Journal of Graph Algorithms and Applications 1-bend Orthogonal Partial Edge Drawing

Recently, a new layout style to avoid edge crossings in straight-line drawings of non-planar graphs received attention. In a Partial Edge Drawing (PED), the middle part of each segment representing an edge is dropped and the two remaining parts, called stubs, are not crossed. To help the user inferring the position of the two end-vertices of each edge, additional properties like symmetry and homogeneity are ensured in a PED. In this paper we explore this approach with respect to orthogonal drawings-a central concept in graph drawing. In particular, we focus on orthogonal drawings with one bend per edge, i.e., 1-bend drawings, and we define a new model called 1-bend Orthogonal Partial Edge Drawing, or simply 1-bend OPED. Similarly to the straight-line case, we study those graphs that admit 1-bend OPEDs when homogeneity and symmetry are required, where these two properties are defined so to support readability and avoid ambiguities. According to this new model, we show that every graph that admits a 1-bend drawing also admits a 1-bend OPED as well as 1-bend homogeneous orthogonal PED, i.e., a 1-bend HOPED. Furthermore , we prove that all graphs with maximum degree 3 admit a 1-bend symmetric and homogeneous orthogonal PED, i.e., a 1-bend SHOPED. Concerning graphs with maximum degree 4, we prove that the 2-circulant graphs that admit a 1-bend drawing also admit a 1-bend SHOPED, while there is a graph with maximum degree 4 that does not admit such a representation.


Introduction
In the field of graph drawing, one of the main objectives for visualization of graphs is to avoid visual clutter for readability.To achieve this, one aspect is to avoid crossing edges, which seriously affect the comprehensibility of a graph drawing [21,25].However, avoiding all crossings is not always possible and a lot of research was focused on minimizing the number of crossings in a drawing [5,7,18].Figure 1: (a) A 1-bend drawing of the 4-dim.cube where one vertex has been removed, taken from [14].(b) A 1-bend OPED drawing of the same graph.Omitted segments are drawn by thin dotted lines.
Other papers study which non-planar graphs can be drawn such that the complexity of the edge crossings in the drawing is controlled.In the case of k-planar drawings, each edge is crossed at most k times (see, e.g., [9,11,16,20]), while in k-quasi planar drawings, no k pairwise crossing edges exist (see, e.g., [1,8,23]), and finally in large angle crossing drawings, any two crossing edges form a large angle (refer to [12]).Besides this theoretical research, many proposals come from information visualization, like edge bundling (see [26] for a survey) or confluent drawings (see, e.g., [10,17]).

Previous Work.
Recently a new layout style was investigated to avoid crossings in a straight-line drawing of a non-planar graph, the resulting drawings were called Partial Edge Drawings (PEDs) [3,4].The idea of PEDs is to subdivide each segment representing an edge into 3 parts and to drop the middle part, relying on the remaining parts called stubs.The beginning of PEDs is dated in 1995, when Becker et al. [2] used stubs the first time to focus on visualizing networked data and not the network itself.They wanted to extract information about overloaded connections between switches and drew the width of edges relative to the severeness of overload, producing a drawing where the underlying network was barely visible.The solution was to draw just a fraction (roughly 10%) of each edge, transforming the drawing into a PED.Later Burch et al. [6] investigated the usefulness of PEDs in a user study regarding directed graphs, observing that certain tasks were answered faster and with less errors when the graphs were drawn as PEDs compared to the straight-line model.A similar approach was re-invented by Rusu et al. [22] by just breaking the edges at their crossing points, inspired by the principles of Gestalt [19], that people intuitively try to complete gaps in drawings.
The first formal definition of the PED model (for straight-line drawings) is given in [4], where different types of PEDs are considered.In its easiest form, a PED only requires the stubs to be crossing free.A PED is homogeneous (HPED), if the ratio of the stub length over the edge length is the same over all the edges, and it is symmetric (SPED), if the two stubs of each edge have the same length.In [3], Bruckdorfer et al. concentrated on SHPEDs, where the length of each stub is a quarter of the total edge length, and proved the existence of SHPEDs for some complete bipartite graphs, k-circulant graphs and graphs with bandwidth k, as well as the non-existence of a SHPED for complete graphs with more than 196 vertices.
Contribution.In this paper we extend PEDs to orthogonal drawings in two dimensions with exactly one bend per edge, called 1-bend drawings.Indeed, orthogonal drawing is a central concept in graph drawing (see [7,18] as a reference), and we see as a natural question to ask how the techniques and results for PEDs carry over to orthogonal drawings.Also, restricting to 1-bend drawings is the first step before extending the model to orthogonal drawings with more than one bend per edge.In a 1-bend drawing vertices are represented as points with integer coordinates and edges as chains of two orthogonal (axis-aligned) segments, hence, each vertex can have degree at most 4, (∆ G ≤ 4).A characterization of the graphs that can be drawn orthogonally with one bend per edge has been recently presented by Felsner, Kaufmann and Valtr in [14], where they adopted the general position model, i.e., no two points can share a coordinate.Indeed, these drawings are also bend minimal, since each edge is represented exactly by one segment parallel to each coordinate axis.In this paper we also adopt the general position model.
Observe that we cannot directly extend the SHPED model from straightline drawings to 1-bend drawings, since bent edges might lead to ambiguous interpretations.In particular, it would be unclear how long one should follow a vertical stub or a horizontal stub to reach the bend, see Figure 2. Therefore, we introduce the new model 1-bend Orthogonal Partial Edge Drawing, or simply 1-bend OPED, where for every edge in a 1-bend drawing we erase the longer segment (we actually draw it as a thin dotted segment).Figure 1(a) illustrates a 1-bend drawing of the 4-dimensional hypercube, while Figure 1(b) illustrates a 1-bend OPED of the same graph.Similarly as for straight-line PEDs, we define two further properties, homogeneity and symmetry.Formal definitions for 1bend homogeneous orthogonal PEDs (1-bend HOPED) and for 1-bend symmetric and homogeneous orthogonal PEDs (1-bend SHOPED) are given in Section 2.
The remainder of the paper is organized as follows.Preliminaries and formal definitions are given in Section 2. We present the main results of the paper in Section 3 to Section 5.In particular, in Section 3 we show that every graph that admits a 1-bend drawing also admits a 1-bend OPED as well as a 1-bend HOPED.Furthermore, in Section 4 we prove that all graphs with maximum degree 3 admit a 1-bend SHOPED.In Section 5 we consider graphs with maximum degree 4 and prove that the 2-circulant graphs that admit a 1-bend drawing also admit a 1-bend SHOPED, while there is a graph with maximum degree 4 that does not admit a 1-bend SHOPED.Conclusions and future work are discussed in Section 6.

