#### Arrangements of Pseudocircles and Circles

Discrete Comput Geom
Arrangements of Pseudocircles and Circles
Ross J. Kang 0
Tobias Müller 0
0 R. J. Kang Radboud University Nijmegen , Nijmegen , The Netherlands
An arrangement of pseudocircles is a finite collection of Jordan curves in the plane with the additional properties that (i) every two curves meet in at most two points; and (ii) if two curves meet in a point p, then they cross at p. We say that two arrangements C = (c1, . . . , cn) and D = (d1, . . . , dn) are equivalent if there is a homeomorphism ϕ of the plane onto itself such that ϕ[ci ] = di for all i ∈ {1, . . . , n}. Linhart and Ortner (Beiträge Algebra Geom 46:351-356, 2005) gave an example of an arrangement of five pseudocircles that is not equivalent to an arrangement of circles, and they conjectured that every arrangement of at most four pseudocircles is equivalent to an arrangement of circles. Here we prove their conjecture. We also consider two related recognition problems. The first is the problem of deciding, given a (combinatorial description of a) pseudocircle arrangement, whether it is equivalent to an arrangement of circles. The second is deciding whether it is equivalent to an arrangement of convex pseudocircles. We prove that both problems are NP-hard, answering questions of Bultena et al. (11th Canadian Conference on Computational Geometry, 1998) and of Linhart and Ortner (Geombinatorics 18:66-71, 2008). We also give an example of an arrangement of convex pseudocircles with the property that its intersection graph (i.e. the graph with one vertex for each pseudocircle and an edge between two vertices if and only if the corresponding pseudocircles intersect) cannot be realised as the intersection graph of a family of circles. This disproves a folklore conjecture communicated to us by Pyatkin.
Arrangements; Pseudocircles; Intersection graphs
1 Introduction and Statement of Results
An arrangement of pseudocircles is a finite list C = (c1, . . . , cn) of Jordan curves in
the plane satisfying the following two conditions:
(i) every two curves intersect in at most two points; and
(ii) if two curves meet in a point p, then they cross at p.
We speak of a simple arrangement of pseudocircles if in addition the following holds:
(iii) no three curves intersect in a point.
Naturally, an arrangement of pseudocircles C = (c1, . . . , cn) is called an
arrangement of circles if each ci is a circle. Arrangements of circles and pseudocircles have
been studied previously by several groups of authors including Agarwal et al. [
1
],
Alon et al. [
2
], Linhart and Ortner [
15–17
] and Linhart and Yang [
18
].
We will say that two arrangements C = (c1, . . . , cn ) and D = (d1, . . . , dn) are
equivalent if there exists a homeomorphism ϕ from the plane onto itself with the
property that ϕ[ci ] = di for all i ∈ {1, . . . , n}. We will say that an arrangement of
pseudocircles is circleable if it is equivalent to an arrangement of circles. A natural
question is whether every arrangement of pseudocircles is circleable. As it turns out the
answer to this question is negative: Edelsbrunner and Ramos [
8
] gave an example of an
arrangement of six pseudocircles that is not circleable. Later, Linhart and Ortner [
16
]
showed that the arrangement of five pseudocircles shown in Fig. 1 is not circleable.
They also conjectured this to be the smallest non-circleable pseudocircle
arrangement. Some evidence in favour of this conjecture was reported in the PhD thesis
of Ortner [
22
]. There, a computer enumeration was implemented which verified the
circleability of all simple arrangements on at most four pseudocircles satisfying the
additional condition that every two pseudocircles intersect. Here we confirm the full
conjecture.
Theorem 1.1 Every arrangement of at most four pseudocircles is circleable.
It should be mentioned that Linhart and Ortner used a more restrictive notion of
arrangement of pseudocircles (corresponding to arrangements that we have called
simple above) and a more general notion of equivalence than ours. So Theorem 1.1
not only proves their conjecture, but also strengthens it slightly.
Theorem 1.1 provides a natural analogue of a celebrated result of Goodman and
Pollack [
11
], stating that every arrangement of up to eight pseudolines is equivalent to
an arrangement of lines. Prior to this an example was already known of an arrangement
of nine pseudolines inequivalent to any arrangement of lines.
Given that not all pseudocircle arrangements are circleable, one may wonder how
easy it is to tell whether a pseudocircle arrangement is circleable. One way to frame
this as a mathematically precise question is by asking the complexity of an appropriate
computational recognition problem. Let us define the combinatorial description of a
pseudocircle arrangement as the associated labelled cell-complex—we describe this
precisely in the next section. The next theorem shows that the computational problem of
deciding, given a combinatorial description of a pseudocircle arrangement C, whether
C is equivalent to a circle arrangement is NP-hard, even if we restrict the input to
simple arrangements of convex pseudocircles.
Theorem 1.2 It is NP-hard to decide, given a combinatorial description of a simple
arrangement of convex pseudocircles, whether the arrangement is circleable.
The proof of this theorem is given in Sect. 5.
It is also natural to consider the analogous problem for convex pseudocircle
arrangements. We say that an arrangement of pseudocircles is convexible if it is equivalent to
an arrangement of convex pseudocircles. Bultena et al. [
4
] have asked about the
complexity of the computational problem of deciding, given a combinatorial description
of a pseudocircle arrangement, whether the arrangement is equivalent to an
arrangement of convex pseudocircles. Later Linhart and Ortner [
17
] asked the indeed weaker
question of whether there exists a pseudocircle arrangement that is not convexible.
The next result answers both questions.
Theorem 1.3 It is NP-hard to decide, given a combinatorial description of an
arrangement of pseudocircles, whether the arrangement is convexible.
The proof of this theorem is given in Sect. 6.
If A = ( A1, . . . , An) is a list of sets, then the intersection graph of A is the
graph G = (V , E ) with vertex set V = {1, . . . , n} and an edge i j ∈ E if and only if
Ai ∩ A j = ∅. A folklore conjecture that was communicated to us by Artem Pyatkin [
23
]
states that every intersection graph of an arrangement of convex pseudocircles is also
the intersection graph of a list of circles. (We do not use the word “arrangement” here
because we do not necessarily want to impose the condition (ii) above.) This
conjecture was apparently inspired by the work of Dobrynin and Mel’nikov [
5–7
] on the
chromatic number of “arrangement graphs” of Jordan curves in the plane (i.e. graphs
whose vertices are the intersection points of the curves and whose edges are the curve
segments between these intersection points). To get a feel for the conjecture observe
for instance that, while all the pseudocircles of the arrangement in Fig. 1 are convex
curves and the arrangement is not equivalent to any arrangement of circles, one can
easily construct a family of five circles in the plane with the same intersection graph.
We are however able to produce a counterexample to the folklore conjecture by adding
a “scaffolding” of additional pseudocircles as in Fig. 2.
Theorem 1.4 The intersection graph of the convex pseudocircles in Fig. 2 cannot be
realised as the intersection graph of a list of circles.
2 Preliminaries
We will write [n] := {1, . . . , n}. If p ∈ R2 and r > 0 then B( p, r ) denotes the open
disk of radius r around p.
Throughout the paper we identify the Euclidean plane R2 with C. Often it will
be convenient to work in the extended complex plane C = C ∪ {∞}, the one-point
compactification of C. Recall that the extended complex plane C is homeomorphic to
the sphere S2. (Stereographic projection provides the canonical example of a
homeomorphism between S2 and C if we adopt the convention that the north pole N ∈ S2 is
mapped to ∞ ∈ C.) Working in the extended complex plane has a number of
advantages. There is for instance no “outer face” that needs separate treatment and we can
view lines as circles passing through the point ∞.
