SSimultaneous Embedding of Colored Graphs ∗ Debajyoti Mondal Department of Computer Science, University of Saskatchewan, Saskatoon, Canada, [email protected]
January 19, 2021
Abstract
A set of colored graphs are compatible, if for every color i , the number of vertices of color i isthe same in every graph. A simultaneous embedding of k compatibly colored graphs, each with n vertices, consists of k planar polyline drawings of these graphs such that the vertices of the samecolor are mapped to a common set of vertex locations.We prove that simultaneous embedding of k ∈ o (log log n ) colored planar graphs, each with n vertices, can always be computed with a sublinear number of bends per edge. Specifically, we showan O (min { c, n − /γ } ) upper bound on the number of bends per edge, where γ = 2 (cid:100) k/ (cid:101) and c isthe total number of colors. Our bound, which results from a better analysis of a previously knownalgorithm [Durocher and Mondal, SIAM J. Discrete Math., 32(4), 2018], improves the bound for k ,as well as the bend complexity by a factor of √ k . The algorithm can be generalized to obtain smalluniversal point sets for colored graphs. We prove that n (cid:100) c/b (cid:101) vertex locations, where b ≥
1, suffice toembed any set of compatibly colored n -vertex planar graphs with bend complexity O ( b ), where c isthe number of colors. Let G be a set of k planar graphs such that each graph G ∈ G has n vertices and these vertices arelabelled without repetition with the numbers 1 , . . . , n . A simultaneous embedding of G is a set of k planarpolyline drawings of the graphs in G such that the vertices with the same label have the same location in R (Fig. 1(a)–(c)). Simultaneous embedding can be used to model multilevel circuit layout [17] and tovisualize different types of relations among a common set of nodes, e.g., different types of code clonesover the files of a software [15].In a colored simultaneous embedding problem (Fig. 1(d)–(f)), the input consists of graphs whosevertices are labelled with the colors 1 , . . . , c , where 1 ≤ c ≤ n , such that for every color q and every pairof graphs G, G (cid:48) in G , the number of vertices of color q in G is the same as that of G (cid:48) , and the locationsof the vertices of color q in the drawing of G are the same as that of G (cid:48) . From both the perspective ofreadability and VLSI applications, a desirable goal is to minimize the bend complexity , i.e., the numberof bends per edge, in the drawing. A b -bend simultaneous embedding consists of drawings with bendcomplexity at most b . Colored simultaneous embedding was first considered by Brandes et al. [2], where they examined classesof graphs that admit simultaneous embedding with bend complexity 0, i.e., each edge is a straight linesegment, which is also known as simultaneous geometric embedding . Although every pair of paths admitssimultaneous geometric embedding, there exist 3 paths that do not admit simultaneous embedding [3],even when colored with 6 colors [2]. However, any number of 3-colored paths can be simultaneously ∗ This work is supported in part by Natural Sciences and Engineering Research Council of Canada (NSERC). a r X i v : . [ c s . C G ] J a n v v v v v v v v v (a) (b) v v v v v (c)(d) (e) (f) Figure 1: (a)–(b) A pair of graphs, and (c) their simultaneous embedding with bend complexity 2. (d)–(f)Illustration for a colored simultaneous embedding of two graphs with bend complexity 1. Here the verticesare colored with red, blue, and black (i.e., large, small, and tiny discs).embedded with bend complexity 0 [2]. A rich body of literature examines simultaneous embedding andits variants (e.g., see the survey by Bl¨asius et al. [1]).The thickness of a graph G is the minimum number t such that G can be decomposed into t planarsubgraphs. By F´ary’s theorem [14], every planar graph (equivalently, thickness-1 graph) has a planarstraight-line drawing. Simultaneous embedding of k planar graphs with small bend complexity can beseen as an extension of F´ary’s theorem for drawing thickness- k graphs on k planar layers. Every pair ofplanar graphs can be simultaneously embedded using bend complexity 2 [12, 9], and thus thickness-2graphs admit 2-bend polyline drawings on 2 planar layers.Durocher and Mondal [10] showed that every thickness- k graph can be drawn on k planar layerswith bend complexity at most O ( √ k · n − /β ), where β = 2 (cid:100) ( k − / (cid:101) . Although the bound seems tobe sublinear for k ∈ o ( n /β )), they claimed a sublinear upper bound only for fixed k . This caseis straightforward to observe from their algorithm, but the case when k is not fixed, requires a carefulanalysis of the dependency between n and k . In this paper, we modify the algorithm to accommodate thecolors, and provide a better analysis of the algorithm leading to a bound of O (min { c, n − /γ } ), where γ = 2 (cid:100) k/ (cid:101) . This bound holds for k ∈ o (log log n ). In addition to the improved bound on k , it also removesthe multiplicative factor √ k from the previously known bound.A point set S is t -bend universal for a class of graphs if every graph in that class admits a t -bendplanar polyline drawing on S . There exists an 1-bend universal point set of size n for planar graphs [13].Note that this can be seen as a universal point set for graphs that are colored with a single color. Forouterplanar graphs, which are compatibly colored with c colors, a (4 c + 1)-bend universal point set isknown [5].Pach and Wenger [16] showed that Ω( n ) bends are sometimes necessary to construct a planar polylinedrawing of a graph if for every vertex, a vertex location is specified in the input. While the graphs herecan be seen as colored with n colors, non-trivial lower bound on the bend complexity has recently beenachieved also for graphs colored with only three colors [7]. We show that every set of k ∈ o (log log n ) planar graphs, each with n vertices and compatibly colored with c colors, can be simultaneously embedded with bend complexity O (min { c, n − /γ } ), where γ = 2 (cid:100) k/ (cid:101) .Our bound results from a better analysis of a previously known algorithm [Durocher and Mondal, SIAMJ. Discrete Math., 32(4), 2018], and improves the previously known bound for k , as well as the bendcomplexity by a factor of √ k . We also examine the potential trade-off between the number of vertexlocations and the bend complexity. We show that for n -vertex planar graphs, which are compatibly2olored with c colors, there exists an O ( b )-bend universal point set of size n (cid:100) c/b (cid:101) , where b ≥ n -vertex graphs on a set of n vertex locations. Section 4 shows a potential trade-off between the bend complexity and the number ofvertex locations allowed to compute a simultaneous embedding. Finally, Section 5 concludes the papersuggesting some open problems. In this section we introduce some notation and preliminary results.A monotone topological book embedding [6] of a planar graph G is a planar drawing of G , where thevertices are represented as points on a horizontal line (cid:96) , and each edge is drawn as an x -monotone polylinebetween their corresponding end points such that it does not cross (cid:96) more than once (Fig. 2(a)–(b)). Theline (cid:96) is the spine of the embedding, and the crossing points on (cid:96) are the division vertices . The pathobtained by connecting the vertices (including the division vertices) on (cid:96) in order (e.g., left to right) isthe spinal path of G . v v v v v (a) v v v v v v v d v v v (d) v v v dv v spine crossing(b)(c) d r Figure 2: (a) A graph G , and (b) its monotone topological book embedding. The spinal path is P = ( v , v , d, v , v , v ), where d is a division vertex. (c)–(d) Illustration for the drawing of G from anuphill drawing of its spinal path.Let Γ be a planar polyline drawing of a path P = ( v , v , . . . , v n ). The drawing Γ is an uphill drawingif for any point r (here r may be a vertex location or an interior point of some edge) in Γ, the upward rayfrom r does not intersect the polyline v , . . . , v i , r , where 1 ≤ i ≤ n (Fig. 2(c)).Two graphs are compatibly colored if for each color q , both graphs have the same number of vertices ofcolor q . Throughout the paper we assume that the graphs are compatibly colored. Let G = { G , . . . , G k } be an instance of the simultaneous embedding problem, where the graphs are colored compatibly with c colors. Consider a monotone topological book embedding of G i and let P i be the spinal path. To colorthe division vertices in all the graphs compatibly, we introduce dummy vertices as necessary (at the endof the spinal paths). We then color all the division vertices with a color different than the input colors.Durocher and Mondal [10] showed that if P i admits an b -bend uphill drawing on a point set S , then G i admits a O ( b )-bend polyline drawing on S . Hence it suffices to consider only the simultaneous embeddingof the spinal paths. Here we give a concise proof for completeness. Lemma 1 (Durocher and Mondal [10]) . If a spinal path admits a b -bend uphill drawing on a point set S ,then the corresponding graph admits an O ( b ) -bend planar polyline drawing on S .Proof. Let B be the axis-aligned bounding box of S . While drawing the spinal path, at each vertex v , draw two b -bend polygonal paths starting at v . One path hits the left and the other hits the right3oundary of B (e.g., see the bold paths starting at d in Fig. 2(d)). Then draw the rest of the spinal pathabove these polygonal paths. Finally, the leftward (rightward) polygonal paths are used to draw theedges that are above (below) the spinal path in the monotone topological book embedding. Remark 2.