Preliminaries and Definitions
Felsner et al. [14] showed that any graph G = (V, E) with ∆ G ≤ 4 has a  of this density condition for each induced subgraph of G comes from the clear observation that in any 1-bend drawing there must be a topmost vertex without any vertex towards the top, a bottommost vertex without any vertex towards the bottom, and similarly for the left and right directions.Notice that, 4-regular graphs cannot guarantee such a condition, hence, Felsner et al. [14] placed one vertex at a point at ∞. Since we consider only the Euclidean plane, we will remove this vertex from the graph.Let Γ be a 1-bend drawing of a graph G.The bounding box R of Γ is the minimum axis-aligned rectangle containing the drawing.Let v be a vertex of G.We denote the four possible anchor points for an edge incident to v by north, south, west and east ports of v. Also, we denote the x-and y-coordinates of v in Γ by x(v) and y(v), respectively.Let e be an edge of G, the segments parallel to the x-axis and to the y-axis in the chain of segments representing e in Γ are called horizontal and vertical segments, respectively, and denoted by e h and e v , respectively.The length of a segment s is denoted by |s|.
Let s v e ⊆ e v , respectively s h e ⊆ e h be subsegments (the stubs of e), so that (e v ∪e h )−(s v e ∪s h e ) is connected.In what follows we introduce three variations of our model for orthogonal partial edge drawings.Each variation is defined as an orthogonal 1-bend drawing (short: 1-bend drawing) in general position, where for every edge e, the two segments e v and e h are replaced by their stubs s v e and s h e (defined above) and no two stubs cross.Furthermore, instead of completely erasing the non-drawn parts, we draw them by thin dotted segments, which help the user to follow the edge correctly.This is just a visualization tool to support the reader, since the graph extracted from the drawing is already unique.
We define a 1-bend Orthogonal Partial Edge Drawing, or simply 1-bend OPED, as a 1-bend drawing in general position where for every edge e, we remove the longer segment and the remaining (shorter) segments do not cross.That is, if |e h | > |e v |, then we have s h e = ∅ and s v e = e v , and otherwise in case of |e h | ≤ |e v |, we have s h e = e h and s v e = ∅.Since we adopt the general position model, each vertical (horizontal) stub uniquely determines a horizontal (vertical) line on which we can find its end-vertex.
An illustration of the 1-bend OPED is given in Figure 3(b).
A 1-bend HOPED is a 1-bend drawing in general position where half of each edge is dropped, while the shorter segment is always entirely drawn.More precisely |s v e |+|s h e | = (|e v |+|e h |)/2 is true for every edge e of G and, in addition, we always draw the shorter segment of an edge completely (as in the 1-bend OPED) and draw on the remaining segment (starting from the bend point or from the end-vertex arbitrarily) as long as we need to reach half of the total edge length (see Figure 3(c)).Therefore, the two stubs may be continuous, forming a unique bent stub.
A 1-bend SHOPED is a 1-bend drawing in general position where we symmetrically remove half of the horizontal segment and half of the vertical segment for each edge, i.e. 2|s v e | = |e v | and 2|s h e | = |e h | for every edge e of G.The dropped parts of e v and e h are connected by definition, i.e., they meet at the bend point and the stubs s v e and s h e are incident to the end vertices of e.An illustration of the 1-bend SHOPED is given in Figure 3(d).

