1. Introduction
Coloring of Rectangles
Let be a family of axisparallel rectangles in the plane. The chromatic number of , denoted by , is the minimum number of colors that can be assigned to the rectangles so that any two intersecting rectangles receive different colors. The clique number of , denoted by , is the maximum size of a set such that any two rectangles in intersect. These two terms are equivalent to the chromatic number and the clique number of the intersection graph of . Since , a natural question is whether can be bounded from above in terms of . This question in various graph classes has received a lot of attention from discrete mathematics community, and it has also played crucial roles in the theory of algorithms and mathematical programming.
In general, it is well known that trianglefree graphs (that is, graphs with clique number ) can have arbitrarily large chromatic number [15]. Classes of graphs that admit a function bounding in terms of for every are called bounded. There has been immense progress in the study of bounded classes of graphs in recent years—see the survey by Scott and Seymour [30] and the references therein. In particular, various classes of geometric intersection graphs are known to be bounded or not bounded—see e.g. [24, 28, 29]. The history of this question for rectangle intersection graphs dates back to 1948, when Bielecki [5] asked whether trianglefree rectangle graphs have bounded chromatic number. Asplund and Grünbaum [4] not only answered this question in the positive but also showed a more general bound of . This bound was later improved by Hendler [21] to . Kostochka [24] constructed rectangle families with , and this remains the best known lower bound. Chalermsook [7] proved the bound for the special case that contains no nested pair of rectangles. Closing or even narrowing down the gap between the linear lower bound and the quadratic upper bound for the general families of rectangles has been a longstanding open problem.
Maximum Weight Independent Set of Rectangles (MWISR)
In MWISR, we are given a family of axisparallel rectangles in the plane together with weights assigned to them, and we aim at finding a maximum weight subfamily (called an independent set or a packing) that contains no two intersecting rectangles. Besides being a fundamental problem in geometric optimization, MWISR is interesting from several perspectives. First, it arises in various applications, including map labeling [3, 14], resource allocation [27], data mining [26, 22, 19], and unsplittable flow routing [6]. Second, it is one of the “somewhat tractable” special cases of the general maximum weight independent set problem: given an vertex graph with weights on the vertices, find a maximum weight subset of the vertices containing no two vertices connected by an edge. This problem for general graphs is NPhard to approximate to within a factor of for every [20, 31], with the best known approximation factor being [17]. In special graph classes defined by intersections of geometric objects (such as disks, squares, and more generally—fat objects), polynomialtime approximation schemes (PTASes) are known [10, 16]. Rectangles are perhaps the simplest natural objects for which the maximum independent set problem is not known to admit a PTAS. MWISR is NPhard [18] and there have been active attempts in the past decade from various groups of researchers on obtaining approximation algorithms. The best known approximation factor is by Chan and HarPeled [11]. In the unweighted case, Chalermsook and Chuzhoy [8] presented an approximation. Recently, a quasipolynomialtime approximation scheme (QPTAS) was presented by Adamaszek, HarPeled, and Wiese [1, 2] (see also an improvement on the unweighted case by Chuzhoy and Ene [13]). Obtaining a PTAS or even a polynomialtime constantfactor approximation for MWISR remains an elusive open problem.
Connections between Coloring and MWISR
These two problems are related through the perspective of mathematical programming. In particular, consider the cliqueconstrained independent set polytope of a graph :
For a graph
and a weight vector
, let and , the latter being the maximum weight of an independent set in with respect to the weights . Clearly, . The integrality ratio (or integrality gap) is the ratio . Since a fractional solution with value can be found efficiently^{1}^{1}1In general graphs, it can be computed via SDP, as a solution optimizing over the Lovász theta body of ., rounding this LP solution is a natural algorithmic paradigm for approximating the maximum weight independent set problem, especially in restricted graph classes.The integrality ratio of has a strong connection to certain Ramseytype bounds. More formally, let be any graph class that is closed under clique replacement operation^{2}^{2}2This holds for various natural graph classes such as perfect graphs and geometric intersection graphs.. When (the unweighted case), proving the upper bound for all , is equivalent to proving the upper bound on the Ramsey numbers^{3}^{3}3The Ramsey number is the minimum integer such that every vertex graph contains a clique of size or an independent set of size . for all graphs in the same graph class . When allowing an arbitrary weight function , proving for all , is equivalent to upper bounding the ratio for all . These connections are constructive [9]. Therefore, one way to design an efficient approximation algorithm for the maximum independent set problem in any graph class is to prove an (algorithmic) upper bound on for graphs in the same graph class.
