Vertex Contact Representations of Paths on a Grid

We study Vertex Contact representations of Paths on a Grid (VCPG). In such a representation, the vertices of G are represented by a family of interiorly disjoint grid-paths on a square grid. Adjacencies are represented by contacts between an endpoint of one grid-path and an interior point of another grid-path. Defining u → v if the path of u ends on the path of v, we obtain an orientation on G from a VCPG. To control the bends of the grid paths the orientation is not enough. We therefore consider pairs (α,ψ): a 2-orientation α and a flow ψ in the angle graph. The 2orientation describes the contacts of the ends of a grid-path and the flow describes the behavior of a grid-path between its two ends. We give a necessary and sufficient condition for such a pair (α,ψ) to be realizable as a VCPG. Using realizable pairs, we show that every planar (2,2)-tight graph admits a VCPG with at most 2 bends per path and that this bound is tight. In a similar way, we show that simple planar (2,1)-sparse graphs have a 4-bend representation and simple planar (2,0)-sparse graphs have 6-bend representation. Submitted: March 2015 Reviewed: July 2015 Revised: October 2015 Reviewed: November 2015 Revised: December 2015 Accepted: December 2015 Final: December 2015 Published: December 2015 Article type: Regular paper Communicated by: C. D. Tóth We acknowledge support by German Science Foundation (DFG) grant Fe 340/7-2. An extended abstract of this paper was presented at 40th International Workshop on GraphTheoretic Concepts in Computer Science, in Le Domaine de Chalès, France [1]. E-mail addresses: aerts@math.tu-berlin.de (Nieke Aerts) felsner@math.tu-berlin.de (Stefan Felsner) 818 N. Aerts and S. Felsner VCPGs


Introduction
Outline of results.In this paper, we consider the question whether a graph G admits a VCPG, i.e., a Vertex Contact representation of Paths on a Grid, where the grid is the square grid.In such a representation, the vertices are represented by a family of interiorly disjoint grid-paths with distinct endpoints.An endpoint of one grid-path coincides with an interior point of another gridpath if and only if the two represented vertices are adjacent.The shared point is not a bend-point of either of the grid-paths.We denote such a contact by proper contact.It follows that a graph admitting such a representation has to be planar.
A VCPG induces a unique orientation of the edges of G: Orienting the edge uv as u → v if the grid-path of u ends on grid-path of v we obtain an orientation of G.As each grid-path has two ends, in the induced orientation each vertex has outdegree at most 2. We denote such an orientation simply 2-orientation.The existence of a 2-orientation implies that the density of graphs admitting a VCPG is rather restricted, in technical terms they have to be (2, 0)-sparse.
Every 2-orientation of a planar graph induces a VCPG (Lemma 8).However, a 2-orientation of G defines the representation of the edges in a VCPG, but not how the grid-paths behave (e.g.how many bends a grid-path has).To control the behavior of the grid-paths between its endpoints, we introduce a flow network in the angle graph (Section 3).
To obtain a full combinatorial description of a VCPG, we consider a pair (α, ψ): a 2-orientation α in the graph and a flow ψ in the angle graph.The main contribution of this paper is a necessary and sufficient condition for such a pair (α, ψ) to be realizable as a VCPG.We will then use such realizable pairs to give bounds on the number of bends needed to represent certain classes of graphs.
When the number of bends of each path is at most k, we call the representation B k -VCPG and when every path has precisely k bends, we speak about strict B k -VCPG.We will show that there are graphs admitting a B 1 -VCPG but no strict B 1 -VCPG and we construct a planar (2,2)-tight graph that does not admit a B 1 -VCPG (Section 4.1).We show that planar (2,2)-tight graphs admit a B 2 -VCPG (Section 4.2) and planar (2,0)-tight graphs admit a B 6 -VCPG (Section 4.3).
Related work.Graphs that can be represented as intersection graphs of gridpaths without bends, i.e., by vertical and horizontal segments are known as Grid Intersection Graphs (GIG).This is a well studied class of bipartite graphs.Contact graphs of vertical and horizontal segments, i.e., B 0 -VCPG graphs or contact 2DIR graphs, are exactly the bipartite planar graphs [6,11].In fact, 2orientations1 of maximal bipartite planar graphs are in bijection with separating decompositions of this graph and separating decomposition induce a segment contact representation by segments in two directions (cf.[5,6,8,11]).
Intersection graphs of grid-paths are called VPG graphs.VPG graphs were introduced by Asinowski et al. [3].When the number of bends of each path is at most k, we denote the representation by B k -VPG.From the result that every planar graph has a contact representation by T-shapes [7], it follows that every planar graph has an intersection representation by grid-paths where each path has at most three bends [3].To obtain the B 3 -VPG, every T-shape is replaced by a grid-path that follows the T-shape in such a way that every point is covered on at least one side.Asinowski et al. conjectured this bound to be tight, i.e., there exists a planar graph for which three bends are necessary.Chaplick and Ueckerdt disproved this by showing that every planar graph admits a B 2 -VPG [4].It is still open to decide whether this is tight.Chaplick and Ueckerdt conjectured that every planar graph admits a B 1 -VPG [4].
In a contact representation by grid-paths (as well as segments, pseudosegments, etc.), the vertices are represented by a family of internally disjoint objects.An endpoint of one object coincides with an interior point of another object if and only if the two represented vertices are adjacent.Kobourov, Ueckerdt and Verbeek show that all planar (2,3)-tight graphs admit an L-contact representation [12], i.e., a strict B 1 -VCPG (a sketch of their algorithm is given in Section 4.1).It follows that every (2,3)-sparse graph also admits a strict B 1 -VCPG.There are graphs that are not (2,3)-sparse but admit a strict B 1 -VCPG, e.g.K 4 .Kobourov et al. stated that every graph that admits a B 1 -VCPG is planar and (2,2)-sparse and they asked whether these conditions are also sufficient.
Chaplick and Ueckerdt show that triangle-free planar graphs, i.e., (2,4)sparse planar graphs, admit a contact representation of {L, L ,I, I }-shapes [4].In other words, triangle-free planar graphs admit a B 1 -VCPG with only two possible orientations of the one-bend paths.Alam et al. investigate which (2,0)sparse graphs have a contact representation of circular arcs [2].
Another related area of research is orthogonal graph drawing.Here one wants to draw a given graph in the classical setting, i.e., the vertices are points in the plane and the edges are grid-paths.In orthogonal graph drawing, there have been many results on minimizing the number of bends.Note that in this setting vertices have degree at most 4, or as a workaround, the vertices can be represented as boxes.An early result of Tamassia gives an algorithm to obtain an orthogonal drawing with minimal bend number of a planar graph in polynomial time [14].The algorithm preserves the embedding of the input graph.Optimizing the number of bends per path has received much attention too.Schäffter [13] gives an algorithm to draw 4-regular graphs in a grid with at most two bends per edge (which is tight when not restricted to planar graphs).For orthogonal drawings without degree restriction, Fößmeier, Kant and Kaufmann have shown that every plane graph has an orthogonal drawing preserving the embedding with at most one bend per edge [9].
Outline of the paper.In Section 2, we define sparse and tight graphs and show that every (2,0)-sparse graph has a 2-orientation.In Section 3, we intro-duce a flow network.We then give the necessary and sufficient condition for a pair, a 2-orientation and a flow, to be realizable as a VCPG.In Section 4, we apply realizable pairs to give bounds on the number of bends in a VCPG.
2 Preliminaries: on (2, )-sparse graphs Let G = (V, E) be a planar graph and let R be a VCPG of G.When we speak about the embedding of G, we mean the embedding R. In R, the vertices of G are represented by internally disjoint grid-paths with distinct endpoints.We denote the grid-path of v ∈ V by p v .Each grid-path has two endpoints.When two grid-paths intersect, the intersection point is an endpoint for precisely one of the grid-paths.An endpoint of p v coincides with an interior point of p u precisely when u and v are adjacent.If an endpoint does not coincide with another grid-path, we call it a free end.The number of free ends is denoted by .A free end in an interior face can be extended to represent an edge, which is considered a dummy edge.As interior free ends can be seen as dummy edges, we assume that R only has free ends in the outer face.Later, in Lemma 2, we will see that free ends in the outer face reduce the number of bends needed to close the outer face.Therefore, the assumption that all free ends are in the outer face is also attractive.If a graph G = (V, E) is given and W ⊂ V , then we let E[W ] be the set of edges induced by W in G.We start with an easy proposition which gives an upper bound on the number of edges in a VCPG.
Proposition 1 If G = (V, E) admits a VCPG then: Proof: Each edge is represented as a proper contact between two grid-paths representing two vertices.For one of the vertices this contact point is an endpoint of its grid-path.This vertex is the representative for the edge.Each vertex can be a representative for at most two edges.Therefore the number of edges induced by a set W ⊆ V is at most twice the number of vertices.
If every grid-path has at most one bend, then there must be at least two free ends in the outer face2 .Therefore, the number of edges of a B 1 -VCPG is at most twice the number of vertices minus two.