1-bend OPEDs and 1-bend HOPEDs
First we consider 1-bend OPEDs and begin with a simple observation.Every graph that admits a 1-bend drawing also admits a 1-bend OPED: it is sufficient to drop all the horizontal (vertical) segments of a 1-bend drawing after stretching the drawing horizontally (vertically) by a factor equal to the length of the longest vertical (horizontal) segment.If homogeneity is required, the technique we use is slightly more difficult and it is presented in the following.
In the remainder of this section we consider 1-bend HOPEDs.We prove that every graph that admits a 1-bend drawing also admits a 1-bend HOPED.Namely, let G be a graph and Γ be a 1-bend drawing of G produced by the algorithm described in [14], we modify the x-coordinates of the vertices of Γ so that for each edge e of G, e v is always shorter than e h , and therefore s v e = e v .Then, we draw s h e always from right to left, as much as we need to reach half of the edge length.Observe that, s v e and s h e might be continuous, forming a unique bent stub.If necessary, we further modify the x-coordinates of the vertices of Γ so that each crossing involving s h e is repaired.
Theorem 1 Every n-vertex graph G with ∆ G ≤ 4 that admits a 1-bend drawing also admits a 1-bend HOPED.
Proof: We start by constructing a 1-bend drawing Γ of G = (V, E) by using the technique in [14].Recall that we adopt the general position model and consider the total ordering of the edges of G defined as follows: e = (u, v) ≺ e = (w, z), if and only if max{x(u), x(v)} < max{x(w), x(z)} and v = z.The only incomparable edges are those with common rightmost end-vertex.Then we break the ties as follows.Let e = (u, v) and e = (w, v) be two edges with common rightmost end-vertex, e ≺ e , if and only if y(u) < y(w).We scan the edges following the order ≺ described above.Namely, let e = (u, v) ∈ E and assume v is the end-vertex of e placed at the point with largest x-coordinate.Consider the point p v (if any) defined by the crossing involving e h with largest JGAA, 18(1) 111-131 (2014) 117 x-coordinate, denoted as x pv .Then, we shift v and all the vertices to the right of v in the following way.For all w ∈ V with x(w) ≥ x(v): x(w) = x(w) + max{δ, δ }, (iff max{δ, δ } > 0) , where Clearly δ > 0 if the vertical segment is at least as long as the horizontal segment of an edge.Thus, the shift operation ensures that the vertical segment is always shorter than the horizontal one (i.e., the vertical segment will be drawn entirely).Similarly δ > 0 if the horizontal distance between u and p v is at least as large as the horizontal distance between p v and v plus the length of the vertical segment.In this case, drawing the vertical stub entirely and the horizontal stub from right to left as much as necessary to reach half of the total edge length would cross p v .Hence, the shift operation ensures that all the crossings involving e h lie on its non-drawn part.It is easy to see that each crossing has been considered exactly once (due to the total order of the edges) and it has been repaired, i.e., it lies on the non-drawn part of the involved horizontal segment.Also, each shift operation does not affect previously repaired crossings.
Conversely, in Section 5 we show that, if symmetry is also required, constructing a 1-bend SHOPED is not always possible.We first construct a 1-bend drawing Γ of the subgraph G = (V = V \ {s, t}, E(V )) and then add s and t in a proper way.

Construction 1.
We assign to vertex v i the coordinates x(v i ) = i and y(v i ) = i, i = 2, . . ., n − 1.We orient the edges e = (v i , v j ) so that e goes from v i to v j , when 2 ≤ i < j ≤ n − 1.Notice that there are only two possible shapes for the edges in Γ .Namely, let e = (v i , v j ), 2 ≤ i < j ≤ n − 1, be a directed edge, it can either leave the east port of v i and enter the south port of v j or leave the north port of v i and enter the west port of v j .In the first case we call e a blue edge, while in the second case we say that e is a red edge.We denote by E B (respectively E R ) the set of blue (respectively red) edges.Due to the properties of an st-numbering, each vertex has at most two incoming edges and at most two outgoing edges.Furthermore, v 2 has only one incoming edge from s and v n−1 has only one outgoing edge to t.We want to find a 2-coloring of the edges E(V ) = E R ∪ E B such that the following two properties hold: 1. ∀v ∈ V , the (at most) two incoming edges receive different colors.
2. ∀v ∈ V , the (at most) two outgoing edges receive different colors.
To this aim, we construct the following undirected graph G * from G .For each vertex v in G there will be two vertices v − and v + in G * .For each edge e = (w, z) in G (oriented from w to z) there will be an edge e * = (w + , z − ) in G * .See Figure 4(b) for an illustration.Thus, G * is a (possibly not connected) bipartite graph (clearly there are no cycles with odd length) with maximum degree 2. Thus, each connected component is either a path or a cycle.It follows that the edges of G * can be colored with two colors in a straightforward way.Namely, let C be a component of G * with m C edges, we define a total ordering of the edges of C, i.e., e 1 ≺ e 2 ≺ • • • ≺ e m C .If C is a path such an ordering is directly defined by the order of its edges along the path (rooted at an arbitrary end-vertex).If C is a cycle, we simply choose an arbitrary edge to be the first one (e 1 ) and remove it from C, the rest of the order is defined by the remaining path.Finally, we color the edges as follows, e i receives color c ei = i mod 2, i = 1, . . ., m C .Since there is a clear one-to-one mapping between edges in G and edges in G * , we can directly color the edges of G with the colors assigned in G * .Let e be an edge of G and let e * be the related edge in G * , we assign to e the red color if c e * = 0 and the blue color if c e * = 1.
We prove now that such a coloring of the edges of G respects the properties 1 and 2 defined above.Let e and e be two incoming (outgoing) edges with respect to the same vertex v.By construction they will belong to the same component C of G * .If C is a path, then e and e will always appear consecutive in any possible order of the edges of C, thus, they will be assigned different colors.If C is a cycle, they may not be consecutive only in such an order where e is the first (last) one and e is the last (first) one.In this case, since m C is even, they will again receive two different colors.
Γ is now defined and we only need to place s and t to construct Γ.

