1 Introduction
The importance of contact and intersection representations of graphs stems not only from their numerous applications including information visualization, chip design, bio informatics and robot motion planning (see for example the references in [2, 10]), but also from the structural and algorithmic insights accompanying the investigation of these intriguing geometric arrangements. From a structural point of view, the certainly most fruitful contact representations (besides the “Kissing Coins” of Koebe, Andrew, and Thurston [17, 1, 24]) are axisaligned segment contact representations: families of interiordisjoint horizontal and vertical segments in where the intersection of any two segments is either empty or an endpoint of at least one of the segments. The corresponding touching graph^{1}^{1}1We use the term touching graphs rather than the more standard contact graph to underline the fact that segments with coinciding endpoints (e.g., two horizontal segments touching a vertical segment in the same point but from different sides, but also nonparallel segments with coinciding endpoint) do not form an edge. has the segments as its vertices and the pairs of segments as its edges for which an endpoint of one segment is an interior point of the other segment, see the left of Fig. 1. It has been discovered several times [16, 22] that any such touching graph is bipartite and planar, and that these two obviously necessary conditions are in fact already sufficient: Every planar bipartite graph is the touching graph of interiordisjoint axisaligned segments in . In fact, edgemaximal segment contact representations endow their associated plane graphs with many useful combinatorial structures such as orientations [10], separating decompositions [4], bipolar orientations [23, 25], transversal structures [13], and Schnyder woods [27].
In this paper we extend axisaligned segment contact representations in to axisaligned rectangle contact representations in . That is, we consider families of axisaligned closed and bounded rectangles in with the property that for all the intersection is a subset of the boundary of at least one of them, i.e., the rectangles are interiorly disjoint. We call such a family a Plattenbau^{2}^{2}2Plattenbau (plural Plattenbauten) is a German word describing a building (Bau) made of prefabricated concrete panels (Platte).. Given a Plattenbau one can consider its intersection graph . However, for us the more important concept is a certain subgraph of , called the touching graph of . There is one vertex in for each rectangle in and two vertices are adjacent if the corresponding rectangles touch, i.e., their intersection is nonempty and contains interior points of one and only one of the rectangles. We say that is a Plattenbau graph if there is a Plattenbau such that . In this case we call a Plattenbau representation of .
Plattenbauten are a natural generalization of axisaligned segment contact representations in and thus Plattenbau graphs are a natural generalization of planar bipartite graphs. While clearly all Plattenbau graphs are tripartite (properly vertex colorable), it is an interesting challenge to determine the exact topological properties in that hold for all Plattenbau graphs, thus generalizing the concept of planarity from to dimensions (for tripartite graphs). We present results towards a characterization of Plattenbau graphs in three directions.
Our Results and Organization of the Paper.
In Section 2 we provide some simple examples of Plattenbau graphs and give some simple necessary conditions for all Plattenbau graphs. We observe that unlike touching graphs of segments, general Plattenbau graphs are not closed under taking subgraphs. We circumvent this issue by introducing the subclass of proper Plattenbauten where for any two touching rectangles the intersection must be a boundary edge of one of . Moreover, we introduce boxed Plattenbauten where every bounded region of is a box, and discuss questions of augmentability.
In Section 3 we show that within planar graphs the necessary condition of colorability is also sufficient for Plattenbau graphs. Thus, the topological characterization of Plattenbau graphs must fully contain planarity (which is not obvious as we consider colorable graphs and not only bipartite graphs).
Theorem 1.1 ()
Every colorable planar graph is the touching graph of a proper Plattenbau.
Along the proof of Theorem 1.1, we obtain a characterization of skeletons of orthogonal surfaces which is implicit already in work of Eppstein and Mumford [6]. Another proof of Theorem 1.1 can be obtained from Gonçalves’ recent proof that colorable planar graphs admit segment intersection representations with segments of 3 slopes [14]. We further comment on these alternative approaches in Section 3.3. A consequence of our approach is a natural partial order  namely a distributive lattice  on the set of orthogonal surfaces with a given skeleton.
In Section 4 we consider proper and boxed Plattenbau graphs as the dimensional correspondence to the edgemaximal planar bipartite graphs, the quadrangulations. We give a complete characterization of these graphs.
Theorem 1.2
A graph is the touching graph of a proper boxed Plattenbau if and only if there are six outer vertices in such that each of the following holds:

is connected and the outer vertices of induce an octahedron.

The edges of admit an orientation such that

the bidirected edges are exactly the outer edges,

each vertex has exactly outgoing edges.


The neighborhood of each vertex induces a spherical quadrangulation in which the outneighbors of induce a cycle.

If is an outer vertex, this cycle bounds a face of .


