Independent Set of Intersection Graphs of Convex Objects in 2D

Transcription

1 Independent Set of Intersection Graphs of Convex Objects in 2D Pankaj K Agarwal abil H Mustafa Department of Computer Science, Duke University, Durham, C , USA pankaj, Abstract he intersection graph of a set of geometric objects is defined as a graph in which there is an edge between two nodes if he problem of computing a maximum independent set in the intersection graph of a set of objects is known to be!#" - complete for most cases in two and higher dimensions We present approximation algorithms for computing a maximum independent set of intersection graphs of convex objects in $&% Specifically, given (' ) a set of ( line segments in the plane with maximum independent set of size ), we present algorithms that find an independent set of size at least *), (5+),6876% in time :;(=< and 4)+->02@ -(5+),676A in time :*(BAC<D>02FEG(5, ('H' ) a set of ( convex objects with maximum independent set of size ), we present an algorithm that finds an independent set of size at least *),+-02G -(5+) 76< in time :;( <JILK 4 M, assuming that can be preprocessed in time K M to answer certain primitive operations on these convex sets, and (' 'H' ) a set of ( rectangles with maximum independent set of size =(, for PORQ, we present an algorithm that computes an independent set of size S;U%(5 All our algorithms use the notion of partial orders that exploit the geometric structure of the convex objects Introduction An independent set of a graph is a subset of pairwise nonadjacent nodes of the graph he maximumindependent-set problem asks for computing a largest independent set of a given graph Given a graph W and an integer XY\[, determining whether there is an independent set in W of size X is known to be ]_^ -complete even for many restricted cases (eg planar graphs [2], bounded-degree graphs [2], geometric graphs [22]) aturally, the attention then turned toward approximating the largest independent set in polynomial time Unfortunately, the existence of polynomial-time algorithms for approximating the maximum independent set efficiently for general graphs is unlikely [4] However, efficient approximation algorithms are known for many restricted classes of graphs For planar graphs, approximation algorithms exist that can compute an independent set of ` Research is supported by SF grants IR , EIA , EIA-0-305, and CCR

2 \ size arbitrarily close to the size of the maximum independent set ote that a graph is planar if and only if there exists a set of unit disks in the plane whose contacts give the edges of the planar graph [] hus a natural direction is to investigate the independent-set problem for the graphs induced by a set of geometric objects he intriguing question there is whether (and what) geometric nature of objects aids in efficient computation of maximum independent set One such family of graphs arising from geometric objects that have been studied are the so-called intersection graphs Given a set of geometric objects in, the intersection graph of, W is defined as follows: each node! #" corresponds to the object $ and %'&(") if $*+,&- A subset 023 is an independent set in W0 if for every pair of nodes & "4, *) & For brevity, we say independent set of when we mean independent set of the intersection graph of In this paper, we present approximation algorithms for the independent-set problem on intersection graphs of line segments and convex objects in the plane Besides the inherent interest mentioned above, independent sets of intersection graphs have found applications in map labeling in computational cartography [2], and frequency assignment in cellular networks [8] For example, in the map-labeling problem, we are given a set of labels of geometric objects, and the goal is to place the maximum number of labels that are pairwise disjoint Computing the maximum independent set of these labels yields a labeling with the maximum number of labeled objects Related work Given, let denote a maximum independent set of W, and let denote it size We will use to denote when is clear from the context We say that an algorithm computes a : -approximation to 5;6 8 if it computes an independent set of size at least 8=<>: For a general graph W with vertices, there cannot be a polynomial-time approximation algorithm with approximation ratio better than A@CB for any D YR[ unless P = PP [4] Currently the best algorithm for a general graph finds an independent set of size E GFIHKJMLAOP<$Q [7], where is the size of a maximum independent set in W However, for intersection graphs of geometric objects, better approximation ratios are possible If R is a set of intervals in, then the maximum independent set of the intersection graph of R can be computed in linear time Computing 5S6 8 is known to be ]_^ -complete if is a set of unit disks or a set of orthogonal segments in [6] For unit disks in, a polynomial time U7 D, - approximation scheme was proposed in [5] For arbitrary disks, independently Erlebach et al [] and Chan [8] presented a polynomial time U7 DI -approximation scheme he above schemes for computing independent set of disks use shifted dissection, which relies heavily on the fact that the geometric objects are disks (or fat objects) A divide-and-conquer technique is used for the case of the intersection graphs of axis-parallel rectangles in the plane, for which Agarwal et al [2] presented a W HKJMLQ -approximation algorithm in time W XHKJML8Q If the rectangles have unit height, they describe a UY DI -approximation scheme with running time W XHKJML B[@O Berman et al [5] show that a HKJML]\Q -approximation can be computed in W 8A time Recently Chan [] improved these algorithms by describing an algorithm that returns a HKJML \ 8 -approximation in time W \ XHKJML where X`ba is any constant Efficient algorithms are known for other classes of graphs as well [3, 6] In other related work [7], it was shown that the problem of recognizing intersection graphs of 2

