Inserting an Edge into a Geometric Embedding

07/31/2018 ∙ by Marcel Radermacher, et al. ∙ KIT Universität Passau 0

The algorithm of Gutwenger et al. to insert an edge e in linear time into a planar graph G with a minimal number of crossings on e, is a helpful tool for designing heuristics that minimize edge crossings in drawings of general graphs. Unfortunately, some graphs do not have a geometric embedding Γ such that Γ+e has the same number of crossings as the embedding G+e. This motivates the study of the computational complexity of the following problem: Given a combinatorially embedded graph G, compute a geometric embedding Γ that has the same combinatorial embedding as G and that minimizes the crossings of Γ+e. We give polynomial-time algorithms for special cases and prove that the general problem is fixed-parameter tractable in the number of crossings. Moreover, we show how to approximate the number of crossings by a factor (Δ-2), where Δ is the maximum vertex degree of G.



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1 Introduction

Crossing minimization is an important task for the construction of readable drawings. The problem of minimizing the number of crossings in a given graph is a well-known -complete problem [8]. A very successful heuristic for minimizing the number of crossings in a topological drawing of a graph is to start with a spanning planar subgraph of and to iteratively insert the remaining edges into a drawing of . The edge insertion problem for a planar graph and two vertices asks to find a drawing  of  with the minimum number of crossings such that the induced drawing  of is planar. The problem comes with several variants depending on whether the drawing  can be chosen arbitrarily or is fixed [9, 10]. In the planar topological case both problems can be solved in linear time. More general problems such as inserting several edges simultaneously [2] or inserting a vertex together with all its incident edges [1] have also been studied.

All these approaches have in common that they focus on topological drawings where edges are represented as arbitrary curves between their endpoints. By contrast, we focus on geometric embeddings, i.e., planar straight-line drawings, and the corresponding rectilinear crossing number. In this scenario we are only aware of a few heuristics that compute straight-line drawings of general graphs [12, 13]. Clearly, if a geometric embedding  of the input graph is provided as part of the input, there is no choice left; we can simply insert the straight-line segment from to into the drawing and count the number of crossings it produces. If, however, only the combinatorial embedding is specified, but one may still choose the outer face and choose the vertex positions so that this results in a straight-line drawing with the given combinatorial embedding, then the problem becomes interesting and non-trivial. We call this problem geometric edge insertion.

Contribution and Outline.

We show several results on the complexity of geometric edge insertion with a fixed combinatorial embedding. Namely, we give a linear-time algorithm for the case that the maximum degree  of is at most 5 (Sec. 3). For the general case, we give a -approximation that runs in linear time. Moreover, we give an efficient algorithm for testing in special cases whether there exists a way to insert the edge so that it does not produce more crossings than when we allow to draw it as an arbitrary curve (Sec. 4). Finally, we give a randomized FPT algorithm that tests in  time whether an edge can be inserted with at most crossings (Sec. 5).

2 Preliminaries

Figure 1: (a) The extended dual (red + blue) of the primal graph (grey) and the red vertices corresponding to and . (b) Labeling induced by the blue path.

Let be a planar graph with a given combinatorial embedding where only the choice of the outer face is free. Additionally, let and be two distinct vertices with . Denote by the graph together with the edge . We want to insert the edge into the embedded graph . That is, we seek a straight-line drawing of (with the given embedding) such that can be inserted into with a minimum number of crossings. In , the edge starts at , traverses a set of faces and ends in . Topologically, this corresponds to a path from to in the extended dual of , i.e., in the dual graph plus and connected to all vertices of their dual faces; see Fig. 1a. The number of crossings in corresponds to the length of the path minus two. However, not all -paths in are of the form for a straight-line drawing of .

Figure 2: Ratio between length of the shortest path and the length of a shortest consistent -path. The solid black edges induce a graph of maximum degree . Red vertices have label , blue vertices have label . (a) The shortest path from to in is not consistent.

A labeling of is a mapping that labels vertices as either left or right. Consider an edge of that is crossed by a path such that and are to the left and to the right of , respectively. The edge is compatible with a labeling if and . A path of and a labeling of are compatible if is compatible with each edge that is crossed by . A path is consistent if there is a labeling of that is compatible with . Eades et al. [4] show the following result.

Proposition 1 (Eades et al. [4], Theorem 1)

An -path in is of the form if and only if it is consistent, where is a geometric embedding of .

In order to minimize the number of crossings of , we look for a consistent -path of minimum length in . Given a path , it is easy to check whether is consistent. Fig. 2 shows that the ratio between the length of a shortest -path and the length of a shortest consistent -path can be arbitrarily large. Thus, our goal is to find short consistent -paths.