Two basic topological facts we shall use frequently in this paper are the Jordan
curve theorem and the Jordan–Schoenflies theorem. The Jordan curve theorem states
that if c is a Jordan curve in the plane (i.e. c is a homeomorphic image of the unit
circle S1), then C \ c has precisely two arcwise connected components, a bounded
one (the “inside”) and an unbounded one (the “outside”). The Jordan–Schoenflies
theorem states that if ϕ : S1 → c is a homeomorphism, then there also exists a
homeomorphism ϕ : C → C that agrees with ϕ on S1. We shall make frequent use of
these basic topological facts without mentioning them explicitly each time.
If C is a circle in the plane with centre p and radius r , then the circle inversion in
C is the map ϕ : C \ { p} → C \ { p} that assigns to a point z = p the unique point
ϕ(z) that satisfies | p − ϕ(z)| = r 2/| p − z| and lies on the “ray” starting from p and
going through z. See Fig. 3 for a depiction. The point ϕ(z) can be written in an explicit
expression as
ϕ(z) =
p · z + r 2 − | p|2 .
z − p
To obtain a map from C to itself, we also define ϕ( p) := ∞ and ϕ(∞) := p. A
circle inversion has the convenient properties that it maps circles not going through
p to circles; it maps circles through p to lines; it maps lines through p to lines;
it maps lines not through p to circles; and it exchanges the inside and outside of
C (in particular, it is the identity on C and swaps p and ∞). Circle inversions are
conformal, meaning that they preserve angles (i.e. if two curves meet at a point p and
make angle α there, then their images also make angle α at ϕ( p)). Note also that ϕ is
a self-homeomorphism of the extended complex plane C. More background on circle
inversions can for instance be found in [
25
].
If C = (c1, . . . , cn) is an arrangement of pseudocircles, then we also call the
intersection points of the curves the vertices of C and we denote the set of vertices by
I (C). A segment of C is the piece of curve connecting two consecutive intersection
points. The faces of C are the connected components of C \ C, each of which is
an open region. There is exactly one unbounded face, also called the outer face. The
unbounded face is homeomorphic to an open disk with its centre removed, while all
bounded faces are homeomorphic to the open disk.
The combinatorial description of a pseudocircle arrangement is the associated
labelled cell-complex. That is, the lists of faces, segments and intersection points
together with a list of the incidences between them and a labelling indicating which is
the infinite face and which segment belongs to which pseudocircle. The
combinatorial description contains all the relevant combinatorial information of a pseudocircle
arrangement, and in particular two pseudocircle arrangements are equivalent if and
only if they have the same combinatorial description. Alternative notions of the
combinatorial description of an arrangement of pseudocircles have been given by Linhart
and Ortner in [
15
] and Goodman and Pollack in [
12
].
Recall that a planar embedding of a planar (multi)graph G is a map A that assigns a
point A(v) ∈ C to each vertex v ∈ V (G), and an arc A(e) ⊆ C to each edge e ∈ E (G)
in such a way that (i) if e ∈ E (G) has endpoints u, v ∈ V (G), then A(u), A(v) are the
endpoints of A(e), and (ii) if e, f ∈ E (G) are edges, then A(e) and A( f ) do not meet
except possibly at common endpoints. We speak of a straight-line embedding if each
arc A(e) is simply the line segment [A(u), A(v)] when e = uv. (Note that straight-line
embeddings only make sense for simple graphs.) We will say that two embeddings
A, B of a planar multigraph G are equivalent (up to the choice of outer face) if there
exists a homeomorphism ϕ : C → C such that ϕ(A(v)) = B(v) for all v ∈ V (G)
and ϕ[A(e)] = B(e) for all e ∈ E (G). (Note that our edges are labelled/named so
that there is no confusion among parallel edges.)
A fact that will turn out to be quite useful to us is Whitney’s unique embeddability
theorem [
26
].
Theorem 2.1 [
26
] If G is a simple, 3-connected planar graph then every two
embeddings of G are equivalent.
Let us remark that the condition that G be simple is essential in the theorem. The
conclusion is clearly false if we allow multiple edges or loops.
The faces of a planar embedding are defined similarly to the faces of an arrangement
of pseudocircles. Every face of a planar embedding is bounded by a closed walk in
G, called a face-walk. Note that a face-walk need not be a cycle. In fact, a connected
planar graph is 2-connected if and only if every face-walk is a cycle (see for instance
Theorem 1.6.1 in [
13
]). It is well known and easy to see that the set of all face-walks
determines the embedding of a connected graph up to equivalence.
Rotation systems provide another way to describe an embedding combinatorially.
Given a multigraph G, a rotation system π assigns to each vertex of v ∈ V (G) a
cyclic permutation π(v) of the edges that are incident with v. (Recall that a cyclic
permutation of a set S = {s1, . . . , sn} is one that can be written in cycle notation as
(si1 . . . sin ), i.e. there is a single cycle containing each element. Put differently, the
orbit of each element is the entire set S.) If A is an embedding of a connected planar
(multi)graph G, then the corresponding rotation system πA assigns to each vertex
v ∈ V (G) the cyclic permutation corresponding to the order in which we encounter
the edges incident with v if we go around the vertex in clockwise fashion. Observe
that if we produce a new embedding B by applying a reflection in a line to A then
πB(v) will be exactly the opposite order of πA(v) for every vertex v ∈ V (G). This
motivates the next definition. We will say that two rotation systems π, τ defined on a
planar multigraph G are equivalent if either π(v) = τ (v) for every vertex v ∈ V (G)
or if π(v) is the reverse order of τ (v) for every v ∈ V (G).
From the rotation system πA, we can retrieve the set of all face-walks on A. (Starting
from some vertex, exit along some edge, and keep going “immediately clockwise”
until traversal of the initial edge in the same direction. Then the edges traversed form
a face-walk. Repeating the procedure for each vertex and edge clearly produces the
set of all face-walks corresponding to the embedding.) Conversely, suppose we are
given the set of face-walks of some embedding. Let us take two edges e, f sharing an
endpoint and lying on a common face F of length at least three. If we declare that e
is the successor of f on the “clockwise face-walk around F ”, then this determines all
the cyclic orders π(v) of a rotation system. And, if we had taken f as the successor of
e on the clockwise walk around F , then we would get the reverse order everywhere.
Since the face-walks determine the embedding up to equivalence as mentioned earlier,
we have the following.
Lemma 2.2 If G is a connected, planar graph and A, B are two embeddings of G,
then A is equivalent to B if and only if the rotation systems πA, πB they define are
equivalent.
A line arrangement is a system L := ( 1, . . . , n) of lines in the plane. If every
two lines intersect and no point lies on more than two lines, then we call L a
simple line arrangement. A pseudoline is the image of a line under a homeomorphism
of the plane. A pseudoline arrangement is a system of pseudolines satisfying the
requirements that every two pseudolines intersect in at most one point, and that when
two pseudolines meet in a point they must cross at that point. A simple pseudoline
arrangement furthermore satisfies the requirement that every two pseudolines
intersect and no point lies on more than two pseudolines. We will say that two pseudoline
arrangements L = ( 1, . . . , n) and M = (m1, . . . , mn) are equivalent if there exists
a homeomorphism ϕ from the plane onto itself with the property that ϕ[ i ] = mi
for all i ∈ {1, . . . , n}. The faces, segments and intersection points and combinatorial
description of a pseudoline arrangement are defined analogously as for pseudocircle
arrangements. Again, it can be seen that two pseudoline arrangements are equivalent
if and only if their combinatorial descriptions coincide. We should mention that
various alternative notions of a combinatorial description of a pseudoline arrangement are
available in the literature such as local sequences, allowable sequences and oriented
matroids of rank 3 (see for instance [
9
]).
