2-Layer Right Angle Crossing Drawings

Transcription

1 DOI /s Layer Right Angle Crossing Drawings Emilio Di Giacomo Walter Didimo Peter Eades Giuseppe Liotta Received: 26 July 2011 / Accepted: 24 October 2012 Springer Science+Business Media New York 2012 Abstract A 2-layer drawing represents a bipartite graph where each vertex is a point on one of two parallel lines, no two vertices on the same line are adjacent, and the edges are straight-line segments. In this paper we study 2-layer drawings where any two crossing edges meet at right angle. We characterize the graphs that admit this type of drawing, provide linear-time testing and embedding algorithms, and present a polynomial-time crossing minimization technique. Also, for a given graph G and a constant k, we prove that it is NP-complete to decide whether G contains a subgraph of at least k edges having a 2-layer drawing with right angle crossings. Keywords Graph drawing Right angle crossings 2-layer drawings Efficient algorithms Testing algorithms NP-hardness 1 Introduction The study of drawings of graphs where any two crossing edges form crossing angles that are not too small is among the emerging topics in Graph Drawing. This interest Work supported in part by MIUR of Italy under project AlgoDEEP prot. 2008TFBWL4. An abstract of this work was presented at the International Workshop on Algorithms and Combinatorics (IWOCA 2011) [7]. E. Di Giacomo W. Didimo ( ) G. Liotta Università di Perugia, Perugia, Italy didimo@diei.unipg.it E. Di Giacomo digiacomo@diei.unipg.it G. Liotta liotta@diei.unipg.it P. Eades University of Sydney, Sydney, Australia peter@cs.usyd.edu.au

2 is motivated by recent experiments of Huang et al. [23, 24], who show that crossing angles guarantee good readability properties only if they are large enough (approximatively larger than π 3 ). These experiments therefore imply that non-planar drawings of graphs should not only be optimized in terms of classical parameters such as the number of edge crossings and the number of bends along the edges, but also in terms of the minimum angle formed by any two crossing edges. In thispaperwe study straight-linerightanglecrossingdrawings (or RAC drawings for short). In a RAC drawing any two crossing edges meet at angle π 2.RAC drawings have been first introduced in [10, 12], where it is proved that straight-line RAC drawings with n vertices have at most 4n 10 edges, which is a tight bound. Straight-line RAC drawings are also studied by Dujmović et al. [15],whogivean alternative proof of the 4n 10 bound. The relationship between straight-line RAC drawings with 4n 10 edges and 1-planar graphs is studied in [18]. Angelini et al. [3] investigate straight-line upward RAC drawings of digraphs. Van Kreveld [30] studies how much better a straight-line RAC drawing of a planar graph can be than any straight-line planar drawing of the same graph. Complete bipartite straight-line RAC drawable graphs are studied in [11]. More references on RAC drawings can be found in [13]. Despite the growing literature about straight-line RAC drawings, no algorithms for computing such drawings have been described so far. Existing papers either establish combinatorial properties of RAC drawings (they typically address Turán-type questions) or study RAC drawings with bends along the edges (see, e.g., [1, 3, 5, 9, 10, 12]). Also, deciding whether a graph admits a straight-line RAC drawing is NP-hard in the general case [4] and it is not even known if this problem is in NP. In this paper we present the first efficient algorithms for straight-line RAC drawings. We focus on bipartite graphs and consider both the problem of deciding whether a bipartite graph admits a straight-line RAC drawing and the problem of computing one in the positive case. We also study how to efficiently compute straight-line RAC drawings of bipartite graphs with the minimum number of edge crossings. We adopt the widely accepted 2-layer drawing paradigm, in which the vertices of each partition set lie on a distinct horizontal layer. This paradigm is motivated by the fact that it clearly conveys the bipartite structure of the graph. Also, it represents a building block for the popular Sugiyama-style framework, adopted for visualizing general graphs on several layers [27, 31]. A limited list of papers on 2-layer drawings of bipartite graphs includes [17, 21, 25, 29]; for more references see also [26]. An overview of our results is given below (n denotes the number of vertices of the input graph). A 2-layer RAC drawing is a 2-layer straight-line drawing with right angle crossings. We characterize 2-layer RAC drawable graphs (Theorem 6). This is the counterpart for RAC drawings of the well-known characterization of 2-layer planar drawings (see, e.g., [19, 22, 28]). Our characterization implies that 2-layer RAC drawable graphs have at most 1.5n 2 edges, which is a tight bound (Corollary 1); also it allows us to describe an O(n)-time algorithm that tests whether a graph admits a 2-layer RAC drawing and, if so, it computes one.

3 We show an O(n 2 log n)-time algorithm to compute a 2-layer RAC drawing with the minimum number of edge crossings (Theorem 7). The algorithm models the optimization problem as the one of computing a flow of minimum cost on a suitable network. We recall that computing a 2-layer drawing of a graph with the minimum number of crossings is NP-hard and that heuristics [17], approximation algorithms [21], FPT algorithms [14, 16], and exact methods [25, 29] for this problem have been described. Finally, we study the complexity of computing the maximum 2-layer RAC drawable subgraph. We prove that for a given bipartite graph G and for a given k, itis NP-complete to decide whether G has a 2-layer RAC drawable subgraph with at least k edges (Theorem 8). This extends to RAC drawings the NP-completeness result for the maximum 2-layer planar subgraph problem [20]. The rest of the paper is organized as follows. Section 2 gives preliminary definitions and proves that the geometric problem of computing a 2-layer RAC drawing reduces to the combinatorial problem of computing a 2-layer RAC embedding. In Sect. 3 an overview of the characterization of 2-layer RAC drawable graphs is given. Sections 4 and 5 present the characterization of biconnected 2-layer RAC drawable graphs and 2-layer RAC drawable trees, respectively. The general characterization of 2-layer RAC drawable graph is given in Sect. 6. The problem of computing a 2-layer RAC drawing with the minimum number of edge crossings is studied in Sect. 7. The NP-hardness result about the maximum 2-layer RAC drawable subgraph problem is presented in Sect. 8. Open problems are listed in Sect Preliminary results: Geometry and Combinatorics Let G = (V 1,V 2,E) be a bipartite graph. A 2-layer drawing of G has a fan crossing if there exist two adjacent edges that are both crossed by a third edge; we say that the three edges involved in the fan crossing form a fan crossing. For a given 2-layer drawing of G, denote by l i the horizontal line on which the vertices of V i are drawn (i = 1, 2). We always assume that l 1 is above l 2. Two 2-layer drawings of G are equivalent if they have the same left-to-right order π i of the vertices of V i (i = 1, 2) along l i.a2-layer embedding is an equivalence class of 2-layer drawings and it is described by a pair of linear orderings (i.e., permutations) γ = (π 1,π 2 ) of the vertices in V 1 and V 2, respectively. If Γ is a drawing within class γ, we also say that γ is the embedding of Γ. Let Γ 1 and Γ 2 be 2-layer drawings of G with the same embedding γ. Two edges e and e cross in Γ 1 if and only if they cross in Γ 2. We say that embedding γ has a crossing at edges e and e. Also, three edges e, e and e form a fan crossing in Γ 1 if and only if they form a fan crossing in Γ 2. Correspondingly, we say that embedding γ has a fan crossing at edges e, e, and e.letγ = (π 1,π 2 ) be a 2-layer embedding of a bipartite graph G. The first vertex of π 1 and the first vertex of π 2 are called the leftmost vertices of γ ; the last vertex of π 1 and the last vertex of π 2 are called the rightmost vertices of γ. The following result proves that the problem of computing a 2-layer RAC drawing of a bipartite graph G = (V 1,V 2,E) can be studied in purely combinatorial terms as

