arXiv:math/0307370v3 [math.CO] 31 Oct 2005
COMBINATORIAL PSEUDO-TRIANGULATIONS
DAVID ORDEN, FRANCISCO SANTOS, BRIGITTE SERVATIUS, AND HERMAN
SERVATIUS
Abstract. We prove that a planar graph is generically rigid in the plane if
and only if it can be embedded as a pseudo-triangulation. This generalizes the
main result of [4] which treats the minimally generically rigid case.
The proof uses the concept of combinatorial pseudo-triangulation, CPT, in
the plane and has two main steps: showing that a certain “generalized Laman
property” is a necessary and sufficient condition for a CPT to be “stretchable”,
and showing that all generically rigid plane graphs admit a CPT assignment
with that property.
Additionally, we propose the study of combinatorial pseudo-triangulations
on closed surfaces.
1. Introduction
The study of pseudo-triangulations in the plane was initiated recently [1, 10],
but it is rapidly becoming a standard topic in Computational Geometry. This
paper continues the program established in [4, 9, 11, 12], showing that pseudotriangulations provide a missing link between planarity and rigidity in geometric
graph theory. Our main result is:
Theorem 1. For a plane graph G, the following conditions are equivalent:
(1) G is generically rigid in the plane.
(2) G can be stretched to become a pseudo-triangulation of its vertex set with
the given topological embedding.
Recall that a plane graph is a graph together with a given (non-crossing) topological embedding in the plane. It is interesting to observe that property (1) is a property of the underlying (abstract) graph, while property (2) in principle depends on
the embedding. That is, our result in particular implies that pseudo-triangulation
stretchability of a plane graph is independent of the embedding. Another consequence is that:
Corollary 1. The class of planar and generically rigid (in the plane) graphs coincides with the class of graphs of pseudo-triangulations.
The implication from (2) to (1) in Theorem 1 follows from [9]. Here we prove
the implication from (1) to (2) in two steps, via the concept of combinatorial
pseudo-triangulation introduced in Section 3. Section 5 proves that generically
rigid plane graphs can be turned into combinatorial pseudo-triangulations with a
certain “generalized Laman” property, and Section 4 proves that combinatorial
Work of D. Orden and F. Santos was partially supported by grant BFM2001–1153 of the
Spanish Dirección General de Enseñanza Superior e Investigación Cientı́fica.
Work of B. Servatius and H. Servatius was partially supported by NSF grant 0203224.
1
2
ORDEN, SANTOS, AND SERVATIUS
pseudo-triangulations with this property can be stretched. In Section 6 we initiate
the study of combinatorial pseudo-triangulations on closed surfaces, which we think
is an interesting topic to be developed further.
2. Pseudo-triangulations and rigidity
Let A be a finite point set in the Euclidean plane, in general position. A pseudotriangle in the plane is a simple polygon with exactly three convex angles. A
pseudo-triangulation of A is a geometric (i.e., with straight edges) non-crossing
graph with vertex set A, containing the convex hull edges of A and in which every
bounded region is a pseudo-triangle. A vertex v in a geometric graph G is called
pointed if all the edges of G lie in a half-plane supported at v or, equivalently,
if one of the angles incident to v is greater than 180◦. A pseudo-triangulation is
called pointed if all its vertices are pointed. The following numerical result has been
stated several times in different forms, and is crucial to some of the nice properties
of pseudo-triangulations:
Lemma 1. Let G be a non-crossing straight-line embedding of a connected graph
in the plane. Let e, x and y denote the numbers of edges, non-pointed vertices
and pointed vertices in G. Then, e ≤ 3x + 2y − 3, with equality if and only if the
embedding is a pseudo-triangulation.
Proof. Let f be the number of bounded faces of the embedding. By Euler’s formula,
x + y + f = e + 1. We now double-count the number of “big” and “small” angles in
the embedding (that is, angles bigger and smaller than 180 degrees, respectively).
The total number of angles equals 2e. The number of big angles equals y, and
the number of small angles is at least 3f (every bounded face has at least three
corners) with equality if and only if the embedding is a pseudo-triangulation. These
equations give the statement.
Recall that a graph is generically rigid in the plane, see [3] or [13], if any generically embedded bar and joint framework corresponding to the graph has no nontrivial infinitesimal motions. Generic rigidity is a property of the graph, and not of
any particular embedding. In fact, edge-minimal generically rigid graphs on a given
number n of vertices are characterized by Laman’s Condition: they have exactly
2n − 3 edges and every subset of k vertices spans a subgraph with at most 2k − 3
edges, see [5]. Generically rigid graphs with |E| = 2n − 3 are also known as Laman
graphs.
The connection between rigidity and pseudo-triangulations was first pointed out
in Streinu’s seminal paper [12] where it is proved that the graphs of pointed pseudotriangulations are minimally generically rigid graphs, i.e. Laman graphs. In [4] it
was shown that a graph G has a realization as a pointed pseudo-triangulation in
the plane if and only if the graph is a planar Laman graph. The following theorem
in [9] extends Streinu’s result to non-minimally rigid graphs and relates the number
of non-pointed vertices to the degree to which a planar rigid graph is overbraced.
Theorem 2. Let G be the graph of a pseudo-triangulation of a planar point set in
general position. Then:
(1) G is infinitesimally rigid, hence rigid and generically rigid.
(2) Every subset of x non-pointed plus y pointed vertices of G, with x + y ≥ 2,
spans a subgraph with at most 3x + 2y − 3 edges.