A useful way to picture a pseudoline arrangement is as a wiring diagram. In a wiring
diagram, each pseudoline is a finite union of line segments. Each pseudoline starts off
and ends with half-infinite horizontal segments, all other segments are of finite length
having slope either 0, 1 or −1, and the endpoints of these line segments all lie on
the integer grid Z2. See Fig. 4 for an example of a wiring diagram. It can be shown
that every pseudoline arrangement is equivalent to (moreover, can be “continuously
deformed” into) a wiring diagram [
10
].
If a pseudoline arrangement is equivalent to a line arrangement, then we say it is
stretchable. The name “stretchability” of course comes from imagining the
pseudolines “stretching” into lines. STRETCHABILITY is the computational problem of
deciding, given a combinatorial description of a pseudoline arrangement as input,
whether it belongs to a stretchable arrangement. Analogously, SIMPLE
STRETCHABILITY is the same computational problem when the input is restricted to
combinatorial descriptions of simple pseudoline arrangements. Our proofs of Theorems 1.2
and 1.3 rely heavily on the following classical result.
Theorem 2.3 (Shor [
24
]) SIMPLE STRETCHABILITY is NP-hard.
This theorem is also a straightforward corollary of a deep topological result by
Mnëv [
20,21
]. Shor’s proof is more direct and uses Pappus’ and Desargues’ theorems
to encode instances of a SAT variant as instances of the simple stretchability problem.
(See also Chap. 8 of [3].)
3 Small Pseudocircle Arrangements are Circleable
Let us say that an arrangement of pseudocircles C = (c1, . . . , cn ) is circleable in the
extended complex plane if there exists an arrangement of circles D = (d1, . . . , dn )
and a homeomorphism ϕ of C onto itself such that ϕ[ci ] = di . When dealing with an
arrangement of pseudocircles in the plane there is always an outer face with a different
topology from the other faces (i.e. the outer face is homeomorphic to a “punctured
disk” while the other faces are homeomorphic to disks), but in the extended complex
plane all faces behave the same. This might lead us to suspect that circleability in the
plane and circleability in the extended complex plane are two different notions. On
the contrary (and fortunately for us) circle inversions provide a short argument that
the two notions coincide.
Lemma 3.1 A pseudocircle arrangement is circleable in the plane if and only if it is
circleable in the extended complex plane.
Proof Any self-homeomorphism of the plane also induces a self-homeomorphism of
the extended complex plane—we just send ∞ to itself. In other words, if an
arrangement is circleable in the ordinary plane it is also circleable in the extended complex
plane.
To see the converse, suppose that C = (c1, . . . , cn ) is circleable in the extended
complex plane, and let D = (d1, . . . , dn ) be an arrangement of circles such that there
is a self-homeomorphism ϕ : C → C with ϕ[ci ] = di for all i ∈ {1, . . . , n}. If it
so happens that ϕ maps ∞ to itself, then we are done: the restriction of ϕ to C is a
self-homeomorphism of C demonstrating that C and D are equivalent.
So suppose ϕ(∞) = p ∈ C. Observe that p does not lie on any of the circles
d1, . . . , dn . Let σ : C → C be the circle inversion in a circle C of centre p and
radius one, say (in fact any radius will do). Then σ maps each circle di to some other
circle ei and it is a self-homeomorphism of the extended complex plane. Observe that
(σ ◦ ϕ)[ci ] = ei for each i ∈ {1, . . . , n}, that σ ◦ ϕ is a self-homeomorphism of C
and that (σ ◦ ϕ)(∞) = ∞. These three properties show that the restriction of σ ◦ ϕ to
C demonstrates that C is circleable in the (ordinary) plane.
For the rest of this section, all action will take place in the extended complex plane.
We will often find it useful to apply a suitable circle inversion so that one or two
circles of an arrangement are mapped to lines. This makes the ensuing case analysis
conceptually simpler. We may always apply another circle inversion (in a circle whose
centre does not lie on any pseudocircle) to convert an arrangement C consisting of lines
and circles into an equivalent arrangement consisting only of circles. We shall often
be informal by referring to pictures and stating that two depicted arrangements are
equivalent. It will always be easy to see that the “combinatorial structures” agree,
keeping in mind that all lines meet in the point at ∞. (It may also be helpful to
think of arrangements as drawn on the sphere S2 by taking the inverse stereographic
projection and adding the north pole to pre-images of the lines.) Readers can check that
equivalences can always be formally justified by repeated invocations of the Jordan–
Schoenflies theorem. Before we proceed with the proof of Theorem 1.1, let us make
one more fairly straightforward observation.
Lemma 3.2 If C = (c1, . . . , cn ) is such that (c1, . . . , cn−1) is circleable and cn
intersects at most one of c1, . . . , cn−1, then C is circleable.
Proof We can assume that c1, . . . , cn−1 are all circles. Let us furthermore assume that
cn intersects precisely one other pseudocircle. (The case when cn intersects no other
pseudocircle is similar and easier.) We can suppose without loss of generality that the
unique pseudocircle that cn intersects is cn−1. By applying a suitable circle inversion
(in a circle whose centre lies on cn−1 but not on any other ci ) we obtain the situation
where c1, . . . , cn−2 are all circles and cn−1 is a line. By applying a suitable isometry if
needed, we can further assume that the origin lies inside cn and that cn−1 is the x -axis
(a.k.a. the real line). See Fig. 5, left, for a depiction.
Let A ⊆ [n − 2] be such that ci lies inside cn for i ∈ A and ci lies outside cn when
i ∈ A. If we apply a dilation by a small enough λ1 > 0 to all ci : i ∈ A and a dilation
by a large enough λ2 > 0 to all ci : i ∈ A, then we can ensure that ci is contained in
a disk of radius 1/2 around the origin for i ∈ A, while ci does not intersect the disk
of radius 2 for i ∈ [n − 2] \ A. (We leave cn−1 and cn intact.) This way we have an
arrangement that is equivalent to the original such that c1, . . . , cn−2 are circles and
cn−1 is a line. See Fig. 5, middle, for a depiction. It is now clear that we can replace cn
by the unit circle and still have an equivalent arrangement (see Fig. 5, right). Applying
a circle inversion in a circle with centre not on any ci gives the sought arrangement of
circles and concludes the proof.
Proof of Theorem 1.1 We first prove the theorem for simple pseudocircle
arrangements, and later extend it to general pseudocircle arrangements.
Let C = (c1, . . . , cn ) be a simple arrangement of pseudocircles on the extended
complex plane with n ≤ 4. By Lemma 3.2, C is circleable if n ≤ 2. Let us now
consider the case n = 3. We may assume c1 and c2 meet, because otherwise each of
them intersects at most one other pseudocircle, and we are done by Lemma 3.2. By
the above, we can apply a homeomorphism of the extended complex plane, to arrive
at an arrangement equivalent to C where both c1, c2 are circles (and c3 is a general
Jordan curve). By applying an inversion in a circle with centre one of the intersection
points of c1, c2, we arrive at the situation where both c1, c2 are lines. Applying another
homeomorphism if necessary, we arrive at the situation where c1 coincides with the
x -axis and c2 coincides with the y-axis (and c3 is still a general Jordan curve). If c3
has the origin in its interior, then c3 must intersect both the positive and the negative
x -axis and both the positive and the negative y-axis. (Recall that two pseudocircles
intersect either never or twice.) Clearly, the arrangement must then be equivalent to
that in Fig. 6, left.
