Multi-Stack Boundary Labeling Problems

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1 Multi-Stack Boundary Labeling Problems Micael A. Bekos 1, Micael Kaufmann 2, Katerina Potika 1 Antonios Symvonis 1 1 National Tecnical University of Atens, Scool of Applied Matematical & Pysical Sciences, Zografou, Atens, Greece mikebekos@mat.ntua.gr, symvonis@mat.ntua.gr, epotik@cs.ntua.gr 2 University of Tübingen, Institute for Informatics, Sand 13, Tübingen, Germany mk@informatik.uni-tuebingen.de Abstract. Te boundary labeling problem was recently introduced in [4] as a response to te problem of labeling dense point sets wit large labels. In boundary labeling, we are given a rectangle wic encloses a set of n sites. Eac site is associated wit an axis-parallel rectangular label l i. Te main task is to place te labels in distinct positions on te boundary of, so tat tey do not overlap, and to connect eac site wit its corresponding label by non-intersecting polygonal lines, so called leaders. Suc a label placement is referred to as legal leader-label placement. In tis paper, we study boundary labeling problems along a new line of researc. We seek to obtain labelings wit labels arranged on more tan one stacks placed at te same side of te enclosing rectangle. We refer to problems of tis type as multi-stack boundary labeling problems. We present algoritms for maximizing te uniform label size for boundary labeling wit two and tree stacks of labels. Te key component of our algoritms is a tecnique tat combines te merging of lists and te bounding of te searc space of te solution. We also present NPardness results for multi-stack boundary labeling problems wit labels of variable eigt. 1 Introduction A common task in te process of information visualization is te placement of extra information, usually in te form of text labels, next to te features of a drawing (diagram, map, tecnical or grap drawing) tey describe. Wen te labels are small and te features tey describe are sparsely distributed in te drawing, it may be feasible to place most labels next to te features tey describe so tat te labels do not overlap wit eac oter and tey do not obscure oter drawing features. Obtaining optimal label placements wit respect to some optimization criterion is, in general, NPard [7]. An extensive bibliograpy about map labeling can be found at [11]. In te case of very large labels (or, equivalently, dense feature sets), usually it is impossible to find a label placement, i.e. place eac label next to te feature it describes. In response to tis problem, Bekos, Kaufmann, Symvonis and Wolff [4] (see also [3]) proposed te boundary labeling model. In tis model te labels are placed on te boundary of a rectangle enclosing all features and eac label is connected to its associated feature wit polygonal lines, called leaders. If te labels are non overlapping and te leaders non intersecting we ave a legal leader-label placement or legal labeling. Te boundary labeling model is a realistic model for medical atlases and tecnical drawings, were certain features of a drawing are explained by blocks of text placed outside te drawing so tat no part of te drawing is obscured. SmartDraw [9] provides boundary labelings in a primitive form based on labeling templates. It does not support any form of automated boundary labeling optimization. Bler [5] supports te boundary labeling process and facilitates te annotation of drawings wit text labels. Sites model features of te drawing. If tey model a point-feature (e.g., a city on a map) tey are naturally represented as points (see points in ectangle of Figures 1, 2 and 5). So, in its simplest

2 2 form, a boundary labeling problem specifies as part of its input a set P of n points = (x i, y i ) on te plane in general position, i.e. no tree points lie on a line and no two points ave te same x- or y-coordinate. Anoter interesting variation is te one wit two candidate points on te plane for eac site (see Figure 3). In practice, several times we want to associate a label wit an area-feature (e.g., a region on a map, a body part on a medical atlas, a macine part on a tecnical drawing). To keep tings simple, we specify tese regions by a closed polygonal line or by a line segment internal to te feature area, and assume tat te site slides along te boundary of te polygon or on te line segment (see Figure 4). Track outing Area Fig. 1: Type-opo leaders. Fig. 2: Type-po leaders. Fig. 3: Sites wit 2 candidate positions. Fig. 4: Sites associated wit vertical line segments. Fig. 5: Tree stacks of labels. In general, eac site as a corresponding axis-parallel rectangular, open label l i of widt w i and eigt i. Te labels are to be placed around an axis-parallel rectangle = [l, r ]x[b, t ] of eigt H = t b and widt W = r l wic contains all sites in P. Wile in te general case te labels are of variable dimensions, it is natural to consider te restricted cases were te labels are of uniform size (eigt and/or widt), or of maximum uniform size. Eac site is connected wit its corresponding label in a simple and elegant way by using polygonal lines, called leaders. In our approac we ave leaders tat consist of a single straigt line segment or a sequence of rectilinear segments. A leader consisting of a straigt line segment is referred to as a type-s leader. Wen a leader is rectilinear, it consists of a sequence of axis-parallel segments eiter parallel (p) or ortogonal (o) to te side of containing te label it leads to. Te type of a leader is defined by an alternating string over te alpabet {p, o}. We focus on leaders of types-opo and po, see Figures 1 and 2, respectively. Furtermore, we assume tat eac type-opo leader as te parallel p-segment outside te bounding