1 Introduction
This paper considers the following problem. Let be a planar bipartite graph such that the vertices in , called fixed vertices, have fixed distinct locations (points) in the plane, while the vertices in , called mobile vertices, can be freely placed. Does admit a crossingfree drawing with at most bends per edge, where is a given nonnegative integer? We assume that each vertex of is drawn in as a distinct point of the plane and that each edge is drawn as a simple polyline. We call an FMbigraph and a drawing with the properties mentioned above a planar bend drawing of . In particular, since edge bends negatively affect the readability of a graph layout (see, e.g., [33, 34]), we are mainly interested in drawings with small values of , ideally with (i.e., straightline drawings). We define the bend number of as the minimum value of for which admits a planar bend drawing.
Besides its intrinsic theoretical interest, our problem is motivated by the following practical scenario. Fixed vertices represent geographic locations and each mobile vertex is an attribute of one or more locations. One wants to place each mobile vertex in the plane and connect it to its associated locations, while guaranteeing a “readable” layout. We interpret readability in terms of planarity and small number of bends per edge. Other criteria can also be studied, like for example angular resolution, edge length, drawing area (see, e.g., [10, 35]).
Contribution. We introduce bend drawings of FMbigraphs with a focus on and . Our results are as follows:
We prove that computing the bend number of an FMbigraph is hard. More generally, deciding whether admits a planar bend drawing is at least as hard as deciding whether an vertex graph admits a planar embedding with given correspondence (mapping) on a set of points, such that each edge has at most bends (Section 2.1). If all fixed vertices of are collinear, the existence of a 0bend drawing can be tested in linear time.
Since it is difficult to discretize the problem in the general case [15, 31], we investigate the case in which each mobile vertex is restricted to lie in the convex hull of its neighbors. This scenario is reasonable in practice, as the user may expect that each attribute is placed in a sort of “barycentric” position with respect to its associated locations. In this setting, we prove that testing the existence of a 0bend drawing is a problem in . With a reduction to a combinatorial problem, which is of its own independent interest (but unfortunately hard in its general form), we obtain polynomialtime solutions when the intersection graph of the convex hulls is a path, a cycle or, more generally, a cactus (Section 2.2).
We finally study 1bend drawings of FMbigraphs in a convention called strip drawing model, inspired by practical labeling scenarios [3]. All fixed vertices are partitioned into a finite set of horizontal strips and each mobile vertex is placed outside these strips. Edges are not allowed to cross any strip, i.e., to intersect both its top and bottom side; see also Fig. 4. For this model we provide polynomialtime testing algorithms (Section 3).
Related Work. Our problem is related to several problems addressed in the literature, but it also has substantial differences from all of them.
Point labeling. A close connection is with the problem of labeling a given set of points in the plane (see, e.g., [29, 38]), because mobile vertices can be regarded as labels for the fixed vertices (points). Similarly to our setting, the manytoone boundary labeling problem [3, 24] assumes that each label can have multiple associated vertices and it is visually connected to them by polyline edges. However, edges can only be drawn as chains of horizontal and vertical segments (which may partially overlap), and the labels are placed outside a single rectangular region that encloses all vertices. Variants of the boundary labeling problem, where each fixed vertex is associated with exactly one label are also studied in the literature (see, e.g., [4, 23, 2]). Note that, in labeling problems, labels are geometric shapes of nonempty area, while we model mobile vertices as points.
Partial drawings. Our problem is a special case of the problem of extending a partial drawing of a (not necessarily bipartite) planar graph to a planar straightline drawing of . This problem is hard in general [31] and polynomialtime solvable for restricted cases [37, 13, 9, 18, 25].
Pointset embedding. In a pointset embedding problem, a planar graph with vertices must be planarly mapped onto a given set of points, with or without a predefined correspondence between the vertices and the points (see, e.g., [1, 8, 11, 12, 22, 30]). Thus, in all settings of the pointset embedding problem, each vertex can only be mapped to a finite set of points. The results in [1, 30] imply that any vertex planar FMbigraph admits a bend drawing with . Indeed, [1, 30] prove that any vertex planar graph can be planarly mapped onto any set of points, with given correspondences, using a linear number of bends per edge (which is also necessary in some cases). Hence, for a given FMbigraph, one can place the mobile vertices anywhere so to realize a planar drawing.
