Coloring TriangleFree Rectangle Overlap Graphs with \(O(\log \log n)\) Colors
Tomasz Krawczyk
Arkadiusz Pawlik
Bartosz Walczak
Recently, it was proved that trianglefree intersection graphs of n line segments in the plane can have chromatic number as large as ( log log n). Essentially the same construction produces ( log log n)chromatic trianglefree intersection graphs of a variety of other geometric shapesthose belonging to any class of compact arcconnected sets in R2 closed under horizontal scaling, vertical scaling, and translation, except for axisparallel rectangles. We show that this construction is asymptotically optimal for intersection graphs of boundaries of axisparallel rectangles, which can be alternatively described as overlap graphs of axisparallel rectangles. That is, we prove that trianglefree rectangle overlap graphs have chromatic number O (log log n), improving on the previous bound of O (log n). To this end, we exploit a relationship between offline coloring of rectangle overlap graphs and online coloring of interval overlap graphs. Our coloring method decomposes the graph into a bounded number of subgraphs with a treelike structure that encodes strategies of the adversary in the online coloring problem. Then, these subgraphs are colored with O (log log n) colors

using a combination of techniques from online algorithms (firstfit) and data structure
design (heavylight decomposition).
1 Introduction
A proper coloring of a graph is an assignment of colors to the vertices of the graph
such that no two adjacent ones are in the same color. The minimum number of colors
sufficient to color a graph G properly is called the chromatic number of G and denoted
by (G). The maximum size of a clique (a set of pairwise adjacent vertices) in a graph
G is called the clique number of G and denoted by (G). It is clear that (G) (G).
Classes of graphs for which there is a function f : N N such that (G) f ( (G))
holds for any graph G in the class are called bounded. A graph is trianglefree if it
does not contain a triangle, that is, if (G) 2.
It was observed in the 1940s that large cliques are not necessary for the chromatic
number to grow. Various classical constructions [14, 19] show that it can be arbitrarily
large even for trianglefree graphs. Kim [11] constructed trianglefree graphs on n
vertices with chromatic number ( n/ log n). An earlier result due to Ajtai et al. [1]
implies that this bound is tight.
In this paper, we focus on the relation between the chromatic number and the number
of vertices in classes of trianglefree graphs arising from geometry. The intersection
graph of a family of sets F is the graph with vertex set F and edge set consisting of
pairs of intersecting members of F . The overlap graph of a family of sets F is the
graph with vertex set F and edge set consisting of pairs of members of F that intersect
but neither contains the other. We always assume that the family of sets F is finite.
It is well known that interval graphs (intersection graphs of intervals in R) are
perfectthey satisfy (G) = (G). The study of the chromatic number of graphs
with geometric representation was initiated in a seminal paper of Asplund and
Grnbaum [2], where they proved that the intersection graphs of families of axisparallel
rectangles in the plane are bounded. In particular, they proved the tight bound of
6 on the chromatic number of trianglefree intersection graphs of axisparallel
rectangles. Gyrfs [7, 8] proved that the class of overlap graphs of intervals in R is
bounded. By contrast, Burling [4] showed that trianglefree intersection graphs
of axisparallel boxes in R3 can have arbitrarily large chromatic number. Pawlik et
al. [15, 16] provided a construction of trianglefree intersection graphs of segments,
and more generally, trianglefree intersection graphs of families of vertically and
horizontally scaled translates of any fixed arcconnected compact set in R2 that is not an
axisparallel rectangle, with arbitrarily large chromatic number. These graphs require
( log log n) colors, where n is the number of vertices. One of the problems posed in
[16] is to determine (asymptotically) the maximum chromatic number that a
trianglefree intersection graph of n segments can have.
We solve the analogous problem for trianglefree intersection graphs of frames,
which are boundaries of axisparallel rectangles. These graphs can be alternatively
defined as overlap graphs of axisparallel rectangles. Therefore, they can be
considered as twodimensional generalizations of interval overlap graphs. We show that the
construction of Pawlik et al. is asymptotically best possible for these graphs.
Theorem 1.1 Every trianglefree overlap graph of n rectangles (intersection graph
of n frames) can be properly colored with O(log log n) colors.
For the completeness of exposition, we also include the proof of the bound from
the other side, which appears in [15].
Theorem 1.2 There are trianglefree overlap graphs of n rectangles (intersection
graphs of n frames) with chromatic number ( log log n).
Note the difference in the behavior of rectangle intersection graphs and rectangle
overlap graphs. The former have chromatic number bounded by a function of their
clique number. The latter can have arbitrarily large chromatic number even when they
are trianglefree.
Theorem 1.1 provides the first asymptotically tight bound on the chromatic number
for a natural class of geometric intersection or overlap graphs that does not allow a
constant bound. So far, best upper bounds were of order O(log n), following from the
results of McGuinness [13] and Suk [18] for intersection graphs of families of shapes
including segments and frames, or polylogarithmic in n, obtained by Fox and Pach [6]
for arbitrary families of curves with bounded clique number. The only known lower
bounds follow from the abovementioned constructions of Burling and Pawlik et al.
We hope that our ideas will lead to improving the bounds for other important classes,
in particular, for segment intersection graphs.
Online coloring is an extensively studied variant of the coloring problem. The
difference between ordinary and online coloring is that in the online setting, the
vertices are introduced one by one and the coloring algorithm must assign colors to
them immediately, knowing only the edges between vertices presented thus far. Our
proof exploits a correspondence between online coloring of interval overlap graphs
and ordinary (offline) coloring of rectangle overlap graphs. We obtain a structural
decomposition of an arbitrary rectangle overlap graph with bounded clique number
into a bounded number of socalled directed families of rectangles. For families whose
overlap graphs are trianglefree, we can further decompose them into socalled clean
families, in which no rectangle is entirely contained in the intersection of two
overlapping rectangles. It turns out that overlap graphs of clean directed families of rectangles
have a particular structure of what we call overlap game graphs, that is, they can be
viewed as encodings of adversary strategies in the online interval overlap graph
coloring problem. We succeed in coloring overlap game graphs with O(log log n) colors
by combining two ideas: heavylight decomposition and firstfit coloring. The
reductions to directed and clean families are purely geometricthey exploit boundedness
results and coloring techniques from [2,7,13,18].
