On the Edge-length Ratio of Outerplanar Graphs

08/31/2017 ∙ by Sylvain Lazard, et al. ∙ Williams College Inria Università Perugia 0

We show that any outerplanar graph admits a planar straightline drawing such that the length ratio of the longest to the shortest edges is strictly less than 2. This result is tight in the sense that for any ϵ > 0 there are outerplanar graphs that cannot be drawn with an edge-length ratio smaller than 2 - ϵ. We also show that every bipartite outerplanar graph has a planar straight-line drawing with edge-length ratio 1, and that, for any k ≥ 1, there exists an outerplanar graph with a given combinatorial embedding such that any planar straight-line drawing has edge-length ratio greater than k.

READ FULL TEXT VIEW PDF
POST COMMENT

Comments

There are no comments yet.

Authors

page 1

page 2

page 3

page 4

This week in AI

Get the week's most popular data science and artificial intelligence research sent straight to your inbox every Saturday.

1 Introduction

The problem of computing a planar straight-line drawing with prescribed edge lengths has been addressed by several authors, partly for its theoretical interest and partly for its application in different areas, including VLSI, wireless sensor networks, and computational geometry (see for example [DBLP:journals/siamdm/BattistaV96, DBLP:conf/infocom/DohertyPG01, DBLP:series/natosec/HeldKRV11, DBLP:conf/usenix/SavareseRL02]). Deciding whether a graph admits a straight-line planar drawing with prescribed edge lengths was shown to be NP-hard by Eades and Wormald for 3-connected planar graphs  [DBLP:journals/dam/EadesW90]. In the same paper, the authors show that it is NP-hard to determine whether a 2-connected planar graph has a unit-length planar straight-line drawing; that is, a drawing in which all edges have the same length. Cabello et al. extend this last result by showing that it is NP-hard to decide whether a 3-connected planar graph admits a unit-length planar straight-line drawing [DBLP:journals/jgaa/CabelloDR07]. In addition, Bhatt and Cosmadakis prove that deciding whether a degree-4 tree has a planar drawing such that all edges have the same length and the vertices are at integer grid points is also NP-hard [DBLP:journals/ipl/BhattC87].

These hardness results have motivated the study of relaxations and variants of the problem of computing straight-line planar drawings with constraints on the edge lengths. For example, Aichholzer et al. [DBLP:conf/cccg/AichholzerHKR14] study the problem of computing straight-line planar drawings where, for each pair of edges of the input graph , it is specified which edge must be longer. They characterize families of graphs that are length universal, i.e. they admit a planar straight-line drawing for any given total order of their edge lengths.

Perhaps one of the most natural variants of the problem in the context of graph drawing is that where, instead of imposing constraints on the edge lengths, one aims at computing planar straight-line drawings where the variance of the lengths of the edges is minimized. See for example  

[DBLP:books/ph/BattistaETT99], where this optimization goal is listed among the most relevant aesthetics that impact the readability of a drawing of a graph. Computing straight-line drawings where the ratio of the longest to the shortest edge is close to also arises in the approximation of unit disk graph representations, a problem of interest in the area of wireless communication networks (see, e.g.  [Chen2011, DBLP:conf/dialm/KuhnMW04]).

Discouragingly, Eades and Wormald observe in their seminal paper that the NP-hardness of computing 2-connected planar straight-line drawings with unit edge lengths persists even when a small tolerance (independent of the problem size) in the length of the edges is allowed. To our knowledge, little progress has been made on bounding the ratio between the longest and shortest edge lengths in planar straight-line drawings. We recall the work of Hoffmann et al. [DBLP:conf/cccg/HoffmannKKR14], who compare different drawing styles according to different quality measures including the edge-length variance.

In this paper we study planar straight-line drawings of outerplanar graphs that bound the ratio of the longest to the shortest edge lengths from above by a constant. We define the planar edge-length ratio of a planar graph as the smallest ratio between the longest and the shortest edge lengths over all planar straight-line drawings of . The main result of the paper is the following.

Theorem 1.1

The planar edge-length ratio of an outerplanar graph is strictly less than . Also, for any given real positive number , there exists an outerplanar graph whose planar edge-length ratio is greater than .

Informally, Theorem 1.1 establishes that is a tight bound for the planar edge-length ratio of outerplanar graphs. The upper bound is proved by using a suitable decomposition of an outerplanar graph into subgraphs called strips, then drawing the graph strip by strip. The lower bound is proved by taking into account all possible planar embeddings of a maximal outerplanar graph whose maximum vertex degree is a function of . Theorem 1.1 naturally suggests some interesting questions that are discussed in Section 3.

We shall assume familiarity with basic definitions of graph planarity and of graph drawing [DBLP:books/ph/BattistaETT99] and introduce only the terminology and notation that is strictly needed for our proofs.