Otherwise, c3 does not contain the origin and it intersects either the positive x -axis
twice or the negative x -axis twice, and similarly for the y-axis. We can assume without
loss of generality that c3 intersects the positive x -axis and the positive y-axis twice.
But now C is equivalent to the arrangement in Fig. 6, right.
This proves the statement for simple pseudocircle arrangements on at most three
curves. Let us thus consider the case when n = 4. Again by Lemma 3.2, we can
assume that each pseudocircle intersects at least two others.
Our first main case in considering simple C is that each pseudocircle intersects
exactly two others. Without loss of generality we can assume c1 does not intersect c3
and c2 does not intersect c4. By the above, (c1, c2, c3) is circleable. Arguing as before,
we can apply a self-homeomorphism of C that sends c1 to the x -axis, c2 to the y-axis
and c3 to a circle. Hence up to symmetries, the arrangement now looks as in Fig. 7,
left; and it is clearly circleable (see Fig. 7, right).
Our second main case is that at least one pseudocircle intersects all three others,
while some pseudocircle intersects only two others. Note that there are exactly two
pseudocircles intersecting all others and two intersecting exactly two others. Without
loss of generality, we can assume c3, c4 do not intersect, while c1, c2 intersect each
other as well as both of c3, c4. Let us first assume that c4 separates the two intersection
points of c1, c2 (i.e. one of the two intersection points lies inside c4 and the other
outside c4). By the previous cases, we can assume c1, c2, c3 are circles. Applying a
circle inversion if needed, we can further assume that c3 is not contained inside c4.
We can now replace c4 by a small circle c4 that lies within c4 and contains the relevant
intersection point of c1, c2, to obtain an arrangement of circles equivalent to C (see
Fig. 8).
By symmetry, we are also done if c3 separates the two intersection points of c1, c2.
Let us therefore consider the case that neither c3 nor c4 separate the two intersection
points of c1, c2. Note that the pseudocircle arrangement (c1, c2, c3) must necessarily
be equivalent (in the extended plane) to the arrangement on the right of Fig. 6. By
applying a suitable circle inversion, we see that this arrangement is equivalent to the
one shown in Fig. 9, left. (We have refrained from drawing it as a circle arrangement
in this figure to avoid the figure getting too “cramped”.)
First we suppose that c4 intersects some segment of this arrangement exactly once.
Then this segment must be one of the segments labelled a, a , b, b , c, c in Fig. 9,
left, because if c4 hits one of the other segments exactly once, then it will separate
the intersection points of c1, c2 or c1, c3 or c2, c3 (here we use that if c4 crosses some
segment exactly once then the two endpoints of the segment are in different components
of C \ c4). Hence, if c4 intersects a segment other a, a , b, b , c, c exactly once, then
we will be done either by a previous case or by contradiction—if c4 separates the
intersection points of c1, c3 or c2, c3, then it intersects c3. Let F be the 4-face whose
boundary contains a, a . Since c4 intersects the boundary of F an even number of
times, c4 intersects a exactly once if and only if it intersects a exactly once. Similarly
c4 intersects b exactly once if and only if it intersects b exactly once; and c4 intersects
c exactly once if and only if it intersects c exactly once. By considering the triangular
faces abc and a b c , we see that, similarly, if c4 intersects one of a, b, c exactly once,
then it intersects precisely two of them exactly once. And, if c4 intersects one of
a , b , c exactly once, then it intersects precisely two of them exactly once. It follows
that, if c4 intersects some segment exactly once, then the situation must be as in Fig. 9,
middle. It is then easily seen that the arrangement C is equivalent to an arrangement
of circles (see Fig. 9, right).
We can thus assume that c4 intersects every segment of the arrangement (c1, c2, c3)
an even number of times. (Yet c4 does not intersect c3, and (c1, c2, c3) is equivalent to
Fig. 9, left.) Note that, since c4 does not intersect c3 and it intersects both c1 and c2,
we must have that it in fact intersects exactly two segments. There must be a face F
of the arrangement (c1, c2, c3) such that c4 intersects two of its bounding segments.
If these segments are consecutive on F , then we can replace c4 by a sufficiently small
circle, close to the point where the segments meet, similar to what we did in Fig. 8.
If c4 intersects two segments of F that are not consecutive, then F must be a 4-face,
and the situation must be as in Fig. 10, left. It is easily seen that in this case we can
find an equivalent arrangement of circles (see Fig. 10, right).
This concludes the analysis of the second main case. Our last main case for
considering when C is simple is that every pair of pseudocircles intersect. We first assume
that (c1, c2, c3) is as in Fig. 6, left. If c4 surrounds the origin (i.e. it separates the origin
from the point ∞), then it must intersect both the positive and the negative x -axis, and
both the positive and the negative y-axis. Figure 11 shows the possible positions of
these intersection points with respect to c3 (up to obvious symmetries).
In the first case, c4 must intersect some segment of c3 twice. By symmetry we can
assume it is the north-east segment. In the second case, the position of the intersection
points with c1, c2 forces c4 to intersect c3 in its north-west and north-east segments.
Since c4 can intersect c3 at most twice, it is easily seen that the third option in Fig. 11 in
fact cannot occur. In the fourth case, c4 necessarily intersects c3 in its north-west and
south-east segments. Observe that in the fifth case, if we apply a circle inversion in a
circle with center the origin then we arrive at a situation equivalent to the second case
(up to relabelling). Similarly, the sixth case is equivalent to the first case. Figure 12
shows that in the first, second and fourth cases the arrangement C is circleable.
Let us thus assume that (c1, c2, c3) is as in Fig. 6, left, c4 intersects all three other
pseudocircles, and it does not separate the two intersection points of c1, c2 (which is
the same as separating the origin from ∞ in the arrangement of Fig. 6, left). Let us
observe that if c4 intersects some segment of the arrangement (c1, c2, c3) exactly once,
then it separates the two intersection points of a pair of pseudocircles. (Note that the
arrangement (c1, c2, c3) is in fact extremely symmetric.) Hence, if c4 intersects some
segment of the arrangement (c1, c2, c3) exactly once, then we can relabel pseudocircles
and apply a previous case. We can therefore assume c4 intersects each segment either
twice or not at all. There must be some face F such that two of its sides are intersected
by c4. Now c4 either intersects the third side of this face, or it intersects a segment of
another face (which must share a side with F ). Hence, up to symmetry, the situation
must be one of those depicted in Fig. 13.
Figure 14 shows that in each of these two cases the arrangement is circleable.
It thus remains to consider the case when (c1, c2, c3) is as in Fig. 6, right. If c4
separates the intersection points of c1, c2, then we see that (c1, c2, c4) is as in Fig. 6,
left. We can then relabel the arrangement and apply a previous case. So we can assume
this is not the case. Similarly, we can assume c4 does not separate the intersection points
of c1, c3 or of c2, c3. Note that this in particular implies that each segment bounding
a 2-face is intersected an even number of times by c4. Hence, if c4 intersects some
segment exactly once, then it is one of the segments a, a , b, b , c or c shown in Fig. 9,
left. Moreover, by earlier arguments, if c4 intersects such a segment exactly once, then
there are precisely four segments that c4 intersects exactly once; and up to relabelling
these segments are a, a , b, b in Fig. 9. Hence we must be in one of the situations in
Fig. 15.
By obvious symmetries, the second and fourth cases in Fig. 15 are the same.
Similarly, the first, third and fifth cases are also identical (by applying a suitable inversion
to switch the inner triangular face and outer triangular face to see the third case is
equivalent to the first case). It is thus again easy to see that in both cases there is an
equivalent arrangement of circles (see Fig. 16).