Construction 2. (adding s and t)
We recall that s and t are connected by an edge, v 2 has coordinates (2, 2) and just one incoming edge from s, as well as v n−1 has coordinates (n − 1, n − 1) and just one outgoing edge to t.Hence, they can be easily connected to s and t, respectively, without causing crossings.Let v i , 2 < i < n, be a possible third vertex connected to s and let v j , 1 < j < n − 1, be a possible third vertex connected to t.We can skip the following consideration for v i (respectively v j ), if s (respectively t) has degree 2. We consider the two free ports of v i and we can assume they are two consecutive ports.If not, we can just toggle the color of one of the two edges incident to v i to match this situation (the colors of the edges in the same component of this edge in G * must be toggled accordingly).Thus, either the north or the south port of v i is free, as well as either the west or the east port of v i is free.We choose which port of v i to use after considering the free ports of v j .Consider the two free ports of v j .Either one between the east or the west port is free or, if both are occupied, then both the north and the south port will be free.In total we have 4 possible cases.
1.If the east port of v j is free, then we set x(t) = x(v j ) + 0.5 and y(t) = n.
While if the west port of v j is free, x(t) = x(v j ) − 0.5 and y(t) = n.Also, we can always assign to s the x-coordinate x(s) = 1.(a) If the north port of v i is free, then we set y(s)=y(v i ) + 0.5, see Figure 5(a).
(b) If the south port of v i is free, then we set y(s) = y(v i ) − 0.5.
2. If the north port of v j is free, then we set x(t) = n and y(t) = y(v j ) + 0.5.While if the south port of v j is free, x(t) = n and y(t) = y(v j ) − 0.5.Also, we can always assign to s the y-coordinate y(s) = 1.
(a) If the east port of v i is free, then we set x(s)=x(v i ) + 0.5, see Notice that, before adding s and t, the vertices were placed in general position, thus there could not be overlaps among edges and vertices.After adding s and t, this property is still maintained due to the introduced fractional coordinates (the grid unit must be halved to get integer coordinates).An example of a 1-bend drawing constructed with this technique is presented in Figure 4(a).
Finally, we observe that constructing an st-numbering of G takes O(n + m) time [13], as well as placing vertices (including s and t), constructing G * and coloring its edges.Thus, since m ≤ 1.5n, the algorithm runs in O(n) time.
Theorem 2 Every biconnected n-vertex graph G with ∆ G ≤ 3 admits a 1-bend SHOPED.Furthermore, such a drawing can be constructed in O(n) time.
Proof: Let Γ be a 1-bend drawing of G constructed by Construction 1 and Construction 2. We adopt the notation used in the proof of Lemma 1.Consider again the subgraph G = (V = V \ {s, t}, E(V )) and the induced subdrawing Γ .In Γ a red edge can be crossed only by red edges, as well as a blue edge can be crossed only by blue edges.Indeed, red edges are all above the diagonal formed by the vertices, while blue edges are all below this diagonal.If a crossing is caused by two red edges, it can be repaired by shifting the rightmost endpoint of the horizontal segment involved in the crossing, so that such a crossing will lie (in a SHOPED) on the non-drawn part of this horizontal segment.In a similar way, if the crossing is caused by two blue edges, it can be repaired by shifting the topmost endpoint of the vertical segment involved in the crossing, so that such a crossing will lie (in a SHOPED) on the non-drawn part of this vertical segment.We will repair the crossings by assigning the vertices new coordinates.

Repair 3.
We assign new coordinates to the vertices v 2 to v n−1 in the following way: Let v i , 2 ≤ i ≤ n − 1 : (x(v i ), y(v i )) = (2 i , 2 i ) are the new coordinates.After that s, t are again placed according to Construction 2. To prove that all crossings are repaired by the assignment of the new coordinates, we consider the vertical segment of the edge between v i and v j , j < i.The length of the stub on this segment is Thus the horizontal segments incident to the vertices v j , j ≤ i − 1 are never crossed by a vertical stub.To prove that vertical segments are never crossed by horizontal stubs we use the same argument for the x-coordinates.
Finally, we need to repair the crossings caused by the outgoing edges of s and by the incoming edges of t.We observe that edges (s, t), (s, v 2 ),(v n−1 , t) are not crossed due to the placement of s, t, v 2 , v n−1 on the bounding box of the drawing, see also Figure 5(a) and Figure 5(b).Thus, only the crossings affecting (s, v i ), 2 < i < n, and (v j , t), 1 < j < n − 1, must be repaired.