Constrained drawings of bipartite graphs. Misue [27] proposed a model and a technique for drawing bipartite graphs such that the vertices of a partition set, called anchors, are evenly distributed on a circle. Anchors are similar to fixed vertices in our setting, but the order of the anchors in Misue’s model can be freely chosen. Extensions to the 3D space and to semibipartite graphs have been subsequently presented [20, 28]. Finally, several papers study how to draw a bipartite graph such that the vertices of each partition set are on a line or within a specific plane region (see, e.g., [5, 6, 14]). In these scenarios, the vertices do not have predefined locations.
Notation. We assume familiarity with graph theory (see, e.g., [17]). For standard definitions on planar graphs and drawings, we point the reader to [21, 36].
We denote an FMbigraph by a pair , where is a bipartite graph and is a function that maps each vertex to a distinct point . A bend drawing of is a bend drawing of such that each vertex is mapped to . In order to study the complexity of computing the bend number of planar FMbigraphs, we introduce the bend FMbigraph decision problem: Given a planar FMbigraph and a nonnegative integer , is there a planar bend drawing of ?
From now on, we assume that is planar. Also, we let , , and . Some proofs are moved to the appendix.
2 Straightline Planar Drawings of FMbigraphs
We show that the bend FMbigraph problem is hard (Theorem 2.1), which implies that it is hard to compute the bend number of a planar FMbigraph. A simple lineartime testing algorithm is given when all fixed vertices are collinear (Theorem 2.3). If each mobile vertex must be placed inside the convex hull of its neighbors, then the bend FMbigraph problem belongs to (Theorem 2.4), and it becomes polynomialtime solvable if the intersection graph of the convex hulls is a cactus (Theorem 2.5).
2.1 hardness and Collinear Fixed Vertices
To prove that bend FMbigraph is hard we use a reduction from the 1bend point set embeddability with correspondence problem (or 1BPSEWC, for short), which has been proven to be hard by Goaoc et al. [15]. Problem 1BPSEWC is defined as follows: Given a planar graph , a set of points in the plane, and a onetoone correspondence between and , is there a planar bend drawing of such that each vertex is mapped to point ?
Theorem 2.1
The bend FMbigraph problem is hard, even if each mobile vertex has degree at most two.
Proof
Let be an instance of 1BPSEWC. Construct (in linear time) an instance of bend FMbigraph as follows: Let and ; for each edge , define a corresponding vertex and two edges , in . Clearly, has a 1bend drawing that respects if and only if has a planar 0bend drawing that respects : The position of a bend along an edge of corresponds to the positions of the mobile vertex in ; if has no bend, is drawn anywhere along segment .
The reduction in Theorem 2.1 can be applied with no change to prove that, for any , problem bend FMbigraph is at least as difficult as problem BPSEWC, which allows up to bends per edge.
Theorem 2.2
The bend FMbigraph problem is at least as hard as the BPSEWC problem, for any .
When all fixed vertices of an FMbigraph are collinear, it can be checked in linear time whether admits a planar 0bend drawing.
collinear
Theorem 2.3
Let be an vertex FMbigraph such that all vertices of are collinear. There exists an time algorithm that tests whether admits a planar 0bend drawing.
Proof (sketch)
Let be the line passing through all fixed vertices. Deciding whether has a planar 0bend drawing coincides with testing the planarity of a graph obtained by augmenting with a cycle that connects all fixed vertices in the order they appear along .