COMBINATORIAL PSEUDO-TRIANGULATIONS
3
Property (2) will be crucial in our proof of Theorem 1. Observe, however, that
it is not a property of the graph G, but a property of the specific straight-line
embedding of G.
Another remarkable connection between planarity, rigidity, and pseudo-triangulations concerns planar rigidity circuits. These are redundantly rigid graphs such that
the removal of any edge leaves a minimally rigid graph. They can, by our results,
be realized as pseudo-triangulations with exactly one non-pointed vertex. Rigidity
circuits, or Laman circuits, have the nice property that the number of faces equals
the number of vertices and that their geometric dual (which exists and is unique) is
also a Laman circuit, see [2]. In [8] we show, using techniques developed here and
Maxwell’s classical theory of reciprocal diagrams [6], that if C is a planar Laman
circuit, then C and its geometric dual C ∗ can be realized as pseudo-triangulations
with the same directions for corresponding edges.
3. Combinatorial pseudo-triangulations in the plane
We now consider a combinatorial analog of pseudo-triangulations. Let G be
a plane graph. We call angles of G the pairs of consecutive edges in the vertex
rotations corresponding to the embedding. Equivalently, an angle is a vertex-face
incidence. By a labelling of angles of G we mean an assignment of “big” or “small”
to every angle of G. Such a labelling is called a combinatorial pseudo-triangulation
labelling (or CPT-labelling, for short) if every bounded face has exactly three angles
labelled “small”, all the angles in the unbounded face are labelled “big”, and no
vertex is incident to more than one “big” angle.
The embedded graph G together with a CPT-labelling of its angles is called a
combinatorial pseudo-triangulation, or CPT. In figures we will indicate the large
angles by an arc near the vertex between the edge pair. Figure 1 shows three
graphs with large angles labelled. The one in the left is not a CPT, because the
exterior face has three small angles. Figure 1b is a CPT, whose bounded faces
a.
b.
c.
Figure 1. A tree with unbounded face a pseudo-triangle (a), the
Mercedes Graph as a CPT (b), and a non-stretchable CPT (c).
are three “triangles” and a hexagonal “pseudo-triangle”. If possible we shall draw
large angles larger than 180◦ , small ones as angles smaller than 180◦ and edges as
straight non-crossing segments, but it has to be observed that this is sometimes not
possible, since there are non-stretchable CPT’s, such as the one in Figure 1c.
Following the terminology of true pseudo-triangulations we say that the interior
faces of a CPT are pseudo-triangles with the three small angles joined by three
pseudo-edges. As in the geometric case, a vertex is called pointed if there is a big
angle incident to it and the CPT is called pointed if this happens at every vertex, see
4
ORDEN, SANTOS, AND SERVATIUS
Figure 1b. The following result and its proof, a straightforward counting argument
using Euler’s formula, are completely analogous to the geometric situation.
Lemma 2. Every combinatorial pseudo-triangulation in the plane with x nonpointed and y pointed vertices has exactly 3x + 2y − 3 edges.
We recall here the main result of [4]:
Theorem 3. Given a plane graph G, the following conditions are equivalent:
(i) G is generically minimally rigid (isostatic),
(ii) G satisfies Laman’s condition,
(iii) G can be labelled as a pointed CPT.
(iv) G can be stretched to a pointed pseudo-triangulation preserving the given
topological embedding.
Throughout the paper, we say that a combinatorial pseudo-triangulation G has
the generalized Laman property or is generalized Laman if every subset of x nonpointed plus y pointed vertices, with x + y ≥ 2, induces a subgraph with at most
3x + 2y − 3 edges. This property is inspired by Theorem 2 and it is crucial to
our proof; see Theorem 4. We call it the generalized Laman property because it
restricts to the Laman condition for the pointed case.
Lemma 3. The generalized Laman property is equivalent to requiring that every
subset of x′ non-pointed plus y ′ pointed vertices of G, with x′ +y ′ ≤ n−2 be incident
to at least 3x′ + 2y ′ edges.
Proof. Using Lemma 2, x plus y vertices satisfy the condition in the definition of
generalized Laman if and only if the cardinalities, x′ and y ′ , of the complementary
sets of vertices satisfy this reformulated one.
Lemma 3 implies that the generalized Laman property forbids vertices of degree 1
and that vertices of degree 2 must be pointed. Moreover, any edge cutset separating
the graph into two components, each containing more than a single vertex, has
cardinality at least 3.
The following is a more detailed formulation of our main result, Theorem 1.
Theorem 4. Given a plane graph G, the following conditions are equivalent:
(i) G is generically rigid,
(ii) G contains a spanning Laman subgraph,
(iii) G can be labelled as a CPT with the generalized Laman property.
(iv) G can be stretched as a pseudo-triangulation (with the given embedding and
outer face).
The equivalence of (i) and (ii) is Laman’s theorem and the fact that (i) and (iii)
follow from (iv) is Theorem 2. We will prove (ii)⇒(iii) (Section 5) and (iii)⇒(iv)
(Section 4).
Note that having the generalized Laman property is not superfluous in the statement, even for pointed CPT’s. Figure 2a shows a combinatorial pointed pseudotriangulation (CPPT) which is not rigid because the innermost three link chain has
a motion or, equivalently, which is not Laman because those four pointed vertices
are incident to only seven edges.
It is also not true that every rigid CPT has the generalized Laman property, as
Figure 2b shows, where the two non-pointed vertices are incident to only five edges.
