#### Toroidal Maps: Schnyder Woods, Orthogonal Surfaces and Straight-Line Representations

Discrete Comput Geom
Toroidal Maps: Schnyder Woods, Orthogonal Surfaces and Straight-Line Representations
Daniel Gonçalves 0
Benjamin Lévêque 0
0 D. Gonçalves
A Schnyder wood is an orientation and coloring of the edges of a planar map satisfying a simple local property. We propose a generalization of Schnyder woods to graphs embedded on the torus with application to graph drawing. We prove several properties on this new object. Among all we prove that a graph embedded on the torus admits such a Schnyder wood if and only if it is an essentially 3-connected toroidal map. We show that these Schnyder woods can be used to embed the universal cover of an essentially 3-connected toroidal map on an infinite and periodic orthogonal surface. Finally we use this embedding to obtain a straight-line flat torus representation of any toroidal map in a polynomial size grid.
Schnyder woods; Toroidal graphs; Embedding
1 Introduction
A closed curve on a surface is contractible if it can be continuously transformed into
a single point. Given a graph embedded on the torus, a contractible loop is an edge
forming a contractible cycle. Two homotopic multiple edges are two edges with the
same extremities such that their union forms a contractible cycle. In this paper, we
will almost always consider graphs embedded on the torus with no contractible loop
and no homotopic multiple edges. We call these graphs toroidal graphs for short and
keep the distinction with graph embedded on the torus that may have contractible
loops or homotopic multiple edges. A map on a surface is a graph embedded on this
surface where every face is homeomorphic to an open disk. A map embedded on the
torus is a graph embedded on the torus that is a map (it may contains contractible
loops or homotopic multiple edges). A toroidal map is a toroidal graph that is a map
(it has no contractible loop and no homotopic multiple edges). A toroidal
triangulation is a toroidal map where every face has size three. A general graph (i.e., not
embedded on a surface) is simple if it contains no loop and no multiple edges. Since
some loops and multiple edges are allowed in toroidal graphs, the class of toroidal
graphs is larger than the class of simple toroidal graphs.
The torus is represented by a parallelogram in the plane whose opposite sides are
pairwise identified. This representation is called the flat torus. The universal cover
G∞ of a graph G embedded on the torus is the infinite planar graph obtained by
replicating a flat torus representation of G to tile the plane (the tiling is obtained by
translating the flat torus along two vectors corresponding to the sides of the
parallelogram). Note that a graph G embedded on the torus has no contractible loop and no
homotopic multiple edges if and only if G∞ is simple.
Given a general graph G, let n be the number of vertices and m the number of
edges. Given a graph embedded on a surface, let f be the number of faces. Euler’s
formula says that any map on a surface of genus g satisfies n − m + f = 2 − 2g,
where the plane is the surface of genus 0, and the torus the surface of genus 1.
Schnyder woods were originally defined for planar triangulations by
Schnyder [
26
].
Definition 1 (Schnyder wood, Schnyder property) Given a planar triangulation G,
a Schnyder wood is an orientation and coloring of the edges of G with the colors 0,
1, 2 where each inner vertex v satisfies the Schnyder property (see Fig. 1 where each
color is represented by a different type of arrow):
• Vertex v has outdegree one in each color.
• The edges e0(v), e1(v), e2(v) leaving v in colors 0, 1, 2, respectively, occur in
counterclockwise order.
• Each edge entering v in color i enters v in the counterclockwise sector from
ei+1(v) to ei−1(v) (where i + 1 and i − 1 are understood modulo 3).
For higher genus triangulated surfaces, a generalization of Schnyder woods has
been proposed by Castelli Aleardi et al. [
3
], with applications to encoding.
Unfortunately, in this definition, the simplicity and the symmetry of the original Schnyder
wood are lost. Here we propose an alternative generalization of Schnyder woods for
toroidal graphs, with application to graph drawings.
By Euler’s formula, a planar triangulation satisfies m = 3n − 6. Thus, there are
not enough edges in the graph for all vertices to be of outdegree three. This explains
why just some vertices (inner ones) are required to satisfy the Schnyder property.
For a toroidal triangulation, Euler’s formula gives exactly m = 3n, so there is hope
for a nice object satisfying the Schnyder property for every vertex. This paper shows
that such an object exists. Here we do not restrict ourselves to triangulations and we
directly define Schnyder woods in a more general framework.
Felsner [
7, 8
] (see also [20]) has generalized Schnyder woods to 3-connected
planar maps by allowing edges to be oriented in one direction or in two opposite
directions. We also allow edges to be oriented in two directions in our definition.
Definition 2 (Toroidal Schnyder wood) Given a toroidal graph G, a (toroidal)
Schnyder wood of G is an orientation and coloring of the edges of G with the colors 0, 1,
2, where every edge e is oriented in one direction or in two opposite directions (each
direction having a distinct color), satisfying the following (see example of Fig. 2):
(T1) Every vertex v satisfies the Schnyder property (see Definition 1)
(T2) Every monochromatic cycle of color i intersects at least one monochromatic
cycle of color i − 1 and at least one monochromatic cycle of color i + 1.
Note that in this definition each vertex has exactly one outgoing arc in each color.
Thus, there are monochromatic cycles and the term “wood” has to be handled with
care here. The graph induced by one color is not necessarily connected, but each
connected component has exactly one directed cycle. We will prove that all the
monochromatic cycles of one color have the same homotopy.
In the case of toroidal triangulations, m = 3n implies that there are too many edges
to have bi-oriented edges. Thus, we can use this general definition of Schnyder wood
for toroidal graphs and keep in mind that when restricted to toroidal triangulations all
edges are oriented in one direction only.
Extending the notion of essentially 2-connectedness [
23
], we say that a toroidal
graph G is essentially k-connected if its universal cover is k-connected. Note that an
essentially 1-connected toroidal graph is a toroidal map. We prove that essentially
3-connected toroidal maps are characterized by the existence of Schnyder woods.
Theorem 1 A toroidal graph admits a Schnyder wood if and only if it is an essentially
3-connected toroidal map.
The dual of a Schnyder wood is the orientation and coloring of the edges of G∗
obtained by the rules represented on Fig. 3.
Our definition supports duality and we have the following results.
Theorem 2 There is a bijection between Schnyder woods of a toroidal map and
Schnyder woods of its dual.
The dual Schnyder wood of the Schnyder wood of Fig. 2 is represented on Fig. 4.
In our definition of Schnyder woods, two properties are required: a local one (T1)
and a global one (T2). This second property is important for using Schnyder woods
to embed toroidal graphs on orthogonal surfaces, as has been done in the plane by
Miller [
20
] (see also [
8
]).
Theorem 3 The universal cover of an essentially 3-connected toroidal map admits a
geodesic embedding on an infinite and periodic orthogonal surface.
A geodesic embedding of the toroidal map of Fig. 2 is represented on Fig. 5. The
black parallelogram represents a copy of the graph of Fig. 2; this is the basic tile that
is used to fill the plane.
A straight-line flat torus representation of a toroidal map G is the restriction to a
flat torus of a periodic straight-line representation of G∞. The problem of finding a
straight-line flat torus representation of a toroidal map was previously solved on
exponential size grids [
21
]. There are several works that represent a toroidal map inside
a parallelogram in a polynomial size grid [
5, 6
], but in these representations the
opposite sides of the parallelogram do not perfectly match. In the embeddings obtained
by Theorem 3, vertices are not coplanar, but we prove that for toroidal triangulations
one can project the vertices on a plane to obtain a periodic straight-line representation
of G∞. This gives the first straight-line flat torus representation of any toroidal map
in a polynomial size grid.
Theorem 4 A toroidal graph admits a straight-line flat torus representation in a
polynomial size grid.
In Sect. 2, we explain how our definition of Schnyder woods in the torus
generalizes the planar case. In Sect. 3, we show that our Schnyder woods are of two
fundamentally different types. In Sect. 4, we study the behavior of Schnyder woods
in the universal cover; we define the notion of regions and show that the existence
of Schnyder woods for a toroidal graph implies that the graph is an essentially
3connected toroidal map. In Sect. 5, we define the angle labeling and the dual of a
Schnyder wood. In Sect. 6, we show how the definition of Schnyder woods can be
relaxed for one of the two types of Schnyder wood. This relaxation is used in the next
sections for proving the existence of a Schnyder wood. In Sect. 7, we use a result of
Fijavz [
13
] on the existence of non-homotopic cycles in simple toroidal triangulations
to obtain a short proof of existence of Schnyder woods for simple triangulations. In
Sect. 8, we prove a technical lemma showing how a Schnyder wood of a graph G
can be derived from a Schnyder wood of the graph G , where G is obtained from
G by contracting an edge. This lemma is then used in Sect. 9 to prove the existence
of Schnyder woods for any essentially 3-connected toroidal maps. In Sect. 10, we
use Schnyder woods to embed the universal cover of essentially 3-connected toroidal
maps on periodic and infinite orthogonal surfaces by generalizing the region
vector method defined in the plane. In Sect. 11, we show that the dual map can also
be embedded on this orthogonal surface. In Sect. 12, we show that, in the case of
toroidal triangulations, this orthogonal surface can be projected on a plane to obtain
a straight-line flat torus representation.
2 Generalization of the Planar Case
Felsner [
7, 8
] has generalized planar Schnyder woods by allowing edges to be
oriented in one direction or in two opposite directions. The formal definition is the
following:
Definition 3 (Planar Schnyder wood) Given a planar map G, let x0, x1, x2 be three
distinct vertices occurring in counterclockwise order on the outer face of G. The
suspension Gσ is obtained by attaching a half-edge that reaches into the outer face
to each of these special vertices. A (planar) Schnyder wood rooted at x0, x1, x2 is an
orientation and coloring of the edges of Gσ with the colors 0, 1, 2, where every edge
e is oriented in one direction or in two opposite directions (each direction having a
distinct color), satisfying the following (see the example of Fig. 6):
(P1) Every vertex v satisfies the Schnyder property and the half-edge at xi is directed
outwards and colored i
(P2) There is no monochromatic cycle.
In the definition given by Felsner [
8
], property (P2) is in fact replaced by “There
is no interior face the boundary of which is a monochromatic cycle”, but the two are
equivalent by results of [
7, 8
].
With our definition of Schnyder woods for toroidal graphs, the goal is to generalize
the definition of Felsner. In the torus, property (P1) can be simplified as every vertex
plays the same role: there are no special outer vertices with a half-edge reaching into
the outer face. This explains property (T1) in our definition. Then if one asks that
every vertex satisfies the Schnyder property, there are necessarily monochromatic
cycles and (P2) is not satisfied. This explains why (P2) has been replaced by (T2) in
our generalization to the torus.
It would have been possible to replace (P2) by “there is no contractible
monochromatic cycles”, but this is not enough to suit our needs. Our goal is to use Schnyder
woods to embed universal covers of toroidal graphs on orthogonal surfaces, as has
been done in the plane by Miller [
20
] (see also [
8
]). The difference is that our surface
is infinite and periodic. In such a representation the three colors 0, 1, 2 correspond to
the three directions of the space. Thus, the monochromatic cycles with different
colors have to intersect each other in a particular way. This explains why property (T2)
is required. Figure 7 gives an example of an orientation and coloring of the edges of
a toroidal triangulation satisfying (T1) but not (T2), as there is no pair of intersecting
monochromatic cycles.
Let G be a toroidal graph given with a Schnyder wood. Let Gi be the directed
graph induced by the edges of color i. This definition includes edges that are
halfcolored i, and in this case, the edges get only the direction corresponding to color i.
Each graph Gi has exactly n edges, so it does not induce a rooted tree (contrarily to
planar Schnyder woods). Note also that Gi is not necessarily connected (for example,
in the graph of Fig. 8, every Schnyder wood has one color whose corresponding
subgraph is not connected). But each component of Gi has exactly one outgoing
arc for each of its vertices. Thus, each connected component of Gi has exactly one
directed cycle that is a monochromatic cycle of color i, or i-cycle for short. Note that
monochromatic cycles can contain edges oriented in two directions with different
colors, but the orientation of an i-cycle is the orientation given by the (half-)edges of
color i. The graph G−1 is the graph obtained from Gi by reversing all its edges. The
i
graph Gi ∪ G−1
i−1 ∪ Gi−+11 is obtained from the graph G by orienting edges in one or
two directions depending on whether this orientation is present in Gi , Gi−−11, or G−1 .
i+1
The following lemma shows that our property (T2) in fact implies that there are no
contractible monochromatic cycles.