A 2-Orientation Representing Edges
Defining u → v if p u ends on p v , we obtain an orientation on G from a VCPG.Every vertex has outdegree at most two in this orientation.The vertices with outdegree less than two are precisely those for which the grid-path has free ends.Such an orientation is denoted by 2-orientation.
A VCPG is not completely described by a plane graph with a 2-orientation.For example, the VCPGs in Figure 1 induce the same orientation on the same plane graph, but the number of bends in the representations is different.It is clear that for a (2, )-tight graph with > 0, not every vertex has outdegree precisely 2. We also assume that the vertices that do not have outdegree 2 are on the outer face.With the next lemma, we show that for 0 ≤ ≤ 2 every (2, )-tight graph has such an orientation.
Lemma 1 Every planar (2, )-tight graph has a 2-orientation.If 0 < ≤ 2, then for every embedding, there exists a 2-orientation such that all vertices with outdegree less than 2 are on the boundary of the outer face.
Proof: Let G = (V, E) be a planar (2, )-tight graph.Suppose there is a subset W of the vertices of G that has less than 2|W | incident edges.Then G[V − W ] must induce at least: edges, which contradicts (2, )-sparseness.Hence, every subset W of the vertices of G has at least 2|W | incident edges.Now we construct a bipartite graph B. The first color class, V 1 , consists of two copies of all but of the vertices of G.The remaining vertices are only added once.The second color class, V 2 , consists of the edges of G.The edge set of B is defined by the incidences in G: two vertices are connected if the corresponding vertex of G is an endpoint of the corresponding edge of G.We will show that (2, )-tightness of G implies that this bipartite graph has a perfect matching.A perfect matching defines a 2-orientation of G.
Let U ⊆ V 1 consist of n duplicate vertices, and m vertices without a copy, so |U | = 2n + m.Let N B (U ) denote the set of neighbors of U in B. The neighbors are all edges in G, except for the edges between two vertices that are not in U .There are at most 2(|V | − n − m) + such edges.Therefore, Hence, Hall's marriage condition is satisfied for each subset the graph B has a perfect matching and, hence, G has a 2-orientation.
To prove the second part of the lemma, we fix an embedding of G.The vertices that are not added twice to V 1 are chosen arbitrarily from the boundary of the outer face.The above argument implies the existence of a perfect matching and thus the desired 2-orientation.
Remark.Lemma 1 holds for all (2, )-tight graphs, not only simple and planar (2, )-tight graphs.In the proof, neither simplicity nor planarity is used.The only difference is that in a non-planar graph the vertices with outdegree less than 2 should be selected differently, because "being on the outer face" is meaningless.