4 the one of choosing a suitable pair (π 1,π 2 ) of permutations of the vertices in V 1 and V 2, disregarding details about the exact coordinates of the vertices. Theorem 1 Let G be a connected graph with n vertices. G is 2-layer RAC drawable if and only if it has a 2-layer embedding without fan crossings. Also, if γ is a 2-layer embedding of G without fan crossings, there exists an O(n)-time algorithm that, taking γ as input, computes a 2-layer RAC drawing of G whose embedding is γ. Proof The necessity is a consequence of the fact that a fan crossing always implies at least one crossing angle smaller than π 2. In order to prove the sufficiency, we shall describe a drawing algorithm that computes a 2-layer drawing Γ with the following properties: (1) The embedding of Γ is γ ; (2) any edge involved in a crossing has either slope 1 or slope 1. These two properties imply that Γ is a 2-layer RAC drawing: Let e 1 and e 2 be two edges that cross each other in γ and assume that these two edges have the same slope in Γ ; either the two edges overlap or they are parallel and never intersect; in both cases the linear order of the vertices on the layers defined by γ would not be respected. Our algorithm strongly relies on the following two properties: Property 1 If G is connected and has a 2-layer embedding γ = (π 1,π 2 ) without fan crossings, then γ has no three mutually crossing edges. Property 2 If a 2-layer embedding γ = (π 1,π 2 ) has no fan crossings, then for every vertex v at most two of its incident edges can be crossed by other edges. Property 1 is proved by contradiction. Indeed, assume that (u 1,u 2 ), (u 1,u 2 ), and (u 1,u 2 ) are three mutually crossing edges in γ, such that π 1(u 1 )<π 1 (u 1 )<π 1(u 1 ) and π 2 (u 2 )<π 2(u 2 )<π 2(u 2 ). Since G is connected, then at least one of u 1 and u 2 has degree two, say for example u 1. This implies that there exists a vertex v 2 adjacent to u 1 (possibly coincident with either u 2 or u 2) such that (u 1,v 2), (u 1,u 2 ), and one of (u 1,u 2 ) and (u 1,u 2 ) form a fan crossing. Property 2 follows by observing that γ defines the left-to-right order of the edges incident to v and only the leftmost and the rightmost edges incident to v can receive a crossing (otherwise there would be a fan crossing in γ ). The drawing algorithm consists of the following steps (see Fig. 1). For each step we also discuss its time complexity. Since each step can be executed in O(n) time, then the whole algorithm runs in linear time. Notice that, since G is straight-line RAC drawable, then G has O(n) edges [10, 12]. We will assume that G has at least two edges, otherwise G is trivially 2-layer RAC drawable. Fig. 1 Example of the drawing algorithm. The bold edges represent the crossing edges (edges in C). The gray vertices are not incident to crossing edges

5 Step 1 Compute the set C of all edges of G that cross in γ, and denote by G the subgraph of G consisting of all edges in C and its end-vertices. In order to compute G in O(n) time, we first observe that since G is connected, every edge is incident to a vertex of degree at least two, that is, the set of vertices of degree at least two covers the whole set of edges of G. For each vertex v of degree at least two, denote by (v, a) and (v, b) the leftmost and the rightmost edge of v in γ, respectively. Clearly, the leftmost and the rightmost edge for all vertices can be detected and stored in linear time. By Property 2, no edge incident to v other than (v, a) and (v, b) can cross other edges. Also, if an edge crosses (v, a) or (v, b), one of its end-vertices must be in between a and b on their layer (note that an edge cannot cross both (v, a) and (v, b), otherwise there would be a fan crossing). If there is no vertex in between a and b (which can be established in constant time), then we can conclude that all edges incident to v are crossing free. Otherwise, let a be the vertex immediately to the right of a on its layer, and let b be the vertex immediately to the left of b on its layer. Observe that a and b may coincide. We claim that if (v, a) and (v, b) do not cross some edges incident to a or b, then they cannot cross any other edge. Indeed, suppose that neither (v, a) nor (v, b) cross an edge incident to a or b : (i) If a = b is the only vertex in between a and b on its layer, then the claim is trivial; (ii) If a b, since a and b cannot be isolated vertices, G would contain the edges (v, a ) and (v, b ), and any edge incident to a vertex in between a and b that crosses either (v, a) or (v, b) would also cross either (v, a ) and (v, b ), thus causing a fan crossing, a contradiction; this proves the claim. It follows that, to establish whether (v, a) and (v, b) cross some other edges, it is sufficient to check if each of them crosses one among the leftmost and the rightmost edges of a and of b. This can be done in constant time for each vertex v. Step 2 Compute the set P of connected components of G. Since by Property 2 at most two edges per vertex cross, each connected component is a path. This step can clearly be executed in O(n) time. Step 3 Iteratively draw each path of P, while maintaining the vertex ordering defined by γ on each layer. Each path is drawn in such a way that each pair of consecutive edges have slopes 1 and 1. To perform Step 3 in O(n) time, define a suitable left-to-right ordering for the paths of P, and draw a path at a time according to this order. More in details, the first path is any path containing a leftmost vertex of γ (there are at most two of these paths). Assume that at the generic step we have added a path Π to the drawing and that e is the leftmost edge of Π. In the next step we consider the embedding γ consisting only of those paths of P that have not been drawn yet, and consider those paths of P that are incident to the leftmost vertices of γ. Since there are at most two leftmost vertices in γ, then we have either one or two of such paths. If there is only one, then it is the next path to be drawn. If there are two of such paths, say Π and Π, then their leftmost edges necessarily cross; denote by e and e the leftmost edges of Π and Π.Itis not possible that both e and e cross e, otherwise there would be three mutually crossing edges in γ (which is not possible by Property 1). If neither e nor e crosses e, then we arbitrarily chose Π or Π as the next path to be drawn. If one of the two edges crosses e, then we chose its path as the next path to be drawn.