The polytope
has played crucial roles from both algorithms and mathematical optimization perspectives; a notable example is its application to finding maximum cliques and independent sets in perfect graphs. It is particularly appealing for rectangle intersection graphs
, which have only maximal cliques. For these graphs, an LP over can be explicitly written and solved by a nearlineartime algorithm [12]. Therefore, it is an interesting question on its own to pinpoint the value of for rectangle graphs.Our Contributions
First, we present the following improvement on the coloring bound of Asplund and Grünbaum [4].
Theorem 1.
Every family of axisparallel rectangles in the plane with clique number is colorable, and an coloring of it can be computed in polynomial time.
Second, via a simple reduction, we obtain the following result for MWISR. We remark that the reduction was used implicitly in the paper of Chalermsook and Chuzhoy [8].
Theorem 2.
There is a polynomialtime approximation algorithm for MWISR, and the integrality ratio of the cliqueconstrained LP for rectangle graphs is at most .
This result improves upon the approximation by Chan and HarPeled [11] for MWISR. It also substantially simplifies and derandomizes the known approximation in the unweighted setting [8]. The bound on the integrality ratio combined with a fast LP solver from [12] imply that an estimate on the value of MWISR can be computed in time.
The main new technical ingredient of this paper is a “hierarchical decomposition” of a family of rectangles, inspired by the work of Kierstead and Trotter [23]. In section 3, we present a small “warmup” result that highlights the main idea behind this decomposition. In section 4, we define the decomposition and use it to prove Theorem 1. We present the proof of Theorem 2 in section 5.
2. Preliminaries
Definitions
A rectangle is a closed set of the form in the plane, where and . The width and the height of such a rectangle are the values and , respectively.
Let be a family of rectangles. The intersection graph of has vertex set and edge set defined as follows: two rectangles are connected by an edge if they intersect, that is, if . A subfamily of is a clique if the intersection of all rectangles in is nonempty. This is the same as to say that every pair of rectangles in intersects, so this notion of a clique corresponds to a clique in the intersection graph of . We say that a clique in contains a point if belongs to every rectangle in . We let denote the clique number of , that is, the maximum size of a clique in . A subfamily of is an independent set if the rectangles in are pairwise disjoint. A coloring of is an assignment of colors to the rectangles in such that the rectangles of any given color form an independent set. These notions correspond to independent sets and colorings of the intersection graph of . We say that is colorable if there is a coloring of using colors. We let denote the chromatic number of , that is, the minimum number such that is colorable.
Let and be two rectangles in the plane that intersect (). We distinguish several possible types of intersections; see Figure 1. If contains at least one corner of or vice versa, then we have a corner intersection between and . Otherwise, we have a crossing intersection between and , and we say that and cross. We have a containment intersection when one rectangle contains the other (which is a particular case of a corner intersection). We have a vertical intersection if one rectangle intersects both the top and the bottom sides of the other.
Let us fix a family of rectangles that is the input to our problem. For each rectangle , let denote the rectangles in that intersect both the bottom and the top sides of , and let denote the rectangles in that cross . Thus, if and the height of is greater than the height of , then the following holds:

there is a vertical intersection between and if and only if ;

there is a crossing intersection between and if and only if .
It is important to observe that if , then .
Preliminary Results
A family of rectangles is sparse if one can fix points in each rectangle so that the intersection of any two crossing rectangles contains at least one of the points . For instance, a family of squares is sparse, because no pair of squares can cross. The following lemma is slightly modified from [27, 7].
Lemma 3.
For each , every sparse family of rectangles with clique number is colorable, and such a coloring can be computed in polynomial time.
Proof.