Lemma 1 The graph Gi ∪ Gi−−11 ∪ Gi−+11 contains no contractible directed cycle.
Proof Suppose there is a contractible directed cycle in Gi ∪ G−1
i−1 ∪ Gi−+11. Let C be
such a cycle containing the minimum number of faces in the closed disk D bounded
by C. Suppose by symmetry that C turns clockwise around D. Then, by (T1), there
is no edge of color i − 1 leaving the closed disk D. So there is an (i − 1)-cycle in D,
and this cycle is C by minimality of C. Then, by (T1), there is no edge of color i
leaving D. So, again by minimality of C, the cycle C is an i-cycle. Thus, all the
edges of C are oriented clockwise in color i and counterclockwise in color i − 1.
Then, by (T1), all the edges of color i + 1 incident to C have to leave D. Thus, there
is no (i + 1)-cycle intersecting C, a contradiction to property (T2).
Let G be a planar map and let x0, x1, x2 be three distinct vertices occurring in
counterclockwise order on the outer face of G. One can transform Gσ into the
following toroidal map G+ (see Fig. 9): Add a vertex v in the outer face of G. Add three
non-parallel and non-contractible loops on v. Connect the three half-edges leaving xi
to v such that there are no two such edges entering v consecutively. Then we have
the following.
Theorem 5 The Schnyder woods of a planar map G rooted at x0, x1, x2 are in
bijection with the Schnyder woods of the toroidal map G+ (where the orientation of
one of the loops is fixed).
Proof ( ⇒) We are given a Schnyder wood of the planar graph G, rooted at x0,
x1, x2. Orient and color the graph G+ as in the example of Fig. 9, i.e., the edges of
the original graph G have the same color and orientation as in Gσ , the edge from
xi to v is colored i and leaving xi , and the three loops around v are colored and
oriented appropriately so that v satisfies the Schnyder property. Then it is clear that
all the vertices of G+ satisfy (T1). By (P2), we know that Gσ has no monochromatic
cycles. All the edges between G and v are leaving G, so there is no monochromatic
cycle of G+ involving vertices of G. Thus, the only monochromatic cycles of G+ are
the three loops around v and they satisfy (T2).
(⇐ ) Given a Schnyder wood of G+, the restriction of the orientation and
coloring to G and the three edges leaving v gives a Schnyder wood of Gσ . The three
loops around v are three monochromatic cycles corresponding to three edges
leaving v; thus, they have different colors by (T1). Thus, the three edges between G and
v are entering v with three different colors. The three loops around v have to leave
v in counterclockwise order 0, 1, 2 and we can assume the colors such that the edge
leaving xi is colored i. Clearly, all the vertices of Gσ satisfy (P1). By Lemma 1, there
are no contractible monochromatic cycles in G+, so Gσ satisfies (P2).
A planar map G is internally 3-connected if there exist three vertices on the outer
face such that the graph obtained from G by adding a vertex adjacent to the three
vertices is 3-connected. Miller [
20
] (see also [
7
]) proved that a planar map admits a
Schnyder wood if and only if it is internally 3-connected. The following results show
that the notion of essentially 3-connected is the natural generalization of internally
3-connected to the torus.
Theorem 6 A planar map G is internally 3-connected if and only if there exist three
vertices on the outer face of G such that G+ is an essentially 3-connected toroidal
map.
Proof ( ⇒) Let G be an internally 3-connected planar map. By definition, there
exist three vertices x0, x1, x2 on the outer face such that the graph G obtained from
G by adding a vertex adjacent to these three vertices is 3-connected. Let G be the
graph obtained from G by adding three vertices y0, y1, y2 that form a triangle and
by adding the three edges xi yi . It is not difficult to check that G is 3-connected.
Since G∞ can be obtained from the (infinite) triangular grid, which is 3-connected,
by gluing copies of G along triangles, G∞ is clearly 3-connected. Thus, G+ is an
essentially 3-connected toroidal map.
(⇐ ) Suppose there exist three vertices on the outer face of G such that G+ is
an essentially 3-connected toroidal map, i.e., G∞ is 3-connected. A copy of G is
contained in a triangle y0y1y2 of G∞. Let G be the subgraph of G∞ induced by
this copy plus the triangle, and let xi be the unique neighbor of yi in the copy of G.
Since G is connected to the rest of G∞ by a triangle, G is also 3-connected. Let us
now prove that this implies that G is internally 3-connected for x0, x1, and x2. This
is equivalent to saying that the graph G , obtained by adding a vertex z connected
to x0, x1, and x2, is 3-connected. If G had a separator {a, b} or {a, z}, with a, b ∈
V (G ) \ {z}, then {a, b} or {a, yi }, for some i ∈ [
0, 2
], would be a separator of G .
This would contradict the 3-connectedness of G . So G is internally 3-connected.
3 Two Different Types of Schnyder Woods
Two non-contractible closed curves are homotopic if one can be continuously
transformed into the other. Homotopy is an equivalence relation, and as we are on the torus
we have the following.
Lemma 2 Let C1, C2 be two non-contractible closed curves on the torus. If C1, C2
are not homotopic, then their intersection is non-empty.
Two non-contractible oriented closed curves on the torus are fully homotopic if
one can be continuously transformed into the other by preserving the orientation. We
say that two monochromatic directed cycles Ci , Cj of different colors are reversal if
one is obtained from the other by reversing all the edges (Ci = Cj−1). We say that two
monochromatic cycles are crossing if they intersect but are not reversal. We define
the right side of an i-cycle Ci , as the right side while “walking” along the directed
cycle by following the orientation given by the edges colored i.
Let G be a toroidal graph given with a Schnyder wood.
Lemma 3 All i-cycles are non-contractible, non-intersecting, and fully homotopic.
Proof By Lemma 1, all i-cycles are non-contractible. If there exist two such
distinct i-cycles that are intersecting, then there is a vertex that has two outgoing edges
of color i, a contradiction to (T1). So the i-cycles are non-intersecting. Then, by
Lemma 2, they are homotopic.
Suppose that there exist two i-cycles Ci , Ci that are not fully homotopic. By the
first part of the proof, cycles Ci , Ci are non-contractible, non-intersecting, and
homotopic. Let R be the region between Ci and Ci situated on the right of Ci . Suppose
by symmetry that C−1 is not an (i + 1)-cycle. By (T2), there exists a cycle Ci+1
i
intersecting Ci and thus Ci+1 is crossing Ci . By property (T1), Ci+1 is entering Ci
from its right side and so it is leaving the region R when it crosses Ci . To enter the
region R, the cycle Ci+1 has to enter Ci or Ci from their left side, a contradiction to
property (T1).
Lemma 4 If two monochromatic cycles are crossing, then they are of different colors
and they are not homotopic.
Proof By Lemma 3, two crossing monochromatic cycles are not of the same color.
Suppose that there exist two monochromatic cycles Ci−1 and Ci+1, of color i − 1
and i + 1, that are crossing and homotopic. By Lemma 1, the cycles Ci−1 and Ci+1
are not contractible. Since Ci−1 = Ci−+11 and Ci−1 ∩ Ci+1 = ∅, the cycle Ci+1 is
leaving Ci−1. It is leaving Ci−1 on its right side by (T1). Since Ci−1 and Ci+1 are
homotopic, the cycle Ci+1 is entering Ci−1 at least once from its right side. This is
in contradiction with (T1).
Let Ci be the set of i-cycles of G. Let (Ci )−1 denote the set of cycles
obtained by reversing all the cycles of Ci . By Lemma 3, the cycles of Ci are
noncontractible, non-intersecting, and fully homotopic. So we can order them as follows:
Ci = {Ci0, . . . , Ciki −1}, ki ≥ 1, such that, for 0 ≤ j ≤ ki − 1, there is no i-cycle in the
region R(Cij , Cij +1) between Cij and Cij +1 containing the right side of Cij
(superscript understood modulo ki ).
We show that Schnyder woods are of two different types (see Fig. 10):
Theorem 7 Let G be a toroidal graph given with a Schnyder wood. Then all i-cycles
are non-contractible, non-intersecting, and fully homotopic, and either:
or
• For every pair of two monochromatic cycles Ci , Cj of different colors i, j , the two
cycles Ci and Cj are not homotopic and thus intersect (we say the Schnyder wood
is of Type 1);
• There exists a color i such that Ci−1 = (Ci+1)−1 and for any pair of monochromatic
cycles Ci , Cj of colors i, j , with j = i, the two cycles Ci and Cj are not homotopic
and thus intersect (we say the Schnyder wood is of Type 2, or Type 2.i if we want
to specify the color i).
Moreover, if G is a toroidal triangulation, then there are no edges oriented in two
directions and the Schnyder wood is of Type 1.
Proof By Lemma 3, all i-cycles are non-contractible, non-intersecting, and fully
homotopic. Suppose that there exist a (i − 1)-cycle Ci−1 and a (i + 1)-cycle Ci+1 that
are homotopic. We prove that the Schnyder wood is of Type 2.i. We first prove that
Ci−1 = (Ci+1)−1. Let Ci−1 be any (i − 1)-cycle. By (T2), Ci−1 intersects an (i +
1)cycle Ci+1. By Lemma 3, Ci−1 (resp. Ci+1) is homotopic to Ci−1 (resp. Ci+1).
So Ci−1 and Ci+1 are homotopic. By Lemma 4, C i+1 are reversal. Thus,
Ci−1 ⊆ (Ci+1)−1 and so by symmetry Ci−1 = (Ci+1i)−−11.anNdowC we prove that for any
pair of monochromatic cycles Ci , Cj of colors i, j , with j = i, the two cycles Ci and
Cj are not homotopic. By (T2), Cj intersects an i-cycle Ci . Since Ci−1 = (Ci+1)−1,
cycle Cj is bi-oriented in color i − 1 and i + 1, and thus we cannot have Cj = Ci−1.
So Cj and Ci are crossing and by Lemma 4, they are not homotopic. By Lemma 3,
Ci and Ci are homotopic. Thus, Cj and Ci are not homotopic. Thus, the Schnyder
wood is of Type 2.i.
If there are no two monochromatic cycles of different colors that are homotopic,
then the Schnyder wood is of Type 1.
For toroidal triangulation, m = 3n by Euler’s formula, so there are no edges
oriented in two directions, and thus only Type 1 is possible.
Note that in a Schnyder wood of Type 1, we may have edges that are in two
monochromatic cycles of different colors (see Fig. 2).
We do not know if the set of Schnyder woods of a given toroidal graph has a kind of
lattice structure as in the planar case [
9
]. De Fraysseix et al. [
15
] proved that Schnyder
woods of a planar triangulation are in one-to-one correspondence with the orientation
of the edges of the graph where each inner vertex has outdegree three. It is possible
to retrieve the coloring of the edges of a Schnyder wood from the orientation. The
situation is different for toroidal triangulations. There exist orientations of toroidal
triangulations where each vertex has outdegree three but there is no corresponding
Schnyder wood. For example, if one considers a toroidal triangulation with just one
vertex, the orientations of edges that satisfy (T1) are the orientations where there are
not three consecutive edges leaving the vertex (see Fig. 11).
Let G be a toroidal graph given with a Schnyder wood. Consider the orientation and
coloring of the edges of G∞ that correspond to the Schnyder wood of G.
Lemma 5 The orientation and coloring of the edges of G∞ satisfy the following:
(U1) Every vertex of G∞ satisfies the Schnyder property
(U2) There is no monochromatic cycle in G∞.
Proof Clearly, (U1) is satisfied. Now we prove (U2). Suppose by contradiction that
there is a monochromatic cycle U of color i in G∞. Let C be the closed curve of
G corresponding to edges of U . If C self-intersects, then there is a vertex of G with
two edges leaving v in color i, a contradiction to (T1). So C is a monochromatic
cycle of G. Since C corresponds to a cycle of G∞, it is a contractible cycle of G,
a contradiction to Lemma 1.
One can remark that properties (U1) and (U2) are the same as in the definition of
Schnyder woods for 3-connected planar graphs (properties (P1) and (P2)). Note that if
the orientation and coloring of the edges of G∞, corresponding to an orientation and
coloring of the edges of G, satisfy properties (U1) and (U2), we do not necessarily
have a Schnyder wood of G. For example, the graph G∞ obtained by replicating the
graph G of Fig. 7 satisfies (U1) and (U2), whereas the orientation and coloring of G
do not make a Schnyder wood as (T2) is not satisfied.