Let P be a set of compatibly colored spinal paths and let S be a point set. If each path in P admits an uphill embedding on S with bend complexity O ( b ) , then the graphs corresponding to P admit asimultaneous embedding with bend complexity O ( b ) . Throughout the paper we consider colors as positive integers. Since the drawing algorithm will heavilyuse results on partitioning an integer sequence , i.e., an ordered list of integers, we often consider a coloredpath as an integer sequence.A sequence (i.e., ordered set) of numbers is monotonic if it is either non-increasing or non-decreasing.A k -tuple is an ordered set of k numbers. Given an ordered set of k -tuples, the i th dimension , where1 ≤ i ≤ k , is the sequence of integers obtained by taking the i th element of every k -tuple in the givenorder. A sequence of k -tuples is monotonic if each dimension of the sequence is monotonic. Here weprove two lemmas on partitioning a sequence of integers and k -tuples, which will be used in our drawingalgorithm. Lemma 3.
Assume that for every sequence of n integers (resp., k -tuples), one can find a monotonicsubsequence of n δ , where < δ < . Then given a sequence S of n integers (resp., k -tuples), where n ≥ /δ , one can partition S into O ( n − δ − δ ) disjoint monotonic subsequence.Proof. To obtain the required partition, we repeatedly extract a monotonic sequence of n δ elements from S .Thus the number of elements in the partition is determined by the recurrence relation T ( n ) = T ( n − n δ )+1.We now show that T ( n ) is upper bounded by dn − δ − δ , where d = 2(1 − δ ) + 1.We choose m ≤ /δ as the base case. First observe that in the base case, we have T ( m ) = m , becausewe can partition S such that each subsequence contains a single element, which is trivially monotonic.Observe that m ≤ /δ ≤ d /δ − δ ) = d (1 − δ ) /δ − δ = dm − δ − δ . This proves the base case.We now assume that the inequality T ( k ) ≤ dk − δ − δ holds for every k from 1 to ( n −
1) and then considerthe general case. We will use the Bernoulli’s inequality, i.e., for real numbers x, r , where 0 ≤ r ≤ x ≥ −
1, we have (1 + x ) r ≤ rx . T ( n ) = T ( n − n δ ) + 1 = (cid:18) d − δ (cid:19) ( n − n δ ) − δ + 1 ≤ (cid:18) d − δ (cid:19) n − δ (1 − (1 − δ ) n δ − ) + 1 , using Bernoulli’s inequality= (cid:18) dn − δ − δ (cid:19) − d + 1 ≤ (cid:18) dn − δ − δ (cid:19) , for any d ≥ T ( n ) ≤ dn − δ − δ , where d = 2(1 − δ ) + 1. Lemma 4.
Given a sequence S of n integer k -tuples, where k ∈ o (log log n ) , one can partition S into O ( n − δk − δ k ) disjoint subsequences of k -tuples such that each subsequence is monotonic.Proof. We first extract n δ k -tuples such that the integer sequence in the first dimension is monotonic.We then repeat the process for every dimension, each time on the newly selected k -tuples. Therefore, onecan find a subsequence of at least n δ κ k -tuples such that they are monotonic.Since a single k -tuple is trivially monotonic, and since the term n δ κ is not a constant for k ∈ o (log log n ),we can apply Lemma 3 with subsequence size n δ κ . Consequently, we can partition S into O ( n − δk − δ k )disjoint subsequences of k -tuples, which are monotonic.We will also use the following properties of ordered sets of real numbers and integers. Theorem 5 (Erd˝os–Szekeres [11]) . Given an ordered set of ( n + 1) distinct real numbers, one can alwayschoose a monotonic subsequence of size at least n + 1 . emma 6. Given an ordered multiset S of n integers, it is always possible to choose a monotonicsubsequence of size at least √ n .Proof. Let I be the set of distinct integers in S . For each integer i ∈ I , let S i be the ordered subsetconsisting of all occurrences of i in S . Let x be the j th element of S i . Add the number j/ ( | S i | + 1) to x . After processing all the elements of I , we obtain a set S (cid:48) of n distinct numbers. By Erd˝os–Szekerestheorem [11], we now can choose a monotonic subsequence of size at least √ n from S (cid:48) . In this section we describe our result on colored simultaneous embedding. By Remark 2, it suffices toconcentrate on computing uphill drawings of the spinal paths with low bend complexity.