Hence we can assume that c4 intersects each segment of (c1, c2, c3) either zero or
two times (and (c1, c2, c3) is as in Fig. 6, right, and every two pseudocircles intersect).
There must be some face F such that c4 intersects two of its sides. If F is a 2-face
then, up to symmetry, the situation must be as in Fig. 17, left.
If we perform a circle inversion in the point p labelled in the figure, then we get
a situation as in Fig. 17, middle, and this is again easily seen to be circleable (see
Fig. 17, right).
Let us now suppose F is a 3-face. If c4 intersects all three sides of F , then the
situation must be as in Fig. 18, left. (Recall that by a suitable circle inversion we can
make any face the outer face.)
It is again easy to see C is circleable (see Fig. 18, right). If, on the other hand, F is
a 3-face and c4 intersects exactly two of its sides, then up to symmetry the situation is
as in Fig. 19, left.
If we perform a circle inversion in a circle with centre the point labelled p in Fig. 19,
left, then we obtain a situation as in Fig. 19, middle, and it again easy to see that C is
circleable (see Fig. 19, right).
Let us then assume that F is a 4-face. If c4 intersects the sides of a 2-face or 3-face
twice, then we are done. Hence it remains to consider the case when c4 intersects three
sides of F . In that case, up to symmetry the situation is as in Fig. 20, left, and the
arrangement is easily seen to be circleable (see Fig. 20, right).
This concludes our case analysis for the case when C is simple. It remains to
consider the case that C is not simple. In this case, there must be a point p common to
at least three pseudocircles. Let us first assume that p lies on all of the pseudocircles.
If we perform an inversion in a circle with centre p, then we in fact get a pseudoline
arrangement L on at most four lines. By the result of Goodman and Pollack [
11
], L
is equivalent to a line arrangement L . Performing an inversion in a circle with centre
not on any line of L yields a circle arrangement C that is equivalent to C.
Let us thus assume that n = 4, that there is a point p common to three pseudocircles,
but that there is no point common to all four pseudocircles. We can assume without loss
of generality that p lies on c1, c2, c3. Applying a circle inversion to an arrangement
in which c1, c2, c3 are circles and c4 is a general Jordan curve (such an arrangement
exists by the last argument), we obtain a situation where c1, c2, c3 are lines and c4 is
a Jordan curve.
Let us first assume that c1, c2, c3 have another point q in common, which implies
that the corresponding line arrangement is the unique non-simple line arrangement of
three lines. Then, depending on whether c4 separates p and q, whether it intersects
only two other pseudolines, or whether it does not separate p, q but intersects the
other three pseudolines, the arrangement must be equivalent (up to relabelling and
symmetries) to one of the arrangements in Fig. 21, and hence is circeable.
Let us therefore assume that p is the only point common to c1, c2, c3, (so that the
corresponding line arrangement is the unique simple line arrangement on three lines).
If furthermore c4 intersects only two other curves, then the arrangement must clearly
be equivalent (up to relabeling and symmetries) to one of the arrangements in Fig. 22.
If p is the only point common to c1, c2, c3 and c4 has at least one of the three
intersection points of the lines in its interior, then the arrangement must be equivalent
(up to relabeling and symmetries) to one of the arrangements in Fig. 23, and hence it
is clearly circleable.
Finally, let us assume p is the only point common to c1, c2, c3, that c4 intersects all
three other pseudolines and that c4 does not have any of the three intersection points
of the corresponding lines in its interior. Then, up to relabelling and symmetries, the
situation must be as in one of the four cases in the top row of Fig. 24. Hence the
arrangement is circleable as well (Fig. 24, bottom row).
This concludes our (lengthy) case analysis, and proves that every pseudocircle
arrangement on at most four pseudocircles is circleable.
4 A Counterexample to the Folklore Conjecture
Before starting with the proof of Theorem 1.4, let us clarify the construction of the
arrangement C = (c1, . . . , c346) depicted in Fig. 2. We start by taking the five curves
c1, . . . , c5 from Fig. 1 and overlay the curves of the arrangement H = (c6, . . . , c346)
shown in Fig. 25, left. The curves from H are placed in such a way that each of the initial
five curves does not intersect the unbounded face of H, and for each of the 8-faces of
H, the corresponding eight curves intersect zero, one or two of c1, . . . , c5 in one of
the ways drawn in Fig. 26. This concludes our description of the construction of C.
The proof of Theorem 1.4 amounts to showing, under the assumption that C can be
realised as the intersection graph of a list of circles, that the pseudocircle arrangement
of Fig. 1 is circleable, a contradiction. The proof also relies on the following
observation, the proof of which is a straightforward perturbation argument. Nevertheless, we
include this argument for completeness.
Lemma 4.1 Every graph that is an intersection graph of a list of circles is also the
intersection graph of a simple arrangement of circles.
Proof For notational convenience, let us denote by S( p, r ) the circle with centre p
and radius r . We will make use of the following three straightforward observations.
(i) If |S( p, r )∩ S(q, s)| = 0, then there exists ε > 0 such that |S( p , r )∩ S(q, s)| = 0
for all r ∈ (r − ε, r + ε) and p ∈ B( p, ε).
(ii) If |S( p, r )∩ S(q, s)| = 2, then there exists ε > 0 such that |S( p , r )∩ S(q, s)| = 2
for all r ∈ (r − ε, r + ε) and p ∈ B( p, ε).
(iii) If |S( p, r ) ∩ S(q, s)| = 1 and r ≤ s, then there exists ε > 0 such that |S( p, r ) ∩
S(q, s)| = 2 for all r ∈ (r, r + ε).
Let us fix a graph G that is an intersection graph of a list of circles. For every list
D = (d1, . . . , dn ) of circles that has G as its intersection graph, let us say that index i
is good if di is not tangent to any other circle of D, and no point of di lies on three or
more circles of D. If i is not good, then we call it bad. Let bad(D) denote the number
of bad indices in D.
Now let D = (d1, . . . , dn ) be an arbitrary list of circles with intersection graph G,
and suppose bad(D) > 0. Let pi be the centre of di and ri its radius. For now, let
us assume that d1, . . . , dn are all distinct (i.e. ( pi , ri ) = ( p j , r j ) if i = j ). We can
assume without loss of generality that the labelling is such that r1 ≤ r2 ≤ . . . ≤ rn.
Let i ∈ {1, . . . , n} be the maximal index such that each of d1, . . . , di are good. (So
in particular di+1 is bad.)
By observations (i–iii) above, there exists ε > 0 such that S( pi+1, r )
intersects precisely the same subset of d1, . . . , di , di+2, . . . , dn as di+1 does for all
r ∈ [ri+1, ri+1 + ε). We can therefore also choose r ∈ [ri+1, ri+1 + ε) such that
the circle di+1 := S( pi+1, r ) intersects the same set of circles as di+1, is not tangent
to any of the other circles, and does not pass through the intersection points of any pair
of the other circles. Hence the list D := (d1, . . . , di , di+1, di+2, . . . , dn ) of circles
has intersection graph G and bad(D ) < bad(D).
Clearly, after finitely many iterations of this procedure, we arrive at a list D of
circles with intersection graph G and bad(D ) = 0. In other words, D is the sought
arrangement of circles with intersection graph G.
This proves the lemma in the case when all circles in the initial list are distinct.
Otherwise, we first remove duplicates, apply the above argument to the thinned list,
and then use (i) and (ii) to replace the duplicates, but each with a centre p that is
distinct from the centres of all the other circles.
We are now ready for the proof of Theorem 1.4.