Recall that the notation Ci = {Ci0, . . . , Ciki −1} denotes the set of i-cycles of G
such that there is no i-cycle in the region R(Cij , Cij+1). As monochromatic cycles
are not contractible by Lemma 1, a directed monochromatic cycle Cij corresponds to
a family of infinite directed monochromatic paths of G∞ (infinite in both directions
of the path). This family is denoted Lij . Each element of Lij is called a monochromatic
line of color i, or i-line for short. By Lemma 3, all i-lines are non-intersecting and
oriented in the same direction. Given any two i-lines L, L , the unbounded region
between L and L is noted R(L, L ). We say that two i-lines L, L are consecutive if
no i-lines are contained in R(L, L ).
Let v be a vertex of G∞. For each color i, vertex v is the starting vertex of a unique
infinite directed monochromatic path of color i, denoted Pi (v). Indeed this is a path
since there is no monochromatic cycle in G∞ by property (U2), and it is infinite (in
one direction of the path only) because every reached vertex of G∞ has exactly one
edge leaving in color i by property (U1). As Pi (v) is infinite, it necessarily contains
two vertices u, u of G∞ that are copies of the same vertex of G. The subpath of
Pi (v) between u and u corresponds to an i-cycle of G and thus is part of an i-line
of G∞. Let Li (v) be the i-line intersecting Pi (v).
Lemma 6 The graph Gi∞ ∪ (Gi∞−1)−1 ∪ (Gi∞+1)−1 contains no directed cycle.
Proof Suppose there is a contractible directed cycle C in Gi∞ ∪ (Gi∞−1)−1 ∪
(Gi∞+1)−1. Let D be the closed disk bounded by C. Suppose by symmetry that C
turns around D clockwise. Then, by (U1), there is no edge of color i − 1 leaving the
closed disk D. So there is an (i − 1)-cycle in D, a contradiction to (U2).
Lemma 7 For every vertex v and color i, the two paths Pi−1(v) and Pi+1(v) only
intersect on v.
i−1 ∪ (Gi∞+1)−1
conProof If Pi−1(v) and Pi+1(v) intersect on two vertices, then G∞
tains a cycle, contradicting Lemma 6.
By Lemma 7, for every vertex v, the three paths P0(v), P1(v), P2(v) divide G∞
into three unbounded regions R0(v), R1(v), and R2(v), where Ri (v) denotes the
region delimited by the two paths Pi−1(v) and Pi+1(v). Let Ri◦(v) = Ri (v) \ (Pi−1(v) ∪
Pi+1(v)) (see Fig. 12).
Lemma 8 For all distinct vertices u, v, we have:
(i) If u ∈ Ri (v), then Ri (u) ⊆ Ri (v).
(ii) If u ∈ Ri◦(v), then Ri (u) Ri (v).
(iii) There exists i and j with Ri (u)
Ri (v) and Rj (v)
Rj (u).
Proof (i) Suppose by symmetry that the Schnyder wood is not of Type 2.(i +1). Then
in G, i-cycles are not homotopic to (i − 1)-cycles. Thus in G∞, every i-line crosses
every (i − 1)-line. Moreover an i-line crosses an (i − 1)-line exactly once and from its
right side to its left side by (U1). Vertex v is between two consecutive monochromatic
(i − 1)-lines Li−1, Li−1, with Li−1 situated on the right of Li−1. Let R be the region
situated on the right of Li−1, so v ∈ R.
Claim 1 For any vertex w of R, the path Pi (w) leaves the region R.
Proof of Claim 1 The i-line Li (w) has to cross Li−1 exactly once and from right to
left; thus, Pi (w) leaves the region R. This proves Claim 1.
The path Pi+1(v) cannot leave the region R as this would contradict (U1). Thus,
by Claim 1 for w = v, we have Ri (v) ⊆ R and so u ∈ R. Moreover, the paths
Pi−1(u) and Pi+1(u) cannot leave region Ri (v) as this would contradict (U1). Thus
by Claim 1 for w = u, the path Pi (u) leaves the region Ri (v) and so Ri (u) ⊆ Ri (v).
(ii) By (i), Ri (u) ⊆ Ri (v), so the paths Pi−1(u) and Pi+1(u) are contained
in Ri (v). Then none of them can contain v as this would contradict (U1). So all
the faces of Ri (v) incident to v are not in Ri (u) (and there is at least one such face).
(iii) By symmetry, we prove that there exists i with Ri (u) Ri (v). If u ∈ Ri◦(v)
for some color i, then Ri (u) Ri (v) by (ii). Suppose now that u ∈ Pi (v) for some i.
By Lemma 7, at least one of the two paths Pi−1(u) and Pi+1(u) does not contain v.
Suppose by symmetry that Pi−1(u) does not contain v. As u ∈ Pi (v) ⊆ Ri+1(v), we
have Ri+1(u) ⊆ Ri+1(v) by (i), and as none of Pi−1(u) and Pi (u) contains v, we
have Ri+1(u) Ri+1(v).
Lemma 9 If a toroidal graph G admits a Schnyder wood, then G is essentially
3connected.
Proof Let u, v, x, y be any four distinct vertices of G∞. Let us prove that there exists
a path between u and v in G∞ \ {x, y}. Suppose by symmetry, that the Schnyder wood
is of Type 1 or Type 2.1. Then the monochromatic lines of color 0 and 2 form a kind
of grid; i.e., the 0-lines intersect all the 2-lines. Let L0, L0 be 0-lines and L2, L2 be
2-lines, such that u, v, x, y are all in the interior of the bounded region R(L0, L0) ∩
R(L2, L2).
By Lemma 7, the three paths Pi (v), for 0 ≤ i ≤ 2, are disjoint except on v. Thus,
there exists i, such that Pi (v) ∩ {x, y} = ∅. Similarly there exists j , such that Pj (u) ∩
{x, y} = ∅. The two paths Pi (v) and Pj (u) are infinite, so they intersect the boundary
of R(L0, L0) ∩ R(L2, L2). Thus, Pi (v) ∪ Pj (u) ∪ L0 ∪ L0 ∪ L2 ∪ L2 contains a path
from u to v in G∞ \ {x, y}.
By Lemma 9, if G admits a Schnyder wood, then it is essentially 3-connected, so
it is a map and each face is a disk.
Note that if (T2) is not required in the definition of Schnyder woods, then Lemma 9
is false. Figure 13 gives an example of an orientation and a coloring of the edges of
a toroidal graph satisfying (T1), such that there are no contractible monochromatic
cycles, but where (T2) is not satisfied as there is a 0-cycle not intersecting any 2-cycle.
This graph is not essentially 3-connected; indeed, G∞ is not connected.
We are given a planar map G, and x0, x1, x2, three distinct vertices occurring in
counterclockwise order on the outer face of G. A Schnyder angle labeling [
7
] of G
with respect to x0, x1, x2 is a labeling of the angles of Gσ satisfying the following:
(L1) The label of the angles at each vertex form, in counterclockwise order,
nonempty intervals of 0’s, 1’s, and 2’s. The two angles at the half-edge at xi have
labels i + 1 and i − 1
(L2) The label of the angles at each inner face form, in counterclockwise order,
nonempty intervals of 0’s, 1’s, and 2’s. At the outer face the same is true in
clockwise order.
Felsner [
8
] proved that, for planar maps, Schnyder woods are in bijection with
Schnyder angle labellings. In the toroidal case, we do not see a simple definition
of Schnyder angle labeling that would be equivalent to our definition of Schnyder
woods. This is due to the fact that, unlike (P2) which is local and can be checked just
by considering faces, (T2) is global. Nevertheless, we have one implication.
The angle labeling corresponding to a Schnyder wood of a toroidal map G is a
labeling of the angles of G such that the angles at a vertex v in the counterclockwise
sector between ei+1(v) and ei−1(v) are labeled i (see Fig. 14).
Lemma 10 The angle labeling corresponding to a Schnyder wood of a toroidal map
satisfies the following: the angles at each vertex and at each face form, in
counterclockwise order, non-empty intervals of 0’s, 1’s, and 2’s.
Proof Clearly, the property is true at each vertex by (T1). To prove that the property
is true at each face, we count the number of color changes around vertices, faces, and
edges. This number of changes is denoted d . For a vertex v there are exactly three
changes, so d(v) = 3 (see Fig. 14). For an edge e, that can be either oriented in one or
two directions, there are also exactly three changes, so d(e) = 3 (see Fig. 14). Now
consider a face F . Suppose we cycle counterclockwise around F ; then an angle
colored i is always followed by an angle colored i or i + 1. Consequently, d(F ) must
be a multiple of 3. Suppose that d(F ) = 0; then all its angles are colored with one
color i. In that case the cycle around face F would be completely oriented in
counterclockwise order in color i + 1 (and in clockwise order in color i − 1). This cycle
being contractible, this would contradict Lemma 1. So d(F ) ≥ 3.
The sum of the changes around edges must be equal to the sum of the changes
around faces and vertices. Thus 3m = e d(e) = v d(v) + F d(F ) = 3n +
F d(F ). Euler’s formula gives m = n + f , so F d(F ) = 3f and this is
possible only if d(F ) = 3 for every face F .
There is no converse to Lemma 10. Figure 7 gives an example of a coloring and
orientation of the edges of a toroidal triangulation not satisfying (T2) but where the
angles at each vertex and at each face form, in counterclockwise order, non-empty
intervals of 0’s, 1’s, and 2’s.
Let G be a toroidal graph given with a Schnyder wood. By Lemma 9, G is an
essentially 3-connected toroidal map, and thus the dual G∗ of G has no contractible
loop and no homotopic multiple edges. Let G be a simultaneous drawing of G and
G∗ such that only dual edges intersect.
The dual of the Schnyder wood is the orientation and coloring of the edges of G∗
obtained by the following method (see Figs. 3 and 4): Let e be an edge of G and
e∗ the dual edge of e. If e is oriented in one direction only and colored i, then e∗
is oriented in two directions, entering e from the right side in color i − 1 and from
the left side in color i + 1 (the right side of e is the right side while following the
orientation of e). Symmetrically, if e is oriented in two directions in colors i + 1 and
i − 1, then e∗ is oriented in one direction only and colored i such that e is entering e∗
from its right side in color i − 1.
Lemma 11 Let G be a toroidal map. The dual of a Schnyder wood of a toroidal map
G is a Schnyder wood of the dual G∗. Moreover we have:
(i) On the simultaneous drawing G of G and G∗, the i-cycles of the dual Schnyder
wood are homotopic to the i-cycles of the primal Schnyder wood and oriented in
opposite directions.
(ii) The dual of a Schnyder wood is of Type 2.i if and only if the primal Schnyder
wood is of Type 2.i.
Proof In every face of G, there is exactly one angle of G and one angle of G∗. Thus
a Schnyder angle labeling of G corresponds to an angle labeling of G∗. The dual of
the Schnyder wood is defined such that an edge e is leaving F in color i if and only
if the angle at F on the left of e is labeled i − 1 and the angle at F on the right of
e is labeled i + 1, and such that an edge e is entering F in color i if and only if at
least one of the angles at F incident to e is labeled i (see Fig. 15). By Lemma 10,
the angles at a face form, in counterclockwise order, non-empty intervals of 0’s, 1’s,
and 2’s. Thus, the edges around a vertex of G∗ satisfy property (T1).
Consider G with the orientation and coloring of primal and dual edges.
Let C be a i-cycle of G∗. Suppose, by contradiction, that C is contractible. Let
D be the disk delimited by C. Suppose by symmetry that C is going anticlockwise
around D. Then all the edges of G that are dual to edges of C are entering D in color
i − 1. Thus, D contains an (i − 1)-cycle of G, a contradiction to Lemma 1. Thus,
every monochromatic cycle of G∗ is non-contractible.
The dual of the Schnyder wood is defined in such a way that an edge of G and
an edge of G∗ of the same color never intersect in G. Thus the i-cycles of G∗ are
homotopic to i-cycles of G. Consider a i-cycle Ci (resp. Ci∗) of G (resp. G∗). The two
cycles Ci and Ci∗ are homotopic. By symmetry, we assume that the primal Schnyder
wood is not of Type 2.(i −1). Let Ci+1 be an (i + 1)-cycle of G. The two cycles Ci
and Ci+1 are not homotopic and Ci is entering Ci+1 on its left side. Thus, the two
cycles Ci∗ and Ci+1 are not homotopic, and by the dual rules Ci∗ is entering Ci+1 on
its right side. So Ci and Ci∗ are homotopic and going in opposite directions.
