Coloring Triangle-Free Rectangle Overlap Graphs with O(loglogn)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$O(\log \log n)$$\end{document} Colors

Recently, it was proved that triangle-free intersection graphs of n\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$n$$\end{document} line segments in the plane can have chromatic number as large as Θ(loglogn)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\Theta (\log \log n)$$\end{document}. Essentially the same construction produces Θ(loglogn)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\Theta (\log \log n)$$\end{document}-chromatic triangle-free intersection graphs of a variety of other geometric shapes—those belonging to any class of compact arc-connected sets in R2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb {R}^2$$\end{document} closed under horizontal scaling, vertical scaling, and translation, except for axis-parallel rectangles. We show that this construction is asymptotically optimal for intersection graphs of boundaries of axis-parallel rectangles, which can be alternatively described as overlap graphs of axis-parallel rectangles. That is, we prove that triangle-free rectangle overlap graphs have chromatic number O(loglogn)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$O(\log \log n)$$\end{document}, improving on the previous bound of O(logn)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$O(\log n)$$\end{document}. To this end, we exploit a relationship between off-line coloring of rectangle overlap graphs and on-line coloring of interval overlap graphs. Our coloring method decomposes the graph into a bounded number of subgraphs with a tree-like structure that “encodes” strategies of the adversary in the on-line coloring problem. Then, these subgraphs are colored with O(loglogn)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$O(\log \log n)$$\end{document} colors using a combination of techniques from on-line algorithms (first-fit) and data structure design (heavy-light decomposition).


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 χ(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 ω(G).It is clear that χ(G) ω(G).The classes for which there is a function f : N → N such that χ(G) f (ω(G)) holds for any graph G from the class are called χ-bounded.A triangle is a clique of size 3.A graph is triangle-free if it does not contain any triangle.
It has been observed in the 1940s that the existence of large cliques is not necessary for the chromatic number to grow.Various classical constructions show that it can be arbitrarily large even for triangle-free graphs.The first such construction was given by Zykov [16], and the one perhaps best known is due to Mycielski [11].Kim [8] constructed triangle-free graphs with chromatic number Θ( n/ log n), which is tight as shown by Ajtai, Komlós and Szemerédi [1].
In this paper, we focus on the relation between the chromatic number and the number of vertices of a graph for classes of triangle-free 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 elements 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 elements of F that intersect but neither contains the other.
T. Krawczyk  The study of the chromatic number of graphs with geometric representation was initiated in the seminal paper of Asplund and Grünbaum [2], where they proved that the intersection graphs of families of axis-aligned rectangles are χ-bounded.In particular, they proved a tight bound of 6 on the chromatic number of triangle-free intersection graphs of axis-aligned rectangles.Gyárfás [5,6] proved that the class of overlap graphs of intervals on a line is χ-bounded.In contrast, Burling [3] showed that triangle-free intersection graphs of axis-aligned boxes in R 3 can have arbitrarily large chromatic number.Pawlik et al. [12,13] provided a construction of triangle-free intersection graphs of segments and, more generally, triangle-free intersection graphs of families of vertically and horizontally scaled translates of any fixed arc-connected compact set in R 2 that is not an axis-aligned rectangle, with arbitrarily large chromatic number.These graphs require Ω(log log n) colors, where n is the number of vertices.One of the problems posed in [12] is to determine (asymptotically) the maximum chromatic number that a triangle-free intersection graph of n segments can have.
We solve the analogous problem for triangle-free intersection graphs of boundaries of axisaligned rectangles, which we call frames.These graphs can be alternatively defined as overlap graphs of axis-aligned rectangles.Therefore, they can be considered as two-dimensional generalizations of interval overlap graphs.We show that the construction of Pawlik et al. is asymptotically best possible for these graphs.

Theorem 1. Every triangle-free 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 [13].
Theorem 2. There are triangle-free overlap graphs of n rectangles (intersection graphs of n frames) with chromatic number Θ(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 the clique number.The latter can have arbitrarily large chromatic number even when they are triangle-free.
Theorem 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 [10] and Suk [15] for intersection graphs of families of shapes including segments and frames, or polylogarithmic in n, obtained by Fox and Pach [4] for arbitrary families of curves with bounded clique number.The only known lower bounds follow from the above-mentioned 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.
