1 Introduction and Preliminary Results
The problem of producing drawings of graphs with geometric constraints is a core topic for Graph Drawing [3, 4, 5, 11, 14, 23, 24, 27, 34, 43, 45]. In this context, a classic question is the one of testing if a planar graph can be drawn planarly and straightline with prescribed edge lengths. The study of such a question is related to several topics in computational geometry [17, 41, 47], rigidity theory [16, 30, 32], structural analysis of molecules [8, 31], and sensor networks [13, 38, 40]. Formally, given a weighted planar graph , i.e., a planar graph equipped with a length function , the Fixed EdgeLength Planar Realization problem (FEPR for short) asks whether there exists a planar straightline realization of (PR for short), i.e., a planar straightline drawing of where the Euclidean length of each edge is . The FEPR problem was first studied by Eades and Wormald [26], who showed its hardness for triconnected planar graphs and for biconnected planar graphs with unit lengths. Cabello, Demaine, and Rote strengthened this result by proving hardness for triconnected planar graphs with unit lengths [12]. Abel et al. [1] proved the completeness of the FEPR problem with unit lengths, solving a problem posed by Schaefer [42].
Since large triconnected minors are essential in the known hardness proofs of the FEPR problem, we study its complexity for trees, which are the maximal graphs with no Kminor. A tree is a graph composed of cycles glued together along edges in a treelike fashion; see Fig. 1, where we show a planar and a nonplanar realization of a weighted tree. Every tree is planar and biconnected, and the class of trees is the class of maximal seriesparallel graphs. There is a vast amount of research on trees in Graph Drawing (e.g., in [18, 25, 28, 33, 39]). The edge lengths of trees have been studied in [9, 10].
In this paper, we first show that the FEPR problem can be solved in linear time for trees with prescribed embedding^{1}^{1}1As in [12], our algorithms adopt the real RAM model, which is customary in computational geometry and supports standard arithmetic operations in constant time.. We note the FEPR problem is hard for general planar graphs with a prescribed embedding [12]. Second, we show that, in the variable embedding setting, the FEPR problem is hard when the number of distinct lengths is at least four, whereas it is lineartime solvable when the number of distinct lengths is one or two. Note that, for general planar graphs, the problem is hard even when all the edges are required to have the same length [26]. Third, we deal with maximal outerplanar graphs. We show that the FEPR problem can be solved in linear time for maximal outerpaths, i.e., the maximal outerplanar graphs whose dual tree is a path, and in cubic time for maximal outerpillars, i.e., the maximal outerplanar graphs whose dual tree is a caterpillar. Finally, we present a slicewise polynomial algorithm for trees, parameterized by the length of the longest path.
Because of space limitations, several proofs are omitted. They can be found in the full version of the paper [2].
1.0.1 Preliminaries.
We assume familiarity with Graph Drawing (see, e.g., [21]). A planar drawing of a graph defines a clockwise order of the edges incident to each vertex of ; the set of such orders for all the vertices is a rotation system for . Two planar drawings of are equivalent if (i) they define the same rotation system for and (ii) their outer faces have the same boundaries. An equivalence class of planar drawings is a plane embedding (or simply an embedding). When referring to a planar drawing of a graph that has a prescribed embedding , we always imply that respects ; sometimes, we explicitly stress this.
An outerplanar drawing is a planar drawing in which all the vertices are incident to the outer face. An outerplane embedding is an equivalence class of outerplanar drawings. An outerplanar graph is a graph that admits an outerplanar drawing. The dual tree of a biconnected outerplanar graph is defined as follows. Consider the (unique) outerplane embedding of . Then has a node for each internal face of and has an edge between two nodes if the corresponding faces of are incident to the same edge of . An outerpath is a biconnected outerplanar graph whose dual tree is a path. A caterpillar is a tree that becomes a path if its leaves are removed. An outerpillar is a biconnected outerplanar graph whose dual tree is a caterpillar.
A tree is recursively defined as follows. A cycle is a tree. Given a tree containing an edge , the graph obtained by adding to a vertex and two edges and is a tree. We observe that the neighbors of any degree vertex are adjacent. The treelike structure of a tree is encoded by means of the decomposition tree rooted at a cycle of . Each node in represents a cycle of , and two nodes are adjacent if their corresponding cycles share an edge; see Fig. 2. The decomposition tree of a tree is easily computed in linear time. We adopt the Euclidean metric and assume that the length function of is such that every cycle satisfies the triangle inequality. This is a necessary condition for the existence of a straightline realization of , i.e., a (not necessarily planar) drawing of in which each edge is represented by a line segment with the prescribed length. We often refer to cycles of , nodes of , and triangles in a straightline realization of interchangeably.