COMBINATORIAL PSEUDO-TRIANGULATIONS
a.
5
b.
Figure 2. Left, a CPPT which is not a Laman graph. Right, a
rigid CPT which is not generalized Laman.
If we do not require the generalized Laman property, it is easy to show that
every rigid graph possesses a CPT labelling. One can start with a minimally rigid
spanning subgraph, which has a CPPT labelling by Theorem 3, and then insert
edges while only relabelling angles of the subdivided face. For details see [7]. At
each step one pointed vertex must be sacrificed. But it is not obvious how to
preserve the generalized Laman condition in this process, even though one starts
out with a Laman graph. In Section 5 we show that this can be done.
4. Generalized Laman CPT’s can be stretched
Here we prove the implication (iii)⇒(iv) of Theorem 4. Our proof is based on a
partial result contained in Section 5 of [4]. To state that result we need to introduce
the concept of corners of a subgraph.
Let G = (V, E) be a CPT. Since G comes (at least topologically) embedded in
the plane, we have an embedding of every subgraph of G. If H is such a subgraph,
every angle in H is a union of one or more angles of G. Also, H comes with a
well-defined outer face, namely the region containing the outer face of G. We say
that a vertex v of H incident to the outer face is a corner of H if either
(1) v is pointed in G and its big angle is contained in the outer face of H, or
(2) v is non-pointed in G and it has two or more consecutive small angles
contained in the outer face of H.
The following statement is Lemma 15 in [4]:
Lemma 4. Let H ⊂ G be a subgraph of a CPT and suppose that it is connected
and contains all the edges interior to its boundary cycle (that is to say, H is the
graph of a simply connected subcomplex of G).
Let e, x, y and b denote the numbers of edges, non-pointed vertices, pointed
vertices and length of the boundary cycle in S, respectively. Then, the number c1
of corners of the first type (big angles in the outer boundary) of H equals
c1 = e − 3x − 2y + 3 + b.
We say that a plane graph has non-degenerate faces if the edges incident to every
face form a simple closed cycle. The following statement is part of Theorem 7 of [4]:
Theorem 5. For a combinatorial pseudo-triangulation G with non-degenerate faces
the following properties are equivalent:
(i) G can be stretched to become a pseudo-triangulation with the given assignment of angles.
6
ORDEN, SANTOS, AND SERVATIUS
(ii) Every subgraph of G with at least three vertices has at least three corners.
With this, in this section we only need to prove that:
Theorem 6. Let G be a generalized Laman CPT. Then:
1. Faces of G are non-degenerate.
2. Every subgraph H of G on at least 3 vertices has at least 3 corners.
Hence, G can be stretched.
Proof. 1. Every face in a plane graph has a well-defined contour cycle. What we
need to prove is that no edge appears twice in the cycle. For the outer face this is
obvious, since all angles in it are big: a repeated edge in the cycle would produce
two big angles at each of its end-points. Hence, assume that there is a repeated
edge a in the contour cycle of a pseudo-triangle of G. This implies that G \ a has
two components, “one inside the other”. Let us call H the interior component. We
will show that the set of vertices of H violates the generalized Laman property by
Lemma 3.
Indeed, let f and e be the number of bounded faces and edges of H. Let x and
y be the numbers of non-pointed and pointed vertices in it. The number of edges
incident to the component is e + 1 (for the edge a). Hence, the generalized Laman
property says that:
e + 1 ≥ 3x + 2y.
On the other hand, twice the number of edges of H equals the number of angles of
G incident to H minus one (because the removal of the edge a merges two angles
into one). The number of small angles is at least 3f and the number of big angles
is exactly y. Hence,
2e + 1 ≥ 3f + y.
Adding these two equalities we get 3e + 2 ≥ 3f + 3y + 3x, which violates Euler’s
formula e + 1 = f + y + x.
2. Observe first that there is no loss of generality in assuming that H is connected
(if it is not, the statement applies to each connected component and the number
of corners of H is the sum of corners of its components) and that H contains all
the edges of G interior to its boundary cycle (because these edges are irrelevant
to the concept of corner). We claim further that there is no loss of generality in
assuming that the boundary cycle of H is non-degenerate. Indeed, if H has an
edge a that appears twice in its boundary cycle, its removal creates two connected
components H1 and H2 , whose numbers of vertices we denote v1 and v2 . We
claim that each Hi contributes at least min{vi , 2} corners to H. Indeed, if vi is
1 or 2, then all vertices of Hi are corners in H. If vi ≥ 3, then Hi has at least
three corners and all but perhaps one are corners in H. Hence, H has at least
min{v1 , 2} + min{v2 , 2} ≥ min{v1 + v2 , 3} corners, as desired.
Hence, we assume that H consists of a simple closed cycle plus all the edges of G
interior to it.
Let y, x, e, and b be the numbers of pointed vertices, non-pointed vertices,
edges and boundary vertices of H, respectively. Let V be the set of vertices of H
which are either interior to H or boundary vertices, but not corners. V consists
of x + y − c1 − c2 vertices, where c1 and c2 are the corners of type 1 and 2 of H,
respectively. Hence, Lemma 3 implies that the number of edges incident to V is at
least
2(y − c1 ) + 3(x − c2 ) = 3x + 2y − 2c1 − 3c2 .
COMBINATORIAL PSEUDO-TRIANGULATIONS
7
(Remark: V certainly does not contain the corners of the whole CPT, whose number
is at least three. This guarantees that we can apply Lemma 3 to V ).