3 line segments, ie, given a graph W, does there exist a set of segments whose intersection graph is W, is ]_^ -hard Another work related to ours is of Pach and ardos [20] Given a set of disjoint objects (which could be line segments or convex shapes) in lying on a sheet of glass, they study the following combinatorial question: by iteratively cutting the glass into two separate sheets with a straight line, and then recursively the cutting the two pieces, how many objects can be separated into different sheets of glass Using decomposition schemes similar to ours, they present various upper and lower bounds on the number of objects that can be separated Our results We first present approximation algorithms for a set of segments in, all of which intersect a common vertical line We show that we can compute in W in W time an independent set of of size at least, and HKJML Q time an independent set of of size at least Using these results, we show that for an arbitrary set independent set of size at least Q< #< a#hkjml a#hkjml a>p<> A in time W a>p<> AA in time W, or HKJML Q of segments in, we can compute an We then extend our results 2 to convex sets amely, for a family of convex sets in, we can compute in W A time an independent set of size at least #<>a#hkjml a>p<> A, assuming that certain primitive operations (namely sidedness and pairwise object intersection queries) on these convex sets can be performed by preprocessing in 2 8 time Finally, for a set of rectangles with maximum independent set of size P, for some, we compute in W time, an independent set of size at least E Q Organization In Section 2 we describe approximation algorithms for a set of line segments in Section 3 describes the algorithm for convex sets, and Section 4 presents approximation algorithms for the case of axis-parallel rectangles 2 Approximation Algorithms for Line Segments Let be a set of line segments in Let denote the - and -coordinates of a point ") Let (resp M ) denote the -coordinate of the left (resp right) endpoint of the segment 0", and let denote the slope of $ 2 A -Approximation Algorithm In this section we assume that all the segments in intersect the -axis We also assume that the segments in are sorted in increasing order of their intersection points with the -axis, and we use All logarithms in this paper are base 2 3

4 (a) (b) Figure : (a) Bold segments form an increasing -monotone sequence, (b) '& in solid (c) & \ as in the proof of Lemma 22 to denote this sorted sequence We call a subsequence * (c) of -monotone (see Figure (a)) if for all "!$# X &% # for all '!(#)% (called increasing -monotone) or Y for all "!*#+% (called decreasing -monotone) Lemma 2 Let R sequence R be a subsequence of pairwise-disjoint segments here exists an -monotone R of size at least, R, PROOF By Dilworth s theorem [0], there is a subsequence R of R such that the slopes of the segments are either monotonically increasing or monotonically decreasing, and the size of R is at least is -monotone, R, Since the segments in R are pairwise disjoint, R We describe an algorithm for computing the longest -monotone subsequence of By Lemma 2, its size is at least 8 Without loss of generality, we describe how to compute the longest increasing -monotone subsequence; the same procedure can compute the longest decreasing - monotone subsequence of, and we return the longer of the two We add a segment to such that it intersects -axis below all the segments of, does not intersect any segment of, # for all - `, and it spans all the other segments of (ie, # [ and Y for all - ) We add another similar segment 0 that intersects the -axis above all the other segments in, does not intersect any segment in, and 0 Y for all - 4