1.0.2 Prescribed Embedding.
First, we deal with trees with a prescribed rotation system or embedding. We start by presenting a geometric tool.
Theorem 1.1.
Let be an vertex weighted tree, be a plane embedding (resp. be a rotation system) for , and be a straightline realization of . There is an time algorithm to test whether is a PR respecting (resp. ).
The proof of Theorem 1.1 is based on an algorithm that: i. tests if respects (resp. ), ii. triangulates the faces of and checks if they are simple polygons [15], and iii. tests if the obtained drawing is a convex subdivision [20].
We now present our prescribed embedding result.
Theorem 1.2.
Let be an vertex weighted tree and be a plane embedding (resp. be a rotation system) for . There is an time algorithm to test whether admits a PR that respects (resp. ) and to construct one, if any.
The proof of Theorem 1.2 is based on: i. computing a decomposition tree rooted at the cycle of with the largest sum of the edge lengths; ii. computing a candidate PR by visiting in preorder while greedily adding to the drawing of each cycle , by exploiting (resp. ) and the containment relationship between and ; and iii. testing whether is a PR of whose plane embedding (resp. rotation system) is (resp. ) by means of Theorem 1.1.
2 NPhardness for 2trees with 4 Edge Lengths
We sketch a reduction from the complete Planar Monotone 3SAT problem [7] (PMS for short) to the FEPR problem with four edge lengths.
Theorem 2.1.
The FEPR problem is hard for weighted trees, even for instances whose number of distinct edge lengths is .
A Boolean CNF formula is an instance of PMS if the variableclause incidence graph of is planar, and each clause of is either positive (it consists of positive literals) or negative (it consists of negated literals). The PMS problem is complete even when comes with a monotone rectilinear representation [7], i.e., a crossingfree drawing of in which i. variables and clauses are boxes, ii. edges are vertical segments, and iii. positive (resp. negative) clauses lie above (resp. below) the horizontal strip containing the variable boxes; see Fig. 3(a).
First, we transform into a representation of that uses segments with slope , , or ; see Fig. 3(b). Then we obtain from a weighted tree that admits a PR if and only if is satisfiable; see Fig. 3(c). The edges of are assigned the lengths , , , and . To obtain we construct gadgets for the variables, the clauses, and the edges of . Our gadgets exploit two main types of triangles: equilateral triangles with sides of length (frame triangles), and isosceles triangles with base of length and two sides of length (transmission triangles). The union of the frame triangles of the gadgets representing variables (gray), edges (yellow), and clauses (green), together with a set of frame triangles connecting the variable gadgets (blue), forms a maximal outerplanar graph. Since this graph is formed by frame triangles, it has a unique PR up to rigid transformations.
Our strategy is to construct from a “rigid” part (mainly formed by the union of the frame triangles of the gadgets), and a part that instead allows for different embedding choices (mainly encoded by the flips of transmission triangles). Consider for example Fig. 4, where we illustrate the PR of a clause gadget, which is the most critical gadget in the construction. Each transmission triangle (in red) has two possible different embeddings. The choice of this embedding influences the choice of the embeddings of the transmission triangles that “conflict” with , that is, that overlaps with in one of their embeddings. These chains of conflict relationships allow for “truth values” that come from the variable gadgets to “move” along the gadgets representing edges. The relevant triangles , , and encode such values: In Fig. 4, and point downward since the corresponding variable is True, and points upward since the corresponding variable is False. A special set of transmission triangles, whose flip depends on the orientation of the relevant triangles, overlap in the pink hexagonal region if all the relevant triangles point upward. Conversely, if at least one relevant triangle points downward, the clause gadget admits a PR.
3 Lineartime Algorithm for 2trees with 2 Edge Lengths
This section is devoted to sketch the proof of the following theorem.
Theorem 3.1.
Let be an vertex weighted tree, where with . There is an time algorithm to test whether admits a PR and to construct one, if any.
Let be a PR of . If , then the existence of implies that is an outerplanar graph, which is a lineartime testable property [19, 36, 46], and that is outerplanar. Since is connected, it has a unique outerplane embedding which can be constructed in linear time [19, 36, 37, 44, 46]. Hence, the problem reduces to the problem of testing whether has a PR that respects , which can be solved in linear time by Theorem 1.2.
Consider now the case in which . W.l.o.g. assume . Also, let . The realization of any cycle of is one of the following types of triangles: i. an equilateral (small equilateral) triangle of side , ii. an equilateral (big equilateral) triangle of side , iii. an isosceles (tall isosceles) triangle with base and two sides of length , and iv. an isosceles (flat isosceles) triangle with base and two sides of length ; refer to Fig. 5.