2 Overview
All rectangles that we consider are axisparallel, that is, their sides are parallel to the
horizontal or the vertical axes. Throughout the paper, we also assume that all rectangles
Fig. 1 a Crossing; be leftward, rightward, downward and upwarddirected intersections, respectively;
fg diagonal intersections; h forbidden configuration in a clean rightwarddirected family
are in general position, that is, no corner of any rectangle lies on the boundary of
another rectangle. We can easily adjust any family of rectangles so as to satisfy this
condition without changing the overlap relation, just by expanding each rectangle in
every direction by a tiny amount inversely proportional to the area of the rectangle.
The boundary of a rectangle is a frame. The filling rectangle of a frame F , denoted by
rect(F ), is the rectangle whose boundary is F .
From now on, we will work with families of frames and their intersection graphs.
We denote the chromatic and the clique numbers of the intersection graph of a
family of frames F by (F ) and (F ), respectively. Triangles, cliques and connected
components of the intersection graph of F are simply called triangles, cliques and
connected components of F . The x coordinates of the left and right sides and the
ycoordinates of the bottom and top sides of a frame F are denoted by ( F ), r (F ),
b(F ), t (F ), respectively. Thus rect(F ) = [ ( F ), r (F )] [b(F ), t (F )].
We distinguish the following types of frame intersections, illustrated in Fig. 1ag:
crossings, leftward, rightward, downward and upwarddirected intersections, and
diagonal intersections. A family of frames F is leftward, rightward, downward or
upwarddirected if the following condition is satisfied:
(F1) for any two intersecting frames in F , their intersection is leftward, rightward,
downward or upwarddirected, respectively.
A family of frames F is directed if it is leftward, rightward, downward or
upwarddirected. Note that in a directed family, we still allow only one of the four types of
directed intersections, we just do not specify which one. A family of frames F is clean
if the following holds:
(F2) no frame in F is enclosed in two intersecting frames in F (see Fig. 1h).
The first step in our proof of Theorem 1.1 is to reduce the problem of coloring
arbitrary trianglefree families of frames to the problem of coloring clean directed
trianglefree families of frames. This is done by the following two lemmas, the first of
which works with any bound on the clique number, not only for trianglefree families.
Lemma 2.1 Every family of frames F with (F ) k can be partitioned into a
bounded number of directed subfamilies, where the bound depends only on k.
Lemma 2.2 Every trianglefree family of frames can be partitioned into two
subfamilies so that every connected component of either subfamily is a clean family of
frames.
The proofs of both lemmas are technical and are postponed to Sect. 6.
The next step is a more abstract description of the structure of intersection graphs
of clean directed families of frames in terms of intervals in R. The family of all closed
intervals in R is denoted by I. The left and right endpoints of an interval I are denoted
by ( I ) and r (I ), respectively. Two intervals overlap if and only if they intersect but
neither contains the other. The overlap graph defined on a family of intervals has
an edge for each pair of overlapping intervals. Again, we can assume without loss
of generality that the intervals representing an overlap graph are in general position,
which means that all their endpoints are distinct. With this assumption, intervals I1
and I2 overlap if and only if ( I1) < ( I2) < r (I1) < r (I2) or ( I2) < ( I1) <
r (I2) < r (I1).
A rooted forest is a forest in which every tree has one vertex chosen as its root. Let
G be a graph, M be a rooted forest with V (M ) = V (G), and : V (G) I. For
u, v V (G), we write u v if u = v and u is an ancestor of v in M . The graph G
is an overlap game graph with metaforest M and representation if the following
conditions are satisfied:
(G1) (( x )) < (( y)) whenever x y;
(G2) x y E (G) if and only if x y or y x and (x ) and (y) overlap;
(G3) there are no x , y, z such that x y z, (x ) and (y) overlap, and (z)
(x ) (y).
In particular, for every set of vertices Q on a common path from a root to a leaf in M ,
the induced subgraph G[Q] of G is isomorphic to the overlap graph of the family of
intervals {(x ) : x Q}. Here, once again, we assume without loss of generality that
the intervals (x ) for x Q are in general position. Since two vertices x , y V (G)
that are not in the ancestordescendant relation of M (x y and y x ) are never
adjacent in G, the graph G is the union of the overlap graphs G[Q] over the vertex
sets Q of all paths from a root to a leaf in M .
Lemma 2.3 A graph is an overlap game graph if and only if it is isomorphic to the
intersection graph of a clean directed family of frames.1
Now, Theorem 1.1 follows from Lemmas 2.1 and 2.2, the fact that each connected
component of a graph can be colored separately, and Lemmas 2.3 and 2.4. Theorem 1.2
follows from Lemma 2.3 and the following result, implicit in [15], which asserts that
the bound in Lemma 2.4 is tight.
Lemma 2.5 There are trianglefree overlap game graphs with chromatic number
( log log n).
1 Lemmas 2.3 and 2.4 remain valid if we drop the cleanliness condition on the families of frames considered
and the condition (G3) in the definition of an overlap game graph. However, the condition (G3) is necessary
for the proof of Lemma 2.4. To derive the analogue of Lemma 2.4 without the condition (G3), we would first
apply the analogue of Lemma 2.3 to turn the graph into a directed family of frames, then apply Lemma 2.2
to partition it into two subfamilies with clean connected components, and then we would apply Lemma 2.3
again to these connected components going back to overlap game graphs, for which we would apply the
original Lemma 2.4.
We prove Lemma 2.3 in Sect. 5 and Lemma 2.4 in Sect. 4. In Sect. 3, we mostly
recall some known results about online colorings of forests, which motivate the proof
of Lemma 2.5 from [15] and the coloring algorithm presented in Sect. 4. We also
include a sketch of the proof of Lemma 2.5, for completeness and to illustrate the idea
behind overlap game graphs.
We complete this outline by explaining the meaning of the word game in the term
overlap game graphs. Let k N. Consider the following overlap game between two
players: Presenter, who builds a family of intervals presenting them one by one, and
Algorithm, who colors them online, that is, each interval is assigned its color right
after it is presented and without possibility of changing the color later. Presenters
moves are restricted by the following rules:
(I1) if an interval I2 is presented after I1, then ( I1) < ( I2);
(I2) no three intervals I1, I2, I3 such that I1 and I2 overlap and I3 I1 I2 are ever
presented.
The coloring constructed by Algorithm has to be a proper coloring of the overlap graph
defined by the intervals presented. Presenter aims to force Algorithm to use more than
k colors, while Algorithm tries to keep using at most k colors.