Suppose the Schnyder wood of G is of Type 1. Then two monochromatic cycles
of G of different colors are not homotopic. Thus, the same is true for
monochromatic cycles of the dual. So (T2) is satisfied and the dual of the Schnyder wood is a
Schnyder wood of Type 1.
Suppose now that the Schnyder wood of G is of Type 2. Assume by symmetry
that it is of Type 2.i. Then all monochromatic cycles of color i and j , with j ∈ {i − 1,
i + 1}, intersect. Now suppose, by contradiction, that there is a j -cycle C∗, with j ∈
{i − 1, i + 1}, that is not equal to a monochromatic cycle of color in {i − 1, i + 1} \ {j }.
By symmetry we can assume that C∗ is of color i − 1. Let C be the (i − 1)-cycle of
the primal that is the first on the right side of C∗ in G. By definition of Type 2.i,
C−1 is an (i + 1)-cycle of G. Let R be the region delimited by C∗ and C situated on
the right side of C∗. Cycle C∗ is not an (i + 1)-cycle, so there is at least one edge
of color i + 1 leaving a vertex of C∗. By (T1) in the dual, this edge is entering the
interior of the region R. An edge of G∗ of color i + 1 cannot intersect C and cannot
enter C∗ from its right side. So in the interior of the region R there is at least one
(i + 1)-cycle Ci∗+1 of G∗. Cycle Ci∗+1 is homotopic to C∗ and going in the opposite
direction (i.e., Ci∗+1 and C∗ are not fully homotopic). If Ci∗+1 is not an (i − 1)-cycle,
then we can define R R the region delimited by Ci∗+1 and C situated on the left
side of Ci∗+1 and as before we can prove that there is an (i − 1)-cycle of G∗ in the
interior of R . So in any case, there is an (i − 1)-cycle Ci∗−1 of G∗ in the interior of
R and Ci∗−1 is fully homotopic to C∗. Let R R be the region delimited by C∗ and
Ci∗−1 situated on the right side of C∗. Clearly, R does not contain C. Thus, by the
definition of C, the region R does not contain any (i − 1)-cycle of G. But R is
nonempty and contains at least one vertex v of G. The path Pi−1(v) cannot leave R , a
contradiction. So (T2) is satisfied and the dual Schnyder wood is of Type 2.i.
By Lemma 11, we have Theorem 2.
6 Relaxing the Definition
Ci = Cj−1.
In the plane, the proof of existence of Schnyder woods can be done without too much
difficulty, as the properties to be satisfied are only local. In the toroidal case, things
are much more complicated, as property (T2) is global. The following lemma shows
that property (T2) can be relaxed a bit in the case of Schnyder woods of Type 1.
Lemma 12 Let G be a toroidal graph given with an orientation and coloring of the
edges of G with the colors 0, 1, 2, where every edge e is oriented in one direction or
in two opposite directions. The orientation and coloring is a Schnyder wood of Type 1
if and only if it satisfies the following:
(T1’) Every vertex v satisfies the Schnyder property.
(T2’) For each pair i, j of different colors, there exists an i-cycle intersecting a j
cycle.
(T3’) There are no monochromatic cycles Ci , Cj of different colors i, j such that
Proof ( ⇒) If we have a Schnyder wood of Type 1, then property (T1’) is
satisfied, as it is equal to property (T1). Property (T1) implies that there always exist
monochromatic cycles of each color, and thus property (T2’) is a relaxation of (T2).
Property (T3’) is implied by definition of Type 1 (see Theorem 7).
(⇐ ) Conversely, suppose we have an orientation and coloring satisfying (T1’),
(T2’), (T3’). We prove several properties.
Claim 2 All i-cycles are non-contractible, non-intersecting, and homotopic.
Proof of Claim 2 Suppose there is a contractible monochromatic cycle. Let C be
such a cycle containing the minimum number of faces in the closed disk D bounded
by C. Suppose by symmetry that C turns around D clockwise. Let i be the color
of C. Then, by (T1’), there is no edge of color i − 1 leaving the closed disk D. So
there is an (i − 1)-cycle in D and this cycle is C by minimality of C, a contradiction
to (T3’).
If there exist two distinct i-cycles that are intersecting, then there is a vertex that
has two outgoing edges of color i, a contradiction to (T1’). So the i-cycles are
nonintersecting. Then, by Lemma 2, they are homotopic. This proves Claim 2.
Claim 3 If two monochromatic cycles are intersecting, then they are not homotopic.
Proof of Claim Suppose by contradiction that there exist C, C , two distinct directed
monochromatic cycles that are homotopic and intersecting. By Claim 2, they are not
contractible and of different color. Suppose C is an (i − 1)-cycle and C an (i +
1)cycle. By (T1’), C is leaving C on its right side. Since C, C are homotopic, the
cycle C is entering C at least once from its right side, a contradiction with (T1’).
This proves Claim 3.
We are now able to prove that (T2) is satisfied. Let Ci be any i-cycle of color i.
We have to prove that Ci intersects at least one (i − 1)-cycle and at least one (i +
1)cycle. Let j be either i − 1 or i + 1. By (T2’), there exists an i-cycle Ci intersecting
a j -cycle Cj of color j . The two cycles Ci , Cj are not reversal by (T3’); thus, they
are crossing. By claim (3), C and Cj are not homotopic. By Claim 2, Ci and Ci are
i
homotopic. Thus, by Lemma 2, Ci and Cj are not homotopic and intersecting.
Thus, (T1) and (T2) are satisfied, and the orientation and coloring are a Schnyder
wood. By (T3’) and Theorem 7 it is a Schnyder wood of Type 1.
Note that for toroidal triangulations, there are no edges oriented in two directions
in an orientation and coloring of the edges satisfying (T1’), by Euler’s formula. So
(T3’) is automatically satisfied. Thus, in the case of toroidal triangulations it is
sufficient to have properties (T1’) and (T2’) to have a Schnyder wood. This is not true in
general, as shown by the example of Fig. 16 that satisfies (T1’) and (T2’) but that is
not a Schnyder wood. There is a monochromatic cycle of color 1 that is not
intersecting any monochromatic cycle of color 2, so (T2) is not satisfied.
7 Existence for Simple Triangulations
In this section we present a short proof of existence of Schnyder woods for simple
triangulations. Sections 8 and 9 contain the full proof of existence for essentially
3-connected toroidal maps.
Fijavz [
13
] proved a useful result concerning the existence of particular
nonhomotopic cycles in toroidal triangulations with no loop and no multiple edges.
(Recall that in this paper we are less restrictive, as we allow non-contractible loops and
non-homotopic multiple edges.)
Theorem 8 ([
13
]) A simple toroidal triangulation contains three non-contractible
and non-homotopic cycles that all intersect on one vertex and that are pairwise
disjoint otherwise.
g : F (G∞) −→ Z, we note that |g| = F ∈F (G∞) g(F ) (when the sum is finite).
Thus, i vi = i (|1(Ri (v) \ Ri (zi (v)))| − |1(Ri (zi (v)) \ Ri (v))|) = | i (1(Ri (v) \
Ri (zi (v)))−1(Ri (zi (v))\Ri (v)))|. Now we compute g = i (1(Ri (v)\Ri (zi (v)))−
1(Ri (zi (v)) \ Ri (v))). We have:
g =
1 Ri (v) \ Ri zi (v)
+ 1 Ri (v) ∩ Ri zi (v)
i
− 1 Ri zi (v) \ Ri (v) − 1 Ri (v) ∩ Ri zi (v)
As Ri (v) \ Ri (zi (v)) and Ri (zi (v)) \ Ri (v) are disjoint from Ri (v) ∩ Ri (zi (v)), we
have:
g =
1 Ri (v) − 1 Ri zi (v)
i
Because the interior of the three regions Ri (v), for i = 0, 1, 2, is disjoint and spans the
whole plane P (by definition), we have i 1(Ri (v)) = 1(∪i (Ri (v))) = 1(P).
Moreover, the regions Ri (zi (v)), for i = 0, 1, 2, are also disjoint and i 1(Ri (zi (v))) =
1(∪i (Ri (zi (v)))) = 1(P \ T ), where T is the bounded region delimited by the lines
L0(v), L1(v), and L2(v). So g = 1(P) − 1(P \ T ) = 1(T ). And thus i vi = |g| =
|1(T )|.
Lemma 22 shows that if the Schnyder wood is of Type 1, then the set of points
are not necessarily coplanar as in the planar case [
12
], but all the copies of a vertex
lie on the same plane (the bounded region delimited by the lines L0(v), L1(v) and
L2(v) has the same number of faces for any copies of a vertex v). Surprisingly, for
Schnyder woods of Type 2, all the points are coplanar.
Lemma 23 The mapping is periodic with respect to S and S .
Proof Let v be any vertex of G∞. Then vitop − vi = N (fi+1(Li+1(vtop)) −
fi+1=(Lvi++1(Sv.))S)im−ilNar(lyf,i−vr1i(gLhti=−1v(v+topS)). − fi−1(Li−1(v))) = N (ci+1 − ci−1)f . So
vtop
For each color i, let γi be the integer such that two monochromatic cycles of G of
respective colors i − 1 and i + 1 intersect exactly γi times, with the convention that
γi = 0 if the Schnyder wood is of Type 2.i. By Lemma 3, γi is properly defined and
does not depend on the choice of the monochromatic cycles. Note that if the Schnyder
wood is of Type 2.i, then γi−1 = γi+1 and if the Schnyder wood is not of Type 2.i, then
γi = 0. Let γ = max(γ0, γ1, γ2). Let Z0 = ((γ1 + γ2)Nf, −γ1Nf, −γ2Nf ), Z1 =
(−γ0Nf, (γ0 + γ2)Nf, −γ2Nf ), and Z2 = (−γ0Nf, −γ1Nf, (γ0 + γ1)Nf ).
Lemma 24 For any vertex u, we have {u + k0Z0 + k1Z1 + k2Z2 | k0, k1, k2 ∈ Z} ⊆
{u + kS + k S | k, k ∈ Z}.
Proof Let u, v be two copies of the same vertex, such that v is the first copy of u
in the direction of L0(u). (That is, L0(u) = L0(v) and on the path P0(u) \ P0(v)
there are not two copies of the same vertex.) Then vi − ui = N (fi+1(Li+1(v)) −
fi+1(Li+1(u)))−N (fi−1(Li−1(v))−fi−1(Li−1(u))). We have |R(L0(v), L0(u))| =
0, |R(L1(v), L1(u))| = γ2f , and |R(L2(v), L2(u))| = γ1f . So v0 − u0 =
N (γ1 + γ2)f , v1 − u1 = −N γ1f , and v2 − u2 = −N γ2f . Thus, v = u + Z0, and
similarly for the other colors. So the first copy of u in the direction of Li(u) is equal
to u + Zi . By Lemma 23, all the copies of u are mapped on {u + kS + k S | k, k ∈ Z},
and so we have the result.
Lemma 25 We have dim(Z0, Z1, Z2) = 2, and if the Schnyder wood is not of
Type 2.i, then dim(Zi−1, Zi+1) = 2.
Proof We have γ0Z0 + γ1Z1 + γ2Z2 = 0 and so dim(Z0, Z1, Z2) ≤ 2. We can
assume by symmetry that the Schnyder wood is not of Type 2.1 and so γ1 = 0. Thus,
Z0 = 0 and Z2 = 0. Suppose by contradiction that dim(Z0, Z2) = 1. Then there exist
α = 0, β = 0, such that αZ0 + βZ2 = 0. The sum of this equation for the coordinates
0 and 2 gives (α + β)γ1 = 0 and thus α = −β. Then the equation for coordinate 0
gives γ0 + γ1 + γ2 = 0, contradicting the fact that γ1 > 1 and γ0, γ2 ≥ 0.
Lemma 26 The vectors S, S are not collinear.
Proof By Lemma 24, the set {u + k0Z0 + k1Z1 + k2Z2 | k0, k1, k2 ∈ Z} is a subset
of {u + kS + k S | k, k ∈ Z}. By Lemma 25, we have dim(Z0, Z1, Z2) = 2, thus
dim(S, S ) = 2.
di(u, zi (u)) < (n − 1) × fij+1.