On the other hand, the edges incident to V are the e − b interior edges of H plus
at most two edges per each boundary non-corner vertex. Hence,
3x + 2y − 2c1 − 3c2 ≤ e − b + 2(b − c1 − c2 ),
or, equivalently,
c2 ≥ 3x + 2y − e − b.
Using Lemma 4 this gives c1 + c2 ≥ 3.
Corollary 2. The following properties are equivalent for a combinatorial pseudotriangulation G:
(i) G can be stretched to become a pseudo-triangulation (with the given assignment of angles).
(ii) G has the generalized Laman property.
(iii) G has non-degenerate faces and every subgraph of G with at least three
vertices has at least three corners.
Proof. The implications (i)⇒(ii), (ii)⇒(iii), and (iii)⇒(i), are, respectively, Theorems 2, 6 and 5.
To these three equivalences, Theorem 7 of [4] adds a fourth one: that a certain
auxiliary graph constructed from G is 3-connected in a directed sense. That property was actually the key to the proof of (iii)⇒(i), in which the stretching of G
is obtained using a directed version of Tutte’s Theorem saying that 3-connected
planar graphs can be embedded with convex faces.
5. Obtaining Generalized Laman CPT-labellings
Theorem 7. The angles of a generically rigid plane graph can be labelled so that
the labelling is a CPT satisfying the generalized Laman condition.
The proof of the theorem proceeds by induction on the number n of vertices. As
base case n = 3 suffices. Consider a generically rigid graph G with more than three
vertices. Since G is generically rigid it contains a Laman spanning subgraph L. Let
v be a vertex of minimal degree in L. Vertices in L have minimum degree at least
two and L has fewer than 2n edges (where n is the number of vertices), so v has
degree two or three:
• If v has degree two in L, then G \ v is generically rigid, because L \ v is
a spanning Laman subgraph of it. By inductive hypothesis, G \ v has a
generalized Laman CPT labelling. Lemma 5 shows that G has one as well.
• If v has degree three in L, then G \ v is either generically rigid (and then we
proceed as in the previous case) or it has one degree of freedom. If the latter
happens there must be two neighbors a and b of v such that if we insert
the edge e = ab into G \ v we get again a generically rigid graph. Since
the plane embedding of G induces a plane embedding of G′ := G \ v ∪ e,
we have by inductive hypothesis a generalized Laman CPT labelling of G′ .
Lemma 6 shows that a and b can be chosen so that there is a CPT labelling
of G′ extending to one of G.
Lemma 5. If G\v is generically rigid, then every generalized Laman CPT labelling
of G \ v extends to one of G.
8
ORDEN, SANTOS, AND SERVATIUS
Proof. Let T be the region of G \ v where v needs to be inserted. This region will
either be a pseudo-triangle (of the CPT labelling of G) or the exterior region of
the embedding. For now, we assume that T is a pseudo-triangle. At the end of
the proof we mention how to proceed in the (easier) case where T is the exterior
region.
Let a, b and c be the three corners of T . There are the following cases:
(a) If there is a pseudo-edge, say ab, containing at least two neighbors of v,
then let a′ and b′ be the neighbors closest to a and b on that pseudo-edge.
We may have a = a′ or b = b′ , but certainly a′ 6= b′ . We separate two
subcases:
(a.1) If all the neighbors of v are on the pseudo-edge ab, then let us label
v as pointed, with the big angle in the face containing c. The vertices
which are not neighbors of v keep the angles they had in G \ v. The
neighbors of v are labelled with both angles small, except that if a 6= a′
(respectively, b 6= b′ ) then a′ (resp. b′ ) gets a big angle in the face
containing a (resp. b). See Figure 3(a.1).
(a.2) If there is a neighbor of v not on ab, let c′ be one of them, chosen
closest to c (it may be c itself). We label v as non-pointed, and the
remaining angles as before except if c′ 6= c, we also put a big angle at
c′ , in the face containing c. See Figure 3(a.2).
(b) If no pseudo-edge contains two neighbors, then v has either two or three
neighbors. The labelling is as in Figure 3(b), with one of the edges from
v removed (and hence v pointed) in the case of only two neighbors. One
of the corners of T may coincide with the corresponding neighbor of v. In
this case that neighbor will get two small angles.
c
c
c
c’
c’
v
v
v
a
a’
b’
b’
(a.1)
a’
b
a
a’
b’
(a.2)
b
a
b
(b)
Figure 3
In all cases, the CPT labelling has the property that every vertex that was nonpointed on G \ v will remain non-pointed in G. This in turn has the following
consequence: let l be the number of neighbors of v that were pointed in G \ v and
are not pointed in G. By the count of edges in a CPT, the degree d of v equals
l + 2 if v is pointed and l + 3 if v is not pointed. This relation follows also from
the above case study. The neighbors of v that keep their status are precisely the
points a′ , b′ and (if it exists) c′ in each case.
This is crucial in order to prove the generalized Laman property, which we now
do. Let S be a subset of vertices of G \ v. Since the subgraph induced by S on G
and G \ v is the same, and since no vertex changed from non-pointed to pointed,
COMBINATORIAL PSEUDO-TRIANGULATIONS
9
the generalized Laman property of S in G \ v implies the same for S in G. But we
also need to check the property for S ∪ v. For this, by Lemma 3 it will be enough if
the number of neighbors of v in S that did not change from pointed to non-pointed
is at most two if v is pointed and at most three if v is non-pointed. This follows
from the above equations d = l + 2 and d = l + 3 respectively.