Every finite strategy of Presenter gives rise to an overlap game graph G with
metaforest M and representation such that the roottoleaf paths in M correspond to the
intervals presented in the possible scenarios of the strategy. Specifically, each root r
of M corresponds to an interval (r ) that can be played in Presenters first move,
and each child of a vertex x of M corresponds to an interval that Presenter can play
right after (x ) at the position represented by x . Conversely, an overlap game graph
G with metaforest M and representation can be interpreted as a nondeterministic
strategy of Presenter, as follows. Presenter starts with an arbitrarily chosen root r of
M presenting (r ), and then, in each move from a position u in M , follows to an
arbitrarily chosen child v of u presenting (v). The key observation is that Algorithm
has a strategy to use at most k colors against such a strategy of Presenter if and only
if (G) k. The proof of Lemma 2.5 constructs a strategy of Presenter forcing
Algorithm to use more than k colors by presenting at most 2k intervals, while the
proof of Lemma 2.4 essentially shows that every such strategy needs to have a double
exponential number of scenarios.
The presented online interpretation of overlap game graphs is exploited in the
proof of Lemma 2.5 presented in the next section. It may also provide a useful insight
into our arguments in Sect. 4, which are formulated using the static definition of an
overlap game graph.
3 Restricted Families of Intervals
Let J be a family of intervals in R in general position. We say that J is restricted if
the following condition is satisfied:
(R) for any three intervals I1, I2, I3 J , if ( I1) < ( I2) < ( I3) and I1 and I2
overlap, then r (I1) < ( I3).
Fig. 2 Three configurations of intervals excluded in restricted families
That is, restricted families exclude the three configurations of intervals presented in
Fig. 2.
For I1, I2 J , we write I1 I2 if ( I1) < ( I2) < r (I1) < r (I2). That is, is
the orientation of the edges of the overlap graph of J from left to right.
Lemma 3.1 Let J be a restricted family of intervals.
(1) For every I1 J , there is at most one I2 J such that I1 I2. In particular,
the overlap graph of J is a forest.
(2) There are no I1, I2, I3, I4 J such that ( I1) < ( I2) < ( I3) < ( I4), I1 I3,
and I2 I4.
Proof Let I1, I2, I3 J , ( I1) < ( I2) < ( I3), and I1 I2. It follows from
(R) that r (I1) < ( I3). Hence the intervals I1 and I3 are disjoint, so I1 I3. This
completes the proof of (1).
For the proof of (2), let I1, I2, I3, I4 J , ( I1) < ( I2) < ( I3) < ( I4),
and I1 I3. It follows from (1) that I1 I2 I3. This and I1 I3 yields
( I1) < ( I2) < r (I2) < ( I3) < r (I1) < r (I3). Therefore, since ( I3) < ( I4), the
intervals I2 and I4 are disjoint, so I2 I4.
The restricted overlap game is a variant of the overlap game in which Presenter
is required to present a restricted family of intervals. Note that the condition that the
family is restricted implies the condition (I2) of the definition of the overlap game.
By Lemma 3.1 (1), the overlap graphs of families of intervals presented in the
restricted overlap game are forests. A wellknown result in the area of online graph
coloring algorithms asserts that Presenter has a strategy to force Algorithm to use
more than k colors on a forest with at most 2k vertices. Such a strategy was first found
by Bean [3] and later rediscovered by Gyrfs and Lehel [9]. Erlebach and Fiala [5]
described a particular version of such a strategy, which can be carried out on forests
represented as intersection graphs of discs or axisparallel squares in the plane. Pawlik
et al. [15] implemented the same strategy on forests represented as overlap graphs of
intervals presented in the restricted overlap game.
We present this strategy first for abstract forests, and then in the restricted overlap
game. Next, we use it to prove Lemma 2.5. Finally, we show that this strategy is
optimal, that is, any forest with fewer than 2k vertices can be properly colored online
using at most k colors.
Proposition 3.2 (Bean [3]; Gyrfs, Lehel [9]) Presenter has a strategy to force
Algorithm to use more than k colors by presenting a forest with at most 2k vertices.
Proof For a tree T of a forest presented online, let r (T ) denote the vertex of T
presented last. The strategy constructs, by induction, a forest of trees T1, . . . , Tm with
Fig. 3 Illustration for the proof of Proposition 3.2
at most 2k 1 vertices in total such that Algorithm is forced to use at least k colors
on the vertices r (T1), . . . , r (Tm ).
The case k = 1 is trivialit is enough to present a single vertex. To obtain a strategy
for k 2, we apply the strategy for k 1 twice, building two forests of trees T1, . . . , Tm
and T1, . . . , Tm with at most 2k 2 vertices in total such that Algorithm is forced to use
at least k 1 colors on r (T1), . . . , r (Tm ) and at least k 1 colors on r (T1), . . . , r (Tm ).
If the two sets of colors are different, then we have already forced Algorithm to use
at least k colors on r (T1), . . . , r (Tm ), r (T1), . . . , r (Tm ). Otherwise, we present one
more vertex v and connect it to r (T1), . . . , r (Tm ), thus merging T1, . . . , Tm into one
tree Tm+1 with r (Tm+1) = v. Algorithm has to color v with a color different from
those used on r (T1), . . . , r (Tm ). Hence it has been forced to use at least k colors on
r (T1), . . . , r (Tm+1). See Fig. 3 for an illustration.
In the final step, after playing the strategy for k claimed above, we present one more
vertex v connected to r (T1), . . . , r (Tm ), on which a (k + 1)st color must be used. This
way, we have forced the use of more than k colors on a tree with at most 2k vertices.
Proposition 3.3 (Pawlik et al. [15]) Presenter has a strategy to force Algorithm to use
more than k colors on a family of at most 2k intervals in the restricted overlap game.
Proof A tree of a restricted family of intervals J is a subfamily T J that is a
connected component of the overlap graph of J . For a tree T of a family of intervals
presented in the restricted overlap game, let R(T ) denote the interval in T presented
last.
The strategy is an adaptation of the strategy described in the proof of Proposition 3.2.
It constructs, by induction, a restricted family of intervals J with trees T1, . . . , Tm of
total size at most 2k 1 such that
(i) R(T1) R(Tm ),
(ii) every interval in J {R(T1), . . . , R(Tm )} lies entirely to the left of r (R(Tm )),
(iii) Algorithm is forced to use at least k colors on the intervals R(T1), . . . , R(Tm ).
