Ordered Level Planarity, Geodesic Planarity and Bi-Monotonicity

08/24/2017 ∙ by Boris Klemz, et al. ∙ 0

We introduce and study the problem Ordered Level Planarity which asks for a planar drawing of a graph such that vertices are placed at prescribed positions in the plane and such that every edge is realized as a y-monotone curve. This can be interpreted as a variant of Level Planarity in which the vertices on each level appear in a prescribed total order. We establish a complexity dichotomy with respect to both the maximum degree and the level-width, that is, the maximum number of vertices that share a level. Our study of Ordered Level Planarity is motivated by connections to several other graph drawing problems. Geodesic Planarity asks for a planar drawing of a graph such that vertices are placed at prescribed positions in the plane and such that every edge is realized as a polygonal path composed of line segments with two adjacent directions from a given set S of directions symmetric with respect to the origin. Our results on Ordered Level Planarity imply NP-hardness for any S with |S|> 4 even if the given graph is a matching. Katz, Krug, Rutter and Wolff claimed that for matchings Manhattan Geodesic Planarity, the case where S contains precisely the horizontal and vertical directions, can be solved in polynomial time [GD'09]. Our results imply that this is incorrect unless P=NP. Our reduction extends to settle the complexity of the Bi-Monotonicity problem, which was proposed by Fulek, Pelsmajer, Schaefer and Štefankovič. Ordered Level Planarity turns out to be a special case of T-Level Planarity, Clustered Level Planarity and Constrained Level Planarity. Thus, our results strengthen previous hardness results. In particular, our reduction to Clustered Level Planarity generates instances with only two non-trivial clusters. This answers a question posed by Angelini, Da Lozzo, Di Battista, Frati and Roselli.

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1 Introduction

In this paper we introduce Ordered Level Planarity and study its complexity. We establish connections to several other graph drawing problems, which we survey in this first section. We proceed from general problems to more and more constrained ones.

Upward Planarity: An upward planar drawing of a directed graph is a plane drawing where every edge  is realized as a -monotone curve that goes upward from  to . Such drawings provide a natural way of visualizing a partial order on a set of items. The problem Upward Planarity of testing whether a directed graph has an upward planar drawing is -complete [12]. However, if the -coordinate of each vertex is prescribed, the problem can be solved in polynomial time [19]. This is captured by the notion of level graphs.

Level Planarity: A level graph  is a directed graph  together with a level assignment where  is a surjective map with  for every edge . Value  is the height of . The vertex set is called the -th level of  and  is its width. The level-width  of  is the maximum width of any level in . A level planar drawing of  is an upward planar drawing of  where the -coordinate of each vertex  is . The horizontal line with -coordinate  is denoted by . The problem Level Planarity asks whether a given level graph has a level planar drawing. The study of the complexity of Level Planarity has a long history [9, 17, 18, 19, 11], culminating in a linear-time approach [19]. Level Planarity has been extended to drawings of level graphs on surfaces different from the plane such as standing cylinder, a rolling cylinder or a torus [1, 4, 5].

An important special case are proper level graphs, that is, level graphs in which  for every edge . Instances of Level Planarity can be assumed to be proper without loss of generality by subdividing long edges [19, 9]. However, in variations of Level Planarity where we impose additional constraints, the assumption that instances are proper can have a strong impact on the complexity of the respective problems [2].

Level Planarity with Various Constraints: Clustered Level Planarity is a combination of Cluster Planarity and Level Planarity. The task is to find a level planar drawing while simultaneously visualizing a given cluster hierarchy according to the rules of Cluster Planarity. The problem is -complete in general [2], but efficiently solvable for proper instances [10, 2].

T-Level Planarity is a consecutivity-constrained version of Level Planarity: every level  is equipped with a tree  whose set of leaves is . For every inner node  of  the leaves of the subtree rooted at  have to appear consecutively along . The problem is -complete in general [2], but efficiently solvable for proper instances [21, 2]. The precise definitions and a longer discussion about the related work are deferred to Appendix 0.C.

Very recently, Brückner and Rutter [7] explored a variant of Level Planarity in which the left-to-right order of the vertices on each level has to be a linear extension of a given partial order. They refer to this problem as Constrained Level Planarity and they provide an efficient algorithm for single-source graphs and show -completeness of the general case.

A Common Special Case - Ordered Level Planarity: We introduce a natural variant of Level Planarity that specifies a total order for the vertices on each level. An ordered level graph  is a triple  where  is a level graph and is a level ordering for . We require that restricted to domain bijectively maps to . An ordered level planar drawing of an ordered level graph  is a level planar drawing of  where for every  the -coordinate of  is . Thus, the position of every vertex is fixed. The problem Ordered Level Planarity asks whether a given ordered level graph has an ordered level planar drawing.