Repair 4.
In case 1 of Construction 2, s is always placed 0.5 grid units above or below v i , thus the vertical segment of the edge (s, v i ) cannot be crossed.In order to fix the crossings on the horizontal segment of (s, v i ) it is enough to shift s on the left, i.e., x(s) = x(s) − (x(v i ) − x(v 2 )) − 1. Similarly in case 2 of Construction 2, s is always placed 0.5 grid units to the left or to the right of v i , thus the horizontal segment of the edge (s, v i ) cannot be crossed.In order to fix the crossings on the vertical segment of (s, v i ) it is enough to shift s to the bottom, i.e., y(s) = y(s) − (y(v i ) − y(v 2 )) − 1.A symmetric argument can be applied to fix the crossings on the edge (v j , t).
A 1-bend SHOPED constructed from the 1-bend drawing in Figure 4(a) is shown in Figure 6(a).
Next we explain how to extend the previous result to any connected graph G with ∆ G ≤ 3. Recall that a cut vertex is a vertex whose removal disconnects G, while a bridge is an edge whose removal disconnects G.We observe for a graph G with ∆ G ≤ 3 that cut vertices are absent in G, if and only if bridges are absent in G. Indeed using the fact that the graphs have maximum degree at most 3, any bridge is incident to at least one cut vertex (and vice versa), justifying the observation.
Lemma 2 Let G be a connected n-vertex graph with ∆ G ≤ 3. We can construct a 1-bend drawing Γ of G in O(n 2 ) time.
Proof: We start by removing all the bridges of G, obtaining a set of k biconnected components C = {C 1 , . . ., C k }, where each component C i ∈ C, 1 ≤ i ≤ k, is either a single vertex or a graph such that ∆ Ci ≤ 3. Next, we define a graph T having one vertex n i for each component C i of G and an edge (n i , n j ), iff C i and C j are connected by a bridge in G. Clearly T is a tree, since a cycle in T would imply a biconnected component comprised by the cycle, inferring a contradiction to the decomposition.Also there is at least one vertex n r in T that represents a component C r , which is either a single vertex or a biconnected graph having two adjacent vertices not incident to any bridge.We choose n r to be the root of T .In the following we describe an algorithm that takes as input G and T and computes a 1-bend drawing Γ of G.We assume that a 1-bend drawing Γ Ci of a component by Construction 1 and 2, where the two poles of the st-numbering will be defined by this construction.
(b) If C i is composed by a single vertex v, then Γ Ci is defined by placing v in the origin and the definition of the two poles can be ignored.
We visit T from the root n r following a breadth first search (bfs) order as follows.
Root n r : Consider C r , we construct a drawing Γ Cr of C r , where in case (a) (i.e., C r is a biconnected graph with ∆ Cr ≤ 3) the st-numbering is defined so that s and t are two adjacent vertices in C r that are not incident to any bridge.Then we set Γ = Γ Cr .
Node n i , i = r: Assume n i is the next vertex of T according to the bfs order.Let Γ be the drawing constructed so far.We first compute a 1-bend drawing Γ Ci of C i , where s = v i and t is an adjacent vertex of s if C i is a biconnected graph with ∆ Cr ≤ 3 (case (a)).If C i is composed by a single vertex v, then v = v i (case (b)).Drawing Γ Ci is now attached to drawing Γ as follows.
Attachment: Let n j be the parent of n i in T and let (v j , v i ) be the bridge in G that corresponds to the edge (n j , n i ) in T .Consider the vertex v j of C j .If C j is a biconnected graph with ∆ Cj ≤ 3, then v j cannot be the s pole in Γ Cj .Furthermore, in this case, since the degree of v j is 2 in C j , v j has either the east port or the north port free in Γ.If C j consists of a single vertex v = v j , then again v j has either the east port or the north port free in Γ.We use this free port of v j to place the bridge e = (v j , v i ) and connect Γ Ci to Γ.In the first case we rotate Γ Ci such that v i is the southernmost vertex and it can be connected by its south port, see Figure 6(b), while in the second case we rotate Γ Ci such that v i is the westernmost vertex and it can be connected by its west port, see Figure 6(c).
Before attaching the drawing Γ Ci , we modify the current drawing Γ in the following way.We assign new x-coordinates to all vertices v with x(v) > x(v j ) by setting x(v) = x(v)+|C i |+1, and we assign new y-coordinates to all vertices v with y(v) > y(v j ) by setting y(v) = y(v) + |C i | + 1.Now we place Γ Ci in this free area, i.e. for each vertex v Ci ∈ C i , x(v Ci ) = x(v j )+x Γ C i (v Ci ), and y(v Ci ) = y(v j ) + y Γ C i (v Ci ), where x Γ C i (v Ci ) and y Γ C i (v Ci ) are the xand y-coordinates of v Ci in Γ Ci , respectively.Then we connect v j with v i .Notice that, by placing Γ Ci in this free area, no edges of Γ \ Γ Ci can cross edges of Γ Ci .
Finally, we observe that finding the bridges of G can be done in linear time [24].Furthermore, constructing a 1-bend drawing Γ Ci for each component However, the time complexity of the technique is dominated by the shift operation required to add the drawing of each component to the current drawing, which takes O(n 2 ).
By iteratively applying Repair 3 and 4 to drawings constructed by Construction 5 we can prove the next theorem.
Theorem 3 Every n-vertex graph G with ∆ G ≤ 3 admits a 1-bend SHOPED.Furthermore, such a drawing can be constructed in O(n 2 ) time.
Proof: Let Γ be a 1-bend drawing of G constructed by Construction 5. We adopt the notation used in the proof of Lemma 2 and we traverse T in the reversed bfs order defined in that proof.Also, we assume that a 1-bend drawing Γ Ci of a component C i ∈ C, 1 ≤ i ≤ k, can be transformed into a 1-bend SHOPED as follows.If C i is a biconnected graph with ∆ Ci ≤ 3, then Γ Ci is repaired by Repair 3 and 4. If C i is composed by a single vertex v, then Γ Ci does not need to be transformed into a 1-bend SHOPED.
Let Γ Ci be the 1-bend drawing of C i .Let Γ Cj be the 1-bend drawing of C j , where n j is the parent of n i in T .Recall that no edges of C j can cross edges of C i in Γ.First, we transform Γ Ci into a 1-bend SHOPED.Then, let s i = max{width(Γ Ci ), height(Γ Ci )}, we assign new x-coordinates to all vertices v of Γ Cj with x(v) > x(v j ) by setting x(v) = x(v)+s i +1, and new y-coordinates to all vertices v with y(v) ≥ y(v j ) by setting y(v) = y(v) The time complexity is dominated by the construction of Γ, which takes O(n 2 ) time, while repairing the components takes