As promised, we now address the case where T is the exterior region of the embedding of G \ v. This case can actually be considered a special case of (a.1) above,
since the exterior region has only “one pseudo-edge”. And, indeed, a labelling similar to the one shown in Figure 3(a.1) works in this case, where the arc a′ b′ now
should be understood as the segment of the boundary of the exterior region of G \ v
that becomes interior in G.
Lemma 6. Let v be a vertex of degree three in a Laman spanning subgraph of G
and let e = ab be an edge between two neighbors of v such that G′ := G \ v ∪ e is
generically rigid.
(1) Every CPT labelling of G′ extends to one of G.
(2) If a and b are consecutive neighbors of v, then every generalized Laman
CPT labelling of G′ extends to a generalized Laman CPT labelling of G.
(3) If a and b are at “distance two” among neighbors of v, that is, if there is a
vertex w such that a, w and b are consecutive neighbors of v, then either
(i) every generalized Laman CPT labelling of G′ extends to a generalized
Laman CPT labelling of G, or
(ii) there is a generically rigid subgraph H of G \ v containing w and with
v lying inside a bounded region of H.
It should be clarified what we mean by “extends” here. We mean that all the
labels of angles common to G and G′ have the same status. We exclude from this
the angles at the end-points of the edge ab, which may change status, even if these
angles could in principle be considered to survive in G, split into two edges av
and vb.
Proof. 1. As in the previous Lemma, we start with a generalized Laman CPT
labelling of G′ = G \ v ∪ e. Let T1 and T2 be the two pseudo-triangles containing e.
(As in the previous Lemma, the case where one of them, say T1 , is the exterior
region can be treated as if T1 was a pseudo-triangle with all neighbors of v in the
same pseudo-edge). We call ai , bi and ci the vertices of Ti , in such a way that
the pseudo-edge from ai to bi contains a and b, in this order. Clearly, a coincides
with at least one of a1 and a2 , and b with one of b1 and b2 . Figure 4 shows the
four possibilities, modulo exchange of all a’s and b’s or all 1’s and 2’s. In all cases
we have drawn the union as a pseudo-quadrilateral even if in parts (a) and (b) a
pseudo-triangle would in principle be possible. But that would actually imply that
G \ v is generically rigid (because it can be realized as a generalized Laman CPT),
in which case we could use Lemma 5.
We will try to extend the labelling independently in T1 and T2 . We concentrate
on one of them, say T1 . Let a′1 and b′1 be the neighbors of v closest to a1 and b1 ,
respectively, on the pseudo-edge a1 b1 . Observe that a′1 will coincide with a1 if a1
is a neighbor, and will coincide with a if a is the only neighbor on the path aa1 . If
not all the neighbors of v in T1 are on the pseudo-edge a1 b1 , let c′1 be one which
is closest to c1 (possibly c1 itself). We assign labels as follows: All non-neighbors
of v keep their labels. All the angles at v and at neighbors of v are labelled small,
10
ORDEN, SANTOS, AND SERVATIUS
b=b 1=b 2
b2
b=b 1=b 2
b1
b=b1
c1
c1
c2
c2
c1
(a)
c2
c1
c2
a=a 1
a=a 1
a=a 1
a=a 1=a 2
b=b 2
a2
a2
a2
(b)
(c)
(d)
Figure 4
with the following exceptions: If a′1 6= a1 , (respectively, b′1 6= b1 or c′1 6= c1 ) the
angle at a′1 on the edge to a1 is big (respectively, the angle at b′1 on the edge to b1
or the angle at c′1 on the edge to c1 ). Also, if all neighbors are on the pseudo-edge
a1 b1 , then the angle at v on the pseudo-triangle a1 b1 c1 is labelled big. Figure 5
schematically shows the two cases. For future reference, observe that the neighbors
of v whose status does not change are precisely a′1 , b′1 and, unless v gets a big
angle, c′1 .
b1
b’1
b1
b’1
b
c1
c’1
v
a’1
b
v
c1
a
a
a’1
a1
a1
(a)
(b)
Figure 5
This clearly produces a pseudo-triangulation of T1 , and we use the same idea in
T2 . Observe also that we have not put big angles where there were none before.
In particular, no vertex other than perhaps v will receive two big angles, and no
vertex that was not pointed in G′ will be pointed in G.
But v itself may actually get two big angles, one on T1 and one on T2 . This will
happen if all neighbors of v in T1 are on the pseudo-edge a1 b1 and all neighbors in
T2 are on the pseudo-edge a2 b2 . In this case, since v has at least three neighbors,
one of a′1 , a′2 , b′1 or b′2 is different from a and b. Without loss of generality, suppose
that a′2 is different from a. In particular a2 6= a and then a1 = a = a′1 (as in parts
(b), (c) or (d) of Figure 4). In this case a has received a small angle both in T1
and in T2 , but it was originally pointed in G′ . We are then allowed to change the
angle of a in T1 to be big, and that of v in the same pseudo-triangle to be small.
Figure 6 shows the change. This finishes the proof of part 1 of the Lemma.