On-line coloring is an intensively studied variant of the coloring problem.The difference between regular and on-line coloring is that in the on-line setting the vertices appear one by one and the coloring algorithm must assign colors to them immediately, knowing only the edges between vertices shown thus far.Our proof exploits a correspondence between on-line coloring of interval overlap graphs and usual (off-line) coloring of rectangle overlap graphs.We obtain a structural decomposition of an arbitrary rectangle overlap graph with bounded clique number, yielding a constant number of so-called directed families of rectangles.For families whose overlap graphs are triangle-free, we can further decompose them into so-called clean families, in which no rectangle is entirely contained in the intersection of two overlapping rectangles.It turns out Figure 1.From left to right: a crossing; a leftward-, rightward-, downward-and upward-directed intersection; two diagonal intersections; forbidden configuration in a clean rightward-directed family 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 on-line interval overlap graph coloring problem.We succeed in coloring overlap game graphs with O(log log n) colors using a mixture of two strategies: heavy-light decomposition of trees (first introduced by Sleator and Tarjan [14]) and first-fit coloring.For frame intersection graphs, we use a result due to McGuinness [10] that intersection graphs of simple triangle-free families of arc-connected compact sets in the plane pierced by a common line have bounded chromatic number, as well as coloring techniques introduced by Gyárfás in his proof that interval overlap graphs are χ-bounded.

Basic ideas
All rectangles that we consider are axis-aligned, that is, their sides are parallel to the horizontal or the vertical axes.Throughout the paper, we also assume that all rectangles 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 with colorings of their intersection graphs.For convenience, we denote the chromatic number and the clique number of a family of frames F by χ(F) and ω(F), respectively.We say that a family of frames F is triangle-free if its intersection graph is triangle-free, that is, ω(F) 2. The x-coordinates of the left and right sides of a frame F and the y-coordinates of the bottom and top sides of R are denoted by ℓ(F ), We distinguish the following types of frame intersections, illustrated in Fig. 1: crossings, leftward-, rightward-, downward-and upward-directed intersections, and diagonal intersections.A family of frames F is leftward-, rightward-, downward-or upward-directed if the following condition is satisfied: (F1) the intersection of any two frames in F is leftward-, rightward-, downward-or upwarddirected, respectively.
A family of frames F is directed if it is leftward-, rightward-, downward-or upward-directed.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 by two intersecting frames in F (see Fig. 1).
The first step in our proof of Theorem 1, explained in Section 6, is to reduce the problem of coloring arbitrary triangle-free families of frames to the problem of coloring clean directed triangle-free 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 triangle-free families: Lemma 3. Every family of frames F with ω(F) k can be partitioned into a bounded number of directed subfamilies, where the bound depends only on k.Lemma 4. Every triangle-free family of frames can be partitioned into two clean subfamilies.
The next step is a more abstract description of the structure of intersection graphs of clean directed families of frames in terms of intervals on R. We denote the family of all closed intervals on R by I.The left and right endpoints of an interval I are denoted by ℓ(I) and r(I), respectively.Again, we assume that we are dealing with intervals in general position, that is, the endpoints of all intervals are distinct.Intervals I and J overlap if ℓ(I) < ℓ(J) < r(I) < r(J) or ℓ(J) < ℓ(I) < r(J) < r(I).The overlap graph defined on a family of intervals has an edge for each pair of overlapping intervals.
Let G be a graph, M be a rooted forest with V (M ) = V (G), and µ : 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 meta-forest M and representation µ if the following conditions are satisfied: (G1) ℓ(µ(x)) < ℓ(µ(y)) whenever x ≺ y, (G2) x and y are adjacent in G if and only if x ≺ y and µ(x) and µ(y) overlap, (G3) there are no x, y, z such that x ≺ y ≺ z, µ(x) and µ(y) overlap, and µ(z) ⊂ µ(x) ∩ µ(y).

Lemma 5. The intersection graph of a clean directed family of frames is an overlap game graph, and vice versa.