Let be an sparse family of rectangles with clique number . It suffices to show that the number of edges in the intersection graph of is strictly less than , because this implies that there is a vertex of degree less than in every induced subgraph, which leads to a coloring by straightforward induction.
For each edge in the intersection graph, if and cross, we give one token to one of the points that lies in . Otherwise, corresponds to a corner intersection, which involves at least two corners (four in case of containment and two otherwise; see Figure 1), and we give half of a token to any two corners involved in the intersection. Clearly, the total number of tokens handed out is equal to the number of edges in the intersection graph. Moreover, for each , each point receives at most tokens, and each corner of a rectangle receives at most tokens (with some corners receiving strictly fewer tokens). Therefore, the number of edges is less than . ∎
Corollary 4.
Every family of rectangles with no crossing intersections and with clique number is colorable, and such a coloring can be computed in polynomial time.
The following result of Asplund and Grünbaum [4] will be used as a subroutine.
Lemma 5.
Every family of rectangles with clique number is colorable, and such coloring can be computed in polynomial time.
Proof.
Let be a family of rectangles with clique number . Let be the strict partial order on defined so that if and only if . Every chain in the poset is a clique in , so the height of is at most . Therefore, can be partitioned into antichains in the poset , that is, families with no crossing intersections. By Corollary 4, each of the families is colorable, so the entire family is colorable. ∎
3. WarmUp
In this section, we present two simple coloring results that capture some of the key ideas behind our general bound.
Proposition 6.
Let be a family of rectangles.

If there are only crossing and containment intersections within , then .

If there are only vertical intersections within , then .
Proof.
Let be a family of rectangles with only vertical intersections, and let . We process the rectangles in in the decreasing order of their heights and put each rectangle into one of the sets as follows. We put a rectangle into where is the maximum integer such that there is an witnessing clique for , that is, a clique in containing at least one rectangle from each of the sets . This is well defined—when a rectangle is being processed, the rectangles in have been already processed and distributed to , and is always at most because the aforesaid clique in together with forms a clique in of size at most . Let denote any witnessing clique for a rectangle where is such that .
First, we show that, for each , there can be no crossing or containment intersection between rectangles from a single set . Suppose for the sake of contradiction that two rectangles form a crossing or containment intersection, where is taller than , that is, . If contains , then is an witnessing clique for , contradicting the assumption that . If and cross, then , and is an witnessing clique for , again contradicting the assumption that . See Figure 2.
If there are only containment and crossing intersections between the rectangles in , then are independent sets. This completes the proof of statement (1).
To prove statement (2), we argue below that is still an independent set and that for , which together imply that . For , if two rectangles intersect, where , then is a witnessing clique for , which contradicts the assumption that . Therefore, is an independent set.
Now, fix . We argue that . Let . We claim that at most one other rectangle in contains the right side of . Suppose for the sake of contradiction that two rectangles contain the right side of , where the right side of lies between the right sides of and . Let be the intersection of the rectangles in . If , then ; this implies that is an witnessing clique for , contradicting the assumption that . Thus assume . This implies that lies between the right sides of and , so . Now, there are two cases. If , then is an witnessing clique for , contradicting the assumption that . If , then is an witnessing clique for , contradicting the assumption that . See Figure 3 for an illustration. We have thus proved that at most one rectangle in other than contains the right side of . By symmetry, at most one rectangle in other than contains the left side of .
Now, we present a coloring algorithm for each set . We process the rectangles in in the decreasing order of their heights and color them greedily. When a rectangle is being processed, at most two other rectangles in have been assigned colors (one intersecting the left side and one intersecting the right side of ), so the greedy coloring uses at most three colors. ∎
Proposition 6 (1) is a strengthening of the observation by Asplund and Grünbaum [4] (used in the proof of Lemma 5) that families with only crossing intersections satisfy .