For every edge of with common neighborhood , the cyclic ordering of around in is the reverse of the cyclic ordering of around in .
A spherical quadrangulation is a graph embedded on the dimensional sphere without crossings with all faces bounded by a cycle. Spherical quadrangulations are connected, planar, and bipartite. We remark that Theorem 1.2 does not give a complete characterization of proper Plattenbau graphs since some proper Plattenbau graphs are not contained in any proper boxed Plattenbau graph as discussed in Section 2.
2 Types of Plattenbauten and Questions of Augmentation
Let us observe some properties of Plattenbau graphs. Clearly, the class of all Plattenbau graphs is closed under taking induced subgraphs. Examples of Plattenbau graphs are , see Fig. 2, and the class of grid intersection graphs, i.e., bipartite intersection graphs of axisaligned segments in the plane [16]. For the latter take the segment intersection representation of a graph, embed it into the plane in and thicken all horizontal segments a small amount into direction and all vertical segments a bit into direction outwards the plane. In particular, is a Plattenbau graph, see Fig. 2. In order to exclude some graphs, we observe some necessary properties of all Plattenbau graphs.
Observation 1
If is a Plattenbau graph, then

the chromatic number of is at most ,

the neighborhood of any vertex of is planar.

the boxicity of , i.e., the smallest dimension such that is the intersection graph of boxes in , is at most .
Proof
Item 1: Each orientation class is an independent set.
Item 2: Let be a vertex of represented by . Let
be the supporting hyperplane of
and the corresponding closed halfspaces. The neighborhood consists of rectangles intersecting and those intersecting . The rectangles in each of these sets have a plane touching graph, since it corresponds to the touching graph of the axisaligned segments given by their intersections with . The neighboring rectangles in are on the outer face in both graphs in opposite order, so identifying them gives a planar drawing of the graph induced by . See Fig. 3 for an illustration.Item 3: A Plattenbau can be transformed into a set of boxes such that the touching graph of is the intersection graph of as follows: First, shrink each rectangle orthogonal to the axis by a small enough in both dimensions different from . As a result, we obtain a set of pairwise disjoint rectangles. Then, expand each such rectangle by in dimension . The obtained set of boxes are again interiorly disjoint and all intersections are touchings. ∎
Note that for Items 2 and 1 of Observation 1 it is crucial that is the touching graph and not the intersection graph. Moreover, Observation 1 allows to reject some graphs as Plattenbau graphs:

is not a Plattenbau graph (by Item 1 of Observation 1),

is not a Plattenbau graph (by Item 2 of Observation 1).