6 Step 4 Add to the drawing the vertices and the edges of G not yet drawn, if any. Each vertex is drawn in such a way that the vertex ordering defined by γ is maintained. This can be easily done because we are working with real coordinates. For example, suppose that u, v, and z are three consecutive vertices of γ on the same layer; also, suppose that u and z are already in the drawing while v is not; it is sufficient to draw v at any point in between u and z. Hence, this step can be computed in O(n) time. A 2-layer embedding without fan crossings is called 2-layer RAC embedding. Based on Theorem 1, the problem of characterizing 2-layer RAC drawable graphs is equivalent to characterizing which graphs have a 2-layer RAC embedding. Note that, a graph is 2-layer RAC drawable if and only if its connected components are 2-layer RAC drawable. Hence, from now on we assume to work on connected graphs. The following notations will be used in the remainder of the paper. Given two vertices a and b of a path, we denote by d(a,b) the distance between a and b on the path, that is, the number of edges from a to b along the path. Given a vertex v, the degree of v is the number of edges incident to v and is denoted as deg(v). 3 Characterization of 2-layer RAC Drawable Graphs: Overview In this section we give an overview of the characterization of 2-layer RAC drawable graphs. In the more general case, a 2-layer RAC drawable graph consists of a set of non-trivial biconnected components 1 and a set of tree components that are partially ordered along the two layers (two components are ordered if their drawings can be separated by a line cutting the two layers). Such an order is not a total order because there can be some overlap between different tree components (two components overlap if an edge of a component crosses an edge in the other component). For example, Fig. 2(b) shows a 2-layer RAC drawing of the graph G depicted in Fig. 2(a). G consists of two non-trivial biconnected components and five tree components (highlighted in Fig. 2(c)) that are partially ordered left-to-right along the two layers in the drawing; as shown in Fig. 2(d) the left-to-right order is not a total ordering because in some cases the tree components overlap as it happens for components T 1 and T 2 and for components T 4 and T 5 in the drawing of Fig. 2(b). In what follows we first characterize biconnected 2-layer RAC drawable graphs (Sect. 4), and then 2-layer RAC drawable trees (Sect. 5). These two characterizations will then be combined to characterize 2-layer RAC drawable graphs. The combination, however, is not straightforward because the drawings of the different components of G must satisfy additional properties in order to be combined together. In particular, it must be possible to order the components along the two layers in such a way that: (i) Each component shares vertices only with the components that immediately precede and follow it and with the components that overlap with it; (ii) the vertices shared by two consecutive components C 1 and C 2 (in this left-toright order) must be the rightmost for C 1 and the leftmost for C 2. For example, the 1 A trivial biconnected component consists of a single edge.

7 Fig. 2 (a) AgraphG. (b) A 2-layer RAC drawing Γ of G. (c) The biconnected components and the tree components of G. (d) The biconnected components and the tree components of G in Γ.(e) AgraphG. (f) G does not admit a 2-layer RAC drawing graph G of Fig. 2(e) has a biconnected component (isomorphic to B 1 in Fig. 2(c)) and a tree component (consisting of edge (3, 9)). Both components are 2-layer RAC drawable, but the biconnected component does not admit a 2-layer RAC drawing with the vertex 3 as the leftmost or the rightmost vertex; as a consequence whatever the position of vertex 9 will be, it is not possible to obtain a 2-layer RAC drawing of G (see Fig. 2(f)). For this reason, Sect.s 4 and 5 contain: (i) the characterizations of biconnected 2-layer RAC drawable graphs and 2-layer RAC drawable trees; (ii) the characterization of biconnected graphs that are 2-layer RAC drawable in such a way that two specified edges are the leftmost and the rightmost ones; and (iii) the characterization of trees that are 2-layer RAC drawable in such a way that two specified leaves are the leftmost and the rightmost ones. 4 Characterization of Biconnected Graphs Let G be a biconnected bipartite graph with at least four edges, and let e and e be two independent edges of G. If there exists a 2-layer RAC embedding γ of G such that the end-vertices of e are the leftmost vertices of γ and the end-vertices of e are the rightmost vertices of γ, we say that γ is a 2-layer RAC embedding of G with respect to e and e. Clearly, edges e and e cannot cross any edge in γ. A 2-layer RAC drawing with embedding γ is called a 2-layer RAC drawing with respect to e and e.ifg admits such a drawing, we also say that G is 2-layer RAC drawable with respect to e and e. Figure 3(b) shows an example of a 2-layer RAC drawing with respect to edges (1, 10) and (5, 6). A biconnected bipartite graph is a ladder if it consists of two paths of the same length u 1,u 2,...,un 2 and v 1,v 2,...,vn 2 plus the edges (u i,v i ) (i = 1, 2,..., n 2 ).