Proof of Theorem 1.4 Let us denote the intersection graph of C by G, with vertex
set {1, . . . , 346}. Let H be the subgraph of G corresponding to H. (Observe that H
can also be obtained by taking the 11 × 11-grid and subdividing each edge. A planar
embedding of H is given in Fig. 25, right.)
Aiming for a contradiction, suppose that D = (d1, . . . , d346) is a list of circles
whose intersection graph is G. By Lemma 4.1, we may assume without loss of
generality that D is also a simple arrangement of circles. For notational convenience, let pi
denote the centre of di and let ri denote its radius, and let us set H := (d6, . . . , d346).
Observe that H has the following property:
• for every three distinct vertices u, v, w of H , there is a uv-path P in H \ w such
that no internal vertex of P is adjacent to w.
This implies that there cannot be three distinct indices i, j, k ∈ {6, . . . , 346} such
that d j is in the interior of di and dk is outside of di . (Otherwise, if i1 . . . im is a j k-path
in H that avoids i , then one of the circles di1 , . . . , dim will intersect di , a contradiction.)
Thus, if there is at least one circle of H in the interior of circle di , then every other
circle of H either intersects di or is in its interior. It follows that, by applying a suitable
circle inversion if needed, we can assume that no di has another d j in its interior (with
i, j ∈ {6, . . . , 346}).
For i ∈ {6, . . . , 346}, let Di := B( pi , ri ) be the (open) disk with centre pi and
radius ri (i.e. the boundary of Di is di ). Observe that
Di ∩ D j = ∅ if and only if di ∩ d j = ∅ for all pairs i, j ∈ {6, . . . 346}.
In other words, we may consider H as a realisation of H as an intersection graph
of disks.
Since H is triangle-free and has minimum degree two, the points pi together with
the line segments [ pi , p j ] for i j ∈ E (H ) constitute a straight-line embedding of
H (see for instance Lemma 5.2 of [
19
]). Observe that, as H is a subdivision of a
3connected planar graph, it also follows from Whitney’s unique embeddability theorem
that every two embeddings of H are equivalent.
In fact, as we will now see, something even stronger is true.
Claim 4.2 The arrangements of pseudocircles H and H are equivalent in the
extended complex plane.
Proof of Claim 4.2 Consider the rotation system corresponding to the straight-line
embedding of H we have obtained. By Lemma 2.2, we know that it is equivalent to
the rotation system provided by the embedding in Fig. 25, right. Applying a reflection
if needed, we can assume that the two rotation systems are in fact identical. (And
clearly D still has all the properties we assumed and derived until now.)
Let us now fix i ∈ {6, . . . , 346} and suppose that i is adjacent to four neighbours
in H . (In other words, ci is one of the square boxes not touching the outer face of
H.) Let i1, i2, i3, i4 denote the four neighbours of i . Without loss of generality, they
appear in this order in the cyclic permutation of i . Since the rotation systems agree, di
intersects each of di1 , . . . , di4 twice, and H is triangle free, we see that there is also a
straightforward correspondence between the intersection points and segments in our
1 2
arrangement H and those in H. See Fig. 27. (For j ∈ {1, 2, 3, 4}, let p(i,i j ), p(i,i j ) be
the two intersection points of ci and ci j , and let q(1i,i j ), q(i,i j ) be the intersection points
2
1
of di and di j . Without loss of generality, the labelling is such that p(i,i j ) comes before
2 1 2
p(i,i j ) if we go around ci in the clockwise direction; and similarly for q(i,i j ), q(i,i j ) and
the clockwise order on di . Now observe that the clockwise order of the intersection
1 2 1 2 1 2 1 2
points on ci is p(i,i1), p(i,i1), p(i,i2), p(i,i2), p(i,i3), p(i,i3), p(i,i4), p(i,i4) and on di the
clockwise order is q(1i,i1), q(2i,i1), q(1i,i2), q(2i,i2), q(1i,i3), q(2i,i3), q(1i,i4), q(2i,i4).) A similar
statement holds if ci is one of the square boxes on the outer face, so we can indeed
label the intersection points of H and H in a consistent way. We can then also label
the segments of H and H such that a segment of H and a segment of H are labelled
the same if and only if their endpoints correspond and they are part of corresponding
pseudocircles.
It is easy to see that in this way we have obtained two embeddings of the same planar
graph (the “arrangement graph” of H, whose vertices are the intersection points—not
Fig. 28 The segments
a, a , b, b
the intersection graph of the pseudocircles) and both embeddings define the same
rotation system. It follows by Lemma 2.2 that there is a self-homeomorphism ϕ :
C → C such that ϕ[di ] = ci for all i ∈ {6, . . . , 346}. This proves the claim.
Let ϕ : C → C be the self-homeomorphism with ϕ[di ] = ci for all i ∈ {6, . . . , 346}
provided by the last claim. Let us set ci := ϕ[di ] for i ∈ {1, . . . , 5}. Then C :=
(c1, . . . , c5, c6, . . . , c346) is an arrangement of pseudocircles equivalent to D (in the
extended complex plane), and
ci ∩ c j = ∅ if and only if ci ∩ c j = ∅ (for all i ∈ {1, . . . , 5}, j ∈ {1, . . . 346}).
(1)
Our next aim in the proof will be to show that, for every i ∈ {1, . . . , 5}, the curve
ci intersects precisely the same segments and faces of the arrangement H as ci . From
this we will then be able to derive that the arrangements (c1, . . . , c5) and (c1, . . . , c5)
are equivalent, which will give us a contradiction since (c1, . . . , c5) is not circleable
by the result of Linhart and Ortner.
Let us thus pick i ∈ {1, . . . , 5} and j ∈ {6, . . . , 346} such that ci ∩ c j = ∅. We first
suppose that c j is one of the non-square rectangles of H. Let j1, j2 ∈ {6, . . . , 346}
be the two neighbours of j in H . Let b be the segment of c j inside c j1 , let b be the
segment of c j inside c j2 , and let a, a be the two other segments (see Fig. 28). By the
construction of C, ci intersects both of the segments a, a exactly once and it does not
intersect b, b (see Fig. 26). We aim to show the same is true for ci .
Let us first note that ci cannot intersect b, since it does not intersect c j1 but it does
intersect some ck lying outside c j1 . Similarly, ci does not intersect b . Since C is
an arrangement of pseudocircles, ci intersects c j at exactly two points. Aiming for a
contradiction, let us assume that both intersection points lie on a (and therefore not on
a ). Let us consider the 8-face F of H whose boundary contains a. Let us first assume
that ci intersects them as in the second (from left) situation in Fig. 26. Let γ be the curve
consisting of the segments bounding F , and let γ be the curve bounding the union of F
with the eight rectangles touching F . Let f, f be the two other segments of these eight
rectangles that ci intersects. See Fig. 29. Observe that ci cannot intersect any of the
segments in Fig. 29 besides a, f, f (otherwise it would intersect a pseudocircle that
ci does not or it would intersect the inside of a pseudocircle that ci does not intersect,
both of which are impossible). Since ci intersects pseudocircles that lie outside of
γ , it follows that ci crosses both f and f . Now recall that two Jordan curves cross
an even number of times—this is a straightforward consequence of the Jordan curve
theorem. Applying this fact to γ and γ , it follows that ci intersects both f and f
at least twice. In particular, ci intersects some pseudocircle ck at least four times,
contradicting the fact that C is an arrangement of pseudocircles. For the other case,
i.e. ci intersects F as in the fifth situation of Fig. 26, we obtain a contradiction in a
completely analogous way. It thus follows that ci intersects each of a, a precisely
once.