The full subdivision of is not a Plattenbau graph (by Item 3 of Observation 1 and [3]).
In particular, some bipartite graphs are not Plattenbau graphs. Together with being a Plattenbau graph, this shows that the class of Plattenbau graphs is not closed under taking subgraphs; an unusual situation for touching graphs, which prevents us from solely focusing on edgemaximal Plattenbau graphs. To overcome this issue, we say that a Plattenbau is proper if for any two touching rectangles an edge of one is contained in the other rectangle. (Note the ambiguity of the term “edge” here, which refers to a maximal line segment in the boundary of a rectangle as well as to a pair of adjacent vertices in a graph.) If is proper, then each edge of the touching graph can be removed by shortening one of the participating rectangles slightly. That is, the class of graphs with proper Plattenbau representations is closed under subgraphs.
We furthermore say that a Plattenbau is boxed if six outer rectangles constitute the sides of a box that contains all other rectangles and all regions inside this box are also boxes. (A box is an axisaligned fulldimensional cuboid, i.e., the Cartesian product of three bounded intervals of nonzero length.) For boxed Plattenbauten we use the additional convention that the edgetoedge intersections of outer rectangles yield edges in the touching graph, even though these intersections contain no interior points. In particular, the outer rectangles of a proper boxed Plattenbau induce an octahedron in the touching graph.
Observation 2
The touching graph of a proper Plattenbau with vertices has at most edges. Equality holds if and only if is boxed.
Proof
For a proper Plattenbau with touching graph there is an injection from the edges of to the edges of rectangles in : For each edge in with corresponding rectangles , take the edge of or that forms their intersection . This way, each of the four edges of each of the rectangles in corresponds to at most one edge in .
Moreover, if contains at least two rectangles of each orientation, the bounding box of contains at least edges of rectangles in its boundary, none of which corresponds to an edge in . Thus, in this case has at most edges. Otherwise, for one of the three orientations, contains at most one rectangle in that orientation. In this case, is a planar bipartite graph plus possibly one additional vertex. In particular, has at most edges, as long as .
Finally, in order to have exactly edges, the above analysis must be tight. This implies that has at least two rectangles of each orientation and its bounding box contains exactly edges of rectangles, i.e., is boxed. ∎
An immediate consequence of Observation 2 is that is a Plattenbau graph which has no proper Plattenbau representation. Contrary to the case of axisaligned segments in , not every proper Plattenbau in can be completed to a boxed Plattenbau; see Fig. 4 for a problematic example. This example can also easily be extended to give a graph that is the touching graph of a proper Plattenbau, but that is not a subgraph of any Plattenbau graph with a proper and boxed Plattenbau representation.
3 Planar Colorable Graphs
Let us recall what shall be the main result of this section:
See 1.1
The proof of this theorem is in several steps. First we introduce orthogonal surfaces and show that the dual graph of the skeleton of an orthogonal surface is a Plattenbau graph (Proposition 1). In the second step we characterize triangulations whose dual is the skeleton of an orthogonal surface (Proposition 2). One consequence of this is a natural very wellbehaved partial order, namely a distributive lattice, on the set of orthogonal surfaces with given skeleton (Corollary 1). We then show that a Plattenbau representation of a colorable triangulation can be obtained by patching orthogonal surfaces in corners of orthogonal surfaces (Section 3.2).
We begin with an easy observation.
Observation 3
Every colorable planar graph is an induced subgraph of a colorable planar triangulation.
Proof (Sketch)
Consider with a plane embedding. It is easy to find a connected colorable which has as an induced subgraph.
Fix a coloring of . Let be a face of of size at least four and be a color such that at least three vertices of are not colored . Stack a vertex inside and connect it to the vertices on that are not colored . The new vertex is colored and the sizes of the new faces within are or . After stacking in a face, the face is either triangulated or there is a color which is not used on the newly created face. A second stack triangulates it. ∎
A plane triangulation is colorable if and only if it is Eulerian. Hence, the dual graph of apart from being connected, cubic, and planar is also bipartite. The idea of the proof is to find an orthogonal surface such that is the skeleton of . This is not always possible but with a technique of patching one orthogonal surface in an appropriate corner of a Plattenbau representation obtained from another orthogonal surface, we shall get to a proof of the theorem.
Consider with the dominance order, i.e., if and only if for . The join and meet of this distributive lattice are the componentwise and . Let be a finite antichain, i.e., a set of mutually incomparable points. The filter of is the set and the boundary of is the orthogonal surface generated by . The left part of Fig. 5 shows an example in . The nine vertices of the generating set are emphasized.
Orthogonal surfaces have been studied by Scarf [15] in the context of test sets for integer programs. They later became of interest in commutative algebra, cf. the monograph of Miller and Sturmfels [21]. Miller [20] observed the connections between orthogonal surfaces, Schnyder woods and the BrightwellTrotter Theorem about the order dimension of polytopes, see also [8].
A maximal set of points of an orthogonal surface which is constant in one of the coordinates is called a flat. A nonempty intersection of two flats is an edge. A point contained in three flats is called a vertex. An edge incident to only one vertex is a ray. We will only consider orthogonal surfaces obeying the following nondegeneracy conditions: (1) The boundary of every bounded flat is a simple closed curve. (2) There are exactly three rays. Note that from (1) it can be deduced that every vertex is contained in exactly three flats.
The skeleton graph of an orthogonal surface consists of the vertices and edges of the surface, in addition there is a vertex which serves as second vertex of each ray. The skeleton graph is planar, cubic, and bipartite. The bipartition consists of the maxima and minima of the surface in one class and of the saddle vertices in the other class. The vertex is a saddle vertex. The dual of is a triangulation with a designated outer face, the dual of .
The generic structure of a bounded flat is as shown in Fig. 6; the boundary consists of two zigzag paths sharing the two extreme points of the flat. The minima of the lower zigzag are elements of the generating set , they are minimal elements of the orthogonal surface . The maxima of the upper zigzag are maximal elements of , they can be considered to be dual generators.
With the following proposition we establish a first connection between orthogonal surfaces and Plattenbau graphs.
Proposition 1
The dual triangulation of the skeleton of an orthogonal surface is a Plattenbau graph and admits a proper Plattenbau representation.
Proof
Choose a point not on on each of the three rays of and call these points the extreme points of their incident unbounded flats.