8 Fig. 3 (a) A bipartite simple cycle C. (b) A 2-layer RAC drawing of C with respect to edges (1, 10) and (5, 6); the drawing has an interlaced embedding. (c) A ladder with extremal edges e and e.(d) A2-layer RAC drawing of the ladder with respect to e and e Figure 3(c) shows an example of a ladder. Note that, a ladder is a maximal bipartite outerplanar graph where all vertices have degree at most three and exactly four vertices u 1,v 1,un,vn 2 2 have degree two. The edges e = (u 1,v 1 ) and e = (u n,vn 2 2 ) are called the extremal edges of the ladder. We will now prove that a biconnected graph G is 2-layer RAC drawable with respect to two independent edges e and e if and only if it is a spanning subgraph of a ladder having e and e as extremal edges (Theorem 2). Figure 3(d) shows a 2-layer RAC drawing of the ladder in Fig. 3(c). Theorem 2 also proves the existence of an efficient algorithm that tests whether G admits a 2-layer RAC drawing with respect to e and e and, if so, it computes such a drawing. Theorem 3 is the counterpart of Theorem 2 without any fixed extremal edges. We give some preliminary definitions and lemmas. Let C be a bipartite simple cycle and let γ be a 2-layer embedding of C. Denote by a 1,a 2,...,an 2 the left-to-right sequence of vertices on one layer and by b 1,b 2,...,bn 2 the left-to-right sequence of vertices on the other layer in γ. We say that γ is an interlaced embedding if the edges of C are (a 1,b 1 ), (a n,bn 2 2 ), (a i,b i+1 ), (a i+1,b i ) for each i {1,..., n 2 1} (see Figs. 3(a) and 3(b) for an example). In other words, if γ is an interlaced embedding of C, then γ is a 2-layer RAC embedding with respect to two edges e and e and each edge distinct from e and e is crossed exactly once. Let γ be an interlaced embedding with three consecutive vertices a 1, b 1, c 1 on one layer and three consecutive vertices a 2, b 2, c 2 on the other layer, such that the edge (b 1,a 2 ) crosses the edge (b 2,a 1 ) and the edge (b 1,c 2 ) crosses the edge (b 2,c 1 ). We say that b 1 and b 2 are opposite vertices of γ. For example, in the embedding of Fig. 3(b), the pairs {9, 2}, {3, 8}, and {7, 4} are pairs of opposite vertices.

9 Lemma 1 Let C be a bipartite simple cycle. C is a 2-layer RAC drawable graph and any 2-layer RAC embedding of C is an interlaced embedding. Proof Clearly, C always admits an interlaced embedding, and therefore it is 2-layer RAC drawable by Theorem 1. On the other hand, suppose γ is a 2-layer RAC embedding of C. We prove that γ is an interlaced embedding. Let a 1,a 2,...,an 2 be the left-to-right sequence of vertices on one layer and let b 1,b 2,...,bn 2 be the left-toright sequence of vertices on the other layer in γ. If edge (a 1,b 1 ) is not in C, then, since a 1 has degree two, C has two edges (a 1,b h ), (a 1,b k ) with h, k > 1. This would imply that any edge (a j,b 1 )(j>1) crosses both the edges (a 1,b h ), (a 1,b k ), causing a fan crossing, a contradiction. It follows that C contains edge (a 1,b 1 ). With a symmetric argument, C contains edge (a n,bn 2 2 ). We now prove by induction on i that C contains edges (a i,b i+1 ) and (a i+1,b i ),fori {1,..., n 2 1}. Fori = 1wehave to prove that C contains edges (a 1,b 2 ) and (a 2,b 1 ). If edge (a 1,b 2 ) is not in C, then, since a 1 has degree two, C has an edge (a 1,b h ) for h>2, other than (a 1,b 1 );but this would imply that the two edges incident to b 2 form a fan crossing with (a 1,b h ), a contradiction. For the same reason, C has the edge (a 2,b 1 ). Assume now that C contains the edges (a i 1,b i ), (a i,b i 1 ) for i>2. If edge (a i,b i+1 ) is not in C, then C would have an edge (a i,b i+h ) for h>1. Also, b i+1 cannot be adjacent to a vertex to the left of a 1, because each of these vertices has already degree two by inductive hypothesis. Hence, the two edges incident to a i+1 would be incident to vertices that are to the right of a i, but in this case they would form a fan crossing with edge (a i,b i+h ). It follows that edge (a i,b i+1 ) is in C. With a symmetric argument it can be proved that edge (a i+1,b i ) is in C. Lemma 2 Let γ be an interlaced embedding of a simple cycle C and let u and v be two non-adjacent vertices of C. The embedding γ with the addition of edge (u, v) is a 2-layer RAC embedding if and only if u and v are opposite vertices of γ. Proof As also observed in the proof of Lemma 1, any interlaced embedding of a cycle is a 2-layer RAC embedding by Theorem 1. Ifu and v are opposite vertices of γ, the addition of edge (u, v) does not create crossings, and therefore the new embedding is still a 2-layer RAC embedding. On the other hand, if u and v are not opposite vertices of γ, the addition of edge (u, v) causes a fan crossing, and therefore the new embedding is not a 2-layer RAC embedding. Lemma 3 Let G be a biconnected bipartite graph. G is a 2-layer RAC drawable graph only if it is outerplanar. Proof If G consists only of one edge the lemma trivially holds. Suppose that G has at least four vertices. We assume that the vertices of G are colored black and white, according to the bipartition. Since G is biconnected and has at least four vertices, it contains a simple cycle. Let C be the longest simple cycle of G. IfG is outerplanar, C contains all vertices of G and placing all the vertices of G on a circumference (in the cyclic order they occur in C) we get a planar drawing of G.IfG is not outerplanar, two cases are possible.

10 Fig. 4 Illustration for the proof of Lemma 3 Case 1: C contains all vertices of G, but placing these vertices on a circumference there are two crossing edges (refer to Fig. 4(a)). Denote by (u, z) and (v, w) two crossing edges of G, where the vertices u, v, z, w are encountered in this order walking on the circumference clockwise. According to Lemma 1 every 2-layer RAC embedding γ of C is an interlaced embedding. Consider any 2-layer RAC embedding of C and assume, without loss of generality, that π 1 (u) < π 1 (v); this implies that π 2 (w) < π 2 (z), because the interlaced embedding reflects the circular order of the vertices in C. It follows immediately that it is not possible to add to γ both (u, z) and (v, w) without creating a fan crossing. By Theorem 1, this implies that G is not 2-layer RAC drawable, a contradiction. Case 2: C does not contain all vertices of G (refer to Fig. 4(b)). According to Lemma 1 every 2-layer RAC embedding γ of C is an interlaced embedding. Let v beavertexofg that is not contained in C. Without loss of generality, assume that v is white. Let w and z be two black vertices of C (note that C contains at least two black vertices and two white vertices). Since G is biconnected, there exist two edge disjoint paths Π vw, Π vz, such that Π vw goes from v to w, Π vz goes from v to z.walk along Π vw from v to w and along Π vz from v to z. Denote by (w,w ) the first edge along Π vw such that w does not belong to C and w belongs to C. Note that, w may coincide with v and that w may coincide with w. Similarly, denote by (z,z ) the first edge along Π vz such that z does not belong to C and z belongs to C (again, z may coincide with v and z main coincide with z). If w is not a leftmost/rightmost vertex of γ then edge (w,w ) would cause a fan crossing with some edges of γ (see, e.g., Case 2.1 of Fig. 4(b)). Analogously, if z is not a leftmost/rightmost vertex of γ then edge (z,z ) would cause a fan crossing. Also, if w is a leftmost vertex of γ and z is a rightmost vertex of γ, then for every placement of v there would be at least one edge in the subpath of Π vw from v to w or at least one edge in the subpath of Π vz from v to z that causes a fan crossing with some edges of γ (see, e.g., Case 2.2 of Fig. 4(b)). Finally, suppose that w and z are both leftmost (or both rightmost) vertices of γ. In this case it would be possible to enhance γ with an embedding γ 1 of the subpath of Π vw going from v to w and with an embedding γ 2 of the subpath of Π vz going from v to z, in such a way that γ = γ γ 1 γ 2 without the edge (w,z )