Completely analogously, if c j is one of the squares rather than one of the non-square
rectangles, then ci intersects precisely the same segments of c j in H (precisely the
same number of times) as does ci . These are the third and fourth situations of Fig. 26.
Hence ci intersects precisely the same faces and segments of H as ci . Let us fix
1 ≤ i < j ≤ 5. Observe that if ci , c j do not intersect then they intersect distinct sets
of edges and faces of H. In that case ci , c j also do not intersect. Now suppose that
ci , c j do intersect. Then there are two distinct faces of H where they cross. Looking
at the rightmost two cases of Fig. 26, we see that if ci , c j (with 1 ≤ i < j ≤ 5) cross
inside a face of H then ci , c j cross (at least once) inside the same face. Since ci , c j
cross at most two times and cross at least once in two distinct faces, it follows that
they cross exactly once inside each of these faces.
Similarly to what we did in the proof of Claim 4.2, we can make a correspondence
between those intersection points and segments of C and those of C and then apply
Lemma 2.2 to see that C and C are in fact equivalent. But this also proves that C
and D are equivalent, and in particular (c1, . . . , c5) is equivalent to (d1, . . . , d5). But
this cannot be, because d1, . . . , d5 are circles and (c1, . . . , c5) is not equivalent to any
arrangement of circles by the result of Linhart and Ortner [
16
]. This last contradiction
proves that G is not the intersection graph of a list of circles, as required.
5 Hardness of Circleability
In overview, the proof of Theorem 1.2 passes through two reductions, embodied in
Lemmas 5.1 and 5.4, relying on an intermediate class of pseudocircle arrangements
we now define. We will say that an arrangement of pseudocircles C = (c1, . . . , cn) is
pencil if there is a point p common to all curves, and every point of C \ { p} is on at
most two of the curves. See Fig. 30 for a depiction of a pencil arrangement with n = 3.
Lemma 5.1 Given a combinatorial description of a simple pseudoline arrangement
L, one can construct, in polynomial time, a combinatorial description of a pencil
pseudocircle arrangement C that is circleable if and only if L is stretchable.
Proof Let L be a simple pseudoline arrangement. Observe that if x is a point not on any
line then circle inversion in a circle with centre x will produce a pencil pseudocircle
arrangement C. The face of L containing x will get mapped to the outer face of C,
and x will be the unique point common to all pseudocircles of the pencil pseudocircle
arrangement C. More generally, there is a one-to-one correspondence between the
faces, segments and intersection points of L and those of C, and the combinatorial
description of C can be easily read off from that of L once we fix the face that the
point x lies in.
It remains to see that L is stretchable if and only if C is circleable. Let us first assume
that L is stretchable. In that case, if we take an arrangement L of lines equivalent to
L, and we pick x as the face corresponding to the face of L that we took x in, then
it is easy to see that circle inversion in a circle with centre x will produce a circle
arrangement C equivalent to C. Similarly, suppose that C is an arrangement of circles
equivalent to C, and let p be the unique point contained in every circle of C . Then
circle inversion in a circle with centre p will produce a line arrangement L that is
equivalent to L.
Lemma 5.2 Every pseudoline arrangement is equivalent to a pseudoline arrangement
in which each pseudoline is a convex curve consisting of finitely many line segments.
Proof Let L be an arbitrary pseudoline arrangement. By a result of [
11
], we may
assume L is represented by a wiring diagram. Without loss of generality, we assume
that all finite line segments of the wiring diagram (i.e. all except the beginning- and
end-segment of each pseudoline) are contained within the square [0, a]2. We first
apply the self-homeomorphism of the plane g(x , y) := (x , y · 2−1000a ), to obtain an
equivalent pseudoline arrangement L . Next, we apply the self-homeomorphism
⎧ (x , y − x − 1)
f (x , y) := ⎨⎪ x , y − ij=0 2−i − (x − j ) · 2−( j+1)
if x < 0,
if j ≤ x < j + 1
for some j ∈ {0, . . . , a − 1},
if x ≥ a.
to obtain a new pseudoline arrangement L . See Fig. 31 for a depiction.
By construction, L is equivalent to L and each of its pseudolines consists of at most
a + 2 line segments, with their endpoints on the vertical lines {x = 0}, . . . , {x = a}.
It remains to see that each pseudoline of L is convex.
To see this, fix some pseudoline ∈ L . Observe that ∩ {x ≤ 0} is a half-infinite
line segment with slope −1. For each 0 ≤ j < a, the segment ∩ { j ≤ x < j + 1}
has slope between −2−( j+1) − 2−1000a and −2−( j+1) + 2−1000a (as it is the image
under f ◦ g of a line segment with slope between −1 and +1). The last segment
(i.e. ∩{x ≥ a}) has slope equal to zero. We conclude the slopes are strictly increasing,
which amounts to the curve being convex.
This last lemma allows us to give a relatively easy proof of the fact that every
pencil pseudocircle arrangement is equivalent to a convex pseudocircle arrangement.
(Observe that this is not true for general pseudocircle arrangements by Theorem 1.3,
which we will prove in the next section.)
Lemma 5.3 Every pencil pseudocircle arrangement is equivalent to an arrangement
of convex polygons.
Proof Let C be an arbitrary pencil pseudocircle arrangement, and let L be the simple
pseudoline arrangement we get by an inversion in a circle whose centre is the unique
point common to all pseudocircles. Let L be a wiring diagram equivalent to L. Observe
that the face of L corresponding to the outer face of C is either the face “above” all the
pseudolines or the face “below” all of them. Without loss of generality, let us assume
it is below.
By Lemma 5.2, there is an equivalent pseudoline arrangement L in which each
line is a convex curve made up of finitely many line segments, with the face of L
below all pseudolines corresponding to the outer face of C. If we apply an inversion
in a circle completely contained within this face, then we get a pencil pseudocircle
arrangement C = (c1, . . . , cn) that is equivalent to C. Moreover, since circle inversions
are conformal and map lines outside the circle to circles inside, it follows that each ci
is a closed convex curve through the origin, consisting of finitely many circular arcs.
We can now replace the circular arcs of each ci by polygonal paths with the same
endpoints (that are sufficiently fine approximations) so that each ci remains convex
and the combinatorial description of C remains unchanged. In this way, we obtain the
required arrangement C of convex polygons that is equivalent to C.
The construction in the next lemma is inspired by a similar construction of Jaggi et
al. [
14
] for oriented matroids.
Lemma 5.4 Given a combinatorial description of a pencil pseudocircle arrangement
C = (c1, . . . , cn), one can construct, in polynomial time, the combinatorial description
of a simple, convex pseudocircle arrangement C := (c1−, c1+, . . . , cn−, cn+) on twice
as many curves that is circleable if and only if C is.
The proof of this last lemma relies on the following geometric fact, whose
straightforward proof we leave to the reader.
Lemma 5.5 Suppose that c, c are two circles in the plane with c inside c, and let the
point p lie inside c but outside c . Then there is a circle c through p such that c lies
inside c and c lies inside c . See Fig. 32.
Proof of Lemma 5.4 Let C = (c1, . . . , cn) be an arbitrary pencil pseudocircle
arrangement, and let p denote the unique point that is on all curves. By Lemma 5.3, we assume
without loss of generality that the curves c1, . . . , cn are convex polygons. We construct
the curves ci− by tracing along ci , very close to the original curve, yet completely inside
ci ; and construct ci+ similarly by tracing along and just outside of ci (see Fig. 33, left).