Lemma 6.Every triangle-free overlap game graph has chromatic number O(log log n). 1ow, Theorem 1 follows from Lemmas 3, 5, and 6.The bound in Lemma 6 is tight as shown by the following result implicit in [13].
Lemma 7.There are triangle-free overlap game graphs with chromatic number Ω(log log n).
We prove Lemma 5 in Section 5 and Lemma 6 in Section 4. In Section 3, we present the proof of Lemma 7 from [13], 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 notion of overlap game graphs.Let k ∈ N. Consider the following overlap coloring game between two players: Presenter, who presents intervals one by one, and Painter, who colors them on-line, that is, each interval is colored right after it is presented and without possibility of changing the color later.Presenter's moves are restricted by the following rules: presented, that is, no two overlapping intervals contain a third one.
The coloring constructed by Painter has to be proper.Presenter aims to force Painter to use more than k colors, while Painter tries to do with at most k colors.Every finite strategy of Presenter (not necessarily winning or deterministic) gives rise to an overlap game graph G with meta-forest M and representation µ such that the root-to-leaf paths in M correspond to the intervals presented on the possible branches of the strategy.Specifically, each root r of M corresponds to an interval µ(r) that can be played in Presenter's first move, and each child of a vertex x of M corresponds to an interval that Presenter can play right after µ(x) at the position represented by x.Conversely, an overlap game graph G with meta-forest M and representation µ represents a non-deterministic strategy of Presenter, as follows.Presenter starts with an arbitrarily chosen root r of M presenting µ(r), and then, in each move from position u in M , follows to an arbitrarily chosen child v of u and presents µ(v).Now, the crucial observation is that Painter has a strategy to use at most k colors against the considered strategy of Presenter if and only if χ(G) k.The proof of Lemma 7 constructs a strategy of Presenter forcing Painter to use at least k colors by presenting O(2 k ) intervals, while the proof of Lemma 6 essentially shows that each such strategy needs to have a double exponential number of branches.
The presented "on-line" interpretation of overlap game graphs is exploited in the proof of Lemma 7 presented in the next section.It may also provide a useful insight into our arguments in Section 4, which are formulated using the static definition of an overlap game graph.

Overlap game graphs with large chromatic number
For a family J of intervals and a point x ∈ R, let J (x) denote the family of those intervals in J that have non-empty intersection with [x, ∞).
Proof of Lemma 7. According to what has been told in the previous section, any strategy of Presenter forcing the use of many colors in the overlap coloring game yields an overlap game graph with large chromatic number.We show that for any k ∈ N and any interval R, Presenter has a strategy Σ(k, R) to present a family of intervals J and a point x ∈ R such that (i) all intervals in J are contained in the interior of R, (ii) the overlap graph of J is triangle-free, (iii) J (x) is a chain of nested intervals, (iv) Painter is forced to use at least k colors on the intervals in J (x).
In the strategy Σ(1, R), Presenter just puts a single interval in the interior of R.This clearly satisfies the conditions above.Now, let k 2, and assume by induction that we have already constructed the strategy Σ(k − 1, R) for every R.The strategy Σ(k, R) looks as follows.First, Presenter plays the strategy Σ(k − 1, R) constructing a family J and a point x ∈ R such that J (x) is nested and Painter is forced to use at least k − 1 colors on the intervals in J (x).Next, Presenter continues with the strategy Σ(k − 1, R ′ ), where R ′ = J (x) ∩ [x, ∞), constructing another family J ′ and a point x ′ ∈ R ′ such that J ′ (x ′ ) is nested and Painter is forced to use at least k − 1 colors on the intervals in J ′ (x ′ ).Suppose that the sets of colors used by Painter on J (x) and J ′ (x ′ ) differ.It follows that Painter has used at least k colors on the family J (x) ∪ J ′ (x ′ ) = (J ∪ J ′ )(x ′ ).Thus Presenter has achieved the goal of the strategy Σ(k, R) with J ∪ J ′ and x ′ .Now, suppose that Painter has used the same sets of colors on J (x) and J ′ (x ′ ).Presenter puts an interval J contained in the interior of J (x) ∩ [x ′ , ∞) and overlapping all intervals in J ′ (x ′ ).The J (x) overlap graph remains triangle-free as the family of neighbors of J, which is J ′ (x ′ ), is nested.The family J (x) ∪ {J} is nested as well, and we have J (x) ∪ {J} = (J ∪ J ′ ∪ {J})(y) for any point y in the interior of J and to the right of all intervals in J ′ (x ′ ).Painter has to color J with a color distinct from those of the intervals in J ′ (x ′ ) and hence is forced to use at least k colors on J (x) ∪ {J}.The presented family J ∪ J ′ ∪ {J} and the point y witness that the goal of the strategy Σ(k, R) has been achieved.One can calculate that Presenter playing according to the strategy Σ(k, R) always presents at most 2 k −1 intervals and each position in Σ(k, R) has at most k children.Hence k = Ω(log log n), where n is the number of vertices of the overlap game graph defined by Σ(k, R).