Lemma 27 If u, v are two distinct vertices such that v is in Li−1(v), u is in Pi−1(v),
both u and v are in the region R(Li+1(u), Li+1(v)), and Li+1(u) and Li+1(v) are
two consecutive (i + 1)-lines with Li+1(u) ∈ Lij+1 (see Fig. 34), then di (zi (v), v) +
Proof Let Qi+1(v) the subpath of Pi+1(v) between v and Li+1(v) (maybe Qi+1(v)
has length 0 if v = zi (v)). Let Qi+1(u) be the subpath of Pi+1(u) between u and
Li+1(u) (maybe Qi+1(u) has length 0 if u = zi (u)). The path Qi+1(v) cannot contain
two different copies of a vertex of G, otherwise Qi+1(v) will correspond to a
noncontractible cycle of G and thus will contain an edge of Li+1(v). So the length of
Qi+1(v) is ≤ n − 1.
The total number of times a copy of a given face of G can appear in the region
R = Ri (zi (v)) \ Ri (v), corresponding to di (zi (v), v), can be bounded as follows.
Region R is between two consecutive copies of Li+1(u). So in R, all the copies of
a given face are separated by a copy of Li−1(v). Each copy of Li−1(v) intersecting
R must intersect Qi+1(v) on a specific vertex. As Qi+1(v) has at most n vertices,
a given face can appear at most n − 1 times in R. Similarly, the total number of
times that a copy of a given face of G can appear in the region Ri (u) \ Ri (zi (u)),
corresponding to di (u, zi (u)), is ≤ (n − 1).
A given face of G can appear in only one of the two gray regions of Fig. 34.
So a face is counted ≤ n − 1 times in the quantity di (zi (v), v) + di (u, zi (u)). Only
the faces of the region R(Cij+1, Cij++11) can be counted, and there is at least one face
of R(Cij+1, Cij++11) (for example, one incident to v) that is not counted. So in total
di (zi (v), v) + di (u, zi (u)) ≤ (n − 1) × (fij+1 − 1) < (n − 1) × fi+1
j .
Clearly, the symmetric of Lemma 27, where the roles of i + 1 and i − 1 are
exchanged, is also true.
The bound of Lemma 27 is somehow sharp. In the example of Fig. 35, the
rectangle represents a toroidal map G and the universal cover is partially
represented. If the map G has n vertices and f faces (n = 5 and f = 5 in the example),
then the gray region, representing the quantity d1(z1(v), v) + d1(u, z1(u)), has size
n(n2−1) = Ω(n × f ).
Lemma 28 Let u, v be vertices of G∞ such that Ri (u) ⊆ Ri (v), then ui ≤ vi .
Moreover, if Ri (u) Ri (v), then vi − ui > (N − n)(|R(Li−1(u), Li−1(v))| +
|R(Li+1(u), Li+1(v))|) ≥ 0.
Proof We distinguish two cases depending on whether the Schnyder wood is of
Type 2.i or not.
• Case 1: The Schnyder wood is not of Type 2.i.
Suppose first that u and v are both in a region delimited by two consecutive lines
of color i − 1 and two consecutive lines of color i + 1. Let Lij−1, Lij−+11, Lij+1, Lij++11 be
these lines such that Lij−+11 is on the positive side of Lij−1, Lij++11 is on the positive side
of Lij+1, and Lk ∈ Lk (see Fig. 36). We distinguish cases corresponding to equality
or not between lines Li−1(u), Li−1(v) and Li+1(u), Li+1(v).
N (fi+1(Li+1(v)) − fi+1(Li+1(u))) = di(v, zi (v)) − di(u, zi (u)) + Nfij+1. Let u
be the intersection of Pi+1(u) with Lij−+11 (maybe u = u ). Let v be the intersection
of Pi+1(v) with Lj+1
i−1 (maybe v = v ). Since Li+1(u) = Li+1(v), we have u = v .
Since u ∈ Ri (v), we have u ∈ Ri(v ) and so u ∈ Pi−1(v ). Then, by Lemma 27,
di(zi (v ), v ) + di (u , zi (u )) < (n − 1)fij+1. If Li−1(u) = Lij−+11, then one can see that
di(v, zi (v)) − di(u, zi (u)) ≥ di(v , zi (v )) − di (u , zi (u )). If Li−1(u) = Lij−1, one
can see that di(v, zi (v)) − di(u, zi (u)) ≥ di(v , zi (v )) − di(u , zi (u )) − fij+1. So
finally, vi − ui = di(v, zi (v)) − di (u, zi (u)) + Nfij+1 ≥ di(v , zi (v )) − di (u , zi (u )) +
Case 1.1: Li−1(u) = Li−1(v) and Li+1(u) = Li+1(v). Then vi − ui =
di(v, zi (v)) − di(u, zi (u)) = di (v, u). Thus clearly, if Ri(u) ⊆ Ri(v), then ui ≤ vi
and if Ri(u) Ri (v), vi − ui > 0 = (N − n)(|R(Li−1(u), Li−1(v))| +
|R(Li+1(u), Li+1(v))|).
Case 1.2: Li−1(u) = Li−1(v) and Li+1(u) = Li+1(v). As u ∈ Ri(v), we have
Li+1(u) = Lij+1 and Li+1(v) = Lij++11. Then vi − ui = di(v, zi (v)) − di (u, zi (u)) +
(N − 1)fij+1 > (N − n)fij+1 = (N − n)(|R(Li−1(u), Li−1(v))| +
|R(Li+1(u), Li+1(v))|) ≥ 0.
Case 1.3: Li−1(u) = Li−1(v) and Li+1(u) = Li+1(v). This case is completely
symmetric to the previous case.
Case 1.4: Li−1(u) = Li−1(v) and Li+1(u) = Li+1(v). As u ∈ Ri(v), we
have Li+1(u) = Lij+1, Li+1(v) = Lij++11, Li−1(u) = Lij−+11, and Li−1(v) = Lij−1.
Then vi − ui = di (v, zi (v)) − di (u, zi (u)) + N (fi+1(Li+1(v)) − fi+1(jLi+1(u))) −
N (fi−1(Li−1(v)) − fi−1(Li−1(u))) = di(v, zi (v)) − di(u, zi (u)) + Nfi+1 + Nfij−1.
Let u be the intersection of Pi+1(u) with Lij−+11 (maybe u = u ). Let u be the
intersection of Pi−1(u) with Lij+1 (maybe u = u ). Let v be the intersection of Pi+1(v)
with Lij−+11 (maybe v = v ). Let v be the intersection of Pi−1(v) with Lij+1 (maybe
v = v ). Since Li+1(u) = Li+1(v), we have u = v . Since u ∈ Ri (v), we have
u ∈ Ri (v ) and so u ∈ Pi−1(v ). Then, by Lemma 27, di(zi (v ), v )+di(u , zi (u )) <
(n−1)fij+1. Symmetrically, di(zi (v ), v )+di (u , zi(u )) < (n−1)fij−1. Moreover,
we have di(v, zi (v)) − di(u, zi (u)) ≥ di(v , zi (v )) − di (u , zi (u )) + di(v , zi (v )) −
di(u , zi (u )) − fij+1 − fij−1. So finally, vi − ui = di(v, zi (v)) − di(u, zi (u)) +
Nfij+1 + Nfij−1 ≥ di(v , zi (v )) − di (u , zi (u )) + di(v , zi (v )) − di(u , zi (u )) +
(N − 1)fij+1 + (N − 1)fij−1 > (N − n)fij+1 + (N − n)fij−1 = (N − n)(|R(Li−1(u),
Li−1(v))| + |R(Li+1(u), Li+1(v))|) ≥ 0.
Suppose now that u and v do not lie in a region delimited by two consecutive lines
of color i − 1 and/or in a region delimited by two consecutive lines of color i + 1.
One can easily find distinct vertices w0, . . . , wr (wi , 1 ≤ i < r chosen at
intersections of monochromatic lines of colors i − 1 and i + 1) such that w0 = u, wr = v,
and for 0 ≤ ≤ r − 1, we have Ri(w ) Ri(w +1) and w , w +1 are both in a region
delimited by two consecutive lines of color i − 1 and in a region delimited by two
consecutive lines of color i + 1. Thus, by the first part of the proof, (w )i − (w +1)i >
(N − n)(|R(Li−1(w +1), Li−1(w ))| + |R(Li+1(w +1), Li+1(w ))|). Thus vi − ui >
(N − n) (|R(Li−1(w +1), Li−1(w ))| + |R(Li+1(w +1), Li+1(w ))|). For any
a, b, c such that Ri (a) ⊆ Ri (b) ⊆ Ri (c), we have |R(Lj (a), Lj (b))| +
|R(Lj (b), Lj (c))| = |R(Lj (a), Lj (c))|. Thus, we obtain the result by summing the
size of the regions.
• Case 2: The Schnyder wood is of Type 2.i.
Suppose first that u and v are both in a region delimited by two consecutive lines
of color i + 1.
Let Lij+1, Lij++11 be these lines such that Lij++11 is on the positive side of Lij+1, and
L . We can assume that we do not have both u and v in Lij++11 (by eventually
chio+o1s∈inLgio+t1her consecutive lines of color i + 1). We consider two cases:
Case 2.1: v ∈/ Lij++11. Then by Lemma 21, Lij+1 = Li+1(u) = (Li−1(u))−1 =
Li+1(v) = (Li−1(v))−1. Then vi − ui = di (v, zi (v)) − di (u, zi (u)) = di (v, u). Thus
clearly, if Ri (u) ⊆ Ri (v), then ui ≤ vi and if Ri (u) Ri (v), then vi − ui > 0 =
(N − n)(|R(Li−1(u), Li−1(v))| + |R(Li+1(u), Li+1(v))|).
Case 2.2: v ∈ Lij++11. Then Lj+1
i+1 = Li+1(v) = (Li−1(v))−1 and di (v, zi (v)) = 0.
By assumption u ∈/ Lij++11 and by Lemma 21, Lij+1 = Li+1(u) = (Li−1(u))−1.
Then vi − ui = di (v, zi (v)) − di (u, zi (u)) + N (fi+1(Li+1(v)) − fi+1(Li+1(u))) −
N (fi−1(Li−1(v)) − fi−1(Li−1(u))) = −di (u, zi (u)) + 2Nfij+1. Let Li and Li
be two consecutive i-lines such that u lies in the region between them and Li
is on the right of Li . Let u be the intersection of Pi+1(u) with Li (maybe
u = u ). Let u be the intersection of Pi−1(u) with Li (maybe u = u ). Then, by
Lemma 27, di (u , zi (u )) < (n − 1)fij+1 and di (u , zi (u )) < (n − 1)fij+1. Thus,
we have di (u, zi (u)) ≤ di (u , zi (u )) + di (u , zi (u )) + fij+1 < (2(n − 1) + 1)fij+1.
So finally, vi − ui > −(2n − 1)fij+1 + 2Nfij+1 > 2(N − n)fij+1 = (N − n) ×
(|R(Li−1(u), Li−1(v))| + |R(Li+1(u), Li+1(v))|) ≥ 0.
If u and v do not lie in a region delimited by two consecutive lines of color i + 1,
then as in case 1, one can find intermediate vertices to obtain the result.
Lemma 29 If two vertices u, v are adjacent, then for each color i, we have
|vi − ui | ≤ 2Nf .
Proof Since u, v are adjacent, they are both in a region delimited by two
consecutive lines of color i − 1 and in a region delimited by two consecutive lines
of color i + 1. Let Lij−1, Lij−+11 be these two consecutive lines of color i − 1 and
Lij+1, Lij++11 these two consecutive lines of color i + 1 where Lk ∈ Lk, Lij−+11 is on
the positive side of Lij−1, and Lij++11 is on the positive side of Lij+1 (see Fig. 36
when the Schnyder wood is not of Type 2.i). If the Schnyder wood is of Type 2.i,
we assume that Lij−+11 = (Lij+1)−1 and Lij−1 = (Lij++11)−1. Let z be a vertex on
the intersection of Lij−+11 and Lij+1. Let z be a vertex on the intersection of Lij−1
and Lij++11. Thus, we have Ri (z) ⊆ Ri (u) ⊆ Ri (z ) and Ri (z) ⊆ Ri (v) ⊆ Ri (z ). So by
Lemma 28, zi ≤ ui ≤ zi and zi ≤ vi ≤ zi . So |vi − ui | ≤ zi − zi = N (fi+1(Lij++11) −
fi+1(Lij+1)) − N (fi−1(Lij−1) − fi−1(Lij−+11)) = Nfij+1 + Nfij−1 ≤ 2Nf .
