangle corner is incident to an angle (measured between tangents) larger than zero and less than π, we call the spherical triangle nonreflex.We say a vertex v inside a spherical triangle S is spherically visible from a point c, if there exists a circular arc connecting c and v that lies entirely in S (see Fig. 8(a)).
Lemma 5 Let S be a nonreflex spherical triangle and let u be a point in the interior of S. Then u is spherically visible from every of the three corners of S. Furthermore, the arc that witnesses the spherical visibility and the boundary arcs of the corresponding corner c have all a distinct tangent at c. Proof.Let the corners of S be v 1 , v 2 , v 3 .We prove the statement for the corner v 1 .Let f be the Möbius transformation that maps the arcs v 1 v 2 and v 1 v 3 to straight-line segments.Clearly, u := f (u) lies inside S := f (S).We set v i := f (v i ).Let s be the ray that starts in v 1 and is pointed towards u .Since Möbius transformations are conformal, the angles at v 2 and v 3 in S are both less than π.It follows that s hits first the vertex u and then the boundary of S without reentering.Therefore, the Möbius function f −1 maps the segment v 1 u to a circular arc that witnesses the spherical visibility of u in S. Clearly, the tangents of v 1 v 2 , v 1 v 3 , and f −1 (s) are all distinct in v 1 because f and f −1 are conformal.Fig. 8 shows an example of a triangle S and its image S . 2 Lemma 6 Let G be a k-fan with outer face f 0 , and let S be a nonreflex spherical triangle.Then G can be drawn with k + 4 circular arcs such that the boundary of S realizes f 0 .Assume now that G is a 2-fan.We extend the arc ending at v 4 (without changing the curvature), such that it reaches inside the spherical triangle spanned by v 1 , v 2 , v 4 .Let the endpoint of the extended arc be v 5 .We can interpolate in between the two arcs between v 1 and v 2 such that we get a circular arc connecting v 1 and v 2 via v 5 .The new arc does not introduce any crossings.See Fig. 8(c) for an illustration.By repeating this argument, we can draw every k-fan with k + 4 arcs. 2 Theorem 3 Every planar 3-tree G = (V, E) can be drawn with 11|E|/18 + 3 arcs as a composite arc-drawing.For every number n, there is a planar 3-tree G = (V, E) with more than n vertices, whose composite arc-drawings require at least |E|/6 arcs.
Proof.Note that we can naturally recurse on a planar 3-tree since when the first vertex v 4 is stacked on the face v 1 v 2 v 3 , the graphs contained in the three interior triangles are planar 3-trees as well.For the drawing algorithm we assume that there were at least two stacking operations, otherwise the bound of the lemma follows directly.Let G f the subgraph of G that is isomorphic to a k-fan and that includes the boundary face, such that k ≥ 2 (see the solid-edge subgraph in Fig. 9).We draw G f as discussed in Lemma 6 including all induced 1-fans of G that would lie inside faces of G f .For every 1-fan we need 2 arcs.This implies that in the worst case there is such a 1-fan for every face in G f , except for v 1 , v 2 , v k+3 .Therefore we have 2k 1-fans, contributing a total of 4k arcs.
The k-fan requires k + 1 arcs for the interior edges.Thus we have 5k + 1 arcs for the 9k interior edges.This shows that the ratio between interior arcs and edges is at most 11/18 (recall that k ≥ 2).For all faces of G f that are not faces of G we still have to fill in a subgraph of G.These subgraphs (including its face in G f ) form planar 3-trees as well.We draw these 3-trees (recursively) with the above strategy.To do so, we draw the boundary face, which is an interior face of G f , as it was drawn for G f .By the above analysis we "save" for every subgraph a ratio of 11/18 of the interior edges.The asserted bound of 11|E|/18 + 3 follows.
The lower bound is due to Lemma 1, since an arbitrarily large planar 3tree with odd degree vertices only can be easily constructed (see Fig. 9 for an example). 2 Figure 9: A planar 3-tree with odd degree vertices only.