5 For [ - #&! 3 segments \ st (S) -# X #"!, (S2) # \ # &, (S3) (S4) * \ \ Y such that C*),&X & A, and & * \ and # &, let _ '& denote the subsequence of See Figure (b) for an illustration of & Let -A!] '& denote the longest increasing - monotone subsequence ofo '& If there is more than one such sequence, we choose the lexicographically minimum one Set -A!]S, -A!, We wish to compute [] ote that by definition of and 0, 0 3 Lemma 22 For all [ &-#"!, -A! -A X X!] 3 () PROOF Let -!] Let be the segment in -!] with the leftmost left endpoint, ie, for all X ote that and & Since each of these two subsequences is -monotone 3 and " '&, -A ]8` and!]8`!" herefore -A! # -A $] ]!] and hence -A! 3 -A X X!] Conversely, let \ " & By definition of \ and \ & (cf ( " \ &, (- ) - # # X # #"!, (-- ) # )' # \ # )($# &, (--- ) ' (- ) ' *G \ ( Y \ Y, and ( * \ & A, &%] ), for all 'b" \ and See Figure (c) As observed in [20], (- ) (- ) imply that the line \ supporting \ does not intersect ' and ( Indeed, (- ) & (-- ) imply that \ does not intersect ' or ( to the right of the -axis, and (--- ) & (- ) imply that \ does not intersect ' or ( to the left of the -axis Since *' and ( lie on the opposite sides of \, they neither intersect each other, nor or & Hence, the segments in -A X,+ X! are pairwise disjoint Moreover, the fact that ' and ( do not intersect or & and -A - imply that ' and ( satisfy -%] for '& Hence \ + \ & '& herefore the sequence -A X \ X!] is an -monotone subsequence of & Hence, -A! `0 -A X 5 X!]

6 W (a) (b) (c) (d) Figure 2: he four types of -monotone sequences described in Lemma 24 his completes the proof of the lemma -A! using a dynamic-programming approach We can compute the set #'& X! have already been -A!] for all -$#! is We can compute in W Q time and -A!] in another W Q time, assuming -A X and computed herefore, the total time spent in computing Putting everything together, we conclude the following heorem 23 Given a set of segments in, all intersecting a common vertical line, an independent set of size at least 8 can be computed in time W 22 A -Approximation Algorithm We now present a faster algorithm at the expense of a larger approximation factor he algorithm again tries to find a large subset of that has a certain special structure, which allows its computation in polynomial time We assume that all the segments of intersect the -axis and are sorted in increasing order by the -coordinates of their left endpoints Let C denote this sequence Let : be the -coordinate of the intersection point of with the -axis Lemma 24 Let R be a subsequence of pairwise-disjoint segments hen there exists a subsequence R, such that R R,, R, `, R,, and it has one of the following properties: For all &-#&%, (L) (L2) (L3) (L4) # # Y Y 0 : # : 0 0 : Y : 0 0 : # : 0 0 : Y : 0 and (Figure 2(a)),, and (Figure 2(b)), and (Figure 2(c)),, and (Figure 2(d)) 6

7 R R PROOF By Dilworth s theorem, there exists a subsequence R R of length at least, R, such that the -coordinates of the right endpoints are either monotonically increasing or monotonically decreasing Again applying Dilworth s theorem to R, one can find a subsequence R of length at least, R, `, R, such that : for " are either monotonically increasing or monotonically decreasing If R is increasing and R increasing (resp decreasing), then the sequence is of type (resp a ) If R is decreasing and R increasing (resp decreasing), then the sequence is of type (resp % ) We refer to a subsequence of pairwise-disjoint segments of that satisfies one of %] property as an -monotone sequence he following property of -monotone sequences allows us to compute them efficiently Lemma 25 Let M be a sequence of segments so that one of the conditions %] is satisfied and * 0 for all - hen is an -monotone sequence PROOF It is clear that for segments of type %], two segments and &, for - #"! intersect without either intersecting 0 or & intersecting herefore if * 0 all -, then the segments are pairwise non-intersecting and hence -monotone, cannot for By Lemma 25, the segments in any sequence satisfying one of %] are pairwise nonintersecting if the adjacent segments do not intersect (See Figure 2) We present an algorithm that, given a sequence of segments, computes the longest -monotone subsequence of each type By Lemma 24, the longest of them is an independent set of size at least 8A We describe an algorithm for computing the longest -monotone subsequence of type he others can be computed analogously Define [& to be the set of segments such that \ " & if (- ) X #"!, (-- ) \ # &, (--- ) : \ # : &, (- ) \ * & Let! be the longest -monotone subsequence of & of type that contains & Set,!, We wish to compute &!] Lemma 26 For "!,! PROOF Let! & Clearly is an -monotone sequence, and " & herefore M@O ` and!] X for all \ " & Conversely, Lemma 25 implies that for any \ " &, the sequence 3 X - & is an -monotone!]8`0 X sequence Hence, X 3!] 7