In the above definitions, the - and -coordinates assigned via  and  merely act as a convenient way to encode total and partial orders respectively. In terms of realizability, the problems are equivalent to generalized versions where  and  map to the reals. In other words, the fixed vertex positions can be any points in the plane. All reductions and algorithms in this paper carry over to these generalized versions, if we pay the cost for presorting the vertices according to their coordinates. Ordered Level Planarity is also equivalent to a relaxed version where we only require that the vertices of each level  appear along  according to the given total order without insisting on specific coordinates. We make use of this equivalence in many of our figures for the sake of visual clarity.

Geodesic Planarity: Let  be a finite set of directions symmetric with respect to the origin, i.e. for each direction , the reverse direction  is also contained in . A plane drawing of a graph is geodesic with respect to  if every edge is realized as a polygonal path  composed of line segments with two adjacent directions from . Two directions of  are adjacent if they appear consecutively in the projection of  to the unit circle. Such a path  is a geodesic with respect to some polygonal norm that corresponds to . An instance of the decision problem Geodesic Planarity is a 4-tuple where  is a graph, and map from  to the reals and  is a set of directions as stated above. The task is to decide whether  has a geodesic drawing, that is,  has a geodesic drawing with respect to  in which every vertex  is placed at .

Katz, Krug, Rutter and Wolff [20] study Manhattan Geodesic Planarity, which is the special case of Geodesic Planarity where the set  consists of the two horizontal and the two vertical directions. Geodesic drawings with respect to this set of direction are also referred to as orthogeodesic drawings [14, 13]. Katz et al. [20] show that a variant of Manhattan Geodesic Planarity in which the drawings are restricted to the integer grid is -hard even if  is a perfect matching. The proof is by reduction from 3-Partition and makes use of the fact the number of edges that can pass between two vertices on a grid line is bounded. In contrast, they claim that the standard version of Manhattan Geodesic Planarity is polynomial-time solvable for perfect matchings  [20, Theorem 5]. To this end, they sketch a plane sweep algorithm that maintains a linear order among the edges that cross the sweep line. When a new edge is encountered it is inserted as low as possible subject to the constraints implied by the prescribed vertex positions. When we asked the authors for more details, they informed us that they are no longer convinced of the correctness of their approach. Theorem 1.2 of our paper implies that the approach is indeed incorrect unless .

Bi-Monotonicity: Fulek, Pelsmajer, Schaefer and Štefankovič [11] present a Hanani-Tutte theorem for y-monotone drawings, that is, upward drawings in which all vertices have distinct -coordinates. They accompany their result with a simple and efficient algorithm for Y-Monotonicity, which is equivalent to Level Planarity restricted to instances with level-width . They propose the problem Bi-Monotonicity and leave its complexity as an open problem. The input of Bi-Monotonicity is a triple  where  is a graph and  and  injectively map from  to the reals. The task is to decide whether  has a bi-monotone drawing, that is, a plane drawing in which edges are realized as curves that are both -monotone and -monotone and in which every vertex  is placed at .

Main results: In Section 3 we study the complexity of Ordered Level Planarity. While Upward Planarity is -complete [12] in general but becomes polynomial-time solvable [19] for prescribed -coordinates, we show that prescribing both -coordinates and -coordinates renders the problem -complete. We complement our result with efficient approaches for some special cases of ordered level graphs and, thereby, establish a complexity dichotomy with respect to the level-width and the maximum degree.

Theorem 1.1

Ordered Level Planarity is -complete, even for maximum degree  and level-width . For level-width or  or proper instances Ordered Level Planarity can be solved in linear time, where  and  are the maximum in-degree and out-degree respectively.

Ordered Level Planarity restricted to instances with  and  is an elementary problem. We expect that it may serve as a suitable basis for future reductions. As a proof of concept, the remainder of this paper is devoted to establishing connections between Ordered Level Planarity and several other graph drawing problems. Theorem 1.1 serves as our key tool for settling their complexity. In Section 2 we study Geodesic Planarity and obtain:

Theorem 1.2

Geodesic Planarity is -hard for any set of directions with even for perfect matchings in general position.

Observe the aforementioned discrepancy between Theorem 1.2 and the claim by Katz et al. [20] that Manhattan Geodesic Planarity for perfect matchings is in . Bi-Monotonicity is closely related to a special case of Manhattan Geodesic Planarity. With a simple corollary we settle the complexity of Bi-Monotonicity and, thus, answer the open question by Fulek et al. [11].

Theorem 1.3

Bi-Monotonicity is -hard even for perfect matchings.