We are now able to prove the following theorem.
Theorem 13 If G is a toroidal map given with a Schnyder wood, then the mapping
of each vertex of G∞ on its region vector gives a geodesic embedding of G.
Proof By Lemmas 23 and 26, the mapping of G∞ on its region vector is periodic
with respect to S, S that are not collinear. For any pair u, v of distinct vertices of G∞,
by Lemma 8(iii), there exists i, j with Ri (u) Ri (v) and Rj (v) Rj (u); thus, by
Lemma 28, ui < vi and vj < uj . So V∞ is a set of pairwise incomparable elements
of R3.
(D1) V∞ is a set of pairwise incomparable elements, so the mapping between
vertices of G∞ and V∞ is a bijection.
(D2) Let e = uv be an edge of G∞. We show that w = u ∨ v is on the surface SV∞ .
By definition, u ∨ v is in DV∞ . Suppose, by contradiction, that w ∈/ SV∞ . Then there
exists x ∈ V∞ with x < w. Let x also denote the corresponding vertex of G∞.
Edge e is in a region Ri (x) for some i. So u, v ∈ Ri (x) and thus by Lemma 8(i),
Ri (u) ⊆ Ri (x) and Ri (v) ⊆ Ri (x). Then by Lemma 28, wi = max(ui , vi ) ≤ xi ,
a contradiction. Thus, the elbow geodesic between u and v is on the surface.
(D3) Consider a vertex v ∈ V and a color i. Let u be the extremity of the arc ei (v).
We have u ∈ Ri−1(v) and u ∈ Ri+1(v), so by Lemma 8(i), Ri−1(u) ⊆ Ri−1(v) and
Ri+1(u) ⊆ Ri+1(v). Thus, by Lemma 8(iii), Ri (v) Ri (u). So, by Lemma 28,
vi < ui , ui−1 ≤ vi−1, and ui+1 ≤ vi+1. So the orthogonal arc of vertex v in the
direction of the basis vector ei is part of the elbow geodesic of the edge ei (v).
(D4) Suppose there exists a pair of crossing edges e = uv and e = u v on the
surface SV∞ . The two edges e, e cannot intersect on orthogonal arcs, so they intersect
on a plane orthogonal to one of the coordinate axes. Up to symmetry we may assume
that we are in the situation of Fig. 37 with u1 = u1, u2 > u2, and v2 < v2. Between
u and u , there is a path consisting of orthogonal arcs only. With (D3), this implies
that there is a bidirected path P ∗ colored 0 from u to u and colored 2 from u to u.
We have u ∈ R2(v), so by Lemma 8(i), R2(u) ⊆ R2(v). We have u ∈ R2(u), so
u ∈ R2(v). If P0(v) contains u , then there is a contractible cycle containing v, u, u
in G1 ∪ G0−1 ∪ G2−1, contradicting Lemma 1, so P0(v) does not contain u . If P1(v)
contains u , then u ∈ P1(u) ∩ P0(u), contradicting Lemma 7. So u ∈ R2◦(v). Thus,
the edge u v implies that v ∈ R2(v). So by Lemma 28, v2 ≤ v2, a contradiction.
Theorems 1 and 13 imply Theorem 3.
One can ask: What is the “size” of the obtained geodesic embedding of
Theorem 13? Of course, this mapping is infinite so there is no real size, but as the object
is periodic, one can consider the smallest size of the vectors such that the mapping
is periodic with respect to them. There are several such pairs of vectors, one is S, S .
Recall that Si = N (ci+1 − ci−1)f and Si = N (ci+1 − ci−1)f . Unfortunately, the size
of S, S can be arbitrarily large. Indeed, the values of ci+1 − ci−1 and ci+1 − ci−1 are
unbounded, as a toroidal map can be artificially “very twisted” in the considered flat
torus representation (independent of the number of vertices or faces). Nevertheless,
we can prove the existence of bounded size vectors for which the mapping is periodic
with respect to them.
Fig. 37 A pair of crossing
elbow geodesics
Lemma 30 If G is a toroidal map given with a Schnyder wood, then the mapping of
each vertex of G∞ on its region vector gives a periodic mapping of G∞ with respect
to non-collinear vectors Y and Y , where the size of Y and Y is in O(γ Nf ). In
general, we have γ ≤ n, and in the case where G is a simple toroidal triangulation
given with a Schnyder wood obtained by Theorem 9, we have γ = 1.
Proof By Lemma 24, the vectors Zi−1, Zi+1 (when the Schnyder wood is not of
Type 2.i) span a subset of S, S (it can happen that this subset is strict). Thus, in
the parallelogram delimited by the vectors Zi−1, Zi+1 (that is, a parallelogram by
Lemma 25), there is a parallelogram with sides Y, Y containing a copy of V . The
size of the vectors Zi is in O(γ Nf ) and so the size of Y and Y is also.
In general, we have γi ≤ n, as each intersection between two monochromatic
cycles of G of color i − 1 and i + 1 corresponds to a different vertex of G and thus
γ ≤ n. In the case of simple toroidal triangulation given with a Schnyder wood
obtained by Theorem 9, we have, for each color i, γi = 1, and thus γ = 1.
As in the plane, one can give weights to faces of G. Then all their copies in G∞
have the same weight, and instead of counting the number of faces in each region one
can compute the weighted sum.
Note that the geodesic embeddings of Theorem 13 are not necessarily rigid.
A geodesic embedding is rigid [
11, 20
] if for every pair u, v ∈ V such that u ∨ v is
in SV , u and v are the only elements of V that are dominated by u ∨ v. The geodesic
embedding of Fig. 5 is not rigid, as the bend corresponding to the loop of color 1 is
dominated by three vertices of G∞. We do not know if it is possible to build a rigid
geodesic embedding from the Schnyder wood of a toroidal map. Maybe a technique
similar to the one presented in [11] can be generalized to the torus.
It has already been mentioned that in the geodesic embeddings of Theorem 13 the
points corresponding to vertices are not coplanar. The problem of building a coplanar
geodesic embedding from the Schnyder wood of a toroidal map is open. In the plane,
there are some examples of maps G [
11
] for which it is not possible to require both
rigidity and coplanarity. Thus, the same is true in the torus for the graph G+.
Another question related to coplanarity is whether one can require that the points
of the orthogonal surface corresponding to edges of the graph (i.e., bends) are
coplanar. This property is related to contact representation by homotopic triangles [
11
]. It
is known that in the plane, not all Schnyder woods are supported by such surfaces.
Kratochvil’s conjecture [
19
], recently proved [
17
], states that every 4-connected
planar triangulation admits a contact representation by homothetic triangles. Can this be
extended to the torus?
When considering not necessarily homothetic triangles, it has been proved [
14
]
that there is a bijection between Schnyder woods of planar triangulations and
contact representations by triangles. This result has been generalized to internally
3connected planar maps [
16
] by exhibiting a bijection between Schnyder woods of
internally 3-connected planar maps and primal-dual contact representations by
triangles (i.e., representations where both the primal and the dual are represented). It
would be interesting to generalize these results to the torus.
11 Duality of Orthogonal Surfaces
Given an orthogonal surface generated by V , let FV be the maximal points of SV ,
i.e., the points of SV that are not dominated by any vertex of SV . If A, B ∈ FV and
A ∧ B ∈ SV , then SV contains the union of the two line segments joining A and B
to A ∧ B. Such arcs are called dual elbow geodesics. The dual orthogonal arc of
A ∈ FV in the direction of the standard basis vector ei is the intersection of the ray
A + λei with SV .
Given a toroidal map G, let G∞∗ be the dual of G∞. A dual geodesic embedding
of G is a drawing of G∞∗ on the orthogonal surface SV∞ , where V∞ is a periodic
mapping of G∞ with respect to two non-collinear vectors, satisfying the following
(see the example of Fig. 38):
(D1*) There is a bijection between the vertices of G∞∗ and FV∞ .
(D2*) Every edge of G∞∗ is a dual elbow geodesic.
(D3*) Every dual orthogonal arc in SV∞ is part of an edge of G∞∗.
(D4*) There are no crossing edges in the embedding of G∞∗ on SV∞ .
Let G be a toroidal map given with a Schnyder wood. Consider the mapping of
each vertex on its region vector. We consider the dual of the Schnyder wood of G.
By Lemma 11, it is a Schnyder wood of G∗. A face F of G∞ is mapped on the point
v∈F v. Let G∞ be a simultaneous drawing of G∞ and G∞∗ such that only dual
edges intersect. To avoid confusion, we denote Ri the regions of the primal Schnyder
wood and Ri∗ the regions of the dual Schnyder wood.
Lemma 31 For any face F of G∞, we have that
v∈F v is a maximal point of SV∞ .
Proof Let F be a face of G∞. For any vertex u of V∞, there exists a color i, such that
the face F is in the region Ri (u). Thus for v ∈ F , we have v ∈ Ri (u). By Lemma 28,
we have vi ≤ ui and so Fi ≤ ui . So F = v∈F v is a point of SV∞ .
Suppose, by contradiction, that F is not a maximal point of SV∞ . Then there is a
point α ∈ SV∞ that dominates F and, for at least one coordinate j , we have Fj < αj .
By Lemma 10, the angles at F form, in counterclockwise order, non-empty intervals
of 0’s, 1’s, and 2’s. For each color, let zi be a vertex of F with angle i. We have that
F is in the region Ri (zi ). So zi−1 ∈ Ri (zi ) and by Lemma 8(i), we have Ri (zi−1) ⊆
Ri (zi ). Since F is in Ri−1(zi−1), it is not in Ri (zi−1) and thus Ri (zi−1) Ri (zi ).
Then by Lemma 28, we have (zi−1)i < (zi )i and symmetrically (zi+1)i < (zi )i . So
Fj−1 = (zj−1)j−1 > (zj )j−1 and Fj+1 > (zj )j+1. Thus, α strictly dominates zj ,
a contradiction to α ∈ SV∞ . Thus, F is a maximal point of SV∞ .
Lemma 32 If two faces A, B are such that Ri∗(B) ⊆ Ri∗(A), then Ai ≤ Bi .
Proof Let v ∈ B be a vertex whose angle at B is labeled i. We have v ∈ Ri∗(B) and
so v ∈ Ri∗(A). In G∞, the path Pi (v) cannot leave Ri∗(A), the path Pi+1(v) cannot
intersect Pi+1(A), and the path Pi−1(v) cannot intersect Pi−1(A). Thus, Pi+1(v)
intersects Pi−1(A) and the path Pi−1(v) cannot intersect Pi+1(A). So A ∈ Ri (v).
Thus, for all u ∈ A, we have u ∈ Ri (v), so Ri (u) ⊆ Ri (v), and so ui ≤ vi . Then
Ai = maxu∈A ui ≤ vi ≤ maxw∈B wi = Bi .
Theorem 14 If G is a toroidal map given with a Schnyder wood and each vertex of
G∞ is mapped on its region vector, then the mapping of each face of G∞∗ on the
point v∈F v gives a dual geodesic embedding of G.
Proof By Lemmas 23 and 26, the mapping is periodic with respect to non-collinear
vectors.
(D1*) Consider a counting of elements on the orthogonal surface, where we count
two copies of the same object just once (note that we are on an infinite and periodic
object). We have that the sum of primal orthogonal arcs plus dual ones is exactly 3m.
There are 3n primal orthogonal arcs and thus there are 3m − 3n = 3f dual orthogonal
arcs. Each maximal point of SV∞ is incident to 3 dual orthogonal arcs and there is no
dual orthogonal arc incident to two distinct maximal points. So there are f maximal
points. Thus by Lemma 31, we have a bijection between faces of G∞ and maximal
points of SV∞ .