2.2 Mobile Vertices at Internal Positions
We now focus on convexhull drawings, in which all fixed vertices are in general position and each vertex lies in the convex hull of its neighbors. With slight abuse of notation, we denote by the convex hull of the neighbors of . Let be the arrangement of lines defined by all pairs of fixed points; see Fig. 2(a). has lines and cells [16]. Lemma 1 allows us to discretize the set of possible positions for the mobile vertices; it implies that all positions of in the same cell of within are equivalent for a planar 0bend drawing of .
cellequiv
Lemma 1
Let be an FMbigraph, , and a cell of inside . Let also and be two points in . Suppose that is a bend drawing of where is at point and let be a bend drawing of obtained from by only moving from point to point . Then is planar if and only if is planar.
Proof
Suppose by contradiction that is planar and is not (the proof for the other direction is symmetric). This implies that while moving along some trajectory from to inside cell , at some point we get a crossing along one of the edges incident to . Let be the point of closest to , such that placing at causes such a crossing (i.e., placing on any point between and implies no crossing). Let be an edge crossed by some other edge , when is placed at . Assume w.l.o.g. that and , and that lies to the right of the oriented edge ; see Fig. 1. Denote by the line through and and by the line through and . Let be the region delimited by lines , and edge that contains . Let be a point of lying between and . Notice that has to lie outside of . Thus crosses the border of . Let us additionally assume that lies arbitrarily close to the border of . We distinguish three cases, based on whether crosses , , or edge .
Case 1. crosses . Since line is part of , lies outside . This is a contradiction to the assumption that lies in .
Case 2. crosses ; see Fig. 1. Since is in the convex hull of its neighbors, there is an edge with and on different sides of (if not, the crossing is resolved). Placing at yields a crossing with , as is arbitrarily close to ; a contradiction to the choice of .
Case 3. crosses . Since lies in the convex hull of its neighbors, there is an edge , where and lie on different sides of the line through edge ; see Fig. 1. Placing at would introduce a crossing between and , as lies arbitrarily close to . This again contradicts the choice of .
Lemma 1 implies that, the bend FMbigraph problem belongs to for convexhull drawings^{1}^{1}1We remark that in a preliminary version of [31], it is claimed membership in for the partial planarity extension problem [32], which would imply membership in also for our problem. That claim, however, lacks a proof in [32] and the author was only able to prove the hardness of the problem in [31] (personal communication).. A nondeterministic algorithm guesses an assignments of the mobile vertices to the cells and, since is planar, checks in time whether the corresponding 0bend drawing is planar (note that ). We summarize this observation in the following theorem.
Theorem 2.4
The bend FMbigraph problem belongs to if each mobile vertex must lie in the convex hull of its neighbors.
Central ingredients to prove that the problem is in fact in for certain input configurations are the CH intersection graph of , the cell graph of , and the skeleton graph of , which we formally define in the following.
The CH intersection graph is defined as the intersection graph [26] of the convex hulls over all .
The cell graph is a clustered graph defined as follows; see Fig. 2(b) for an example. Each mobile vertex is associated with a cluster ; the vertices of , called cell vertices, are the cells of that intersect with (and in fact are contained in ). The vertices of are defined by the disjoint union of the vertices of all clusters^{2}^{2}2Cells in the intersection of two convex hulls correspond to different vertices of ., that is, . For a cell vertex of , we denote by the cell corresponding to in . For a pair of mobile vertices and such that , a cell vertex is adjacent to a cell vertex if and only if placing in and in produces no crossing among the edges incident to and ; see Fig. 2(c). Note that has vertices and edges. Also, by definition, for each pair of mobile vertices and such that , and can be positioned within their convex hulls without creating edge crossings if and only if there exist two adjacent cell vertices and in .
The skeleton graph is created by selecting exactly one cell vertex, called a skeleton vertex, from each cluster of , such that for every pair of mobile vertices and with , the skeleton vertices of and are adjacent in . Graph is the subgraph of induced by the skeleton vertices. Note that might not exist. If exists, then it is isomorphic to . The following characterization is an immediate consequence of our definitions.
fromgeometrytotopology
Lemma 2
An FMbigraph admits a planar 0bend convexhull drawing if and only if cell graph has a skeleton.