Ordered Level Planarity is an immediate and very constrained special case of Constrained Planarity. Further, in Appendix 0.C we establish Ordered Level Planarity as a special case of both Clustered Level Planarity and T-Level Planarity by providing the following reductions.

Theorem 1.4

Ordered Level Planarity with maximum degree  and level-width reduces in linear time to T-Level Planarity with maximum degree and level-width .

Theorem 1.5

Ordered Level Planarity with maximum degree  and level-width reduces in quadratic time to Clustered Level Planarity with maximum degree , level-width  and clusters.

Angelini, Da Lozzo, Di Battista, Frati and Roselli [2] propose the complexity of Clustered Level Planarity for clustered level graphs with a flat cluster hierarchy as an open question. Theorem 1.5 answers this question by showing that -hardness holds for instances with only two non-trivial clusters.

2 Geodesic Planarity and Bi-Monotonicity

In this section we establish that deciding whether an instance of Geodesic Planarity has a geodesic drawing is -hard even if  is a perfect matching and even if the coordinates assigned via  and  are in general position, that is, no two vertices lie on a line with a direction from . The -hardness of Bi-Monotonicity for perfect matchings follows as a simple corollary. Our results are obtained via a reduction from Ordered Level Planarity.

Lemma 1

Let  with be a finite set of directions symmetric with respect to the origin. Ordered Level Planarity with maximum degree  and level-width  reduces to Geodesic Planarity such that the resulting instances are in general position and consist of a perfect matching and direction set . The reduction can be carried out using a linear number of arithmetic operations.

Proof Sketch. In this sketch, we prove our claim only for the classical case that contains exactly the four horizontal and vertical directions. Our reduction is carried out in two steps. Let be an Ordered Level Planarity instance with maximum degree  and level-width . In Step (i) we turn into an equivalent Geodesic Planarity instance . In Step (ii) we transform into an equivalent Geodesic Planarity instance where  is a perfect matching and the vertex positions assigned via and are in general position.

Step (i): In order to transform into we apply a shearing transformation. We translate the vertices of each level by units to the right, see Figure 1(a) and Figure 1(b). Clearly, every geodesic drawing of  can be turned into an ordered level planar drawing of . On the other hand, consider an ordered level planar drawing  of . Without loss of generality we can assume that in  all edges are realized as polygonal paths in which bend points occur only on the horizontal lines  through the levels  where . Further, we may assume that all bend points have -coordinates in the open interval . We shear  by translating the bend points and vertices of level by units to the right for , see Figure 1(b). In the resulting drawing , the vertex positions match those of . Furthermore, all edge-segments have a positive slope. Thus, since the maximum degree is  we can replace all edge-segments with -geodesic rectilinear paths that closely trace the segments and we obtain a geodesic drawing  of , see Figure 1(c).

Fig. 1: (a), (b) and (c): Illustrations of Step (i). (d) The two gadget squares of each level. Grid cells have size . (e) Illustration of Step (ii). Turning a drawing of into a drawing of (f) and vice versa (g).

Step (ii): In order to turn  into the equivalent instance we transform into a perfect matching. To this end, we split each vertex by replacing it with a small gadget that fits inside a square  centered on the position  of , see Figure 1(e). We call  the square of  and use , , and to denote the top-right, top-left, bottom-right and bottom-left corner of , respectively. We use two different sizes to ensure general position. The size of the gadget square is if and it is if . The gadget contains a degree-1 vertex for every edge incident to . In the following we explain the gadget construction in detail, for an illustration see Figure 1(d). Let  be an edge incident to . We create an edge  where  is a new vertex which is placed at if  is located to the top-right of  and it is placed at  if  is located to the bottom-left of . Similarly, if  is incident to a second edge , we create an edge  where  is placed at or depending on the position of . Finally, we create a blocking edge  where  is placed at  and  is placed at . The thereby assigned coordinates are in general position and the construction can be carried out in linear time.

Assume that  has a geodesic drawing . By construction, all blocking edges have a top-left and a bottom-right endpoint. On the other hand, all other edges have a bottom-left and a top-right endpoint. As a result, a non-blocking edge  can not pass through any gadget square , except the squares  or  since  would have to cross the blocking edge of . Accordingly, it is straight-forward to obtain a geodesic drawing of : We remove the blocking edges, reinsert the vertices of  according to the mappings  and  and connect them to the vertices of their respective gadgets in a geodesic fashion. This can always be done without crossings. Figure 1(f) shows one possibility. If the edge from  passes to the left of , we may have to choose a reflected version. Finally, we remove the vertices  and  which now act as subdivision vertices.