Before going into the proofs of parts 2 and 3 we make the following observations
about the CPT labelling that we have just constructed:
(a) The number of neighbors of v that do not change status is three if v is
pointed and four if v is non-pointed. This can be proved with a case study
COMBINATORIAL PSEUDO-TRIANGULATIONS
b1
b’1
b1
b’1
b
v
c1
11
b
c1
a= a1
v
a= a1
(a)
(b)
Figure 6
using our explicit way of labelling, but it also follows from global counts of
small angles in the CPT’s G and G′ .
(b) Assume now that G′ has the generalized Laman property, and let us try to
prove the property for G. Every subset S of vertices of G not containing
v satisfies the generalized Laman count in this CPT labelling: indeed, the
subgraph induced by G on S is contained in the one induced by G′ , and
no vertex changed from non-pointed to pointed. Hence, the Laman count
translates from G′ to G.
(c) But if we try to prove the generalized Laman property for S ∪ v, we encounter a problem: Suppose that a and b do not both belong to S, so that
the subgraph induced by G on S is the same as that induced by G′ . Suppose
also that S is tight in G′ , meaning that the subgraph induced has exactly
the number of edges permitted by the generalized Laman count. Then, we
need to prove that if v is pointed (respectively, non-pointed) at most two
(respectively, three) of the neighbors of v in S keep their pointedness status.
But, globally, we know that three (respectively, four) of the neighbors of v
keep their status, so we cannot finish the proof.
However, from this analysis we get very precise information on the cases
where G happens not to have the generalized Laman property. Namely, if
S ∪ v fails to satisfy the generalized Laman count, S has the following four
properties:
• S does not contain both a and b. Otherwise, the deleted edge ab allows
for one extra edge to be inserted and the count is satisfied.
• S is tight in G′ , otherwise again an extra edge is allowed to be inserted.
This implies that if G′ is embedded as a pseudo-triangulation (via
Theorem 6), then the subgraph G′ |S induced by S is itself a pseudotriangulation. Indeed, let eS , xS and yS be the numbers of edges,
non-pointed vertices and pointed vertices of G′ |S . Let x and y be the
numbers of vertices of S which are non-pointed and pointed in G′ , we
have
eS = 3x + 2y − 3 ≥ 3xS + 2yS − 3.
′
Hence, G |S is a pseudo-triangulation by Lemma 1.
• In particular, G′ |S is generically rigid. Since G′ |S does not contain ab,
this implies that S moves rigidly in the 1-degree of freedom (1-dof )
mechanism G \ v.
12
ORDEN, SANTOS, AND SERVATIUS
• S contains all the neighbors of v that did not change their status when
we extended the CPT labelling. In particular, at least one of a or b
did change its status.
2. We now suppose that a and b are consecutive neighbors of v. Suppose that
T1 is the pseudo-triangle in which there is no other neighbor. Certainly, T1 does
not impose any change of status for a or b. So, if a or b change their status, this
must be because of what happens in the pseudo-triangle T2 . There are two cases:
• If our CPT-labelling of T2 only changes the status of one of a or b, we can
restore its status by the same type of change that we used when v got two
big angles: in the pseudo-triangle T1 we change the angle of v from big to
small, and that of the neighbor that changed status in T2 (which, clearly,
had a big angle on T2 and hence a small angle in T1 ), from small to big.
We now have a CPT-labelling where neither a nor b changed status, hence
the generalized Laman property holds.
• If our CPT-labelling of T2 changes the status of both a and b, let a′ and
b′ be the neighbors of v in the pseudo-edge a2 b2 that do not change their
status by the CPT-labelling of T2 . We can try to apply the trick of the
previous case at vertex a and at vertex b. If the first one fails, we have a
CPT-labelling where a, a′ and b′ did not change their status and without
the generalized Laman property. By our final remarks in the proof of part 1,
this implies that a, a′ and b′ move rigidly in the 1-dof mechanism G \ v.
Similarly, if the second fails, b, b′ and a′ move rigidly. Hence, if both fail,
a, b, a′ and b′ move rigidly in G \ v, which contradicts our initial choice of
the edge e = ab.
3. Applying the construction of part 1 to the edge e, it turns out that w must be
one of the points that does not change status (that is, one of the points a′i , b′i or c′i
of Figure 5). Suppose that the resulting CPT is not generalized Laman, for some
generalized Laman CPT of G′ . We want to prove that there is a rigid component
of G \ v that includes w and contains v in the interior of a cycle.
By the remarks at the end of the proof of part 1, if the generalized Laman
property fails in G, then there is an induced subgraph S of G \ v that contains
all the neighbors of v that did not change status (in particular, contains w), and
which is itself a pseudo-triangulation (with respect to the CPT labelling of G′ ).
If this pseudo-triangulation already contains v in the interior of a pseudo-triangle,
then the claim is proved. But it may happen that v is in the exterior of this
sub-pseudo-triangulation (see a schematic picture in Figure 7).
The crucial point now is that in the construction of part 1 the vertices that do not
change status cannot all lie on the same pseudo-edge of the pseudo-quadrilateral of
G \ v containing v. Indeed, one of them is either a′1 or a′2 (this is the point marked
a′ in Figure 7), and lies on one of the two pseudo-edges “on the a side”. Another
one is either b′1 or b′2 (marked b′ in Figure 7), and lies on the two pseudo-edges “on
the b-side”.