Proof
A planar 0bend drawing immediately defines a skeleton. Conversely, if has a skeleton , a planar 0bend drawing is obtained by placing each in the cell corresponding to the skeleton vertex of in . Since crossings may only occur between edges incident to mobile vertices and such that , is planar.
The characterization of Lemma 2 allows us to translate the geometric problem of finding a 0bend convexhull drawing of an FMbigraph bigraph to a purely combinatorial problem on a support clustered graph constructed from . Unfortunately, however, this combinatorial problem is hard in its general form, as Theorem 2.6 shows. Nonetheless, we are able to solve it efficiently when is a cactus (Theorem 2.5), which includes the special cases in which is a cycle or a tree. The next two lemmas are base cases for Theorem 2.5.
Lemma 3
Let be an FMbigraph such that is a path. There exists a polynomialtime algorithm that tests whether has a planar 0bend convexhull drawing.
Proof
By Lemma 2, it is enough to test whether has a skeleton. Let be the mobile vertices in the order their convex hulls appear along path . Call a cell vertex of active if and only if the subgraph of induced by has a skeleton containing , where . Thus, has a skeleton if and only if there is an active cell vertex in . A simple algorithm that tests this condition works as follows. Initially mark all cell vertices of as active, and then propagate this information forward to the cell vertices of , that is, for each , mark each cell vertex of as active if it has an active neighbor in . The time complexity is bounded by the number of vertices and edges in .
Lemma 4
Let be an FMbigraph such that is a simple cycle. There exists a polynomialtime algorithm that tests whether has a planar 0bend convexhull drawing.
Proof
Let be the mobile vertices in the cyclic order their convex hulls appear along . Our approach is similar to the one of Lemma 3, but now it is not enough to propagate the information about the active vertices only from to . Indeed, there might be vertices of that cannot close a cycle with an active vertex of .
In the first phase, our refined algorithm starts by marking all cell vertices of as active and then propagates this information forward to , as in the case of a path. However, all cell vertices of a cluster that are not marked as active at the end of this phase are now definitely removed from (along with their incident edges), as they cannot occur in any skeleton. Then, the algorithm cleans all vertex marks and executes a backward propagation phase from to (symmetric to the previous one), where all the remaining vertices in are initially marked as active. As before, all vertices that are not marked as active at the end of this phase are definitely removed from . Now, the algorithm removes from all vertices with no neighbor in and from all vertices with no neighbor in , as these vertices cannot occur in a skeleton of . Finally, for each pair of adjacent vertices and in , the algorithm checks whether there exists a path from to that passes through each exactly once . If exists, both and are marked as confirmed. At the end, every vertex in that is not confirmed is removed from , as it cannot occur in a skeleton. Conversely, by construction, every remaining vertex in has an adjacent vertex such that is a skeleton (simple cycle) of . Thus the test is positive if and only if is not empty.
It is immediate to see that the whole testing algorithm works in polynomial time in the size of and, if the test is positive, a skeleton of can be easily reconstructed visiting from any vertex .
We now extend the previous result to the case that is a cactus, which also covers the case of a tree. We recall that a cactus is a connected graph in which any two simple cycles share at most one vertex. A cactus is an outerplanar graph and can always be decomposed into a tree where each node corresponds to either a single vertex or a simple cycle (refer to Fig. 3(a)).
cactusalgorithm
Theorem 2.5
Let be an FMbigraph such that is a cactus. There exists a polynomialtime algorithm that tests whether has a planar 0bend convexhull drawing.
Proof
By Lemmas 3 and 4, the statement holds when is a path or a cycle. In the general case, our testing algorithm decomposes into its tree (as in Fig. 3(b)), roots at any node, and visits bottomup. More precisely, each vertex of corresponds to a convex hull of a mobile vertex , and it has a onetoone correspondence with a cluster of . Thus, each node of corresponds to either a single cluster of or to a cycle of clusters of . Note that when in two cycles share a vertex (cluster of ), we replicate such a vertex in both nodes of that correspond to the two cycles. For example, in Fig. 3(b) cluster inside and cluster inside correspond to the same vertex of . Once the root of has been chosen, we define the anchor of as the cluster that connects to its parent node in (lightgray in Fig. 3(b)). During the bottomup visit of , two cases are possible when a node is visited:

is a leaf. If contains a single cluster (i.e., its anchor), then all its cell vertices are marked as active; if contains a cycle of clusters, then the active cell vertices of its anchor are computed as for in the proof of Lemma 4.