8 A naive approach takes W Q time to compute each!, provided that X, for all X #"!, have already been computed his yields an W time algorithm We now describe a more sophisticated approach by exploiting geometry to compute each! in W HKJML Q amortized time Let & & We compute!] sequentially for!, maintaining a data structure that stores all the segments of & Given the segment & 0, the data structure returns X Once we have computed!], we insert the segment & together with its weight!] into the data structure ote that after we have inserted!], the data structure stores all the segments in the set4& 0 We now describe, a three-level data structure, that stores a set of weighted segments For a query segment intersecting the -axis, it returns the maximum weight of a segment in so that 2 +*, and intersects the -axis above he first-level is a balanced binary search tree on the -coordinates of the right endpoints of the segments in Let denote the canonical subset of segments stored in the subtree rooted at " 8 For each node, the second-level data structure is a balanced binary search tree on :, " ] Let denote the set of segments stored in the subtree rooted at " ] Finally, for the set of segments, we construct a segment-intersection data structure as described in [] It stores a family of canonical subsets of in a tree-like structure he total size of the data structure is W HKJML Q, and it can be constructed in W HKJML Q time For a query segment, we can report in W HKJML Q time the segments of not intersecting as a union of W HKJML Q canonical subsets For each canonical subset, we store the maximum weight of a segment in he overall data structure can be constructed in time W HKJML Q For a query segment, we wish to report, where the maximum is taken over all segments of so that 2, *, and the intersection point of with the -axis lies below that of We query the first-level tree with the right endpoint of and identify a set of W HKJMLQ nodes so that is the set of segments whose right endpoints lie to the left of 2 ext, for each ", we compute a set 2 of W HKJMLYQ nodes st * is the set of segments of that intersect the -axis below does For each " O, we compute the maximum weight of a segment in that does not intersect, using the third-level data structure We return he total time spent is W HKJML Q Finally, we can use the standard dynamization techniques by Bentley and Saxe [4] to handle insertions in the data structure Since the data structure can be constructed in W HKJML Q time, the amortized insertion time is W HKJML 0 Q However, in our applications, we know in advance all the segments that we want to insert they are the segments of So we can somewhat simplify the data structure as follows Set the weight of all segments to [, and construct on all the segments of When we wish to insert a segment, we update its weight and update the weight of appropriate canonical subsets at the third level of Omitting all the details, we conclude the following heorem 27 Given a set of segments in, all intersecting a common vertical line, one can compute an independent set of size at least 8 in time W HKJML ]Q 8

9 ` ` ` & & & & 23 Independent Set for Arbitrary Segments Let be a set of arbitrary segments in We describe a recursive algorithm for computing an independent set of Let be the vertical line passing through the median -coordinate of the right endpoints of segments in, ie, at most P<>a segments have their right endpoints on each side of We partition into three sets, P and 6 (resp ) is the subset of segments that lie completely to the left (resp right) of, andy6 is the subset of segments whose interiors intersect We compute an independent set R 6 of 6, and recursively compute an independent set R (resp R ) of (resp ) Since the segments in do not intersect any segments in, R + R is an independent set of +) We return either R 6 or R +GR, whichever is larger Suppose our algorithm computes an independent set of size at least #, where b 8 Let,,,, and Q6 6 Suppose the algorithm for computing an independent set of 6 returns a set of size at least 6 hen # ` 2 P 6 A where functions, it can be argued that # 8` Q60`, P 3P<>a, 4`, and `3 Since and are sub-linear P<>a We now show that the solution to the above recurrence is # 8` Expanding (2), we obtain # 8` where is the depth of recursion and 6 6 A (2) a#hkjml a>p<> P<>a A P<>a P<>a & A A Moreover, for all!, a & ` A (3) We claim that ` a#hkjml a>p<> Indeed if it were not true, then 0` a#hkjml a>p<> hen for! HKJML a>p<>, P<>a Q<>a, while & & 2 Y ' ' Q<>a, thereby contradicting (3) he proof then follows