Proposition 6 (2) is closely related to the result of Kierstead and Trotter [23] that families of intervals on the real line with clique number can be colored online using at most colors. Specifically, by a correspondence described in [25], for every deterministic strategy of the adversary in the online coloring problem for intervals, there is an ‘equivalent’ family of axisparallel rectangles (with only vertical intersections) whose chromatic number is equal to the number of colors forced by that strategy against any online coloring algorithm. The decomposition of the family into sets used in the proof above is an adaptation of a decomposition used by Kierstead and Trotter in their proof of the upper bound of for the online problem. Kierstead and Trotter [23] also showed a deterministic strategy of the adversary forcing any online coloring algorithm to use at least using on a family of intervals with clique number . That strategy gives rise to a family of axisparallel rectangles with only vertical intersections and with [25, Proposition 3.1], which shows that the bound in Proposition 6 (2) is sharp.
4. An Coloring Algorithm
Let be a family of rectangles and let . Thus . We show how to construct a coloring of using colors (in polynomial time), which yields Theorem 1.
The argument consists of two steps. In the first step, we construct a “hierarchical decomposition” of similar to the decomposition into sets used in the proof of Proposition 6, but defined with a “divide and conquer” approach rather a simple linear induction. This modification is essential to make it work with the second step—a “clique reduction” argument, which is an adaptation of an argument used before in [7, 8].
Step 1: Hierarchical Decomposition
For , let denote the set of binary words of length , and let denote the empty binary word, so that and for . For each , by induction, we construct a partition of and a nonempty set of witness points for every rectangle .
For , let , and for every rectangle , let be the set of intersection points of the top side of with the left and right sides of the rectangles in (which includes the two top corners of ). See Figure 4 for an illustration.
Now, let , and let . We partition the set into two subsets and , and we define the witness sets for the rectangles , as follows. We consider the rectangles in the order decreasing by height, so that all rectangles in are considered before . For each rectangle (in that order), if some witness point belongs to at least rectangles from that have been already added to , then we add to and let be the set of all such witness points ; otherwise, we add to and let .
Proposition 7.
We remark the following basic properties of this (hierarchical) decomposition.

For each , the family forms a partition of .

For each and each , we have .

For each rectangle , we have . Moreover, if , then .
Next, we prove some crucial properties of this decomposition. We will use them in the design and analysis of our coloring algorithms.
Lemma 8.
For each , each , each rectangle , every witness point in belongs to fewer than rectangles in .
Proof.
The proof goes by induction on . For the base case when , every point in belongs to fewer than rectangles in , because has clique number at most . For the induction step, let and or , where . For the sake of contradiction, suppose a witness point belongs to at least rectangles in . If , then should be taken to instead of , because is a witness point in that belongs to at least rectangles in with height greater than the height of . If , then by the fact that is a witness point in , the point additionally belongs to at least rectangles in , so it belongs to at least rectangles in , contradicting the induction hypothesis. ∎
Lemma 9.
For each , each , any two rectangles such that , and any witness point , there is a witness point that belongs to and to all rectangles in containing .
Proof.
The proof goes by induction on . For the base case when , the claim follows from the fact that when : if is the intersection point of the top side of with the left or right side of some rectangle , then we let be the intersection point of the top side of with the same side of . See Figure 5 for an illustration.
For the induction step, let and or , where . Let be such that . If , then the claim follows directly from the induction hypothesis, as and . Now, suppose . Let be a witness point in . Thus belongs to at least rectangles in . By the induction hypothesis, since , there is a witness point that belongs to all rectangles in containing . In particular, belongs to at least rectangles in and thus in , as . This shows that . ∎
Corollary 10.
For each , each , and any two rectangles such that , there is a witness point that belongs to .
Proof.
Immediate from Lemma 9, as the witness set is always nonempty. ∎
Lemma 11.
For each and each rectangle , every clique in has size at most .
Proof.
For the sake of contradiction, suppose that there is a clique of size greater than in . Let be a point on the top side of that lies in all rectangles in . Let be the rectangles from with leftmost left sides. Let be the rectangles from with rightmost right sides. Let be a rectangle in , which exists as . By Corollary 10 applied to and , there is a witness point such that . If is to the left of , then belongs to all rectangles in (as they contain and their left sides are more to the left than the left side of ), which contradicts Lemma 8. An analogous contradiction is reached for if is to the right of . ∎
Step 2: Clique Reduction
For , an covering of a rectangle is a clique that contains , at least rectangles intersecting the top side of , and at least rectangles intersecting the bottom side of (not necessarily different from those intersecting the top side).