The smallest overlap game graph with chromatic number k that we know is obtained by playing according to the strategy Σ(k − 1, R) and finally putting one additional interval inside [x, ∞) that overlaps all intervals in J (x).

Coloring triangle-free overlap game graphs
For the purpose of this entire section, let G be an n-vertex triangle-free overlap game graph with meta-forest M and representation µ.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.
The relation ≺ defines an orientation of the edges of G: we write x → y if x is adjacent to y and x ≺ y.We classify each vertex v of G as either primary or secondary as follows: , where u is the parent of v, then v is secondary; otherwise v is primary.
Let P (u) ⊂ V (G) be the set of vertices v for which u is the first primary vertex on the path from v to the root of M (including u itself).Clearly, P (u) is independent in G and µ(v) ⊂ µ(u) for v ∈ P (u).Let P (u) → P (v) denote that u ≺ v and there are x ∈ P (u) and y ∈ P (v) such that x → y.
We are going to show that any proper k-coloring of the primary vertices of G can be transformed into a proper 2k-coloring of the whole graph G.This can be done with the help of the following lemma.Lemma 8. Let S be an independent set of primary vertices in G, and let v ∈ S.There is at most one vertex u ∈ S such that P (u) → P (v).
We now show how to color the vertices of G with 2k colors.Let S be a color class in a proper k-coloring of the primary vertices of G. Consider all the sets P (u) for u ∈ S. Each of these sets is independent, and by Lemma 8, the edges between the sets form a bipartite graph.Therefore, we need just two colors for the vertices in u∈S P (u) and 2k colors in total for the whole G.
It suffices to show that the chromatic number of the graph induced on the primary vertices is O(log log n).The primary vertices induce an overlap game graph witnessed by the restriction of µ and ≺.From now on, we simply assume that there are no secondary vertices in G.
Let P be the set of (primary) vertices lying on a path from the root to a leaf of M .We say that two edges x → y and z → t with x, y, z, t ∈ P overlap if either x ≺ z ≺ y ≺ t or z ≺ x ≺ t ≺ y.The following lemma essentially means that the set P induces a forest in G with no overlapping edges.
To continue the proof we need to introduce the idea of heavy-light decomposition due to Sleator and Tarjan [14].Let T be a rooted tree, and let T v denote the subtree of T rooted at a vertex v.For every internal vertex u of T , let s(u) be a child v of u such that the number of vertices in T v is as large as possible.The edges of T connecting u to s(u) are called heavy.Clearly, there is a unique root-to-leaf path in T consisting of heavy edges only-call it the heavy path of T .Remove the heavy path from the tree, obtaining a forest.Continue by removing heavy paths from each tree in the forest, until there is nothing left to remove.The resulting vertex cover of T by paths is called the heavy-light decomposition of T .We will call each path in this cover a heavy path of T .The heavy-light decomposition has the following crucial property.

Lemma 10. If a root-to-leaf path in T intersects k heavy paths, then T has at least
Fix a heavy-light decomposition of M .Form an auxiliary graph G ′ by removing the edges of G that connect two vertices in different heavy paths.By Lemma 9 (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 C 1 and C 2 be the coloring classes in a proper two-coloring of G ′ , and fix i ∈ {1, 2}.For any x, y ∈ C i , the following holds: ( * ) If x → y, then x and y are in different heavy paths.