The case k = 1 is trivialit is enough to present a single interval. To obtain a
strategy for k 2, we first apply the strategy for k 1, building a restricted
family of intervals J with trees T1, . . . , Tm of total size at most 2k1 1 such that
(i)(iii) are satisfied. Let x be the maximum of ( R(Tm )) and the right endpoints
of all intervals in J {R(T1), . . . , R(Tm )}. We apply the strategy for k 1 again,
this time playing entirely inside the interval (x , r (R(Tm ))), building a restricted
family of intervals J with trees T1 , . . . , Tm of total size at most 2k1 1 such that
Fig. 4 Illustration for the proof of Proposition 3.3
(i)(iii) are satisfied. Let x be the maximum of ( R(Tm )) and the right endpoints of
all intervals in J {R(T1 ), . . . , R(Tm )}. It is clear that the family J J with trees
T1, . . . , Tm , T1 , . . . , Tm satisfies (i) and (ii). If the two sets of at least k 1 colors used
on R(T1), . . . , R(Tm ) and R(T1 , . . . , Tm ) differ, then we have already forced
Algorithm to use at least k colors on R(T1 ), . . . , R(Tm ), R(T1 ), . . . , R(Tm ), so (iii) is also
satisfied. Otherwise, we present one more interval J with ( J ) (x , r (R(Tm ))) and
r ( J ) (r (R(T1 )), r (R(Tm ))). This way, J overlaps the intervals R(T1 , . . . , Tm ) and
no other intervals, and thus merges T1 , . . . , Tm into one tree Tm+1 with R(Tm+1) = J .
Again, the family J J { J } with trees T1, . . . , Tm+1 satisfies (i) and (ii). Moreover,
Algorithm has to color J with a color different from those used on R(T1 ), . . . , R(Tm ).
Hence it has been forced to use at least k colors on R(T1), . . . , R(Tm+1), so (iii) is
also satisfied. See Fig. 4 for an illustration.
In the final step, after playing the strategy for k claimed above, we present one more
interval J such that ( J ) (x , r (R(Tm ))) and r ( J ) > r (R(T1)), where x is the
maximum of ( R(Tm )) and the right endpoints of all intervals in J {R(T1), . . . , R(Tm )}.
It follows that a (k + 1)st color must be used on J . This way, we have forced the use of
more than k colors on a family of at most 2k intervals in the restricted overlap game.
Proof of Lemma 2.5 (sketch) According to what has been told in Sect. 2, any strategy
of Presenter in the overlap game gives rise to an overlap game graph encoding this
strategy. Since the overlap graph of the family J presented by the strategy described
in the proof of Proposition 3.3 is trianglefree, the overlap game graph G obtained
from this strategy is trianglefree as well. Since Algorithm is forced to use at least
k colors on J , the graph G has chromatic number at least k (actually, its chromatic
number is exactly k). One can calculate that the number of vertices of G is less than
23 22k2 . See [15] for more details.
To show that any forest with fewer than 2k vertices can be properly colored online
using at most k colors, we use the algorithm called Firstfit. It colors each vertex with
the least positive integer that has not been used on any neighbor presented before.
Proposition 3.4 (folklore) Firstfit uses at most k colors on any forest with fewer than
2k vertices presented online in any order.
Proof Let F be a forest with fewer than 2k vertices presented online. For u, v V (F ),
let u v denote that uv E (F ) and u has been presented before v. Let f (u)
denote the color chosen by Firstfit for each vertex u. For every v V (F ), the colors
1, . . . , f (v) 1 have been chosen for some vertices u with u v. Let r be a vertex
with maximum color, and let T be a minimal subforest of F that satisfies the following
conditions (see Fig. 5):
T contains the vertex r ,
for any v V (T ), there are u1, . . . , u f (v)1 V (T ) such that u j v and
f (u j ) = j for 1 j f (v) 1.
It follows that there is a directed path in T from every vertex to r , so T is actually
a tree. Moreover, for every u V (T ) {r }, there is a unique v V (T ) such that
u v. Indeed, if there were two such vertices v1 and v2, then there would be two
different paths in T from u to r , which cannot exist in a tree. Let n j denote the number
4 Coloring TriangleFree Overlap Game Graphs
For the purpose of this entire section, let G be an nvertex trianglefree overlap game
graph with metaforest M and representation . Our goal is to prove that G has
chromatic number O(log log n). Since different components of M are not connected
by edges of G, they can be colored independently using the same set of colors. Thus it
is enough to consider each component of M separately, and therefore we can assume
without loss of generality that M is a single tree.
For any set S of vertices lying on a common roottoleaf path of M , the family
of intervals {(v) : v S} excludes the two configurations of intervals presented in
Fig. 2a, by the condition (G3) on G, and 2c, by the assumption that G is trianglefree,
but it may contain the configuration in Fig. 2b. Our first goal is to reduce the problem
to the case that all three configurations in Fig. 2 are excluded.
We classify each vertex of G as either primary or secondary according to the
following inductive rule. The root of M is primary. Now, let x be a vertex other than
the root of M , and suppose that all vertices v with v x have been already classified.
If there are primary vertices u and v with u v x such that (u), (v) and (x )
form the configuration in Fig. 2b, that is, (v) (x ) and (u) overlaps both (v)
and (x ), then x is secondary. Otherwise, x is primary. It clearly follows that for
any set P of primary vertices lying on a common roottoleaf path of M , the family
of intervals {(v) : v P} excludes all three configurations in Fig. 2, that is, it is
restricted.
r
5
Fig. 5 Example of a tree T with f (r) = 5
For every primary vertex v, let S(v) be the set consisting of v and all vertices x
with v x for which there is a primary vertex u with u v such that (u), (v)
and (x ) form the configuration in Fig. 2b. Hence every primary or secondary vertex
belongs to some S(v).
Lemma 4.1 For every primary vertex v, the set S(v) consists of those and only those
vertices x for which v is the last primary vertex on the path from the root to x in M .
Moreover, every S(v) is an independent set in G.
Proof Let v be a primary vertex, and let y S(v). It follows that there is a primary
vertex u with u v y such that (u), (v) and (y) form the configuration in
Fig. 2b. Let x be any vertex with v x y. We claim that (v) (x ) (y) (see
Fig. 6). Indeed,
if r ((x )) < r ((u)), then (u), (v) and (x ) form the configuration in Fig. 2a,
which is excluded by the condition (G3) on G;
if r ((u)) < r ((x )) < r ((y)), then (u), (x ) and (y) form the configuration
in Fig. 2c, which contradicts the assumption that G is trianglefree;
if r ((x )) > r ((v)), then (u), (v) and (x ) form the configuration in Fig. 2c,
which again contradicts the assumption that G is trianglefree;
hence r ((y)) < r ((x )) < r ((v)), that is, (v) (x ) (y). It follows that
(u), (v) and (x ) also form the configuration in Fig. 2b, so x is secondary and
x S(v). This yields the first statement of the lemma. It also follows that (x ) (y)
whenever x , y S(v) and x y, which proves that S(v) is an independent set in G.