Lemma 12.
Every clique in of size greater than is an covering of one of its rectangles.
Proof.
Let be a clique in of size greater than . Let be the rectangles in with topmost top sides. Let be the rectangles in with bottommost bottom sides. Let be a rectangle in , which exists as . The rectangles in and witness that is an covering of . ∎
For each and each , let be the set of rectangles such that there is a covering of in . Observe that when and that it is easy to compute the sets in polynomial time. By Lemma 12, at least one rectangle from every clique in of size greater than belongs to , so .
Lemma 13.
For each and each , the set is sparse.
Proof.
Let be a rectangle in . We define three points as follows. Let be the leftmost point and be the rightmost point in the witness set . Recall that these points lie on the top side of . Choose a clique that forms a covering of in , and let be a point in the intersection of all rectangles in (which include ).
We verify that these points capture all intersecting pairs of rectangles, as in the definition of sparseness. Consider two crossing rectangles such that . We claim that at least one point of lies in . Suppose, for the sake of contradiction, that this is not the case.
First, observe that cannot lie to the left of : due to Corollary 10, this would mean that some witness point in lies to the left of , contradicting to the choice of . For the same reason, cannot lie to the right of , so it must lie between and .
Since , the point must lie either above or below . Assume that it lies below (see Figure 6; the other case is analogous, by symmetry). Since , there is a covering of in . Let be the rectangles in that intersect the top side of . Thus , by the definition of covering. Since lies above and below every point on the top side of , we have . Lemma 11 yields , so . Each rectangle in fully contains the left or the right side of (or both). Let be the rectangles in containing the left side of , and let be those containing the right side of , so that . It follows that for some . This contradicts Lemma 8, because is a clique in of size at least containing the witness point . ∎
Finally, we present the coloring algorithm. It proceeds in rounds. Initially, we have . At round , we obtain by removing from . After the last round, we are left with the set . Let and be the set of rectangles removed in round . Hence, the families and form a partition of . We argue that each family in this partition can be colored using colors.
For each and each , by Lemma 13, the family is sparse, and since (where is the prefix of in ) and (by the remark before Lemma 13), the maximum size of a clique in is at most . Therefore, by Lemma 3, each set is colorable. Since , by coloring each set where using a separate bunch of colors, we color using colors. Finally, for each , we have , so (by the remark before Lemma 13) , and therefore (by Lemma 5) the set can be colored using colors. Again, using a separate bunch of colors on each set where , we color the family using colors.
5. An Approximation Algorithm for MWISR
In this section, we present a reduction from MWISR to the coloring problem, which leads to a polynomialtime approximation algorithm for MWISR. Similar reductions from the maximum weight independent set problem to the coloring problem have been already used in the literature of approximation algorithms [7, 8, 27]. We also show that the reduction can be made deterministic via a derandomization trick similar to the one used by Chan and HarPeled [11].
Let be a family of rectangles, and for each , let be the weight associated with . Assume that for each (otherwise can be disregarded). Let be the family of inclusionmaximal cliques in . Thus , because the intersection of every such clique is a rectangle whose top left corner is the intersection point of the top side of some rectangle in with the left side of some (possibly the same) rectangle in . Consider the following cliqueconstrained LP relaxation of the maximum weight independent set problem:
maximize  
subject to  
Let be an optimal fractional solution to the LP, and let be the optimum value, which is therefore an upper bound on the maximum weight of an independent set in . Let .
Claim 14.
There is an integral vector such that
Moreover, such a vector can be computed by a polynomialtime deterministic algorithm.
Proof.
If for some , then it is enough to set and for . Therefore, assume henceforth that for every .
For each , let , let
be a random variable in
such that , and let . It follows that for . The LP inequality for a clique yieldswhich implies
where the strict inequality is the following form of the Chernoff bound for a sum of independent zeroone random variables : . For each clique , let a random variable be defined as follows:
so that . Let a random variable be defined as follows:
It follows that
Therefore, there is a choice of where
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