A Combinatorial Characterization of VCPGs 3.1 A Feasible Flow Representing Bends
From here on, we consider graphs that are (2, )-tight, plane and 2-connected.Note that any (2, )-tight graph can easily be extended to a 2-connected (2, )tight graph by adding an appropriate number of degree 2 vertices.
To describe the behavior of the bends of a grid-path, we introduce a flow network.The construction is inspired by Tamassia's flow network [14].A flow network N is a graph that has two distinct sets of vertices that are the sources and the sinks of the network.The sources and the sinks in our case, as well as in the network designed by Tamassia, are the faces of the graph (see Figure 2).For a face of a graph G, the need of a convex angle of a bend is presented as having a negative demand3 , i.e., it is a source.Similarly, needing a concave angle of a bend is presented as having a positive demand, i.e., it is a sink.
The network will be the angle graph of G. Formally, the angle graph arises from G by taking the union of the vertices and faces of G as the vertices of A(G) and the pairs (v, f ), v ∈ V (G), f ∈ F (G), such that v is a vertex on f in G, as the edges of A(G) (see Figure 4).The angle graph A(G) is a plane bipartite graph with one color class consisting of vertex-vertices and the other of the face-vertices.Edges of A(G) represent the angles4 of G.The angle graph of a 2-connected plane graph is a maximal bipartite graph, i.e., a quadrangulation.The outer face has special properties in a VCPG, therefore, we consider graphs with a fixed embedding.
The sources and sinks of the network are faces of G.A path in the network from a source to a sink has to traverse vertices of G, as the vertices will be represented as grid-paths with bends.
A directed path in the network going from a source to a sink passing through a vertex v represents a bend in the grid-path p v representing vertex v.A path f 1 → v → f 2 represents a bend of p v where the convex angle belongs to f 1 and the concave angle to f 2 .The difference between the network needed for VCPGs and the network defined by Tamassia for orthogonal drawings is that in the case of VCPGs the paths, resp.flow, goes through a vertex and in the case of orthogonal drawings the flow goes through an edge.In other words, the underlying graph of the network defined by Tamassia is the dual graph and not the angle graph (see Figure 3).We proceed with the formal introduction of the encoding of the bends of a VCPG.
A flow ψ is a weighted directed graph, with the network N as underlying graph.The weight of an edge denotes the amount of flow that is routed over this edge.Flow conservation holds relative to correction terms given by the positive or negative demands.An example of a flow ψ in the angle graph of K 4 is shown on the right side of Figure 4.A face-vertex of A(G) can be a source or a sink, depending on the size of the face in G (see Proposition 2 below).The vertex-vertices of A(G) are neither sources nor sinks.The capacities of edges are unbounded.From a VCPG R we construct a flow ψ as follows.For each bend of p v , such that the convex angle of this bend lies in f 1 and the concave angle lies in f 2 , add a unit of flow f 1 → v → f 2 .Note that the matching between concave and convex corners at faces sharing a bend of an edge forces us to adopt a strange classification of convex and concave angles at the outer face, e.g., the outer face of a rectangle has four concave angles.Proof: Following the boundary of an interior region of a VCPG and adding the changes in direction, one should obtain 2π.Each edge is represented as a proper contact and therefore changes the direction with π/2.A convex angle at a vertex changes the direction with π/2 as well and a concave angle at a vertex changes the direction with −π/2.For a face f , let c(f ) be the number of concave angles at vertices minus the number of convex angles at vertices.For an interior face f we obtain: The value c(f ) also counts the incoming minus the outgoing flow in ψ and therefore is the demand of f .By rearranging (2), we obtain i.e., the desired result.
For the outer face f ∞ , the bends have to be counted differently.For the outer face, an edge implies a change in direction of −π/2.A convex angle contributes a −π/2 change of direction and a concave angle a change of π/2.A free end gives a change in direction of π.Together we obtain and after rearranging: where is the number of free ends.
A valid flow for the network N satisfies the flow conservation law at each vertex that is not a source or a sink, i.e., the net flow is zero at each vertex v ∈ S ∪ T .At a sink, the sum of the incoming flow minus the sum of the outgoing flow is at most the demand of the sink.At a source, the sum of the incoming flow minus the sum of the outgoing flow is at least the size of the demand 5 .A valid flow ψ is feasible if the demand of every sink is satisfied.We next show that the demands of the sinks add up to the absolute value of the sum of the demands of the sources, hence, in a feasible flow the negative demand of each source is also satisfied.
The total value of a feasible flow is equal to the number of interior 3-faces (the number of sources), which in turn is equal to the sum of the demands of all sinks.We take the sum of all demands.
We have used Euler's formula to replace |F | and the fact that the graph is (2, ) tight for the last equation.

The number of bends prescribed by the flow is
where ψ(v) denotes the amount of flow that goes through the vertex v.In order to relate to a VCPG it is necessary that ψ is an integral feasible flow for the network N .Note that since the demands are integral the existence of a (minimum cost) integral feasible flow is guaranteed.In the sequel, we will omit the word 'integral' as we only consider integral feasible flows.
A flow does not uniquely define a VCPG either.The two VCPGs in Figure 5 induce the same feasible flow, but the edges are represented in a different way.The obvious question is whether every feasible integral flow in A(G) belongs to a VCPG.Unfortunately this is not the case (see Figure 6).However, using both the orientation and the feasible flow, we obtain a characterization of VCPGs.This is the subject of the following section.

Realizable Pairs
Starting with a VCPG, a 2-orientation α and a feasible flow ψ can be obtained, as described in the previous sections.In this section we identify a necessary property of a pair, (α, ψ) that comes from a VCPG.Not every pair (α, ψ) on G induces a VCPG of G.We call a pair (α, ψ) realizable when it does.We will prove that the necessary property is also sufficient, hence, realizable pairs are in bijection to VCPGs.Our proof method is algorithmic, it shows how one can construct a VCPG (the geometric setting) from a realizable pair (the combinatorial setting).

A Property of Realizable Pairs
We will deduce a property of realizable pairs and show that this property is necessary and sufficient for a pair, a 2-orientation α and a feasible flow ψ, to be realizable.The property depends on α and on ψ.First we define the property for a vertex v (see Figure 7 ] denote the angle graph induced by the closed neighborhood of a vertex v, i.e. induced by v and all its neighbors in A(G).Let n 1 , n 2 be the neighbors of v along the outgoing edges of v in α.If v has outdegree 0, then n 1 , n 2 are its neighbors on the outer face.If v has outdegree 1, then n 1 is its neighbor along the outgoing edge.The vertex n 2 is the neighbor of v on the outer face, chosen such that the units of • there is a decomposition of the flow into non-crossing paths, and, • every path of such a decomposition crosses the path n 1 , v, n 2 .
When the pair (α, ψ) satisfies the realizability condition at each vertex we say that the pair is realizable.Let us look at the feasible flow ψ from Figure 6 in the context of the realizability condition.The graph is (2,0)-tight and therefore all vertices have outdegree 2 in a 2-orientation that comes from a VCPG.A 2-orientation α such that (α, ψ) is a realizable pair, is such that u → v and w → v. Then v has outdegree at most 1.This shows that there is no 2-orientation α such that (α, ψ) is realizable, and hence, due to Lemma 2, ψ does not come from a VCPG.
We now show that a pair that comes from a VCPG satisfies the realizability condition.
Proof: First note that a VCPG of G describes an embedding of G.If there is a grid-path with one free end, then before proceeding, we reduce all unnecessary bends, i.e., if a grid-path has bends between its last neighbor and its free end, these bends are removed.The 2-orientation α that is induced by the VCPG has an edge u → v if and only if the grid-path of u ends on the grid-path of v. Consider the grid-path that represents a vertex v.If this path has no bends, the realizability condition is satisfied at this vertex.Suppose the path has k bends.Draw an arrow from the face containing a convex corner to the face in which the associated concave corner lies.Now the set of arrows represents the flow ψ(v).This flow is non-crossing through v and every unit of flow is cut by the the grid-path of v.When these arrows are introduced for all bends of all grid-paths, the flow given by these arrows satisfies the demand of each face.Contract the strictly interior steps of the grid-path to a vertex.Every unit of the non-crossing flow through v is now cut by the outermost two segments of the grid-path, which correspond to the outgoing edges of v, or to the outgoing edge and the location of the last incoming edge before the free end of the grid-path.Hence, the realizability condition is satisfied at each vertex.Therefore, the pair (α, ψ) obtained from the VCPG is realizable.