Note: Some proofs have been moved to the Appendix.

2 Proof of Theorem 1

It suffices to establish the result for maximal outerplanar graphs. To show that the edge-length ratio of a maximal outerplanar graph is always less than , we imagine decomposing the dual of into a set of disjoint paths, which we call chains. Each chain corresponds to some sequence of pairwise-adjacent triangles of . The set of chains inherits a tree structure from , and we use this structure to direct an algorithm that draws each of the chains proceeding from the root of this tree down to its leaves. We formally define a chain to be a sequence of triangles of where , and such that (i) consists of an outer edge of whose vertices are labeled with , along with a third vertex labeled , (ii) for each , the vertices of are labeled by so that and share the edge having vertices labeled and , and (iii) for each , the vertices of are labeled by so that and share the edge having vertices labeled and . Note that this definition prohibits fans (consecutive triangles all sharing a common vertex) containing more than triangles, except for the vertex labeled , which has four incident triangles on the chain.

The decomposition into chains is constructed by first selecting an edge on the outer face of some outerplanar topological embedding of . The edge is incident with a unique triangle of . Label each vertex of with , and label the third vertex of the triangle with . There is now a unique maximal chain in containing this labeled triangle. The edges of can be partitioned into two sets: and where consists of edges of whose vertex labels differ by and consists of all edges of whose vertex labels differ by , along with .

Removing the edges of from produces a set of -connected components in 1-1 correspondence with the edges of : Each component contains exactly one element of which lies on its outer face. For each edge , let be the component of containing . We can then recursively decompose each by choosing the (unique) maximal chain in containing the one triangle (if any) of that is incident with . We call the set of chains so constructed a chain decomposition of . A chain decomposition produces a decomposition of the edges of into sets and , where is the union of the edges in each and is the union of all of the edges in each . Note that there is a single chain for each edge in , and that the collection of chains produced naturally form a tree: The root of the tree is the chain and its children are the chains ; the chain decomposition is entirely determined by the choice of external edge .

The drawing algorithm proceeds by first drawing the root chain of the chain decomposition tree of and then recursively drawing the chain decomposition trees of each . The algorithm depends on a specific method for drawing a single chain. To describe it, we need a few definitions. First, given a line segment

in the plane and a direction (unit vector)

not parallel to , denote by the half-infinite strip bounded by and the two infinite rays in direction that have their sources at the endpoints of . Finally, given a chain , the edges of are called external edges of ; note that each external edge is incident to exactly one triangle of .

Lemma 1

Given a chain with vertices, an external edge of , a segment of length in the plane, and a direction such that the (smaller) angle between and is , there exists a planar straight-line drawing of such that: (i) the drawing is completely contained within the strip ; (ii) no external edge of is parallel to , and the strips are all empty; (iii) each external edge has length and all other edges have lengths greater than ; (iv) each external edge forms an angle less than with . Moreover, such a drawing can be computed in -time in the real RAM model.

Proof

Let be the triangle of containing . is either adjacent to zero, one, or two triangles of . We handle these three cases in turn. If is the only triangle in , then we simply draw in as an isosceles triangle with drawn as and with its third vertex drawn so that its two edges have length , where .

Assume now that is adjacent to a triangle of . Denote the vertices of as follows: , where and is not incident with . The vertex of not in is denoted by , and, subsequently, the vertex of each not in is denoted by . We draw as previously, but with more careful positioning of . To determine where to position , we draw edge of as a unit-length segment in direction . As long as is positioned within but outside of the disks of radius centered at , and , the edges from each of these vertices to will have length greater than (see top half of Figure 1). By placing close to , the edges and will have lengths less than . Also, since , edge will have length less than .

Figure 1: Drawing a chain

Assuming that have been drawn for some , is drawn by positioning one unit distant from in direction . The result is that each is congruent to and so the edge-length ratio of is less than . At this point, all of the unit-length segments, except for , lie on the two rays in direction emanating from and . By rotating these rays a very small amount towards one another, we can preserve the lengths of the unit-length segments while ensuring that all of the remaining segments have lengths in the range . See the top of Figure 1.

Finally, suppose that has two adjacent triangles. Starting with , label the other triangles in so that the labels of adjacent triangles differ by . Thus, for example, is adjacent to and . The vertices in the will be labeled as in the previous case: the unique vertex in not in is labeled . The vertices in the will be similarly labeled: the unique vertex in not in is labeled .

Draw each as in the previous case. Now draw the in a similar fashion: Place vertex one unit distant from in direction . Then, as above, all of the unit length edges of the triangles will lie on the two rays in direction emanating from and , and these two rays can be rotated slightly towards each other while maintaining the length of the unit-length edges and ensuring that the other edges still have lengths in the range . See the bottom of Figure 1. This can clearly be done so that all external edges form angles less than with .