The two extreme points of a flat of span a rectangle . Note that the other two corners of are and . We claim that the collection of rectangles is a weak rectangle contact representation of the dual triangulation of the skeleton of . Here weak means that the contacts of pairs of rectangles of different orientation can be an edge to edge contact. If and share an edge of the skeleton, then since one of the ends of is a saddle point of and thus extreme in two of its incident flats, it is extreme for at least one of and . This shows that is contained in the boundary of at least one of the rectangles , i.e., the intersection of the open interiors of the rectangles is empty.
Let and be two flats which share no edge. Let and be the supporting planes. If is contained in an open halfspace defined by , then and , the other two corners of , are also in , hence and . If intersects and intersects , then consider the line . This line is parallel to one of the axes, hence it intersects in a closed interval . If and are the intervals obtained by intersecting with and respectively, then one of them equals and the other is an edge of the skeleton of , i.e., is an edge of .
It remains to expand some of the rectangles to change weak contacts into true contacts. Let be an edge such that the contact of and is weak. Select one of and , say . Now expand the rectangle with a small parallel shift of the boundary segment containing . This makes the contact of and a true contact. The expansion can be taken small enough as to avoid that new contacts or intersections are introduced. Iterating over the edges we eventually get rid of all weak contacts. See Fig. 7 for an illustration. ∎
Recall that we aim at realizing , the dual of the colorable triangulation as the skeleton of an orthogonal surface. Since is Eulerian its dual is bipartite. Let (black) and (white) be the bipartition of the vertices of such that the dual of the outer face of is in . The critical task is to assign two extreme vertices to each face of which does not contain . This has to be done so that each vertex in (except ) is extremal for exactly two of the faces.
To solve the assignment problem we will work with an auxiliary graph . The faces of which do not contain are the interior vertices of , we denote this set with . As the vertices of are the facial triangles of , we think of as representing the black triangles of . We also let , this is the set of bounded black triangles of . The vertices of are the edges of correspond to the incidence relation in and respectively, i.e., with and is an edge if vertex is a corner of the black triangle . A valid assignment of extreme vertices is equivalent to an orientation of such that each vertex has outdegree two and each vertex has indegree two, i.e., the outdegrees of the vertices are prescribed by the function with for and for . Since it is readily seen that the sum of the values of all vertices equals the number of edges of .
Orientations of graphs with prescribed outdegrees have been studied e.g. in [9], there it is shown that the following necessary condition is also sufficient for the existence of an orientation. For all and and
() 
Here and denote the set of edges induced by , and the set of edges in the cut defined by , respectively.
Inequality () does not hold for all triangulations and all . We next identify specific sets violating the inequality, they are associated to certain badly behaving triangles, we call them babets. In Proposition 2 we then show that babets are the only obstructions for the validity of ().
Let be a separating triangle of such that the faces of bounding from the outside are white. Let be the set of vertices inside and let be the collection of black triangles of which have all vertices in . We claim that is violating (). If and , then . The triangulation whose outer boundary is has triangles, half of them, i.e., , are black and interior. The right side of () is counting the number of incidences between and black triangles. Black triangles in have incidences in total. There are black triangles have an incidence with a corner of and 3 of them have incidences with two corners of . Hence the value on the right side is . This shows that the inequality is violated. A separating triangle of with white touching triangles on the outside is a babet.
Proposition 2
If has no babet, then there is an orientation of whose outdegrees are as prescribed by .
Before we prove Proposition 2 and Theorem 1.1, let us briefly summarize the procedure.
First, we construct a Plattenbau representation in the babetfree case (Section 3.1) based on an auxiliary graph arising from the bipartition of , and a Schnyder wood for . We then find an orthogonal surface based on the Schnyder wood and show that the skeleton of is , which together with Proposition 1 gives a Plattenbau for . Then (Section 3.2), in case contains some babets, we cut the triangulation along an innermost babet, find orthogonal surfaces for the orthogonal surfaces for the inside and outside, and patch the former into a saddle point of the latter.
Now, let us start with the proof of Proposition 2.
Proof
Suppose that there is an violating inequality (). We are going to modify in several steps always maintaining the property that the inequality is violated. At the end we will be able to point to a babet in .
Suppose there is a with neighbors in . Let when going from to the left side of () is loosing while on the right side we loose the edges of which are incident to but not to . Since the set is violating. From now on we assume that every has 3 neighbors in , in particular .
Now the left side of () equals and for the right side we have and contains no edge incident to . We define the notation indicates that we only have to care of boundary edges of . The assumption that Inequality () is violated then becomes or equivalently
() 
We can assume that the subgraph of induced by is connected, otherwise a connected component would also violate. Let and . The set is a set of black triangles in the triangulation and is the set of vertices of these triangles. Let be the plane embedding of all edges of triangles of as seen in
. Classify the faces of
as black triangles, white triangles and big faces, and let their numbers be , and , respectively. We consider the outer face of a big face independent of its size, therefore, . Consider the triangulation obtained by stacking a new vertex in each big face and connecting it to all the angles of the face, i.e., the degree of the vertex stacked into face equals the length of the boundary of . Note that may have multiedges but every face of is a triangle so that Euler’s formula holds. Let be the sum of degrees of the stack vertices. Since has vertices it has faces. However, we also know that has faces. Counting the edges incident to the triangles of we obtain . Using this to eliminate we obtain , i.e., . With ( ‣ 3) this implies .Claim 1
is divisible by .
Proof of Claim. Actually we prove that each is divisible by 3. Let be a collection of triangles in a colorable triangulation and let be the boundary of , i.e., is the set of edges incident to exactly one triangle from . Let and be the black and white triangles in and let and be the edges of which are incident to black and white triangles of . Double counting the number of edges in we get , hence, . In our case and depending on the chosen either or .
Let be the boundary cycle of a big face of which is not the outer face. We have seen that . We are interested in the contribution of to , i.e., in the number of incidences of vertices of with black triangles in the inside. For convenience we can use multiple copies of a vertex to make simple. The claim is that the contribution of to is at least . If this is shown we can add all the inner black triangles of to and all the inner vertices to , the left side of the violator inequality is thereby reduced by and the right side is reduced by at least , i.e., violators are preserved.