We color the vertices from C i with positive integers one at a time, going from the root of M towards the leaves.We choose the color of a vertex v ∈ C i to be the least positive integer not occurring as a color of any vertex w with w ∈ C i and w → v (every such vertex has been colored before v).This coloring strategy is known as first-fit.Lemma 11.If first-fit assigns a color k to some vertex v ∈ C i , then the path P in M from the root to v intersects at least k−2 heavy paths.
Proof.Let f (w) denote the color chosen by first-fit for each vertex w ∈ C i .The colors 1, . . ., f (v) − 1 have been chosen for vertices w ∈ P ∩ C i with w → v, so there are at least f (v) − 1 such vertices.Let F be a minimal subset of P ∩ C i that satisfies the following conditions (see Fig. 6): • for any w ∈ F , there are w 1 , . . ., w f (w)−1 ∈ F such that w j → w and f (w j ) = j for 1 j f (w) − 1.
By Lemma 9 (1), the set F induces a directed tree T in G with the root v and edges directed towards v.By the minimality of F , each vertex w of T has exactly f (w) − 1 children (which are closer in M to the root of M ).It follows that T has exactly 2 k−1 vertices and 2 k−2 internal vertices.
Let w be an internal vertex of T , and let u ∈ F be the vertex that precedes w in the order ≺ on F .We claim that u is a child of w in T .Suppose it is not.Since u precedes w in the order ≺ and therefore is not the root of T , it has a parent p in T .Since w is an internal vertex of T , it has a child c in T .We know that w ≺ p and c ≺ u, as u and w are consecutive in the order ≺ on F .It follows that c ≺ u ≺ w ≺ p and the edges c → w and u → p overlap, which contradicts Lemma 9 (2).We have shown that there is an edge between u and w in T , so they must lie on different heavy paths.Consequently, any two ≺-consecutive internal vertices of T lie on different heavy paths, which shows that P intersects at least 2 k−2 heavy paths.
Proof of Lemma 6.As noted previously, we may assume that G consists of primary vertices only and is connected.Suppose that k is the maximal color used by first-fit on a vertex v ∈ C i .By Lemma 11, the path in M from the root to v intersects at least 2 k−2 heavy paths.This implies, by Lemma 10, that n 2 2 k−2 − 1.Therefore, first-fit uses at most O(log log n) colors on C i .We color C 1 and C 2 by first-fit using two separate sets of colors, obtaining a proper coloring of G with O(log log n) colors.Proof of Lemma 5. Let F be a directed family of rectangular frames.We assume without loss of generality that F is rightward-directed.Define a map µ : F → I so that µ(F ) is the interval obtained by projecting F on the x-axis.Thus we have ℓ(µ(F )) = ℓ(F ) and r(µ(F )) = r(F ).
For F ∈ F, let L(F ) be the subfamily of F consisting of such We define a rooted forest M on F as follows.If L(F ) is empty, then F is a root of M .Otherwise, the parent of F in M is the member F ′ of L(F ) with greatest ℓ(F ′ ).We show that the intersection graph of F is an overlap game graph with meta-forest M and representation µ.To this end, we argue that the conditions (G1)-(G3) from the definition of an overlap game graph are satisfied by the intersection graph of F.
It follows directly from the definition of parent that . This already shows (G1) and the right-to-left implication in . However, such a configuration is forbidden in a clean family of frames.This contradiction shows (G3).Now, let F 1 and F 2 be two intersecting members of F. By the assumption that F is rightward-directed, µ(F 1 ) and µ(F 2 ) overlap, and we have F 1 ∈ L(F 2 ) or F 2 ∈ L(F 1 ).Therefore, in order to prove the left-to-right implication in (G2), it remains to show that F 1 ∈ L(F 2 ) implies F 1 ≺ F 2 .To this end, we use induction on the increasing order of ℓ(F 2 ).There is nothing to prove when F 1 is the parent of F 2 , so assume the other case.Let F ′ 2 be the parent of F 2 .We have . This and the induction hypothesis yield F 1 ≺ F ′ 2 and thus F 1 ≺ F 2 .It remains to prove that each overlap game graph is (isomorphic to) the intersection graph of a directed family of frames.Let G be an overlap game graph with meta-tree M and representation µ.For every vertex v of G we define a rectangular frame , and therefore F (u) and F (v) intersect if and only if µ(u) and µ(v) overlap.Moreover, when F (u) and F (v) intersect, this intersection is rightward-directed.A straightforward argument shows that no frame is enclosed by any two intersecting frames.Consequently, {F (v) : v ∈ V (G)} is a clean rightward-directed family of rectangular frames and its intersection graph is isomorphic to G.