However, we need to ensure that can be placed so that both triangle and triangle can simultaneously satisfy the required edge-length conditions: Namely, that edges , , and are all unit-length, while edges , , , and all have lengths in the range . However, it is relatively simple to show that can always be successfully placed if the (smaller) angle between and is less than . [The angle is the angle opposite an edge of length in an isosceles triangle having side lengths , , and .] Finally, the computation of the locations of the vertices can each be computed in constant time in the real RAM model, giving a run-time linear in the size of the chain.∎

We are now ready to prove the following lemma. For a planar straight-line drawing , we denote with the ratio of a longest to a shortest edge in .

Lemma 2

A maximal outerplanar graph with vertices admits a planar straight-line drawing with that can be computed in time assuming the real RAM model of computation .

Proof

We call the drawing computed as in Lemma 1 a U-strip drawing of and adopt the same notation as in Lemma 1. Recall that in a chain decomposition of a graph, the external edges of the chains are exactly the edges of . A drawing of is computed as follows.

  1. Compute a chain decomposition tree for ; let be the root of the tree.

  2. Select a line segment of length in the plane and an initial direction not parallel to such that .

  3. Apply Lemma 1 to compute a U-strip drawing of .

  4. Each edge is drawn as a segment of length , not parallel to , that forms an angle with that is less than , so draw the subtree of rooted at in the empty strip .

The result is an outerplanar straight-line drawing in which all edges of (long edges) have length while all edges in (short edges) have length strictly greater than . If we assume that the input is provided to the algorithm in the form of a doubly-connected edge list [MULLER1978217], then a chain decomposition tree for can be computed in linear time. Also, since by Lemma 1 each chain can be drawn in time proportional to its length, the algorithm runs in time in the real RAM model. ∎

The following lemma can be proved by means of a packing argument and elementary geometry (see Appendix A1 for details).

Lemma 3

For any there exists a maximal outerplanar graph whose planar edge length ration is greater than .

We conclude the section by observing that Lemmas 2 and 3 imply Theorem 1.1.

3 Additional Remarks and Open Problems

The upper and the lower bound of Theorem 1.1 suggest some questions that we find worth investigating. One question is whether better bounds on the planar edge-length ratio can be established for subfamilies of outerplanar graphs (for example, it is easy to show that trees have unit-length drawings). A second question is whether an edge length variance bounded by a constant can be guaranteed for drawings of outerplanar graphs where not all vertices lie in a common face. By a variant of the approach used to prove Lemma 2 and by using some simple geometric observations, the following results can be proved (see Appendix A1 and A2 for details).

Theorem 3.1

The planar edge-length ratio of a bipartite outerplanar graph is .

The plane edge-length ratio of a planar embedding of a graph is the minimum edge-length ratio taken over all embedding-preserving planar straight-line drawings of .

Theorem 3.2

For any given , there exists an embedded outerplanar graph whose plane edge-length ratio is at least .

We conclude this paper by listing some open questions that we find interesting to study: (i) Study the edge-length ratio of triangle-free outerplanar graphs. For example, it is not hard to see that if all faces of an outerplanar graph have five vertices, a unit edge length drawing may not exist; however, the planar edge length ratio for this family of graphs could be smaller than the one established in Theorem 1.1. (ii) Extend the result of Theorem 1.1 to families of non-outerplanar graphs. For example it would be interesting to study whether the planar edge-length ratio of 2-trees is bounded by a constant. (iii) Study the complexity of deciding whether an outerplanar graph admits a straight-line drawing where the ratio of the longest to the shortest edge is within a given constant. This problem is interesting also in the special case that we want all edges to be unit length.

References

Appendix

A1: Proof of Lemma 3

Proof

Let the length of the longest edge be . We show that, for any value , there exists a a maximal outerplanar graph such that in any planar straight-line drawing of the length of the shortest edge must be smaller than . Let us rewrite as , where . In any planar straight-line drawing of a maximal outerplanar graph such that the longest edge has length and the shortest edge has length at least , the area of every triangular face cannot become arbitrarily small, but it has a lower bound that depends on the value . More precisely, by observing that the minimum area of a triangular face is obtained when one of its sides has length while the other two have length , by Heron’s formula we have that the area of any triangular face under these assumptions is at least .

Figure 2: An example of the graph in the proof of Lemma 3 when .