10 R (a) (b) (c) a Figure 3: (a) Sequences of type (bold) and (dashed) (b) Sequence of type, (c) #'& (solid) If we can compute the independent set of6 in time Q, then the running time of the algorithm is W Q HKJMLYQ If Q ` 0_B, then the running time is W QA Hence we conclude the following heorem 28 Let be a set of % segments, all of which intersect a common vertical line Suppose we can compute an independent set of of size at least 0A in time % hen for any set of segments in, we can compute an independent set of size at least #<>a#hkjml a>p<> A where 4 8 he running time is W QA if Q ` 0_B, and W Q HKJMLQ otherwise Corollary 2 For a set of segments in, one can compute an independent set of size at least (- ) Q<>a#HKJML a>p<> A U in time W and (-- ) Q<>a#HKJML a>p<> A in time W HKJML Q 3 Independent Set for Convex Objects We now describe how the results of the previous section can be extended to find an independent set in a set of convex objects in Let be a set of convex objects in, and let 56 8 be a maximum independent set of As for segments, we describe an algorithm for the case in which all objects in intersect the -axis We can then use the approach in Section 23 to handle the general case Define (resp ) to be the smallest (largest) -coordinate of all the points ") Let : be the maximum -coordinate of the intersection of with the -axis Again assume that is sorted in increasing order of the -coordinates of the leftmost endpoints An application of Dilworth s theorem similar to Lemma 24 gives the following Lemma 3 Given a set R (C), where, R #, `, R, of pairwise-disjoint convex objects, there exists a subsequence and R 0, and : # : 0 (Figure 3(a)), or has one of the following structure: For all +- #&% 0

11 W W (C2) (C3) # Y 0 and : Y : 0 (Figure 3(a)), or 0 (Figure 3(b)) Sequences satisfying condition or a can be computed using a dynamic programming approach similar to the one in Section 22 We outline the algorithm for computing a longest subsequence of type For - #! such that *4 & so that (- ) : # : \ # : &, (-- ) (--- ) * \ & A # and & * \ \ #, let '& \ # denote the subsequence of segments \ & A, and See Figure 3(c) Let -A!] & denote the longest subsequence of type of '& Set, -A!, hen we can prove the following Lemma 32 For all &-#"!, -A! -A X X!] 3 -!] (4) We can compute -A!] using a dynamic-programming approach Assuming -A X and X!] have already been computed, we only need to compute the set '& Suppose we can preprocess in time 2 8 so that we have the following information at our disposal: (^ ) [ = : for each ", (^0a ) whether * & U for each & " 2 hen the set '& can be computed in time Q Hence, we can compute Q in W 8A time Plugging this procedure into the recursive scheme of Section 23 we obtain the following heorem 2 33 Let be a set of convex objects in so that (^ 8 time hen an independent set of size at least 2 Q<>a#HKJML a>p<> A A 4 Independent Set for Axis-Parallel Rectangles )-(^0a ) can be computed in can be computed in time Let be a set of rectangles in, and let ^ denote the size of the largest clique in the intersection graph of Without loss of generality, we can assume that no rectangle completely contains any another rectangle We first consider two special cases of rectangle intersection graphs the piercing and non-piercing intersection subgraphs, derived by defining the following partial order among the rectangles Given two rectangles and, if and only if intersects both vertical edges of, and intersects both horizontal edges of (See Figure 4(a)) Clearly, if, then and intersect, and neither contains any vertex of the other ote that is a transitive relation: implies If, we say that and pierce, and that is pierced by