11 is an interlaced embedding of a new cycle C, which contains more vertices than C (see, e.g., Case 2.3 of Fig. 4(b)); this contradicts the hypothesis that C is the longest simple cycle in the graph. It follows that C cannot exist. Lemma 4 Every ladder is 2-layer RAC drawable with respect to its extremal edges. Proof Let G be a ladder. Since by definition of ladder G is a maximal bipartite outerplanar graph and all its vertices have degree at most three, then G has a planar embedding where all vertices are on the external boundary, all internal faces of G are quadrilateral, and there are no two internal edges incident to the same vertex. Hence, a 2-layer RAC drawing of G can be constructed so that the external boundary has an interlaced embedding γ and each internal edge connects two opposite vertices of γ. Embedding γ is such that the end-vertices of the extremal edges of G are the leftmost and the rightmost vertices, respectively. Theorem 2 Let G be a biconnected graph and let e and e be two independent edges of G. G is 2-layer RAC drawable with respect to e and e if and only if it is a spanning subgraph of a ladder with extremal edges e and e. Also, if G has n vertices, there exists an O(n)-time algorithm that tests whether G admits a 2-layer RAC drawing with respect to e and e and, if so, it computes such a drawing. Proof If G is a spanning subgraph of a ladder with extremal edges e and e, then G is 2-layer RAC drawable by Lemma 4. Suppose vice-versa that G has a 2-layer RAC drawing Γ with respect to e and e. By Lemma 3 G is outerplanar. Since G is an outerplanar graph, it admits a planar embedding where all vertices are on the external boundary C. By Lemma 1 Γ is such that cycle C has an interlaced embedding γ ;by Lemma 2 every internal edge of Γ connects two opposite vertices of γ. By connecting all pairs of non-adjacent opposite vertices u, v of γ (if any), we obtain a ladder that contains G as a spanning subgraph and with extremal edges e and e. Concerning the time complexity, the algorithm first checks in O(n) time if G is a bipartite graph and if the vertex degree of G is at most three. Then, it tests in O(n) time whether G admits an outerplanar embedding, by applying the following standard technique: (i) Temporarily adds to G a dummy vertex u and a dummy edge (u, v) for every vertex v of G. (ii) run on the new graph an O(n)-time planarity testing and embedding algorithm (see, e.g., [6]); if the testing algorithm finds a planar embedding, then removing the dummy vertex and edges we get an outerplanar embedding of G, otherwise G is not outerplanar, and the whole test is negative. Suppose that G is a bipartite outerplanar graph with vertex degree at most three, and that the outerplanar embedding of G has been found; since G is biconnected, it has only one outerplanar embedding. Denote by C the external boundary of this embedding. To complete the test, the algorithm has to verify if the two following conditions hold: (a) C consists of two paths of the same length u 1,u 2,...,un 2, v 1,v 2,...,vn 2 such that e = (u 1,v 1 ) and e = (u n 2,vn 2 ). (b) Any edge of G that does not belong to C connects a pair of vertices {u i,v i } (2 i n 2 1). If (a) and (b) are not both true, the test is negative. Otherwise, the test is positive and a 2-layer RAC embedding γ of G with respect to e and e is immediately obtained

12 from the outerplanar embedding of G, using the circular ordering of the vertices along C (see, e.g., Fig. 3(a) and Fig. 3(b)). A 2-layer RAC drawing with embedding γ can be computed in linear time with the constructive technique of Theorem 1. Theorem 3 Let G be a biconnected graph. G is 2-layer RAC drawable if and only if it is a spanning subgraph of a ladder. Also, if G has n vertices, there exists an O(n)- time algorithm that tests whether G is 2-layer RAC drawable and, if so, it computes a 2-layer RAC drawing of G. Proof Let G be a biconnected graph. If G is a spanning subgraph of a ladder then G is 2-layer RAC drawable by Lemma 4. Suppose vice-versa that G is a 2-layer RAC drawable graph. By Lemma 3 G is outerplanar. If G is already a ladder we are done. Assume that G is not a ladder and let Γ be any 2-layer RAC drawing of G. Since G is an outerplanar graph, it admits a planar embedding where all the vertices are on the external boundary C. By Lemmas 1 and 2, Γ is such that cycle C has an interlaced embedding γ and every internal edge of G connects two opposite vertices of γ.by connecting all pairs of non-adjacent opposite vertices u, v of γ we obtain a ladder that contains G as a spanning subgraph. According to the characterization above, a testing algorithm has to check if G is bipartite and if G can be augmented to a ladder (i.e., to a maximal outerplanar graph with vertex degree at most three), by adding some extra edges. Testing if G is bipartite, outerplanar, and with vertex degree at most three is done as described in the proof of Theorem 2. Suppose that G is a bipartite outerplanar graph with vertex degree at most three, and that the outerplanar embedding of G has been found; since G is biconnected, it has only one outerplanar embedding. To complete the test, the algorithm has to decide whether it is possible to add extra edges so that all the internal faces of G are split into faces of degree four while maintaining the graph bipartite and with vertex degree at most three. This can be done as follows. While there exists an internal face f of degree greater than four, consider an internal edge e = (u, v) on the boundary of f.let(u, w) and (v, z) be the edges of the boundary of f that are adjacent to e. Observe that w and z have different colors. If both w and z have degree two then split f into two faces, by adding the new edge (v, z); otherwise, G cannot be augmented to a ladder, the algorithm stops and the test is negative. The test is positive when every internal face has degree four and the algorithm never stopped during the edge augmentation. It is not difficult to see that the edge augmentation can be performed in O(n) time. Also, if the test is positive, a 2-layer RAC embedding γ of G is immediately obtained from the outerplanar embedding of G, usingthe circular ordering of the vertices along the external boundary. A 2-layer RAC drawing with embedding γ can be computed in linear time with the constructive technique of Theorem 1. 5 Characterization of Trees In this section we characterize 2-layer RAC drawable trees. As explained in Sect. 3, along with the characterization of 2-layer RAC drawable trees (Theorem 5), we give