Reduction to clean directed families of frames
The goal of this section is to prove Lemmas 3 and 4. 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 do not necessarily agree).Let ξ(F) denote the minimum size of such a partition of a family of frames F. It is enough to bound ξ(F), because we can gather the connected components of a partition class that have the same direction, thus turning each partition class into four directed families.This way we will obtain the bound of 4ξ(F) on the size of a partition of F into directed families.
Two families of frames F 1 and F 2 are mutually independent if there are no two intersecting frames F 1 ∈ F 1 and F 2 ∈ F 2 .The following properties of ξ (analogous to the properties of chromatic number) are straightforward: We will use them implicitly in the rest of this section.Also, whenever we consider a partition of a family into a number of subfamilies, we allow these subfamilies to be empty.
Recall that two frames cross when both vertical sides of one frame intersect both horizontal sides of the other frame.A family of frames F is non-crossing if it contains no pair of crossing frames.
Lemma 12 (implicit in [2]).Every family of frames F with ω(F) k can be partitioned into k non-crossing subfamilies.
Proof.Define a partial order < on F so that F 1 < F 2 whenever F 1 and F 2 cross so that the vertical sides of F 1 intersect the horizontal sides of F 2 .The graph of crossing pairs of frames in F is the comparability graph of <, so it has chromatic number equal to its clique number, which is at most k.Hence the lemma follows.
Let F be a family of frames.For L ⊂ F, a frame Lemma 13.Every family of frames F has a partition P into layers.Moreover, P can be partitioned into two subclasses (possibly empty) each consisting of mutually independent layers.Proof.When F is not connected, we can construct the layers and the bipartition for each component independently.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 L d = ∅.We claim that P = {L 0 , . . ., L d } is a partition of F into layers.
We have |L 0 | = 1, so L 0 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 F 0 F 1 . . .F j be a path from F 0 = R to F j = F of length j.Let i be greatest such that F i is external to L j (such i exists, as R is external to L j ).Since F i intersects F i+1 , which is not external to L j , F i must intersect some frame F ′ ∈ L j .This witnesses a path from R to F ′ of length i + 1, so i + 1 = j.Therefore, F i 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.Therefore, the families {L j : j is odd} and {L j : j is even} consist of mutually independent layers.Corollary 14.For m 1, every family of frames F has a partition P into m-fold layers.Moreover, P can be partitioned into 2 m subclasses (possibly empty) each consisting of mutually independent m-fold layers.
Proof.We proceed by induction on m.For m = 1, this is exactly the statement of Lemma 13.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 2 m−1 subclasses P 1 , . . ., P 2 m−1 (possibly empty) each consisting of mutually independent layers.We apply Lemma 13 to construct a partition P L of each (m − 1)-fold layer L ∈ P into layers, which are now m-fold layers of F, and a partition of P L into two subclasses P L 1 and P L 2 (possibly empty) each consisting of mutually independent layers.We set P = L∈P P L .The desired partition of P is formed by the families Theorem 15 (Asplund,Grünbaum [2]).Let R be a family of axis-aligned rectangles with ω(R) k.It follows that χ(R) α k , where α k depends only on k.
The precise bound of Theorem 15 in [2] is 4k 2 − 3k.This was improved to 3k 2 − 2k − 1 by Hendler [7].However, we are going to apply Theorem 15 to non-crossing families R only, for which the bound in [2] is 4k − 3.
A curve is x-semimonotone if its intersection with every vertical line is a connected interval (possibly empty).
Theorem 16 (Suk [15]).Let V be a vertical line.Let  The actual theorem in [15] 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 graphs.The statement of Theorem 16 for k = 2 but without the x-semimonotonicity assumption has been proved by McGuinness [10].Lasoń et al. [9] have generalized Theorem 16 removing the x-semimonotonicity assuption for any k.