The relation defines an orientation of the edges of G: we write x y if x y
E (G) and x y. We write S(u) S(v) if u v and there are x S(u) and
y S(v) such that x y. We are going to show that any proper kcoloring of the
primary vertices of G can be transformed into a proper 2kcoloring of the whole G.
This is done with the help of the following lemma.
Lemma 4.2 Let I be an independent set of primary vertices in G, and let v I . There
is at most one vertex u I such that S(u) S(v).
Proof Suppose there are two vertices u1, u2 I such that u1 u2 v, S(u1)
S(v), and S(u2) S(v). It follows that there are vertices x1 S(u1), x2 S(u2),
and y1, y2 S(v) with x1 y1 and x2 y2. Hence, by Lemma 4.1, we have
u1 x1 u2 x2 v yi for i {1, 2}. Since (x1) (u1), (x2) (u2),
(y1) (v), (y2) (v), and u1, u2 and v are independent in G, it follows that
(u1), (u2) and (v) are nested (see Fig. 7). Clearly, (x1) and (x2) overlap (v),
and (x1) overlaps (u2). If r ((x1)) < r ((x2)), then (x1), (x2) and (v) form
the configuration in Fig. 2c, which contradicts the assumption that G is trianglefree. If
r ((x1)) > r ((x2)), then (x1), (u2) and (x2) form the configuration in Fig. 2a,
which is excluded by (G3).
We now show how to color the vertices of G with 2k colors. Let I be a color class in
a proper kcoloring of the primary vertices of G. Consider all the sets S(u) for u I .
By Lemma 4.1, each of them is independent. By Lemma 4.2, the edges between the
sets form a bipartite graph. Therefore, we need two colors for the vertices in uI S(u)
and hence 2k colors for the whole G.
It remains to show that the chromatic number of the subgraph of G induced on
the primary vertices is O(log log n). Clearly, this subgraph is itself an overlap game
graph. Therefore, from now on, we simply assume that all vertices of G are primary.
This means that for every set P of vertices lying on a common roottoleaf path of
M , the family of intervals {(v) : v P} is restricted. The following is an immediate
consequence of Lemma 3.1.
Lemma 4.3 Let P be a set of vertices lying on a common roottoleaf path of M .
(1) For every u P, there is at most one v P such that u v. In particular, the
graph G[ P] is a forest.
(2) There are no u1, u2, v1, v2 P such that u1 u2 v1 v2, u1 v1, and
u2 v2.
Now, the simplest idea would be to color the vertices by Firstfit, processing them
from left to right in the order . Then, by Lemma 4.3 (1) and Proposition 3.4, the
number of colors used would be logarithmic in the maximum length of a
roottoleaf path in M . This is not enough when M contains paths longer than O(log n). To
overcome this problem, we need to introduce the concept of heavylight decomposition
due to Sleator and Tarjan [17].
Let T be a rooted tree. We call an edge uv of T , where v is a child of u, heavy if
the subtree of T rooted at v contains more than half of the vertices of the subtree of T
rooted at u. Otherwise, we call the edge uv light. The resulting partition of the edges
of T into heavy and light edges is called the heavylight decomposition of T . Since
every vertex u of T has a heavy edge to at most one of its children, the heavy edges
induce in T a collection of paths, called heavy paths. The heavylight decomposition
has the following crucial property, easily proved by induction.
Lemma 4.4 (Sleator, Tarjan [17]) If there is a path in T starting at the root and
containing at least k 1 light edges, then T has at least 2k 1 vertices.
Fix a heavylight decomposition of M . Form an auxiliary graph G by removing
from G the edges connecting pairs of vertices in different heavy paths. By Lemma 4.3
(1), the vertices on each heavy path induce a forest in G. Hence G is a forest and can
be properly colored with two colors. Let C1 and C2 be the color classes in a proper
twocoloring of G . Fix i {1, 2}. It follows that for any x , y Ci , if x y, then x
and y lie on different heavy paths. Color the induced subgraph G[Ci ] of G by Firstfit,
processing the vertices from left to right in the order . That is, the color assigned to
a vertex v Ci is the least positive integer not assigned to any u Ci with u v.
Lemma 4.5 If Firstfit assigns a color k to some vertex r Ci , then the path Mr in
M from the root to r contains at least 2k2 1 light edges.
Proof Let f (u) denote the color chosen by Firstfit for each vertex u Ci . Since
G[Ci ] is a forest, we can find, as in the proof of Proposition 3.4, a minimal subtree T
of G[Ci ] that satisfies the following conditions (see Fig. 5):
T contains the vertex r ,
for any v V (T ), there are u1, . . . , u f (v)1 V (T ) such that u j v and
f (u j ) = j for 1 j f (v) 1.
Clearly, the entire T lies on the path Mr . Moreover, as in the proof of Proposition 3.4,
T contains exactly 2k1 vertices, 2k2 of which have color 1 and 2k2 have color
greater than 1. Let v be a vertex of T with color greater than 1, and let u V (T ) be
the vertex directly preceding v in the order on V (T ). We claim that u is a child of v
in T . Suppose it is not. Since u is not the greatest vertex of T , we have u = r , and
therefore u has a parent p in T . Since v has color greater than 1, it has a child c in T .
It follows that c u v p, as u and v are consecutive in the order on V (T ). But
we also have c v and u p, which contradicts Lemma 4.3 (2). We have shown
that there is an edge between u and v in T , so they must lie on different heavy paths.
Consequently, each vertex of T with color greater than 1 belongs to a different heavy
path. This shows that the path Mr contains at least 2k2 1 light edges.
Proof of Lemma 2.4 As noted before, we may assume that G consists only of primary
vertices and M is a tree. Let k be the maximum color used by Firstfit on a vertex
r Ci . By Lemma 4.5, the path Mr in M from the root to r contains at least 2k2 1
light edges. This implies, by Lemma 4.4, that n 22k2 1. Therefore, Firstfit uses
at most O(log log n) colors on Ci . We color C1 and C2 by Firstfit using two separate
sets of colors, obtaining a proper coloring of G with O(log log n) colors.