Let and be two triangles realizing two different cycles in a PR of . We say that is drawn inside if all the points of are points of and at least one vertex of is an interior point of . Let be the decomposition tree of . We have the following.
Lemma 1.
If is rooted at the cycle with the largest sum of edge lengths and is drawn inside , then is a leaf triangle that shares a side with .
By Lemma 1, we assume that is rooted at a cycle with the largest sum of edge lengths. The framework of is the subgraph obtained, in linear time, as follows: For each leaf triangle that can be drawn inside its parent or a sibling triangle , we remove from the vertex that does not share with , along with the two edges incident to . Note that i. is a tree which may contain any type of triangle, ii. by Lemma 1, no triangle of is drawn inside any other triangle in any PR of , and iii. is rooted at a triangle of the framework. We test in linear time if is outerplanar and, in such case, we compute in linear time its unique outerplanar embedding . Exploiting Theorem 1.2, we test if admits a PR respecting . In the negative case admits no PR, otherwise we denote by the obtained PR of . Hereafter, we assume that exists.
Refer to Fig. 6, where we show an example of a PR of a weighted tree. Let be the set of leaf triangles that were removed from to obtain . Observe that is formed by small equilateral and flat isosceles triangles. We show how to extend to a PR of by embedding the triangles of , if possible. Let denote a triangle in . The triangle has a side in that is incident to either two internal faces, or to an internal face and the outer face of ; hence has exactly two embedding choices. We say that has an internal embedding if it is embedded inside an internal face of , and an outer embedding otherwise. We say that induces a framework conflict if it has an outer embedding and is not planar; e.g., see the (dashed) outer embedding of in Fig. 6. Let and be two triangles of . We say that and induce an internal conflict if both have an internal embedding and is not planar; e.g., see the (dashed) internal embedding of and the (solid) internal embedding of . On the other hand, we say that and induce an external conflict if both have an outer embedding and is not planar; e.g., see the (dashed) outer embeddings of and .
Lemma 2.
Let and be two leaf triangles of that can both be drawn inside some triangle . The triangles and induce an internal conflict if at least one of the following properties holds true:

and share an edge;

and is a tall isosceles;

.
The weighted tree shown in Fig. 6 is such that , that is . By Property b of Lemma 2, two flat isosceles triangles can be drawn inside a big equilateral triangle without inducing conflicts, and any pair of leaf triangles (i.e., either two flat isosceles triangles or a flat isosceles triangle and a small equilateral triangle) induce an internal conflict inside a tall isosceles triangle. On the other hand, by Property a of Lemma 2 and the fact every triangle in has two embedding choices, for a PR of to exist, it must hold that no three triangles in share the same edge with a triangle of . If satisfies this requirement, then we say that is consistent.
Lemma 3.
If is consistent, then there are pairs of triangles sharing an edge with the same triangle of that induce an internal conflict.
Proof.
By Lemma 1, two triangles of that induce an internal conflict share an edge with a common triangle of . Since is consistent, there exist at most triangles in incident to . Hence, at most pairs of triangles sharing an edge with can induce an internal conflict. ∎
3.0.1 Extending to a PR of .
Next, we show how to test whether there is a choice of embeddings for the triangles in that yields a PR of . We distinguish two cases, based on whether or .
Case . In this case there are no flat isosceles triangles, and hence the setting is much simpler than the one depicted in Fig. 6. Every leaf triangle in is a small equilateral triangle whose parent is a tall isosceles triangle. We act as follows. For any two triangles that share an edge , we embed and in on opposite sides of . We embed every other leaf triangle in inside its parent triangle. At the end of this process we obtain, in linear time, a straightline realization of , and a plane embedding of .
Lemma 4.
has a PR if and only if is planar.
Proof sketch.
First, if two leaf triangles share an edge , then they lie on opposite sides of in any PR, since they would overlap otherwise. Second, each isosceles triangle may contain only one leaf triangle, since it has only one side of length . Hence, embedding a leaf triangle inside an isosceles triangle that shares an edge with cannot cause crossings, since does not induce internal conflicts. Therefore, a crossing in can only be caused by framework and external conflicts, which are, however, unavoidable. ∎
By Lemma 4, in order to test whether admits a PR, we can apply Theorem 1.1 to test in time whether is a PR of with embedding .