rectangle, routed in te so-called track routing area. We consider type-o leaders to be of type-opo and of type-po as well. A furter refinement of te labeling model as to do wit te sides of te enclosing rectangle containing te labels. Labels can be placed on one or more sides of te enclosing rectangle (in Figures 1 and 2 all labels are placed on te east side of te enclosing rectangle). Finally, in order to allow for greater numbers of larger labels, we migt ave te labels arranged in more tan one stack at eac side of te enclosing rectangle. Tis paper is devoted to te case of multi-stack labelings. Figure 5 sows a labeling were te labels occupy tree stacks to te east side of te enclosing rectangle. Notice tat in te case of multiple stacks of labels (say m stacks), a leader of type-opo can ave its p segment eiter in between te rectangle and te first stack (called first track routing area) or between te i-t and te (i + 1)-t stack, were i < m (called (i + 1)-t track routing area). Eac leader tat connects a site to a label, touces te label on a point on its side tat faces te enclosing rectangle. Te point were te leader touces te label is called port. We can assume eiter fixed ports, i.e. te leader is only allowed to use a fixed set of ports on te label side (a typical case is were te leader uses te middle point of te label side) or sliding ports were te leader can

3 touc any point of te label s side. Te labelings in Figures 1, 2, 3 and 4 use sliding ports, wile in Figure 5 it uses fixed ports. Keeping in mind tat we want to obtain simple and easy to visualize labelings, te following criteria can be adopted from te areas of VLSI and grap drawing: minimizing te total number of bends of te leaders, minimizing te total leader lengt, minimizing te maximum leader lengt. An additional criterion tat we consider is te maximization of te label size for uniform size labels. Tis is a quite common optimization criterion in te map labeling literature. In tis paper, we seek to obtain labelings wit labels of maximum uniform size arranged on more tan one stacks of labels at te same side of te enclosing rectangle Tis paper is structured as follows: Section 2, reviews preliminary results required for te development of our algoritms. In Section 3, we present algoritms for obtaining multi-stack labelings of maximum uniform label eigt for te cases of two and tree stacks of labels arranged at te same side of. In Section 4, we present several NP -ardness results for non-uniform sized labels placed in two stacks. We conclude in Section 5 wit open problems and future work Previous Work Most of te known results on boundary labeling wit point sites were presented in [4] (see also [3]). A legal labeling, on one side wit type-opo (type-po) leaders can be acieved in O(n log n) time (in O(n 2 ) time, respectively), wereas on all four sides wit type-opo leaders in O(n log n) time. Te minimization of te total leader lengt wen uniform sized labels can be placed on two opposite sides of wit eiter type-opo and type-po leaders needs O(n 2 ) time. Te same problem on two opposite sides of wit type-opo leaders and non-uniform label sizes O(n 2 H) is needed. Te problem of minimizing te total number of leader bends on one side wit type-opo leaders can be solved in O(n 2 ) time. An algoritm for minimizing te total leader lengt on four sides wit type-opo leaders in O(n 2 log 3 n) time is presented for points in [1] and for polygons in [2]. 2 Preliminaries Trougout te paper we use lists tat contain pairs of integers. Given a pair (a, b) of integers, a and b are referred to as te first and te second coordinate of te pair, respectively. Inspired by an idea of Stockmeyer [10] wic was subsequently used by Eades et. al. [6], we manage to keep te lengt of eac list bounded by pruning pairs tat cannot occur in an optimal solution. Definition 1. A list L of pairs of integers is sorted if te pairs it contains are lexicograpically sorted in decreasing order wit respect to teir first coordinate and in increasing order wit respect to teir second coordinate. Definition 2. Let (a, b) and (c, d) be pairs of integers. (a, b) dominates (c, d) a c and b d. Suppose we ave to solve a problem were te searc space of te solution consists of pairs of integers, and let f be a monotone function computing a minimization objective on pairs from te solution searc space. Tus, if (a, b) and (c, d) represent possible solutions, ten te pair (a, b) can never be involved in an optimal solution and may be safely removed from te solution set. Given a list L of pairs of integers, a pair (a, b) L tat does not dominate any oter pair in L is called an atom (wit respect to L). In our algoritms we maintain lists (of pairs) tat contain only atoms. A frequently performed operation is te merging of two lists of atoms, resulting in a new list of atoms. Te merging algoritm resembles te merging step of merge sort algoritm and can be considered to be folklore. It supports te following lemmas:

4 4 Lemma 1. k sorted lists L 1, L 2,..., L k, k 2, of atoms can be merged in O((k 1) k i=1 L i ) time into a new sorted list L of at most k i=1 L i atoms. Lemma 2. Let A and B be two finite set of integers and let L = {(a, b) a A and b B} be a list of atoms. Ten, L min( A, B ). Finally, we present some notation and terminology tat we use in te description of our algoritms. We say tat a pair (a, b) obeys te boundary conditions, if a H and b H, were H is te eigt of te enclosing rectangle. We also define operator H :, were: { a + b, if a + b H a H b =, oterwise 3 Label Size Maximization 3.1 Two Stacks of Labels on te Same Side We consider boundary labeling wit type-opo leaders, were te labels are placed at two stacks on te same side (say, te east side) of te enclosing rectangle. We assume tat all labels ave te same size (widt and eigt) and we seek to maximize te eigt of all labels. Our approac is as follows: Given labels of eigt, we propose an algoritm, tat determines weter tere exists a legal labeling. To determine te maximum value of, we apply a binary searc on all possible discrete values for eigt. We assume te more general case of sliding labels wit sliding ports. Te type-opo leaders connecting sites to labels tat are on te second stack can ave teir bends (or equivalently teir p segments) eiter in te first or in te second track routing area. Observe tat, in any legal one-side labeling wit type-opo leaders, te vertical order of te sites is identical to te vertical order of teir corresponding labels on bot stacks. Tis, togeter wit te assumption tat no two sites sare te same y-coordinate, guarantees tat leaders do not intersect. So, we assume tat te sites are sorted according to increasing y-coordinate. For a fixed label eigt, we propose a dynamic programming algoritm tat outputs a boolean value, wic indicates weter tere exists a legal label placement, wen all sites are associated wit labels of eigt. Our algoritm maintains a table T of size (n + 1) (n + 1). Eac entry T [i, k], i k of table T describes different possible placements for te subproblem consisting only of te first i sites, suc tat k out of te i leaders ave teir bends in te second track routing area. Assuming tat we ave placed te labels for te first i 1 sites, we try to place te i-t label, wic corresponds to te i-t site. We distinguis two cases based on weter te label is placed on te first or second stack. Additionally, if l i is to be placed in te second stack, ten we ave to ceck weter tis can be done wit a leader tat as its bends in te first or second track routing area. Obviously, suc placements can be obtained from placements of te first i 1 sites wit eiter k or k 1 leaders, tat bend in te second track routing area. For eac possible placement, we are interested in te eigt of two stacks. Tis information is captured by a pair (a, b), were a (b) is te igest occupied Y -coordinate of te first (respectively second) stack. Tus, entry T [i, k] contains a list of atoms, eac corresponding to a different placement. List T [i, k] is empty, wen it is impossible to place te first i labels, using k leaders tat ave teir bends in te second track routing area. Label l i is placed at te first stack: Let T 1 [i, k] be a list of pairs (a, b), were a (b) is te igest occupied Y -coordinate of te first (respectively second) stack, wen te labels of te first i sites ave been placed, te i-t site as its label at te first stack and k out of te i leaders ave teir bends in te second track routing area. T 1 [i, k] can be computed based on entry T [i 1, k] (see Figure 6a), as follows: T 1 [i, k] = {(a H, b) : (a, b) T [i 1, k]}

5 5 a b a b (a) Placing l i in 1st stack; bend in 1st track routing area. (b) Placing l i in 2nd stack; bend in 1st track routing area. a b a b (c) Placing l i in 2nd stack; bend in 1st track routing area. (d) Placing l i in 2nd stack; bend in 2nd track routing area. Fig. 6: Different placements obtained for te label of site i. In Figures 6a, 6b and 6c: (a, b) T [i 1, k], wereas in Figure 6d: (a, b) T [i 1, k 1]. Label l i is placed at te second stack - bend at te first track routing area: Let T 21 [i, k] be a list of pairs (a, b), were a (b) is te igest occupied Y -coordinate of te first (respectively second) stack, wen te labels of te first i sites ave been placed, te i-t site is connected wit a label placed at te second stack using a leader tat as its bends at te first track routing area and k out of te i leaders ave teir bends in te second track routing area. Again, T 21 [i, k] can be computed based on entry T [i 1, k]. Tis computation is a little bit more complicated. If for some pair (a, b) T [i 1, k] it olds tat a b (i.e. te occupied area of te first stack is lower or equal tan te occupied area of te second stack), ten a pair (b, b H ) is added in T 21 [i, k] (see Figure 6b). Oterwise, pair (a, max{b H, a}) is added in T 21 [i, k] (see Figure 6c). So, we conclude tat T 21 [i, k] can be computed by using te following formula: were: T 21[i, k] = A 21[i, k] B 21[i, k], A 21 [i, k] = {(b, b H ) : (a, b) T [i 1, k] s.t. a b} B 21[i, k] = {(a, max{b H, a}) : (a, b) T [i 1, k] s.t. a > b} Label l i is placed at te second stack - bend at te second track routing area: Similarly, let T 22 [i, k] be a list of pairs (a, b), were a (b) is te igest occupied Y -coordinate of te first (respectively second) stack, wen te labels of te first i sites ave been placed, te i-t site is connected wit a label placed at te second stack using a leader tat as its bends at te second track routing area and k out of te i leaders ave teir bends in te second track routing area. T 22 [i, k] can be computed based on entry T [i 1, k 1] (see Figure 6d), as follows: T 22 [i, k] = {(y i, b H ) : (a, b) T [i 1, k 1] s.t. a < y i } All pairs (, a), (a, ) can be removed from lists T 1 [i, k], T 21 [i, k] and T 22 [i, k], in linear to teir lengt time, since tey do not capture possible placements. Te implied lists are merged into list T [i, k] of atoms, based on Lemma 1. We can easily sow tat T [i, k] 2 T [i 1, k] +3. Tis implies tat T [n, k] = O(2 n ), n k. Also, by Lemma 2, we ave tat T [n, k] H. However, by employing te following Lemma 3, we can improve on bot of tese bounds. Its correctness can easily be sown inductively, by proving tat te distinct values tat bot coordinates of te pairs in T [i, k] can receive are drawn from te sets {0,, 2,..., i}, {y 1, y 2,..., y i }, and i j=1 {y j +, y j +2,..., y j +(i 1)}.