So, if v is exterior to S, then the boundary of S contains a concave chain connecting two different pseudo-edges of the pseudo-quadrilateral. Together with the
opposite part of the pseudo-quadrilateral this produces a pseudo-triangle in G \ v
that contains v in its interior and all the vertices that did not change status on
its boundary. This pseudo-triangle, together with everything in its exterior, is a
COMBINATORIAL PSEUDO-TRIANGULATIONS
13
b’
v
b
w
S
a
a’
Figure 7
pseudo-triangulation, hence generically rigid (here, we are assuming that G′ and,
in particular, G \ v = G′ \ e has been stretched, via Theorem 6).
Corollary 3. Let v be a vertex of degree three in a Laman spanning subgraph of G.
Then, there is an edge e = ab between two neighbors of v such that G′ := G \ v ∪ e
is generically rigid and every generalized Laman CPT labelling of G′ extends to a
generalized Laman CPT labelling of G.
Proof. Let w1 , . . . , wk be the cyclic list of neighbors of v in G. If there is an edge
e = wi wi+1 (with indices meant modulo k) between consecutive neighbors of v that
restores rigidity in G \ v ∪ e, then part 2 of the previous Lemma gives the statement
for that e. Hence, in the rest of the proof we suppose that this is not the case.
We now look at edges of the form e = wi−1 wi+1 . At least one of them must
restore rigidity in G \ v ∪ e: if not, let G0 be the graph consisting of the 2k edges
wi wi+1 and wi−1 wi+1 . Our hypothesis is that G \ v ∪ G0 is still a 1-dof mechanism.
But, since G0 is rigid and contains all neighbors of v, this implies that G ∪ G0 ,
hence G, is a 1-dof mechanism too, a contradiction.
Let S be the set of neighbors of v in G with the property that wi−1 wi+1 restores
rigidity. In the rest of the proof we show that there is a wi ∈ S such that e =
wi−1 wi+1 is as claimed in the statement. We argue by contradiction, so assume the
claim is not true. Part 3 of the previous Lemma says that then for each wi ∈ S there
is a generically rigid subgraph of G \ v, let us denote it Hwi , such that wi ∈ Hwi
and v lies in the interior of a bounded face of Hwi . We now claim that the same
holds for every S ′ ⊂ S: there is a generically rigid subgraph HS ′ of G \ v such
that (1) S ′ ⊂ HS ′ and (2) v lies in the interior of a bounded face of HS ′ . Indeed,
after we know this for one-element subsets we just need to show that from HS1 and
HS2 we can construct HS1 ∪S2 . We consider first the union of the two graphs HS1
and HS2 . It clearly contains both S1 and S2 , and v lies in the interior of a face:
the intersection of the faces F1 of HS1 and F2 of HS2 that contain v. The only
problem is that HS1 ∪ HS2 may not be generically rigid. Since HS1 and HS2 clearly
intersect (the boundaries ∂F1 and ∂F2 of F1 and F2 are two cycles around v that
must intersect because they both contain neighbors of v in G), if their union is not
generically rigid then they intersect in a single point. This point must actually be
in the two cycles ∂F1 and ∂F2 . That is, the cycles “are tangent and one is inside
the other”. Since S1 ⊂ ∂F1 and S2 ⊂ ∂F2 consist only of neighbors of v in G, this
implies that one of S1 and S2 (say S1 ) consists of a single point wi , which is the
intersection point. In particular, S1 ⊂ HS2 and we can take HS1 ∪S2 = HS2 .
14
ORDEN, SANTOS, AND SERVATIUS
So, taking S ′ = S, the conclusion is that all the wi ’s such that wi−1 wi+1 restores
rigidity lie in a rigid subgraph of G\v. Our final claim is that under these conditions
all neighbors of v move rigidly in the 1-dof mechanism G \ v, which (as above)
contradicts the fact that G is rigid. Indeed, assume without loss of generality that
the vertex w1 is such that wk w2 restores rigidity. To seek a contradiction, let wi+1
be the first neighbor of v (in the order w1 , . . . , wk ) which does not move rigidly with
w1 and w2 , and let wj+1 be the first neighbor after wi which does not move rigidly
with wi wi+1 . In particular, both wi−1 wi+1 and wj−1 wj+1 restore rigidity and, by
the above conclusion, the three vertices w1 , wi and wj lie in a rigid subgraph. This
subgraph has two vertices in common with both {w1 , . . . , wi } and {wi , . . . , wj },
which lie respectively in two rigid subgraphs by the choice of wi and wj . Hence, all
{w1 , . . . , wj } lies in a rigid subgraph, in contradiction with the choice of wi .
6. Pseudo-triangulations on closed surfaces
This section contains a couple of observations on the concept of pseudo-triangulations (and combinatorial ones) on closed surfaces. We believe it would be interesting
to develop this concept further.
Let G be a graph embedded on some closed surface of genus g. Every closed
surface can be realized as the quotient of the sphere, the Euclidean plane, or the
hyperbolic plane, by a discrete group of isometries, so in each case there is a well
defined notion of distance and angle. Similar to the situation in the plane, a pseudotriangulation of the surface is a graph embedding with geodesic arcs such that
every face has exactly three angles smaller than 180◦ . A combinatorial pseudotriangulation is a (topological) embedding together with an assignment of “big”
and “small” to angles such that every face has exactly three small angles and every
vertex has at most one big angle. One difference with the situation in the plane is
that now there is no “outer” face.
Proposition 1. Let G be embedded on a surface S of genus g. If G possesses a
combinatorial pseudo-triangular assignment then the numbers e, x and y of edges,
non-pointed vertices and pointed vertices satisfy e = 3x+2y−6+6g if S is orientable,
and e = 3x + 2y − 6 + 3g if S is non-orientable.