12 ^ (a) (b) (c) Figure 4: (a), (b) he set (solid) and the set R to vertices, and rectangles in R to edges (dashed), (c) Mapping rectangles in 4 Piercing and on-piercing Rectangle Intersection Graphs Given a set, the relation partitions the edges in the intersection graph W into two sets ( and : given an edge C +" W(O, +")X if or, and " otherwise Piercing Intersection Graphs Define the piercing intersection graph of as the directed graph W 0, ie, each vertex in W ] corresponds to a rectangle in, and there is a directed edge C between two vertices and if is pierced by, ie, C 8" W ] if and only if Lemma 4 Given a set of rectangles in, let W be the piercing graph of hen ^ W ]8`, and the optimal independent set in W can be computed in polynomial time PROOF It follows from the transitivity of that W is a transitive graph It is a well-known fact that all transitive graphs are perfect graphs, ie, for every induced subgraph of W, the size of the maximum clique equals the chromatic number herefore the vertices of W can be partitioned into W subsets, each of which is an independent set in Ẅ Since the largest subset has at least P<^ W vertices, we conclude that ^ W PF W ` By a classical result of Grötschel et al [3], a largest independent set of a perfect graph can be computed in polynomial time W F on-piercing Intersection Graphs We now consider the case of pairwise non-piercing rectangles Let be a set of rectangles such that no two rectangles pierce (although they could intersect); the intersection graph of is called the non-piercing intersection graph and denoted as W First, note that computing the optimal independent set of remains P-hard since the construction in [22] proving the P-hardness for intersection graphs of rectangles uses only non-piercing rectangles We call a subset r-maximal if it satisfies the following four conditions: (A) is an independent set, 2

13 " R " " " (A2) Every ") 7 intersects some (A3) For each, there is at most one rectangle "G5S6 not intersect any other rectangle in, and, 8 such that intersects and does (A4) For every pair of disjoint rectangles, there are at most two rectangles 7"4 S such that Y*, and intersect both and, and no other rectangle in Lemma 42 Let be an -maximal set hen,, ` 8=<>:, where : M PROOF Let R be the set of rectangles in the maximum independent set We will charge each rectangle in R to a rectangle in such that each rectangle in is charged at most a constant : times he lemma then follows Let be a rectangle in R By maximality condition a, intersects at least one rectangle in, and since the rectangles in are pairwise non-piercing, does not pierce any rectangle in We charge to a rectangle in as follows First, if ", we charge to itself Clearly, each rectangle in *GR receives only one unit of charge Otherwise, we charge as follows: C only intersects one rectangle "4 Charge to receives at most one unit of charge By condition, each rectangle in C2 intersects in a corner (ie, contains a vertex of ) Charge to Since R and are independent sets, each rectangle in receives at most four units of charge Let R R be the set of rectangles of R that have not yet been charged hus far each rectangle in has received at most 5 charges, so therefore, R R,,, We now bound the number of rectangles in R R has the following property: each rectangle in R intersects exactly two rectangles in (see Figure 4(b)); C and C2 above deal with all other intersection possibilities We map each rectangle in and R to a vertex and an edge, respectively, by constructing a bipartite multi-graph W Ö where each vertex in corresponds to a rectangle in, and there is an edge between & and \ if there is a rectangle " that intersects both and Clearly each rectangle in R maps to an edge in W We now show that W is a planar multi-graph as follows Replace each rectangle with its center :, and each rectangle ")R with a polygonal curve (consisting of three piecewise-linear segments) as illustrated in Figure 4(c) It is clear, given the independence property of the rectangles in R and, that the above embedding is non-intersecting Hence W is a planar multi-graph For a planar graph, the number of edges is at most thrice the number of vertices Since W is a planar multi-graph with at most two edges between any pair of vertices by condition (A4), we have, R,,,,,,, Hence, combining the bounds,, R,, R,, RR,,,,, M,, 3