6 6 Lemma 3. List T [n, k], n k contains O(n 2 ) pairs. To prove te correctness of our algoritm, consider a pair (a, b) T [i, k] tat dominates pair (c, d) T [i, k]. Assume, for te sake of contradiction, tat pair (a, b) yields a solution and pair (c, d) does not. Tat means tat, for at least one pair out of {(y i, b + ), (b, b + ), (a, max{b +, a}), (a +, b)} te boundary condition olds wile te boundary condition does not old for any of te pairs {(y i, d + ), (d, d + ), (c, max{d +, c}), (c +, d)}. Tis is impossible since a c and b d. Terefore (a, b) can never be involved in an optimal solution and can be discarded. Tis implies tat eac list T [i, k] sould only contain atoms. Eac of te (n + 1) (n + 1) entries of T is computed in O(n 2 ) time. Tus, our algoritm terminates after O(n 4 ) time. For a fixed label eigt, te algoritm outputs a boolean value, wic indicates weter tere exists a legal label placement. Tis is done by identifying weter tere exists a non-empty list T [n, j], wit 0 j n. By using an extra table of te same size as T, our algoritm can easily be modified, suc tat it also computes te label and leader positions. Teorem 1. Given a rectangle of integer eigt H and a set P of n points (sites) in general positions, tere exists an O(n 4 log H) time algoritm tat produces a legal multi-stack labeling wit two stacks of labels and wit type-opo leaders suc tat te uniform integer eigt of te labels is maximum. Proof. In order to solve te size maximization problem (i.e. to determine te maximum eigt of te labels), we can simply apply a binary searc on all possible discrete values for eigt. To complete te proof, observe tat H n 2H n. 3.2 Sample Labelings Figures 7 and 8 are produced from te algoritm of Section 3.1 and depict two regional maps of UK and Italy, respectively. Te labels occupy two stacks to te east side of te enclosing rectangle. In bot labelings te label size is maximum. Fig. 7: A regional map of UK. Fig. 8: A regional map of Italy. 3.3 Tree Stacks of Labels on te Same Side In tis section, we extend te algoritm of Section 3.1 to support an additional stack of labels. We consider te case, were leaders connected to labels of te i-t stack are restricted to bend in te i-t track routing area. Te objective, again, is to maximize te uniform eigt of all labels.

7 7 a a b a b m (a) Placing l i in 1st stack. (b) Placing l i in 2nd stack. (c) Placing l i in 3rd stack. Fig. 9: Different placements obtained for te label of te i-t site. In Figure 9a: a T [i 1, k, m], in Figure 9b: (a, b) T [i 1, k 1, m], wereas in Figure 9c: (a, b) T [i 1, k, m 1] Teorem 2. Given a rectangle of integer eigt H and a set P of n points (sites) in general positions, tere exists an O(n 4 log H) time algoritm tat produces a legal multi-stack labeling wit tree stacks of labels and wit type-opo leaders suc tat te uniform integer eigt of te labels is maximum and te leaders connected to labels of te i-t stack are restricted to bend in te i-t track routing area. Proof. We use dynamic programming algoritm employing a table T of size (n+1) (n+1) (n+1). For eac i k + m, entry T [i, k, m] contains a list of pairs (a, b), were a (b) is te Y -coordinate of te first (second) stack, tat is needed to place te first i labels, wen m labels are placed in te tird stack, k labels are placed in te second stack and i k m labels are placed in te first stack. Note tat te eigt of te tird stack is m, since all leaders connected to labels of te tird stack are restricted to bend in te tird track routing area. List T [i, k, m] is empty, wen it is impossible to route te first i labels using k labels in te second stack and m in te tird stack. Tis implies tat table entries T [i, k, m], were i < k + m, contain empty lists. Following similar arguments as in Section 3.1, entry T [i, k, m] can be computed based on te following recurrence relation: were: T [i, k, m] = Merge{T 1 [i, k, m], T 2 [i, k, m], T 3 [i, k, m]} (1) T 1 [i, k, m] = {(a H, b) : (a, b) T [i 1, k, m]} T 2 [i, k, m] = {(y i, b H ) : (a, b) T [i 1, k 1, m] s.t. a < y i } T 3[i, k, m] = {(y i, y i) : (a, b) T [i 1, k, m 1], s.t. m H and (a, b) < (y i, y i)} List T 1 [i, k, m] of Equation 1 captures placements of te i-t label in te first stack (see Figure 9a). Similarly, list T 2 [i, k, m] of Equation 1 captures placements of te i-t label in te second stack. Since we assumed tat leaders connected to labels of te second stack are restricted to bend in te second track routing area, tis is possible only for pairs (a, b) T [i 1, k 1, m] wit a y i (see Figure 9b). Finally, list T 3 [i, k, m] of Equation 1 captures placements of te i-t label in te tird stack. Tis is possible only for pairs (a, b) T [i 1, k, m 1] wit (a, b) (y i, y i ) (see Figure 9c). To compute