Case . In this case there might be flat isosceles triangles in which might or might not need to be embedded inside a framework triangle; in Fig. 6 such triangles are shaded blue. Also, more than one leaf triangle can be drawn inside the same framework triangle and this might or might not induce internal conflicts. Recall that the triangles in are small equilateral triangles and/or flat isosceles triangles.
We construct a 2SAT formula with a Boolean variable for each triangle . The values of are associated with the two possible embeddings of . If two triangles in induce a conflict for certain embeddings, then contains a clause that is True if and only if, at least one of the variables representing the triangles does not have the value corresponding to the embedding generating the conflict. Further, for each triangle that induces a framework conflict, contains a clause that is True if and only if, the variable representing the triangle does not have the value corresponding to an outer embedding. We test in time if is satisfiable [6]. In the positive case, we obtain a PR of from by embedding each triangle in according to the value of the corresponding variable. We reject the instance in the negative case.
It only remains to prove that the number of conflicts (and hence the size of ) is in and that such conflicts can be found in time. Detecting internal conflicts is fairly easy: i. by Lemma 1, triangles inducing an internal conflict are “close” in (they share an edge with a common framework triangle); ii. by Lemma 3, there exist leaf triangles sharing an edge with the same framework triangle; and iii. since is consistent, the maximum degree of is bounded by a constant. Hence, by traversing we compute in time the set of pairs of leaf triangles that induce an internal conflict.
Efficiently detecting external and framework conflicts is more challenging. Let be the subset of composed of those triangles that are incident to external edges of . We give an outer embedding to every triangle in . This results in a (possibly nonplanar) straightline realization of the graph . We now construct a boundeddegree graph whose nodes are associated with sets of vertices of so that the following properties hold: (a) Each node is associated with degree vertices of that belong to triangles in , (b) if two triangles in induce an external conflict, then their degree vertices are associated either with the same node or with adjacent nodes, and (c) if a triangle intersects an edge of (inducing a framework conflict), then the degree vertex of and the endvertices of are associated either with the same node or with adjacent nodes. After constructing , the external and framework conflicts can be detected with a lineartime traversal of .
The graph is defined as follows. Assume that the bottomleft corner of the bounding box of lies on the origin of the Cartesian axes. Consider a square grid covering the plane whose grid cells have side length ; see Fig. 7. Assign a label to each vertex of . Then has a node for each label assigned to at least one vertex of , and two distinct nodes and are connected if and only if and . Note that has edges since it has at most nodes and maximum degree 8.
We now prove that satisfies Property (a). The number of degree vertices of that belong to triangles in and are associated with a node of , is upper bounded by the number of framework triangles that i. are contained in the union of the grid cell and its surrounding grid cells, and ii. share an edge with a triangle in . Note that is actually the number of big equilateral and tall isosceles triangles in such nine cells. Since the area of a big equilateral or tall isosceles triangle is at least the area of a small equilateral triangle, then is upper bounded by the ratio between the area of cells, which is , and the area of a small equilateral triangle, which is . Therefore, since , we have that .
We next sketch an algorithm to construct in time. The vertex set of is constructed by removing repetitions from the set of labels computed for the vertices in . To this aim, we compute a total order of the vertices of such that, for any two vertices and with and , we have if and only if i. or ii. and . Since is connected and any edge of has length at most , then for any label . Hence, we compute in time with counting sort. Since vertices with the same label are consecutive in , repetitions can be removed with a linear scan of .
The edge set of consists of four disjoint subsets , , , and . These sets contain the edges that connect nodes of corresponding to grid cells that are adjacent horizontally, vertically, along the main, and the minor diagonal, respectively; see Fig. 8. We appropriately define four orders , , , and of the nodes of such that nodes that are connected by an edge in , , , and are consecutive in the corresponding order. We compute the four sets of edges with a linear scan of the orders , , , and .
4 Maximal Outerplanar Graphs
In this section we study the FEPR problem for weighted outerplanar trees, i.e., for weighted maximal outerplanar graphs. We prove the following theorems.
Theorem 4.1.
Let be an vertex weighted maximal outerpath. There is an time algorithm to test whether admits a PR and to construct one, if any.
Theorem 4.2.
Let be an vertex weighted maximal outerpillar. There is an time algorithm to test whether admits a PR and to construct one, if any.
Let be a weighted tree and be an edge of . An outer realization of is a PR of such that is incident to the outer face. An outer realization of is optimal if, for every outer realization of , there is a rigid transformation of such that the segment representing coincides with the one in and such that the interior of is a subset of the interior of .