5 Clean Directed Families of Frames = Overlap Game Graphs
Proof of Lemma 2.3 First, we show that every overlap game graph is isomorphic to
the intersection graph of a clean directed family of frames. Let G be an overlap game
graph with metaforest M and representation . We define frames F (v) for all vertices
v of G so that
b(F (v)) < b(F (v1)) < t (F (v1)) < < b(F (vk )) < t (F (vk )) < t (F (v)),
where v1, . . . , vk are the children of v in M in any order.
See Fig. 8 for an illustration. The numbers b(F (v)) and t (F (v)) can be computed
by performing a depthfirst search over the forest M and recording, for each vertex
v, the times at which the subtree of M rooted at v is entered and left, respectively.
Clearly, if u and v are two vertices unrelated in M , then t (F (u)) < b(F (v)) or
t (F (v)) < b(F (u)), so F (u) and F (v) do not intersect. If u v, then we have
b(F (u)) < b(F (v)) < t (F (v)) < t (F (u)), and therefore F (u) and F (v) intersect
if and only if (u) and (v) overlap. Moreover, when F (u) and F (v) intersect, this
intersection is rightwarddirected. If there are three vertices u, v, w such that F (u)
and F (v) intersect and F (w) is enclosed in both F (u) and F (v), then u v w,
(u) and (v) overlap, and (w) (u) (v), which is excluded by the condition
(G3) of the definition of an overlap game graph. Consequently, {F (v) : v V (G)} is
a clean rightwarddirected family of frames, and its intersection graph is isomorphic
to G.
It remains to show that the intersection graph of every clean directed family of
frames is an overlap game graph. Let F be a directed family of frames. We can assume
without loss of generality that F is rightwarddirected. Define a mapping : F I
so that (( F )) = ( F ) and r ((F )) = r (F ), that is, (F ) is the interval obtained
by projecting F on the x axis.
For F F , let L(F ) be the subfamily of F consisting of those F for which
( F ) < ( F ) < r (F ) and b(F ) < b(F ) < t (F ) < t (F ). We define a rooted forest
M on F as follows. If L(F ) = , then F is a root of M . Otherwise, the parent of F in
M is the member F of L(F ) with greatest ( F ). We show that the intersection graph
of F is an overlap game graph with metaforest M and representation . To this end,
we argue that the conditions (G1)(G3) of the definition of an overlap game graph are
satisfied by the intersection graph of F .
It follows directly from the definition of parents that F1 F2 implies ( F1) < ( F2)
and b(F1) < b(F2) < t (F2) < t (F1). This already shows (G1) and the righttoleft
implication in (G2). Suppose there are F1, F2, F3 F such that F1 F2 F3, (F1)
and (F2) overlap, and (F3) (F1) (F2). We have ( F1) < ( F2) < ( F3) <
r (F3) < r (F1) < r (F2) and b(F1) < b(F2) < b(F3) < t (F3) < t (F2) < t (F1), a
configuration that is forbidden in a clean family of frames. This contradiction shows
(G3). Now, let F1 and F2 be two intersecting members of F . By the assumption
that F is rightwarddirected, (F1) and (F2) overlap, and we have F1 L(F2)
or F2 L(F1). Therefore, in order to prove the lefttoright implication in (G2), it
remains to show that F1 L(F2) implies F1 F2. To this end, we use induction
on the increasing order of ( F2). There is nothing to prove when F1 is the parent of
F2. Hence assume that F1 L(F2), but F1 is not the parent of F2. Let F2 be the
parent of F2. We have ( F1) < ( F2) < ( F2) < r (F1) and, since F is
rightwarddirected, b(F1) < b(F2) < t (F2) < t (F1). Thus F1 L(F2). This and the induction
hypothesis yield F1 F2 and thus F1 F2.
6 Reduction to Clean Directed Families of Frames
The goal of this section is to prove Lemmas 2.1 and 2.2. To make the description
simpler, we are going to partition a family of frames with bounded clique number into
a bounded number of subfamilies with the property that the connected components
of each subfamily are directed, but the directions of different components may be
different. Let (F ) denote the minimum size of such a partition of a family of frames
F . It is enough to bound (F ), because we can gather the connected components of
each partition class that have the same direction, thus turning each partition class into
at most four directed families. This way we will obtain the bound of 4(F ) on the size
of a partition of F into directed families.
Two families of frames F1 and F2 are mutually disjoint if there are no two
intersecting frames F1 F1 and F2 F2. The following properties of (analogous to the
properties of chromatic number) are straightforward:
(i) (F1 Fm ) (F1) + + (Fm ),
(ii) if Fi and F j are mutually disjoint for i = j , then (F1 Fm ) =
max{(F1), . . . , (Fm )}.
We will use them implicitly in the rest of this section. For convenience, whenever
we consider a partition of a family into a number of subfamilies, we allow these
subfamilies to be empty.
Recall that two frames are said to cross if both vertical sides of one intersect both
horizontal sides of the other. A family of frames F is noncrossing if it contains no
two crossing frames.
Lemma 6.1 (implicit in [2]) Every family of frames F with (F )
tioned into k noncrossing subfamilies.
Proof Define a partial order < on F so that F1 < F2 whenever both vertical sides of F1
intersect both horizontal sides of F2. The graph G+ of crossing pairs of frames in F is
the comparability graph of <, so it is perfect. Hence (G+) = (G+) (F ) k,
and the lemma follows.
A frame E is external to a family of frames F if E FF rect(F ). A subfamily
L of a family of frames F is a layer of F if L = 1 or every frame in L intersects
some frame in F L external to L. A family L F is an mfold layer of F if there
is a chain of families L = Lm Lm1 L0 = F such that Li+1 is a layer of
Li for 0 i m 1. Note that every subfamily of a layer (an mfold layer) of F is
also a layer (an mfold layer) of F .
Lemma 6.2 Every family of frames F has a partition P into layers. Moreover, P
can be partitioned into two subclasses each consisting of mutually disjoint layers.
Proof When F is not connected, the layers and the bipartition can be constructed
for each component separately. Thus assume, without loss of generality, that F is
connected. Choose any frame R F that is external to F {R}. For j N, let L j
consist of those frames in F whose distance to R in the intersection graph of F is j .
Let d be greatest such that Ld = . We claim that P = {L0, . . . , Ld } is a partition
of F into layers.
We have L0 = 1, so L0 is a layer. To show that L j with j 1 is a layer, we find,
for each F L j , a frame in F L j external to L j and intersecting F . Let F0 F1 . . . Fj
be a path from F0 = R to Fj = F of length j . Let i be greatest such that Fi is external
to L j (such i exists, as R is external to L j ). Since Fi intersects Fi+1, which is not
external to L j , Fi must intersect some frame F L j . This witnesses a path from R
to F of length i + 1, so i + 1 = j . Therefore, Fi is the requested frame external to
L j and intersecting F .