entry T [i, k, m], we first remove all pairs (, a), (a, ) from lists T 1 [i, k, m], T 2 [i, k, m] and T 3 [i, k, m] and ten we merge te implied lists to T [i, k, m] of atoms, based on Lemma 1. Lemma 4. For n k + m, T [n, k, m] n + 1. Proof. Lists T 2 [i, k, m] and T 3 [i, k, m] contain pairs of numbers wit te same first coordinate. Tis implies tat tey contribute at most one non-dominating pair, wile list T 1 [i, k, m] contains at most i elements, since T [i 1, k, m] i. Tus, T [i, k, m] i + 1. Eac of te (n+1) (n+1) (n+1) entries of T is computed in O(n) time. Tus, our algoritm terminates after O(n 4 ) time. For a fixed label eigt, te algoritm outputs a boolean value, wic indicates weter tere exists a legal label placement. Tis is done by identifying weter

8 8 tere exists a non-empty list T [n, i, j], wit 0 i + j n. By using an extra table of te same size as T, our algoritm can easily be modified, suc tat it also computes te label and leader positions. In order to solve te size maximization problem (i.e. to determine te maximum eigt of te labels), we can simply apply a binary searc on all possible discrete values for eigt. To complete te proof, observe tat H n 3H n. 4 Computational Complexity of Multi-Stack Labeling Problem In tis section, we investigate te computational complexity of several multi-stack boundary labeling problems wit eiter type-opo or po leaders and labels of arbitrary size, wic can be placed at two stacks on te same side of te enclosing rectangle. Witout loss of generality, we assume tat te labels are located to te east side of te enclosing rectangle. We consider several different type of sites. In te most applicable case, site s i is associated wit a point = (x i, y i ) on te plane. However, we also consider te cases, were site s i is associated wit eiter two candidate points p 1 i = (x1 i, y1 i ) and p2 i = (x2 i, y2 i ) on te plane (see Figure 3) or wit a vertical line segment, so tat te site slides along te boundary of te proposed line segment (see Figure 4). Te assumed models are quite general, since we allow sliding labels wit sliding ports. 4.1 Line sites wit type-opo leaders at two stacks on one side. We focus on type-opo leaders, were eac site s i can slide along a line segment parallel to te y-axis and is associated wit a label l i of eigt i. We seek to find a legal labeling. Teorem 3. Given a rectangle of eigt H, a set P of n line segments (sites) tat are parallel to te y-axis and a label of eigt i for eac site s i P, it is NP -ard to place all labels at two stacks on one side of wit non-intersecting type-opo leaders. Proof. We reduce te Partition problem [8] to our problem. Te Partition problem is defined as follows: Given positive integers a 1, a 2,..., a m, is tere a subset I of J = {1, 2,..., m} suc tat i I a i = i J I a i? We will reduce an instance A of Partition to an instance B of our problem, suc tat instance A can be partitioned if and only if tere exists a legal labeling for instance B. Te reduction we propose is somewat immediate. Our site set P = {s 1, s 2,..., s m } consists i J a i (H is te eigt of te of m (parallel to y-axis) line segments of identical lengt H = 1 2 enclosing rectangle ; see Figure 10). Eac site s i is also associated wit a label l i of eigt a i. Bot stacks contribute 2H eigt, wic is equal to te sum of all label eigts. It is clear tat if tere exists a labeling for instance B, te labels sould be partitioned into two sets suc tat te sum of te label eigts of eac set sums up to H. Terefore, te indices of te sites wose labels lie in te first stack imply te desired partition I of J. Suppose now tat tere exists a subset I of J = {1, 2,..., m} suc tat i I a i = i J I a i. Witout loss of generality, we furter suppose tat I J I. For eac site s i wit i I, we coose to place its label to te first stack. Te remaining labels are placed to te second stack. To complete te reduction, we ave to describe ow to connect eac site wit its corresponding label suc tat te implied labeling is legal. Te leaders of te sites wose labels lie in te first stack are of type-o, i.e. orizontal line segments wit no bends. In tis case, te ports of bot sites and labels can be cosen arbitrarily. Since te labeling is tigt (i.e. te sum of te label eigts on eac stack is equal to H), we use te fact tat te labels are open and use te gaps between tem as corridors to route te leaders, wose labels lie in te second stack. Since we assumed tat I J I, tere exist enoug corridors to route all leaders, adopting te following scenario: Te leader wic corresponds to te lowest label tat as not been routed yet, can use te lowest available corridor. In tis case te site ports are defined based on te corridors, wereas te label ports can be cosen arbitrarily again.