Lemma 7.
Let Q be a set of k spinal paths, each with n vertices. Assume that the paths are compatiblycolored with n α colors, where ≤ α ≤ . Then one can simultaneously draw Q such that each path isdrawn uphill with bend complexity O (min { n α , n − (1 / k } ) .Proof. We consider the following two cases.
Case 1 ( ≤ α ≤ / : In this case we construct the vertex locations on a horizontal line and assignthe vertices of the same color a subset of contiguous locations. We now show that every path admitsan uphill drawing with bend complexity O ( n α ) on these vertex locations. Let P = { x , x , . . . , x n } bea path in Q and let C ( x i ), where 1 ≤ i ≤ n , be the contiguous vertex locations for the color of x i . Wecompute an uphill drawing of P , as follows. We map the vertex x to the leftmost vertex location of C ( x ). For each i from 2 to n , we map x i to the leftmost available vertex location of C ( x i ). We thendraw the edge ( x i − , x i ) with an x -monotone polyline L such that L does not create any edge crossingwith the drawing of x . . . x i − , and the unmapped vertex locations lie above L .Fig. 3 illustrates the construction, where the mapped locations for each color are shown in shadedrectangles. Since we always map a new vertex v to the leftmost available location in set C ( v ), the mappedlocations for each color occupy a contiguous subset of vertex locations. We can use this property, to routean edge from one color set C ( v ) to another color set C ( w ), using O ( b ) bends, where b is the number ofdistinct colors between these two sets. We need O (1) bends to skip each intermediate color set (e.g., seethe bold edge in Fig. 3(f), where the intermediate color sets are skipped by ‘jumping’ over the shadedrectangles). Since there are at most n α colors, one can construct L with O ( n α ) bends. Case 2 ( / < α ≤ ): We first label the vertices of each spinal path with unique integers from 1 to n such that for every label (cid:96) , the vertices with label (cid:96) in all the paths have the same color (Fig. 4(a)).We then create a set I of n integer k -tuples, each of size k . The i th entry of the j th k -tuple, where1 ≤ i ≤ k and 1 ≤ j ≤ n , is the position of the j th label in the i th spinal path. Fig. 4(b) illustrates theconstruction of I for the paths of Fig. 4(a). Note that the colors of the vertices do not play any role inthis construction.By Lemma 6, for any multiset of n integers, we can find a monotonic subsequence of length at least n / . By Lemma 4, one can partition I into O (cid:16) n − (1 / k − (1 / k (cid:17) ∈ O ( n − (1 / k ) disjoint subsequences, whereeach subsequence is monotonic. Fig. 4(c)–(d) illustrate such a partition for the sequence I of Fig. 4(b).We now construct the vertex locations on a horizontal line and assign the vertices of the samesubsequence a subset of contiguous locations. Fig. 5(b) illustrates the point set determined by themonotonic sequences of Fig. 4(c). The monotonic sequences are ( X , X , X ) , ( X , X ) , ( X , X ) and( X , X ). We now show that every path admits an uphill drawing with O ( n − (1 / k ) bends per edge onthese vertex locations.Let P = { x , x , . . . , x n } be a path in Q and let C (cid:48) ( x i ), where 1 ≤ i ≤ n , be the contiguous vertexlocations for the subsequence containing x i . We compute an uphill drawing of P , as follows. We firstmap the vertex x to its corresponding vertex location X . For each i from 2 to n , we first map x i to itscorresponding vertex location X i , and then draw the edge ( x i − , x i ) with an x -monotone polyline L suchthat L does not create any edge crossing with the drawing of x . . . x i , and the unmapped locations lieabove L .The monotonicity of the set C (cid:48) ( x i ) ensures that the mapped locations of C (cid:48) ( x i ) remain consecutive.Therefore, one can route the edge from one monotonic sequence C (cid:48) ( x i − ) to another monotonic sequence C (cid:48) ( x i ), using O ( b ) bends, where b is the number of distinct monotonic sequences between these two5 x x x x x (a) x x x x x x x (b)(c) (d) x x x x x x x x x x (f) x x (e) Figure 3: (a) A spinal path P . (b)–(f) Construction of an uphill drawing of P . X : (1 , , X : (2 , , X : (3 , , X : (4 , , X : (5 , , X : (6 , , X : (7 , , X : (8 , , X : (9 , , (b) (d) x x x x x x x x x x x x x x x x x x x x x x x x x x x X : (2 , , X : (8 , , x x x x x x x x x x x x x x x x x x x x x x x x x x x (a)(c) X : (1 , , X : (3 , , X : (6 , , X : (5 , , X : (9 , , X : (4 , , X : (7 , , P : P : P : P : P : P : Figure 4: (a) A set of spinal paths. (b) Construction of I . (c) Partition of the paths into monotonicsequences. (d) Illustration for the monotonicity for the first two elements of the partition.6 a) (b)(c) (d)(f)(e) x x x x x x x x x P x x x x x x x x x x x x x x x x x x x x X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X Figure 5: (a) The spinal path P of Fig. 4, and (b)–(f) it’s uphill drawing on the point set determined bythe monotonic sequences.monotonic sequences. Fig. 5(b)–(f) illustrate the computation of an uphill drawing for the path ofFig. 