Let be a directed acyclic graph. A path is a directed path if for each , . It is undirected if for each , either or . We refer to the number of edges of a path as the length of . Two paths and are edge-disjoint if they do not share an edge. Two paths and of an embedded graph are non-crossing if at each common vertex , the edges of and incident to do not alternate in the cyclic order around in the graph induced by and . We denote by the subpath of a path from to .

3 Bounded Degree

The shortest -path of the graph in Fig. 2a is not consistent. Note that the maximum vertex degree is . In this section, we show that every shortest -path in graphs of bounded degree is consistent, and that in each planar graph with vertex degree at most , there is a shortest -path that is consistent. Finally, we prove that there is a consistent -path of length in a graph with maximum vertex degree  and a shortest -path of length in .

Let be an -path in and let be an edge of . An endpoint of the primal edge of is left of if it is locally left of on (Fig. 1b). A vertex of is left (right) of if is left (right) of an edge of . We now consider a labeling extended by two more labels . We define the labeling  induced by as follows. Each vertex that is left and right of gets the label . The remaining vertices that are either left or right of get labels and , respectively. Vertices neither left nor right of get the label . Obviously, there is a labeling of compatible with if and only if does not use the label .

Theorem 3.1 ()

Let be a planar embedded graph of degree at most . Then every shortest -path in is consistent.

Theorem 3.2

Let be a planar embedded graph with maximum degree . Then there is a shortest -path in that is consistent.

Figure 3: Inconsistent path around (a) a degree-4 vertex and (b,c) a degree-5 vertex.

Let be a shortest -path in . We call an edge of good if the vertices left and right of it do not have label in the labeling induced by .

If  is not consistent, then let denote the last edge of  that is not good. Then an endpoint of the primal edge corresponding to has label . Without loss of generality, we may assume that lies left of . Since , there is an edge of that has to its right. By the choice of , it follows that lies before on . We now distinguish cases based on the degree of .

If , then we find that enters or leaves a face twice, which contradicts the assumption that it is a shortest -path.

If , we denote the edges around in clockwise order as such that crosses . Moreover, we denote the faces incident to in clockwise order as where is the starting face of .

Since no face has two incoming or two outgoing edges of , it follows that crosses and crosses ; see Fig. 3a. Let be the path obtained from by replacing the subpath  by the edge that crosses . Since is a shortest path, it follows that . By construction, it is . Observe that lies inside the region bounded by and a curve connecting and that crosses . The only vertex inside this region whose label changed is . Therefore, the path consists of good edges, and we have thus increased the length of the suffix of the shortest path that consists of good edges.

Now assume that . We denote the edges around as in clockwise order such that crosses . We further denote the faces incident to in clockwise order as such that starts in . Since no face has two incoming or two outgoing edges, it follows that either crosses from to or crosses from to .

If crosses , then we consider the path obtained from by replacing the subpath by the edge that crosses ; see Fig. 3b. As above, it follows that and is a cutvertex and that consists of good edges.

If crosses , then we obtain by replacing by the single edge that crosses ; see Fig. 3c. As above, we find that and is a cutvertex and that consists of good edges.

Thus, in all cases, we increase the length of the suffix of the shortest path consisting of good edges. Eventually, we thus arrive at a shortest path whose edges are all good and that hence is consistent.

Theorem 3.3

Let a planar embedded graph with maximum vertex-degree and let be a shortest -path in with . Then there is a consistent path of length at most .

Figure 4: Inconsistent path around a degree vertex.

Let be an -path in . Assume that is not consistent. Then there is a shortest prefix of that is not consistent; refer to Fig. 4. Let be a vertex incident to the primal edge of with . Without loss of generality let be the faces around in counterclockwise order, i.e., lies left of .

Since is not consistent, there is a second edge of that crosses a primal edge incident to . Let be the last edge of that crosses a primal edge incident to . Since is the shortest inconsistent prefix of , lies right of , i.e., for some with . Moreover, let be the first vertex in clockwise order from that lies on the path . Note that such a vertex exists, since at the latest satisfies the condition.

Let be the path . We obtain a path from by replacing by , i.e., . Note that, since is the first vertex in clockwise order on , is a simple path. Since does not contain the edges and , and contains at least one edge, the path has length at most . We claim that the prefix is consistent.

Then, since is a subpath of and is consistent, it follows that we have decreased the maximum length of a suffix of the path whose removal results in an inconsistent path. Since this suffix has initially length at most , we inductively find a consistent -path of length at most .