13 the characterization of 2-layer RAC drawable trees where two given vertices u and v are required to be a leftmost vertex and a rightmost vertex, respectively (Theorem 4). The latter characterization will be used in Sect. 6 to characterize 2-layer RAC drawable graphs. In Theorem 4 we focus on the case when u and v are two leaves for two reasons: (i) This is the only case needed for the characterization of 2-layer RAC drawable graphs that contain both non-trivial biconnected components and tree components. (ii) As Lemma 6 proves, any 2-layer RAC drawable tree admits a 2-layer RAC drawing where a leftmost vertex and a rightmost vertex are leaves. The characterization given in Theorem 5 is an immediate consequence of Lemma 6 and Theorem 4. Roughly speaking, a 2-layer RAC drawing of a tree consists of a zig-zag path between the two layers with some suitable sub-structures attached to its vertices. In order to define the different types of sub-structures we define a sort of simplified version of the tree which we call the weighted contraction of the tree. Section 5.1 gives some preliminary definitions, and presents and proves Lemma 6. Weighted contractions are defined in Sect The sufficiency of the characterization is given in Sect. 5.3, while the necessity is proved in Sect Section 5.5 presents and proves Theorems 4 and Basic Properties of 2-layer RAC Drawable Trees We start with definitions and lemmas that can be stated for more general connected graphs. Let γ = (π 1,π 2 ) be a 2-layer embedding of a bipartite graph G and let G be a subgraph of G consisting of a vertex u V i attached to three vertices v 1,v 2,v 3 V j (i, j {1, 2} and i j) where π j (v 1 )<π j (v 2 )<π j (v 3 ). We say that G is a degree- 3blockerin γ and that u is the apex of the degree-3 blocker. An edge (w, z) of G with w V i and z V j is called a traversing edge of G in γ if either π i (w) < π i (u), π j (v 2 )<π j (z) or π i (u) < π i (w), π j (z) < π j (v 2 ). Refer to Fig. 5(a). Let γ = (π 1,π 2 ) be a 2-layer embedding of a bipartite graph G and let G be a subgraph of G consisting of a path v 1,v 2,v 3,v 4, such that v 1,v 3 V 1, v 2,v 4 V 2 and either π 1 (v 1 )<π 1 (v 3 ) and π 2 (v 4 )<π 2 (v 2 ) or π 1 (v 3 )<π 1 (v 1 ) and π 2 (v 2 )< π 2 (v 4 ). We say that G is a butterfly blocker in γ. An edge (w, z) of G with w V 1 and z V 2 is called a traversing edge of G in γ if either π 1 (w) < π 1 (v 3 ), π 2 (v 2 ) π 2 (z) or π 1 (v 3 )<π 1 (w), π 2 (z) π 2 (v 2 ). Refer to Fig. 5(b). If G is either a degree- 3 blocker or a butterfly blocker, every traversing edge of G causes a fan crossing. Hence, by Theorem 1 the following property holds. Fig. 5 (a) Thebold edges form a degree-3 blocker. The remaining edges are traversing edges. (b) The bold edges form a butterfly blocker. The remaining edges are traversing edges

14 Property 3 Let γ = (π 1,π 2 ) be a 2-layer RAC embedding of a bipartite graph G and let G G be either a degree-3 blocker or a butterfly blocker in γ. There is no traversing edge of G in γ. Let γ = (π 1,π 2 ) be a 2-layer embedding of a bipartite graph G. A path Π of G is monotone in γ if, for one of the two possible orientations of Π, every two vertices u, v in Π V i are such that π i (u) < π i (v) (i {1, 2}) if and only if u appears before v in Π. Lemma 5 Let γ = (π 1,π 2 ) be a 2-layer RAC embedding of a connected bipartite graph G. Let u and v be a leftmost vertex and a rightmost vertex in γ, respectively. Every path of G from u to v is monotone in γ. Proof Let Π be a path from u to v in G (Π exists because G is connected) and walk on Π from u to v. IfΠ is not monotone in γ, there exist three consecutive vertices u,z,v along Π, such that u,v V i (i {1, 2}) and π i (v )<π i (u ). Assume that u, z, and v are the first three vertices along Π that are in this situation. This implies that there is a subpath G = v 1,v 2 = u,v 3 = z, v 4 = v such that π h (v 1 )<π h (v 3 ) and π i (v 4 )<π i (v 2 ), where h, i {1, 2} and h i. Hence, G is a butterfly blocker in γ. Since v 1 is encountered before v 2 and v 3 walking along Π from u to v, v 2 v and v 3 v. Also, since v is a rightmost vertex of γ, either v V h and π h (v 3 )<π h (v) or v V i and π i (v 2 )<π i (v). In both cases, the subpath of Π from v 4 to v contains a traversing edge of G, and therefore causes a fan crossing. This contradicts the assumption that γ is a 2-layer RAC embedding of G. Let T beatreeandletu and v be two leaves of T. If there exists a 2-layer RAC embedding γ where u and v are a leftmost vertex and a rightmost vertex of γ, respectively, we say that γ is a 2-layer RAC embedding with respect to u and v. A 2-layer RAC drawing with embedding γ is called a 2-layer RAC drawing with respect to u and v. IfT admits such a drawing, we also say that T is 2-layer RAC drawable with respect to u and v. The next lemma guarantees that in order to test the 2-layer RAC drawability of a tree we can concentrate only on those 2-layer RAC embeddings that have two leaves as leftmost and rightmost vertices. Lemma 6 Let T be a 2-layer RAC drawable tree. There exist two leaves u and v such that T is 2-layer RAC drawable with respect to u and v. Proof Since T is a 2-layer RAC drawable tree, it admits a 2-layer RAC embedding γ = (π 1,π 2 ). If a leftmost vertex in γ is a leaf and a rightmost vertex of γ is a leaf, γ is the desired 2-layer RAC embedding. Otherwise both the leftmost vertices are not leaves and/or both the rightmost vertices are not leaves. We prove that it is possible to modify the 2-layer RAC embedding γ to obtain a new 2-layer RAC embedding γ that satisfies the statement. Let u 1 V 1 and u 2 V 2 be the leftmost vertices in γ and assume that they are not leaves. It follows that they must be adjacent because otherwise u 1 would be adjacent