5(a). The mapped locations for each monotonic sequence are enclosed in shaded rectangles.We need O (1) bends to skip each intermediate monotonic sequence (e.g., the intermediate monotonicsequences can be skipped by ‘jumping’ over the shaded rectangles). Since there are at most O ( n − (1 / k )monotonic sequences, we can construct L with O ( n − (1 / k ) bends.Lemma 7 and Remark 2 together implies that any set of k graphs, compatibly colored with c colors,admits a simultaneous embedding with bend complexity O (min { c, n − (1 / k } ). This can be furtherimproved by the following observations (Obs 1-2). These observations are made in [10], which generalizea standard technique of computing simultaneous geometric embedding of two paths [4] to drawing anynumber of paths with bends. Obs1.
The vertex locations of an uphill drawing can be moved vertically and the edges can be redrawnsuch that the new drawing is also uphill with the same bend complexity. Fig. 6 (a)–(b) illustratesuch an example.
Obs2.
Half of the spinal paths can be drawn along the x-axis, and the other half along the y-axis.While Obs 1 is straightforward to verify, we give a brief overview of Obs 2 to make the paper self-contained.Given a set Q of k spinal paths, first partition them into two sets Q and Q such that Q contains (cid:100) k/ (cid:101) spinal paths, and Q the rest of the spinal paths. Let I j , where 1 ≤ j ≤
2, be the set of integer k -tuples constructed for Q j , and let M j be the partition of I j into monotonic sequences. Now construct apoint set along the x-axis using M , and another point set along the y-axis using M (Fig. 6(c)). Thefinal point set is constructed by taking for each label, the point with x and y -coordinates equal to thepositions of the label along the x-axis and y-axis, respectively. Fig. 6(c) depicts the final point set S onan integer grid. We have two monotonic sequences, one along x-axis and the other along y-axis. We now7 x x x x x x x x x (b)(a) x x x x x x x x x x x x x x x x x x (c) Figure 6: (a)–(b) Illustration for computing an uphill drawing when the points are not on a line. (c)Construction of the point set S .draw Q along the x-axis, and Q along the y-axis. The bend complexity of the drawing of Q does notinterfere with bend complexity of the drawing of Q . Therefore, the bend complexity of the resultingdrawing are determined by only half of the spinal paths.By Lemma 7 and Remark 2, we now have the following theorem. Theorem 8.
Given k compatibly colored planar graphs, each with n vertices, one can compute asimultaneous embedding of these graphs with bend complexity O (min { c, n − /γ } ) , where c is the numberof distinct colors and γ = 2 (cid:100) k/ (cid:101) . In this section we show that allowing more than n vertex locations help improve the bend complexity. Inparticular, n (cid:100) c/b (cid:101) vertex locations, where b ≥
1, suffice to embed any set of compatibly colored n -vertexplanar graphs with bend complexity O ( b ), where c is the number of colors.Let Q be a set of compatibly colored spinal paths, each with n vertices. We first partition the colorsinto at most b disjoint sets C , C , . . . C b such that each contains at most (cid:100) c/b (cid:101) colors. For each 1 ≤ j ≤ b ,let N j be the number of vertices, whose color belong to C j (in the spinal path). Then n = (cid:80) ≤ j ≤ b N j .8y a chain of C j , we denote a point set of | C j | points, each colored by a distinct color from C j . We nowcreate a point set S of size n (cid:100) c/b (cid:101) along the x-axis, as follows.The points in S are arranged into b subsets S , . . . , S b , each contains a set of contiguous points of S . The set S j consists of N j chains of C j , and thus N j | C j | = N j (cid:100) c/b (cid:101) points. Fig. 7 illustrates such anexample with 5 colors. We now show that the uphill drawing of a spinal path can be computed with O ( b )bends per edge. S S S (c) x x x x x x x x x x x (a)(b) x x x x x x x Figure 7: (a) A spinal path P to illustrate N j for different colors. (b) Construction of S , with a mappingfor the vertices of P . (c) Illustration for an uphill drawing of P on S .For each C j , we map the corresponding vertices of the spinal path from left to right on the chains of S j . We now construct the drawing following the same approach as in the proof of Lemma 7, where wetreat each chain as a single vertex location (i.e., from each chain, we use only one point for mapping). Ateach step of the construction, the mapped chains of each subset S j remain consecutive. Since there are atmost b such subsets, the edges can be routed with O ( b ) bends per edge. By Remark 2, we now have thefollowing theorem. Theorem 9.