1-bend SHOPEDs for graphs with maximum degree 4
We first present a class of graphs with maximum degree 4, the 2-circulant graphs, that admit a 1-bend SHOPED, and show afterwards that there is a graph that does not admit a 1-bend SHOPED.
Recall that the k-circulant graph C k n with n > 2k vertices is the simple graph whose vertex set is V = {v 0 , . . ., v n−1 } and whose edge set is Theorem 4 Every 2-circulant n-vertex graph that admit a 1-bend drawing also admits a 1-bend SHOPED.Furthermore, such a drawing can be constructed in O(n) time.
Proof: We start by observing that one vertex of a 2-circulant n-vertex graph, v t ∈ V (0 ≤ t ≤ n − 1), has to be removed in order to match the necessary density condition for 1-bend drawings, i.e. |E(S)| ≤ 2|S| − 2 for all S ⊂ V .W.l.o.g., let t = 0, otherwise the vertices of C 2 n can be easily renumbered so to match this condition.
We first construct a 1-bend drawing Γ of ) adopting a similar strategy as in Construction 1. Namely, we assign to Again, there can be only two possible shapes of edges in Γ according to such a placement of the vertices.Indeed, let e = (v i , v j ), so that j = i + 1, i = 2, . . ., n − 3, we call e a red edge, e ∈ E R ⊂ E(V ), and we draw it so that it leaves the north port of v i and enters the west port of v j .While, if e = (v i , v j ), so that j = i + 2, i = 2, . . ., n − 4, we call e a blue edge, e ∈ E B ⊂ E(V ), and we draw it so that it leaves the east port of v i and enters the south port of v j .Thus, red edges are never crossed, while each blue edge receives at most two crossings (one involving the vertical segment and one involving the horizontal segment).
To construct a 1-bend drawing Γ of G, the addition of v 1 and v n−1 to Γ can be managed by adopting the same strategy used to add s and t in Construction 2 Finally, we can apply Repair 3 and 4 (restricted to blue edges) to turn Γ into a 1-bend SHOPED.Now, we show that there exists a graph with maximum degree 4 that admits a 1-bend drawing but not a 1-bend SHOPED.To this end, we need to introduce some additional notation.Consider a 1-bend SHOPED Γ.We say that two horizontal (vertical) stubs s h e , s h e (s v e , s v e ) overlap, if there exists a vertical (horizontal) line l such that s h e ∩ l = ∅ and s h e ∩ l = ∅ (s v e ∩ l = ∅ and s v e ∩ l = ∅).Two horizontal (vertical) stubs overlap by u units if there are two vertical (horizontal) lines l,l with the above property and such that their horizontal (vertical) distance is u.Also, we call ≺ x and ≺ y the total vertex orderings induced by the projection of the x-and y-coordinates of the vertices in Γ, respectively (we remark that we adopt the general position model).
Recall from [14] (see also Section 2) that that any graph G = (V, E) with ∆ G ≤ 4 has a 1-bend drawing if and only if |E(S)| ≤ 2|S|−2 for all S ⊂ V .This necessary and sufficient condition is satisfied by any of the following three cases: (i) there are four vertices of degree at most three; (ii) there are two vertices of degree at most two; (iii) there are one vertex of degree at most two and two vertices of degree at most three.Since a 4-regular graph cannot guarantee such a condition, one vertex must be removed.Furthermore, if we are in case (i) each side of the bounding box of any possible 1-bend drawing contains one of the four vertices of degree three.In case (ii) the two vertices of degree two must be placed at the opposite corners of the bounding box.While in case (iii) the degree two vertex must be placed at one corner of the bounding box and the other two vertices of degree three must be placed on the two opposite sides.We call these vertices lying on the sides of the bounding box by external vertices.All the other vertices must lie in the interior of the bounding box and we refer to them as internal vertices.
Consider a graph G = (V , E ) so that V ={v 1 , . . ., v 6 } and E ={(v 1 , v 2 ), (v 2 , v 3 ), (v 4 , v 5 ), (v 5 , v 6 )}, see Figure 7(a).We prove that there exist two total vertex orderings ≺ x , ≺ y such that in every 1-bend drawing of G at least two stubs cross each other.Recall that ≺ x and ≺ y are orders, where the x-and y-order of the vertices coincide with these two orderings.
Proof: Let Γ be a 1-bend drawing of G respecting the properties defined in the statement of this lemma (see Figure 7(a)), and consider the stubs (in the 1-bend SHOPED model) of the segments representing the edges of G in Γ .First, we observe that the two horizontal stubs s h v1v2 and s h v2v3 overlap, because of , that is: By definition x(v 2 ) − x(v 3 ) = 2|s h v2v3 |, which implies: Since x(v 3 ) − x(v 1 ) ≥ 1, the two stubs must overlap by at least half of a grid unit.In a similar way, the two vertical stubs s v v4v5 , s v v5v6 overlap, because of v 5 ≺ y v 4 ≺ y v 6 .Assume the stub s v v5v6 does not cross the stub s h v1v2 .Due to the overlap between s v v5v6 and s v v4v5 , and since v 3 ≺ x v 4 ≺ x v 5 , it follows that s v v4v5 will cross either s h v1v2 or s h v2v3 , or s v v5v6 will cross s h v2v3 .