We sketch the proof of Theorem 4.1; the proof of Theorem 4.2 uses similar ideas. Let be an vertex weighted maximal outerpath; see Fig. 9. Let be the dual tree of the outerplane embedding of ; since is an outerpath, is a path . For , let be the cycle of bounding the internal face of dual to and let be the unique, up to rigid transformation, PR of . For , let be the edge of dual to .
Let be such that has maximum edge length sum. Let and be the subgraphs of composed of the cycles and , respectively. Since the length of is maximum, the restrictions of any PR of to and are outer and outer realizations, respectively. We prove that (resp. ) admits an outer (resp. outer) realization if and only if it admits an optimal (resp. optimal) realization. The core of the proof of Theorem 4.1 is an time algorithm, called OuterChecker, that constructs an optimal (resp. an optimal) realization of (resp. of ) and its plane embedding, if any such a realization exists. If OuterChecker concludes that both and admit a PR, then and (as well as their embeddings) can be combined in four ways with (see Fig. 10) and each resulting straightline realization can be tested for planarity in time, by Theorem 1.1.
We describe how OuterChecker works on . A key observation is that the restriction of any optimal realization of to the graph composed of the cycles is an optimal realization of . This allows OuterChecker to work by induction on to decide whether has an optimal realization . If , the graph is the cycle whose unique PR is optimal. If , an optimal realization of is constructed, if it exists, by combining and so that coincides in the two realizations. Three things might happen. First, if “fits” inside , as in Fig. 11(left), then the resulting PR is optimal. Else, if “fits” outside , as in Fig. 11(middle), once cycles and lie on different sides of , then the resulting PR is optimal. Otherwise, admits no optimal realization, as in Fig. 11(right).
A naive implementation of OuterChecker takes time. Indeed, for each of the inductive steps, one can check in time whether fits inside and/or outside using Theorem 1.1. We achieve total running time avoiding a planarity test at each step. For , we compute a “candidate” straightline realization of , and only test for planarity the final realization . By “candidate” we mean that, if admits an optimal realization, then is such a realization. In order to do that, OuterChecker dynamically maintains the boundary of the convex hull of , which is guaranteed to actually be the boundary of the convex hull of if is planar. We compute by suitably exploiting a lineartime algorithm by Melkman [35], which incrementally computes the convex hull of a point set spanned by a planar path, provided that the points are given in the order of the path.
After constructing (which comes with a plane embedding), we test its planarity in time using Theorem 1.1. If the test is successful, is an optimal PR of , otherwise no optimal PR of exists. Each step of OuterChecker takes time, except for the computation of the boundary . However, the computation of the boundaries takes time in total [35]. Hence, the overall running time of OuterChecker is in .
5 2Trees with Short Longest Path
In this section, we sketch a proof of the following theorem.
Theorem 5.1.
Let be an vertex weighted tree and let be the length of a longest path of . There is an time algorithm to test whether admits a PR and to construct one, if any.
Theorem 5.1 is actually a corollary of a stronger theorem, which relates to SPQtrees; these are a specialization for trees of the wellknown SPQRtrees [22, 29]. The SPQtree of is a tree that represents a recursive decomposition of into subgraphs along separation pairs. Each node of corresponds to a subgraph of , which is joined to the rest of the graph via two vertices and . Assume that is rooted at the neighbor of an edge of with maximum length and let be the height of . We design an time algorithm that tests whether admits a PR and, in the positive case, constructs such a realization. Then Theorem 5.1 follows, as we can prove that .
The time algorithm performs a visit of . When visiting a node , the algorithm either concludes that admits no PR, or constructs a set of “optimal” PRs of . Here, “optimal” means that, for every PR of , there is a PR whose interior is a subset of the interior of , after a suitable rigid transformation. The main ingredient needed for bounding the running time of the algorithm is the following. Suppose that consists of a “parallel” composition of graphs . Then “few” of the permutations of need to be considered when constructing . Namely, we can sort by increasing length of the edge paths between and they contain. Then, in any PR of , the graph is either “to the left” or “to the right” of all the graphs ; further, whether a PR of is optimal only depends on the choice of the “leftmost” and “rightmost” graphs among (and on their drawings, which are taken from ), and not on the permutation of the remaining graphs, as long as planarity is ensured.
6 Open problems
Our results on the FEPR problem when is a tree motivate the study of several open questions:

Determine the computational complexity of the FEPR problem for weighted trees with prescribed edge lengths (we proved it is lineartime solvable for and hard for ).

Determine if it is possible to improve our algorithm for general trees to an algorithm.

Study the computational complexity of the FEPR problem for general maximal outerplanar graphs.

Study the computational complexity of the FEPR problem for graphs with treewidth and for degenerate planar graphs; both these classes generalize the one of trees.
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