9 9 Fig. 10: Instance to wic Partition is reduced. 4.2 Two candidate points wit type-opo leaders at two stacks on one side. We will sow tat te problem remains NP -ard even if we restrict ourselves in sites, wic may ave two candidate points, i.e. leader of site s i connects eiter point p 1 i = (x1 i, y1 i ) or point p2 i = (x2 i, y2 i ) wit label l i. To sow NP -ardness, we reduce te following variant of Partition to our problem. Te NP -ardness of tis problem follows easily from te Even Odd Partition problem (see [8]). Lemma 5 (Partition). Given 2m non-negative integers a 1, a 2,... a 2m, te problem of finding a subset I of J = {1, 2,... 2m} suc tat te following tree conditions are satisfied is NP -ard. 1. I contains exactly one of {2i 1, 2i} for i = 1, 2,... m 2. i I a i = i J I a i 3. i I&i k a i < i J I&i k a i for k = 2, 4,... 2m 2 Teorem 4. Given a rectangle of eigt H, a set P of n sites, eac associated wit two candidate points, and a label of eigt i for eac site s i P, it is NP -ard to place all labels at two stacks on one side of wit non-intersecting type-opo leaders. Proof. We will reduce an instance A of Partition to an instance B of our problem, suc tat A can be partitioned if and only if tere exists a legal labeling for instance B. Let C be a very large number, e.g. C = (2m + 1) 2 i J a i. Set P = {s 1, s 2,..., s 2m } consists of 2m sites. Site s i is associated wit p 1 i = (x i, yi 1) and p2 i = (x i, yi 2). Consecutive sites s 2i 1 and s 2i, i = 1, 2,..., m, form m parallelograms r i, i = 1, 2,..., m, suc tat y2i 1 1 < y1 2i < y2 2i 1 < y2 2i and y2 2i 1 y1 2i = a2i 1+a2i 2 +1 (see Figure 11). We assume tat parallelogram r i 1 is placed lower tan r i. Te vertical distance between two consecutive parallelograms is C, wereas te vertical distance between te bottommost (topmost) parallelogram r 1 (r m ) and te bottommost (topmost respectively) side of te enclosing rectangle is C/2. Te eigt of te enclosing rectangle is H = m(c + 1) i J a i. Te corresponding label l i of site s i, as eigt i = C + a i + 1, tus i J i = 2m(C + 1) + i J a i. Observe, tat bot stacks contribute 2H eigt, wic is equal to te sum of all label eigts. Te construction ensures tat te same number of labels are placed at te two stacks. For te sake of contradiction, suppose tat te number of labels of te first stack is m + δ, were δ 1. Ten, te sum of te corresponding label eigts is at least (m + δ)(c + 1), wic is greater tan H, since δc > i J a i, yielding to a contradiction. If tere exists a legal labeling L for instance B, te labels sould be partitioned into two sets of equal cardinality, suc tat te sum of te label eigts of eac set sums up to H. Since te labeling L is tigt, we use te fact tat te labels are open and use te gaps between te labels of te first stack as corridors for te leaders of te sites routed at te second stack. Observe, tat due to te large label eigts and te point locations of te sites, all leaders sould bend in te first track routing area. Two consecutive sites s 2i 1 and s 2i, i = 1, 2,... m can not ave teir labels bot at te first stack (consequently at te second stack), because at least one corridor is lost and terefore at least one label at te second stack can not be routed. To avoid leader crossings, te order of indices sould be preserved at bot stacks, i.e. label l i will be stacked lower tan l j if

10 10 s 2m s 2m 1 (r,t ) s 2 s 2i s 2m s 2m 1 (r,t ) C H 1 + a 2i+a 2i 1 2 (l,b ) ic C 2 s 2i s 2 s 2i 1 W s 1 H (l,b ) W s 2i 1 3iC 2 s 1 Fig. 11: Instance to wic Partition is reduced. Fig. 12: Instance to wic Partition is reduced. i < j. To connect all sites wit teir labels, it must eiter old i I&i k i < i J I&i k i or i I&i k i > i J I&i k i for all k = 2, 4,... 2m 2, wic is equivalent to te tird condition of Partition, since all labels are augmented by C + 1. Terefore, te indices of te sites wose labels lie on te first or on te second stack respectively imply te desired partition I of J. Suppose tat tere exists a subset I of J = {1, 2,..., 2m} of instance A suc tat all tree conditions of Partition are satisfied. If i I, ten te label of site s i is placed at te first stack preserving te order of indices, i.e. label l i will be stacked lower tan l j, if i < j. Te remaining labels (i J I) are placed to te second stack in te same manner. Tis ensures tat all labels of te second stack can be it by a leader. Suppose for te sake of contradiction tat a label of te second stack, say te k-t from te bottom, can not be it by a