In order to remove the assumption that the x-order and the y-order of the vertices cannot be changed, we augment G adding new edges and vertices obtaining a new graph G, such that in every 1-bend drawing of G the x-order and the y-order of the vertices in V is the same up to rotation.Let K 4 be the complete graph with 4 vertices.Let K 4 be a copy of K 4 and let a, b, c, d (in clockwise order traversing the bounding box) be its 4 vertices.We connect v 1 to a and b and v 3 to c and d.See also Figure 7(b).In a similar way we add one more copy of K 4 between v 4 and v 6 .We call the subgraph induced by the pair v 1 , v 3 and its copy of K 4 as K * 13 , analogously we call K * 46 the subgraph induced by the pair v 4 , v 6 and its copy of K 4 .Furthermore, we add the edge (v 3 , v 4 ) and (v 2 , v 5 ).Finally, we add the vertices v 7 and v 8 and the edges (v 1 , v 7 ), (v 5 , v 7 ) and (v 2 , v 8 ), (v 6 , v 8 ).The final graph G is shown in Figure 7(d), it clearly admits a 1-bend drawing.
Theorem 5 There exists a graph that does not admit a 1-bend SHOPED.Proof: Consider the graph G shown in Figure 7(d).We will evaluate all the possible configurations in terms of x-order and y-order for the vertices of G.For every feasible configuration we will then apply Lemma 3 to prove that there is no 1-bend SHOPED.
Let Γ be a 1-bend SHOPED of G. Consider the vertices v 7 and v 8 , they must be the external vertices of Γ and they must be placed at the corners of one of the two diagonals of the bounding box R of Γ.Consider the bounding-box R * of the subdrawing Γ * induced by the graph G * = (V * = V \ {v 7 , v 8 }, E(V * )) (see Figure 7(c)).Vertices v 1 , v 2 , v 5 , v 6 must be the external vertices of Γ * and they must lie one on each side of R * .For the same reason, vertices v 3 , v 4 are internal vertices (as well as all the vertices in the two copies of K 4 ) of Γ * .Also, consider the bounding-box R 13 of the subdrawing Γ 13 induced by the graph K * 13 .Vertices v 1 and v 3 must be the external vertices of Γ 13 and must be placed at the corners of one of the two diagonals of R 13 .W.l.o.g., let v 7 be at the top-left corner of R. Hence, either v 1 is the westernmost vertex and v 5 is the northernmost vertex in Γ * or vice versa.Assume to be in the former case, since the latter case can be proved with symmetric arguments.It follows that v 2 is either the southernmost or the easternmost vertex, while v 6 is either the easternmost or the southernmost vertex in Γ * .
We start the analysis looking at the vertex v 1 .We already know that v 1 ≺ x v 3 and consider two possible cases for the relative y-order of these two vertices.
1. Assume v 3 ≺ y v 1 .In this case, since the only free port of v 1 is the north one, we have v 1 ≺ y v 2 .This implies that v 2 cannot be the southernmost vertex, but it must be the easternmost vertex and v 6 the southernmost instead.Hence, the edge (v 2 , v 3 ) must leave the south port of v 2 and enter the east port of v 3 .It follows that the only free port of v 3 is the south one, and, due to the edge (v 3 , v 4 ), v 4 ≺ y v 3 .Now consider two further subcases, either v 4 ≺ x v 3 or v 3 ≺ x v 4 .We prove by contradiction that v 3 ≺ x v 4 .
(a) Let v 4 ≺ x v 3 (see Figure 8(a)), this implies that v 6 ≺ x v 4 .Consider the edge (v 5 , v 6 ).It must leave the west port of v 6 and enter the south port of v 5 .Also, consider the edge (v 4 , v 5 ), it must leave the north port of v 4 and enter the east port of v 5 .Finally, consider the edge (v 2 , v 5 ), again it must leave the north port of v 2 and enter the west port of v 5 , thus, it must be bent at least three times, a contradiction.
(b) Let v 3 ≺ x v 4 , thus v 4 ≺ x v 6 .Consider the edge (v 5 , v 6 ), it must leave the east port of v 6 and enter the south port of v 5 , thus v 6 ≺ x v 5 .Also, consider the edge (v 4 , v 5 ), it must leave the north port of v 4 and enter the west port of v 5 .Finally, consider the edge (v 2 , v 5 ), it can be placed so that it leaves the east port of v 5 and enters the north port of v 2 .Notice that, v 6 ≺ y v 4 ≺ y v 3 ≺ y v 1 ≺ y v 2 ≺ y v 5 , as well as v 1 ≺ x v 3 ≺ x v 4 ≺ x v 6 ≺ x v 5 ≺ x v 2 .Thus, by Lemma 3, there is no 1-bend SHOPED for the subdrawing Γ induced by the subgraph G .
2. Assume v 1 ≺ y v 3 , as depicted in Figure 8(b).In this case, since the only free port of v 1 is the south one, we have v 2 ≺ y v 1 .Now consider the edge (v 2 , v 3 ), it must leave the north port of v 2 and enter the east port of v 3 .Also, consider the edge (v 2 , v 5 ), it must leave the east port of v 2 and enter the south port of v 5 .This implies that v 2 is the southernmost vertex and v 6 is the easternmost vertex.Thus, consider the edge (v 5 , v 6 ), it must leave the east port of v 5 and enter and north port of v 6 .Now, since v 4 must lie on the opposite extreme of the diagonal of R 46 with respect to v 6 , we have that v 4 ≺ y v 6 .Thus, since the edge (v 4 , v 5 ) must use the west port of v 5 and the west or south port of v 4 , it must have at least two bends, which implies that this configuration is not feasible and that v 3 ≺ y v 1 .