Let be an integer such that and consider a fan graph with a vertex of degree . is constructed by adding triangular faces to as follows: for each edge of not incident to , add a new vertex adjacent to both vertices of . See Figure 2 for an illustration when . We call these non-fan triangles pendant triangles. Observe that in any planar straight-line drawing of independently of whether all vertices of appear on a common face or not, we have that the drawing has at least area-disjoint pendant triangles, since any pendant triangle that contains another triangle must contain the entire graph. Also, since the longest edge in the drawing has length and since every vertex has graph theoretic distance at most from vertex , we have that lies inside a disk of radius centered at . The number of area disjoint triangles that can be packed inside such that every triangle has area at least must have . Since , it follows that any planar straight-line drawing of the maximal outerplanar graph where the length of the longest edge is requires at least one edge having length shorter than . ∎

A2: Proof of Theorem 2

The proof adopts a variant of the approach used to prove Lemma 2. Given a unit-length line segment in the plane and two directions and (unit vectors) in the same half-plane determined by , consider the region bounded by and the two infinite rays in directions and having their sources at and , respectively. Define to be the sum of the angles and , where and are in the half-plane of containing and . We call a wedge and the wedge angle. We will show that any bipartite outerplanar graph admits a unit-length drawing within any wedge such that .

To simplify the construction, we first add edges to until it is a maximal (bipartite) outerplanar graph and then show that all such graphs admit the desired type of drawing. Every face of , with the exception of its outerface, is now a quadrilateral. We call any edge on the non-quadrilateral face an external edge of . If itself is a quadrilateral, then every edge of is external. We are now ready to prove Theorem 3.1.

Proof

Let be a maximal bipartite outerplanar graph, an external edge of , a unit length segment, and two directions in the same half-plane of such that has . We construct an outerplanar straight-line drawing of within such that is drawn as and every edge of has length .

The proof proceeds by induction on the number of internal faces of . If , then can clearly be drawn within as a rhombus. So assume now that the result holds for some and that has internal faces.

Let and consider the unique quadrilateral face containing . We draw within so that is identified with , and denote by the line segment drawn for each edge . Now select directions and for rays emanating from and , respectively, such that for , each pair is in the same half-plane determined by , and has an angle greater than . Since , such directions exist. Now separate into its (at most) three 2-connected components , where has designated edge . By induction, each admits an outerplanar straight-line drawing in the corresponding wedge so that is identified with the segment . ∎

A3: proof of Theorem 3

In this section we construct a family of embedded outerplanar graphs having unbounded plane edge-length ratios. To prove the correctness of the construction we use the following fact that can be proved by using elementary geometry.

Lemma 4

Let be a triangle with longest edge , shortest edge , and containing two points and such that the triangles and are area-disjoint. Then at least one of the triangles and has perimeter at least shorter than the perimeter of .

Proof

Let be the perimeter of , where are the lengths of the edges opposite the vertices , respectively. Consider the segment from to the (unique) point on that divides into two triangles and of equal perimeter , where is the length of segment .

The difference in perimeter between and (or ) is since . ∎

We are now ready to prove Theorem 3.2

Proof

We construct a family of outerplanar graphs having planar embeddings such that as , where denotes the plane edge-length ratio of the planar embedding of the graph . Each will have a set of distinguished edges, and for , each will contain as a subgraph. is a single triangle, with two distinguished edges. is constructed from by adding, for each distinguished edge of , a new vertex that is adjacent to both vertices of ; the newly added edges incident to each are the distinguished edges of . There is only planar embedding of ; the embedding of is obtained by putting the two vertices of in the inner face of . For , the embedding of is defined by placing each new vertex in the (unique) triangular face of having on its boundary.

Assume that for each there is a planar straight-line drawing of that preserves the embedding described above, and that, for some for all . We show that this assumption leads to a contradiction. To simplify our notation slightly, we will assume that some edge of has length , since we can always scale the drawing to make this so; thus the smallest edge-length in any of the is at least .

Consider a triangular face created in the construction of for some . It consists of a distinguished edge of along with the vertex in adjacent to . The two edges and incident with are distinguished edges of , so in each of them will form a triangle with some new vertex, say and , respectively, and both new vertices will be in the face .

Now consider the line segment from to that bisects the perimeter of , as in the proof of Lemma 4, forming two triangles and , where contains on its boundary. Now either contains or contains ; whichever contains then contains the entire triangular face formed by and . Assume, w.l.o.g., that contains . We consider two cases.

  1. If is the longest edge of , then by Lemma 4, the perimeter of is smaller than that of by at least half the length of the shortest side of . But no edge in any of the is shorter than , and so the perimeter of is smaller than that of by at least .

  2. If is not the longest edge of , then one of or must be. If is the longest edge of , then after the construction of , must be the longest edge of the (new) triangle formed by and (because is contained in , which has as its longest edge). Thus, in the construction of , one of the new triangles formed, by similar application of Lemma 4, will have perimeter that is at least shorter than the perimeter of , which itself has perimeter shorter than that of .

In each of the two cases above we have identified a triangle, either in or in , with perimeter at least shorter than that of . Repeating this process times results in a triangle that has perimeter at least shorter than that of . But the perimeter of is at most , so eventually the quantity becomes negative—a contradiction. ∎