leader. Tis implies tat te sum of te label eigts of te k 1 bottommost labels of te second stack is greater tan te corresponding sum of te k bottommost labels of te first stack, wic is a contradiction due to te tird condition of Partition, and te facts tat all labels are augmented by C + 1 and C > i J a i. A legal labeling can be obtained as follows: Take te lowest site wic as not been routed. If its label is to be placed at te second stack, use te lowest available corridor for its leader, else route it at te first stack wit a type-o leader. Since two consecutive sites s 2i 1 and s 2i form a parallelogram r i, we can determine in constant time wic point of bot s 2i 1 and s 2i will be used, suc tat teir leaders do not intersect. 4.3 Type-po leaders at two stacks on one side. We focus on type-po leaders, were eac site s i corresponds to a point = (x i, y i ) of te plane and is associated wit a label l i of eigt i. Te labels are located to te east side of te enclosing rectangle at two stacks. We seek to find a legal labeling. Teorem 5. Given a rectangle of eigt H, a set P of n and a label of eigt i for eac site s i P, it is NP -ard to place all labels at two stacks on one side of wit non-intersecting type-po leaders. Proof. As in te proof of Teorem 4, we will reduce an instance A = {a 1, a 2,... a m } of Partition to an instance B of our problem, suc tat A can be partitioned if and only if tere exists a legal labeling for B. Witout loss of generality, we furter suppose tat a 1 < a 2. Let C be a very large number, e.g. C = (2m + 1) 2 i J a i. Our point set P = {s 1, s 2,..., s 2m } consists of 2m sites. Site s i is associated wit a label l i of eigt i = C + a i. Te eigt of te enclosing rectangle is H = mc i J a i. Bot stacks contribute 2H eigt, wic is equal to te sum of all label eigts. As sown in proof of Teorem 4, te construction ensures tat te same

11 number of labels are placed at te two stacks. In te construction we propose te odd and even indexed sites lie on two separate lines. A detailed description of teir exact positions is given below. Sites s 2i 1, i = 1, 2,... m 1 lie on te line, wic is defined from te points (r, b ) and (l + W/2, t ), suc tat: y 1 = b and y 2i 1 = b + (2i 1)C/2, i = 2, 3,... m 1. Sites s 2i, i = 1, 2,... m lie on te line, wic is defined from points (l, t C) and (l +W/2, t ), suc tat: y 2 = t C and y 2i+2 y 2i = C m, i = 1, 2,..., m 1. Site s 2m 1 is placed at x 2m 1 = x 2m+x 2m 3 2 and suc tat 0 < (y 2m y 2m 1 ) 0 Suc a construction is depicted in Figure 12 and can be obtained in linear time. Te last condition (i.e. 0 < (y 2m 1 y 2m ) 0) ensures tat no orizontal leader can lie between s 2m 1 and s 2m. Suppose tat tere exists a subset I of J = {1, 2,..., 2m} of instance A, suc tat all conditions of Partition are satisfied. If i I, ten te label of site s i is placed at te first stack preserving te order of indices, i.e. if i < j, label l i will be stacked lower tan l j. Te remaining labels are placed at te second stack in te same manner. Following similar arguments as in proof of Teorem 4, we can sow tat tere exists no label wic can not be it by a leader. A legal labeling can be obtained as follows: Take te lowest site wic as not been routed. If its label is to be placed at te second stack, use te lowest available corridor for its leader, else route it at te first stack. In te oter direction, suppose tat tere exists a legal labeling L for instance B. Ten, te labels sould be partitioned into two sets of equal cardinality, suc tat te sum of te label eigts of eac set sums up to H. Since te labeling L is tigt, we use te fact tat te labels are open and use te gaps between te labels of te first stack as corridors for te leaders of te sites routed at te second stack. We first focus on te case, were label l 1 is placed at te first stack. Ten it olds, for all k = 1, 2,..., m 1, tat te sum of te label eigts of te k bottommost labels of te first stack is less tan te sum of te eigts of te k bottommost labels of te second stack, oterwise at least one label of te second stack would be unable to be connected to its site. Label l 1 can only be te bottommost label of te first stack, oterwise at least one corridor (te one wic will be used for te leader of te bottommost label of te second stack) will be lost. Similar claim can be proved for label l 2, wic corresponds to te leftmost site s 2, i.e. tat l 2 is te bottommost label of te second stack. For te sake of contradiction, suppose tat l 2 is at te first