is an internal node. Let be the children of in and denote by the cluster of connected to the anchor of . Note that may coincide with some if . Also, the anchor of and may correspond to the same vertex of .

For each , if the anchor of differs from in , remove from all cell vertices that are not connected to an active cell vertex of the anchor of in , as they cannot occur in any skeleton of . On the other hand, if the anchor of and coincide in , remove from all vertices of the anchor of that are not marked as active.

Now, if contains a single cluster (i.e., its anchor), then all its remaining cell vertices are marked as active; if contains a cycle of clusters, then the active cell vertices of its anchor are computed as in Lemma 4. At this point, if the anchor of contains no active vertex, the algorithm stops and the instance is rejected, as a skeleton does not exist.

Once the bottomup visit of ends, the test is positive if and only if the anchor of the root node of has an active cell vertex , and in this case one can reconstruct a skeleton of starting from and visiting topdown. In particular, during the topdown visit, for each node of , any active vertex in the anchor of can be arbitrarily selected, as it is connected to the parent node of by construction. Also, if corresponds to a simple cycle of clusters, the construction of a cycle that connects these clusters is done as in Lemma 4.
Concerning the time complexity, the above algorithm takes polynomial time in the size of . Indeed, the number of clusters that may occur in multiple nodes of (i.e., those that are shared by multiple cycles of clusters) is at most the number of cycles in . Therefore, the total number of clusters over all nodes of is linear in the number of clusters of . This also implies that the total number of cell vertices over all clusters of is linear in the number of cell vertices in . Finally, each node of is visited twice (once in the bottomup visit and once in the topdown visit), and in each visit of the algorithm has a running time that is polynomial in the number of cell vertices in the clusters of .
Finally, we show the following completeness result on a combinatorial generalization of our problem. We reuse the terminology of the bend FMbigraph problem to emphasize the analogies.
skeletonhardness
Theorem 2.6
Let be a graph, where is a set of disjoint clusters of cells. Also, let be a graph, where each is a cell of a cluster and only if . It is complete to test if there is a subset of skeleton vertices, containing exactly one cell from each cluster in such that the induced subgraph is isomorphic to .
Proof (sketch)
The problem is clearly in . The hardness proof is by reduction from 3Sat. For a boolean 3Sat formula create a cluster for each variable in and a cluster for each clause of . In each clause cluster is adjacent to the three clusters of the variables occurring in the clause. Each variable cluster consists of two cells in , one for the positive literal and one for its negation . Also, each clause cluster contains three cells, one for each literal. Finally, connect each literal cell of a clause to the corresponding cell of its variable cluster and to all four cells of the other two variables of . It can be seen that has a satisfying truth assignment iff there exists a subset of skeleton vertices in that induces a subgraph isomorphic to .
3 1bend Drawings in the Strip Drawing Model
Our model for bend FMbigraphs is inspired by the boundary labeling approach [3], where mobile vertices are regarded as labels that must be connected to the fixed vertices. In the boundary labeling model, the fixed vertices are inside a single rectangular region and each label is either to the left or to the right of this region. Our model allows for multiple rectangular regions (corresponding to horizontal strips); each mobile vertex is placed outside of these regions, either below or above each of them. To avoid long edges and make the drawing more readable, edges are not allowed to traverse regions.
More formally, our model is called the strip model and is defined as follows. Let be an FMbigraph and assume that the vertices of all have distinct coordinates (this condition is always achievable by a suitable rotation of the plane). For the sake of simplicity, for a vertex , we do not distinguish between and its fixed position . Let be a toptobottom sequence of (closed) disjoint horizontal strips of the plane that partition , i.e., each has a finite height and infinite width, each vertex of lies in one , and for . Since the strips are disjoint, there is always a nonempty region of the plane between two consecutive strips, which does not contain fixed vertices. Also, there are no fixed vertex above and below . Any point that is not inside a strip is called a free point. For a vertex , the strip that contains is called the strip of .