15 to at least two vertices to the right of u 2 and u 2 would be adjacent to at least two vertices to the right of u 1. But this would imply a fan crossing. By the same reasoning, both u 1 and u 2 have degree 2. Let v be a rightmost vertex of γ. By Lemma 5, the path Π 1 from u 1 to v is monotone, and the path Π 2 from u 2 to v is monotone. One among Π 1 and Π 2 contains edge (u 1,u 2 ) and properly contains the other path (otherwise T would contain a cycle, which is impossible). Assume, without loss of generality, that Π 2 contains (u 1,u 2 ) (hence, Π 1 Π 2 ), and let e i be the edge incident to u i and distinct from (u 1,u 2 ) (i {1, 2}). Clearly, edge e 1 belongs to Π 2.LetC be the connected component obtained by removing e 2 that does not contain u 1 and let T = C {e 2 }. Observe that T shares with Π 2 only vertex u 2, otherwise there would be a cycle in T, which is impossible. This implies that, since T is embedded to the right of u 1 and u 2 (because they are the leftmost vertices), it cannot contain a degree-3 vertex, because otherwise it would contain a degree-3 blocker G and one edge of Π 2 would be a traversing edge of G. Thus T is a path. (Note also that T is a monotone path in γ ; if not, it would contain a butterfly blocker G and one edge of Π 2 would be a traversing edge of G.) Let a 1,a 2,...,a h be the vertices of T that belong to V 1 in the left-to-right order they appear in π 1 and let b 1,b 2,...,b k be the vertices of T \{u 2 } that belong to V 2 in the left-to-right order they appear in π 2. We construct from γ a 2-layer RAC embedding γ = (π 1,π 2 ) by mirroring the monotone path T with respect to edge (u 1,u 2 ). More precisely, π 1 in such that a 1,a 2,...,a h are to the left of u 1 and π 1 (a h )<π 1 (a h 1 )< <π 1 (a 1 )<π 1 (u 1 ), while all the other vertices of V 1 remain to the right of u 1 in the same relative order they have in π 1. Analogously, π 2 is such that b 1,b 2,...,b k are to the left of u 2 and π 2 (b k )<π 2 (b k 1 )< <π 2 (b 1 )< π 2 (u 2 ), while all the other vertices of V 2 remain to the right of u 2 in the same relative order they have in π 2. It is easy to verify that this mirroring does not create any new crossing and therefore γ is a 2-layer RAC embedding. Also, one of the two leftmost vertices of γ (i.e., either a h or b k ) is the end-vertex of the path T and therefore it is a leaf. At this point, if one of the rightmost vertices of γ is a leaf, then γ is the desired 2-layer RAC embedding, otherwise we can symmetrically modify it so that a rightmost vertex of the resulting 2-layer RAC embedding is a leaf. 5.2 Weighted Contractions We now give some definitions that will be used to characterize 2-layer RAC drawable trees. A chain of T is a maximal path of T such that all its internal vertices have degree two in T.Aweighted contraction of T is a weighted tree obtained from T by replacing each chain of length k>1 with a single edge of weight k. We denote by wc(t ) the weighted contraction of T and by ω(a,b) the weight of an edge (a, b) of wc(t ). An edge of T that also belongs to wc(t ) has weight 1. Note that, the vertices of wc(t ) are a subset of the vertices of T and every vertex of wc(t ) has the same degree in T and in wc(t ). A path between two leaves u, v of T is called a spine of T and is denoted by s uv (T ). The path between u and v in wc(t ) is denoted by sc uv (T ) and is called a spine of wc(t ). For a given spine, we call its vertices spine vertices; the remaining vertices are called non-spine vertices. Figure 6(a) shows a tree T and its weighted

16 Fig. 6 (a) A tree T and its weighted contraction wc(t ); the gray vertices are internal vertices of chains and disappear in wc(t ); the path between u and v in T (edges in bold) isaspine.(b) A2-layerRAC embedding of T with respect to u and v contraction wc(t ). The path between leaves u and v (edges in bold) is a spine; Figure 6(b) shows a 2-layer RAC embedding with respect to the end-vertices u and v of the spine. Our characterization of a 2-layer RAC drawable tree is expressed in terms of properties of its weighted contraction. In order to do that, we look at what type of subtrees are attached to the vertices of the spine. For a given spine sc uv (T ) of wc(t ), we define three kinds of subtrees in wc(t ), called k-fence, y-tree, and star-tree, respectively. Also, for a subtree T of any of the types above, we give the definition of feasibility of T, which expresses the possibility of representing the non-contracted version of T in a 2-layer RAC embedding with respect to the end-vertices, u and v, of the spine. We will prove that a tree T is 2-layer RAC drawable with respect to u and v if and only if every subtree attached to sc uv (T ) is either a feasible maximal k-fence, or a feasible y-tree, or a feasible star-tree. The three kinds of subtrees are defined as follows: k-fence (refer to Figs. 7(a), 7(b), and 7(c)). Let z 0,z 1,z 2,...,z k,z k+1 (k 2) be a sequence of spine vertices such that: (i) deg(z i ) = 3 (i {1,...,k}); (ii) ω(z i,z i+1 ) = 1 (i {1,...,k 1}); deg(w i ) = 1, where w i is the non-spine vertex adjacent to z i (i {1,...,k}). The weighted subtree T induced by the vertices of {z i i = 1,...,k} {w i i = 1,...,k} is called a k-fence of wc(t ). The subsequence z 1,z 2,...,z k is the chain root of T, each w i is a leaf of T. A k-fence is maximal if z 0,z 1,z 2,...,z k,z k+1 is a maximal sequence satisfying (i) and (ii). A 2-fence is feasible if either ω(z 1,w 1 ) ω(z 0,z 1 ) + 1 and ω(z 2,w 2 ) ω(z 2,z 3 ) + 1orω(z 2,w 2 ) ω(z 0,z 1 ) and ω(z 1,w 1 ) ω(z 2,z 3 ). A 3-fence is feasible if one of the following conditions holds: the subtree induced by z 1, z 2, w 1, w 2 is a feasible 2-fence and ω(z 3,w 3 ) ω(z 3,z 4 ) + 1; the subtree induced by z 2, z 3, w 2, w 3 is a feasible 2-fence and ω(z 1,w 1 ) ω(z 0,z 1 ) + 1; Ak-fence (k>3) is feasible if the following conditions holds: the subtree induced by z 1, z 2, w 1, and w 2 is a feasible 2-fence; the subtree induced by z k 1, z k, w k 1, and w k is a feasible 2-fence; If k 5, then ω(z i,w i ) 2 (i {3,...,k 2}).