stack. Obviously it can not be te bottommost label, since tis is label l 1. If l 2 is te topmost label and te proposed leader is of type-o, ten several crossings wit leaders of even indexed sites or s 2m 1 exist, a contradiction since L is legal. type-po, ten te o segment of tis leader eiter lies between s 2m 1 and s 2m (a contradiction since 0 < (y 2m 1 y 2m ) 0) or between s 2i 2 and s 2i, and ence several crossing wit leaders of even indexed sites exist, a contradiction since L is legal. Suppose tat l 2 is te k-t bottommost label (of te first stack), were 1 < k < m. In tis case labeling L as to contain crossings, because only k 1 sites can be connected to te (k 1)-t bottommost labels of bot stacks (totally 2k 2 labels) witout crossings. Terefore, l 2 sould be at te second stack. If l 2 is te topmost label of tis stack, ten its leader would overlap wit site s 2m. Using similar arguments as in te case of l 2 being te k-t bottommost label at te first stack, we conclude tat label l 2 can only be te bottommost label of te second stack. Inductively, we can sow tat two sites s 2i 1 and s 2i, i = 1, 2,... m can not ave teir labels bot at te same stack and tat label l i sould be stacked lower tan l j if bot l i and l j belong to te same stack and i < j. So, te indices of te sites wose labels lie at te first stack imply te desired partition I of J, since we ave supposed tat a 1 < a 2 and l 1 lies in te first stack. Similarly, 11

12 12 we can prove tat if l 1 is placed at te second stack, ten te desired partition I of J is formed from te indices of te labels at te second stack. Since, eac point site can be tougt as a line site of zero lengt, Corollary 1 follows immediately. Corollary 1. Given a rectangle of eigt H, a set P of n lines (sites) and a label of eigt i for eac site s i P, it is NP -ard to place all labels at two stacks on one side of wit non-intersecting type-po leaders. Following similar arguments as in proofs of Teorems 4 and 5, one can sow tat te problem remains NP -ard even if we restrict ourselves in sites, wic may ave two candidate points, i.e. leader of site s i connects eiter point p 1 i = (x1 i, y1 i ) or point p2 i = (x2 i, y2 i ) wit label l i. Teorem 6. Given a rectangle of eigt H, a set P of n sites, eac associated wit two candidate points, and a label of eigt i for eac site, it is NP -ard to place all labels at two stacks on one side of wit non-intersecting type-po leaders. 5 Open Problems and Future Work For multi-stack labeling problems we presented results only for te label size maximization problem and te legal leader-label placement. No results are known regarding te total leader lengt minimization and te minimization of te total number of bends. Anoter line of researc is to design good approximation algoritms tat solve te problems, tat are proved to be N P -ard. eferences 1. M. A. Bekos, M. Kaufmann, K. Potika, and A. Symvonis. Boundary labelling of optimal total leader lengt. In Proc. 10t Panellenic Conference On Informatics (PCI2005), LNCS 3746, pages 80 89, M. A. Bekos, M. Kaufmann, K. Potika, and A. Symvonis. Polygon labelling of minimim leader lengt. In Asia Pacific Symposium on Information Visualisation (APVIS2006), CPIT 60, pages 15 21, M. A. Bekos, M. Kaufmann, A. Symvonis, and A. Wolff. Boundary labeling: Models and efficient algoritms for rectangular maps. Computational Geometry: Teory and Applications. To appear. 4. M. A. Bekos, M. Kaufmann, A. Symvonis, and A. Wolff. Boundary labeling: Models and efficient algoritms for rectangular maps. In Proc. 12t Int. Symposium on Grap Drawing (GD 04), LNCS 3383, pages 49 59, M. A. Bekos and A. Symvonis. Bler: A boundary labeller for tecnical drawings. In Proc. 13t Int. Symposium on Grap Drawing (GD 05), LNCS 3843, pages , P. Eades, T. Lin, and X. Lin. Minimum size -v drawings. In Advanced Visual Interfaces (Proceedings of AVI 92), volume 36 of World Scientific Series in Computer Science, pages , M. Formann and F. Wagner. A packing problem wit applications to lettering of maps. In Proc. 7t Annuual ACM Symposium on Computational Geometry (SoCG 91), pages , M.. Garey and D. S. Jonson. Computers and Intractability: A Guide to te Teory of NP- Completeness. W. H. Freeman, New York, NY, 1979, SmartDraw-7. Product web site: ttp:// 10. L. Stockmeyer. Optimal orientations of cells in slicing floorplan designs. Information and Control, 57(2 3):91 101, A. Wolff and T. Strijk. Te Map-Labeling Bibliograpy. ttp://i11www.ira.uka.de/maplabeling/bibliograpy, 1996.