Composite drawings of 3-connected planar graphs
Let G = (V, E) be a planar 3-connected graph with n vertices.We order the vertices of G with respect to some canonical order as defined by Kant [9].In particular, let • G k is 2-connected and internally 3-connected, • for each k ∈ {2, . . ., m − 1} either V k = {z} where z ∈ C k and z has a neighbor in j>k V j , or V k = {z 1 , . . .z t } where each z i has a neighbor triangulation G = (V, E) with more than n vertices whose composite segmentdrawings need at least |E|/6 + 1 segments.
Proof.As a first step we count the number of arcs used in the drawing.We introduced three types of arcs.(i) Every path π i will be realized as a single arc.Such an arc will be added m − 2 times (for every transition . (ii) When processing a singleton set V i = {u} we might draw additional arcs for edges (u, u ) that are not on π i .We call these arcs short arcs and denote the number of short arcs by s. (iii) For the boundary face we need two arcs.In total we have thus m + s arcs.
For the further analysis we introduce a function Φ who assigns every graph G i a potential.Let the number of edges in G i that are realized as short edges in the drawing be s i , and the number of interior faces in G i be f i .Then we define Φ(G i ) := f i − s i .Although we used the reversed canonical order in the drawing algorithm, we consider the forward canonical order for the analysis.Clearly, we have Φ(G 1 ) = 0.When we add a vertex set in the canonical order the potential changes.In case we add a singleton, we introduce a number of faces, say f new , but we also add f new − 1 edges which will be drawn as short arcs.Hence, we get Φ(G i ) = Φ(G i−1 )+1.In case our set is not a singleton set we add only one face, and no short arc edge.Again we have Φ(G i ) = Φ(G i−1 ) + 1.As a consequence Φ(G) = m − 1, and therefore f − s = m − 1, for f being the number of interior faces in G.
The number of arcs can now be expressed as m+s = f +1.The ratio between arcs and edges is maximized when a graph with |E| edges has a maximal number of vertices, that is, when G 2 Note that we have freedom how to place the vertices horizontally.The algorithm can be easily updated such that all vertices have distance at least one while keeping the largest distance between any two points at most n.From this perspective the drawing has a good vertex resolution.

Future work
In this paper we presented the first algorithms for composite drawings.For all graph classes except for trees there is a gap between the lower and upper bound on the number of necessary arcs.We are interested in tightening these gaps, but we think that new methods are required for a substantial improvement.
This paper concentrates on the combinatorial question, i.e., how small can the visual complexity be.On the other hand, drawings with very low visual complexity might violate other criteria for readable drawings.We addressed this issue in Theorem 1 by combining classical graph drawing criteria (grid size) with low visual complexity.We would like to extend this result for more complicated graph classes in order to construct more readable drawings with low visual complexity.
It is ongoing research to evaluate with empirical user studies our hypothesis that a graph with low visual complexity is easier to percept by the viewer.Our hope is that we can show that drawings with small visual complexity are easier to memorize and we think this might be especially applicable to drawings of graphs with a small number of vertices.
Finally, we would like to point out that we are interested in small decompositions of planar graphs into edge-disjoint simple paths.This graph-theoretic question might yield better lower bounds.Although this problem seems elementary, only partial results are known.If the graph is a triangulation, it can be decomposed into edge-disjoint simple paths that all have exactly three edges [8].The same is true for cubic bridge-less graphs [4].We would like to see a similar bound for general planar 3-connected graphs.

Figure 1 :
Figure 1: A drawing with low visual complexity of the graph of the dodecahedron (a).The drawing uses 10 circular arcs, which is the best possible.A drawing of the icosahedron graph that has not the lowest possible visual complexity (b).

Figure 2 :
Figure2: A tree with its heavy paths (indicated by the thick edges).The numbers refer to the depth of the heavy paths.

Figure 4 :
Figure 4: A snapshot during the execution of the greedy strategy for the box displacement.The safe boxes up to B 4 have already been placed.When placing the last box B 5 we avoid the restricted strip S. The boundaries of the strip S after round 1 and 2 are drawn as dashed lines.

Figure 5 :
Figure 5: The special case where u has only one child connected by a light edge.

Figure 6 :
Figure 6: An example where the analysis of Lemma 4 is tight.

Figure 7 :
Figure 7: (a) A composite arc-drawing of a series-parallel graph, obtained by the method explained in the proof of Theorem 2. (b) A series-parallel graph with only one even-degree vertex.

Figure 8 :
Figure 8: (a) A spherical triangle spanned by v 1 , v 2 , v 3 with interior point u.(b) The image of the same spherical triangle under a Möbius transformation that turns two boundary arcs into straight-lines.(c) Construction in the proof of Lemma 6.

Figure 11 :
Figure 11: The graph of the cube drawn with the technique based on the reverse canonical order of the vertices.The faces C k and the sets V k are highlighted.The perturbation of the straight-line segments was not necessary in this example.

Table 1 :
Combinatorial results obtained in the paper.The lower bounds in the third and fifth row are presented as slightly simplified expressions.

2
Related Work.Dujmović et al. [6] studied the complexity of composite segment-drawings.They presented their results in a slightly different form, namely, the bounds on the number of segments are expressed in terms of |V |, instead in terms of |E|.We are however convinced that a bound in terms of |E| gives a more universal expression since a graph with fewer edges tends to require fewer segments or arcs.The results of Dujmović et al. are presented in Tab. 1.Our results for composite arc-drawings are an improvement over the (straight-line segment) bounds of Dujmović et al.. None of the drawings of Dujmović et al. fulfilled additional aesthetic quality criteria.In fact, they stated the problem of designing algorithms with small area as an open problem.From this perspective, Theorem 1 gives the first algorithm that constructs composite drawings on a small polynomial grid.