On the other hand, let  be a geodesic planar drawing of . Without loss of generality, we can assume that each edge  passes only through the squares of  and . Furthermore, for each  we can assume that its incident edges intersect the boundary of  only to the top-right of or to the bottom-left of , see Figure 1(g). Thus, we can simply remove the parts of the edges in the interior of the gadget squares and connect the gadget vertices to the intersection points of the edges with the gadget squares in a geodesic fashion.

The bit size of the numbers involved in the calculations of our reduction is linearly bounded in the bit size of the directions of . Together with Theorem 1.1 we obtain the proof of Theorem 1.2. The instances generated by Lemma 1 are in general position. In particular, this means that the mappings  and  are injective. We obtain an immediate reduction to Bi-Monotonicity. The correctness follows from the fact that every -geodesic rectilinear path can be transformed into a bi-monotone curve and vice versa. Thus, we obtain Theorem 1.3.

3 Ordered Level Planarity

To show -hardness of Ordered Level Planarity we reduce from a 3-Satisfiability variant described in this paragraph. A monotone 3-Satisfiability formula is a Boolean 3-Satisfiability formula in which each clause is either positive or negative, that is, each clause contains either exclusively positive or exclusively negative literals respectively. A planar 3SAT formula is a Boolean 3-Satisfiability formula with a set of variables and a set of clauses such that its variable-clause graph is planar. The graph  is bipartite, i.e. every edge in  is incident to both a clause vertex from  and a variable vertex from . Furthermore, edge  if and only if a literal of variable occurs in . Planar Monotone 3-Satisfiability is a special case of 3-Satisfiability where we are given a planar and monotone 3-Satisfiability formula  and a monotone rectilinear representation  of the variable-clause graph of . The representation  is a contact representation on an integer grid in which the variables are represented by horizontal line segments arranged on a line . The clauses are represented by E-shapes turned by such that all positive clauses are placed above  and all negative clauses are placed below , see Figure 1(a). Planar Monotone 3-Satisfiability is -complete [6]. We are now equipped to prove the core lemma of this section.

(a)
(b)
(c)
Fig. 2: (a) Representation  of  with negative clauses , and and positive clauses and and (b) its modified version in Lemma 2. (c) Tier .
Lemma 2

Planar Monotone 3-Satisfiability reduces in polynomial time to Ordered Level Planarity. The resulting instances have maximum degree  and all vertices on levels with width at least 3 have out-degree at most and in-degree at most .

Proof Sketch. We perform a polynomial-time reduction from Planar Monotone 3-Satisfiability. Let  be a planar and monotone 3-Satisfiability formula with . Let  the variable-clause graph of . Let  be a monotone rectilinear representation of . We construct an ordered level graph such that  has an ordered level planar drawing if and only if  is satisfiable. In this proof sketch we omit some technical details such as precise level assignments and level orderings.

Overview: The ordered level graph  has  levels which are partitioned into four tiers , , and . Each clause  is associated with a clause edge  starting with  in tier  and ending with  in tier . The clause edges have to be drawn in a system of tunnels that encodes the 3-Satisfiability formula . In  the layout of the tunnels corresponds directly to the rectilinear representation , see Figure 1(c). For each E-shape there are three tunnels corresponding to the three literals of the associated clause. The bottom vertex  of each clause edge  is placed such that  has to be drawn inside one of the three tunnels of the E-shape corresponding to . This corresponds to the fact that in a satisfying truth assignment every clause has at least one satisfied literal. In tier  we merge all the tunnels corresponding to the same literal. We create variable gadgets that ensure that for each variable  edges of clauses containing  can be drawn in the tunnel associated with either the negative or the positive literal of  but not both. This corresponds to the fact that every variable is set to either true or false. Tiers  and  have a technical purpose.

We proceed by describing the different tiers in detail. Recall that in terms of realizability, Ordered Level Planarity is equivalent to the generalized version where  and  map to the reals. For the sake of convenience we will begin by designing  in this generalized setting. It is easy to transform  such that it satisfies the standard definition in a polynomial-time post processing step.

Tier 0 and 2, clause gadgets: The clause edges end in tier . It is composed of  levels each of which contains precisely one vertex. We assign . Observe that this imposes no constraint on the order in which the edges enter .

Tier  consists of a system of tunnels that resembles the monotone rectilinear representation  of , see Figure 1(c). Intuitively it is constructed as follows: We take the top part of , rotate it by  and place it to the left of the bottom part such that the variables’ line segments align, see Figure 1(b). We call the resulting representation . For each E-shape in  we create a clause gadget, which is a subgraph composed of vertices that are placed on a grid close to the E-shape, see Figure 3. The red vertex at the bottom is the lower vertex  of the clause edge  of the clause  corresponding to the E-shape. Without loss of generality we assume the grid to be fine enough such that the resulting ordered level graph can be drawn as in Figure 1(c) without crossings. Further, we assume that the -coordinates of every pair of horizontal segments belonging to distinct E-shapes differ by at least . This ensures that all vertices on levels with width at least 3 have out-degree at most and in-degree at most as stated in the lemma.