To prove the claim consider the interior triangulation of the simple cycle . Suppose . Then on we find an ear, this is a vertex which has no incidence to a black triangle of , i.e., its degree in is 2. If is an ear and , are the neighbors of on , then is an edge and there is a black triangle in . An ear is reducible if has a neighbor on which is only incident to a single black triangle in . Assume that is such a neighbor of , then the second neighbor of on has an edge to . If is the unique common neighbor of and , then we delete and and identify the edges and . This results in a cycle with . We call this an ear reduction with center . Fig. 8 shows sketches of reducible ears. Note that in one case (depending on whether belongs to or not) and in the other .
If is a reducible edge but and share several neighbors, then there is an extreme common neighbor with the property that the cycle encloses all the common neighbors of and . In this case we perform an ear reduction with center , indeed, due to the mod 3 parity exactly one of the interior triangles of and is black. Again we get a a cycle with and .
After a series of reductions we obtain a cycle which has no reducible ear. Let each vertex with at least two incidences with black triangles discharge to each neighbor, then all vertices have a weight of at least 1. Hence and . This completes the proof of the claim.
We have already seen that the claim implies that we may assume that , i.e., the violator only has a single big face, the outer face . We write , the condition for violation becomes .
Let be the boundary cycle of the unique big face. While the outer face of has to be contained in the big face we prefer to think of a drawing of such that the big face is the interior of . It will be crucial, however, that in the inner triangulation of inherited from there is a special black triangle . The goal is to find a babet in the interior of , we use induction on .
In the case the triangle has incidences with black triangles from the inside. The unique configuration with this property is shown on the left in Fig. 9. Since the black triangle is not yet in the picture one of the triangles must be separating. If the separating triangle were not the central one it would lead to an increase of . Hence the white central triangle is separating, i.e., a babet.
Now let . Assuming a black incident triangle for every vertex of we obtain . Therefore, on we find an ear. As above we aim for an ear reduction. Let be a reducible ear with neighbors and such that is only incident to a single black triangle inside and is the second common neighbor of and . If is the unique common neighbor of and , then the reduction leads to a decrease of by 2 or 3 and a decrease of by 1, whence the reduced cycle remains violating.
If is reducible but and share several neighbors, then there we aim at a reduction whose center is the extreme common neighbor of and . If this cycle does not enclose the black triangle we can apply the reduction with center .
The described reductions may decrease until it is 1 whence there is a babet. In fact we will complete the proof by showing that when no reduction is possible and , then the violator inequality is not fulfilled.
We first discuss the case where is reducible but and share several neighbors and is enclosed in the cycle where is the extreme common neighbor of and . We show that in this case we can find a cycle of length 6 which is a violator, i.e., .
Suppose is an interior of in this case we take the cycle add a new common neighbor for and an ear over the black triangle containing . This yields a 6 cycle as shown in Fig. 10 (left 1). In the interior of we replace the triangles inside by the 3 triangles of a reducible ear, see Fig. 10 (left 2) and refer to the cycle with the simplified interior as . Now we compare with and and use the bound previously shown in the claim for , i.e., . Taking into account that on each of and we see two incidences with black triangles which are not counted in we get . Hence , whence is also a violator.
If is a vertex on the cycle we add a new ear either over or over depending on which is incident to a black triangle. This yields a 6cycle as shown in Fig. 10 (right 1). In the interior of we replace the triangles inside by the 3 triangles of a reducible ear, see Fig. 10 (right 2) and refer to the cycle with the simplified interior as . Taking into account that on and we see three incidences with black triangles which are not counted in we get . Hence , whence is also a violator.
If and there are no reducible ears, then both neighbors of each ear vertex have two incidences with black triangles. Let each vertex with at least two incidences with black triangles discharge to the neighbors, then all vertices have a weight of at least 1. We get and the example is not a violator.
For the case we consider circular sequences with such that there is an inner Eulerian triangulation of a 6gon with only white triangles touching the 6gon and black incident triangles at vertex of the 6gon. A vertex with is an ear. Clearly, the circular sequence has no 00 subsequence. With two ears which share a neighbor of degree 1, i.e., with a 010 subsequence, we have the graph from Fig. 11, this triangulation does not occur in our setting since it does not contain a black which is vertex disjoint from the outer 6gon. Making any of the four triangles of separating so that it can accommodate in its interior would make .
With two ears which are not adjacent we have . Again there is no in and making a triangle separating would make . It remains to look at sequences without 00, and 010, and 101 but . The sequence 201110 is the unique sequence with these properties and there is a unique corresponding triangulation . As with and there is no in and making a triangle separating would make . ∎
3.1 Plattenbau Representations in the BabetFree Case
Let be a colorable triangulation which has no babet. Due to Proposition 2 we find the orientation of . This orientation can be represented on the dual of as a collection of cycles such that every face which is not incident to contains exactly one fragment of a cycle which leaves the face in distinct vertices of and every vertex of is covered by one of the cycles (Eppstein and Mumford [6] refer to this structure as cycle cover). The connected components of can be twocolored in white and pink such that the two sides of each cycle have distinct colors, the twocoloring is unique if we want the unbounded region to be colored white. This twocoloring of the plane induces a partition of the vertices of into where consists of all vertices living in white regions and consists of the vertices in pink regions, see Fig. 12 for an example. An important property of the partition is that every vertex of is adjacent to vertices from both classes and . This follows form the fact that a cycle from contains , whence two of the edges of are on one side and the third is on the other side of .
The partition of the vertices of corresponds to the partition into minima, saddle points, maxima of the vertices of the orthogonal surface with skeleton . To construct this orthogonal surface we first define a connected planar graph whose vertex set is , the edges of are in bijection with and each bounded face of contains exactly one vertex of . This graph will be decorated with a Schnyder wood, i.e., an orientation of the edges which obeys the following rules:

On the outer face of there are three special vertices colored red, green, and blue in clockwise order. For vertex is equipped with an outward oriented halfedge of color .

Every edge is oriented in one or in two opposite directions. The directions of edges are colored such that if is bioriented the two directions have distinct colors.

Every vertex has outdegree one in each color . The edges leaving in colors occur in clockwise order. Each edge entering in color enters in the sector bounded by the two with colors different from (see Fig. 13).

There is no interior face whose boundary is a directed cycle in one label.
The construction of and the Schnyder wood is in several steps. The first step is to define a coloring of the edges of . Let be colored with such that on the outer face these colors appear clockwise in the given order. Note that this implies that all white triangles see in clockwise order and all black triangles see in counterclockwise order. The coloring of induces a coloring of the edges: for edge use the unique color which is not used for and . This edge coloring of can be copied to the dual edges. This yields an edge coloring of such that each vertex in is incident to edges of colors in clockwise order. Delete from but keep the edges incident to as half edges at their other endpoint, we denote the obtained graph as . The neighbors of are the special vertices for the Schnyder wood.
Next we introduce some new edges. For every which is adjacent to only one vertex of we identify the unique face which is adjacent to but not to . Connect to a vertex on the boundary of which belongs to . Note that the edge is intersected by a cycle from and that both faces obtained by cutting via contain the vertex . This shows that we can add edges to all vertices of which are adjacent to only one vertex in without introducing crossings. The color of is copied to the new edge . Fig. 14 exemplifies the coloring of together with the additional edges.
Now remove all the edges incident to vertices in . In the remaining graph all the vertices of are of degree 2. We remove these ‘subdivision’ vertices and melt the two edges into one. The result is . We claim that orienting the three edges of which come from an edge of as outgoing we obtain a Schnyder wood of . Indeed a, b, and c follow directly or from what we have already said. For d we need a little argument. Consider a monochromatic edge and let be one of the faces of containing this edge on the boundary. The other edge on the boundary of which contains is either incoming in the same color or outgoing in a different color. This shows that there is no monochromatic directed facial cycle containing . Now consider a face of which has no monochromatic edge. Note that is a face of the graph obtained from by deleting all the edges incident to vertices in . Let be a vertex in the interior of such that in there is an edge with being on the boundary of . Let be the color of the edge . Vertex has two neighbors on . By looking at the colors of edges of we see that at and the incident edge on which is different from has color . In the Schnyder wood we therefore see these outgoing in color at and one of them being oriented clockwise, the other counterclockwise on the boundary of . This shows that supports no monochromatic directed cycle.
For the following we rely on the theory of Schnyder woods for connected planar graphs, see e.g. [7] or [12] or [9]. In the Schnyder wood on for every vertex and every color there is a directed path of color from to . The three paths pairwise only share the vertex . The three paths of partition the interior of into 3 regions. For we let be the region bounded by and . With
we associate the region vector
where is the number of faces of in region . Fig. 14 illustrates regions and region vectors.Let be the generating set for an orthogonal surface . In slight abuse of notation we identify region vectors with their corresponding vertices and say that is generated by . The minima of are the vertices of . Moreover, supports the Schnyder wood in the sense that every outgoing edge at in corresponds to an edge of the skeleton of such that the direction of the skeleton edge is given by the color of the edge. In fact from the clockwise order of directions of skeleton edges at minima, saddle points and maxima it can be concluded that the skeleton of is . With Proposition 1 we obtain a Plattenbau representation of .
Since the set of orientations of a fixed planar graph carries the structure of a distributive lattice [9], and we have shown a correspondence of such a set with the orthogonal surfaces with given skeleton, we obtain:
Corollary 1
Let be a planar cubic bipartite graph with specified vertex . The set of orthogonal surfaces with skeleton and vertex at infinity carries a distributive lattice structure.
3.2 Plattenbau Representations in the Presence of Babets
Let be a colorable triangulation suppose that contains babets. Being separating triangles babets can be nested, let be the family of basic babets of , i.e., of babets which are not contained in the interior of another babet. Let be the triangulation obtained from by cleaning all the babets, i.e., removing the interior vertices and their incident edges from all babets . Clearly is colorable and babetfree. Triangles which have been babets are black. With the method from the previous subsection we get an orthogonal surface for . In this representation triangles which have been babets correspond to saddle points. For later reference let be the saddle point corresponding to .
For each babet let the inside triangulation of in . Clearly, is colorable, hence, its triangles can be colored black and white with the outer face being black (note that this coloring of is obtained from the coloring of triangles in by exchanging black and white). Assuming that has no babet we obtain an orthogonal surface for .