14 " " ^ ` " Lemma 43 Given a set of rectangles in such that no two rectangles pierce, let W denote the corresponding (non-piercing) intersection graph hen ^ W ;F W ], and a : -approximation to the maximum independent set can be computed in polynomial time, PROOF First, by urán s heorem [], there exists an independent set of size at least <,, Second, for each edge %X &, due to the properties of non-piercing graphs, either contains a vertex of & or vice versa Let : denote the number of times rectangle s vertex is contained in another rectangle If contains a vertex of &, charge the edge % to vertex & by incrementing :U&, otherwise increment :I Clearly, :,,, and hence there exists a rectangle \ whose vertices are contained in more than,, <$ rectangles, and therefore at least one vertex contained in more than,, <% rectangles hese rectangles share a common point, and thus form a clique of size greater than,, <% Hence, W 2F W ] `,, F We describe an iterative algorithm to compute a -maximal set, which, by Lemma 42, is a constant-factor approximation of the maximum independent set of W In the beginning of the - -th iteration the algorithm maintains a set of - pairwise-disjoint segments In the - -th iteration, the algorithm checks whether conditions (A2) (A4) are satisfied If the answer is yes, we return, as it is a -maximal set Otherwise,,, % Condition (A2) violated hen there exists a set 0" P Set 0 +) % that does not intersect any 2 Condition (A3) violated hen there exist two disjoint rectangles " some but no other rectangle in Set that intersect 3 Condition (A4) violated hen there exist three pairwise-disjoint rectangles, " that all intersect, but no other rectangle in Set 0 +, O It is clear that is a set of pairwise-disjoint segments he process can continue for at most iterations, and the final set & is obviously a -maximal set Since each of the conditions (A2) (A4) can be checked in polynomial time, the total running time is polynomial 42 General Rectangle Intersection Graphs Combining Lemma 4 and Lemma 43, we attain the following Lemma 44 Given a set of rectangles in, an independent set of size at least some constant % : % %, can be computed in polynomial time PROOF By Lemma 4, compute the maximum independent set, say of ote the following: ' ), for, in the piercing graph 4

15 ^ ` ` % X -U,, ` 8 as an independent set in is an independent set in piercing graph of, and --U ^ 0, since a clique in is a clique in By Lemma 43, (QF^ 0`,, <% Hence, using -A and --A above, it follows that (7`,, <%M^ (7` 8=<%M^ 8 Since the intersection graph of is non-piercing, Lemma 43 gives an algorithm which returns an independent set of size at least 0=<>: ` =<% : ^ 8 From now on, let heorem 45 Given with polynomial time 8# 2, for some 8X 2, one can compute an independent set of size ' ( in PROOF he algorithm repeatedly extracts large cliques (one can compute the maximum clique in rectangle intersection graphs in polynomial time [6]) until a good independent set is found, as follows Set, and let be the set of rectangles in the - -th iteration For now assume that the value of is known At the - -th iteration, if there exists a clique of size at least a <, remove from, ie, set 0, and reiterate If no such clique exists, compute the maximum independent set, say, in the piercing graph of, and return the maximum independent set in Assume the algorithm stops after! iterations, ie ^ & a < ote that! P< a < # 2P<>a+ X<>a Since at most one rectangle from the independent set can be in a clique, each iteration removes at most one rectangle from the optimal independent set of, hence _&M` X! From Lemma 44, one can thus compute an independent set of size at least & X! X X<>a ^ & a < # a < < # yielding the desired result ote that we do not know the value of, but can run the above algorithm for all the possible values, and return the maximum 5 Conclusions In this paper we have presented algorithms for approximating the maximum independent set in the intersection graphs of convex objects in the plane he approximation ratio is better if the convex objects are line segments All the algorithms described in this paper have the same overall approach hey first prove the existence of a large independent subset with some special (separator-like) properties hey then show that this subset can be computed exactly from among the entire set (we used dynamicprogramming) One approach toward improving these results is to show the existence of independent subsets of larger size, which are still computable in polynomial time We leave it as an open problem whether the approximation ratios can be improved In particular, is it possible to design a -approximation algorithm for the case of general convex objects (all intersecting a vertical line) Similarly, is it possible to approximate the independent set of line segments better than For axis-parallel rectangles, devising an algorithm with approximation ratio HKJMLQ remains an intriguing open problem We have made a step toward crossing the logarithmic bound in this paper, but a general-case constant-factor algorithm remains elusive 5

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