The clause gadget (without the clause edge) has a unique ordered level planar drawing in the sense that for every level  the left-to-right sequence of vertices and edges intersected by the horizontal line  through  is identical in every ordered level planar drawing. This is due to the fact that the order of the top-most vertices , , , , and is fixed. We call the line segments , and the gates of . Note that the clause edge  has to intersect one of the gates of . This corresponds to the fact the at least one literal of every clause has to be satisfied.

(a)
(b)
Fig. 3: (a) The E-shape and (b) the clause gadget of clause . The thick gray lines represent the gates of .

The subgraph  induced by  (without the clause edges) has a unique ordered level planar drawing. In tier  we bundle all gates that belong to one literal together by creating two long paths for each literal. These two paths form the tunnel of the corresponding literal. All clause edges intersecting a gate of some literal have to be drawn inside the literal’s tunnel, see Figure 1(c). To this end, for  we use  () to refer to the left-most (right-most) vertex of a negative clause gadget placed on a line segment of  representing . The vertices  and  are the first vertices of the paths forming the negative tunnel  of the negative literal of variable . Analogously, we use  () to refer to the left-most (right-most) vertex of a positive clause gadget placed on a line segment of  representing . The vertices  and  are the first vertices of the paths forming the positive tunnel  of the positive literal of variable . If for some  the variable  is not contained both in negative and positive clauses, we artificially add two vertices   and or  and  on the corresponding line segments in order to avoid having to treat special cases in the remainder of the construction.

Tier 1 and 3, variable gadgets: Recall that every clause edge has to pass through a gate that is associated with some literal of the clause, and, thus, every edge is drawn in the tunnel of some literal. We need to ensure that it is not possible to use tunnels associated with the positive, as well as the negative literal of some variable simultaneously. To this end, we create a variable gadget with vertices in tier  and tier  for each variable. The variable gadget of variable  is illustrated in Figure 3(a). The variable gadgets are nested in the sense that they start in  in the order , from bottom to top and they end in the reverse order in , see Figure 5. We force all tunnels with index at least  to be drawn between the vertices  and . This is done by subdividing the tunnel edges on this level, see Figure 3(b). The long edge  has to be drawn to the left or right of  in . Accordingly, it is drawn to the left of  or to the right of  in . Thus, it is drawn either to the right (Figure 3(b)) of all the tunnels or to the left (Figure 3(c)) of all the tunnels. As a consequence, the blocking edge  is also drawn either to the right or the left of all the tunnels. Together with the edge  it prevents clause edges from being drawn either in the positive tunnel  or negative tunnel  of variable  which end at level  because they can not reach their endpoints in  without crossings. We say  or  are blocked respectively.

(a)
(b)
(c)
Fig. 4: (a) The variable gadget of  in (b) positive and (c) negative state.

The construction of the ordered level graph  can be carried out in polynomial time. Note that maximum degree is  and that all vertices on levels with width at least 3 have out-degree at most 1 and in-degree at most 1 as claimed in the lemma.

Correctness: It remains to show that  has an ordered level planar drawing if and only if  is satisfiable. Assume that  has an ordered level planar drawing . We create a satisfying truth assignment for . If  is blocked we set  to true, otherwise we set  to false for . Recall that the subgraph  induced by the vertices in tier  has a unique ordered level planar drawing. Consider a clause  and let  be the indices of the variables whose literals are contained in . Clause edge  has to pass level  through one of the gates of . More precisely, it has to be drawn between either and , and or and if  is negative or between either and , and or and if  is positive, see Figure 1(c). First, assume that  is negative and assume without loss of generality that it traverses  between  and . In this case clause edge  has to be drawn in . Recall that this is only possible if  is not blocked, which is the case if  is false, see Figure 3(c). Analogously, if  is positive and  traverses w.l.o.g. between  and , then  is true, Figure 3(b). Thus, we have established that one literal of each clause in  evaluates to true for our truth assignment and, hence, formula  is satisfiable.

Fig. 5: Nesting structure of the variable gadgets.

Now assume that  is satisfiable and consider a satisfying truth assignment. We create an ordered level planar drawing  of . It is clear how to create the unique subdrawing of . The variable gadgets are drawn in a nested fashion, see Figure 5. For  we draw edge  to left of  and  and edge  to right of  and . In other words, the pair  is drawn between all such pairs with index smaller than . Recall that the vertices , , , and are located on higher levels than the according vertices of variables with index smaller than  and that and are located on lower levels than the according vertices of variables with index smaller than .