By choosing the curves ci−, ci+ sufficiently close to the original curves, we can ensure
that each intersection point other than p lies in a unique 4-face of the arrangement
C := (c1−, c1+, . . . , cn−, cn+). The 4-face containing the intersection point of ci , c j
other than p is then one of the two regions of the area between ci− and ci+ and between
c −j and c +j. Observe that, since C was an arrangement of convex polygons, we can
certainly ensure that the curves of C are convex too. We can also ensure that the
arrangement C is simple. Moreover, while placing ci−, ci+, we can inductively ensure
that, for all j, k < i , the area containing p that is between c −j, c +j and between ck−, ck+
is also completely contained between ci− and ci+ (see Fig. 33, right).
This last step guarantees that this procedure uniquely determines the combinatorial
description of a simple, convex pseudocircle arrangement. Having access to an explicit
representation of C in the plane is not necessary to determine C . Indeed, it is not hard
to see that the combinatorial description of C can be produced in polynomial time
directly from the combinatorial description of C. (Notice for instance that for every
face of C we get a face in C , for every segment of C we get a face and some segments
bounding it, for every intersection point other than p we get a 4-face, etc., and the
incidences between the faces, segments and intersection points of C are completely
determined by the combinatorial description of C.)
It remains to be seen that C is circleable if and only if C is. First assume that C is
circleable, and let D = (d1, . . . , dn ) be an equivalent arrangement of circles. Then it
is not hard to see that we can pick circles di−, di+, concentric with di for each i , such
that the arrangement D = (d1−, d1+, . . . , dn−, dn+) is equivalent to C .
Conversely, suppose that D = (d1−, d1+, . . . , dn−, dn+) is an arrangement of circles
equivalent to C . Let p be an arbitrary point inside the face of D that corresponds to
the face of C that contains p. By Lemma 5.5, there are circles d1, . . . , dn such that
p ∈ di and di lies inside di+ and outside di− for each i . Clearly D = (d1, . . . , dn )
is a pencil circle arrangement. To see that it is equivalent to C, we can for instance
make use of rotation systems again. Let ϕ be a self-homeomorphism of the plane
satisfying ϕ[di−] = ci−, ϕ[di+] = ci+ for all i , and let the pseudocircle arrangement
E = (e1, . . . , en ) be defined by ei := ϕ[di ]. Then every ei contains the point ϕ( p ),
which lies in the same face of the arrangement C as p. And, for each pair i = j , the
other intersection point of ei and e j lies in the same face of C as the intersection point
of ci and c j not equal to p. Moreover, for each i, ei intersects exactly the same faces
and segments of C as ci does. It follows that the planar multigraphs defined by the
intersection points and segments of E and of C define the same rotation system, and
so must be equivalent. It then also follows that D and C are equivalent, which shows
C to be circleable.
For completeness we collect our findings into an explicit proof.
Proof of Theorem 1.2 Given a combinatorial description of an arbitrary simple
pseudoline arrangement L, we can transform it, in polynomial time, into a
combinatorial description of a pencil pseudocircle arrangement C that is circleable if and only if L
is stretchable using Lemma 5.1. Using Lemma 5.4, we can then produce, in polynomial
time, a combinatorial description of a simple, convex pseudocircle arrangement C that
is circleable if and only if C is circleable. Thus, if there is a polynomial-time algorithm
to decide if an arbitrary simple arrangement of convex pseudocircles is circleable, then
there must also be a polynomial-time algorithm to solve SIMPLE
STRETCHABILITY. By Theorem 2.3, it follows that our computational problem is NP-hard.
6 Hardness of Convexibility
Proof of Theorem 1.3 Let us consider an arbitrary simple arrangement L =
( 1, . . . , n ) of n pseudolines. We may assume that L is given as a wiring diagram.
From L, we shall construct an arrangement C = (c1−, c1+, . . . , cn−, cn+) of 2n
pseudocircles with the property that C is convexible if and only if L is stretchable. For each of
the “wires” i , we construct a pair ci−, ci+ of pseudocircles that trace along i , with ci−
just below and ci+ just above it. We convert the curves ci−, ci+ into Jordan curves, using
vertical and horizontal line segments, as illustrated in Fig. 34, right. Observe that this
can be done in such a way that ci+ and ci− each intersect both c −j and c +j exactly twice
if j = i ; ci− and ci+ do not intersect; and the unique intersection point of i , j (i = j )
lies in a unique 4-face of the arrangement C. As in the last section, having access to
an explicit representation of L in the plane is not actually necessary to determine C:
the combinatorial description of C can be produced in polynomial time directly from
the combinatorial description of L.
Fig. 34 The construction of the pseudocircle arrangement C (right) from a simple pseudoline arrangement
L (left)
Let us now see how C is convexible if and only if L is stretchable. First assume
that L is stretchable. Then there is a representation of L as an arrangement of lines
in the plane. By performing an appropriate linear transformation if needed, we can
assume without loss of generality both that the lines have arbitrarily small slopes
and that the initial vertical ordering of lines “agrees” with the corresponding initial
ordering of pseudolines as given by the wiring diagram. Therefore, using a minor
modification of the procedure that produced C from the wiring diagram, we can produce
a representation of C as an arrangement of convex quadrilaterals (see Fig. 35).
This proves that C is convexible if L is stretchable. To complete the proof, we
now show that L is stretchable under the assumption that C is convexible. Let
D = (d1−, d1+, . . . , dn−, dn+) be a convex pseudocircle arrangement and ϕ be some
self-homeomorphism of the plane such that ϕ[di−] = ci−, ϕ[di+] = ci+ for all
i ∈ {1, . . . , n}. We then fix a line arrangement M = (m1, . . . , mn) where mi is
an arbitrary line separating di− and di+. The existence of M is guaranteed by multiple
invocations of the hyperplane separation theorem. Let L = ( 1, . . . , n) be defined
by setting i := ϕ[mi ]. See Fig. 36. Then L will be a pseudoline arrangement with
the property that i separates ci− and ci+ for all i . It therefore follows that, for each
i = j , the pseudolines i , j intersect inside exactly the same 4-face of C where i , j
intersect (and i and j intersect the same sides of this 4-face as i and j ,
respectively). Using rotation systems for example, it can now easily be seen that L must in
fact be equivalent to L . But then L is also equivalent to M, so L is stretchable. This
concludes the proof of the theorem.
7 Open Problems
Theorem 1.3 proves the existence of an arrangement of pseudocircles that is not
convexible. A natural question is for an analogue of Theorem 1.1 for convexibility.
Question 7.1 What is the least n such that there exists a non-convexible pseudocircle
arrangement on n pseudocircles?
By Theorem 1.1, we must have n ≥ 5 in this last question.
Theorem 1.4 proved the existence of a pseudocircle arrangement such that no circle
arrangement has the same intersection graph. We made no attempt to minimize the
number of pseudocircles (our example had no less than n = 346 pseudocircles), but
the question of what is the smallest example seems very natural.
Question 7.2 What is the smallest n for which there exists a graph on n vertices that
is the intersection graph of a convex pseudocircle arrangement, but not of a circle
arrangement?
Another natural problem is to resolve the computational complexity of determining,
given a combinatorial description of a pseudocircle arrangement as input, whether
there is a circle arrangement with the same intersection graph.
Acknowledgments We thank Artem Pyatkin for bringing the folkore conjecture to our attention and for
helpful pointers to the literature. We thank the anonymous referees for careful proofreading and suggestions
which have improved the presentation. Part of the work of R. J. Kang in this paper was done while this
author was at Durham University, supported by EPSRC Grant EP/G066604/1. He is now supported by a
VENI Grant from Netherlands Organisation for Scientific Research (NWO). Part of the work of T. Müller
in this paper was done while this author was supported by a VENI Grant from Netherlands Organisation
for Scientific Research (NWO).
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