Proof. The number of angles equals twice the number of edges. There are 2e − y
small angles. The number of small angles equals three times the number of faces,
f , which together with Euler’s formula x + y − e + f = 2 − 2g in the orientable case,
or x + y − e + f = 2 − g in the non-orientable case yields the desired relationship
between x, y and e.
The tree in Figure 1a is an example of a pointed pseudo-triangulation of the
sphere. A triangular prism which can realized as a pointed pseudo-triangulation
in the plane, cannot be realized as a pointed pseudo-triangulation of the sphere
since, by Proposition 1, any CPT labelling of the prism for the sphere must have
three non-pointed vertices. The graph of a cube, which has no CPT labelling for
the plane, since it has too few edges to be rigid, can be realized as a pseudotriangulation of the sphere, with two non-pointed vertices, see Figure 8 in which
the pointed vertices are placed on the equator, and the two non-pointed vertices are
at the poles. In this geometric realization the large angles are exactly 180◦, so that
pseudo-triangles are actual triangles, although the complex is not a triangulation
since it is not regular.
COMBINATORIAL PSEUDO-TRIANGULATIONS
15
Figure 8. A pseudo-triangular embedding of a cube on the sphere.
In Figure 9 we see the well known embedding of the one-skeleton of the octahedron into the torus with all square faces. In Figure 10 we have modified this
Figure 9. The octahedron graph as a square tiling of the torus.
Figure 10. The octahedron graph as a pointed pseudotriangulation of the torus.
construction to embed the octahedron graph as a pointed pseudo-triangulation of
the torus. The octahedron graph, which is overbraced as a framework in the plane,
has no pointed pseudo-triangular embedding in either the plane or the sphere.
16
ORDEN, SANTOS, AND SERVATIUS
Acknowledgements
This research was initiated at the Workshops on Topics in Computational Geometry organized by Ileana Streinu at the Bellairs Research Institute of McGill University in Barbados, Jan. 2002 and 2003, and partially supported by NSF Grant
CCR-0203224. We also thank G. Rote and an anonymous referee for comments
that helped us improve the presentation.
References
[1] B. Chazelle, H. Edelsbrunner, M. Grigni, L. Guibas, J. Erschberger, M. Sharir, Ray
shooting in polygons using geodesic triangulations, Algorithmica 12, no. 1, 54–68, 1994.
[2] P. Christopher, B. Servatius, Construction of self-dual graphs, American Mathematical
Monthly, 99, 2, 153–158, 1992.
[3] J. Graver, B. Servatius and H. Servatius, “Combinatorial Rigidity”, AMS, Graduate
Studies in Mathematics vol.2, 1993.
[4] R. Haas, D. Orden, G. Rote, F. Santos, B. Servatius, H. Servatius, D. Souvaine,
I. Streinu, and W. Whiteley, Planar minimally rigid graphs and pseudo-triangulations,
Comput. Geom., 31, no. 1-2, 31–61, 2005.
[5] G. Laman, On graphs and rigidity of plane skeletal structures, J. Eng. Math., 4, 331-340,
1970.
[6] J. C. Maxwell, On reciprocal figures and diagrams of forces, Phil. Mag. Series 4, 27, 250–
261, 1864.
[7] D. Orden, Two problems in geometric combinatorics: Efficient triangulations of the hypercube; Planar graphs and rigidity, Ph.D. Thesis, Universidad de Cantabria, 2003.
[8] D. Orden, G. Rote, F. Santos, B. Servatius, H. Servatius, W. Whiteley Non-crossing
frameworks with non-crossing reciprocals, Discrete and Computational Geometry (special
issue in honor of Lou Billera), 32:4 (2004), 567–600.
[9] D. Orden and F. Santos, The polytope of non-crossing graphs on a planar point set, Discrete
and Computational Geometry 33:2 (2005), 275–305.
[10] M. Pocchiola and G. Vegter, Topologically sweeping visibility complexes via pseudotriangulations, Discrete and Computational Geometry, 16, 419–453, 1996.
[11] G. Rote, F. Santos, and I. Streinu Expansive motions and the polytope of pointed
pseudo-triangulations, in “Discrete and Computational Geometry – The Goodman-Pollack
Festschrift”, (B. Aronov, S. Basu, J. Pach, M. Sharir, eds.), Algorithms and Combinatorics
vol. 25, Springer Verlag, Berlin, 699–736, 2003.
[12] I. Streinu, A combinatorial approach to planar non-colliding robot arm motion planning,
In “Proc. 41st Ann. Symp. on Found. of Computer Science” (FOCS 2000), Redondo Beach,
CA, 443–453, 2000.
[13] W. Whiteley, Rigidity and Scene Analysis, in “Handbook of Discrete and Computational
Geometry”, (J. E. Goodman, J. O’Rourke, eds.), 893–916, 1997.
Departamento de Matemáticas, Universidad de Alcalá. Facultad de Ciencias, Apdo.
de Correos 20, E-28871 Alcalá de Henares (Madrid), SPAIN
E-mail address: david.orden@uah.es
Departamento de Matemáticas, Estadı́stica y Computación, Universidad de Cantabria.
Facultad de Ciencias, Avenida de los Castros s/n, E-39005 Santander, SPAIN
E-mail address: santosf@unican.es
Dept. of Math. Sci., Worcester Polytechnic Institute, Worcester MA 01609
E-mail address: bservat@wpi.edu, hservat@wpi.edu