Realizable Pairs Are in Bijection with VCPGs
Now we are ready for the main result.
Theorem 1 The realizable pairs are in bijection with VCPGs.
The remainder of this section is dedicated to showing that given a realizable pair (α, ψ), there exists a VCPG, with the same vertices on the outer face as in the chosen embedding of G (this embedding is implied by the pair (α, ψ) since ψ lives on the angle graph), and such that: (a) The grid-path of u ends on the grid-path of v if and only if the edge uv is oriented from u to v in α; and (b) The grid-path of v has precisely ψ(v) bends.
We will show how to construct a VCPG given a realizable pair.Note that an embedding can be derived from A(G) (in which the flow ψ is defined).Consider a realizable pair (α, ψ).The construction consists of four steps, which we first outline here.
Step 1: First, we expand the vertices that have k units of flow going through them, to a path6 of length k.We obtain a bipartite graph.
Step 2: We introduce help-edges and vertices in the bipartite graph to construct a quadrangulation (see Figure 8(b)).The orientation α is extended to a 2orientation of the quadrangulation.
Step 3: We then find a segment contact representation of the quadrangulation.It has been shown that the 2-orientations of maximal bipartite planar graphs are in bijection with separating decompositions of this graph (e.g.[5]).In turn, a separating decomposition induces a segment contact representation (cf.[11,6,8]).Hence we can construct a segment contact representation where the representation of the edges is in bijection with the given 2-orientation.An example is shown in Figure 8(c).
Step 4: Last, we will show that the extra edges that have been introduced to make a quadrangulation of the bipartite graph can be deleted in order to obtain a VCPG of G (see Figure 8(d)).
Let us describe the steps in more detail.
Step 1.Let (α, ψ) be a realizable pair for G.We expand all vertices with non-zero flow.The plane graph we obtain is denoted by G.For every vertex v for which ψ(v) = 0, expanding v consists of the following steps (see Figure 7): 1. Expand v to a circle.We will call this the bag of v.
2. Inside the circle, add a path with ψ(v) + 1 vertices between the two outgoing edges of v.If v has outdegree 1 or 0, then v is on the outer face.The path is added between the edge on the outer face and the outgoing edge, or between the two edges on the outer face, respectively.The edges of the path are called path-edges, the inner vertices of the path by path-vertices, and all the vertices that belong to a bag are called bag-vertices.
3. Each incoming edge of v, is now connected to a path-vertex.This is done such that the flow between two faces only crosses a single edge and this is an edge of the path.Step 2. Now we add help-edges to extend G to a quadrangulation.We denote the quadrangulation by G Q .We will also orient the new edges to obtain a 2orientation of G Q .In order to explain how the help-edges are added, we need the following lemma.Proof: Let ψ + (f ) (respectively ψ − (f )) denote the incoming (respectively outgoing) flow in face f .Let f be the face corresponding to f in the original graph G.
As ψ is a feasible flow, the demand c(f ) of f is the difference between incoming and outgoing flow: Now we use that the size of the extended face f is the size of f plus the incoming and the outgoing flow.
Hence we find Using (| f | − 4)/2 help-edges, we can quadrangulate f (see Figure 9 and the proof of Lemma 4).The help-edges should be added in such a way that every bag (vertex expansion) gets as many help-edges as the value of the flow going into f through this vertex.Informally, a concave corner arises from two segments that both end in one point.In the segment contact representations that we will use, there are only proper contacts or free ends.Each help-edge represents a segment of a concave corner that proceeds into the face, and, this part will later be removed.Such a help-edge will be oriented as outgoing from the bag.
Later, we will construct a segment contact representation of the quadrangulation.For this, it is necessary that every interior vertex has outdegree precisely 2. Therefore, the help-edges must be added in such a way that this is possible for all vertices.Each interior bag-vertex should finally get assigned two outgoing arcs (which are not edges in the original graph).The vertices on the boundary of the bag, i.e., the vertices incident to the outgoing edges of the vertex in the original graph, need one additional outgoing arc.To begin with, the help-edges are added along the flow from a vertex into a face.This gives the required number of new edges leaving a bag (a bag with k vertices and 2 outgoing edges in α has k − 1 internal vertices and flow value k − 1, hence, adding one out-edge for each unit of flow is the right amount).
Lemma 4 Each inner face f of G can be quadrangulated in such a way that each bag through which k units of flow enter f gets k new outgoing arcs.The outer face of G can be quadrangulated using four help-vertices (v t , v r , v b , v l ), in such a way that each bag through which k units of flow enter the outer face gets k new outgoing arcs.
An example of quadrangulating an interior face is depicted in Figure 9.The flow ψ is given by the red arrows.First, half-arcs are added, the green solid arcs.Then these half-arcs are subsequently connected in such a way that they close one 4-face (green dashed lines).The quadrangulation of the outer face is based on the same idea.
Proof of Lemma 4: Add an outgoing half-arc into f from each vertex that comes clockwise after a unit of incoming flow, see Figure 9.This ensures that every bag gets the correct number of new outgoing arcs.Now we will show that extending these half-arcs allows us to quadrangulate f .There are (| f | − 4)/2 vertices that get a half-arc, thus (| f | + 4)/2 without a half-arc.Hence, there exists a vertex with a half-arc which is followed by two consecutive vertices u, v without a half-arc.We will extend this half-arc and show that afterwards there is again a half-arc which is followed by two consecutive vertices without a half-arc.Consider the half-arc clockwise before u, v and connect it to the vertex after u, v.We have constructed a 4-face and completed one half-arc.The other face has | f |−2 vertices and (| f |−2+4)/2 half-arcs.Therefore, there must again be a vertex with a half-arc that is followed by two vertices without a half-arc and the step can be repeated on this face.In every step the size of the face is reduced by two and one half-arc is extended, there are (| f | − 4)/2 steps.Since it follows that the result is a quadrangulation of f .The way the half-arcs are added ensures that that each bag through which k units of flow enter f gets k new outgoing arcs.To take care of the outer face we add a quadrilateral around the graph and quadrangulate the new inner face between the graph and the quadrangle.
We distinguish four cases (see Figure 10), a detailed description of the process in each of the cases is given below.We add a quadrilateral around the graph and construct an inner face, of size 2k + 4 such that the amount of incoming flow is k (see Figure 10).Then we can use the same method as for the interior faces.Recall that we consider (2, )-tight graphs.Hence the graph has 2|V | − edges and there are precisely free ends in the outer face of the representation.
(a) Note that = 2. Add a quadrangle around the graph with vertices v l , v t , v r , v b in clockwise order.If s is expanded, we label the vertices in the expansion in counterclockwise order, following the boundary of f ∞ , by s 1 , . . ., s k .If s is not expanded then we label s = s 1 .Add the arcs (s 1 , v t ) and (s 1 , v b ).Now the bounded face f * containing s 1 , v t , v r , v b on its boundary has the following properties: The same method as for an inner face can be used: first, half-arcs are added and then these are extended to arcs by consecutively closing 4-faces.
(b) Note that = 2. Add a quadrangle around the graph with vertices v l , v t , v r , v b in clockwise order.If s and t are expanded, we label the expansion vertices in the respective bags such that s 1 and t 1 have no outgoing arc under α and they are end vertices of the expansion path.If s respectively t is not expanded, we label s = s 1 respectively t = t 1 .Add the arc (s 1 , v t ) and let f * be the inner face that is bounded by the quadrangle.Let ψ t1,s1 denote the incoming flow to f * between t 1 and s 1 clockwise around f ∞ .Assign the label q to the vertex at distance 2ψ t1,s1 + 3 from t 1 walking counterclockwise around f * .Add the arc (t 1 , q).Now we have obtained two faces f u , f d , for which the incoming flow The same method as for an inner face can be used: first, half-arcs are added and then these are extended to arcs by consecutively closing 4-faces.
(c) Note that = 1.Add a quadrangle around the graph with vertices v l , v t , v r , v b in clockwise order.The vertex s has outdegree precisely 1 under α.Label the vertex in the bag of s that has no outgoing edge and is an end vertex of the extension path s 1 or if s is not expanded we label s = s 1 .Add the arc (s 1 , v t ).Let f * be the inner face that is bounded by the quadrangle.The incoming flow of f * is of size 1  2 | f ∞ | + 1 and the size is given by |f The same method as for an inner face can be used: first, half-arcs are added and then these are extended to arcs by consecutively closing 4-faces.
(d) Note that = 0. Add a quadrangle around the graph with vertices v l , v t , v r , v b in clockwise order and let f * be the inner face that is bounded by the quadrangle.We will use one unit of flow to connect f ∞ to the quadrangle.First add a half-arc into f * from each vertex clockwise after a unit of incoming flow.Choose any half-arc and connect it to v t .We obtain a face f with The same method as for an inner face can be used: first, half-arcs are added and then these are extended to arcs by consecutively closing 4-faces.
In the resulting graph each inner face f of G is quadrangulated and the outer face of G is quadrangulated using four help-vertices (v t , v r , v b , v l ).Due to the introduction of the half-arcs according to the incoming flow in a face, we have ensured that each bag through which k units of flow enter the face gets k new outgoing arcs.
To obtain a 2-orientation of the quadrangulation G Q , the edges that are strictly inside the bags, and the four boundary edges need to be oriented.The orientation of the original edges is inherited from α, the help-edges are already Suppose one of the bag-vertices has outdegree 3 or more.At most one of the arcs is an edge of the original graph, as otherwise the vertex would not have been expanded and no new outgoing arc is added.When an outgoing arc is added to a vertex, the vertex is located clockwise after a path-edge, i.e., an edge that was added during the expansion of the vertex, and this path-edge is crossed by a unit of flow routed into the adjacent face for which the vertex is clockwise after the edge.If a vertex has two added arcs, it must be incident to two faces, one on each side of the path and such that the vertex is clockwise next of the path-edge with respect to this face.Hence, this vertex is incident to two path-edges and is not an end vertex of the path.Therefore, it cannot have an outgoing arc that is an edge of the original graph.It follows that every vertex has outdegree at most 2.
Orient v l v t , v r v t , v l v b and v r v b towards v t and v b , respectively.The vertices v t and v b are the two poles of the 2-orientation.The orientation can be completed in a greedy way.
Proof: Let u 1 , u 2 , . . ., u s be the vertices of a bag in the ordering given by the path.Think of the path as being horizontal.We know that u 1 and u s have outgoing edges that come from the 2-orientation α or an outgoing edge that connects to the surrounding quadrilateral.In addition, a unit of flow that goes through the edge u i , u i+1 into the "upper face" implies an outgoing edge at u i .We orient such an edge as u i+1 → u i so that the flow through the edge contributes a single outgoing edge at each of the endpoints.A unit of flow that goes through the edge u j , u j+1 into the "lower face" implies an outgoing edge at u j+1 .We orient such an edge as u j → u j+1 so that again the flow through the edge contributes a single outgoing edge at each of the endpoints.Taking into account the special outgoing edges at u 1 and u s we have a total of 2s outgoing edges from the bag.Also, every vertex in the bag has outdegree at most 2. Therefore, every vertex of the path now has outdegree 2.
Step 3. From G and the realizable pair (α, ψ) we constructed a quadrangulation G Q with a 2-orientation α (v t and v b are the only two vertices with outdegree 0 instead of outdegree 2).We construct a segment contact representation, the vertices of the two color classes of G Q become horizontal and vertical segments and the edges are proper contacts between the segments satisfying α.
Step 4. It remains to show that this segment representation of G Q can be transformed to a VCPG of G that realizes (α, ψ).The segment contact representation of G Q contains a paths p v for each vertex that reflects the behaviour prescribed by the pair (α, ψ).If the bag b v of v contains the path u 1 , u 2 , . . ., u s , then path p v starts and ends at the contact points of the segments representing u 1 and u s with the segments representing their special outneighbors.The bends of p v are at the contact points of segments of pairs u i , u i+1 of adjacent vertices in the bag.
The collection of these paths p v for v ∈ V does not immediately induce the VCPG.In the example in Figure 11 (a), the path of 1, belonging to vertex b does not end on the part of 10 that will belong to vertex a.To be precise, the path of 1 ends on a part of 10 that belongs to a help-edge.In general it might be that a grid-path that is supposed to end on another grid-path ends on a part that belongs to a help-edge.The following lemma shows that we can change the order of segments ending on different sides of a segment.It follows that the segment contact representation can be changed such that all contacts are as we want them to be.
Lemma 7 Given a vertical segment s in a segment contact representation, suppose a horizontal segment s ends on the left of s and s r ends on the right of s such that the endpoint of s on s is below the endpoint of s r on s.Then there is an equivalent segment contact representation, i.e., with the same corresponding 2-orientation, where the endpoint of s on s is above the endpoint of s r on s.
Proof: Consider a cutline consisting of an interval on s and two horizontal rays.The first ray starts just below s and points to the left.The second ray starts just above s r and points to the right.The base-points of both rays on s are chosen such that the rays only intersect vertical segments.The interval on s is the interval between the two base-points.Now cut the graph along the cutline and shift the half containing s up until s is above s r .Vertical segments that are intersected by the rays now have disconnected parts, connect these parts vertically through the slab opened by the shift.This results in a new, equivalent, segment contact representation where s is higher than s r .Figure 11 shows an example where s = 10, s = 9 and s r = 1.
Clearly, there is a symmetric construction that allows to change the order of two segments touching a horizontal segment one from above and the other from below.We call this operation a shift operation.
Let s be a vertical segment and let s 1 , . . ., s a and s r1 , . . ., s r b be the segments touching s from the left resp.from the right, where each of the two sets have increasing order of y-coordinates.Let a merge of the segments be a permutation of the a + b segments where the subpermutation of the segments touching from the left is s 1 , . . ., s a and the order of the segments touching from the right is s r1 , . . ., s r b .We can realize any merge of the a + b segments touching s as the order of their y coordinates.A procedure that achieves this is working upwards on s.Lemma 7 allows to shift the needed next segment from the left or the right chain of segments below all the still untouched segments.Again there is a symmetric statement for horizontal segments.Identification.For each vertex v we will identify a set of parts of segments that form a grid-path p v .Basically the identified set represents the path edges and the original outgoing edges of v.This grid-path will represent the vertex in the VCPG.We call the parts of the segments that are selected the highlighted parts (see Figure 11).For a vertex v that is not expanded, we take the whole segment as p v .For each bag b v , select the segments representing the bag-vertices u 1 , . . ., u s .For each bag-vertex u i , use the part of the segment between u i−1 and u i+1 for p v .For the end vertices of the bag, u 1 (respectively u s ) take the part of the segment between u 2 (respectively u s−1 ) and the outgoing neighbor of u 1 (respectively u s ) that is not in the bag nor reached via a help-edge.Together these pieces form the grid-path p v associated with v.When = 1, 2 we have added arcs while quadrangulating the outer face that do not correspond to the flow nor to original edges.These edges represent the free ends of grid-paths.We cut such segments right after the last neighbor ending on this segment to get the free end of p v .
Shift operations.It may occur that for an arc of G, say v → u, the endpoint from v on u appears on a non-highlighted part of a segment that contributes to the path p u of color u.This may occur because of the added help-edges.This is to be repaired with a shift operation as described in Lemma 7. We have to show that such a shift operation can indeed be used for the repair.
Let u i be the vertex of the bag b u corresponding to the segment s i on which p v ends.Let u i−1 and u i+1 be the neighbors of u i in the bag b u .For i = 1, we also let u 0 be the special outneighbor of u 1 , i.e., the outneighbor that is not introduced because of the flow unit associated with u 1 , u 2 .Similarly, u s+1 is the special outneighbor of u s .
In the following, it is crucial that when walking clockwise around s i , we see the contacts with other segments in the same order as we see around u i in G Q the edges to neighbors.
If u i−1 ← u i → u i+1 , then the complete segment s i belongs to p u .Hence, the end-point of p v on s i is already on p u .
Our second case is For ease of reference, we assume that s i is vertical with the segment of u i+1 to its right.Figure 12(a) shows a generic situation.The figure shows G Q and the flow associated to u i , u i+1 in the upper sketch, and the representation with segment contacts in the lower sketch.With a and b we have included two possible neighbors of u that are attached to u i .The path p b already meets p u , a contact of p a with p u can be created with the shift operation.It is crucial that on the right side of s i above u i+1 there can be no contact with p v .This is because the flow associated to u i , u i+1 and the help-edge leaving u i that is associated to this flow point into the same face of G.The case u i−1 → u i → u i+1 is the same as the previous, just the numbering of the vertices in the bag is reversed.
We are left with the case u i−1 → u i ← u i+1 .Again we assume that s i is vertical with the segment of u i+1 to its right.The segment of u i−1 is to the left of s i because around u i in G Q vertices u i−1 and u i+1 are separated by the outgoing edges.As above on the right side of s i above u i+1 and on the left side of s i below u i−1 , there can be no contact with a path p v .Figure 12  Deletion.After the shift operations, all edges v → u of G are represented by an endpoint of p v on p u .We delete all parts of the segments that are not used by paths p v .
Conclusion.It follows from the three steps that each edge v → u of G is represented by a contact of the corresponding paths p v and p u .Each vertex v is represented by a path consisting of intervals on ψ(v) + 1 segments, hence, it is a grid-path with ψ(v) bends.The result is a VCPG of G that agrees with α and ψ.
With the four steps we have obtained a VCPG from a realizable pair.This completes the proof of Theorem 1.Using this construction, there is no control over the number of bends of a grid-path.A realizable pair precisely gives the number of bends of each vertex.Adding costs to the vertices in the flow network, it is possible to request few bends for certain vertices, and allow for more bends at others.The characterization also gives a certificate that the total number of bends is minimized.If all the costs are set to 1, and there is a realizable pair that attains a minimum flow, then this is a certificate.Even if this is not the case, the flow value gives a lower bound on the number of bends needed in any representation.We do not know an example where the minimum flow value deviates from the minimum number of bends in a representation, i.e., it might be that there is always a realizable pair (α, ψ) where ψ is a minimum cost flow.
In the following sections, we will show some results on the number bends for a vertex, we will try to locally minimize the number of bends for particular graph classes.
The following proposition is an application of the VCPG of a (2, )-sparse graph G.
Proof: Let G be a (2, )-sparse graph.Let B be the bipartite graph consisting of two copies of each vertex of G in one color class, and the edges of G in the second color class.The edges of B are the incidences between the edges and vertices in G. From the sparsity condition, it follows that B has a matching in which all the edge-vertices are matched: Hence, the graph G has a 2-orientation and by Lemma 8 G admits a VCPG.Let R be a VCPG of G. Choose special free ends, which we do not consider in the following.
Within a face f that contains a free end of the grid-path p v which represents v, the following step is taken: • If the f is not a 3-face, p v is extended such that it ends on the grid-path of a non-neighbor of v. • If the f is a 3-face, we add a new segment perpendicular to p v at the free end.This segment is extended on both sides to make contacts with the other two vertices of the 3-face.
This takes care of all but of the free ends, and the resulting graph must be (2, )-tight.To obtain a (2, 0)-tight supergraph, the remaining free ends removed with the same technique.The graph that is represented by the resulting VCPG is a tight graph that contains G as a subgraph.
Lemma 9 is part of the PhD thesis of the first author.In a recent paper Alam et al. [2] prove a slightly stronger result.They show that for ∈ {0, 1, 2, 3} there is a (2, )-tight graph that has G as a spanning subgraph.