Clearly, any two intersecting frames belong to one layer or two layers with
consecutive indices. Hence the families {L j : j is odd} and {L j : j is even} consist of
mutually disjoint layers.
Corollary 6.3 Every family of frames F has a partition P into mfold layers.
Moreover, P can be partitioned into 2m subclasses each consisting of mutually disjoint
mfold layers.
LPi P1L and
LPi P2L for 1
Proof We proceed by induction on m. For m = 1, this is exactly the statement of
Lemma 6.2. For m 2, we apply induction hypothesis to construct a partition P of
F into (m 1)fold layers and a partition of P into 2m1 subclasses P1, . . . , P2m1
each consisting of mutually disjoint layers. We apply Lemma 6.2 to construct a partition
PL of each (m 1)fold layer L P into layers, which are now mfold layers of F ,
and a partition of PL into two subclasses P1L and P2L each consisting of mutually
disjoint layers. We set P = LP PL. The desired partition of P is formed by the
families i 2m1.
Theorem 6.4 (Asplund, Grnbaum [2]) Let R be a family of axisparallel rectangles
with (R) k. It follows that (R) k , where k depends only on k.
The precise bound of Theorem 6.4 in [2] is 4k2 3k. This was improved to 3k2
2k 1 by Hendler [10]. However, we are going to apply Theorem 6.4 only to
noncrossing families R, for which the bound in [2] is 4k 3.
A curve is xsemimonotone if its intersection with every vertical line is connected
or empty.
Theorem 6.5 (Suk [18]) Let V be a vertical line. Let C be a family of x semimonotone
curves contained in the right halfplane delimited by V such that V C  = 1 for
any C C, C1 C2 1 for any distinct C1, C2 C, and (C) k. It follows that
(C) k , where k depends only on k.
The actual theorem in [18] concerns families of x monotone curves, that is, curves
whose intersection with every vertical line is a single point or empty. The generalization
to x semimonotone curves above comes from the fact that we can always perturb
a family of x semimonotone curves to make them x monotone without changing
their intersection graph. Theorem 6.5 for k = 2 but without the x semimonotonicity
assumption was proved by McGuinness [13]. Lason et al. [12] generalized Theorem 6.5
removing the x semimonotonicity assumption for any k.
For the rest of this section, assume that k and k are as in the statements of
Theorems 6.4 and 6.5. Define constants k and k,m for 1 m k by induction on k
and m, as follows:
1 = 1, k,1 = 0, k,m = 4(m 1)(k4 + 2k1 + k,m1), k = 2k k k,k .
Lemma 6.6 Let F be a noncrossing family of frames with (F ) k, let 1 m k,
let B1, . . . , Bkm+1 be pairwise intersecting frames, and let L F be a family of
frames enclosed in each of B1, . . . , Bkm+1 such that L {B1, . . . , Bkm+1} is an
mfold layer of F . Suppose that (L ) k1 holds for any subfamily L L with
(L ) k 1. It follows that (L) k,m .
Proof The proof goes by induction on m. If m = 1, then the assumption that L
{B1, . . . , Bk } is a layer of F yields L = . Indeed, if there was some F L, then
there would be a frame E F external to L {B1, . . . , Bk } and intersecting F . Hence
{B1, . . . , Bk , E } would be a (k + 1)clique in F . We conclude that (L) = 0 = k,1.
Now, assume that 2 m k and the lemma holds for m 1. We show that
(L) 4(m 1)(k4 + 2k1 + k,m1) = k,m .
Since L {B1, . . . , Bkm+1} is an mfold layer of F , there is an (m 1)fold
layer M of F such that L {B1, . . . , Bkm+1} is a layer of M. Let S be the family
of those frames in M that intersect all of B1, . . . , Bkm+1 and do not lie inside
any frame in M intersecting all of B1, . . . , Bkm+1. It follows that (S) m 1
and two frames in S intersect if and only if their filling rectangles intersect. Since
L {B1, . . . , Bkm+1} is a layer of M, for every F L, there is a frame E M that
is external to L {B1, . . . , Bkm+1} and intersects F , which implies that E or some
frame enclosing E belongs to S. Hence every frame in L intersects or is enclosed in
some frame in S.
We partition S into four families S1, S2, S3 and S4 so that every frame in S1, S2,
pSa3rtoitrioSn4Linitnetrosefcotusrtfhaemlielfite,srLig1h,t,Lb2o,tLto3manodr Lto4pssoidtheaotfevBe1r,yrfersapmecetiivneLlyi. iNnteexrts,ewctes
or is enclosed in some frame in Si . We have (L) (L1) + (L2) + (L3) + (L4).
We show that (Li ) (m 1)(k4 + 2k1 + k,m1). In the following, we assume
without loss of generality that i = 1, that is, we are to bound (L1).
We partition L1 into three families X , Y and Z as follows. Fix a frame F L1.
If there is a frame in S1 that encloses F , then we put F in X . Otherwise, if there is a
frame in S1 that encloses the entire top or bottom side of F , then we put F in Y . If
neither of the above holds, then we put F in Z .
The intersection graph of S1 is isomorphic to the intersection graph of the filling
rectangles of the frames in S1, so it is an interval graph, as all these rectangles intersect
the vertical line containing the left side of B1. Hence (S1) = (S1) m 1.
Therefore, we can partition S1 into m 1 families S1, . . . , Sm1 each consisting of
frames with pairwise disjoint filling rectangles. We also partition each of X , Y and Z
into m 1 families X1, . . . , Xm1, Y1, . . . , Ym1 and Z1, . . . , Zm1, respectively, so
that
if F X j , then F is enclosed in a frame in S j ,
if F Y j , then the bottom or top side of F is enclosed in a frame in S j ,
if F Z j , then F intersects a frame in S j and neither of the above holds.
We show that (X j ) k,m1, (Y j )
Once this is achieved, we will have
m 1.
m1
j =1
m1
j =1
m1
j =1
First, we show that (X j ) k,m1. We partition X j into S j  families XS for S
S j so that if F XS , then F is enclosed in S. The families XS for S S j are mutually
disjoint, as the filling rectangles of the S S j are pairwise disjoint. Let S S j . It
follows that S intersects all of B1, . . . , Bkm+1. Since XS { B1, . . . , Bkm+1, S} is
a subfamily of M, it is an (m 1)fold layer of F . The induction hypothesis yields
(XS ) k,m1. Hence (X j ) = maxSS j (XS ) k,m1.