A bend drawing of within is defined as follows; see Fig. 4: Each vertex is mapped to a distinct free point. Each edge , with and consists of a segment from to a point on the boundary of the strip of and of a vertical segment ; all points of but are free points, while is completely inside the strip of . No edge intersects the boundary of a strip twice and no two edges cross in a free point.
Note that in the strip model two distinct edges and , where , share their vertical segments if these segments are incident to both from below or both from above. This overlap does not create ambiguity and reduces the visual complexity caused by the edges. Figure 4 shows a bend drawing of a bigraph within a given set of three strips (gray regions), with fixed vertices in black. Also note that, if an FMbigraph has no bend drawing for a set of strips, splitting an element of into two strips may lead to a feasible solution; see Fig. 5. Conversely, splitting a strip may transform a positive instance into a negative one; see Fig. 5. We prove the following.
onebendstripmodel
Theorem 3.1
Let be an vertex FMbigraph and let be a set of horizontal strips that partition . There exists an time algorithm that tests whether admits a bend drawing within .
Proof (sketch)
Call a fixed vertex black, a mobile vertex with all neighbors in the same strip white, and the remaining vertices gray. A gray vertex with neighbors in two consecutive strips must lie between them, while each white vertex can lie either above or below the strip of its neighbors. If a grey vertex has neighbors that are not in two consecutive strips, the instance is immediately rejected.
Let be the sequence of strips. For each strip , let be the lefttoright sequence of black vertices inside . Also, let be the set of mobile vertices connected to some vertex of . Arranging the vertices of above or below so to avoid crossings between their incident edges equals to assigning each of them either above or below the halfplane determined by a horizontal line that contains , in this lefttoright order. Hence, testing if a bend drawing within exists generalizes testing the existence of a bend drawing when all fixed vertices are collinear. As in Theorem 2.3, this problem is reduced to testing planarity of a graph suitably defined by augmenting . Namely, for each , add a cycle connecting all edges of in their lefttoright order; then, subdivide edge of with three vertices , in this order from to , and call the subdivision of ; finally, for each and , connect to . Graph is planar iff it has a planar embedding where is inside (). A mobile vertex between and corresponds to placing above and below . If is in the outer face of then is below , and if is inside then is above . Since the size of is linear in the size of and graph planarity testing is lineartime solvable, the statement holds.
The next result immediately follows by iterating the technique in the proof of Theorem 0..3 over all possible ways of partitioning into strips.
Corollary 1
Let be an vertex FMbigraph and let be a constant. There is an time algorithm that tests if has a bend drawing within , for some set of strips that partition .
4 Conclusions and Open Problems
We introduced FMbigraphs, showed that the bend FMbigraph problem is hard in the general case, and gave polynomialtime algorithms for in some interesting restricted cases. Several open research questions remain:
Q1. We could solve the bend FMbigraph problem for convexhull drawings if the CH intersection graph is a cactus and show that it is complete in a nongeometric setting. Can we solve the problem for larger classes of convexhull drawings in polynomial time or extend the completeness to our geometric setting? More generally, for which other layout constraints or subfamilies of FMbigraphs does the bend FMbigraph problem become tractable?
Q2. Our focus was on proving the existence of polynomialtime algorithms under certain layout constraints, but some of the algorithms have high time complexity. Thus, finding more efficient algorithms is of interest.
Q3.
We focused on crossingfree drawings of FMbigraphs. Relaxing the planarity requirement (e.g., for a given maximum number of permitted crossings per edge) is an interesting variant, as well as, designing heuristics or exact approaches for crossing/bend minimization.
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Appendix
collinear
Theorem 0..1
Let be an vertex FMbigraph such that all vertices of are collinear. There exists an time algorithm that tests whether admits a planar 0bend drawing.