B 1 -VCPGs
In a B 1 -VCPG, each vertex is represented by a grid-path with at most one bend.Kobourov et al. have shown that every (2,3)-tight graph admits a strict B 1 -VCPG, i.e., every vertex has precisely one bend [12].To obtain a strict B 1 -VCPG, the grid-paths of the two vertices on the outer face, get a bend in the outer face, i.e., both the convex and the concave angle of these bends are in the outer face.The representation obtained by the algorithm of Kobourov et al. has the special property that every interior face has precisely one convex angle of a vertex.The authors call this property 'proper' and they show that plane Laman graphs are precisely the graphs that admit a proper strict B 1 -VCPG.For readers familiar with this work, the angular tree can be seen as a flow in the angle graph (see Figure 14).The 2-orientation obtained from splitting along the angular tree, together with the flow given by the angular tree, is a realizable pair.Kobourov   In this section we will answer this question by showing a graph that admits B 1 -VCPG but no strict B 1 -VCPG and that not all planar (2,2)-tight graphs admit a B 1 -VCPG.We will also show that every (2,2)-tight planar graph that has an edge e such that the removal of e leaves a (2,3)-tight graph, has a strict B 1 -VCPG (not a proper strict B 1 -VCPG).
There are graphs that are not (2,3)-sparse but admit a B 1 -VCPG, for example K 4 .In this section, we give some insight into the class of graphs that admit a B 1 -VCPG.As mentioned before, it is known that plane (2,3)-tight graphs are precisely the graphs that admit a proper, strict B 1 -VCPG [12].It follows that planar (2,3)-sparse graphs admit a strict B 1 -VCPG.The complete graph on four vertices, K 4 , shows that there are graphs that have strict B 1 -VCPG but no proper strict B 1 -VCPG.We will show that there are graphs that have a B 1 -VCPG but no strict B 1 -VCPG, i.e., there exist graphs that can be represented when segments are allowed.We will also show that not all planar (2,2)-tight graphs admit a B 1 -VCPG.
Lemma 10 There are graphs admitting a B 1 -VCPG but no strict B 1 -VCPG.
Proof: The graph G in Figure 15 admits a B 1 -VCPG but no strict B 1 -VCPG.Every embedding of the graph G contains a subgraph embedded as the subgraph H shown in the right part of the figure.We now concentrate on a B 1 -VCPG representation of H.To satisfy the demands of all 3-faces, the six boundary vertices of H in total have six bends.If x had a bend, then there would be a unit of flow traversing each of the two 4-faces incident to x.This implies that a boundary vertex would get one additional unit of flow.Hence, at least one of the boundary vertices will have two bends.Therefore, x has to be represented as a segment in a B 1 -VCPG of this graph.
Proof: Suppose a graph has a B 1 -VCPG, then, by Theorem 1, there must be a realizable pair (α, ψ) such that 0 ≤ ψ(v) ≤ 1 for all vertices v. Consider the graph from Figure 16.Suppose that there is a plane embedding and a flow ψ which satisfies 0 ≤ ψ(v) ≤ 1 for all vertices v.Each embedding has two K 4 subgraphs whose faces are bounded by triangles sharing only vertex m (the grey-colored K 4 subgraphs).These K 4 subgraphs both have demand −3.So, there must be 6 units of flow going out of the two K 4 subgraphs.However, there are only 5 vertices bounding the two K 4 subgraphs.We conclude that there is no feasible flow such that there is at most one unit of flow going through each vertex.A Laman-plus-one graph G is a (2, 2)-tight graph such that there exists an edge e in G for which G − e is (2, 3)-tight, i.e. a Laman graph [10].In the following, we extend the result of Kobourov et al. from planar (2, 3)-tight graphs to planar Laman-plus-one graphs.There are two bends of vertices that are completely in the outer face (the bends of s 1 and s 3 in Figure 14(c)).The idea is to use these bends for the additional edge. of G such that ψ * satisfies all demands of ψ.In the second step, we route the auxiliary flow on the edges of A(G).
The demand of the faces is given by c(f ) = |f | − 4 and for the outer face it is c(f ∞ ) = |f ∞ | (Proposition 2 with = 2).We claim that there is a flow ψ * satisfying all demands such that the flow value in each dual edge is at most one.This is shown by checking the size of all cuts.Let H be a subgraph induced by a subset F H of bounded faces of G. Let b be the number of boundary edges of H.We have to show that b Let v H , e H and f * H be the number of vertices, edges and bounded faces of H, respectively.Using Euler's formula (v H −e H +f * H = 1) and the upper bound on the edges in H, we obtain the following.
The vertices V H of H are partitioned into boundary vertices B H and inner vertices I H .We have Therefore, we obtain the following: Putting this together we obtain: It follows from the max-flow min-cut theorem, that there is a flow in the dual graph that satisfies all demands and respects the capacity bound of 1 in each edge.Since all constraints are integral we can assume that the flow is integral, i.e., a 0, 1-flow.From Lemma 1, we know that there is a 2-orientation α of G such that the only vertices with outdegree less than 2 are incident to the outer face.Using α, we construct the flow ψ in the angle graph as follows: If there is a flow from f 1 to f 2 crossing edge u → v, then we add f 1 → u and u → f 2 to ψ.
Since the flow in the dual graph is edge-disjoint and each vertex has at most two outgoing edges we have ψ(v) ≤ 2 for all vertices v.At each vertex, the flow cuts off the outgoing edge.Hence, the realizability condition is satisfied at each vertex.We conclude that the pair (α, ψ) is realizable.