For  if  is positive we draw the long edge  to the right of  and  and, accordingly, we have to draw all tunnels left of  and  (except for , which has to be drawn to the left of  and end to the right of ), see Figure 3(b). If  is negative we draw the long edge  to the left of  and  and, accordingly, we have to draw all tunnels right of  and  (except for , which has to be drawn to the right of  and end to the left of ), see Figure 3(c). We have to draw the blocking edge  to the right of  if  is positive and to the left of  if  is negative.

It remains to describe how to draw the clause edges. Let be a clause. There is at least one true literal in . Let  be the index of the corresponding variable. We describe the drawing of clause edge  from bottom to top. We start by drawing  in the tunnel  () if  is positive (negative). After the variable gadget of  the edge  leaves its tunnel and is drawn to the left (right) of all gadgets of variables with higher index, see Figure 5.

We obtain -hardness for instances with maximum degree . In fact, we can restrict our attention to instances level-width . To this end, we split levels with width  into  levels containing exactly two vertices each.

Lemma 3

An instance  of Ordered Level Planarity with maximum degree  can be transformed in linear time into an equivalent instance  of Ordered Level Planarity with level-width  and maximum degree . If in  all vertices on levels with width at least 3 have out-degree at most 1 and in-degree at most 1, then . Otherwise, .

The reduction in Lemma 2 requires degree-2 vertices. With , the problem becomes polynomial-time solvable. In fact, even if one can easily solve it as long as the maximum in-degree and the maximum out-degree are both bounded by 1. Such instances consists of a set of -monotone paths. We write , meaning that must be drawn to the left of , if and have vertices and that lie adjacent on a common level. If is acyclic, we can draw  according to a linear extension of , otherwise there exists no solution.

Lemma 4

Ordered Level Planarity restricted to instances with maximum in-degree  and maximum out-degree  can be solved in linear time.

For  Ordered Level Planarity is solvable in linear time since Level Planarity can be solved in linear time [19]. Proper instances can be solved in linear-time via a sweep through every level. The problem is obviously contained in . The results of this section establish Theorem 1.1.

Acknowledgements: We thank the authors of [20] for providing us with unpublished information regarding their plane sweep approach for Manhattan Geodesic Planarity.

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Appendix 0.A Omitted Proofs in Section 2

Lemma 1.  Let  with be a finite set of directions symmetric with respect to the origin. Ordered Level Planarity with maximum degree  and level-width  reduces to Geodesic Planarity such that the resulting instances are in general position and consist of a perfect matching and direction set . The reduction can be carried out using a linear number of arithmetic operations.

Proof

We first prove our claim for the classical case that contains exactly the four horizontal and vertical directions. Afterwards, we discuss the necessary adaptations for the general case. Our reduction is carried out in two steps. Let be an Ordered Level Planarity instance with maximum degree  and level-width . In Step (i) we turn into an equivalent Geodesic Planarity instance . In Step (ii) we transform into an equivalent Geodesic Planarity instance where  is a perfect matching and the vertex positions assigned via and are in general position.

Step (i): In order to transform into we apply a shearing transformation. We translate the vertices of each level by units to the right, see Figure 1(a) and Figure 1(b). Clearly, every geodesic drawing of  can be turned into an ordered level planar drawing of . On the other hand, consider an ordered level planar drawing  of . Without loss of generality we can assume that in  all edges are realized as polygonal paths in which bend points occur only on the horizontal lines  through the levels  where . Further, we may assume that all bend points have -coordinates in the open interval . We shear  by translating the bend points and vertices of level by units to the right for , see Figure 1(b). In the resulting drawing , the vertex positions match those of . Furthermore, all edge-segments have a positive slope. Thus, since the maximum degree is  we can replace all edge-segments with -geodesic rectilinear paths that closely trace the segments and we obtain a geodesic drawing  of , see Figure 1(c).

Step (ii): In order to turn  into the equivalent instance we transform into a perfect matching. To this end, we split each vertex by replacing it with a small gadget that fits inside a square  centered on the position  of , see Figure 1(e). We call  the square of  and use , , and to denote the top-right, top-left, bottom-right and bottom-left corner of , respectively. We use two different sizes to ensure general position. The size of the gadget square is if and it is if . The gadget contains a degree-1 vertex for every edge incident to . In the following we explain the gadget construction in detail, for an illustration see Figure 1(d). Let  be an edge incident to . We create an edge  where  is a new vertex which is placed at if  is located to the top-right of  and it is placed at  if  is located to the bottom-left of . Similarly, if  is incident to a second edge , we create an edge  where  is placed at or depending on the position of . Finally, we create a blocking edge  where  is placed at  and  is placed at . The thereby assigned coordinates are in general position and the construction can be carried out in linear time.