For the rest of this section, assume that α k and β k are as in the statements of Theorems 15 and 16.Define constants γ k and δ k,m for 1 m k by induction on k and m, as follows: We partition S into four families S 1 , S 2 , S 3 and S 4 so that every frame in S 1 , S 2 , S 3 or S 4 intersects the left, right, bottom or top side of B 1 , respectively.Next, we partition L into four families L 1 , L 2 , L 3 and L 4 so that every frame in L i intersects or is enclosed by some frame in S i .We have ξ(L) ξ(L 1 )+ξ(L 2 )+ξ(L 3 )+ξ(L 4 ).We show that ξ(L i ) (m−1)(β 4 k +2γ k−1 +δ k,m−1 ).In the following, we assume without loss of generality that i = 1, that is, we are to bound ξ(L 1 ).
We partition L 1 into three families X , Y and Z as follows.Fix a frame F ∈ L 1 .If there is a frame in S 1 that encloses F , then we put F in X .Otherwise, if there is a frame in S 1 that encloses the entire top or bottom side of F , then we put F in Y.If neither of the above hold, then we put F in Z.
The intersection graph of S 1 is isomorphic to the intersection graph of the filling rectangles of the frames in S 1 , so it is an interval graph, as all these rectangles intersect the vertical line containing the left side of B 1 .Hence ξ(S 1 ) = ω(S 1 ) m−1.Therefore, we can partition 1 into m − 1 families S 1 , . . ., S m−1 each consisting of frames with pairwise disjoint filling rectangles.We also partition each of X , Y and • if F ∈ Y j , then the bottom or top side of F is enclosed by a frame in S j , • if F ∈ Z j , then F intersects a frame in S j and none of the above holds.
We show that ξ(X j ) δ k,m−1 , ξ(Y j ) 2γ k−1 and ξ(Z j ) β 4 k for 1 j m − 1.Once this is achieved, we will have ξ(L 1 ) First, we show that ξ(X j ) δ k,m−1 .We partition X j into |S j | families X S for S ∈ S j so that if F ∈ X S , then F is enclosed by S. The families X S for S ∈ S j are mutually independent.Choose any S ∈ S j .Since X S ∪ {B 1 , . . ., B m−1 , S} is a subfamily of M, it is an (m − 1)-fold layer of F. Therefore, by the induction hypothesis, ξ(X S ) δ k,m−1 .This yields ξ(X j ) = max S∈S j ξ(X S ) δ k,m−1 .Now, we show that ξ(Y j ) 2γ k−1 .We partition Y j into two families Y b j and Y t j so that if F ∈ Y b j , then the bottom side of F is enclosed by some frame in S j , while if F ∈ Y t j , then the top side of F is enclosed by some frame in S j .We show that ω(Y Let F be the frame in K with minimum 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| ω(F) − 1 k − 1.The proof that ω(Y t j ) k − 1 is analogous.Finally, we show that ξ(Z j ) β 4 k .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 by 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 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 .
All short lower (upper) traces can be connected to the left side of B 1 by pairwise disjoint x-monotone curves inside the frames in S j , thus forming a family of x-monotone 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 16, there are proper colorings φ ℓ and φ u of all short lower and short upper traces, respectively, with β k colors.To prove that ξ(Z j ) β 4 k , it is enough to show that ξ(Z ′ ) β 2 k for any family Z ′ ⊂ Z j of frames whose short lower traces have the same color in φ ℓ and whose short upper traces have the same color in φ 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 ψ ℓ and ψ u of all long lower and long upper traces, respectively, with β k colors.The intersection of any two frames in Z ′ whose lower long traces are disjoint and whose upper long traces are disjoint is leftward-directed.Therefore, the frames in Z ′ whose long lower traces have the same color in ψ ℓ and whose long upper traces have the same color in ψ u form a leftward-directed family.This shows that ξ(Z ′ ) β 2 k .Lemma 18.Let F be a non-crossing family of frames with ω(F) k.It follows that ξ(F) γ k .
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.