B k -VCPGs for k > 2
For (2,2)-tight graphs, using a flow in the dual gives a tight bound.We have seen before that there are planar (2, 2)-tight graphs which do not admit a B 1 -VCPG (Lemma 11).On the other hand, using the dual graph, we could show that two bends per vertex are sufficient.Unfortunately this is not the case for (2,1)-and (2,0)-tight graphs.
Proof: claim that there is a flow in the dual graph such that each edge has capacity 2. Similarly to the proof of Theorem 4, we obtain that a simple planar (2, 1)-tight graph admits a B 4 -VCPG.
Again, let H be a subgraph induced by a subset F H of bounded faces of G and let b be the number of boundary edges of H. From Equations ( 5) and ( 6) we obtain: There are no loops, therefore, b > 1.It follows that the demands can be satisfied using b edges with capacity 2.
We have only been able to show that the lower bound is 2.An example is given by the octahedron with an interior edge removed.We conjecture that every simple planar (2, 1)-tight graph admits a B 2 -VCPG.
Proof: Let G be a simple planar (2, 0)-tight graph.Again, let H be a subgraph induced by a subset F H of bounded faces of G and let b be the number of boundary edges of H. From Equations ( 5) and ( 6) we obtain: As there are no loops, we can satisfy all the demands in the dual graph by using every edge at most thrice.As in the proof of Theorem 4, this induces a realizable pair, in which every vertex has at most 6 units of flow.This yields a B 6 -VCPG.