The construction of the orthogonal surface works with the assumption that the vertices of the outer face, i.e., of the triangle are colored in clockwise order. The same assumption for the full triangulation implies that the vertices of are colored in clockwise order.
The goal is to patch at the saddle point to so that the flats corresponding to a vertex of in the two orthogonal surfaces are coplanar. Let , , and be the red, green and blue flat at in , they represent the vertices of with their color in . At exactly one of the three flats has a concave angle, we assume that this flat is , the other cases are completely symmetric. The point is an interior point of the rectangle spanned by the extreme points of . The point is the apex of a convex corner whose sides coincide locally with , and . In this corner we see the colors of the flats in clockwise order as . Hence, we can patch and appropriately scaled down copy of into this corner such that the red outer flat of becomes part of , while the green outer flat of becomes part of and the blue outer flat of becomes part of . With the technique of Proposition 1 we obtain a rectangle contact representation of the subgraph of induced by all the vertices of which are represented by flats of and . See Fig. 17 for an illustration.
Repeating the procedure for further babets in and for babets which may occur in triangulations we eventually obtain an rectangle contact representation of . This completes the proof of Theorem 1.1.
3.3 Comments on Related Work
Eppstein and Mumford [6] study orthogonal polytopes and their graphs.
They define corner polyhedra as polytopes obtained from what we call an orthogonal surface by restricting the surface to the bounded flats and connecting their boundary to the origin , this replaces the 3 unbounded flats of an orthogonal surface by three flats closing the polytope. Eppstein and Mumford show that the skeleton graphs of corner polyhedra are exactly the cubic bipartite connected graphs with the property that every separating triangle of the planar dual graph has the same parity. This is equivalent to our characterization of these graphs as duals of colorable triangulations with admit a choice of the outer face such that there are no babets, see Proposition 2.
A major part of their proof is devoted to the construction of a rooted cycle cover, respectively, to the investigation of necessary and sufficient conditions for the existence of such a cycle cover. The proof is based on a set of operations that allow any connected Eulerian triangulation to be reduced to smaller ones. In contrast show the equivalent existence of an orientation with a counting argument. Given a cycle cover Eppstein and Mumford provide a construction an appropriate orthogonal surface from the combinatorial data by combining the coordinates obtained from plane drawings of three orthogonal projections. In contrast we refer to the established theory of Schnyder woods and their relation to orthogonal structures to get the results.
As a more general class than corner polyhedra Eppstein and Mumford define xyz polyhedra as orthogonal polytopes with the property that each axisparallel line through a vertex contains exactly one other vertex. They characterize the skeletons of as cubic bipartite connected graphs, i.e., as the duals of colorable triangulations. Modulo the ‘reflection’ of the three outer flats this correspond to our Theorem 1.1. The main step in the proof is the gluing of an orthogonal surface into a corner of another orthogonal surface, see Section 3.2 and also Fig. 29 in [6].
A recent paper of Gonçalves [14] can be used as a basis for yet another proof of Theorem 1.1, i.e., a proof of the characterization of xyz polyhedra. Gonçalves uses a system of linear equations to construct a TCscheme for a given colorable triangulation. The TCscheme comes very close to a segment contact representation with segments of 3 slopes for the input graph, however, there can be degeneracies: segments may degenerate to points (this relates to babets) and segments ending on two sides of another segment may have coinciding endpoints. The TCscheme can be transformed into an orthogonal surface. First adjust the directions to have slopes , , and , then add an orthogonal peak in each gray triangle^{3}^{3}3This refers to the two color classes of the triangles of the TCscheme, not to the two classes of triangles of the original triangulation. and an orthogonal valley in each white triangle, and extend the outer flats. This yields an orthogonal surface. If there are no degeneracies the orthogonal surface properly represents the triangulation via flat contacts. Degeneracies of the TCscheme translate into corners of degree 6 in the orthogonal surface, they can be resolved by shifting flats (c.f. [12] for details on flat shifting). Finally as in the other two approaches babets have to be recovered by patching their orthogonal surface into corners of the surface.
A nice aspect of this approach is that the partition of white triangles of the triangulation into peak and valley triangles is done by solving the linear system, no need of computing an orientation or a cycle cover for this task.
4 Proper Boxed Plattenbauten and Octahedrations
In this section we characterize the touching graphs of proper boxed Plattenbauten, that is, we prove Theorem 1.2.
First, in any proper Plattenbau any two touching rectangles have a proper contact, i.e., a boundary edge of one rectangle, say , is completely contained in the other rectangle . We denote this as and remark that this orientation has already been used in the proof of Observation 2.