By Corollary 14, F has a partition P into k-fold layers, and moreover, P can be partitioned into 2 k subclasses P 1 , . . ., P 2 k each consisting of mutually independent k-fold layers.We show that ξ(L) α k δ k,k for any k-fold layer L ∈ P. Once this is done, we will have ξ( P i ) α k δ k,k (because each P i consists of mutually independent k-fold layers) and thus ξ(F) Let L ∈ P. Let R be the family of those frames in L that are not enclosed by any other frame in L. Every frame in L R is enclosed by some frame in R. Hence L R can be partitioned into |R| families L R for R ∈ R so that for every frame in L R is enclosed by R. Since L R ∪ {R} is a k-fold layer of F, it follows from Lemma 17 that ξ(L R ) δ k,k .The intersection graph of R is isomorphic to the intersection graph of the filling rectangles of the frames in R, so Theorem 15 yields χ(R) α k .Hence R can be partitioned into α k subfamilies R 1 , . . ., R α k each consisting of frames with pairwise disjoint filling rectangles.For 1 j α k , the families R j and L R for R ∈ R j are mutually independent, so ξ(R j ∪ R∈R j L R ) = max{ξ(R j ), max R∈R j ξ(L R )} δ k,k .Therefore, ξ(L) Lemma 3. As it has been explained at the beginning of this section, it is enough to bound ξ(F) for families of frames F with bounded clique number.Let F be a family of frames with ω(F) k.By Lemma 12, we can partition F into k non-crossing families F 1 , . . ., F k .By Lemma 18, we have ξ(F i ) γ k for 1 i k.Hence ξ(F) k i=1 ξ(F i ) kγ k .Proof of Lemma 4. Let F be a triangle-free family of frames.By Lemma 13, F has a partition P into layers, and P can be further split as P = P 1 ∪ P 2 so that each of P 1 and P 2 consists and A. Pawlik were supported and B. Walczak was partially supported by Ministry of Science and Higher Education of Poland Grant 884/N-ESF-EuroGIGA/10/2011/0 under ESF EuroGIGA project GraDR.B. Walczak was partially supported by Swiss National Science Foundation Grant 200020-144531.

Figure 6 . 4 5.
Figure 5 C be a family of x-semimonotone curves contained in the right half-plane delimited by V such that |V ∩ C| = 1 for any C ∈ C, |C 1 ∩ C 2 | 1 for any distinct C 1 , C 2 ∈ C, and ω(C) k.It follows that χ(C) β k , where β k depends only on k.
1 ), and γ k = 2 k α k δ k,k .We are going to prove the bound ξ(F) kγ k for families of frames F with ω(F) k.Lemma 17.Let F be a non-crossing family of frames with ω(F) k, let 1 m k, let B 1 , . . ., B k−m+1 be pairwise intersecting frames, and let L ⊂ F be a family of frames enclosed by each of B 1 , . . ., B k−m+1 such that L ∪ {B 1 , . . ., B k−m+1 } is an m-fold layer of F. Suppose that for any subfamilyL ′ ⊂ L with ω(L ′ ) k − 1 we have ξ(L ′ ) α k−1 .It follows that ξ(L) δ k,m .Proof.The proof goes by induction on m.If m = 1, then the assumption that L ∪ {B 1 , . . ., B k } is a layer of F yields L = ∅ and thus ξ(L) = 0 = δ k,1 .Indeed, if F ∈ L, then there would be a frame F ′ external to L ∪ {B 1 , . . ., B k } and intersecting F .Hence {B 1 , . . ., B k , F } would be a (k + 1)-clique in F.Now, assume that 2m k and the lemma holds for m − 1.We show that ξ(L)4(m − 1)(β 4 k + 2γ k−1 + δ k,m−1 ) = δ k,m .Since L ∪ {B 1 , . . ., B k−m+1 } is an m-fold layer of F, there is an (m − 1)-fold layer M of F such that L ∪ {B 1 , . . ., B k−m+1 } is a layer of M.Let S be the family of those frames in M that intersect all of B 1 , . . ., B k−m+1 and are not enclosed by any other frame in M intersecting all of B 1 , . . ., B k−m+1 .It follows that ω(S) m − 1, and two frames in S intersect if and only if their filling rectangles intersect.Every frame in L intersects or is enclosed by some frame in S.