Given a set of compatibly colored planar graphs, each with n vertices, there exists an O ( b ) -bend universal point set of n (cid:100) c/b (cid:101) points for these graphs, where c is the number of distinct colorsin the input and b ≥ . In this paper we show that colored simultaneous embedding of k ∈ o (log log n ) compatibly colored graphs,each with n vertices, can be computed on n vertex locations and with sublinear bend complexity. Therunning time of our approach is polynomial in n and k , which is determined by the time for partitioningan ordered set of n integer k -tuples into monotonic subsequences (Lemma 4). We also show that anynumber of n -vertex compatibly colored graphs can be embedded on a set of n (cid:100) c/b (cid:101) points with bendcomplexity O ( b ), where c is the number of colors and b ≥ √ n in any integer sequence was the key to improvethe bend complexity in Section 3. However, in a colored simultaneous embedding, the correspondingsequences are integer multisets. Hence we attempted to generalize Erd˝os–Szekeres theorem [11] to computelarge monotonic sequences in a multiset of positive integers. We were able to prove the existence of amonotonic subsequence of size max {√ n, √ c + nc − } (Lemma 10), which is larger than √ n only whenthe number of distinct integers c is smaller than √ n . It would be interesting to investigate multisetswhere the number of distinct integers is larger than √ n .9 emma 10. Let S be an ordered multiset of n integers, where n = kc + 1 , c is the number of distinctintegers in S , and k is a positive integer. Assume that c is a perfect square. Then S contains a monotonicsubsequence of size at least max {√ n, √ c + nc − } .Proof. If S does not contain c distinct integers, then we can apply the same technique as in the proof ofLemma 6 to transform S into a multiset that satisfies this property. Hence it suffices to prove the claimwhen S contains c distinct integers. Since the √ n lower bound is guaranteed by Lemma 6, we focus onthe other term.We apply an induction on k . If k = 1, then we have c + 1 integers, and by Erd˝os-Szekeres theorem [11],we can find a monotonic subsequence of size at least √ c . Since n = c + 1, we have √ c ≥ √ c + nc −
2. Wemay thus assume that k >
1, and for each integer smaller than k , the claim holds.By Lemma 6, we can find a monotonic subsequence L of size √ c from the first c + 1 integers and deletethe last element (cid:96) of the subsequence. We choose the earliest possible subsequence, i.e., a monotonicsubsequence that minimizes the position of the last element. Later, we will refer to this last element (cid:96) asthe representative of the monotonic sequence L . Note that we can repeat this step for at least ( n − c )steps, and remove a set X = { x , . . . , x n − c } of ( n − c ) integers. The set X is ordered, i.e., x i , where1 ≤ i ≤ ( n − c ), denotes the element deleted at step i .Let q be an integer with the maximum frequency in X , and let X q ⊆ X be the set consisting of all q .Since there are at most c distinct integers in X , | X q | ≥ (cid:100) ( n − c ) /c (cid:101) . Consider the earliest integer p in X q .The integer p is a representative of a monotonic subsequence Q of size √ c . The subsequence consisting of Q followed by X q \ { p } is the required subsequence of size at least √ c + (cid:108) ( n − c ) c (cid:109) − ≥ √ c + nc − . Our work raises several natural directions for future research, e.g., (a) Prove a tight bound on thelength of the monotonic subsequence that is guaranteed to exist in every multiset of c distinct numbers.(b) Prove a non-trivial lower bound on the bend complexity for colored simultaneous embedding. (c)Improve our results on the interplay between the bend complexity and the number of vertex locations. Acknowledgement.
The author thanks the anonymous reviewers for their helpful comments, which improved the presentationof the paper.
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