It remains to prove that is consistent. Since is the shortest inconsistent prefix of , the prefix is consistent. Therefore, is right of . By construction, is right of . Thus, we have . The only vertices of with can be neighbors of , as otherwise would not be consistent.

Consider the region enclosed by the path and that contains ; refer to Fig. 4. The prefix lies outside of and the path lies entirely in . Moreover, in case that is crossed by , lies outside of . On the other hand, if crosses an edge , then lies inside . Thus, in both cases we immediately get that . Therefore, the prefix is consistent.

4 Consistent Shortest -paths

In Section 3 we showed that every shortest -path in the extended dual of a graph with vertex degree at most is consistent. For every graph of maximum degree , there is a shortest -path that is consistent. On the other hand, Fig. 2 shows that, starting from degree , there are graphs whose shortest -paths are not consistent. In this section we investigate the problem of deciding whether contains a consistent shortest -path. As a consequence of Proposition 1 this problem is in .

In Lemma 1 we show that finding a consistent -path in is closely related to finding two edge-disjoint paths in . Especially, we are interested in two edge-disjoint paths where the length of one is minimized. Eilam-Tzoreff [5] proved that this problem is in general -complete. In planar graphs the sum of the length of two vertex-disjoint paths can be minimized efficiently [11]. In general directed graphs the problem is -hard [7]. Finding two edge-disjoint paths in acyclic directed graphs is -complete [6].

The closest relative to our problem is certainly the work of Eilam-Tzoreff. In fact their result can be modified to show that it is -hard to decide whether a graph contains two edge-disjoint -paths such that one of them is a shortest path. We study this problem in the planar setting with the additional restriction that and lie on a common face of the subgraph of that contains all shortest paths from to .

Lemma 1

An -path in is consistent if and only if there is an -path in that is edge-disjoint from and that does not cross .

Figure 5: (a) The green regions are right of (blue) and the blue left of . (b) The outer region that is not bounded by maximal subpaths of and .

The paths and define a set of regions in the plane. Since and are non-crossing, each region is bounded by one maximal subpath of and one maximal subpath of (Fig. 5). We label each region with either or , depending on whether lies left or right of the unique maximal subpath of on its boundary. We define a labeling of by giving each vertex the label of the region that contains it. We claim that is compatible with .

Since and are edge-disjoint, every primal edge connects vertices of the same or adjacent regions. Moreover, by construction, vertices of adjacent regions have different labels. Thus all vertices left of have label and all vertices right of have label . That is is compatible with , i.e., is consistent.

Figure 6: (a) The line through the segment induces a path in . (b) Modification of the undirected path edge-disjoint from .

Conversely, assume that is consistent. By Proposition 1 there is a straight-line drawing of such that the segment intersects the same edges as and in the same order (Fig. 6a). Let be the line that contains the segment . Each edge of intersects at most once. Thus, the complement of in defines a path from to in that is edge-disjoint from and does not cross .

Thus, we now consider the problem of finding a consistent shortest -path as an edge-disjoint path problem in . Our proof strategy consists of three steps. Step 1) We first show that the problem is equivalent to finding two edge-disjoint paths and in a directed graph such that is directed and is undirected. Step 2) We modify such that is a path in a specific subgraph and lies in the subgraph . These two graphs may share an edge set such that each edge in can be an edge of or of . Moreover, we find pairs of edges and in such that the path in (the path in ) contains either or . Step 3) Finally, we use these properties to reduce our problem to -SAT.

We begin with Step 1. A directed graph is -friendly if contains a consistent shortest -path if and only if contains a directed -path and an undirected -path that is edge-disjoint from and does not cross . We obtain an -friendly graph from as follows. Denote by the directed acyclic graph that contains all shortest paths from to in . If an edge is an edge of , we add it to . For all remaining edges , we add a subdivision vertex to and add the directed edges to in this direction. We claim that is -friendly.

Let be a consistent shortest -path in . By Lemma 1 there is a path in that is edge-disjoint from and does not cross . By construction corresponds to a directed path in and corresponds to an undirected path in . Conversely, due to the directions of the edges , every directed -path in is a directed path in , and therefore it is a shortest -path in . If there is an undirected path that is edge-disjoint from and does not cross , we obtain a path from by contracting edges incident to split vertices . Hence, is -friendly.

We consider the following special case, where and lie on a common face of the subgraph of . Without loss of generality, let be the outer face of and let lie on the outer face of . We denote by and the upper and lower -path of on the boundary of . A vertex of is an interior vertex if does not lie on . An edge of is an interior edge if and are interior vertices. An edge of is a chord if both its endpoints lie on but is not an edge on the boundary of .