Obtaining Better Bounds
In the previous section, we have used an auxiliary flow ψ * in the dual graph to obtain a feasible flow ψ in the angle graph.This way, we are certain that there exists a 2-orientation such that the pair is realizable and, hence, is in bijection to a VCPG.Even stronger, the choice of the 2-orientation does not matter.Unfortunately, it neither minimizes the total number of bends nor does it minimize the number of bends per vertex when a 2-orientation is given.We illustrate this with the following example.In Figure 18, two VCPGs of the octahedron are shown, the left of which does not relate to a feasible flow in the dual graph.Any feasible flow in the dual graph will lead to a VCPG with at least 12 bends in total, while 8 is the minimum.Any feasible flow in the dual graph together with the chosen orientation, ensures that c has no bends incident to the outer face in the VCPG.To close the outer cycle at least 7 bends are needed, which will be divided over a and b, hence, there will be a vertex with 4 bends while 3 is the minimum.It has been fruitful to use the dual graph for bounding the number of bends locally at each vertex, but what if we want to globally minimize the number of bends.Ideally, there would always exists a 2-orientation such that the minimum cost feasible flow is realizable.However, for the graph in Figure 19, a minimum cost flow is given (the value of the flow is equal to the number of 3-faces).For this flow, there does not exist a 2-orientation such that the pair is realizable.Suppose there is such a 2-orientation.The realizability condition requires the orientations a → c and b → c.Then c has only one outgoing edge, but the flow requires the free ends to be in the outer face, contradiction.Note that rerouting a unit of flow that uses a or b to go through c leaves a flow for which there does exist a 2-orientation such that the pair is realizable.