17 Fig. 7 (a) A feasible 2-fence. (b) A feasible 3-fence. (c) A feasible 5-fence. (d) A feasible y-tree. (e) A feasible star-tree y-tree (refer to Fig. 7(d)). Let z be a degree-3 spine vertex of wc(t ) that does not belong to any maximal k-fence and such that: (i) deg(w) = 3, where w is the nonspine vertex adjacent to z: and (ii) the vertices a,b z adjacent to w have degree one. The weighted subtree T induced by z, w, a, b is called a y-tree of wc(t ). Vertex z is the root, w is the internal vertex, and a,b are the leaves of T. Denote by z 1,z 2 the spine vertices adjacent to z. TreeT is feasible if: (i) ω(z,w) = 1; (ii) either ω(w,a) ω(z 1,z) and ω(w,b) ω(z,z 2 ) or ω(w,b) ω(z 1,z) and ω(w,a) ω(z,z 2 ). star-tree (refer to Fig. 7(e)). Let z be a spine vertex of wc(t ) that does not belong to either a maximal k-fence or a y-tree and such that: (i) deg(z) 3; (ii) every nonspine vertex w i adjacent to z has degree one (i = 1,...,deg(z) 2). The weighted subtree T induced by z and all vertices w i is called a star-tree of wc(t ).Vertexz is the root of T. Denote by z 1,z 2 the spine vertices adjacent to z.treet is feasible if: (i) There exist at most two vertices w j, w h such that 1 <ω(z,w j ) ω(z 1,z)+1 and 1 <ω(z,w h ) ω(z,z 2 )+1; (ii) for every vertex w i / {w j,w h }, ω(z,w i ) = 1. A spine s uv (T ) is called a feasible spine if every vertex z sc uv (T ) distinct from u and v is such that one of the three conditions holds: (i) z belongs to the chain root of a feasible maximal k-fence; (ii) z is the root of exactly one feasible y-tree; (iii) z is the root of exactly one feasible star-tree. In what follows we prove that a tree T is 2-layer RAC drawable with respect to u and v if and only if s uv (T ) is a feasible spine (Theorem 4). We prove separately the sufficiency condition (Lemma 7) and the necessity condition (Lemma 9). The proof of the sufficiency condition describes a linear time algorithm that starting from a feasible spine s uv (T ) constructs a 2-layer RAC embedding of T with respect to u and v. 5.3 Sufficiency Lemma 7 Let T be a tree and let u, v be two leaves of T. Tree T is 2-layer RAC drawable with respect to u and v if s uv (T ) is a feasible spine.

18 Fig. 8 Illustration for the proof of Lemma 7: (a) A 2-layer RAC embedding of the spine; (b) A2-layer RAC embedding of the non-contracted version of the y-tree of Fig. 7(d); (c) A 2-layer RAC embedding of the non-contracted version of the star-tree of Fig. 7(e); (d) (e) Two different 2-layer RAC embeddings of the 2-fence of Fig. 7(a); (f) A 2-layer RAC embedding of the 3-fence of Fig. 7(b); (g) A2-layerRAC embedding of the 5-fence of Fig. 7(c) Proof We describe an algorithm to construct a 2-layer RAC embedding γ of T with respect to u and v. This is sufficient by Theorem 1. The algorithm works as follows. First it embeds s uv (T ) as a monotone path. Then for each vertex z s uv (T ) sc uv (T ) the algorithm distinguishes between the following cases: Case 1: z is the root of a feasible y-tree T.Letw be the internal vertex of T and let a,b be its leaves. Also, let z 1,z 2 be the two vertices of sc uv (T ) adjacent to z. Without loss of generality assume that ω(w,a) ω(z,z 1 ) and that ω(w,b) ω(z,z 2 ) (if not, we can exchange the labels of a and b). The algorithm embeds the non-contracted version of T asshowninfig.8(b): The path Π 1 from w to a in T and the path Π 2 from w to b in T are embedded to the left of z and to the right of z, respectively, so that their vertices alternate with the vertices of the spine. Since ω(w,a) ω(z,z 1 ) and that ω(w,b) ω(z,z 2 ), no edge of Π i traverses a blocker (i = 1, 2). Hence, the embedding of the non-contracted version of T does not create fan crossings. Also, note that if z 1 coincides with u, it remains a leftmost vertex of γ. Analogously, if z 2 coincides with v, it remains a rightmost vertex of γ. Case 2: z is the root of a feasible star-tree T.Letw 1,...,w k be the vertices of T adjacent to z and let z 1,z 2 be the two vertices of sc uv (T ) adjacent to z. Without loss of generality, assume that ω(z,w 1 ) ω(z,z 1 ) + 1, ω(z,w k ) ω(z,z 2 ) + 1, and ω(z,w i ) = 1 (i = 2,...,k 1) (if not, we can reorder the vertices). The algorithm embeds the non-contracted version of T as shown in Fig. 8(c): The path Π 1 from z to w 1 in T and the path Π 2 from z to w k in T are embedded to the left and to the right of z, respectively, so that their vertices alternate with the vertices of the spine. Since ω(z,w 1 ) ω(z,z 1 )+1 and that ω(z,w k ) ω(z,z 2 )+1, no edge of Π i traverses a blocker (i = 1, 2). The remaining edges (z, w i )(i= 2,...,k 1) are embedded between the leftmost and the rightmost edges incident to z, so that they do not cross any other edge of γ. Hence, the embedding of the noncontracted version of T does not create fan crossings. Also, note that if z 1 coincides with u, it remains a leftmost vertex of γ. Analogously, if z 2 coincides with v, it remains a rightmost vertex of γ.