Lemma 2

For a directed -path and an undirected -path , that are edge-disjoint and non-crossing, there is an undirected -path that is edge-disjoint from , does not cross , and that does not use interior vertices of .


Since and are non-crossing, there are two distinct vertices on or on , say , such that the inner vertices of lie in the interior of ; refer to Fig. 6b. Moreover, since and are non-crossing, the region enclosed by and does not contain a vertex of in its interior. Therefore, we obtain by iteratively replacing pieces in the form of by .

Figure 7: (a) The red directed path can be circumvented with the blue directed path via vertex . (b) The red path consists of avoidable edges.

This finishes Step 1, and we continue with Step 2. In the following, we iteratively simplify the structure of while preserving -friendliness of . Due to Lemma 2, the graph , obtained from contracting an edge of , is -friendly, if is an interior edge. This may generate a separating triangle . Let be a vertex in the interior of and let be a directed -path that contains . Then, contains at least two vertices of . Hence, can be rerouted using an edge of . Thus, the graph after removing all vertices in the interior of is -friendly. After contracting all interior edges of , each neighbor of an interior vertex of lies either on or on . The remaining edges are edges on and chords.

Consider three vertices that lie in this order on ( and two interior vertices and , with ; refer to Fig. 7a. Note that and can coincide. Then, every directed -path that contains also contains and . Hence, can be rerouted through the edges and as a consequence of Lemma 2, the graph is -friendly. Analogously, if contains the edge , remains -friendly. We call such edges circumventable.

We refer to edges of a subpath () as avoidable if there exists an interior vertex with (Fig. 7b). If there exists a directed path that uses an avoidable edge it can be rerouted by replacing the corresponding path with the edges . Thus, we can split the edge with a vertex  and we direct the resulting edges from towards and , respectively, and remove the edge from . Finally, we iteratively contract edges incident to vertices with in- and out-degree , and we iteratively remove vertices of degree at most , except for and .

Figure 8: (a) Interior partners decoded by color of -edge connected component of . (b) Split a vertex on the boundary of .

Since all interior edges of are contracted, circumventable interior edges are removed and avoidable edges are replaced, each -edge connected component of is an outerplanar graph whose weak dual (excluding the outer face) is a path; compare Fig. 8a. Each face of , with , contains at least one edge of and one edge on . Moreover, every directed -path contains either or . We refer to the edge sets and as interior partners.

Property 1

Choosing a directed -path in is equivalent to choosing for each face of one of the interior partners or such that the following condition holds. Let be two adjacent faces that are separated by a chord that ends at () such that is right of (left of ), then the choice of () implies the choice of ().

In the following, we modify the exterior of , i.e., , with the aim to obtain an analog property for the choice of the undirected path. We refer to edges of as exterior edges. A vertex in is an exterior vertex.

Since the undirected path is not allowed to cross the directed path, we split each cut vertex into an upper copy and a lower copy . We reconnect edges of and incident to to and , respectively. Exterior edges incident to that are embedded to the right of are reconnected to . Likewise, edges embedded to the left of are reconnected to . Note that this operation duplicates bridges of . Thus, we forbid the undirected path to traverse these duplicates. Observe that after this operation the outer face of is bounded by a simple cycle.

Let be a vertex on that is incident to an exterior edge. In this case, we insert a vertex to and we remove each exterior edge from and insert as a replacement edges and ; see Fig. 8b. We refer to the edge as a barrier. Since the barrier is directed from to , the modification preserves the -friendliness of . We now exhaustively contract exterior edges that are not barrier edges, and remove vertices in the interior of separating triangles.

Figure 9: (a) If the undirected path contains , it can be rerouted to use vertex . (b) The color coding of the faces indicate the exterior partners.

Recall that and lie on a common face of the subgraph of and lies on the outer face of . Let be an exterior vertex such that its neighbor comes before its neighbor on , refer to Fig. 9a. Let be a vertex between and on that is connected to a vertex such that the edge () lies in the interior of the region bounded by and . Consider a directed -path in and an undirected -path in that is edge-disjoint from , that does not cross and that contains . Due to Lemma 2 we can assume, that does not contain an interior vertex of . Thus, it contains and . We obtain a new path by replacing the subpath by . Since are exterior edges, and are edge-disjoint and non-crossing. Thus, the graph () is -friendly. After removing all such edges, for any two neighbors and of an exterior vertex , the paths and each contains either and . Hence, the region bounded by and contains a second exterior vertex if and only if contains either or .