Assume that  has a geodesic drawing . By construction, all blocking edges have a top-left and a bottom-right endpoint. On the other hand, all other edges have a bottom-left and a top-right endpoint. As a result, a non-blocking edge  can not pass through any gadget square , except the squares  or  since  would have to cross the blocking edge of . Accordingly, it is straight-forward to obtain a geodesic drawing of : We remove the blocking edges, reinsert the vertices of  according to the mappings  and  and connect them to the vertices of their respective gadgets in a geodesic fashion. This can always be done without crossings. Figure 1(f) shows one possibility. If the edge from  passes to the left of , we may have to choose a reflected version. Finally, we remove the vertices  and  which now act as subdivision vertices.

On the other hand, let  be a geodesic planar drawing of . Without loss of generality, we can assume that each edge  passes only through the squares of  and . Furthermore, for each  we can assume that its incident edges intersect the boundary of  only to the top-right of or to the bottom-left of , see Figure 1(g). Thus, we can simply remove the parts of the edges in the interior of the gadget squares and connect the gadget vertices to the intersection points of the edges with the gadget squares in a geodesic fashion.

The general case: It remains to discuss the adaptations for the case that 

is an arbitrary set of directions symmetric with respect to the origin. By applying a linear transformation we can assume without loss of generality that

and are adjacent directions in . Accordingly, all the remaining directions point into the top-left or the bottom-right quadrant. Further, by vertical scaling we can assume that no direction projects to on the unit square. Observe that if we do not insist on a coordinate assignment in general position, the reduction for the restricted case discussed above is already sufficient. In order to guarantee general position we have to avoid points that lie on a line with a direction from . This requires some easy but a bit technical modifications of our construction.

Note that since no direction of  points to the top-right or bottom-left quadrant, every pair of conflicting vertices from  that defines a line parallel to one of the directions in  has to belong to one or both of the gadgets of two vertices  with . Let  and  be the flattest and steepest slope of respectively. In order to guarantee general position we apply the following two changes.

(1) We increase the horizontal distance in the mapping  between each pair of vertices  and  with  and  and  in order to ensure that there can not be any conflicting vertices , in  such that  belongs to the gadget of  and  belongs to the gadget of  . It suffices to translate  and its square to the right such that  is to the right of the line with direction  through , see Figure 6(a).

(2) In order to ensure that there are no conflicting vertices , that belong to the same gadget square , we change the offset to the gadget square corners from  and  to  and  where is chosen small enough such that the gadget vertices are placed above the line with direction  through , below the line with direction  through , below the line with direction  through  and above the line with direction  through , see Figure 6(b). ∎

Fig. 6: Adaptations (1) (a) and (2) (b) for the general case.

Appendix 0.B Omitted Proofs in Section 3

Lemma 2. Planar Monotone 3-Satisfiability reduces in polynomial time to Ordered Level Planarity. The resulting instances have maximum degree  and all vertices on levels with width at least 3 have out-degree at most and in-degree at most .

Proof

We perform a polynomial-time reduction from Planar Monotone 3-Satisfiability. Let  be a planar and monotone 3-Satisfiability formula with . Let  the variable-clause graph of . Let  be a monotone rectilinear representation of . We construct an ordered level graph such that  has an ordered level planar drawing if and only if  is satisfiable.

Overview: The ordered level graph  has  levels which are partitioned into four tiers , , and . Each clause  is associated with a clause edge  starting with  in tier  and ending with  in tier . The clause edges have to be drawn in a system of tunnels that encodes the 3-Satisfiability formula . In  the layout of the tunnels corresponds directly to the rectilinear representation , see Figure 1(c). For each E-shape there are three tunnels corresponding to the three literals of the associated clause. The bottom vertex  of each clause edge  is placed such that  has to be drawn inside one of the three tunnels of the E-shape corresponding to . This corresponds to the fact that in a satisfying truth assignment every clause has at least one satisfied literal. In tier  we merge all the tunnels corresponding to the same literal. We create variable gadgets that ensure that for each variable  edges of clauses containing  can be drawn in the tunnel associated with either the negative or the positive literal of  but not both. This corresponds to the fact that every variable is set to either true or false. Tiers  and  have a technical purpose.

We proceed by describing the different tiers in detail. Recall that in terms of realizability, Ordered Level Planarity is equivalent to the generalized version where  and  map to the reals. For the sake of convenience we will begin by designing  in this generalized setting. It is easy to transform  such that it satisfies the standard definition in a polynomial-time post processing step.

Tier 0 and 2, clause gadgets: The clause edges end in tier . It is composed of  levels each of which contains precisely one vertex. We assign . Observe that this imposes no constraint on the order in which the edges enter .