Proof
Assume w.l.o.g. that all vertices of lie on a horizontal line . In any planar 0bend drawing of , we can assume that every vertex is either above or below . Indeed, if lies on then has degree either one or two, and in this last case it is incident to two consecutive vertices of along : We can always slightly move above or below without changing the planar embedding of the drawing. Hence, deciding whether has a planar 0bend drawing is equivalent to deciding whether there exists an assignment of each mobile vertex to one of the two half planes determined by that avoids edge crossings. This problem coincides with testing the planarity of a graph obtained by augmenting with a cycle that connects all fixed vertices in the order they appear along : A vertex inside (outisde of) the cycle corresponds to a vertex above (below) . Since the size of is linear in the size of and since the graph planarity testing problem is lineartime solvable [7, 19], the statement follows.
skeletonhardness
Theorem 0..2
Let be a graph, where is a set of disjoint clusters of cells. Also, let be a graph, where each is a cell of a cluster and only if . It is complete to test if there is a subset of skeleton vertices, containing exactly one cell from each cluster in such that the induced subgraph is isomorphic to .
Proof
The problem is clearly in . The proof of the hardness is by reduction from 3Sat. For a boolean 3Sat formula create a cluster for each variable in and a cluster for each clause of . In each clause cluster is adjacent to the three clusters of the variables occurring in the clause. Each variable cluster consists of two cells in , one for the positive literal and one for its negation . Also, each clause cluster contains three cells, one for each literal. Finally, connect each literal cell of a clause to the corresponding cell of its variable cluster and to all four cells of the other two variables of .
We now show that has a satisfying truth assignment iff there exists a subset of skeleton vertices in that induces a subgraph isomorphic to . Assume that we know a satisfying variable assignment of . We select the “true” cells of the variable clusters and one satisfied literal of each clause. The induced subgraph of this set of cells is isomorphic to as the satisfied literal cell of each clause covers all three edges of to its three adjacent variable clusters. Conversely, if we have a subset of skeleton vertices of that induce a subgraph isomorphic to , then setting the literals of the set of selected literal cells to true satisfies . Otherwise, in an unsatisfied clause, none of the three clause cells would connect to the selected cells of all three adjacent variable clusters, which contradicts the skeleton property.
onebendstripmodel
Theorem 0..3
Let be an vertex FMbigraph and let be a set of horizontal strips that partition . There exists an time algorithm that tests whether admits a bend drawing within .
Proof
Call black a fixed vertex, white a mobile vertex with all neighbors in the same strip, and gray the remaining vertices. A gray vertex with neighbors in two consecutive strips must lie between them, while each white vertex can lie either above or below the strip of its (black) neighbors. If a grey vertex has neighbors that are not in two consecutive strips, the instance is immediately rejected.
Let be the sequence of strips. For each strip , let be the lefttoright sequence of black vertices inside . Also, let be the set of mobile vertices connected to some vertex of . Arranging the vertices of above or below so to avoid crossings between their incident edges equals to assigning each of them either above or below the halfplane determined by a horizontal line that contains , in this lefttoright order. Hence, testing if a bend drawing within exists generalizes testing the existence of a bend drawing when all fixed vertices are collinear. As in Theorem 2.3, this problem is reduced to testing planarity of a graph suitably defined by augmenting . Namely, for each , add a cycle connecting all edges of in their lefttoright order; then, subdivide edge of with three dummy vertices , in this order from to , and call the subdivision of ; finally, for each and , connect to . See Fig. 6 for an illustration; in the figure, the dummy vertices are represented by small squares.
Denote by the graph obtained from by removing all mobile vertices. It is immediate to see that is planar if and only if it has a planar embedding such that coincides with the external face of . In such an embedding for , the edges force all cycles to be nested in such a way that is inside (). A mobile vertex between and corresponds to placing above and below . If is in the outer face of then is below , and if is inside then is above . Since the size of is linear in the size of and graph planarity testing is lineartime solvable, the statement holds.