Secondly, in any proper boxed Plattenbau there are six rectangles that are incident to the unbounded region. We refer to them as outer rectangles and to the six corresponding vertices in the touching graph for as the outer vertices. The corners incident to three outer rectangles are the outer corners, and the inner regions/cells of will be called rooms.
Whenever we have specified some vertices of a graph to be outer vertices, this defines inner vertices, outer edges, and inner edges as follows: The inner vertices are exactly the vertices that are not outer vertices; the outer edges are those between two outer vertices; the inner edges are those with at least one inner vertex as endpoint. We shall use these notions for a Plattenbau graph, as well as for some planar quadrangulations we encounter along the way.
Let us start with the necessity of Items a to d in Theorem 1.2.
Proposition 3
Every touching graph of a proper boxed Plattenbau satisfies Items a to d in Theorem 1.2.
Proof
Item a follows directly from the definition of boxed Plattenbauten. Also, for Item b simply orient each edge towards the endvertex whose rectangle has interior points in the intersection. This way, any edge between two outer vertices is bidirected, which is in accordance with Item b. Now, Item c follows from Observation 1 Item 2 together with the fact that edgemaximal planar bipartite graphs are quadrangulations. Indeed, if the rectangles on one side of a given rectangle would not induce a quadrangulation, then there would be a rectangle with a free edge and would not be boxed. Let us finally argue Item d. The common neighbors of two vertices lie on the circle that is the intersection of the spheres with centers and , respectively. Since and are on different sides of the intersection, see the vertices on the circle in opposite order. See Fig. 19 for an illustration. ∎
Next, we prove the sufficiency in Theorem 1.2, i.e., for every graph satisfying Items a to d we find a proper boxed Plattenbau with touching graph .
Fix a graph with six outer vertices and edge orientation fulfilling Items a to d. For each vertex denote by the spherical quadrangulation induced by given in Item c. By Item c, the outneighbors of vertex induce a cycle in , which we call the equator of . The equator splits the spherical quadrangulation into two hemispheres, each being a plane embedded quadrangulation with outer face with the property that each vertex of is contained in exactly one hemisphere. The vertices of are the outer vertices of either hemisphere. Note that one hemisphere (or even both) may be trivial, namely when the equator bounds a face of .
We proceed with a number of claims.
Claim 2
In each hemisphere, each inner vertex has exactly two outgoing edges and no outer vertex has an outgoing inner edge.
Proof of Claim. Let be any inner vertex of a hemisphere of . We shall first show that has at least two outneighbors in that hemisphere. We have , i.e., is a neighbor of but not an outneighbor. Hence, the edge is directed from to and lies on the equator of . As such, has two neighbors on , i.e., . It follows that are outneighbors of , since , and , since . In other words, has at least two outneighbors in and, by planarity of , these are in the same hemisphere as , as desired.
Finally, each hemisphere of is a plane quadrangulation with outer cycle , and as such has exactly inner edges for inner vertices. As each inner vertex has at least two outgoing edges, the edge count gives that each inner vertex has exactly two outgoing edges. Moreover, each inner edge is outgoing at some inner vertex, i.e., no outer vertex of the hemisphere has an outgoing inner edge.
Each equator edge of induces together with a triangle in , and we call these four triangles the equator triangles of .
Claim 3
Every triangle in is an equator triangle.
Proof of Claim. Let be a triangle in with vertices . First, we shall show that is not oriented as a directed cycle, i.e., has a vertex of outdegree two in . Without loss of generality let be directed from to . If is directed from to , we are done. So assume that is directed from to . Then is an inner vertex of a hemisphere of . Moreover has an outgoing edge to , and Claim 2 implies that is an inner vertex of the same hemisphere of . In particular, is directed from to and has outdegree two in , as desired.
Now let be the vertex in with outdegree two and be its outneighbors in . Then and lie on the equator of and are connected by an edge in . Thus is an equator triangle.
Clearly, a vertex forms a triangle with two vertices and if and only if is an edge and is a common neighbor of and . Equivalently, is adjacent to in , which in turn is equivalent to being adjacent to in . Hence, the set of all common neighbors (and thus also the set of all triangles sharing edge ) is endowed with the clockwise cyclic ordering around in , as well as with the clockwise cyclic ordering around in . By Item d, these two cyclic orderings are reversals of each other.
Let us define for a triangle in with vertices the two sides of as the two cyclic permutations of , which we denote by and . So triangle has the two sides and . We define a binary relation on the set of all sides of triangles in as follows.
(1) 
Note that by d comes immediately before in the clockwise ordering around if and only if comes immediately before in the clockwise ordering around . Thus also implies , i.e., is a symmetric relation and as such encodes an undirected graph on the sides of triangles.
Claim 4
Each connected component of is a cube. The corresponding subgraph in is an octahedron.
Proof of Claim. Consider any fixed vertex of . Then is an edge of contained in . As is a quadrangulation, vertex has degree at least two in . Hence, there exists a unique vertex in such that according to Eq. 1. Moreover, and are both neighbors of in and hence nonadjacent in , i.e., . Symmetrically, we find with and
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