Let V∞∗ be the maximal points of SV∞ . Let DV∗∞ = {A ∈ R3 | ∃B ∈ V∞∗ such that
A ≤(DB2}*.)NLoetteeth=at AthBe bboeuanndaerdygoefoDf VG∞∞i∗s.SWVe∞ s.how that w = A ∧ B is on the
sur∗
face SV∞ . By definition, w is in DV∗∞ . Suppose, by contradiction, that w ∈/ SV∞ . Then
there exists C, a maximal point of SV∞ with w < C. By the bijection (D1*) between
maximal points and vertices of G∞∗, the point C corresponds to a vertex of G∞∗,
also denoted C. Edge e is in a region Ri∗(C) for some i. So A, B ∈ Ri∗(C) and thus,
by Lemma 8(i), Ri∗(A) ⊆ Ri∗(C) and Ri∗(B) ⊆ Ri∗(C). Then by Lemma 32, we have
Ci ≤ min(Ai , Bi ) = wi , a contradiction. Thus, the dual elbow geodesic between A
and B is also on the surface.
(D3*) Consider a vertex A of G∞∗ and a color i. Let B be the extremity
of the arc ei (A). We have B ∈ Ri∗−1(A) and B ∈ Ri∗+1(A), so by Lemma 8(i),
aRni∗d−1A( Bi+)1⊆≤ RBi∗i−+11(.AA)s aAndanRdi∗+B1(aBre)d⊆istRini∗c+t1m(Aa)x.imTahlupsobinytsLoefmSmVa∞3, 2th,eAyia−r1e
≤incBoim−1parable, and thus Bi < Ai . So the dual orthogonal arc of vertex A in the direction of
the basis vector ei is part of edge ei (A).
(D4*) Suppose there exists a pair of crossing edges e = AB and e = A B of G∞∗
on the surface SV∞ . The two edges e, e cannot intersect on orthogonal arcs, so they
intersect on a plane orthogonal to one of the coordinate axes. Up to symmetry we
may assume that we are in the situation A1 = A1, A0 > A0, and B0 < B0. Between
A and A there is a path consisting of orthogonal arcs only. With (D3*), this implies
that there is a bidirected path P ∗ colored 2 from A to A and colored 0 from A to A.
We have A ∈ R0(B), so by Lemma 8(i), R0(A) ⊆ R0(B). We have A ∈ R0(A), so
A ∈ R0(B). If P2(B) contains A , then there is a contractible cycle containing A,
A , B in G1∗ ∪ G0∗−1 ∪ G2∗−1, contradicting Lemma 1, so P2(B) does not contain A .
If P1(B) contains A , then A ∈ P1(A) ∩ P2(A), contradicting Lemma 7. So A ∈
R◦(B). Thus, the edge A B implies that B ∈ R0(B). So by Lemma 32, B0 ≥ B0,
0
a contradiction.
Theorems 13 and 14 can be combined to obtain a simultaneous representation of
a Schnyder wood and its dual on an orthogonal surface. The projection of this
3dimensional object on the plane of the equation x + y + z = 0 gives a representation
of the primal and the dual where edges are allowed to have one bend and two dual
edges have to cross on their bends (see the example of Fig. 39).
Theorem 15 An essentially 3-connected toroidal map admits a simultaneous flat
torus representation of the primal and the dual where edges are allowed to have
one bend and two dual edges have to cross on their bends. Such a representation is
contained in a (triangular) grid of size O(n2f ) × O(n2f ) in general and O(nf ) ×
O(nf ) if the map is a simple triangulation. Furthermore, the length of the edges are
in O(nf ).
Proof Let G be an essentially 3-connected toroidal map. By Theorem 1 (or
Theorem 9 if G is a simple triangulation), G admits a Schnyder wood (where
monochromatic cycles of different colors intersect just once if G is simple). By Theorems 13
and 14, the mapping of each vertex of G∞ on its region vector gives a primal and
dual geodesic embedding. Thus, the projection of this embedding on the plane of the
equation x + y + z = 0 gives a representation of the primal and the dual of G∞ where
edges are allowed to have one bend and two dual edges have to cross on their bends.
By Lemma 30, the obtained mapping is a periodic mapping of G∞ with respect to
non-collinear vectors Y and Y where the size of Y and Y is in O(γ Nf ), with γ ≤ n
in general and γ = 1 in case of a simple triangulation. Let N = n. The embedding
gives a representation in the flat torus of sides Y, Y where the size of the vectors Y
and Y is in O(n2f ) in general and in O(nf ) if the graph is simple and the Schnyder
wood is obtained by Theorem 9. By Lemma 29, the lengths of the edges in this
representation are in O(nf ).
12 Straight-Line Representation of Toroidal Maps
The geodesic embedding obtained by the region vector method can be used to obtain
a straight-line representation of a toroidal map (see Fig. 40). For this purpose, we
have to choose N bigger than previously. Note that Fig. 40 is the projection of the
geodesic embedding of Fig. 5 obtained with the value of N = n. In this particular
case this gives a straight-line representation, but in this section we only prove that
such a technique works for triangulations and for N sufficiently large. To obtain a
straight-line representation of a general toroidal map, one first has to triangulate it.
Let G be a toroidal triangulation given with a Schnyder wood and V ∞ the set of
region vectors of vertices of G∞. The Schnyder wood is of Type 1 by Theorem 7.
Recall that γi is the integer such that two monochromatic cycles of G of colors i − 1
and i + 1 intersect exactly γi times.
Lemma 33 For any vertex v, the number of faces in the bounded region delimited by
the three lines Li (v) is strictly less than (5 min(γi ) + max(γi ))f .
Proof Suppose by symmetry that min(γi ) = γ1. Let Li = Li (v) and zi = zi (v). Let T
be the bounded region delimited by the three monochromatic lines Li . The boundary
of T is a cycle C oriented clockwise or counterclockwise. Assume that C is oriented
counterclockwise (the proof is similar if oriented clockwise). The region T is on the
left sides of the lines Li . We have zi−1 ∈ Pi (zi+1).
We define, for j, k ∈ N, monochromatic lines L2(j ), L0(k) and vertices z(j, k) as
follows (see Fig. 41). Let L2(1) be the first 2-line intersecting L0 \ {z1} while
walking from z1, along L0 in the direction of L0. Let L0(1) be the first 0-line of color 0
intersecting L2 \ {z1} while walking from z1, along L2 in the reverse direction of L2.
Let z(1, 1) be the intersection between L2(1) and L0(1). Let z(j, 1), j ≥ 0, be the
consecutive copies of z(1, 1) along L0(1) such that z(j + 1, 1) is after z(j, 1) in the
direction of L0(1). Let L2(j ), j ≥ 0, be the 2-line of color 2 containing z(j, 1). Note
that we may have L2 = L2(0), but in any case L2 is between L2(0) and L2(1). Let
z(j, k), k ≥ 0, be the consecutive copies of z(j, 1) along L2(j ) such that z(j, k + 1) is
after z(j, k) in the reverse direction of L2(j ). Let L0(k), k ≥ 0, be the 0-line
containing z(1, k). Note that we may have L0 = L0(0), but in any case L0 is between L0(0)
and L0(1). Let S(j, k) be the region delimited by L2(j ), L2(j + 1), L0(k), L0(k + 1).
All the regions S(j, k) are copies of S(0, 0). The region S(0, 0) may contain several
copies of a face of G, but the number of copies of a face in S(0, 0) is equal to γ1. Let
R be the unbounded region situated on the right of L0(1) and on the right of L2(1). As
P0(v) cannot intersect L0(1) and P2(v) cannot intersect L2(1), vertex v is in R. Let
P (j, k) be the subpath of L0(k) between z(j, k) and z(j + 1, k). All the lines L0(k)
are composed only of copies of P (0, 0). The interior vertices of the path P (0, 0)
cannot contains two copies of the same vertex, otherwise there will be a vertex z(j, k)
between z(0, 0) and z(1, 0). Thus, all interior vertices of a path P (j, k) correspond
to distinct vertices of G.
The Schnyder wood is of Type 1, thus 1-lines are crossing 0-lines. As a line L0(k)
is composed only of copies of P (0, 0), any path P (j, k) is crossed by a 1-line. Let
The bound of Lemma 33 is somehow sharp. In the example of Fig. 42, the
rectangle represents a toroidal triangulation G and the universal cover is partially
represented. For each value of k ≥ 0, there is a toroidal triangulation G with n = 4(k + 1)
vertices, where the gray region, representing the region delimited by the three
monochromatic lines Li (v), contains 4 j2k=+11 +3(2k + 2) = Ω(n × f ) faces.
Figure 42 represents such a triangulation for k = 2.
For planar graphs the region vector method gives vertices that all lie on the same
plane. This property is very helpful in proving that the positions of the points on
P give straight-line representations. In the torus, things are more complicated, as
our generalization of the region vector method does not give coplanar points. But
Lemmas 22 and 33 show that all the points lie in the region situated between the two
planes of equations x + y + z = 0 and x + y + z = t , with t = (5 min(γi ) + max(γi ))f .
Note that t is bounded by 6nf by Lemma 30 and this is independent from N . Thus,
from “far away” it looks like the points are coplanar and by taking N sufficiently
large, non-coplanar points are “far enough” from each other to enable the region
vector method to give straight-line representations.
Let N = t + n.
Lemma 34 Let u, v be two vertices such that ei−1(v) = uv, Li = Li (u) = Li (v),
and such that both u, v are in the region R(Li , Li ) for L an i-line consecutive to Li .
i
Then vi+1 − ui+1 < |R(Li , Li )| and ei−1(v) is going counterclockwise around the
closed disk bounded by {ei−1(v)} ∪ Pi (u) ∪ Pi (v).
Proof Let y be the first vertex of Pi (v) that is also in Pi (u). Let Qu (resp. Qv ) be
the part of Pi (u) (resp. Pi (v)) between u (resp. v) and y.
Let D be the closed disk bounded by the cycle C = (Qv)−1 ∪ {ei−1(v)} ∪ Qu.
If C is going clockwise around D, then Pi+1(v) is leaving v in D and thus has to
intersect Qu or Qv . In both cases, there is a cycle in Gi+1 ∪ (Gi )−1 ∪ (Gi−1)−1,
a contradiction to Lemma 6. So C is going clockwise around D.
As Li (u) = Li (v) and Li−1(u) = Li−1(v), we have vi+1 − ui+1 = di+1(v, u),
and this is equal to the number of faces in D. We have D R(Li , Li ). Suppose D
contains two copies of a given face. Then, these two copies are on different sides of a
1-line. By property (T1), it is not possible to have a 1-line entering D. So D contains
at most one copy of each face of R(Li , Li ).
Lemma 35 For any face F of G∞, incident to vertices u, v, w (given in
counterclockwise order around F ), the cross product v−→w ∧ −→vu has strictly positive coordinates.
Proof Consider the angle labeling corresponding to the Schnyder wood. By
Lemma 10, the angles at F are labeled in counterclockwise order 0, 1, 2. As
−→uv ∧ u−→w = v−→w ∧ −→vu = w−→u ∧ w−→v, we may assume that u, v, w are such that u is
in the angle labeled 0, vertex v in the angle labeled 1, and vertex w in the angle
labeled 2. The face F is either a cycle completely directed into one direction or it has
two edges oriented in one direction and one edge oriented in the other. Let
⎛ (w1 − v1)(u2 − v2) − (w2 − v2)(u1 − v1) ⎞
−→
X = v−→w ∧ −→vu = ⎝ −(w0 − v0)(u2 − v2) + (w2 − v2)(u0 − v0) ⎠
(w0 − v0)(u1 − v1) − (w1 − v1)(u0 − v0)
By symmetry, we consider the following two cases:
• Case 1: The edges of the face F are in counterclockwise order e1(u), e2(v), e0(w)
(see Fig. 43(a)).
We have v ∈ P1(u), so v ∈ R0(u) ∩ R2(u) and u ∈ R1◦(v) (as there are no edges
oriented in two directions). By Lemma 8, we have R0(v) ⊆ R0(u), R2(v) ⊆ R2(u),
and R1(u) R1(v). In fact, the first two inclusions are strict as u ∈/ R0(v) ∪ R2(v).
So by Lemma 28, we have v0 < u0, v2 < u2, u1 < v1. We can prove similar
inequalities for the other pairs of vertices and we obtain w0 < v0 < u0, u1 < w1 < v1,
v2 < u2 < w2. By just studying the signs of the different terms occurring in the
values of the coordinates of −→X, it is clear that −→X has strictly positive
coordinates. (For the first coordinates, it is easier if written in the following form: X0 =
(u1 − w1)(v2 − w2) − (u2 − w2)(v1 − w1).)