Tier  consists of a system of tunnels that resembles the monotone rectilinear representation  of , see Figure 1(c). Intuitively it is constructed as follows: We take the top part of , rotate it by  and place it to the left of the bottom part such that the variables’ line segments align, see Figure 1(b). We call the resulting representation . For each E-shape in  we create a clause gadget, which is a subgraph composed of vertices that are placed on a grid close to the E-shape, see Figure 3. The red vertex at the bottom is the lower vertex  of the clause edge  of the clause  corresponding to the E-shape. Without loss of generality we assume the grid to be fine enough such that the resulting ordered level graph can be drawn as in Figure 1(c) without crossings. Further, we assume that the -coordinates of every pair of horizontal segments belonging to distinct E-shapes differ by at least . This ensures that all vertices on levels with width at least 3 have out-degree at most and in-degree at most as stated in the lemma.

Technical Details: In the following two paragraphs, we describe the construction of the clause gadgets in detail.

For every  where  is negative we create its 11-vertex clause gadget as follows, see Figure 3. Let  be the three vertical line segments of the E-shape representing  in  where  is left-most and  right-most. Let  be the lower endpoints and  be the upper endpoints of , respectively. We place the tail  of the clause edge  of  at . We create new vertices at , , , , , , and at which are the lattice points one unit to the right of , respectively. To simplify notation, we identify these new vertices with their locations on the grid. We add edges , , , , and to .

As stated above, we can assume without loss of generality that the grid is fine enough such that the resulting ordered level graph can be drawn as in Figure 1(c) without crossing. It suffices to assume that the horizontal and vertical distance between any two segment endpoints of  is at least 3 (unless the endpoints lie on a common horizontal or vertical line).

Gates and Tunnels: The clause gadget (without the clause edge) has a unique ordered level planar drawing in the sense that for every level  the left-to-right sequence of vertices and edges intersected by the horizontal line  through  is identical in every ordered level planar drawing. This is due to the fact that the order of the top-most vertices , , , , and is fixed. We call the line segments , and the gates of . Note that the clause edge  has to intersect one of the gates of . This corresponds to the fact the at least one literal of every clause has to be satisfied.

The subgraph  induced by  (without the clause edges) has a unique ordered level planar drawing. In tier  we bundle all gates that belong to one literal together by creating two long paths for each literal. These two paths form the tunnel of the corresponding literal. All clause edges intersecting a gate of some literal have to be drawn inside the literal’s tunnel, see Figure 1(c). To this end, for  we use  () to refer to the left-most (right-most) vertex of a negative clause gadget placed on a line segment of  representing . The vertices  and  are the first vertices of the paths forming the negative tunnel  of the negative literal of variable . Analogously, we use  () to refer to the left-most (right-most) vertex of a positive clause gadget placed on a line segment of  representing . The vertices  and  are the first vertices of the paths forming the positive tunnel  of the positive literal of variable . If for some  the variable  is not contained both in negative and positive clauses, we artificially add two vertices   and or  and  on the corresponding line segments in order to avoid having to treat special cases in the remainder of the construction.

Tier 1 and 3, variable gadgets: Recall that every clause edge has to pass through a gate that is associated with some literal of the clause, and, thus, every edge is drawn in the tunnel of some literal. We need to ensure that it is not possible to use tunnels associated with the positive, as well as the negative literal of some variable simultaneously. To this end, we create a variable gadget with vertices in tier  and tier  for each variable. The variable gadget of variable  is illustrated in Figure 3(a). The variable gadgets are nested in the sense that they start in  in the order , from bottom to top and they end in the reverse order in , see Figure 5. We force all tunnels with index at least  to be drawn between the vertices  and . This is done by subdividing the tunnel edges on this level, see Figure 3(b). The long edge  has to be drawn to the left or right of  in . Accordingly, it is drawn to the left of  or to the right of  in . Thus, it is drawn either to the right (Figure 3(b)) of all the tunnels or to the left (Figure 3(c)) of all the tunnels. As a consequence, the blocking edge  is also drawn either to the right or the left of all the tunnels. Together with the edge  it prevents clause edges from being drawn either in the positive tunnel  or negative tunnel  of variable  which end at level  because they can not reach their endpoints in  without crossings. We say  or  are blocked respectively.

Technical Details: In the following two paragraphs, we describe the construction of the variable gadgets in detail.

Tier  has  layers each of which contains precisely one vertex. We refer to the vertex in layer  as  and to the vertex in layer  as  for . Tier  has  levels. In each of the levels  and  where  we create one vertex. These vertices are called , and  respectively. In level  we create two vertices  and  in this order. We add the edges ,