Now, we show that (Y j ) 2k1. We partition Y j into two families Y bj and Y tj
so that if F Y bj, then the bottom side of F is enclosed in some frame in S j , while if
F Y tj , then the top side of F is enclosed in some frame in S j . We show that (Y bj)
k 1 and (Y tj ) k 1. Then, it will follow that (Y j ) (Y bj) + (Y tj ) 2k1.
To see that (Y bj) k 1, let K be a clique in Y bj. Let F be the frame in K with
maximum b(F ), and let S be the frame in S j enclosing the bottom side of F . It follows
that F and S intersect. Moreover, every other frame in K {F } intersects or encloses
the bottom side of F and thus intersects S as well. Hence K {S} is a clique in F .
This implies K (F ) 1 k 1. The proof that (Y tj ) k 1 is analogous.
Finally, we show that (Z j ) k4. The definition of Z j and the assumption that
F is noncrossing imply that the right side of no frame in Z j intersects or is enclosed
Fig. 9 Short (blue) and long (blue + red) lower traces
in any frame in S j . For each frame F Z j , we define four curves, the short lower,
short upper, long lower and long upper trace of F , as follows. The short lower (upper)
trace of F starts at the bottom (top) right corner of F and follows along the bottom
(top) side of F and possibly further along F until it reaches an intersection point with
a frame S S j on either the bottom (top) or the left side of F . The long lower (upper)
trace of F starts at the top (bottom) right corner of F , follows along the right side of
F to the bottom (top) corner of F , and then continues along the short lower (upper)
trace until the intersection point with a frame S S j . See Fig. 9 for an illustration.
All short lower (upper) traces can be connected to the left side of B1 by
pairwise disjoint x monotone curves inside the frames in S j , thus forming a family of
x semimonotone curves with the same intersection graph. Any two of these curves
intersect in at most one point, because so do any two short lower (upper) traces.
Therefore, by Theorem 6.5, there are proper colorings and u of all short lower and short
upper traces, respectively, with k colors. To prove that (Z j ) k4, it is enough to
show that (Z ) k2 for any family Z Z j of frames whose short lower traces
have the same color in and whose short upper traces have the same color in u .
Let Z be such a family. Since the short lower (upper) traces of any two frames in Z
are disjoint, their long lower (upper) traces intersect in at most one point. Consequently,
the same argument as for short traces yields proper colorings and u of all long
lower and long upper traces, respectively, with k colors. The intersection of any two
frames in Z whose lower long traces are disjoint and whose upper long traces are
disjoint is leftwarddirected. Therefore, the frames in Z whose long lower traces have
the same color in and whose long upper traces have the same color in u form a
leftwarddirected family. This shows that (Z ) k2.
Proof The proof goes by induction on k. The statement is trivial for k = 1, so assume
that k 2 and the statement holds for k 1. Let F be a noncrossing family of
frames with (F ) k. By Corollary 6.3, F has a partition P into kfold layers,
and moreover, P can be partitioned into 2k subclasses P1, . . . , P2k each consisting
of mutually disjoint kfold layers. We show that (L) k k,k for any kfold layer
L P. Once this is done, we will have (Pi ) k k,k (because each Pi consists
of mutually disjoint kfold layers) and thus
RR j
i=1
Let L P. Let R be the family of those frames in L that do not lie inside any
frame in L. Every frame in L R is enclosed in some frame in R. Hence L R
can be partitioned into R families LR for R R so that every frame in LR is
enclosed in R. Since LR {R} is a kfold layer of F , it follows from Lemma 6.6 that
(LR ) k,k . The intersection graph of R is isomorphic to the intersection graph of
the filling rectangles of the frames in R, so Theorem 6.4 yields (R) k . Hence
R can be partitioned into k subfamilies R1, . . . , Rk each consisting of frames with
pairwise disjoint filling rectangles. For 1 j k , the families R j and LR for
R R j are mutually disjoint, so
Proof of Lemma 2.1 Let F be a family of frames with (F ) k. As it has been
explained at the beginning of this section, it is enough to bound (F ) in terms of
k. By Lemma 6.1, we can partition F into k noncrossing familieks F1, . . . , Fk . By
Lemma 6.7, we have (Fi ) k for 1 i k. Hence (F ) i=1 (Fi ) kk .
Proof of Lemma 2.2 Let F be a trianglefree family of frames. By Lemma 6.2, F has
a partition P into layers, which can be further split as P = P1 P2 so that each of
P1 and P2 consists of mutually disjoint layers. We are going to show that each layer
in P is a clean family of frames. This will show that the subfamilies P1 and P2
of F satisfy the conclusion of the lemma.
Choose any layer L P. Suppose there are three frames F1, F2, F3 L such
that F1 and F2 intersect and F3 is enclosed in both F1 and F2. By the definition of a
layer, there is a frame E F L external to L and intersecting F3. Clearly, E must
intersect both F1 and F2, thus creating a triangle in the intersection graph of F . This
contradiction shows that L is indeed a clean family of frames.
7 Problems
The authors of [16] asked for the maximum possible chromatic number of a
trianglefree intersection graph of n segments. In this paper, we resolved a similar question
for frames. The following problems ask whether segment graphs behave similarly to
frame graphs with respect to containment of overlap game graphs.
Problem 1 Can every trianglefree segment intersection graph be decomposed into a
bounded number of overlap game graphs?
Problem 2 Does every trianglefree segment intersection graph with chromatic
number k contain an overlap game graph with chromatic number at least ck as an induced
subgraph, for some absolute constant c > 0?
The positive answer to either question would yield the bound of ( log log n) on
the chromatic number of trianglefree segment intersection graphs, while the negative
answer would help us understand the limitations of our methods.
Most results of this paper concern trianglefree graphs, but similar statements are
likely to hold for Kk free graphs.
Problem 3 What is the maximum possible chromatic number of a Kk free overlap
graph of n rectangles (intersection graph of n frames)?
Acknowledgments T. Krawczyk and A. Pawlik were supported and B. Walczak was partially supported
by Ministry of Science and Higher Education of Poland Grant 884/NESFEuroGIGA/10/2011/0 within
ESF EuroGIGA project GraDR. B. Walczak was partially supported by Swiss National Science Foundation
Grant 200020144531.
Open Access This article is distributed under the terms of the Creative Commons Attribution License
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the source are credited.