• Case 2: The edges of the face F are in counterclockwise order e0(v), e2(v), e0(w)
(see Fig. 43(b)).
As in the previous case, one can easily obtain the following inequalities: w0 <
v0 < u0, u1 < w1 < v1, u2 < v2 < w2 (the only difference with case 1 is between u2
and v2). Exactly as in the previous case, it is clear that X0 and X2 are strictly positive.
But there is no way to reformulate X1 to have a similar proof. Let A = w2 − v2,
B = u0 − v0, C = v0 − w0, and D = v2 − u2, so X1 = AB − CD and A, B, C, D are
all strictly positive.
Vertices u, v, w are in the region R(L1, L1) for L1 a 1-line consecutive to L1. We
consider two cases depending on equality or not between L1(u) and L1(v).
Subcase 2.1: L1(u) = L1(v).
We have X1 = A(B − D) + D(A − C).
We have B − D = (u0 + u2) − (v0 + v2) = (v1 − u1) + ( ui − vi ).
Since u ∈ P0(v), we have L0(u) = L0(v). Suppose that L2(u) = L2(v); then by
Lemma 22, we have ui = vi , and thus B − D = v1 − u1 > 0. Suppose now
that L2(u) = L2(v). By Lemmas 22 and 33, ui − vi > −t . By Lemma 28,
v1 − u1 > (N − n)|R(L2(u), L2(v))| ≥ N − n. So B − D > N − n − t ≥ 0.
We have A − C = (w0 + w2) − (v0 + v2) = (v1 − w1) + ( wi − vi ) > wi −
vi . Suppose that L1(v) = L1(w); then by Lemma 22, we have vi = wi and
thus A − C = v1 − w1 > 0. Then X1 > 0. Suppose now that L1(v) = L1(w). By
Lemma 34, D = v2 − u2 < |R(L1, L1)|. By Lemma 28, A = w2 − v2 > (N − n) ×
|R(L1, L1)|. By Lemmas 22 and 33, wi − vi > −t , so A − C > −t . Then
X1 > (N − n − t )|R(L1, L1)| > 0.
Subcase 2.2: L1(u) = L1(v).
We have X1 = B(A − C) + C(B − D).
Suppose that L1(w) = L1(v). Then L1(w) = L1(u). By Lemma 34 e0(w) is going
counterclockwise around the closed disk D bounded by {e0(w)} ∪ P1(w) ∪ P1(u).
Then v is inside D and P1(v) has to intersect P1(w) ∪ P1(u), so L1(v) = L1(u),
contradicting our assumption. So L1(v) = L1(w).
By Lemma 28, B = u0 − v0 > (N − n)|R(L1, L1)|. We have A − C =
(w0 + w2) − (v0 + v2) = (v1 − w1) + ( wi − vi ). By Lemma 22, we have
vi = wi and thus A − C = v1 − w1 > 0. By (the symmetric of) Lemma 34,
C = v0 − w0 < |R(L1, L1)|. By Lemmas 22 and 33, B − D = (u0 + u2) − (v0 + v2) =
(v1 − u1) + ( ui − vi ) > −t . So X1 > (N − n − t )|R(L1, L1)| > 0.
Let G be an essentially 3-connected toroidal map. Consider a periodic
mapping of G∞, embedded graph H (finite or infinite), and a face F of H . Denote
(f1, f2, . . . , ft ) the counterclockwise facial walk around F . Given a mapping of
the vertices of H in R2, we say that F is correctly oriented if for any triplet
1 ≤ i1 < i2 < i3 ≤ t , the points fi1 , fi2 , and fi3 form a counterclockwise triangle.
Note that a correctly oriented face is drawn as a convex polygon.
Lemma 36 Let G be an essentially 3-connected toroidal map given with a periodic
mapping of G∞ such that every face of G∞ is correctly oriented. This mapping gives
a straight-line representation of G∞.
Proof We proceed by induction on the number of vertices n of G. Note that the
theorem holds for n = 1, so we assume that n > 1. Given any vertex v of G,
let (u0, u1, . . . , ud−1) be the sequence of its neighbors in counterclockwise order
(subscript understood modulo d ). Every face being correctly oriented, for every
i ∈ [0, d − 1] the oriented angle (oriented counterclockwise) (v−→ui , v−−u−i→+1) < π . Let
the winding number kv of v be the integer such that 2kvπ = i∈[0,d−1](v−→ui , v−−u−i→+1).
It is clear that kv ≥ 1. Let us prove that kv = 1 for every vertex v.
Claim 4 For any vertex v, its winding number kv = 1.
Proof of Claim 4 In a flat torus representation of G, we can sum up all the angles by
grouping them around the vertices or around the faces.
(v−→ui , v−−u−i→+1) =
v∈V (G) ui∈N(v)
F ∈F (G) fi ∈F
(f−−if−i→−1, −f−if−i→+1)
The faces being correctly oriented, they form convex polygons. Thus, the angles of a
face F sum at (|F | − 2)π .
2kvπ =
(|F | − 2)π
v∈V (G)
v∈V (G)
v∈V (G)
F ∈F (G)
kv = m − f
|F | − f
v∈V (G) kv = n, and thus kv = 1 for every vertex v. This
Let v be a vertex of G that minimizes the number of loops whose ends are on v.
Thus, either v has no incident loop, or every vertex is incident to at least one loop.
Assume that v has no incident loop. Let v be any copy of v in G∞ and
denote its neighbors (u0, u1, . . . , ud−1) in counterclockwise order. As kv = 1, the points
u0, u1, . . . , ud−1 form a polygon P containing the point v and the segments [v , ui ]
for any i ∈ [0, d − 1]. It is well known that any polygon admits a partition into
triangles by adding some of the chords. Let us call O the outerplanar graph with outer
boundary (u0, u1, . . . , ud−1), obtained by this “triangulation” of P . Let us now
consider the toroidal map G = (G \ {v}) ∪ O and its periodic embedding obtained from
the mapping of G∞ by removing the copies of v. It is easy to see that in this
embedding every face of G is correctly oriented (including the inner faces of O, or the
faces of G that have been shortened by an edge ui ui+1). Thus by the induction
hypothesis, the mapping gives a straight-line representation of G ∞. It is also a
straightline representation of G∞ minus the copies of v where the interiors of each copy of
the polygons P are pairwise disjoint and do not intersect any vertex or edge. Thus,
one can add the copies of v on their initial positions and add the edges with their
neighbors without intersecting any edge. The obtained drawing is thus a straight-line
representation of G∞.
Assume now that every vertex is incident to at least one loop. Since these loops
are non-contractible and do not cross each other, they form homothetic cycles. Thus,
G is as depicted in Fig. 44, where the dotted segments stand for edges that may be
in G but not necessarily. Since the mapping is periodic, the edges corresponding to
Fig. 44 The graph G if every
vertex is incident to a loop
loops of G form several parallel lines, cutting the plane into infinite strips. Since for
any 1 ≤ i ≤ n, kvi = 1, a line of copies of vi divides the plane, in such a way that their
neighbors which are copies of vi−1 and their neighbors which are copies of vi+1 are
in distinct half-planes. Thus, adjacent copies of vi and vi+1 are on two lines bounding
a strip. Then one can see that the edges between copies of vi and vi+1 are contained
in this strip without intersecting each other. Thus, the obtained mapping of G∞ is a
straight-line representation.
A plane is positive if it has equation αx + βy + γ z = 0 with α, β, γ ≥ 0.
Theorem 16 If G is a toroidal triangulation given with a Schnyder wood, and V ∞
the set of region vectors of vertices of G∞, then the projection of V ∞ on a positive
plane gives a straight-line representation of G∞.
Proof Let α, β, γ ≥ 0 and consider the projection of V ∞ on the plane P of the
equation αx + βy + γ z = 0. A normal vector of the plane is given by the vector
−→n = (α, β, γ ). Consider a face F of G∞. Suppose that F is incident to vertices
u, v, w (given in counterclockwise order around F ). By Lemma 35, (−→uv ∧ u−→w).−→n
is positive. Thus, the projection of the face F on P is correctly oriented. So by
Lemma 36, the projection of V ∞ on P gives a straight-line representation of G∞.
Theorems 1 and 16 imply Theorem 4. Indeed, any toroidal graph G can be
transformed into a toroidal triangulation G by adding a linear number of vertices and
edges and such that G is simple if and only if G is simple (see for example the proof
of Lemma 2.3 of [
21
]). Then by Theorem 1, G admits a Schnyder wood. By
Theorem 16, the projection of the set of region vectors of vertices of G ∞ on a positive
plane gives a straight-line representation of G ∞. The grid where the representation is
obtained can be the triangular grid, if the projection is done on the plane of equation
x + y + z = 0, or the square grid, if the projection is done on one of the planes of
equations x = 0, y = 0, or z = 0. By Lemma 30, and the choice of N , the obtained
mapping is a periodic mapping of G∞ with respect to non-collinear vectors Y and
Y where the size of these vectors is in O(γ 2n2) with γ ≤ n in general and γ = 1 if
the graph is simple and the Schnyder wood obtained by Theorem 9. By Lemma 29,
the lengths of the edges in this representation are in O(n3) in general and in O(n2) if
the graph is simple. When the graph is not simple, there is a non-contractible cycle of
length 1 or 2 and thus the size of one of the two vectors Y , Y is in O(n3). Thus, the
grid obtained in Theorem 4 has size in O(n3) × O(n4) in general and O(n2) × O(n2)
if the graph is simple.
The method presented here gives a polynomial algorithm to obtain flat torus
straight-line representations of any toroidal maps in polynomial size grids. Indeed, all
the proofs lead to polynomial algorithms, even the proof of Theorem 8 [
13
], which
uses results from Robertson and Seymour [
25
] on disjoint paths problems.
It would be nice to extend Theorem 16 to obtain convex straight-line
representations for essentially 3-connected toroidal maps.
13 Conclusion
We have proposed a generalization of Schnyder woods to toroidal maps with
application to graph drawing. Along these lines, several questions were raised. We recall
some briefly:
• Does the set of Schnyder woods of a given toroidal map have a kind of lattice
structure?
• Does any simple toroidal triangulation admit a Schnyder wood where the set of
edges of each color induces a connected subgraph?
• Is it possible to use Schnyder woods to embed the universal cover of a toroidal map
on rigid or coplanar orthogonal surfaces?
• Which toroidal maps admit (primal-dual) contact representation by (homothetic)
triangles in a flat torus?
• Can geodesic embeddings be used to obtain convex straight-line representations
for essentially 3-connected toroidal maps?
The guideline of Castelli Aleardi et al. [
3
] to generalize Schnyder woods to higher
genus was to preserve the tree structure of planar Schnyder woods and to use this
structure for efficient encoding. For that purpose they introduce several special rules
(even in the case of genus 1). Our main guideline while working on this paper was
that the surface of genus 1, the torus, seems to be the perfect surface to define
Schnyder woods. Euler’s formula gives exactly m = 3n for toroidal triangulations. Thus,
a simple and symmetric object can be defined by relaxing the tree constraint. For
genus 0, the plane, there are not enough edges in planar triangulations to have
outdegree three for every vertex. For higher genus (the double torus, . . . ) there are too
many edges in triangulations. An open problem is to find what would be the natural
generalization of our definition of toroidal Schnyder woods to higher genus.
The results presented here motivated Castelli Aleardi and Fusy [
4
] to develop
direct methods to obtain straight-line representations for toroidal maps. They manage to
generalize planar canonical ordering to the cylinder to obtain straight-line
representations of simple toroidal triangulations in grids of size O(n) × O(n2), thus improving
the size of our grid, which is O(n2) × O(n2) in the case of a simple toroidal map.
It should be interesting to investigate further the links between the two methods, as
canonical orderings are strongly related to Schnyder woods.
Planar Schnyder woods appear to have many applications in various areas like
enumeration [
1
], compact coding [
24
], representation by geometric objects [
14, 16
],
graph spanners [2], graph drawing [
7, 18
], etc. In this paper we use a new definition
of Schnyder wood for graph drawing purposes, and it would also be interesting to see
if it can be used in other computer science domains.
Acknowledgements The authors thank Nicolas Bonichon, Luca Castelli Aleardi, and Eric Fusy for
fruitful discussions about this work. They also thank student Chloé Desdouits for developing a software to
visualize orthogonal surfaces.
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