Conclusion
We have shown that there exist planar (2, 2)-tight graphs that do not admit a B 1 -VCPG.However, the only type of (2, 2)-tight planar graph that we found not to have a B 1 -VCPG has at least one vertex which is the intersection of "many" critical subsets.Do all planar (2,2)-tight graphs that have no such vertex admit a B 1 -VCPG?
We have also obtained bounds for simple (2,1)-tight and (2,0)-tight planar graphs.However, we believe that these bounds are not tight.A lower bound of The bounds that we have shown do not depend on a chosen 2-orientation (i.e. the bounds hold for every 2-orientation).It would be interesting to find a sufficient condition on a flow such that, when satisfied, there exists a 2orientation such that the pair is realizable.For (2,3)-tight graphs, the algorithm of Kobourov, Ueckerdt and Verbeek, takes a particular flow and a particular network as an input.Is there a way to construct a realizable pair simultaneously for all (2,0)-sparse graphs?

Figure 1 :
Figure 1: Two VCPGs of the octahedron, which induce the same 2-orientation (shown in the middle) but the number of bends differs.

Figure 2 :
Figure 2: Faces with their demand and possible representations by grid-paths with appropriate numbers of convex and concave angles.

Figure 3 :
Figure 3: An example of a feasible flow for K 4 (inner faces have demand −1, the outer face has demand 3) with the dual graph (left) or the angle graph (right) as underlying graph.

Figure 4 :Proposition 2
Figure 4: The angle graph (black) of the octahedron (grey) and the angle graph with the feasible flow induced by the VCPG on the left of Figure 1.In the flow, the edges that get weight zero are omitted, the edges that have weight larger than 1 are labeled with their weight.

Figure 5 :
Figure 5: A feasible flow does not uniquely define a VCPG.On the left two VCPGs, and on the right the feasible flow induced by both of the VCPGs.The difference is the orientation of the cycle e, a, b, f (in red).

Figure 6 :
Figure6: A feasible flow ψ that does not belong to a VCPG, i.e. there exists no 2-orientation α such that the pair (α, ψ) is realizable.

Figure 7 :
Figure 7: (a) The vertex v in G and the local flow.(b) The expansion of v: The flow through v is decomposed into disjoint paths and each of these paths cuts the path consisting of the outgoing edges of v. (c) A close look at the grid-path of a vertex v in a VCPG.
After all the expansions have been done we obtain a graph where all faces have even length.Each face gets ||f | − 4|+2k extra vertices due to the expansion step, where k is the amount of flow crossing through the face.The resulting faces in G have size |f | + ||f | − 4| + 2k, for |f | = 3 this is 4 + 2k and for |f | > 3 this is 2|f | − 4 + 2k, both are even.So, all faces in G have even length and therefore G is a bipartite graph.

Figure 8 :Lemma 3
Figure 8: From a realizable pair to a VCPG: (a) a plane (2,0)-tight graph with a realizable pair (ψ in red); (b) expanding the vertices according to the flow (in blue) and extending the bipartite graph to a quadrangulation (in green); (c) a segment contact representation of the quadrangulation with the segments that belong to the original graph highlighted; (d) a VCPG.

Figure 9 :
Figure 9: Adding help-edges in a face.The flow ψ is depicted by red arcs.The green half-arcs together with the dashed extensions represent the help-edges.
(a) There exists a vertex s in the original graph G which has no outgoing arcs under α.(b) There exist two vertices s, t in the original graph G which both have precisely one outgoing arc under α.(c) There exists precisely one vertex s which has precisely one outgoing arc under α.(d) All vertices have outdegree 2 under α.

Figure 10 :
Figure 10: Adding a quadrilateral around the graph.

Lemma 5
oriented.Each bag b v contains |b v | − 1 = ψ(v) edges which are not yet oriented.The outdegree of each bag b v in G Q is precisely |b v | + 1 and the outdegree of each vertex is at most 2. Proof: The bag b v of a vertex v has ψ(v) + 1 vertices.According to the flow, ψ(v) outgoing arcs are added at the introduction of the help-edges.If a bag b v comes from a vertex v which has outdegree 2, the bag has outdegree 2 + ψ(v) = |b v | + 1.If a bag b v comes from a vertex v which has outdegree 1 in α, then the quadrangulation step has assigned another outgoing arc to this bag.Therefore the bag has outdegree |b v | + 1.If a bag b v comes from a vertex v which has outdegree 0 in α, then the quadrangulation step has assigned two outgoing arcs to this bag.Therefore the bag has outdegree |b v | + 1.

Figure 11 :Theorem 2
Figure 11: Cutting open, shifting and extending shows that neighbors on each side can be moved independently: (a) a segment contact representation where the highlighted part of 1 does not end on the highlighted part of 10, and a cutting line (dashed); (b) the segment contact representation is cut and the top is pulled upwards extending the vertical segments that are cut.
(b) shows a generic situation.With a and b we have included two possible neighbors of u that are attached to u i .Taking the segments of a and b as the reference segments for a shift operation we transform the representation such that p a and p b both end on p u .

Figure 12 :
Figure 12: Using shift operations, the missing contacts from p a and p b to p u can be established.

Figure 13 :
Figure 13: A VCPG drawn by approximating a straight-line drawing of a graph and using a 2-orientation.

Figure 14 :
Figure 14: An angular tree, splitting to obtain a 2-orientation and a proper, strict B 1 -VCPG.

Figure 15 :
Figure 15: Every embedding of the graph G contains a subgraph embedded as H.In a B 1 -VCPG corresponding to the embedding of H vertex x must be represented by a segment.

Figure 18 :
Figure 18: Two VCPGs of the octahedron that induce the same 2-orientation.The VCPG on the left does not relate to a flow in the dual graph as the rightmost bend on the bottom (of the grid-path of vertex b) cannot be transformed into a flow in the dual graph.

Figure 19 :Conjecture 1 Conjecture 2
Figure19: A minimum feasible flow for which there is no 2-orientation such that the pair is realizable.