Hence, the dual of , with the dual vertex of removed, is a caterpillar , refer to Fig. 9b. In case that or is incident to an exterior vertex , we can assume that the undirected path contains the edge (). Thus, for simplicity, we now assume that neither nor is connected to an exterior vertex. Let and be the vertices in whose primal faces are incident to and , respectively. Then every undirected -path in from to traverses the primal faces of the simple path from to in . Let be a primal face of a vertex on . Since we inserted the barrier edges to , every face contains at least one edge of and one edge of . Therefore, every undirected -path in either contains or . We refer to the sets and as exterior partners.

Property 2

Choosing an undirected -path in is equivalent to choosing for each face of one of the exterior partners or .

This finishes Step 2, and we proceed to Step 3. The problem of finding a directed -path and an undirected -path in reduces to a -SAT instance as follows. For each exterior and interior partner we introduce variables and , respectively, where and correspond to the faces of the partners. If is true, contains the edge of , otherwise it contains . The conditions on the choice of in Property 1 can be formulated as implications. Let an be exterior partners and let and be interior partners. In case that , either can contain edges of or can contains edges of but not both. Thus, and are not allowed to be true at the same time, i.e., . Hence, we have the following Theorem.

Theorem 4.1

If and lie on a common face of , it is decidable in polynomial time whether has a directed -path and an undirected -path that are edge-disjoint and non-crossing.

Corollary 1

If and lie on a common face of , it is decidable in polynomial time whether contains a consistent shortest -path.

5 Parametrized Complexity of Short Consistent -Paths

In this section we show that edge insertion can be solved in FPT time with respect to the minimum number of crossings of a straight-line drawing of where is drawn without crossings and has the specified embedding. Let  be an arbitrary labeling of . Observe that  defines a directed subgraph of  by removing each edge whose dual edge has endpoints with the same label and by directing all other edges such that the endpoint of its primal edge left of has label  and its other endpoint has label . We denote this graph by . Obviously, a shortest -path in  is compatible with , and thus a corresponding drawing exists. Clearly, given the labeling  a shortest -path in  can be computed in linear time by a BFS.

Now assume that the length of a shortest consistent path in  is . We propose a randomized FPT algorithm with running time  for finding a shortest consistent path in , based on the color-coding technique [3].

The algorithm works as follows. First, we pick a random labeling of  by labeling each vertex independently with  or

with probability 

. We then compute a shortest path in . We repeat this process times and report the shortest path found in all iterations.

Clearly the running time is . Moreover, each reported path is consistent, and therefore the algorithm outputs only consistent paths. It remains to show that the algorithm finds a path of length  with constant probability.

Consider a single iteration of the procedure. If the random labeling  is compatible with , then the algorithm finds a path of length . Therefore the probability that our algorithm finds a consistent path of length  is at least as high as the probability that is compatible with the random labeling . Let  denote the vertices of  that are left and right of , respectively. Clearly it is . A random labeling  is consistent with if it labels all vertices in  with and all vertices in with . Since vertices are labeled independently with probability , it follows that  is consistent with .

Therefore, the probability that no path of length  is found in iterations is at most , which is monotonically increasing and tends to . Thus the algorithm succeeds with a probability of . The success probability can be increased arbitrarily to , by repeating the algorithm  times. The probability that each iteration fails is then bounded from above by . E.g., to reach a success probability of , it suffices to do repetitions. The algorithm can be derandomized with standard techniques [3].

Theorem 5.1

There is a randomized algorithm that computes a consistent path of length  if one exists with a success probability of . The running time of is .

6 Conclusion

We have shown that the problem of finding a short consistent -paths in is tractable in special cases and fixed-parameter tractable in general. Whether has a short consistent -path is equivalent to the question of whether has two edge-disjoint and non-crossing -paths, where the length of one path is minimized. Surprisingly, this is related to yet another purely graph theoretic problem: does a directed graph have two edge-disjoint paths where one is directed and the other is only undirected? By the result of Eilam-Tzoreff [5] the former problem is in general -hard. For planar graphs the computational complexity of these problems remains an intriguing open question.

In this paper, we only considered planar graphs with a fixed combinatorial embedding. Allowing for arbitrary embeddings opens new perspectives on the problem and is interesting future work.


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Appendix 0.A Proof of Theorem 3.1

See 3.1


Let be a shortest path in . Assume that is not consistent. Then there is a vertex that left and right of . Let be the first edge of that crosses a primal edge incident to . If the degree of is at most , then contains either a loop or a double edge, contradicting the assumption that is a shortest path. Therefore, assume that the degree of is . Without loss of generality, let and be the faces around in clockwise order (Fig. 10b). Since is left and right of , contains either the edge or . Thus, contains either or twice. This contradicts the assumption that is a shortest path.

Figure 10: Inconsistent path around a degree-3 vertex.