Plattenbauten: Touching Rectangles in Space

07/15/2020 ∙ by Stefan Felsner, et al. ∙ 0

Planar bipartite graphs can be represented as touching graphs of horizontal and vertical segments in ℝ^2. We study a generalization in space, namely, touching graphs of axis-aligned rectangles in ℝ^3. We prove that planar 3-colorable graphs can be represented as touching graphs of axis-aligned rectangles in ℝ^3. The result implies a characterization of corner polytopes previously obtained by Eppstein and Mumford. A by-product of our proof is a distributive lattice structure on the set of orthogonal surfaces with given skeleton. Moreover, we study the subclass of strong representations, i.e., families of axis-aligned rectangles in ℝ^3 in general position such that all regions bounded by the rectangles are boxes. We show that the resulting graphs correspond to octahedrations of an octahedron. This generalizes the correspondence between planar quadrangulations and families of horizontal and vertical segments in ℝ^2 with the property that all regions are rectangles.

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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 axis-aligned segment contact representations: families of interior-disjoint 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 graph111We 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 non-parallel 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 interior-disjoint axis-aligned segments in . In fact, edge-maximal 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].

Figure 1: An axis-aligned segment contact representation (left) and a Plattenbau (right) together with the respective touching graphs.

In this paper we extend axis-aligned segment contact representations in  to axis-aligned rectangle contact representations in . That is, we consider families  of axis-aligned 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 Plattenbau222Plattenbau (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 non-empty 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 axis-aligned 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 edge-maximal 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:

  1. is connected and the outer vertices of  induce an octahedron.

  2. The edges of  admit an orientation such that

    • the bidirected edges are exactly the outer edges,

    • each vertex has exactly outgoing edges.

  3. The neighborhood  of each vertex  induces a spherical quadrangulation  in which the out-neighbors of  induce a -cycle.

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

  4. 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 axis-aligned 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.

Figure 2: Plattenbau representations of  (left) and  (right).
Observation 1

If  is a Plattenbau graph, then

  1. the chromatic number of  is at most ,

  2. the neighborhood of any vertex of  is planar.

  3. 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 axis-aligned 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.

Figure 3: Left: A -rectangle (depicted in blue) in a Plattenbau with its touching rectangles intersecting its upper halfspace . Right: The resulting crossing-free embedding on the upper hemisphere.

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:

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 edge-maximal 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 axis-aligned full-dimensional cuboid, i.e., the Cartesian product of three bounded intervals of non-zero length.) For boxed Plattenbauten we use the additional convention that the edge-to-edge 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 axis-aligned 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.

Figure 4: A proper Plattenbau that cannot be augmented to a boxed Plattenbau.

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 well-behaved 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.

Figure 5: An orthogonal surface and its skeleton (vertex  omitted).

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 Brightwell-Trotter 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 non-empty 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 non-degeneracy 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 zig-zag paths sharing the two extreme points of the flat. The minima of the lower zig-zag are elements of the generating set , they are minimal elements of the orthogonal surface . The maxima of the upper zig-zag are maximal elements of , they can be considered to be dual generators.

Figure 6: Generic flat  and the spanned rectangle .

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 .

Figure 7: Replacing flats by rectangles and expanding in order to avoid weak contacts.

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 out-degrees 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 babet-free 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 multi-edges 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 .

Figure 8: Two sketches of reducible ears. The right one is the special case where is an ear.

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.

this triangle is a babetit contains .
Figure 9: Illustration for the case .

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., .

Figure 10: Illustration for the subconfigurations generated from a cycle enclosing .

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 6-cycle 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 6-gon with only white triangles touching the 6-gon and black incident triangles at vertex of the 6-gon. 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 6-gon. Making any of the four triangles of separating so that it can accommodate in its interior would make .

Figure 11: Three triangulations with and .

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 Babet-Free 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 two-colored in white and pink such that the two sides of each cycle have distinct colors, the two-coloring is unique if we want the unbounded region to be colored white. This two-coloring 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 .

\begin{picture}(10904.0,8755.0)(11161.0,-14361.0)\end{picture}
Figure 12: Example of a with a cycle cover of and the induced partition 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:

  1. 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 half-edge of color .

  2. 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.

  3. 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).

  4. There is no interior face whose boundary is a directed cycle in one label.

\begin{picture}(3208.0,2878.0)(1163.0,-3210.0)\end{picture}
Figure 13: Illustration for the vertex condition c.

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.

\begin{picture}(10904.0,8755.0)(11161.0,-14361.0)\end{picture}
Figure 14: The coloring of the edges of obtained from the -coloring of . The dashed arrows are 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.

\begin{picture}(10904.0,9275.0)(11161.0,-14696.0)\end{picture}
Figure 15: A Schnyder wood with a shading indicating the regions of the yellow vertex. Vertices are labeled with their region vector.

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 .

\begin{picture}(14402.0,13202.0)(48600.0,-24362.0)\end{picture}
Figure 16: The orthogonal surface obtained from the region vectors given in Fig. 15.

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 babet-free. 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.

Figure 17: The different places to patch an orthogonal surfaces into the saddle points of another orthogonal surface..

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 axis-parallel 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 TC-scheme for a given -colorable triangulation. The TC-scheme 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 TC-scheme can be transformed into an orthogonal surface. First adjust the directions to have slopes , , and , then add an orthogonal peak in each gray triangle333This refers to the two color classes of the triangles of the TC-scheme, 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 TC-scheme 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.

\begin{picture}(42984.0,15302.0)(-17874.0,-14012.0)\end{picture}
Figure 18: A -colored triangulation, a TC-scheme of the triangulation and the corresponding orthogonal 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 end-vertex 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 edge-maximal 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. ∎

Figure 19: Two rectangles (green and blue) and their common neighborhood.

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 out-neighbors 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 out-neighbors in that hemisphere. We have , i.e., is a neighbor of but not an out-neighbor. 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 out-neighbors of , since , and , since . In other words, has at least two out-neighbors 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 out-degree 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 out-degree two in , as desired.

Now let be the vertex in with out-degree two and be its out-neighbors 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 non-adjacent in , i.e., . Symmetrically, we find with and