Conclusion and Future Work
We defined a new layout style for orthogonal drawings with one bend per edge, called 1-bend Orthogonal Partial Edge Drawing, extending the already existent PED model for straight-line drawings.We studied those graphs that admit such a representation when homogeneity or both symmetry and homogeneity are required.In the former case, we proved that every graph that admits a 1bend drawing also admits a 1-bend HOPED.In the latter case, we proved that all graphs with maximum degree 3 and the 2-circulant graphs that admit a 1bend drawing, also admit a 1-bend SHOPED.Furthermore we proved that there is a graph with maximum degree 4 that does not admit a 1-bend SHOPED.The complexity of the decision problem is still open, i.e., deciding whether a graph with maximum degree 4 admits a 1-bend SHOPED.We formalized a related question as an integer linear program (ILP), i.e. a test of a 1-bend drawing, whether it admits a 1-bend SHOPED inside a square of given size without changing the relative horizontal and vertical order of the vertices, but the complexity of this question is also still open.Also, it would be of interest to study 1-bend SHOPEDs where a few crossings among stubs are allowed, so to enlarge the family of graphs that admit these representations.To extend our model, one may consider graphs with degree greater than 4, by following similar approaches as in [15] for representing vertices, or orthogonal drawings with more than one bend per edge.

Figure 2 :
Figure 2: (a) A 1-bend drawing of a graph with four edges when applying the SHPED model of straight-line drawings, i.e. half of the edge is removed in such a way that a quarter of the edge remains incident to an end-vertex and a quarter to the other end-vertex.It is ambiguous which vertices are connected, since (b), (c) and (d) are different graphs that can be extracted from (a) according to the SHPED model of straight-line drawings.Such an ambiguity can be resolved by using the new definitions for 1-bend SHOPEDs introduced in Section 2. Indeed, (d), (e) and (f) are unique 1-bend SHOPEDs of the graphs (d), (b) and (c), resp.

Figure 3 :
Figure 3: (a) K 4 has no planar 1-bend drawing, (b) but admits a 1-bend OPED, (c) a 1-bend HOPED and also (d) a 1-bend SHOPED.Observe that in (b) and (c) the bend point is part of the stubs, while in (d) it is not.

Figure 4 :
Figure 4: (a) A 1-bend drawing of a 14-vertex biconnected cubic graph G described in Construction 1 and 2. (b) The graph G * taken from (a) already colored as described in Construction 1 (red edges are represented by dashed segments, while blue edges by dashed-dotted segments).

Figure 5 :
Figure 5: (a) A 1-bend drawing according to the case 1a of the proof of Lemma 1.(b) A 1-bend drawing according to the case 2a of the proof of Lemma 1.

Fig- ure 5
(b).(b)If the west port of v i is free, then we set x(s) = x(v i ) − 0.5.

Figure 6 :
Figure 6: (a) A 1-bend SHOPED constructed from the drawing in Figure 4(a) (for the sake of readability only part of the drawing is shown).(b)-(c) Illustration of the technique described in the proof of Lemma 2 to attach the drawing Γ Ci to Γ when the east port (b) or the north port (c) of v j is free.
The specified index of a vertex is calculated modulo n.Notice that, ∆ C k n = 4 implies k = 2, hence, each vertex has exactly two neighbors with smaller indices and two neighbors with larger indices.Greater values of k are not realizable when ∆(C k n ) = 4. Extending the techniques in Section 4, we can prove the next theorem.

Figure 7 :
Figure 7: (a) A drawing Γ of G that cannot be redrawn as a 1-bend SHOPED if the vertical and horizontal order of the vertices cannot be changed.The relative order of each pair of vertices in G can be constrained by the gadget in (b).(c) An illustration of the graph G * as defined in the proof of Theorem 5. (d) A graph G that admits no 1-bend SHOPED.

Figure 8 :
Figure 8: An illustration of the case 1 (a) and 2 (b) in the proof of Theorem 5.The red dash dotted edge indicates an edge, which cannot be drawn with 1 bend.